<?xml version='1.0' encoding='UTF-8'?><?xml-stylesheet href="http://www.blogger.com/styles/atom.css" type="text/css"?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearchrss/1.0/' xmlns:georss='http://www.georss.org/georss' xmlns:gd='http://schemas.google.com/g/2005' xmlns:thr='http://purl.org/syndication/thread/1.0'><id>tag:blogger.com,1999:blog-2267127033347614408</id><updated>2011-11-27T16:48:18.537-08:00</updated><category term='INFOSYS'/><category term='Programming Puzzles'/><title type='text'>Puzzles Asked in Microsoft , Google, Yahoo, Amazone Interviews</title><subtitle type='html'>Puzzles,Programming Puzzles, Word puzzles, Sudoku puzzles, Brain teasers,Logic Problems,Download Puzzles, puzzles Asked in Top companies interviews, prize winning puzzles. Google - Microsoft - amazon, yahoo interview puzzles</subtitle><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/posts/default'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default?max-results=100'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/'/><link rel='hub' href='http://pubsubhubbub.appspot.com/'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><generator version='7.00' uri='http://www.blogger.com'>Blogger</generator><openSearch:totalResults>98</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>100</openSearch:itemsPerPage><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-3457121565842188420</id><published>2009-09-09T23:03:00.001-07:00</published><updated>2009-09-09T23:03:54.259-07:00</updated><title type='text'>Develop a plan that allows as many people as possible to live.</title><content type='html'>&lt;p&gt;One hundred persons will be lined up single file, facing north. Each person will be assigned either a red hat or a blue hat. No one can see the color of his or her own hat. However, each person is able to see the color of the hat worn by every person in front of him or her. That is, for example, the last person in line can see the color of the hat on 99 persons in front of him or her; and the first person, who is at the front of the line, cannot see the color of any hat.Beginning with the last person in line, and then moving to the 99th person, the 98th, etc., each will be asked to name the color of his or her own hat. If the color is correctly named, the person lives; if incorrectly named, the person is shot dead on the spot. Everyone in line is able to hear every response as well as hear the gunshot; also, everyone in line is able to remember all that needs to be remembered and is able to compute all that needs to be computed.Before being lined up, the 100 persons are allowed to discuss strategy, with an eye toward developing a plan that will allow as many of them as possible to name the correct color of his or her own hat (and thus survive). They know all of the preceding information in this problem. Once lined up, each person is allowed only to say “Red” or “Blue” when his or her turn arrives, beginning with the last person in line. &lt;/p&gt;&lt;p&gt;Develop a plan that allows as many people as possible to live.&lt;/p&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-3457121565842188420?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/3457121565842188420/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=3457121565842188420' title='4 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3457121565842188420'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3457121565842188420'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2009/09/develop-plan-that-allows-as-many-people.html' title='Develop a plan that allows as many people as possible to live.'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>4</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-8671484199659807627</id><published>2009-08-30T21:42:00.000-07:00</published><updated>2009-08-30T21:42:00.153-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>What was his salary to begin with?</title><content type='html'>&lt;span style="font-size:85%;"&gt;A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, &amp;amp; now his salary is Rs. 22702.68&lt;br /&gt;&lt;br /&gt;What was his salary to begin with?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Rs.22176&lt;br /&gt;&lt;br /&gt;Assume his salary was Rs. X&lt;br /&gt;&lt;br /&gt;He earns 5% raise. So his salary is (105*X)/100&lt;br /&gt;&lt;br /&gt;A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100) which is Rs. 22702.68&lt;br /&gt;&lt;br /&gt;Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68&lt;br /&gt;X = 22176&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-8671484199659807627?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/8671484199659807627/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=8671484199659807627' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/8671484199659807627'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/8671484199659807627'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/what-was-his-salary-to-begin-with.html' title='What was his salary to begin with?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-3868665566489423262</id><published>2009-08-29T21:41:00.000-07:00</published><updated>2009-08-29T21:41:00.804-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>How much would you have earned with this business after 50 years?</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;&lt;br /&gt;If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, &amp;amp; so on.&lt;br /&gt;&lt;br /&gt;How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;Rs.333,062,500&lt;br /&gt;&lt;br /&gt;To begin with, you want to know the total number of days: 365 x 50 = 18250.&lt;br /&gt;&lt;br /&gt;By experimentation, the following formula can be discovered, &amp;amp; used to determine the amount earned for any particular day: 1 + 2(x-1), with x being the number of the day. Take half of the 18250 days, &amp;amp; pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, &amp;amp; so on, &amp;amp; you will see that if you add these pairs together, they always equal Rs.36500.&lt;br /&gt;&lt;br /&gt;Multiply this number by the total number of pairs (9125), &amp;amp; you have the amount you would have earned in 50 years.&lt;br /&gt;&lt;br /&gt;Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11.....upto 18250 terms)&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-3868665566489423262?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/3868665566489423262/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=3868665566489423262' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3868665566489423262'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3868665566489423262'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/how-much-would-you-have-earned-with.html' title='How much would you have earned with this business after 50 years?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-1329714617592705773</id><published>2009-08-29T04:55:00.000-07:00</published><updated>2009-08-29T04:55:00.256-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Puzzle 2:   Find sum of digits of D.</title><content type='html'>&lt;span style="font-size:85%;"&gt;Find sum of digits of D.&lt;br /&gt;Let&lt;br /&gt;A= 19991999&lt;br /&gt;B = sum of digits of A&lt;br /&gt;C = sum of digits of B&lt;br /&gt;D = sum of digits of C&lt;br /&gt;&lt;br /&gt;(HINT : A = B = C = D (mod 9))&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;The sum of the digits od D is 1.&lt;br /&gt;&lt;br /&gt;Let E = sum of digits of D.&lt;br /&gt;It follows from the hint that A = E (mod 9)&lt;br /&gt;Consider,&lt;br /&gt;A = 19991999&lt;br /&gt;&lt; 20002000 =" 22000" 10002000 =" 1024200" 106000 =" 106800" 9 =" 61200" 9 =" 45" 9 =" 18" 1999 =" 1" 19991999 =" 1" e="1.&lt;/font"&gt;&lt;br /&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-1329714617592705773?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/1329714617592705773/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=1329714617592705773' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1329714617592705773'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1329714617592705773'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/puzzle-2-find-sum-of-digits-of-d.html' title='Puzzle 2:   Find sum of digits of D.'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-2761392859046414232</id><published>2009-08-28T21:40:00.000-07:00</published><updated>2009-08-28T21:40:00.189-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>how long did Vipul study in candle light?</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;&lt;br /&gt;Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.&lt;br /&gt;&lt;br /&gt;For how long did Vipul study in candle light?&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;Vipul studied for 3 hours in candle light.&lt;br /&gt;&lt;br /&gt;Assume that the initial lenght of both the candle was L and Vipul studied for X hours.&lt;br /&gt;&lt;br /&gt;In X hours, total thick candle burnt = XL/6&lt;br /&gt;In X hours, total thin candle burnt = XL/4&lt;br /&gt;&lt;br /&gt;After X hours, total thick candle remaining = L - XL/6&lt;br /&gt;After X hours, total thin candle remaining = L - XL/4&lt;br /&gt;&lt;br /&gt;Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.&lt;br /&gt;(L - XL/6) = 2(L - XL/4)&lt;br /&gt;(6 - X)/6 = (4 - X)/2&lt;br /&gt;(6 - X) = 3*(4 - X)&lt;br /&gt;6 - X = 12 - 3X&lt;br /&gt;2X = 6&lt;br /&gt;X = 3&lt;br /&gt;&lt;br /&gt;Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-2761392859046414232?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/2761392859046414232/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=2761392859046414232' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2761392859046414232'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2761392859046414232'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/how-long-did-vipul-study-in-candle.html' title='how long did Vipul study in candle light?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-4320269299279446224</id><published>2009-08-27T21:39:00.000-07:00</published><updated>2009-08-27T21:39:00.363-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>What part of the contents of the container is left at the end of the second day?</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day.&lt;br /&gt;&lt;br /&gt;What part of the contents of the container is left at the end of the second day?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;Assume that contents of the container is X&lt;br /&gt;&lt;br /&gt;On the first day 1/3rd is evaporated.&lt;br /&gt;(1 - 1/3) of X is remaining i.e. (2/3)X&lt;br /&gt;&lt;br /&gt;On the Second day 3/4th is evaporated. Hence,&lt;br /&gt;(1- 3/4) of (2/3)X is remaining&lt;br /&gt;i.e. (1/4)(2/3)X = (1/6) X&lt;br /&gt;&lt;br /&gt;Hence 1/6th of the contents of the container is remaining&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-4320269299279446224?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/4320269299279446224/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=4320269299279446224' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/4320269299279446224'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/4320269299279446224'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/what-part-of-contents-of-container-is.html' title='What part of the contents of the container is left at the end of the second day?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-1722152319919698737</id><published>2009-08-26T21:38:00.000-07:00</published><updated>2009-08-26T21:38:00.297-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Find a number of 9 digits has the following properties ?</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;&lt;br /&gt;A number of 9 digits has the following properties:&lt;br /&gt; The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2&lt;=n&lt;=9.&lt;br /&gt; Each digit in the number is different i.e. no digits are repeated.&lt;br /&gt; The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.&lt;br /&gt;Find the number.&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;The answer is 381654729&lt;br /&gt;&lt;br /&gt;One way to solve it is Trial-&amp;amp;-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.&lt;br /&gt;&lt;br /&gt;The other way to solve this problem is by writing a computer program that systematically tries all possibilities&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-1722152319919698737?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/1722152319919698737/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=1722152319919698737' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1722152319919698737'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1722152319919698737'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/find-number-of-9-digits-has-following.html' title='Find a number of 9 digits has the following properties ?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-3591997751488625796</id><published>2009-08-25T21:37:00.000-07:00</published><updated>2009-08-25T21:37:00.269-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>How many medals were awarded ?</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;&lt;br /&gt; In a sports contest there were m medals awarded on n successive days (n &gt; 1).&lt;br /&gt;1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.&lt;br /&gt;2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.&lt;br /&gt;3. On the nth and last day, the remaining n medals were awarded.&lt;br /&gt;How many days did the contest last, and how many medals were awarded altogether?&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;Total 36 medals were awarded and the contest was for 6 days.&lt;br /&gt;&lt;br /&gt;On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals&lt;br /&gt;On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals&lt;br /&gt;On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals&lt;br /&gt;On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals&lt;br /&gt;On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals&lt;br /&gt;On day 6: Medals awarded 6 &lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-3591997751488625796?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/3591997751488625796/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=3591997751488625796' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3591997751488625796'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3591997751488625796'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/how-many-medals-were-awarded.html' title='How many medals were awarded ?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-7943038019803811780</id><published>2009-08-24T22:25:00.000-07:00</published><updated>2009-08-24T22:25:00.828-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Coin weighing puzzle.</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;&lt;br /&gt;There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.&lt;br /&gt;1. Take 8 coins and weigh 4 against 4.&lt;br /&gt;o If both are not equal, goto step 2&lt;br /&gt;o If both are equal, goto step 3&lt;br /&gt;&lt;br /&gt;2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.&lt;br /&gt;o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.&lt;br /&gt; If both are equal, L4 is the odd coin and is lighter.&lt;br /&gt; If L2 is light, L2 is the odd coin and is lighter.&lt;br /&gt; If L3 is light, L3 is the odd coin and is lighter.&lt;br /&gt;&lt;br /&gt;o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2&lt;br /&gt; If both are equal, there is some error.&lt;br /&gt; If H1 is heavy, H1 is the odd coin and is heavier.&lt;br /&gt; If H2 is heavy, H2 is the odd coin and is heavier.&lt;br /&gt;&lt;br /&gt;o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4&lt;br /&gt; If both are equal, L1 is the odd coin and is lighter.&lt;br /&gt; If H3 is heavy, H3 is the odd coin and is heavier.&lt;br /&gt; If H4 is heavy, H4 is the odd coin and is heavier.&lt;br /&gt;&lt;br /&gt;3. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.&lt;br /&gt;o If both are equal, there is some error.&lt;br /&gt;o If X is heavy, X is the odd coin and is heavier.&lt;br /&gt;o If X is light, X is the odd coin and is lighter. &lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-7943038019803811780?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/7943038019803811780/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=7943038019803811780' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/7943038019803811780'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/7943038019803811780'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/coin-wighing-puzzle.html' title='Coin weighing puzzle.'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-2957030956284592046</id><published>2009-08-23T22:24:00.000-07:00</published><updated>2009-08-23T22:24:00.605-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>ow many zeroes will be needed?</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;&lt;br /&gt;A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;The sign-maker will need 192 zeroes.&lt;br /&gt;&lt;br /&gt;Divide 1000 building numbers into groups of 100 each as follow:&lt;br /&gt;(1..100), (101..200), (201..300), ....... (901..1000)&lt;br /&gt;&lt;br /&gt;For the first group, sign-maker will need 11 zeroes.&lt;br /&gt;For group numbers 2 to 9, he will require 20 zeroes each.&lt;br /&gt;And for group number 10, he will require 21 zeroes.&lt;br /&gt;&lt;br /&gt;The total numbers of zeroes required are&lt;br /&gt;= 11 + 8*20 + 21&lt;br /&gt;= 11 + 160 + 21&lt;br /&gt;= 192&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-2957030956284592046?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/2957030956284592046/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=2957030956284592046' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2957030956284592046'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2957030956284592046'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/ow-many-zeroes-will-be-needed.html' title='ow many zeroes will be needed?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-9032015695457932652</id><published>2009-08-22T22:23:00.000-07:00</published><updated>2009-08-22T22:23:00.202-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Tractors Puzzle</title><content type='html'>&lt;span style="font-family:trebuchet ms;font-size:85%;"&gt;There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.&lt;br /&gt;&lt;br /&gt;Find the tractors each originally had?&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.&lt;br /&gt;&lt;br /&gt;It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X &amp;amp; Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)&lt;br /&gt;&lt;br /&gt;Similarly, before Y gave tractors to X &amp;amp; Z, they had 6 &amp;amp; 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)&lt;br /&gt;&lt;br /&gt;Again, before X gave tractors to Y &amp;amp; Z, they had 21 &amp;amp; 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)&lt;br /&gt;&lt;br /&gt;Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-9032015695457932652?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/9032015695457932652/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=9032015695457932652' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/9032015695457932652'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/9032015695457932652'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/tractors-puzzle.html' title='Tractors Puzzle'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-4118936702869398987</id><published>2009-08-21T22:22:00.000-07:00</published><updated>2009-08-21T22:22:00.149-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>RelationShip puzzle 2</title><content type='html'>&lt;span style="font-size:85%;"&gt;Puzzle:&lt;br /&gt;&lt;br /&gt;Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not necessarily in the same order.&lt;br /&gt;Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge. No wife partners her husband and Cam does not play bridge.&lt;br /&gt;&lt;br /&gt;Who is married to Cam?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;Carrie is married to Cam.&lt;br /&gt;&lt;br /&gt;"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."&lt;br /&gt;&lt;br /&gt;It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina.&lt;br /&gt;&lt;br /&gt;As Cam does not play bridge, Billy's husband must be Laurie.&lt;br /&gt;&lt;br /&gt;Hence, Carrie is married to Cam.&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-4118936702869398987?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/4118936702869398987/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=4118936702869398987' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/4118936702869398987'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/4118936702869398987'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/relationship-puzzle-2.html' title='RelationShip puzzle 2'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-827809641188390852</id><published>2009-08-20T22:21:00.000-07:00</published><updated>2009-08-20T22:21:00.279-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Relationship problem</title><content type='html'>&lt;span style="font-family:trebuchet ms;font-size:85%;"&gt;Puzzle:&lt;br /&gt;One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and another is a Doctor.&lt;br /&gt; If the Doctor is a male, then the Engineer is a male.&lt;br /&gt; If the Engineer is younger than the Doctor, then the Engineer and the Doctor are not blood relatives.&lt;br /&gt; If the Engineer is a female, then she and the Doctor are blood relatives.&lt;br /&gt;Can you tell who is the Doctor and the Engineer?&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.&lt;br /&gt;&lt;br /&gt;Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.&lt;br /&gt;&lt;br /&gt;Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr. Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.&lt;br /&gt;&lt;br /&gt;Now from 2, Mr. Bajaj's mother can not be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further.&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-827809641188390852?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/827809641188390852/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=827809641188390852' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/827809641188390852'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/827809641188390852'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/relationship-problem.html' title='Relationship problem'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-3966809009548588733</id><published>2009-08-19T22:19:00.000-07:00</published><updated>2009-08-19T22:19:00.327-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Max Number of Edges on Cube</title><content type='html'>&lt;span style="font-size:85%;"&gt;&lt;br /&gt;Puzzle:&lt;br /&gt;If you take a marker &amp;amp; start from a corner on a cube, what is the maximum number of edges you can trace across if you never trace across the same edge twice, never remove the marker from the cube, &amp;amp; never trace anywhere on the cube, except for the corners &amp;amp; edges?&lt;br /&gt;&lt;br /&gt;Answer : 9&lt;br /&gt;&lt;br /&gt;To verify this, you can make a drawing of a cube, &amp;amp; number each of its 12 edges. Then, always starting from 1 corner &amp;amp; 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube.&lt;br /&gt;&lt;br /&gt;There is no need to start from other corners or edges of the cube, as you will only be repeating the same combinations. The process is a little more involved than this, but is useful for solving many types of spatial puzzles.&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-3966809009548588733?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/3966809009548588733/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=3966809009548588733' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3966809009548588733'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/3966809009548588733'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/max-number-of-edges-on-cube.html' title='Max Number of Edges on Cube'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-6511866986195383983</id><published>2009-08-19T04:56:00.000-07:00</published><updated>2009-08-19T04:56:00.503-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Puzzle 3: Distance Covered.</title><content type='html'>&lt;span style="font-size:85%;"&gt;There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.&lt;br /&gt;&lt;br /&gt;In the mean time the whole platoon has moved ahead by 50m.&lt;br /&gt;&lt;br /&gt;The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.&lt;br /&gt;Submitted&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;The last person covered 120.71 meters.&lt;br /&gt;&lt;br /&gt;It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.&lt;br /&gt;&lt;br /&gt;Let's assume that when the last person reached the first person, the platoon moved X meters forward.&lt;br /&gt;Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.&lt;br /&gt;Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.&lt;br /&gt;&lt;br /&gt;Now, as the ratios are equal,&lt;br /&gt;(50+X)/X = X/(50-X)&lt;br /&gt;(50+X)*(50-X) = X*X&lt;br /&gt;&lt;br /&gt;Solving, X=35.355 meters&lt;br /&gt;&lt;br /&gt;Thus, total distance covered by the last person&lt;br /&gt;= (50+X) + X&lt;br /&gt;= 2*X + 50&lt;br /&gt;= 2*(35.355) + 50&lt;br /&gt;= 120.71 meters&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.&lt;br /&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-6511866986195383983?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/6511866986195383983/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=6511866986195383983' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/6511866986195383983'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/6511866986195383983'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/puzzle-3-distance-covered.html' title='Puzzle 3: Distance Covered.'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-7365166091622128627</id><published>2009-08-17T04:52:00.000-07:00</published><updated>2009-08-16T21:22:50.559-07:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Puzzle 1: Bullets problem</title><content type='html'>&lt;span style="font-size:85%;"&gt;Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided.&lt;br /&gt;&lt;br /&gt;Answer: 18&lt;br /&gt;&lt;br /&gt;Explanation:&lt;br /&gt;&lt;br /&gt;Assume that initial there were 3*X bullets.&lt;br /&gt;So they got X bullets each after division.&lt;br /&gt;All of them shot 4 bullets. So now they have (X - 4) bullets each.&lt;br /&gt;But it is given that,after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X&lt;br /&gt;&lt;br /&gt;Therefore, the equation is&lt;br /&gt;3 * (X - 4) = X&lt;br /&gt;3 * X - 12 = X&lt;br /&gt;2 * X = 12&lt;br /&gt;X = 6&lt;br /&gt;Therefore the total bullets before division is = 3 * X = 18&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-7365166091622128627?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/7365166091622128627/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=7365166091622128627' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/7365166091622128627'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/7365166091622128627'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/06/puzzle-1-bullets-problem.html' title='Puzzle 1: Bullets problem'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-2344613873045261952</id><published>2008-10-22T22:31:00.001-07:00</published><updated>2008-12-23T21:42:41.353-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='INFOSYS'/><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>INFOSYS Puzzles</title><content type='html'>&lt;meta equiv="Content-Type" content="text/html; 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  &lt;w:lsdexception locked="false" priority="37" name="Bibliography"&gt;   &lt;w:lsdexception locked="false" priority="39" qformat="true" name="TOC Heading"&gt;  &lt;/w:LatentStyles&gt; &lt;/xml&gt;&lt;![endif]--&gt;&lt;style&gt; &lt;!--  /* Font Definitions */  @font-face 	{font-family:"Cambria Math"; 	panose-1:2 4 5 3 5 4 6 3 2 4; 	mso-font-charset:0; 	mso-generic-font-family:roman; 	mso-font-pitch:variable; 	mso-font-signature:-1610611985 1107304683 0 0 159 0;}  /* Style Definitions */  p.MsoNormal, li.MsoNormal, div.MsoNormal 	{mso-style-unhide:no; 	mso-style-qformat:yes; 	mso-style-parent:""; 	margin:0in; 	margin-bottom:.0001pt; 	mso-pagination:widow-orphan; 	font-size:12.0pt; 	font-family:"Times New Roman","serif"; 	mso-fareast-font-family:"Times New Roman"; 	mso-fareast-language:EN-US;} .MsoChpDefault 	{mso-style-type:export-only; 	mso-default-props:yes; 	font-size:10.0pt; 	mso-ansi-font-size:10.0pt; 	mso-bidi-font-size:10.0pt;} @page Section1 	{size:8.5in 11.0in; 	margin:1.0in 1.25in 1.0in 1.25in; 	mso-header-margin:.5in; 	mso-footer-margin:.5in; 	mso-paper-source:0;} div.Section1 	{page:Section1;} --&gt; &lt;/style&gt;&lt;!--[if gte mso 10]&gt; &lt;style&gt;  /* Style Definitions */  table.MsoNormalTable 	{mso-style-name:"Table Normal"; 	mso-tstyle-rowband-size:0; 	mso-tstyle-colband-size:0; 	mso-style-noshow:yes; 	mso-style-priority:99; 	mso-style-qformat:yes; 	mso-style-parent:""; 	mso-padding-alt:0in 5.4pt 0in 5.4pt; 	mso-para-margin:0in; 	mso-para-margin-bottom:.0001pt; 	mso-pagination:widow-orphan; 	font-size:10.0pt; 	font-family:"Times New Roman","serif";} &lt;/style&gt; &lt;![endif]--&gt;  &lt;p class="MsoNormal" style="text-align: center;" align="center"&gt;INFOSYS &lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;1. I bought a suit and sold it for 10% profit. If i had bought it for 10% less and sold it for 20% profit, i would have sold it for 25np less. Find the actual cost? (4 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;2. x power 1/3 - x power 1/9 = 60. Find the value of x? (4 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;3. Three squares are chosen at random in a chess board. Find the probability that the chosen squares form a diagonal. (6 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;4. There are two stations A and B 99Kms apart. At each other station in between them, there is a sign post indicating the distance from A &amp;amp; B.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;On how many posts, there are only two digits. (E.g. 33, 66) (6 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;5. Allen, Betty, C, Gim, Ohio are a professional dancer, Oracle Professional, Harpic player and one is in waiting.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;He got married. Now, if only the Oracle Professional speaks truth, &lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;Given some conditions:&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;A : B is not a Professional dancer&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;B : &lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;C : G is not Oracle Professional&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;G :&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;O : I will marry&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;Find the professions of each. (6 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;6. Two dice are thrown. The numbers shown on the dice are not added but multiplied.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;Some conditions are given&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;I&lt;span style=""&gt;        &lt;/span&gt; &lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;II&lt;span style=""&gt;      &lt;/span&gt;3 less than first throw&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;III&lt;span style=""&gt;     &lt;/span&gt;2 less than half of I&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;IV&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;V&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;Find the scores of these five throws. (8 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;7. There are two families Smiths and H family. and four children Rohan, Praveen, Praneeth, and Latif.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;Some relation between their ages is given&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;1) Praneeth is the oldest&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;2) Youngest of H family is 2 years less than the youngest of Smiths family.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;3)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;4)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;Find their ages (8 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;8. T, U, V, W, X, Y, Z are the persons available. Out of them, 5 persons should be chosen.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;1) T will not work with X&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;2) If V is there, X should also be there&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;3) If U is there, Y should also be there&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;4)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;5)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;   &lt;/span&gt;Four questions are given on their arrangement (Multiple Choice Questions) (8 Marks)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;Total Marks : 50&lt;/p&gt;  &lt;p class="MsoNormal"&gt;Total Time&lt;span style=""&gt;   &lt;/span&gt;: 60 mins&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;And One Essay. – 15mins&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-2344613873045261952?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/2344613873045261952/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=2344613873045261952' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2344613873045261952'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2344613873045261952'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/10/infosys-puzzles_8171.html' title='INFOSYS Puzzles'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-1082360872996842522</id><published>2008-10-22T22:29:00.000-07:00</published><updated>2008-12-23T21:42:41.353-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='INFOSYS'/><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>INFOSYS  Puzzles</title><content type='html'>INFOSYS TECHNOLOGIES LIMITED.&lt;br /&gt;&lt;br /&gt;PAPER 1:&lt;br /&gt;&lt;br /&gt;Part 1.&lt;br /&gt;&lt;br /&gt;(1) 9 cards are there. u have to arrange them in a 3*3 matrix.&lt;br /&gt;&lt;br /&gt;cards are of 4 colors. they are red, yellow, blue, green.&lt;br /&gt;&lt;br /&gt;conditions for arrangement: one red card must be in first row&lt;br /&gt;&lt;br /&gt;or second row.2 green cards should be in 3rd column. Yellow&lt;br /&gt;&lt;br /&gt;cards must be in the 3 corners only. Two blue cards must be in&lt;br /&gt;&lt;br /&gt;the 2nd row. Atleast one green card in each row.&lt;br /&gt;&lt;br /&gt;Solution:&lt;br /&gt;&lt;br /&gt;Yello Red Gren&lt;br /&gt;&lt;br /&gt;Blu Blu Gren&lt;br /&gt;&lt;br /&gt;Yello Gren Yello&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2. 4 cards are placed on a table, each card has two colors. U&lt;br /&gt;&lt;br /&gt;don't know the color of the back side of eachcard.4 persons A&lt;br /&gt;&lt;br /&gt;B C and D are sitting on the table before the cards. They can&lt;br /&gt;&lt;br /&gt;see Red, Green Red and blue .Out of the 4 poeple 2 always lie.&lt;br /&gt;&lt;br /&gt;They see the color on the reverse side and give the following&lt;br /&gt;&lt;br /&gt;comment&lt;br /&gt;&lt;br /&gt;A: Yello/green&lt;br /&gt;&lt;br /&gt;B: Neither Blue/nor Green&lt;br /&gt;&lt;br /&gt;c: Blue/Yello&lt;br /&gt;&lt;br /&gt;D: Blue/ Yello&lt;br /&gt;&lt;br /&gt;find out the color on the other side of the 4 cards.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3.Red and brown tribes [FROM BARRONS GRE] Conditions to&lt;br /&gt;&lt;br /&gt;get married with each other.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4. Venn diagram regarding Rich, muscular, soft-skinned,&lt;br /&gt;&lt;br /&gt;employed, etc.,( Refer BARRONS GRE GUIDE)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;PART 2.&lt;br /&gt;&lt;br /&gt;1. SAKUNTALA DEVI'S PUZZLE BOOK : PUZZLES TO PUZZLE YOU.&lt;br /&gt;&lt;br /&gt;problem no: 3. ( Brothers and Sisters)&lt;br /&gt;&lt;br /&gt;A family I know has several children. Each boy in this&lt;br /&gt;&lt;br /&gt;family has as many sisters as brothers but each girl has&lt;br /&gt;&lt;br /&gt;twice as many brothers as sisters. How many brothers&lt;br /&gt;&lt;br /&gt;and sisters are there?&lt;br /&gt;&lt;br /&gt;ans: 4 boys and 3 girls.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2. No. of animals is 11 more than the no. of birds. If the&lt;br /&gt;&lt;br /&gt;no. of birds were the no. of animals and no. of animals&lt;br /&gt;&lt;br /&gt;were the no. of birds( ie., interchanging no.s of animals&lt;br /&gt;&lt;br /&gt;and birds.), the total no. of legs get reduced by one fifth&lt;br /&gt;&lt;br /&gt;(1/5). How many no. of birds and animals were there?&lt;br /&gt;&lt;br /&gt;ans: birds:11,animals:22&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3. In a soap company a soap is manufactured with 11 parts.&lt;br /&gt;&lt;br /&gt;For making one soap you will get 1 part as scrap. At the&lt;br /&gt;&lt;br /&gt;end of the day u have 251 such scraps. From that how many&lt;br /&gt;&lt;br /&gt;soaps can be manufactured? ans: 22 + 2+ 1 = 25.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4. 2 * * |&lt;br /&gt;&lt;br /&gt;3 * * | No. 7 does not occur in this&lt;br /&gt;&lt;br /&gt;---------------- |&lt;br /&gt;&lt;br /&gt;5 * * | multiplication.&lt;br /&gt;&lt;br /&gt;* 4 * |&lt;br /&gt;&lt;br /&gt;* * 3 | Find the product.&lt;br /&gt;&lt;br /&gt;---------------- |&lt;br /&gt;&lt;br /&gt;* * * * * |&lt;br /&gt;&lt;br /&gt;---------------- |&lt;br /&gt;&lt;br /&gt;--------------------------------------------------&lt;br /&gt;&lt;br /&gt;ans 2 8 1&lt;br /&gt;&lt;br /&gt;3 2 2&lt;br /&gt;&lt;br /&gt;-----&lt;br /&gt;&lt;br /&gt;5 6 2&lt;br /&gt;&lt;br /&gt;5 6 2 0&lt;br /&gt;&lt;br /&gt;8 4 3 0 0&lt;br /&gt;&lt;br /&gt;---------&lt;br /&gt;&lt;br /&gt;9 0 4 8 2&lt;br /&gt;&lt;br /&gt;---------&lt;br /&gt;&lt;br /&gt;5. There is a 5digit no. 3 pairs of sum is eleven each.&lt;br /&gt;&lt;br /&gt;Last digit is 3 times the first one.&lt;br /&gt;&lt;br /&gt;3 rd digit is 3 less than the second.&lt;br /&gt;&lt;br /&gt;4 th digit is 4 more than the second one.&lt;br /&gt;&lt;br /&gt;Find the digit. ans : 25296.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;6. There are five thieves, each loot a bakery one after the&lt;br /&gt;&lt;br /&gt;other such that the first one takes 1/2 of the total no.&lt;br /&gt;&lt;br /&gt;of the breads plus 1/2 of a bread. Similarly 2nd, 3rd,4th&lt;br /&gt;&lt;br /&gt;and 5fth also did the same. After the fifth one no. of&lt;br /&gt;&lt;br /&gt;breads remained are 3. Initially how many breads were there?&lt;br /&gt;&lt;br /&gt;ans : 31.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;7.ESCALATOR PROBLEM OF SAKUNTALA DEVI 'PUZZLES TO PUZZLE'book.&lt;br /&gt;&lt;br /&gt;Problem No: Problem 27( Down the escalator)&lt;br /&gt;&lt;br /&gt;ans : the no of steps in the stair way : 46.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;8.Harbour line and Main line Problem of Sakuntala Devi Puzzle&lt;br /&gt;&lt;br /&gt;book. Ans : 4/5.&lt;br /&gt;&lt;br /&gt;( More Puzzles book)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;9.There are some chicken in a poultry. They are fed with corn&lt;br /&gt;&lt;br /&gt;One sack of corn will come for 9 days.The farmer decides to&lt;br /&gt;&lt;br /&gt;sell some chicken and wanted to hold 12 chicken with him.&lt;br /&gt;&lt;br /&gt;He cuts the feed by 10% and sack of corn comes for 30 days.&lt;br /&gt;&lt;br /&gt;So initially how many chicken are there?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;10.Two people X &amp;amp; Y walk on the wall of a godown in opposite&lt;br /&gt;&lt;br /&gt;direction. They meet at a point on one side and then go&lt;br /&gt;&lt;br /&gt;ahead. X after walking for some time, walks in opposite&lt;br /&gt;&lt;br /&gt;direction for 15 mtrs.Then again he turns back and walks&lt;br /&gt;&lt;br /&gt;in the original direction. What distance did Y walk before&lt;br /&gt;&lt;br /&gt;they met again, if X walks 11 mtrs by the time Y walks&lt;br /&gt;&lt;br /&gt;8 mtrs.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;11.Problem from SAKUNTALA DEVI 'PUZZLES TO PUZZLE U'.&lt;br /&gt;&lt;br /&gt;Problem no: 23( Walking back to happiness.)&lt;br /&gt;&lt;br /&gt;The walking time : 55 mins.&lt;br /&gt;===========================================================================================&lt;br /&gt;&lt;br /&gt;PAPER 3:&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;INFOSYS - 99 AT JU, CALCUTTA.&lt;br /&gt;&lt;br /&gt;(1). ALL TALLMEN ARE ANDSOME,FAIRSKIMED,LEAN,NULSCULAR&lt;br /&gt;&lt;br /&gt;RICH AND EMPLOYED.&lt;br /&gt;&lt;br /&gt;1. ALL HANDSOME PERSONS ARE FAIRSKIMED&lt;br /&gt;&lt;br /&gt;2. SOME NULSCULAR ARE FAIR SKIMED AND HANDSOME&lt;br /&gt;&lt;br /&gt;3. ALL LEANS ARE NUSCULAR PERSONS&lt;br /&gt;&lt;br /&gt;4. NO FAIRSKIMED PERSON WHO IS NOT HANDSOME IS&lt;br /&gt;&lt;br /&gt;RICH&lt;br /&gt;&lt;br /&gt;5. NEITHER FAIRSKIMED NOR NULSCULAR ARE EMPLOYED.&lt;br /&gt;&lt;br /&gt;THERE ARE 4 QUESTIONS BASED ON THE ABOVE.&lt;br /&gt;&lt;br /&gt;8 MARKS&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;(2). PROBLEM&lt;br /&gt;&lt;br /&gt;(3). A SOFTWARE ENGINEER STARTS FROM HOME AT 3PM FOR&lt;br /&gt;&lt;br /&gt;EVENING WALK.HE WALKSPEED OF 4KMPH ON LEVEL&lt;br /&gt;&lt;br /&gt;GROUND AND THEN AT A SPEED OF 3 KMPH ON THE UPHILL&lt;br /&gt;&lt;br /&gt;AND THEN DOWN THE HILL AT A SPEED OF 6KMPH TO THE&lt;br /&gt;&lt;br /&gt;LEVEL GROUND AND THEN AT A SPEED OF 4KMPH TO THE&lt;br /&gt;&lt;br /&gt;HOME AT 9PM. WHAT IS THE DISTENCE ON ONE WAY?&lt;br /&gt;&lt;br /&gt;4MARKS&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;(4). A BAG CONTAINS CERTAIN NUMBER OF FILES. EACH FILE IS&lt;br /&gt;&lt;br /&gt;NUMBERED WITH ONE DIGIT OF 0 TO 9. SUPPOSE THE&lt;br /&gt;&lt;br /&gt;PERSON WANT TO GET THE NUMBER BETWEEN 1 TO&lt;br /&gt;&lt;br /&gt;2000 (OR 7000 CHECK ) . HOW MANY MIN NO. OF FILES&lt;br /&gt;&lt;br /&gt;SHOULD BE PRESENT IN THE BAG.&lt;br /&gt;&lt;br /&gt;3MARKS&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;(5). A + B + C +D = D + E + F + G = G + H + I =17.&lt;br /&gt;&lt;br /&gt;IF A = 4 WHAT ARE THE VALUES OF D AND G. EACH&lt;br /&gt;&lt;br /&gt;LETTER TAKEN ONLY ONE OF THE DIGIT FROM 1 TO 9.&lt;br /&gt;&lt;br /&gt;8MARKS&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;ANS : A = 4 ,B = 2, C =6, D = 5, E = 3, F = 8,&lt;br /&gt;&lt;br /&gt;G = 1, H = 7, I = 9.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;(6). SIX PERSONS A,B,C,D,E &amp;amp;F WENT TO SOLIDER CINIMA.&lt;br /&gt;&lt;br /&gt;THERE ARE SIX CONSEUTIVE SEATS. A SITS IN THE&lt;br /&gt;&lt;br /&gt;FIRST SEAT FOLLOWED BY B, FOLLOWED BY C AND SOON. IF A TAKEN ONE OF THE SIX SEATS, THEN&lt;br /&gt;&lt;br /&gt;B SHOULD SIT ADJACENT TO A. C SHOULD SIT&lt;br /&gt;&lt;br /&gt;ADJACENT A OR B. D SHOULD SIT ADJACENT TO A, B,ORC AND SOON. HOW MANY&lt;br /&gt;&lt;br /&gt;POSSIBILITIES ARE THERE?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;(7). SUPPOSE THERE ARE 4 GRADES A, B, C, D. (A IS THE BEST&lt;br /&gt;&lt;br /&gt;AND D IS THE WORST) 4 PERSONS JACK,JEAN,POUL,LUCY&lt;br /&gt;&lt;br /&gt;WROTE THEN FINAL EXAM AND MADE THE STATEMENTS&lt;br /&gt;&lt;br /&gt;LIKE THIS.&lt;br /&gt;&lt;br /&gt;1. JACK: IF I WILL GET A THEN LUCY WILL GET D&lt;br /&gt;&lt;br /&gt;LUCY: IF I WILL GET C THEN JACK WILL GET D .&lt;br /&gt;JACK GRADE IN BETTER THEN POUL GRADE.&lt;br /&gt;&lt;br /&gt;JEAN: IF JEAN DOESNOT GET A, THEN LACK WILL NOT&lt;br /&gt;GET A.&lt;br /&gt;&lt;br /&gt;4. POUL: IF JACK GET A , THEN JEAN WILL NOT GET B.&lt;br /&gt;&lt;br /&gt;LUCY WILL GET C, I WON'T EITHER A OR B.&lt;br /&gt;&lt;br /&gt;IF ALL THE ABOVE STATEMENTS ARE TRUE, THEN WHICH PERSON WILL GET WHICH GRADES.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;(8). EACH MAN DANCES WITH 3 WOMEN. EACH WOMEN DANCES WITH&lt;br /&gt;&lt;br /&gt;3 MEN. AMONG EACHG PAIR OF MEN THEY HAVE EXACTLY TWO WOMEN IN COMMAN.&lt;br /&gt;FIND THE NO. OF MEN &amp;amp; WOMEN.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;ESSAY.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1. INTERNET REVOLUTION&lt;br /&gt;&lt;br /&gt;2. ROLE OF MEDIA FOR YOUNG.&lt;br /&gt;&lt;br /&gt;===============================================================================================&lt;br /&gt;PAPER 4:&lt;br /&gt;&lt;br /&gt;THIS IS INFOSYS 99 FROM IIT BHU&lt;br /&gt;&lt;br /&gt;q1)A frog jumps 3 ft comres bacfk 2ft in a day in how many day&lt;br /&gt;&lt;br /&gt;it will come out of 30ft deep well (2 marks )ns ans 28 day&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;q2) A-B=C&lt;br /&gt;&lt;br /&gt;D/E=F&lt;br /&gt;&lt;br /&gt;G+H=I&lt;br /&gt;&lt;br /&gt;C.F=I ANS A=9,B=5,C=4,F=2.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Q3)when the asctual time pass 1hr wall clock ic 10 min behind it&lt;br /&gt;&lt;br /&gt;when 1 hr is shown by wall clock, table clock shows 10 min ahead&lt;br /&gt;&lt;br /&gt;of ih when table clock shows 1 hr the alarm closck goes 5minbehind&lt;br /&gt;&lt;br /&gt;it,when alarm clock goes 1 hr wrist watch is 5 min ahead of it assuming&lt;br /&gt;&lt;br /&gt;that all clocks are correcrt with actual tinme at 12 noon what will&lt;br /&gt;&lt;br /&gt;be time shown by wrist watch after 6 hr&lt;br /&gt;&lt;br /&gt;ans---5:47:32.5&lt;br /&gt;&lt;br /&gt;(n X 60 )50/60 X 70/60 X 55/60 X 65/60&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;q4)a soft. engr just returned from US has eaten too mucgh fat &amp;amp;put&lt;br /&gt;&lt;br /&gt;a lot of weight everfy sunday he starts walking 4 km/hr on level&lt;br /&gt;&lt;br /&gt;ground then up at 3 km\hr then back down hill at 6km\hr then again&lt;br /&gt;&lt;br /&gt;on level grounggd at 4km\hr till he reaches his destination if he&lt;br /&gt;&lt;br /&gt;returned home at 9 p.m. e what distance did he covered&lt;br /&gt;&lt;br /&gt;ans 24km&lt;br /&gt;&lt;br /&gt;it is aslo asked in some parer fingd half of the dist covered by hoim&lt;br /&gt;&lt;br /&gt;in this case ans 12km&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;q5 ] one family promblem was there in wich varios relations are givern&lt;br /&gt;&lt;br /&gt;ans is correcr 3 families&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;q6] one question was about cards 4 persons play yjthe game of cards&lt;br /&gt;&lt;br /&gt;cards are of 4 colour yellow green blue,red fing colours of playing&lt;br /&gt;cards which are hiddeden ans remember the sequence Yellow,yellow,&lt;br /&gt;green,blue,probably frob gre&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;q7] one very easy question just solve two eq, about ages&lt;br /&gt;&lt;br /&gt;i rembemer only ans boy=10yrsor girl =4 yrs or vice versa pl. check it&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;q8]ans the questions from facts 8marks&lt;br /&gt;&lt;br /&gt;the member of certain tribe are divided into 3 casts abhor,dravid magar&lt;br /&gt;&lt;br /&gt;1 an abhor woman cant marry dravid nman&lt;br /&gt;&lt;br /&gt;2 a nmagar man cant marry a dravid man&lt;br /&gt;&lt;br /&gt;3 a son takes the caste of his father a dauhgter takes cast of her mother&lt;br /&gt;&lt;br /&gt;5 all marriages except those mentioned are not permirtted 6 there&lt;br /&gt;&lt;br /&gt;are no childern born out of a wedlock&lt;br /&gt;&lt;br /&gt;1]aliertnates cant rem u acan previos question&lt;br /&gt;&lt;br /&gt;some ans are&lt;br /&gt;&lt;br /&gt;q]an abhor woman a alternative istrue ---can have a dra, grandson&lt;br /&gt;&lt;br /&gt;q]it can be refer from conditions described that a magar man&lt;br /&gt;&lt;br /&gt;ans is 2 &amp;amp;3 only iii choice --cant have a dravid mother &amp;amp; may have abhou mother&lt;br /&gt;&lt;br /&gt;q]an abhor marreies &amp;amp; has thr e childern allof are members aof&lt;br /&gt;&lt;br /&gt;abhor acsateeich of ffoloewing is true&lt;br /&gt;&lt;br /&gt;ans i &amp;amp; ii only --i is if c were married toa magar man&lt;br /&gt;&lt;br /&gt;ii-- if c were married to abhor man,they have a male child&lt;br /&gt;&lt;br /&gt;one more q] there are 2 scales of temp A &amp;amp; B it was given A 14 to 133&lt;br /&gt;&lt;br /&gt;B 36 to 87 find the temp when temp oa a is eq. to temp of b&lt;br /&gt;&lt;br /&gt;ans 52.5&lt;br /&gt;&lt;br /&gt;let t=mx+c c=-70 m=51/119&lt;br /&gt;&lt;br /&gt;a=a.51/119 -70 ; a=52.5&lt;br /&gt;=============================================================================================&lt;br /&gt;PAPER 5:&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1. There are 4 married couples out ofwhich 3 a group&lt;br /&gt;&lt;br /&gt;is needed . But her should not be his of her spouse .How nmany&lt;br /&gt;&lt;br /&gt;groups are possible ? Ans 32&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2.In the 4 digits 1,2,3,4 how many 4 digited numbers are&lt;br /&gt;&lt;br /&gt;possible which are divisable by 4? Repetetions are allowed&lt;br /&gt;&lt;br /&gt;Ans 64&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3. Two men are goingalong a trackf rail in the opposite&lt;br /&gt;&lt;br /&gt;direction.One goods train crossed the first person in 20 sec.&lt;br /&gt;&lt;br /&gt;After 10 min the train crossed the other person who is commingin&lt;br /&gt;&lt;br /&gt;opposite directionin 18 sec .Afterthe train haspassed, when thetwo&lt;br /&gt;&lt;br /&gt;persons will meet? Approx 72min check it once.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4. Theno. of children adults . Theno .of adults the&lt;br /&gt;&lt;br /&gt;no .of boys .The no.of boys no. of girls .The no.of girls no.of&lt;br /&gt;&lt;br /&gt;family conditions&lt;br /&gt;&lt;br /&gt;1.No family is without a child&lt;br /&gt;&lt;br /&gt;2 Every girl has atleast one brotherand sister .&lt;br /&gt;&lt;br /&gt;Ans c&gt;a&gt;b&gt;g&gt;f; 9 6 5 4 3 .&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;6.There are4 boys Anand ,Anandya ,Madan and Murali with&lt;br /&gt;&lt;br /&gt;nicmnames perich ,zomie ,drummy and madeena not in thesame&lt;br /&gt;&lt;br /&gt;order Some com=nditons&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Ans Anand : Perich&lt;br /&gt;&lt;br /&gt;Anandya: drummy&lt;br /&gt;&lt;br /&gt;Madan : Zombie&lt;br /&gt;&lt;br /&gt;murali: Madeena&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;7.Thereare2diomans ,1 spadeand1 club and 1ace and also 1king ,1&lt;br /&gt;&lt;br /&gt;jack and 1 aceare arranged in a straight line&lt;br /&gt;&lt;br /&gt;1.The king is at third place&lt;br /&gt;&lt;br /&gt;2.Theleft of jack is a heart and itsright is king&lt;br /&gt;&lt;br /&gt;3. No two red colours arein consequtive.&lt;br /&gt;&lt;br /&gt;4.The queensareseperated by two cards. Write the orderor which&lt;br /&gt;&lt;br /&gt;suits&lt;br /&gt;&lt;br /&gt;(hearts ,clubs )and names(jacks queensetc.)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;8. Writeeach statementis true or false 8M&lt;br /&gt;&lt;br /&gt;1.The sum of the1st three statements and the2nd false statement&lt;br /&gt;&lt;br /&gt;givesthe true statement.&lt;br /&gt;&lt;br /&gt;2.The no.oftrue statements &gt;falsestatement&lt;br /&gt;&lt;br /&gt;3. The sum of2nd true statement and 1st falsestatement gives&lt;br /&gt;&lt;br /&gt;the first&lt;br /&gt;&lt;br /&gt;true statement.&lt;br /&gt;&lt;br /&gt;4. Thereareatmost 3 falsestatements&lt;br /&gt;&lt;br /&gt;5.There is no two consequtive true statements&lt;br /&gt;&lt;br /&gt;6.If this containsonly 1-5 statements ,theanswer of this is&lt;br /&gt;&lt;br /&gt;sameasthe an answer of the following question&lt;br /&gt;&lt;br /&gt;9.Question on Venn diagram.&lt;br /&gt;&lt;br /&gt;All handsome are also fair skinned&lt;br /&gt;&lt;br /&gt;Sme musularsare fair skinned&lt;br /&gt;&lt;br /&gt;Some musculars are also handsome&lt;br /&gt;&lt;br /&gt;All lean are also muscular&lt;br /&gt;&lt;br /&gt;Some lean are also fair skinned.&lt;br /&gt;&lt;br /&gt;All rich man inot fair skinned but all rich manare handsome&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;10 Some questions follows.&lt;br /&gt;&lt;br /&gt;There are 3 pileseach containe 10 15 20 stones. There are&lt;br /&gt;&lt;br /&gt;A,B,C,D,F,G and h persons .One man can catch upto four stones&lt;br /&gt;from any pile.The last manwho takeswill win. If first A starts&lt;br /&gt; next B. and so on who will win?&lt;br /&gt;&lt;br /&gt;Ans May be F&lt;br /&gt;&lt;br /&gt;Essay writing&lt;br /&gt;&lt;br /&gt;1 Intrnet revolution&lt;br /&gt;&lt;br /&gt;2.Media for youth&lt;br /&gt;=============================================================================================&lt;br /&gt;&lt;br /&gt;PAPER 6:&lt;br /&gt;&lt;br /&gt;MARKS: 50&lt;br /&gt;&lt;br /&gt;I. Rohith can catch either the harbour line train or main line&lt;br /&gt;train from his nearest railway station kandala to reach his office.&lt;br /&gt;Both the trains have a frequency of 10 min. In the morning office Hours.&lt;br /&gt; The harbour line trains arrive evary 10min. Starting from 5:00a.m&lt;br /&gt;and the main line trains arrive at every 10min. Starting from 5:02a.m&lt;br /&gt;&lt;br /&gt;ii. In a certain department store the position of&lt;br /&gt;&lt;br /&gt;Buyer,cashier,clerk, floorwalkar, &amp;amp; manager are held, though not necessarily&lt;br /&gt; Respectively ,by m$ the cashier &amp;amp; the manager were roommates in college.&lt;br /&gt;the buyer is bachelor, evans &amp;amp; miss ames have only business contacts&lt;br /&gt;with each other mrs conroy was greatly dosappointed when her&lt;br /&gt; husband told her That the manager had refuge to give him a&lt;br /&gt;raise. davis is going to be the best man when the clerk&lt;br /&gt; &amp;amp; the cashier Are married .&lt;br /&gt;&lt;br /&gt;what position does each person held.?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;iii. tanveer and kunal went down a descending escalator .&lt;br /&gt;The escalator Was going down at a contant speed . Kunal&lt;br /&gt;walked down the steps and Had to cover 50 steps to reach&lt;br /&gt;the bottom.tanveer on the other hand , ran Down the steps&lt;br /&gt;of the escalator and hence had to take 75 steps to reach&lt;br /&gt;The bottom . Actually , tanveer took 3 steps in the same&lt;br /&gt;time it took Kunal to take 1 step . How many steps would&lt;br /&gt; be visible when the ecalator Is stopped and not operating.?&lt;br /&gt;&lt;br /&gt;iv. In a four team foot-ball tournament , all the teams played each&lt;br /&gt;&lt;br /&gt;Other in three rounds of matches are shown in the table -a . Some of&lt;br /&gt;&lt;br /&gt;The results of the tournament are shown in the table _b . Using the&lt;br /&gt;&lt;br /&gt;Clues given below, please fill -in the blank columns in the result&lt;br /&gt;&lt;br /&gt;Table-b (goals for &amp;amp; goals for against ? ).&lt;br /&gt;&lt;br /&gt;note : two points for win , one point for draw &amp;amp; zero points for&lt;br /&gt;&lt;br /&gt;Defeat are awarded .&lt;br /&gt;&lt;br /&gt;clues :&lt;br /&gt;&lt;br /&gt;East zone won the tournament despite scoring one less goal than The runners -up.&lt;br /&gt;2. North zone scored an odd number of goals in their first round Game&lt;br /&gt;3. South zone , who failed to score in their final match , were&lt;br /&gt;&lt;br /&gt;Beaten by a two -goal margin in the first round.&lt;br /&gt;&lt;br /&gt;4. East zone lost their match aginst west zone.&lt;br /&gt;&lt;br /&gt;5. All four teams scored goals in the second round matches&lt;br /&gt;&lt;br /&gt;6. West zone scored the same number of goals against east&lt;br /&gt; zone as North zone scored aginst them.&lt;br /&gt;&lt;br /&gt;7. East zone scored four goals in round two match.&lt;br /&gt;&lt;br /&gt;table - a ( matches played )&lt;br /&gt;&lt;br /&gt;round 1 north zone vs south zone&lt;br /&gt;&lt;br /&gt;west zone vs east zone.&lt;br /&gt;&lt;br /&gt;round 2 south zone vs west zone&lt;br /&gt;&lt;br /&gt;east zone vs north zone&lt;br /&gt;&lt;br /&gt;round 3 south zone vs east zone&lt;br /&gt;&lt;br /&gt;west zone vs north zone&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;table - b ( results&lt;br /&gt;&lt;br /&gt;played won draw lost golas for goals&lt;br /&gt;&lt;br /&gt;Points&lt;br /&gt;&lt;br /&gt;against&lt;br /&gt;&lt;br /&gt;$&lt;br /&gt;&lt;br /&gt;east zone 3 - - - ? 3&lt;br /&gt;&lt;br /&gt;4&lt;br /&gt;&lt;br /&gt;north zone 3 - - - ? ?&lt;br /&gt;&lt;br /&gt;4&lt;br /&gt;&lt;br /&gt;west zone 3 - - - 4 3&lt;br /&gt;&lt;br /&gt;3&lt;br /&gt;&lt;br /&gt;south zone 3 - - - 2 5&lt;br /&gt;&lt;br /&gt;1&lt;br /&gt;&lt;br /&gt;v. at six o'clock the wall clock struck 6 times . Checking with my&lt;br /&gt;&lt;br /&gt;Watch , i noticed that the time between the first &amp;amp; last strokes was 30&lt;br /&gt;&lt;br /&gt;Seconds. How long will the clock take to stike 12 at mid night (ans 66&lt;br /&gt;&lt;br /&gt;Seconds)&lt;br /&gt;&lt;br /&gt;vi. Plese answer the q's with reference to the facts given below ( 8&lt;br /&gt;&lt;br /&gt;Marks )&lt;br /&gt;&lt;br /&gt;1 . The members of a certain tribe are divided into three castes&lt;br /&gt;&lt;br /&gt;-abhor , dravid , &amp;amp; magar castes&lt;br /&gt;&lt;br /&gt;2. An abhor women cannot marry a dravid man&lt;br /&gt;&lt;br /&gt;3. A magar man cannot marry a dravid women.&lt;br /&gt;&lt;br /&gt;4. A son takes the caste of his father , a daughter takes the&lt;br /&gt;&lt;br /&gt;Caste of&lt;br /&gt;&lt;br /&gt;her mother .&lt;br /&gt;&lt;br /&gt;5. All marriages except those mentiond above are permitted&lt;br /&gt;&lt;br /&gt;6 . There are no children born out of wed lock.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;q's&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1) an abhor woman marries and has three children , all of whom are Member$&lt;br /&gt;&lt;br /&gt;i) if she were to be married to a magar man , they have&lt;br /&gt;&lt;br /&gt;No male$&lt;br /&gt;&lt;br /&gt;ii) if she were to be married to abhor man ,they may have&lt;br /&gt;&lt;br /&gt;Male chi$&lt;br /&gt;&lt;br /&gt;iii) if her fourth child is male , he may be a dravid,.&lt;br /&gt;&lt;br /&gt;a) i only&lt;br /&gt;&lt;br /&gt;b) iii only&lt;br /&gt;&lt;br /&gt;c) i &amp;amp; ii only&lt;br /&gt;&lt;br /&gt;d) ii &amp;amp; iii only&lt;br /&gt;&lt;br /&gt;e) i , ii &amp;amp; iii&lt;br /&gt;&lt;br /&gt;2) it can be inferred from the condition described that a magar man&lt;br /&gt;&lt;br /&gt;i) cannot have a dravid grand mother&lt;br /&gt;&lt;br /&gt;ii) cannot have a dravid mother&lt;br /&gt;&lt;br /&gt;iii ) may have a abhor mother&lt;br /&gt;&lt;br /&gt;a) i only&lt;br /&gt;&lt;br /&gt;b) ii only&lt;br /&gt;&lt;br /&gt;c) i &amp;amp; ii only&lt;br /&gt;&lt;br /&gt;d) ii &amp;amp; iii only&lt;br /&gt;&lt;br /&gt;e) i , ii &amp;amp; iii&lt;br /&gt;&lt;br /&gt;3) when two dravids are married , it is impossible for them to have&lt;br /&gt;&lt;br /&gt;i) any female abhor descendents&lt;br /&gt;&lt;br /&gt;ii ) an abhor great_ grand daughter \&lt;br /&gt;&lt;br /&gt;iii) an abhor grand daughter&lt;br /&gt;&lt;br /&gt;a)i only&lt;br /&gt;&lt;br /&gt;b) iii only&lt;br /&gt;&lt;br /&gt;c) i &amp;amp; ii only&lt;br /&gt;&lt;br /&gt;d) ii &amp;amp; iii only&lt;br /&gt;&lt;br /&gt;e) i , ii &amp;amp; iii&lt;br /&gt;&lt;br /&gt;4) an abhor woman&lt;br /&gt;&lt;br /&gt;a)can have a dravid grandson&lt;br /&gt;&lt;br /&gt;b) cannot have a magar son&lt;br /&gt;&lt;br /&gt;c) cannot have a magar grand son&lt;br /&gt;&lt;br /&gt;d) can have a magar daughter -in - law&lt;br /&gt;&lt;br /&gt;e) cannot have a dravid grand daughter&lt;br /&gt;&lt;br /&gt;vii. ( 2 marks )&lt;br /&gt;&lt;br /&gt;in certain community there are thousand married couples .&lt;br /&gt;&lt;br /&gt;Two&lt;br /&gt;&lt;br /&gt;thirds of the husbands who are taller than their wives are also&lt;br /&gt;&lt;br /&gt;Heavier&lt;br /&gt;&lt;br /&gt;and three quarters of the husbands who are heavier than their&lt;br /&gt; wives Are also taller . If there are 120 wives who are taller&lt;br /&gt;and heavier&lt;br /&gt;&lt;br /&gt;Than their husbands, how many husbands are taller and heavier than their wives ?&lt;br /&gt;&lt;br /&gt;viii. Both the gupthas and sinhas have two young sons whose ages&lt;br /&gt; are Under eleven. The names of the boys, whose ages roundedoff to&lt;br /&gt;&lt;br /&gt;The nearest year are all different, are rajesh, praveen,lalith and&lt;br /&gt;&lt;br /&gt;Prathap. taking the ages of the ages of the boys only to the nearest&lt;br /&gt;&lt;br /&gt;Year, the following statements are true.&lt;br /&gt;&lt;br /&gt;rajesh is three years younger than his brother is.&lt;br /&gt;&lt;br /&gt;praveen is the oldest.&lt;br /&gt;&lt;br /&gt;prathap is 5yrs older than the younger sinhas boy.&lt;br /&gt;&lt;br /&gt;lalith is half as old as one of the guptha boys.&lt;br /&gt;&lt;br /&gt;the total ages of the boys in each family differ by the&lt;br /&gt;&lt;br /&gt;Same&lt;br /&gt;&lt;br /&gt;amount today as they did five years ago. --- 7 marks.&lt;br /&gt;&lt;br /&gt;ix. A long division sum ---- 7 marks&lt;br /&gt;&lt;br /&gt;xx)xxxxxxxxx(xxxxxxx&lt;br /&gt;&lt;br /&gt;xx&lt;br /&gt;&lt;br /&gt;---&lt;br /&gt;&lt;br /&gt;xxx&lt;br /&gt;&lt;br /&gt;xx&lt;br /&gt;&lt;br /&gt;----&lt;br /&gt;&lt;br /&gt;xxx&lt;br /&gt;&lt;br /&gt;xx&lt;br /&gt;&lt;br /&gt;---&lt;br /&gt;&lt;br /&gt;xx&lt;br /&gt;&lt;br /&gt;xx&lt;br /&gt;&lt;br /&gt;--&lt;br /&gt;&lt;br /&gt;xxx&lt;br /&gt;&lt;br /&gt;xxx&lt;br /&gt;&lt;br /&gt;-----&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;in the complete solution there are four 5's.&lt;br /&gt;&lt;br /&gt;find the missing digits.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;x. Following services are operated by asian airlines between the two&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;are located in different countries with different time zones.&lt;br /&gt;&lt;br /&gt;As is is normally done, the time shown is the local time- viz ist&lt;br /&gt;&lt;br /&gt;&amp;amp; tst. regular flight supersonicflig&lt;br /&gt;&lt;br /&gt;arrive alexandria 17:10 tst 15:40 tst&lt;br /&gt;&lt;br /&gt;depart " " 20:50 tst 22:50 tst&lt;br /&gt;&lt;br /&gt;arrive rampur 23:40 ist ? Ist&lt;br /&gt;&lt;br /&gt;the arrival time of supersonic flight into rampur from&lt;br /&gt;&lt;br /&gt;Alexandr$ arrival time of the super sonic flight assuming each&lt;br /&gt;&lt;br /&gt;Service- regular and supersonic maintains its own constant speed&lt;br /&gt;&lt;br /&gt;Of flig$&lt;br /&gt;&lt;br /&gt;don't answer for all the puzzules. 8 are the safe one.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;PAPER 7:&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1)A,B,C,D,E related.4 of them made these statements each.&lt;br /&gt;&lt;br /&gt;i)C is my son in law's brother.&lt;br /&gt;&lt;br /&gt;ii)B is my father's brother.&lt;br /&gt;&lt;br /&gt;iii)E is my mother in law.&lt;br /&gt;&lt;br /&gt;iv)A is my brother's wife.&lt;br /&gt;&lt;br /&gt;who made these statements?(person mentioned is one of A,B,C,D,E)(10 mar ks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2)e means belong.&lt;br /&gt;&lt;br /&gt;All members of E e D.&lt;br /&gt;&lt;br /&gt;All members of D e A.&lt;br /&gt;&lt;br /&gt;Not all members of D e E.&lt;br /&gt;&lt;br /&gt;Not all members of A e D.&lt;br /&gt;&lt;br /&gt;All members of C e both A and B.some questions are asked about&lt;br /&gt;relatio n.use venn diagram.(5 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3)complete the table.&lt;br /&gt;&lt;br /&gt;Played won lost draw goals goals&lt;br /&gt;&lt;br /&gt;for against&lt;br /&gt;&lt;br /&gt;A 2 2 1&lt;br /&gt;&lt;br /&gt;B 2 1 2 4&lt;br /&gt;&lt;br /&gt;C 2 3 7&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;A,B,C are 3 hockey teams.(2 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4) A says Party was held on :Thursday ,May 8th.&lt;br /&gt;&lt;br /&gt;B says Party was held on :Tuesday,May 10th.&lt;br /&gt;&lt;br /&gt;C says party was held on :Friday ,June 8th.&lt;br /&gt;&lt;br /&gt;Given April 1 st was Tuesday.one of A,B,C says 1 correct.one says 1&lt;br /&gt;&lt;br /&gt;wrong.and one was completely wrong of date,Month and day.&lt;br /&gt;Find the Day the party held. (5marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;5) A ship is away from the shore by 180 miles.A plane is travelling at&lt;br /&gt;&lt;br /&gt;10 times speed of the ship.How long from the shore will they meet?&lt;br /&gt;&lt;br /&gt;(2marks)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;6) Every station in N railroad issue everyother station's ticket.&lt;br /&gt;&lt;br /&gt;some stations are added.Now they have to issue 46 more tickets.&lt;br /&gt;&lt;br /&gt;say the No.of stations after and before added.(5 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;7) 3 persons say these statements.&lt;br /&gt;&lt;br /&gt;A says either Democratic or liberal wins the elections.&lt;br /&gt;&lt;br /&gt;B says Democratic wins.C says neither democratic nor liberal wins&lt;br /&gt;&lt;br /&gt;the election.of these only one is wrong.who wins the election?&lt;br /&gt;&lt;br /&gt;(5 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;8) A clock showing 6 o'clock takes 30 secs to strike 6 times.How long&lt;br /&gt;&lt;br /&gt;will it take to strike 12 at midnight?Ans.66 secs.(2marks)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;9) Only boys aged &gt; 16 wear coats.&lt;br /&gt;&lt;br /&gt;Boys aged &gt; 15 go to watch football.some more statements are given.&lt;br /&gt;&lt;br /&gt;What can be said about those who are watching football ? (age and&lt;br /&gt;&lt;br /&gt;costume)&lt;br /&gt;&lt;br /&gt;(5 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;10) There are 3 societies A,B,C having some tractors each.&lt;br /&gt;&lt;br /&gt;A Gives B and C as many tractors as they already have.&lt;br /&gt;&lt;br /&gt;After some days B gives A and C as many tractors as they have.&lt;br /&gt;&lt;br /&gt;After some days C gives A and B as many tractors as they have.&lt;br /&gt;&lt;br /&gt;Finally each has 24 tractors.what is the original No.of&lt;br /&gt;&lt;br /&gt;tractors each had in the beginning?&lt;br /&gt;&lt;br /&gt;Ans.A -39.&lt;br /&gt;&lt;br /&gt;B- 21.&lt;br /&gt;&lt;br /&gt;C- 12.(7 marks).&lt;br /&gt;&lt;br /&gt;11) 4,5 statements.From that find the answer.(7 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;PAPER 8:&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1a) BE * BE = ACB&lt;br /&gt;&lt;br /&gt;A,B,C,E ARE NON ZERO NUMBERS FIND B,E.&lt;br /&gt;&lt;br /&gt;ANS) B=1 E=9&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2) A,B,C,D,E ARE HAVING NUMERICAL VALUES. THERE ARE SOME CONDITIONS GIVEN a) A=C &lt;===&gt; B!=E&lt;br /&gt;&lt;br /&gt;b) DIFFERENCE BETWEEN A AND C AS SAME AS DIFFERENCE BETWEEN C&lt;br /&gt;AND B AS SAME AS DIFFERENCE BETWEEN A AND D&lt;br /&gt;&lt;br /&gt;c) C&lt;a&gt;D&lt;br /&gt;&lt;br /&gt;THEN FIND A,B,C,D,E&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3) THERE ARE SIX CARDS IN WHICH IT HAS TWO KING CARDS. ALL CARDS ARE&lt;br /&gt; TURNED DOWN AND TWO CARDS ARE OPENED&lt;br /&gt;&lt;br /&gt;a) WHAT IS THE POSSOBILITY TO GET AT LEAST ONE KING.&lt;br /&gt;&lt;br /&gt;b) WHAT IS THE POSSIBILITY TO GET TWO KINGS.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4) A PERSON WENT TO A SHOP AND ASKED FOR CHANGE FOR 1.15PAISE.&lt;br /&gt;BUT HE SAID THAT HE COULD NOT ONLY GIVE CHANGE FOR ONE RUPEE.&lt;br /&gt;BUT ALSO FOR 50P,25P,10P AND 5P. WHAT WERE THE COINS HE HAD?&lt;br /&gt; ans) 1--&gt;50 4---&gt;10P 1---&gt;25P&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;5) THERE ARE 3 NURSES AND THEY WORK ALTOGETHER ONLY ONCE IN A&lt;br /&gt; WEEK.NO NURSE IS CALLED TO WORK FOR 3 CONSECUTIVE DAYS.&lt;br /&gt;NURSE 1 IS OFF ON TUESEDAY,THURSDAY AND SUNDAY.&lt;br /&gt;&lt;br /&gt;NURSE 2 IS OFF ON SATURDAY.&lt;br /&gt;&lt;br /&gt;NURSE 3 IS OFF ON THURSDAY,SUNDAY.&lt;br /&gt;&lt;br /&gt;NO TWO NURSES ARE OFF MORE THAN ONCE A WEEK.&lt;br /&gt;&lt;br /&gt;FIND THE DAY ON WHICH ALL THE 3 NURSES WERE ON WORK.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;6) THERE ARE 5 PERSONS A,B,C,D,E AND EACH IS WEARING A BLOCK&lt;br /&gt;OR WHITE CAP ON HIS HEAD. A PERSON CAN SEE THE CAPS OF&lt;br /&gt;THE REMAINING 4 BUT CAN'T SEE HIS OWN CAP. A PERSON WEARING&lt;br /&gt; WHITE SAYS TRUE AND WHO WEARS BLOCK SAYS FALSE.&lt;br /&gt;&lt;br /&gt;i) A SAYS I SEE 3 WHITES AND 1 BLOCK&lt;br /&gt;&lt;br /&gt;ii) B SAYS I SEE 4 BLOCKS&lt;br /&gt;&lt;br /&gt;iii) E SAYS I SEE 4 WHITES&lt;br /&gt;&lt;br /&gt;iiii) C SAYS I SEE 3 BLOCKS AND 1 WHITE.&lt;br /&gt;&lt;br /&gt;NOW FIND THE CAPS WEARED BY A,B,C,D AND E&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;7) THERE ARE TWO WOMEN, KAVITHA AND SHAMILI AND TWO MALES SHYAM,&lt;br /&gt;ARAVIND WHO ARE MUSICIANS. OUT OF THESE FOUR ONE IS A PIANIST,&lt;br /&gt; ONE FLUTIST, VIOLINIST AND DRUMMER.&lt;br /&gt;&lt;br /&gt;i) ACROSS ARAVIND BEATS PIANIST&lt;br /&gt;&lt;br /&gt;ii) ACROSS SHYAM IS NOT A FLUTIST&lt;br /&gt;&lt;br /&gt;iii) KAVITHA'S LEFT IS A PIANIST&lt;br /&gt;&lt;br /&gt;iiii) SHAMILI'S LEFT IS NOT A DRUMMER&lt;br /&gt;&lt;br /&gt;FLUTIST AND DRUMMER ARE MARRIED.&lt;br /&gt;&lt;br /&gt;8) 1/3 ED OF THE CONTENTS OF A CONTAINER EVAPORATED ON THE&lt;br /&gt;1 ST DAY.3/4 TH OF THE REMAINING CONTENTS OF THE CONTAINER&lt;br /&gt;EVAPORATED THE SECOND DAY. WHAT PART OF THE CONTENTS OF&lt;br /&gt; THE CONTAINER ARE LEFT AT THE END OF THE SECOND DAY.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;9) A MAN COVERED 28 STEPS IN 30 SECONDS BUT HE DECIDED&lt;br /&gt;TO MOVE FAST AND COVERED 34 STEPS IN 18 SECONDS.&lt;br /&gt;HOW MANY STEPS ARE THERE ON THE ESCALATOR WHEN STATIONARY.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;10) ALL FAIR SKINNED, RICH, HANDSOME, MUSCULAR, LEAN AND EMPLOYED ARE TALL MEN&lt;br /&gt;&lt;br /&gt;1) ALL LEAN MEN ARE MUSCULAR.&lt;br /&gt;&lt;br /&gt;2) NO FAIRSKINNED PERSON WHO IS NOT RICH IS HANDSOME.&lt;br /&gt;&lt;br /&gt;3) SOME MUSCULAR MEN ARE HANDSOME.&lt;br /&gt;&lt;br /&gt;4) ALL HANDSOME ARE FAIRSKINNED.&lt;br /&gt;&lt;br /&gt;5) NO PERSON WHO IS NEITHER FAIR SKINNED NOR MUSCULAR IS ENPLYED.&lt;br /&gt;&lt;br /&gt;6) we unable to recall this condition and question also incomplete.&lt;br /&gt;&lt;br /&gt;------------------------------------------------------------------------&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;PAPER 9:&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;This paper contains INFOSYS interview questions.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1&gt; S: Satch J:Jute ANS&lt;br /&gt;&lt;br /&gt;person1: Most of us are satch J&lt;br /&gt;&lt;br /&gt;person2:Most of us are jute S&lt;br /&gt;&lt;br /&gt;person3:Two of us are satch J&lt;br /&gt;&lt;br /&gt;person4: Three of us are jute J&lt;br /&gt;&lt;br /&gt;person5:I am satch J&lt;br /&gt;&lt;br /&gt;we have to find who is satch and who is jute.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2&gt; Four persons are three to cross a bridge they have one torch light&lt;br /&gt;&lt;br /&gt;one person A can cross in 1 min&lt;br /&gt;&lt;br /&gt;person B can cross in 2 min&lt;br /&gt;&lt;br /&gt;person C can cross in 5 min&lt;br /&gt;&lt;br /&gt;person D can cross in 10 min&lt;br /&gt;&lt;br /&gt;They have to cross bridge with in 17 min.&lt;br /&gt;&lt;br /&gt;At a time only two persons can cross&lt;br /&gt;&lt;br /&gt;I A&amp;amp;B ----&gt; 2&lt;br /&gt;&lt;br /&gt;II A &lt;----1&lt;br /&gt;&lt;br /&gt;III C&amp;amp;D ----&gt;10&lt;br /&gt;&lt;br /&gt;IV B &lt;----2&lt;br /&gt;&lt;br /&gt;V A&amp;amp;B ----&gt;2&lt;br /&gt;&lt;br /&gt;__________&lt;br /&gt;&lt;br /&gt;Total 17&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3&gt;7 hours hour glass&lt;br /&gt;&lt;br /&gt;11 hours hour glass&lt;br /&gt;&lt;br /&gt;you have to boil one egg for 15 hours&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4&gt; What is the maximum number of slices can you obtain by cutting&lt;br /&gt;&lt;br /&gt;a cake with only 4 cuts?&lt;br /&gt;&lt;br /&gt;ANS: 16&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;5&gt; Three are three boxes ,&lt;br /&gt;&lt;br /&gt;In one box Two white balls&lt;br /&gt;&lt;br /&gt;In two box 2 black balls&lt;br /&gt;&lt;br /&gt;In three box 1 white &amp;amp;1 black&lt;br /&gt;&lt;br /&gt;The lables on the boxes are not correct.Then you have to open&lt;br /&gt;&lt;br /&gt;one box and to find the colour of the balls in all boxes.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Solution: Open the box labled black&amp;amp; white&lt;br /&gt;&lt;br /&gt;If white balls are there then the box labled with white balls&lt;br /&gt;&lt;br /&gt;contain black balls and labled with black balls contain one&lt;br /&gt;&lt;br /&gt;black and one white ball and viceversa if two black ballsare there.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;PAPER 10:&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1)A begger collects cigarette stubs and makes&lt;br /&gt; one ful cigarette with every 7 stubs. Once he gets 49 stubs .&lt;br /&gt;How many cigarettes can he smoke totally.?&lt;br /&gt;&lt;br /&gt;Ans. 8&lt;br /&gt;&lt;br /&gt;2). A soldiar looses his way in a thick jungle at random walks&lt;br /&gt; from his camp but mathematically in an interestingg fashion.&lt;br /&gt;First he walks one mile east then half mile to north.&lt;br /&gt;Then 1/4 mile to west, then 1/8 mile to south and so&lt;br /&gt; on making a loop. Finally hoe far he is from his camp&lt;br /&gt;and in which direction. ans: in north and south directions&lt;br /&gt;&lt;br /&gt;1/2 - 1/8 + 1/32 - 1/128 + 1/512 - and so on&lt;br /&gt;&lt;br /&gt;= 1/2/((1-(-1/4))&lt;br /&gt;&lt;br /&gt;similarly in east and west directions&lt;br /&gt;&lt;br /&gt;1- 1/4 + 1/16 - 1/64 + 1/256 - and so on&lt;br /&gt;&lt;br /&gt;= 1/(( 1- ( - 1/4))&lt;br /&gt;&lt;br /&gt;add both the answers&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3). how 1000000000 can be written as a product of two factors&lt;br /&gt; neither of them containing zeros&lt;br /&gt;&lt;br /&gt;Ans 2 power 9 x 5 ppower 9 ( check the answer )&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4). Conversation between two mathematcians:&lt;br /&gt;&lt;br /&gt;first : I have three childern. Thew pproduct of their ages&lt;br /&gt; is 36. If you sum their ages . it is exactly same as&lt;br /&gt; my neighbour's door number on my left. The sacond&lt;br /&gt;mathematiciaan verfies the door number and says that&lt;br /&gt;the not sufficient. Then the first says " o.k one more&lt;br /&gt;clue is that my youngest is the youngest" Immmediately&lt;br /&gt; the second mathematician answers . Can you aanswer&lt;br /&gt;&lt;br /&gt;the questoion asked by the first mathematician?&lt;br /&gt;&lt;br /&gt;What are the childeren ages? ans 2 and 3 and 6&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;5). Light glows for every 13 seconds . How many times did it between 1:57:58 and 3:20:47 am&lt;br /&gt;&lt;br /&gt;ans : 383 + 1 = 384&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;6). 500 men are arranged in an array of 10 rows and 50 columns.&lt;br /&gt;ALL tallest among each row aare asked to fall out.&lt;br /&gt; And the shortest among THEM is A. Similarly after&lt;br /&gt;resuming that to their originaal podsitions that the&lt;br /&gt;shorteest among each column are asked to fall out.&lt;br /&gt;&lt;br /&gt;And the longest among them is B . Now who is&lt;br /&gt;&lt;br /&gt;taller among A and B ? ans A&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;7). A person spending out 1/3 for cloths , 1/5 of the remsaining&lt;br /&gt; for food and 1/4 of the remaining for travelles is&lt;br /&gt; left with Rs 100/- . How he had in the begining ?&lt;br /&gt;&lt;br /&gt;ans RS 250/-&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;8). there are six boxes containing 5 , 7 , 14 , 16 ,18, 29&lt;br /&gt;balls of either red or blue in colour. Some boxes&lt;br /&gt; contain only red balls and others contain only blue .&lt;br /&gt;One sales man sold one box out of them and then he says "&lt;br /&gt;I have the same number of red balls left out as that of blue ".&lt;br /&gt;Which box is the one he solds&lt;br /&gt;&lt;br /&gt;out ? Ans : total no of balls = 89 and (89-29 /2 = 60/2 = 30&lt;br /&gt;&lt;br /&gt;and also 14 + 16 = 5 + 7 + 18 = 30&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;9) A chain is broken into three pieces of equal lenths&lt;br /&gt;conttaining 3 links each. It is taken to a backsmith to&lt;br /&gt;join into a single continuous one . How many links are&lt;br /&gt;to to be opened to make it ?&lt;br /&gt;&lt;br /&gt;Ans : 2.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;10). Grass in lawn grows equally thickand in a uniform rate.&lt;br /&gt;It takes 24 days for 70 cows and 60 for 30 cows .&lt;br /&gt;How many cows can eat away the same in 96 days.?&lt;br /&gt;&lt;br /&gt;Ans : 18 or 19&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;11). There is a certain four digit number whose fourth digit&lt;br /&gt; is twise the first digit.Third digit is three more than&lt;br /&gt;second digit.Sum of the first and fourth digits twise the&lt;br /&gt;third number.What was that number ?&lt;br /&gt;&lt;br /&gt;Ans : 2034 and 4368&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1. From a vessel on the first day, 1/3rd of the liquid&lt;br /&gt; evaporates. On the second day 3/4th of the remaining&lt;br /&gt; liquid evaporates. what fraction of the volume is&lt;br /&gt; present at the end of the II day.&lt;br /&gt;&lt;br /&gt;2. an orange galss has orange juice. and white glass has apple juice.&lt;br /&gt;Bothe equal volume 50ml of the orange juice is taken and&lt;br /&gt; poured into the apple juice. 50ml from the white glass is&lt;br /&gt; poured into the orange glass. Of the two quantities, the&lt;br /&gt;amount of apple juice in the orange glass and the amount of&lt;br /&gt;orange juice in the white glass, which one is greater and by how much?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3. there is a 4 inch cube painted on all sides. this is cut&lt;br /&gt;into no of 1 inch cubes. what is the no of cubes which have&lt;br /&gt;no pointed sides.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4. sam and mala have a conversation. sam says i am&lt;br /&gt;vertainly not over 40. mala says i am 38 and you are&lt;br /&gt;atleast 5 years older than me. Now sam says you are atleast 39.&lt;br /&gt;all the sattements by the two are false. How hold are they realy.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;5. ram singh goes to his office in the city, every day from his&lt;br /&gt;suburbun house. his driver mangaram drops him at the railway&lt;br /&gt; station in the morning and picks him up in the evening.&lt;br /&gt; Every evening ram singh reaches the station at 5 o'clock.&lt;br /&gt;mangaram also reaches at the same time. one day ramsingh&lt;br /&gt;started early from his office and came to the station at 4 o'clock.&lt;br /&gt;not wanting to wait for the car he starts walking home.&lt;br /&gt; Mangaram starts at normal time, picks him up on the way&lt;br /&gt;and takes him back&lt;br /&gt;&lt;br /&gt;house, half an hour early. how much time did ram singh walk.&lt;br /&gt;&lt;br /&gt;6. in a railway station, there are tow trains going. One in the&lt;br /&gt; harbour line and one in the main line, each having a&lt;br /&gt; frequency of 10 minutes. the main line service starts&lt;br /&gt;at 5 o'clock. the harbour line starts at 5.02a.m. a man&lt;br /&gt;goes to the station every day to catch the first train.&lt;br /&gt;what is the probability of man catchinhg the first train7.&lt;br /&gt;some people went for vaction. unfortunately it rained for&lt;br /&gt;13 days when they were there. but whenever it rained in&lt;br /&gt;the morning, they had clean afternood and vice versa.&lt;br /&gt;In all they enjoyed 11 morning and 12 afternoons. how many days&lt;br /&gt;did they stay there totally&lt;br /&gt;&lt;br /&gt;8. exalator problem repeat 9. a survey was taken among&lt;br /&gt;100 people to firn their preference of watching t.v. programmes.&lt;br /&gt; there are 3 channels. given no of&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;people who watch&lt;br /&gt;&lt;br /&gt;at least channel 1&lt;br /&gt;&lt;br /&gt;" " 2&lt;br /&gt;&lt;br /&gt;" " 3&lt;br /&gt;&lt;br /&gt;no channels at all&lt;br /&gt;&lt;br /&gt;atleast channels 1and 3&lt;br /&gt;&lt;br /&gt;" " 1 and 2&lt;br /&gt;&lt;br /&gt;" " 2 and 3&lt;br /&gt;&lt;br /&gt;find the no of people who watched all three.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;10. albert and fernandes they have two leg swimming race.&lt;br /&gt;both start from opposite and of the pool. On the first leg,&lt;br /&gt; the boys pass each other at 18 mt from the deep end of the&lt;br /&gt; pool. during the II leg they pass at 10 mt from the shallow&lt;br /&gt; end of the pool. Both go at const speed. but one of them is faster.&lt;br /&gt;each boy rests for 4 sec to see at the end of the i leg.&lt;br /&gt;what is the length of the pool.11. T H I S Each alphabet stands for one&lt;br /&gt;&lt;br /&gt;I S digit, what is the maximum value T&lt;br /&gt;&lt;br /&gt;-------------- can take&lt;br /&gt;&lt;br /&gt;X F X X&lt;br /&gt;&lt;br /&gt;X X U X&lt;br /&gt;&lt;br /&gt;--------------&lt;br /&gt;&lt;br /&gt;X X N X X&lt;br /&gt;&lt;br /&gt;--------------&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1. an escalator is descending at constant speed. A walks down&lt;br /&gt;and takes 50 steps to reach the bottom. B runs down and takes&lt;br /&gt;90 steps in the same time as A takes 10 steps. how many steps&lt;br /&gt; are visible when the escalator is not operating.&lt;br /&gt;&lt;br /&gt;2. every day a cyclist meets a train at a particular crossing.&lt;br /&gt;the road is straignt before the crossing and both are travelling&lt;br /&gt; in the same direction. cyclist travels with a speed of 10 Kmph.&lt;br /&gt; One day the cyclist comes late by 25 min. and meets the train&lt;br /&gt; 5km before the crossing. what is the seppd of the train.&lt;br /&gt;&lt;br /&gt;3. five persons muckerjee, misra, iyer, patil and sharma, all&lt;br /&gt; take then first or middle names in the full names. There are&lt;br /&gt;4 persons having I or middle name of kumar, 3 persons with&lt;br /&gt; mohan, 2 persons withdev and 1 anil.&lt;br /&gt;&lt;br /&gt;--Either mukherjee and patil have a I or middle name of dev&lt;br /&gt; or misra and iyer have their I or middle name of dev&lt;br /&gt;&lt;br /&gt;--of mukherkjee and misre, either both of them have a first&lt;br /&gt;or middle name of mohan or neither have a first or middle name&lt;br /&gt; of mohan--either iyer of sharma has a I or middle name of kumar hut not both.who has the I or middle name of anil&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4. reading comprehension&lt;br /&gt;&lt;br /&gt;5. a bird keeper has got pigeon, M mynas and S sparrows. the&lt;br /&gt;&lt;br /&gt;keeper goes for lunch leaving his assistant to watch the birds.&lt;br /&gt;&lt;br /&gt;a. suppose p=10, m=5, s=8 when the bird keeper comes back, the&lt;br /&gt;&lt;br /&gt;assistant informs the x birds have escaped. the bird keeper&lt;br /&gt;&lt;br /&gt;exclaims oh no! all my sparrows are gone. how many birds flew&lt;br /&gt;&lt;br /&gt;away.&lt;br /&gt;&lt;br /&gt;b. when the bird keeper come back, the assistand told him that&lt;br /&gt;&lt;br /&gt;x birds have escaped. the keeper realised that atleast2 sparrows&lt;br /&gt;&lt;br /&gt;have escaped. what is minimum no of birds that can escape.&lt;br /&gt;&lt;br /&gt;6. select from the five alternatives A,B,C,D,E&lt;br /&gt;&lt;br /&gt;AT THE end of each question ,two conditions will be given.&lt;br /&gt;&lt;br /&gt;the choices are to filled at follows.&lt;br /&gt;&lt;br /&gt;a. if a definete conclusion can be drawn from condition 1&lt;br /&gt;&lt;br /&gt;b. if a definete conclusion can be drawn from condition 2&lt;br /&gt;&lt;br /&gt;c. if a definete conclusion can be drawn from condition 1 and 2&lt;br /&gt;&lt;br /&gt;d. if a definete conclusion can be drawn from condition 1 or 2&lt;br /&gt;&lt;br /&gt;e. no conclusion can be drawn using both conditions&lt;br /&gt;&lt;br /&gt;1. person 1 says N&lt;5&lt;br /&gt;&lt;br /&gt;person says n&gt;5&lt;br /&gt;&lt;br /&gt;person 3 says 3N&gt;20&lt;br /&gt;&lt;br /&gt;person 4 says 3n&gt;10&lt;br /&gt;&lt;br /&gt;person 5 says N&lt;8&lt;br /&gt;&lt;br /&gt;whaT IS value of N&lt;br /&gt;&lt;br /&gt;a) 1. no of persons who speak false being less than no of persons&lt;br /&gt;&lt;br /&gt;who tells the truth.&lt;br /&gt;&lt;br /&gt;2. person 2 is telling the truth.&lt;br /&gt;&lt;br /&gt;b) 1. no of persong telling the truth is greater than no of&lt;br /&gt;&lt;br /&gt;persons telling lies&lt;br /&gt;&lt;br /&gt;2. person 5 is telling the truth.&lt;br /&gt;&lt;br /&gt;7. there are N coins on a table. there are two players A&amp;amp;B.&lt;br /&gt;&lt;br /&gt;you can take 1 or 2 coins at a time. the person who takes the&lt;br /&gt;&lt;br /&gt;last coin is the loser. a always starts first&lt;br /&gt;&lt;br /&gt;--1. if N=7&lt;br /&gt;&lt;br /&gt;a) A can always win by taking two coins in his first chanse&lt;br /&gt;&lt;br /&gt;b) B can win only if A takes two coins in his first chance.&lt;br /&gt;&lt;br /&gt;c) B can always win by proper play&lt;br /&gt;&lt;br /&gt;d) none of the above&lt;br /&gt;&lt;br /&gt;--2. A can win by proper play if N is equal to&lt;br /&gt;&lt;br /&gt;a) 13 b) 37 c) 22 d) 34 e) 48 ans. e.&lt;br /&gt;&lt;br /&gt;--3. B can win by proper play if N is equal to&lt;br /&gt;&lt;br /&gt;a) 25 b)26 c) 32 d) 41 e) none&lt;br /&gt;&lt;br /&gt;--4. if N&lt;4, can A win by proper play always&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;8. Two turns have vertain pecular charcteristics. One of them&lt;br /&gt;&lt;br /&gt;always lies on Monday, Wednesday, Friday. \the other always lies&lt;br /&gt;&lt;br /&gt;on Tuesdays, thursdays and saturdays. On the other days they tel&lt;br /&gt;&lt;br /&gt;the truth. You are given a conversation.&lt;br /&gt;&lt;br /&gt;person A-- today is sunday my name is anil&lt;br /&gt;&lt;br /&gt;person B-- today is tuesday, my name is bill&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;answers for selected questions&lt;br /&gt;&lt;br /&gt;2. equal 1. 150&lt;br /&gt;&lt;br /&gt;3. 8 2. 60 kmph&lt;br /&gt;&lt;br /&gt;4. 37(M),41(S) 3. Mukherjee&lt;br /&gt;&lt;br /&gt;5. 45 min. 8. today is tuesday&lt;br /&gt;&lt;br /&gt;6. 0.8&lt;br /&gt;&lt;br /&gt;7. 18&lt;br /&gt;&lt;br /&gt;11. T max value = 4&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-1082360872996842522?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/1082360872996842522/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=1082360872996842522' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1082360872996842522'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1082360872996842522'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/10/infosys-puzzles_4838.html' title='INFOSYS  Puzzles'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-6810541582325217038</id><published>2008-10-22T22:28:00.001-07:00</published><updated>2008-12-23T21:42:41.353-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='INFOSYS'/><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>INFOSYS Puzzles</title><content type='html'>&lt;meta equiv="Content-Type" content="text/html; 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	mso-header-margin:.5in; 	mso-footer-margin:.5in; 	mso-paper-source:0;} div.Section1 	{page:Section1;} --&gt; &lt;/style&gt;&lt;!--[if gte mso 10]&gt; &lt;style&gt;  /* Style Definitions */  table.MsoNormalTable 	{mso-style-name:"Table Normal"; 	mso-tstyle-rowband-size:0; 	mso-tstyle-colband-size:0; 	mso-style-noshow:yes; 	mso-style-priority:99; 	mso-style-qformat:yes; 	mso-style-parent:""; 	mso-padding-alt:0in 5.4pt 0in 5.4pt; 	mso-para-margin:0in; 	mso-para-margin-bottom:.0001pt; 	mso-pagination:widow-orphan; 	font-size:10.0pt; 	font-family:"Times New Roman","serif";} &lt;/style&gt; &lt;![endif]--&gt;  &lt;p class="MsoNormal"&gt;1.In a competition, Tom Dick &amp;amp; Harry had 9,2 and 2 points respectively. Harry came 2&lt;sup&gt;nd&lt;/sup&gt; in athletics, who came first in high jump?&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;2.A ship was on a dock .A rung was hanging for any one to climb on the ship. Miss Gary was supposed to climb the ship at 11a.m when 3 steps were visible .The distance between each step was 3 inches. The height of the tide was rising at 4 inches per hour. How many steps will Miss Gary have to climb when she reaches the dock at noon?&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;3.4 people are standing at the corners of a square with sides of 5km .The speed of 2&lt;sup&gt;nd&lt;/sup&gt; person is double that of 1&lt;sup&gt;st&lt;/sup&gt;. The speed of 3&lt;sup&gt;rd&lt;/sup&gt; person is double that of 2&lt;sup&gt;nd&lt;/sup&gt; .The speed of 4&lt;sup&gt;th&lt;/sup&gt; person is double that of 3&lt;sup&gt;rd&lt;/sup&gt;.What is the distance between 2&lt;sup&gt;nd&lt;/sup&gt; and 4&lt;sup&gt;th&lt;/sup&gt; person after an hour?&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;4.There was a river with a semi-circular bridge to cross it. The height of the bridge at the center was 40metres and at a distance of 20 meters from the bank it was 10 meters. What is the length of the bridge?&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;5.In a 5 digit nos., the 2&lt;sup&gt;nd&lt;/sup&gt; digit is twice the 3&lt;sup&gt;rd&lt;/sup&gt; ….&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;                  &lt;/span&gt;&lt;span style=""&gt;             &lt;/span&gt;The 1&lt;sup&gt;st&lt;/sup&gt; digit is thrice the 5&lt;sup&gt;th&lt;/sup&gt;..&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;…..similar relations like that….determine the no.?&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;6.In a day according to ram, prob. of raining was ½.his success rate was 50% and according to shyam, prob. of raining was ¼.his success rate was 50%. What is the prob. That it will rain.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;7.A man can carry food sufficient for a man for 4 days. He has to cross a desert and he will need food sufficient for 6 days. How many supporters he should engage to successfully cross the desert?&lt;/p&gt;  &lt;p class="MsoNormal"&gt;(hint: supporter must return after half way)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;8.Questions on series….&lt;/p&gt;  &lt;p class="MsoNormal"&gt;Eg. &lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;    &lt;/span&gt;0,10,100,…….&lt;/p&gt;  &lt;p class="MsoNormal"&gt;Options were:- 6,7,8,9.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;9.questions like 4 person are sitting around a table………….few instructions were given like A is sitting diagonal to D,B is sitting beside C……&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt; &lt;/span&gt;Who is sitting to the right of A?&lt;/p&gt;  &lt;p class="MsoNormal"&gt;(the above is just an indicative eg.)&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;o:p&gt; &lt;/o:p&gt;&lt;/p&gt;  &lt;p class="MsoNormal"&gt;10.5 people were staying together(a,b,c,d,e) .their professions were…..&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;    &lt;/span&gt;doctor was staying in room beside b.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt;    &lt;/span&gt;c was&lt;span style=""&gt;  &lt;/span&gt;engg. By profession….&lt;/p&gt;  &lt;p class="MsoNormal"&gt;&lt;span style=""&gt; &lt;/span&gt;Some more statements.&lt;/p&gt;  &lt;p class="MsoNormal"&gt;Who was staying in the 3&lt;sup&gt;rd&lt;/sup&gt; room?&lt;/p&gt;  &lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-6810541582325217038?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/6810541582325217038/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=6810541582325217038' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/6810541582325217038'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/6810541582325217038'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/10/infosys-puzzles_22.html' title='INFOSYS Puzzles'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-192313353295625070</id><published>2008-10-22T22:24:00.000-07:00</published><updated>2008-12-23T21:42:41.353-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='INFOSYS'/><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>INFOSYS  Puzzles</title><content type='html'>      &lt;br /&gt;                 -----------------------------&lt;br /&gt;                 INFOSYS TECHNOLOGIES LIMITED.&lt;br /&gt;                 -----------------------------&lt;br /&gt;                 Question Paper&lt;br /&gt;                 --------------&lt;br /&gt;                   Part 1.&lt;br /&gt;                   ------&lt;br /&gt;       (1) 9 cards are there. u have to arrange them in a 3*3 matrix.&lt;br /&gt;       cards are of 4 colors.they are red,yellow,blue,green.&lt;br /&gt;       conditions for arrangement: one red card must be in first row&lt;br /&gt;       or second row.2 green cards should be in 3rd column.Yellow&lt;br /&gt;       cards must be in the 3 corners only. Two blue cards must be in&lt;br /&gt;       the 2nd row. Atleast one green card in each row.&lt;br /&gt;       Solution:&lt;br /&gt;       Yello     Red        Gren&lt;br /&gt;       Blu       Blu        Gren&lt;br /&gt;       Yello     Gren       Yello&lt;br /&gt;       2. 4 cards are placed on a table, each card has two colors. U&lt;br /&gt;       don't know the color of the back side of eachcard.4 persons A&lt;br /&gt;       B C and D are sitting on the table before the cards. They can&lt;br /&gt;       see Red, Green Red and blue .Out of the 4 poeple 2 always lie.&lt;br /&gt;       They see the color on the reverse side and give the following&lt;br /&gt;       comment&lt;br /&gt;       A: Yello/green&lt;br /&gt;       B: Neither Blue/nor Green&lt;br /&gt;       c: Blue/Yello&lt;br /&gt;       D: Blue/ Yello&lt;br /&gt;       find out the color on the other side of the 4 cards.&lt;br /&gt;&lt;br /&gt;       3.Red and brown tribes [FROM BARRONS GRE] Conditions to&lt;br /&gt;       get married with each other.&lt;br /&gt;&lt;br /&gt;       4. Venn diagram regarding Rich, muscular, soft-skinned,&lt;br /&gt;       employed, etc.,( Refer BARRONS GRE GUIDE)&lt;br /&gt;&lt;br /&gt; ------------------------------------------------------------------------&lt;br /&gt;                         PART 2.&lt;br /&gt;                         -------&lt;br /&gt;         1. SAKUNTALA DEVI'S PUZZLE BOOK : PUZZLES TO PUZZLE YOU.&lt;br /&gt;            problem no: 3. ( Brothers and Sisters)&lt;br /&gt;            A family I know has several children. Each boy in this&lt;br /&gt;            family has as many sisters as brothers but each girl has&lt;br /&gt;            twice as many brothers as sisters. How many brothers&lt;br /&gt;            and sisters are there?&lt;br /&gt;            ans: 4 boys and 3 girls.&lt;br /&gt;         2. No. of animals is 11 more than the no. of birds. If the&lt;br /&gt;            no. of birds were the no. of animals and no. of animals&lt;br /&gt;            were the no. of birds( ie., interchanging no.s of animals&lt;br /&gt;            and birds.), the total no. of legs get reduced by one fifth&lt;br /&gt;            (1/5). How many no. of birds and animals were there?&lt;br /&gt;            ans: birds:11,animals:22&lt;br /&gt;         3. In a soap company  a soap is manufactured with 11 parts.&lt;br /&gt;            For making one soap you will get 1 part as scrap. At the&lt;br /&gt;            end of the day u have 251 such scraps. From that how many&lt;br /&gt;            soaps can be manufactured? ans: 22 + 2+ 1 = 25.&lt;br /&gt;         4.        2 * *            |&lt;br /&gt;                   3 * *            | No. 7 does not occur in this&lt;br /&gt;             ----------------       |&lt;br /&gt;                   5 * *            | multiplication.&lt;br /&gt;                 * 4 *              |&lt;br /&gt;               * * 3                | Find the product.&lt;br /&gt;             ----------------       |&lt;br /&gt;               * * * * *            |&lt;br /&gt;             ----------------       |&lt;br /&gt;            --------------------------------------------------&lt;br /&gt;            ans   2 8 1&lt;br /&gt;                  3 2 2&lt;br /&gt;                  -----&lt;br /&gt;                  5 6 2&lt;br /&gt;                5 6 2 0&lt;br /&gt;              8 4 3 0 0&lt;br /&gt;              ---------&lt;br /&gt;              9 0 4 8 2&lt;br /&gt;              ---------&lt;br /&gt;           5.  There is a 5digit no. 3 pairs of sum is eleven each.&lt;br /&gt;              Last digit is 3 times the first one.&lt;br /&gt;              3 rd digit is 3 less than the second.&lt;br /&gt;              4 th digit is 4 more than the second one.&lt;br /&gt;              Find the digit.&lt;br /&gt;              ans : 25296.&lt;br /&gt;           6. There are five thieves, each loot a bakery one after the&lt;br /&gt;             other such that the first one takes 1/2 of the total no.&lt;br /&gt;             of the breads plus 1/2 of a bread. Similarly 2nd, 3rd,4th&lt;br /&gt;             and 5fth also did the same. After the fifth one no. of&lt;br /&gt;             breads remained are 3. Initially how many breads were there?&lt;br /&gt;             ans : 31.&lt;br /&gt;           7.ESCALATOR PROBLEM OF SAKUNTALA DEVI 'PUZZLES TO PUZZLE'book.&lt;br /&gt;             Problem No: Problem 27( Down the escalator)&lt;br /&gt;             ans : the no of steps in the stair way : 46.&lt;br /&gt;           8.Harbour line and Main line Problem of Sakuntala Devi Puzzle&lt;br /&gt;             book. Ans : 4/5.&lt;br /&gt;             ( More Puzzles book)&lt;br /&gt;           9.There are some chicken in a poultry. They are fed with corn&lt;br /&gt;             One sack of corn will come for 9 days.The farmer decides to&lt;br /&gt;             sell some chicken and wanted to hold 12 chicken with him.&lt;br /&gt;             He cuts the feed by 10% and sack of corn comes for 30 days.&lt;br /&gt;             So initially how many chicken are there?&lt;br /&gt;           10.Two people X &amp;amp; Y walk on the wall of a godown in opposite&lt;br /&gt;             direction. They meet at a point on one side and then go&lt;br /&gt;             ahead. X after walking for some time, walks in opposite&lt;br /&gt;             direction for 15 mtrs.Then again he turns back and walks&lt;br /&gt;             in the original direction. What distance did Y walk before&lt;br /&gt;             they met again, if X walks 11 mtrs by the time Y walks&lt;br /&gt;             8 mtrs.&lt;br /&gt;           11.Problem from SAKUNTALA DEVI 'PUZZLES TO PUZZLE U'.&lt;br /&gt;              Problem no: 23( Walking back to happiness.)&lt;br /&gt;              The walking time : 55 mins.&lt;br /&gt;&lt;br /&gt;           With this The paper has been completed. For the lack time&lt;br /&gt;           I have'nt typed those probs, which are in Sakuntala Devi&lt;br /&gt;           Puzzle books. I wish all of you the best.&lt;br /&gt;           Send me the mail immediately after the written test will&lt;br /&gt;           be over. thank you.&lt;br /&gt;                                                     -Sarma. B. Y.&lt;br /&gt;&lt;br /&gt;======================================================================================================&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Infosys 2000 -  I.I.T.,Madras&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1. If 1/4 of the time from midnight plus 1/2 of the time from now&lt;br /&gt;to midnight is the present&lt;br /&gt; time,then what is the present time ?&lt;br /&gt;&lt;br /&gt;2. In a 10 digit number, if the 1st digit number is the number of&lt;br /&gt; ones,2nd digit number is&lt;br /&gt; the number of twos, and ... so on. 10th digit is the number of&lt;br /&gt; zeroes,then find the number&lt;br /&gt;&lt;br /&gt;3. A train blows a siren one hour after starting from the station.&lt;br /&gt; After that it travels at 3/5th of its speed it reaches the next&lt;br /&gt; station 2 hours behind schedule. If it had a problem 50 miles&lt;br /&gt; farther from the previous case,it would have reached 40 minutes&lt;br /&gt; sooner.Find the distance between the two stations .&lt;br /&gt;&lt;br /&gt;4. An army 50 miles long marches at a constant rate. A courier&lt;br /&gt;  standing at the rear moves forward and delivers the message&lt;br /&gt;  to the first person and then turns back and reaches the rear&lt;br /&gt; of the army as the army completes 50 miles. Find the distance&lt;br /&gt; travelled by the courier.&lt;br /&gt;&lt;br /&gt;5. A person grows cabbage, he uses a larger square this year than&lt;br /&gt; previous year and produces cabbages than previous year.what is&lt;br /&gt; the no. of cabbages produced this year&lt;br /&gt;6. Olympic race :&lt;br /&gt; 4 contestants  : Alan,charlie, Darren ,Brain.&lt;br /&gt; There are two races and average is taken to decide the winner.&lt;br /&gt; one person comes at the same position in both the race. Charlie&lt;br /&gt; always come before darren. Brian comes first once.&lt;br /&gt; Alan comes third atleast once. Find the positions. Alan never comes last.&lt;br /&gt; Charlie &amp;amp; darren comes 2nd atleast once.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;8. Problem finding who is husband,wife &amp;amp; son from 4 set of families.&lt;br /&gt;9. Rank the persons from set of conditions.&lt;br /&gt;&lt;br /&gt;=========================================================================================================&lt;br /&gt;&lt;br /&gt;/* THIS FILE INCLUDES INFOSYS PAPERS FOR YEAR 2000 HELD AT IISc AND NAGPUR.&lt;br /&gt;*/&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;INFOSYS TEST @ IISC - 2000&lt;br /&gt;&lt;br /&gt;1:there is a robery and four persons are suspected out of them one is&lt;br /&gt;actual&lt;br /&gt;thief,&lt;br /&gt;these are the senteces said by each one of them!&lt;br /&gt;A says D had done&lt;br /&gt;B says A had done&lt;br /&gt;C says i dddnt done&lt;br /&gt;D  B lied when he said that i am thief&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;out of these only one man is true remaining are false&lt;br /&gt;ans C is thef,D is true!&lt;br /&gt;maks 3&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2  how many four digit numbers divisible by four  can be formed  using&lt;br /&gt;&lt;br /&gt;1,2,3,4;repetions are not allowed!&lt;br /&gt;ans 6;&lt;br /&gt;marks 4&lt;br /&gt;3  a vender solds two things at same cost 12 RS with one item at 25%profit&lt;br /&gt;and other at 20%loss,by this transaction he made profit or loss by how&lt;br /&gt;much&lt;br /&gt;ans loss,60paise&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;marks 3&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4 two friends A,B are running up  hill and then to get down!&lt;br /&gt;length if road is 440 yards A on his return journy met B goin up at&lt;br /&gt;20 yards from top,A has finished the rase .5 minuit earlier than&lt;br /&gt;B,then how much time A had taken to complete the rase.&lt;br /&gt;&lt;br /&gt;ans 6.3 minuits&lt;br /&gt;marks 4&lt;br /&gt;&lt;br /&gt;5 barons question diagnostic test question 8-12,5 men and 5 women went for&lt;br /&gt;foot ball match.........&lt;br /&gt;&lt;br /&gt;marks 8&lt;br /&gt;&lt;br /&gt;6 write a five digit number ,&lt;br /&gt;which will be having two prime numbers,&lt;br /&gt;and some two more conditions,like 1st digit greater than 2nd etc&lt;br /&gt;its easy(remember one is not a prime number,most people don mistake&lt;br /&gt;taking 1&lt;br /&gt;as prime number)&lt;br /&gt;&lt;br /&gt;ans 71842&lt;br /&gt;&lt;br /&gt;5 marks&lt;br /&gt;&lt;br /&gt;7  two employs were there&lt;br /&gt;employ one says to employ two your work ezxperience twice as me&lt;br /&gt;emp2 exactly&lt;br /&gt;emp1 but two years before you said that your work experience is&lt;br /&gt;thrice as me&lt;br /&gt;emp2 yes its also true&lt;br /&gt;what are their work experience&lt;br /&gt;ans 4,8&lt;br /&gt;marks 3&lt;br /&gt;&lt;br /&gt;8   ther are four persons A,B,C,D and for languages english ,french&lt;br /&gt;,german&lt;br /&gt;italian,&lt;br /&gt;conditions&lt;br /&gt;1 only one language is spoken by more than two men&lt;br /&gt;2 A dont know english&lt;br /&gt;3 a man can speak either french or german but not both&lt;br /&gt;4 all man cannot spek in a group(no common language)&lt;br /&gt;5 A can mediate when B and C want to speak with each other&lt;br /&gt;6 each men  can speak two languages&lt;br /&gt;ans&lt;br /&gt;A      french italian&lt;br /&gt;B      english french&lt;br /&gt;C      german  italian&lt;br /&gt;D      german  italian&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;marks 7&lt;br /&gt;&lt;br /&gt;9   there are 3 women ,they having three jewells,named diamond emerald&lt;br /&gt;,ruby&lt;br /&gt;&lt;br /&gt;3 women A,B,C&lt;br /&gt;3 thiefs D,E,F&lt;br /&gt;each thef had taken one jewel from each of the women&lt;br /&gt;following conditions&lt;br /&gt;one who had taken diamond is the bachelor and most dangerous&lt;br /&gt;D 's brother in law E who is less dangerous than the thief who had&lt;br /&gt;--------------------------------------------------------------------&lt;br /&gt;stolen emerald&lt;br /&gt;--------------&lt;br /&gt;(this is the key from this e had stolen ruby)&lt;br /&gt;D did nt stolen from B&lt;br /&gt;one more condition is there&lt;br /&gt;&lt;br /&gt;marks 7&lt;br /&gt;&lt;br /&gt;10 ther were five persons out of which two persons heir went white&lt;br /&gt;and three persons hair is black&lt;br /&gt;conditions&lt;br /&gt;if A 's hir is white then B'hair is white&lt;br /&gt;A,s hair is not blakck if C's hair is not black&lt;br /&gt;two more conditions&lt;br /&gt;but we can easily solve it from first codition&lt;br /&gt;ans is Aand B's hair is white&lt;br /&gt;marks 5&lt;br /&gt;&lt;br /&gt;essay&lt;br /&gt;in this age of super computer computer will repalce man!&lt;br /&gt;&lt;br /&gt;what is contribution of industry towards humn growth&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;interview questions!&lt;br /&gt;&lt;br /&gt;1 two time glases 7,4 how you measure 9&lt;br /&gt;2  four persons have to cross the bridge  they are having one torch&lt;br /&gt;ligt&lt;br /&gt;four persons take 1,2,5,10 minuits respectively ,&lt;br /&gt;when two persons are going the will take the time of the slowest&lt;br /&gt;person&lt;br /&gt;whats the time taken to cross allof them&lt;br /&gt;&lt;br /&gt;ans 17minuits&lt;br /&gt;3   a non uniform rope burns for one hour how you will measure&lt;br /&gt;half an hour&lt;br /&gt;ans lit it from both sides!&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;BEST OF LUCK!&lt;br /&gt;&lt;br /&gt;------------------------------------------------------------------------&lt;br /&gt;&lt;br /&gt;INFOSYS TEST @ NAGPUR - 2000&lt;br /&gt;questions&lt;br /&gt;1. there were three suspects for a robbery that happend in a bank, tommy,&lt;br /&gt;joy and bruceEach of them were saying that I haven't done anything and the&lt;br /&gt;other two has done it.police found that tommy was lying .who is the thief.&lt;br /&gt;3M (MARKS).&lt;br /&gt;&lt;br /&gt;2. three clocks where set to true time .First run with the exact time&lt;br /&gt;..second slows one minute/day. third gains one minute/day. after how many&lt;br /&gt;days they will show true time.  3M.&lt;br /&gt;&lt;br /&gt;3. A,B,C,D,E are some numbers.if AB * CD = EEE and CD - A =CC&lt;br /&gt;THEN what is AB * D =? IN SAME CODE .5M.&lt;br /&gt;&lt;br /&gt;4. Joe started from bombay towards pune and her friend julie in opposite&lt;br /&gt;direction.they meet at a point .distance travelled by joe was 1.8 miles&lt;br /&gt;more&lt;br /&gt;than that of julie.after spending some both started there way.joe reaches&lt;br /&gt;in&lt;br /&gt;2 hours while julie in 3.5 hours.&lt;br /&gt;Assuming both were travelling with constant speed.Wath is the distance&lt;br /&gt;between the two cities.   5M.&lt;br /&gt;&lt;br /&gt;5. In a six level building,a person lives .suppose the persons are&lt;br /&gt;A,B,C,D,E,F. --- and there were few clues like---&lt;br /&gt;A cann't live above third level,&lt;br /&gt;B cann't live above A and below C.-- i am recalling each.did u understood&lt;br /&gt;the question--&lt;br /&gt;u have to find person living at each level.  5M&lt;br /&gt;&lt;br /&gt;6. There were some containers of quantity 1,3,4,5,6,12,15,22,24,38&lt;br /&gt;liters.each was filled with some liquid except one.the liquids are milk,&lt;br /&gt;water and oil.quantity of each was like this&lt;br /&gt;water = 2* milk&lt;br /&gt;oil = 2* water&lt;br /&gt;find out which container was empty and cantainers filled with milk and&lt;br /&gt;oil.&lt;br /&gt;6M&lt;br /&gt;&lt;br /&gt;7. there were few diamonds.&lt;br /&gt;1st thief takes half of the diamonds +2&lt;br /&gt;2nd thief takes half of the diamonds +2&lt;br /&gt;3rd thief takes half of the diamonds +2&lt;br /&gt;4th thief takes half of the diamonds +2&lt;br /&gt;when 5 th thief arives there were no diamonds. find total no. of diamonds.&lt;br /&gt;(ans 60)  6M.&lt;br /&gt;&lt;br /&gt;8. there were five hunters A,B,C,D,E and five animals A,B,C,D,E. Hunter&lt;br /&gt;having the same name with the animal didn't kill it. Each hunter has&lt;br /&gt;missed&lt;br /&gt;some animal.&lt;br /&gt;A animal was hunt by the hunter whose name matches with animal hunt by&lt;br /&gt;hunter B.&lt;br /&gt;C animal was hunt by the hunter whose name matches with animal hunt by&lt;br /&gt;hunter D.&lt;br /&gt;E has hunt C and missed D .find out animals hunted by A,B,C.  6m.&lt;br /&gt;&lt;br /&gt;9. Five students A,B,C,D,E.&lt;br /&gt;when conversation started B,C were speaking english.When D join them they&lt;br /&gt;shifted to Spanish.A,E knows French .B,E knows Italian.Portoguese was&lt;br /&gt;known&lt;br /&gt;to there of them.Spanish was the most common language between them.&lt;br /&gt;one of them knows five languages.&lt;br /&gt;one of them knows four languages.&lt;br /&gt;one of them knows there languages.&lt;br /&gt;one of them knows two languages.&lt;br /&gt;one of them knows only one languages.&lt;br /&gt;&lt;br /&gt;A knows ?  4 options&lt;br /&gt;B knows ?  4 options&lt;br /&gt;C knows ?  4 options&lt;br /&gt;E knows ?  4 options       8M&lt;br /&gt;------------------------------------------------------------------------&lt;br /&gt;&lt;br /&gt;WISH U ALL THE BEST.........&lt;br /&gt;MAY ALL YOUR DREAMS COME TRUE !!!&lt;br /&gt;&lt;br /&gt;=================================================================================================&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Infosys Technologies ltd&lt;br /&gt;Aptitude Test  9th july 2000&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1.A boy picks up the phone and asks" who are you?"&lt;br /&gt;The voice from the other side answers" I am your&lt;br /&gt;mother's mother in law".What is the relation of the&lt;br /&gt;boy with the fellow speaking at the other end.(3marks).&lt;br /&gt;&lt;br /&gt;2.Imagine a rectangle.Its length=2*width.A square of 1&lt;br /&gt;inch is cut on all corners so that the remaining portion&lt;br /&gt;forms a boxwhen folded.the volume of the box is_____ cubic&lt;br /&gt;inches.find the original dimensions of the box.  (3 marks).&lt;br /&gt;&lt;br /&gt;3.2 persons are doing part time job in a company9say a and b).&lt;br /&gt;THe company is open for all the 7 days of the week.'A' works&lt;br /&gt;every second day.'B' works every 3rd day.If 'A'works on 1st june&lt;br /&gt;and 'B' works on 2nd june.Find out the date on which both 'A'&lt;br /&gt;and 'B' will work together.  (4 marks).&lt;br /&gt;&lt;br /&gt;4.Consider a pile of Diamonds on a table.A thief enters and steals&lt;br /&gt;1/2 of th e total quanity and then again 2 extra from the remaining.&lt;br /&gt;After some time a second thief enters and steals 1/2 of the remaining+2.&lt;br /&gt;Then 3rd thief enters and steals 1/2 of the remaining+2.Then 4th thief&lt;br /&gt;enters and steals 1/2 of the remaining+2.When the 5th one enters&lt;br /&gt;he finds 1 diamond on the table.Find out the total no. of diamonds&lt;br /&gt;originally on the table before the 1st thief entered.  (4 marks).&lt;br /&gt;&lt;br /&gt;5.Imagine 4 persons A,B,C,D.(It is a strength determining game).&lt;br /&gt;A found it hard,but could pull 'c' and 'd' to his side.&lt;br /&gt;AC and BD pairs on opposite sides found themselves equally balanced.&lt;br /&gt;When A and B exchanged thier positions to form pairs AD and BC ,BC pair&lt;br /&gt;could&lt;br /&gt;&lt;br /&gt;win and pull AD to thier side.Order the 4  persons in Ascending order&lt;br /&gt;according to thier  strengths.  (3 marks).&lt;br /&gt;&lt;br /&gt;6.Consider a beauty contest.3 persons participate.Their names are&lt;br /&gt;Attractive,&lt;br /&gt;Delectable,Fascinating.THere 3 tribes Pukkas,Wottas,Summas.&lt;br /&gt;Pukkas - Always speak truth.&lt;br /&gt;Wottas - Always speak lies.&lt;br /&gt;Summas - Speak truth and lies alternatively.&lt;br /&gt;Each of the 3 persons make 2 statements.&lt;br /&gt;&lt;br /&gt;Attractive - (1)&lt;br /&gt;(2)&lt;br /&gt;&lt;br /&gt;Delectable - (1)   i don't remember&lt;br /&gt;(2)&lt;br /&gt;&lt;br /&gt;Fascinating- (1)&lt;br /&gt;(2)The person who speaks truth is the least beautiful&lt;br /&gt;&lt;br /&gt;&gt;From the statements they give and the character of the 3 tribal types,&lt;br /&gt;find out which person belongs to which tribe.Also find out the persons&lt;br /&gt;in the Ascending order of  their beauty.   (8 marks).&lt;br /&gt;&lt;br /&gt;7.There are 5 positions-Clerk,Buyer,Cashier,Manager,Floorwalker.&lt;br /&gt;THere are 5 persons- Mrs.Allen,Mrs.CLark,Twain,Ewing,Bernett.&lt;br /&gt;Conditions:  1.clerk and cashier lunch time 11.30.to12.30&lt;br /&gt;2.others 12.30 to 1.30&lt;br /&gt;3.Mrs.Allen and Bernett play durind lunch     time.&lt;br /&gt;4.clerk and cashier share Bachler rooms.&lt;br /&gt;5.Ewing and Twain are not in good terms because&lt;br /&gt;one day when Twain retuned early from lunch&lt;br /&gt;he saw Ewing already sitting fro lunch and reported&lt;br /&gt;about him to the manager.&lt;br /&gt;Find out which person holds which post.   (8 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;8.There are 8 courses to be handled by faculty in 2 semesters.4 in 1st&lt;br /&gt;semester and 4 in 2nd semester THe candiadates hired for the post are&lt;br /&gt;k,l,m,n,o.The courses are Malvino,Shakespeare,Joyce,Chauncer...........&lt;br /&gt;Some conditions will be given like.,&lt;br /&gt;1.L and N handle Shakespeare and malvino.&lt;br /&gt;2.M and O handle Malvino and Joyce.&lt;br /&gt;3.  .......&lt;br /&gt;4.&lt;br /&gt;..&lt;br /&gt;..     i don't remember&lt;br /&gt;..&lt;br /&gt;8.  ......&lt;br /&gt;Only one author can handle a course in 1 semester.&lt;br /&gt;then 3 questions were asked based on these information and&lt;br /&gt;3 options for each question were given  (8 marks).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;9.  I don't remember&lt;br /&gt;&lt;br /&gt;10.I don't remember.&lt;br /&gt;&lt;br /&gt;===================================================================================================&lt;br /&gt;&lt;br /&gt;INFOSYS&lt;br /&gt;  --------&lt;br /&gt;&lt;br /&gt;  written test.&lt;br /&gt;&lt;br /&gt;  There are passages from A to M.Each one is having four questions.&lt;br /&gt;&lt;br /&gt;  A.My father had no brothers.but his three.....&lt;br /&gt;  1.My father was only child. (F)&lt;br /&gt;  2.Three of my aunts have sons. (C)&lt;br /&gt;  3.I have six cousins on my mothers side (C)&lt;br /&gt;  4.I have only one uncle. (F)&lt;br /&gt;  B.Senoir manegers in a leading company..........&lt;br /&gt;  1.Some senoir managers said that more people will want to buy new&lt;br /&gt;  cars in the future. (F)&lt;br /&gt;  2.Managers told workers that Japanese workers are taking jobs&lt;br /&gt;  away form India Workers in the car industry (F)&lt;br /&gt;  3.The managers issued their warning after a rise in interest rates(T).&lt;br /&gt;  4.The increased rate of interest will mean that Japanese firms&lt;br /&gt;  will cease to operate in this country.&lt;br /&gt;  C.Reaserchers in Mumbai have found that............&lt;br /&gt;  1.Injecting eather into the gallbladdre dissolves most gallastones.(T)&lt;br /&gt;  2.Surgery is the only treatment for calcium based gallastones(T)&lt;br /&gt;  3.Only a few hundred people suffer from calcium based gallastones(C)&lt;br /&gt;  4.It takes more than one day for the eather to dissolve calcium gallastones.&lt;br /&gt;  (F)&lt;br /&gt;&lt;br /&gt;  Here is a small trick for D,E,F,G&lt;br /&gt;  D. T F T F&lt;br /&gt;&lt;br /&gt;  E.F F T C&lt;br /&gt;&lt;br /&gt;  F.F F T C&lt;br /&gt;&lt;br /&gt;  G.T F T F&lt;br /&gt;&lt;br /&gt;  D and G are same and E and F are same.Be carefull.&lt;br /&gt;  -------------------------------------------------&lt;br /&gt;&lt;br /&gt;  H. Every form of art is proteted by copyright...........(Tagore)&lt;br /&gt;  1.Tagore plays are not protected by copyright(F)&lt;br /&gt;  2.Tagore's desecendents should be asked to make some contribution&lt;br /&gt;  to the arts.(T)&lt;br /&gt;  31.Instead of buying a ticket,theatregoers should pay a fee to a&lt;br /&gt;  trust for the benifit of the arts. (F)&lt;br /&gt;  32.More people could go to the theatre if copyright were&lt;br /&gt;  abolished (C)&lt;br /&gt;  I. Hacking is a crime..............&lt;br /&gt;&lt;br /&gt;  33.Most hackers are authorised to breaks into networks&lt;br /&gt;  34.Computers are only vulnerable to the unauthorised manipulation&lt;br /&gt;&lt;br /&gt;  of thier data via another computer if they are networked. (C)&lt;br /&gt;  35.The main reason why it is relatively easy. (T)&lt;br /&gt;  36.Hackers do not work for the firms whose networks they break into(C)&lt;br /&gt;  J. In this only two questions&lt;br /&gt;&lt;br /&gt;  37.F&lt;br /&gt;  38.T&lt;br /&gt;  K.Electronics......&lt;br /&gt;&lt;br /&gt;  39.F&lt;br /&gt;  40. to 42 all are True&lt;br /&gt;  L.&lt;br /&gt;  All are true.(43 to 46)&lt;br /&gt;&lt;br /&gt;  M. All are true (47 to 50)&lt;br /&gt;  No of candidates appeared : 220&lt;br /&gt;  For the Interview : 90.&lt;br /&gt;&lt;br /&gt;==================================================================================================&lt;br /&gt;&lt;br /&gt;&gt; The Questions are follows&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; 1. Number of null pointers in any binary tree = n+1&lt;br /&gt;&gt;&lt;br /&gt;&gt; 2. max(t1,t2,...tn)   = pipelining&lt;br /&gt;&gt;&lt;br /&gt;&gt; 3. 50%   -DBETXXXXXX    - density&lt;br /&gt;&gt;&lt;br /&gt;&gt; 4. print (Head(T))&lt;br /&gt;&gt;    Traverse(left(T))&lt;br /&gt;&gt;    print (Head(T))&lt;br /&gt;&gt;    Traverse(right(T))    - ans: none of the above&lt;br /&gt;&gt;&lt;br /&gt;&gt; 5. Boolean expn Evalvate&lt;br /&gt;&gt;   &lt;br /&gt;&gt; 6. Common subexpn :   - ans : a + e&lt;br /&gt;&gt;&lt;br /&gt;&gt; 7. LRU :  1, 2, 3.&lt;br /&gt;&gt;&lt;br /&gt;&gt; 8. Tr. Delay - 10000 bits ans. 10.01&lt;br /&gt;&gt;&lt;br /&gt;&gt; 9. Grammar of Number of shift / reduce operator  : ans. 4&lt;br /&gt;&gt;&lt;br /&gt;&gt; 10. CPU scheduling  9,8 ?&lt;br /&gt;&gt;&lt;br /&gt;&gt; 11. if even x/2&lt;br /&gt;&gt;     else p(p(3x+1))&lt;br /&gt;&gt;&lt;br /&gt;&gt;    2^k + 1:  3 . 2^(k-1)  clarify this with sans&lt;br /&gt;&gt;&lt;br /&gt;&gt; 12. allocation  ans: (ii) only&lt;br /&gt;&gt;&lt;br /&gt;&gt; 13.  swapping :  ans: reference only&lt;br /&gt;&gt;&lt;br /&gt;&gt; 14. Compiler - related Qn.&lt;br /&gt;&gt;&lt;br /&gt;&gt; 15. LAN frames - ?  related Qn.&lt;br /&gt;&gt;&lt;br /&gt;&gt; 16. parameter passing (35,20)&lt;br /&gt;&gt;&lt;br /&gt;&gt; 17. sliding window protocol&lt;br /&gt;&gt;          -   BUFFER SIZE large&lt;br /&gt;&gt;&lt;br /&gt;&gt; 18.  kernel mode   -  deallocate resource&lt;br /&gt;&gt;&lt;br /&gt;&gt; 19.  logic circuit&lt;br /&gt;&gt;       ans .  Minimum OR = 3&lt;br /&gt;&gt;&lt;br /&gt;&gt; 20.  Combinatorics  related&lt;br /&gt;&gt;&lt;br /&gt;&gt; 21. priority scheduling&lt;br /&gt;&gt;&lt;br /&gt;&gt; 22. cobegin&lt;br /&gt;&gt;       begin  x = y; x= x+1;  y= x&lt;br /&gt;&gt;       begin x =y; z= z+1; y= z&lt;br /&gt;&gt;     coend&lt;br /&gt;&gt;&lt;br /&gt;&gt;    ans.  Number of values possi = 2&lt;br /&gt;&gt;&lt;br /&gt;&gt; 23. 2 bits flip / 2 bits exchange&lt;br /&gt;&gt;     &lt;br /&gt;&gt;      ans : the word with one '1'&lt;br /&gt;&gt;&lt;br /&gt;&gt; 24.  any addr&lt;br /&gt;&gt;       K^+ v(a) + 2I - 2a&lt;br /&gt;&gt;     &lt;br /&gt;&gt;  clarify with SANS.&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&lt;br /&gt;=================================================================================================&lt;br /&gt;&lt;br /&gt;                        Infosys&lt;br /&gt;&lt;br /&gt;                        --------&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1) There are two balls touching each other circumferencically.&lt;br /&gt;&lt;br /&gt;The radius of the big ball is 4 times the diameter of the small&lt;br /&gt;&lt;br /&gt;ball.The outer small ball rotates in anticlockwise direction&lt;br /&gt;&lt;br /&gt;circumferencically over the bigger one at the rate of 16 rev/sec.&lt;br /&gt;&lt;br /&gt;The bigger wheel also rotates anticlockwise at Nrev/sec. what is&lt;br /&gt;&lt;br /&gt;'N' for the horizontal line from the centre of small wheel always&lt;br /&gt;&lt;br /&gt;is horizontal.&lt;br /&gt;&lt;br /&gt;2)     1 2 3 4&lt;br /&gt;&lt;br /&gt;     + 3 4 5 5&lt;br /&gt;&lt;br /&gt;     ----------&lt;br /&gt;&lt;br /&gt;       4 6 8 9&lt;br /&gt;&lt;br /&gt;     - 2 3 4 5&lt;br /&gt;&lt;br /&gt;     ----------&lt;br /&gt;&lt;br /&gt;       2 3 4 4&lt;br /&gt;&lt;br /&gt;     + 1 2 5 4&lt;br /&gt;&lt;br /&gt;     ------------&lt;br /&gt;&lt;br /&gt;       3 6 9 8&lt;br /&gt;&lt;br /&gt; Q)  Strike off any digit from each number in seven rows (need not&lt;br /&gt;&lt;br /&gt; be at same place) and combine the same operations with 3 digit numbers&lt;br /&gt;&lt;br /&gt; to get the same addition. After this strike off another digit from all&lt;br /&gt;&lt;br /&gt; and add all the No.s to get the same 2 digit No. perform the same&lt;br /&gt;&lt;br /&gt; process again with 1 digit No.s. Give the ' no.s in 7 rows at&lt;br /&gt;&lt;br /&gt; each stage.&lt;br /&gt;&lt;br /&gt;3) there is a safe with a 5 digit No. The 4th digit is 4 greater than&lt;br /&gt;&lt;br /&gt;second digit, while 3rd digit is 3 less than 2nd digit. The 1st digit&lt;br /&gt;&lt;br /&gt;is thrice the last digit. There are 3 pairs whose sum is 11. Find&lt;br /&gt;&lt;br /&gt;the number.  Ans) 65292.&lt;br /&gt;&lt;br /&gt;4) there are 2 guards Bal and Pal walking on the side of a wall of a&lt;br /&gt;&lt;br /&gt;wearhouse(12m X 11m) in opposite directions. They meet at a point and&lt;br /&gt;&lt;br /&gt;Bal says to Pal " See you again in the other side". After a few moments&lt;br /&gt;&lt;br /&gt;of walking Bal decides to go back for a smoke but he changes his&lt;br /&gt;&lt;br /&gt;direction again to his previous one after 10 minutes of walking in&lt;br /&gt;&lt;br /&gt;the other(opposite) direction remembering that Pal will be waiting&lt;br /&gt;&lt;br /&gt;for to meet.If Bal and Pal walk 8 and 11 feet respectively, how&lt;br /&gt;&lt;br /&gt;much distance they would have travelled before meeting again.&lt;br /&gt;&lt;br /&gt;5)          xxx)xxxxx(xxx&lt;br /&gt;&lt;br /&gt;                3xx&lt;br /&gt;&lt;br /&gt;                -------&lt;br /&gt;&lt;br /&gt;                 xxx&lt;br /&gt;&lt;br /&gt;                 x3x&lt;br /&gt;&lt;br /&gt;                 ------&lt;br /&gt;&lt;br /&gt;                  xxx&lt;br /&gt;&lt;br /&gt;                  3xx&lt;br /&gt;&lt;br /&gt;                  ------&lt;br /&gt;&lt;br /&gt;   Q) Find the 5 digit No.&lt;br /&gt;&lt;br /&gt;    Hint: 5 is used atleast once in the calculation.&lt;br /&gt;&lt;br /&gt;6) Afly is there 1 feet below the ceiling right across a wall length&lt;br /&gt;&lt;br /&gt;is 30m at equal distance from both the ends. There is a spider 1 feet&lt;br /&gt;&lt;br /&gt;above floor right across the long wall eqidistant from both the ends.&lt;br /&gt;&lt;br /&gt;If the width of the room is 12m and 12m, what distance is to be&lt;br /&gt;&lt;br /&gt;travelled by the spider to catch the fly? if it takes the shortest&lt;br /&gt;&lt;br /&gt;path.&lt;br /&gt;&lt;br /&gt;7) Ramesh sit around a round table with some other men. He has one&lt;br /&gt;&lt;br /&gt;rupee more than his right person and this person in turn has 1 rupee&lt;br /&gt;&lt;br /&gt;more than the person to his right and so on, Ramesh decided to give&lt;br /&gt;&lt;br /&gt;1 rupee to his right &amp;amp; he in turn 2 rupees to his right and 3 rupees&lt;br /&gt;&lt;br /&gt;to his right &amp;amp; so on. This process went on till a person has&lt;br /&gt;&lt;br /&gt;'no money' to give to his right. At this time he has 4 times the&lt;br /&gt;&lt;br /&gt;money to his right person. How many men are there along with Ramesh&lt;br /&gt;&lt;br /&gt;and what is the money with poorest fellow.&lt;br /&gt;&lt;br /&gt;8)Question related to probabilities of removing the red ball from a&lt;br /&gt;&lt;br /&gt;basket,given that two balls are removed from the basket and the other&lt;br /&gt;&lt;br /&gt;ball is red. The basket contains blue,red,yellow balls.&lt;br /&gt;&lt;br /&gt;9)Venkat has 1boy&amp;amp;2daughters.The product of these children age is 72.&lt;br /&gt;&lt;br /&gt;The sum of their ages give the door numberof Venkat.Boy is elder of&lt;br /&gt;&lt;br /&gt;three.Can you tell the ages of all the three.&lt;br /&gt;&lt;br /&gt;                           ANALYTICAL&lt;br /&gt;&lt;br /&gt;                           ----------&lt;br /&gt;&lt;br /&gt;1)L:says all of my other 4 friends have money&lt;br /&gt;&lt;br /&gt;M:says that P said that exact one has money&lt;br /&gt;&lt;br /&gt;N:says that L said that precisely two have money&lt;br /&gt;&lt;br /&gt;O:says that M said that 3 of others have money.&lt;br /&gt;&lt;br /&gt;P:Land N said that they have money.&lt;br /&gt;&lt;br /&gt;   all are liers.Who has money&amp;amp;who doesn't have?&lt;br /&gt;&lt;br /&gt;2)A hotel has two,the east wing and the west wing.some east wing rooms&lt;br /&gt;&lt;br /&gt;but not all have an ocean view(OV).All WW have a harbour view(HV).The&lt;br /&gt;&lt;br /&gt;charge for all rooms is identical, except as follows&lt;br /&gt;&lt;br /&gt;* Extra charge for all HV rooms on or above the 3rd floor&lt;br /&gt;&lt;br /&gt;* Extra charge for all OV rooms except those without balcony&lt;br /&gt;&lt;br /&gt;* Extra charge for some HV rooms on the first two floor&amp;amp;some EW rooms&lt;br /&gt;&lt;br /&gt;without OV but having kitchen facilities. (GRE modrl Test 3-question&lt;br /&gt;&lt;br /&gt;                                                 1J-22)&lt;br /&gt;&lt;br /&gt;3)Post man has a data of name surname door no.pet name of 4 families.&lt;br /&gt;&lt;br /&gt;But only one is correct for each family.There are a set of statements&lt;br /&gt;&lt;br /&gt;&amp;amp;questions.&lt;br /&gt;&lt;br /&gt;4)4 couples have a party.Depending on the set of statements,find who&lt;br /&gt;&lt;br /&gt;insulted whom and who is the host of the party.&lt;br /&gt;&lt;br /&gt;5)5 women given some of their heights(tall,medium,short)Hair( long,&lt;br /&gt;&lt;br /&gt;plainted),stards(Black or Brown), sari,2 medium,2-short.Tall-&gt;no&lt;br /&gt;&lt;br /&gt;sari.Plainted-&gt;medium.Answer the combinations.&lt;br /&gt;&lt;br /&gt;1) A person has to go both Northwards&amp;amp;Southwards in search of a job.&lt;br /&gt;&lt;br /&gt;He decides to go by the first train he encounters.There are trains for&lt;br /&gt;&lt;br /&gt;every 15 min both southwards and northwards.First train towards south&lt;br /&gt;&lt;br /&gt;is at 6:00 A.M. and that towards North is at 6:10 .If the person arrives&lt;br /&gt;&lt;br /&gt;at any random time,what is the probability that he gets into a train&lt;br /&gt;&lt;br /&gt;towards North.&lt;br /&gt;&lt;br /&gt;2) A person has his own coach&amp;amp;whenever he goes to railway station he&lt;br /&gt;&lt;br /&gt;takes his coach.One day he was supposed to reach the railway station&lt;br /&gt;&lt;br /&gt;at 5 O'clock.But he finished his work early and reached at 3 O'clock.&lt;br /&gt;&lt;br /&gt;Then he rung up his residence and asked to send the coach immediately.&lt;br /&gt;&lt;br /&gt;He came to know that the coach has left just now to tje railway station.&lt;br /&gt;&lt;br /&gt;He thought that the coach has left just now to the railway station.He&lt;br /&gt;&lt;br /&gt;thought that he should not waste his time and started moving towards&lt;br /&gt;&lt;br /&gt;his residence at the speed of 3mi/hr.On the way,he gets the coach and&lt;br /&gt;&lt;br /&gt;reaches home at 6 o'clock.How far is his residence from railway&lt;br /&gt;&lt;br /&gt;station.&lt;br /&gt;&lt;br /&gt;3)Radha,Geeta&amp;amp;Revathi went for a picnic.After a few days they forgot the&lt;br /&gt;&lt;br /&gt;date,day and month on which they went to picnic.Radha said that it was&lt;br /&gt;&lt;br /&gt;onThursday,May 8 and Geeta said that it was Thursday May 10.Revathi&lt;br /&gt;&lt;br /&gt;said Friday Jun 8.Now one of them told all things wrongly,others one&lt;br /&gt;&lt;br /&gt;thing wrong and the last two things wrongly.If April 1st is tuesday&lt;br /&gt;&lt;br /&gt;what is the right day,date and month?&lt;br /&gt;&lt;br /&gt;-------------------------------------------------------------------------&lt;br /&gt;     I am sending mainly c&lt;br /&gt;paper and some questions.Rao&lt;br /&gt;also will send somethig.There are 15 c q's all are discriptive.&lt;br /&gt;&lt;br /&gt;int a=2;&lt;br /&gt;f1(a++);&lt;br /&gt;}&lt;br /&gt;f1(int c)&lt;br /&gt;{&lt;br /&gt;printf("%d", c);&lt;br /&gt;}&lt;br /&gt;1)fallacy &lt;br /&gt;f()&lt;br /&gt;{&lt;br /&gt;int a;&lt;br /&gt;void c;f2(&amp;amp;c,&amp;amp;a);&lt;br /&gt;2)a=0;&lt;br /&gt;b=(a=0)?2:3;&lt;br /&gt;a) What will be the value of b? why&lt;br /&gt;b) If in 1st stmt a=0 is replaced by -1, b=?&lt;br /&gt;c) If in second stmt a=0 is replaced by -1, b=?&lt;br /&gt;3)char *a[2]&lt;br /&gt;int const *p;&lt;br /&gt;int *const p;&lt;br /&gt;struct new { int a;int b; *var[5] (struct new)&lt;br /&gt;4)f()&lt;br /&gt;{&lt;br /&gt;int a=2;&lt;br /&gt;f1(a++);&lt;br /&gt;}&lt;br /&gt;f1(int c)&lt;br /&gt;{&lt;br /&gt;printf("%d", c);&lt;br /&gt;}&lt;br /&gt;c=?&lt;br /&gt;5)f1()&lt;br /&gt;{&lt;br /&gt;f(3);}&lt;br /&gt;f(int t)&lt;br /&gt;{&lt;br /&gt;switch(t);&lt;br /&gt;{&lt;br /&gt;case 2: c=3;&lt;br /&gt;case 3: c=4;&lt;br /&gt;case 4: c=5;&lt;br /&gt;case 5: c=6;&lt;br /&gt;default: c=0;}&lt;br /&gt;value of c?&lt;br /&gt;6)Fallacy&lt;br /&gt;int *f1()&lt;br /&gt;{&lt;br /&gt;int a=5;&lt;br /&gt;return &amp;a;&lt;br /&gt;}&lt;br /&gt;f()&lt;br /&gt;int *b=f1()&lt;br /&gt;int c=*b;&lt;br /&gt;}&lt;br /&gt;7)a)Function returning an int pointer&lt;br /&gt;b)Function ptr returning an int ptr&lt;br /&gt;c)Function ptr returning an array of integers&lt;br /&gt;d)array of function ptr returning an array of integers&lt;br /&gt;(See Scham series book)&lt;br /&gt;8)fallacy&lt;br /&gt;int a;&lt;br /&gt;short b;&lt;br /&gt;b=a;&lt;br /&gt;9)Define function ?Explain about arguments?&lt;br /&gt;10)C passes  By value or By reference?&lt;br /&gt;11)Post processed code for&lt;br /&gt;abc=1;&lt;br /&gt;b=abc1; (1 or 2 blank lines are given)&lt;br /&gt;strcpy(s,"abc");&lt;br /&gt;z=abc;&lt;br /&gt;12)difference between my-strcpy and strcpy ?check&lt;br /&gt;13)f()&lt;br /&gt;{&lt;br /&gt;int *b;&lt;br /&gt;*b=2;&lt;br /&gt;}&lt;br /&gt;14)Function which gives a pointer to a binary trees const an integer value&lt;br /&gt;at each code, return function of all the nodes in binary tree.(Study)Check&lt;br /&gt;15)Calling refernce draw the diagram of function stack illustrating the&lt;br /&gt;variables in the -----then were pushed on the stack at the point when&lt;br /&gt;function f2 has been introduced&lt;br /&gt;type def struct&lt;br /&gt;{ double x,double y} point;&lt;br /&gt;main( int argc, char *arg[3])&lt;br /&gt;{double a;&lt;br /&gt;int b,c;&lt;br /&gt;f1(a,b);}&lt;br /&gt;f1(double x, int y)&lt;br /&gt;{&lt;br /&gt;point p;&lt;br /&gt;stack int n;&lt;br /&gt;f2(p,x,y)}&lt;br /&gt;f2(point p, double angle)&lt;br /&gt;{ int i,j,k,int max)&lt;br /&gt;}&lt;br /&gt;_____________________________________________________________&lt;br /&gt;1)Least no. when divide by [7 gives remainder 6,6gives 5,5 gives 4 and&lt;br /&gt;soon     ans;419&lt;br /&gt;2)What compilation do (ans source code to obj)&lt;br /&gt;3)Artficial language is provided which of the language (Lisp) check&lt;br /&gt;4)241 change its equivalent octal ?&lt;br /&gt;5)for cube and sphere 3 views are similarly draw one such figure?&lt;br /&gt;6)Write a program to exchange two variaables without temp&lt;br /&gt;7)Fortran cannot have value by reference&lt;br /&gt;8)4,6,8,__&lt;br /&gt;9)success is to failure, joy is to&lt;br /&gt;10)MEANING OF JOLLY?&lt;br /&gt;11)opposite to essential?&lt;br /&gt;12)"Raw" means&lt;br /&gt; 13)To be good "Wrestler " one should have?&lt;br /&gt;14)"Command" opposite?&lt;br /&gt;15)genuine opposite?&lt;br /&gt;16)Two proverbs are goven&lt;br /&gt;17)Sum of two consecutive nos is 55, larger one is?&lt;br /&gt;18)A person goes 4/5 of his usual speed reaches 10min lateto his&lt;br /&gt;destinaton, time taken?&lt;br /&gt;19)80% pass in english, 70%pass in maths , 10%fail in both , 144 pass in&lt;br /&gt;both . How many all appeared to the test?&lt;br /&gt;20)To get a parabola if you cut a section of?&lt;br /&gt;21)Bird is flying 120km/hr b/w B to R. two trians at B to R at 60 kmph&lt;br /&gt;The distance trvelled by the bird before it is killed.Ans.120&lt;br /&gt;22)meaning of inert&lt;br /&gt;        If any are there  rao will send you. Prepare well&lt;br /&gt;for the interview. Mostly on graphics , geometry .Prepare questions like&lt;br /&gt;(for interview)Prove  some of the angles in a triangle are 180.Angle in a&lt;br /&gt;half circle is 90.How will you measure hight of building when you  are at&lt;br /&gt;the top of the building and if you have stone with you.&lt;br /&gt;&lt;br /&gt;=================================================================================================&lt;br /&gt;&lt;br /&gt;From satthi@hotmail.comSat Sep 12 21:11:49 1998&lt;br /&gt;Date: Sun, 06 Sep 1998 09:52:12 PDT&lt;br /&gt;From: satish kandanuru &lt;satthi@hotmail.com&gt;&lt;br /&gt;To: vsr@ece, anju_ksr@mailcity.com&lt;br /&gt;Subject: hai..anju&lt;br /&gt;&lt;br /&gt; hai anji ,&lt;br /&gt;     &lt;br /&gt;      yesterday my self &amp;amp; pran sent infosys paper which is not clear.&lt;br /&gt;&lt;br /&gt;       &lt;br /&gt;        now i am sending original &amp;amp; complete paper&lt;br /&gt;&lt;br /&gt;                               &lt;br /&gt;                                MARKS: 50&lt;br /&gt;       &lt;br /&gt;        I. ROHITH CAN CATCH EITHER THE HARBOUR LINE TRAIN OR MAIN LINE&lt;br /&gt;TRAIN&lt;br /&gt;           FROM HIS NEAREST RAILWAY STATION KANDALA TO REACH HIS OFFICE.&lt;br /&gt;BOTH&lt;br /&gt;           THE TRAINS HAVE A FREQUENCY OF 10 min. IN THE MORNING OFFICE&lt;br /&gt;HOURS.&lt;br /&gt;           THE HARBOUR LINE TRAINS ARRIVE EVARY 10min. STARTING FROM&lt;br /&gt;5:00a.m.  &lt;br /&gt;           AND THE MAIN LINE TRAINS ARRIVE AT EVERY 10min. STARTING FROM&lt;br /&gt;5:02a.m&lt;br /&gt;  &lt;br /&gt;         &lt;br /&gt;          II. IN  A CERTAIN DEPARTMENT STORE THE POSITION OF&lt;br /&gt;BUYER,CASHIER,CLERK&lt;br /&gt;  &lt;br /&gt;   , FLOORWALKAR, &amp;amp; MANAGER ARE HELD, THOUGH NOT NECESSARILY&lt;br /&gt;RESPECTIVELY ,BY M$&lt;br /&gt;       &lt;br /&gt;        THE CASHIER &amp;amp; THE MANAGER WERE ROOMMATES IN COLLEGE.&lt;br /&gt;        THE BUYER IS BACHELOR,&lt;br /&gt;        EVANS &amp;amp; MISS AMES HAVE ONLY BUSINESS CONTACTS WITH EACH OTHER&lt;br /&gt;        MRS CONROY WAS GREATLY DOSAPPOINTED WHEN HER HUSBAND TOLD HER&lt;br /&gt;THAT THE&lt;br /&gt;        MANAGER HAD REFUGE TO GIVE HIM A RAISE.&lt;br /&gt;        DAVIS IS GOING TO BE THE BEST MAN WHEN THE CLErk &amp;amp;  THE CASHIER&lt;br /&gt;ARE MAR&lt;br /&gt;        RIED .&lt;br /&gt;&lt;br /&gt;        WHAT POSITION DOES EACH PERSON HELD.?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;   III.&lt;br /&gt;      TANVEER AND KUNAL WENT DOWN A DESCENDING ESCALATOR . THE ESCALATOR&lt;br /&gt;WAS&lt;br /&gt;      GOING DOWN AT A CONTANT SPEED . KUNAL WALKED DOWN THE STEPS AND&lt;br /&gt;HAD TO&lt;br /&gt;      COVER 50 STEPS TO REACH THE BOTTOM.TANVEER ON THE OTHER HAND , RAN&lt;br /&gt;DOWN&lt;br /&gt;      THE STEPS OF THE ESCALATOR AND HENCE HAD TO TAKE 75 STEPS TO REACH&lt;br /&gt;THE&lt;br /&gt;      BOTTOM . ACTUALLY , TANVEER TOOK 3 STEPS IN THE SAME TIME IT TOOK&lt;br /&gt;KUNAL&lt;br /&gt;      TO TAKE 1 STEP . HOW MANY STEPS WOULD BE visible when the ecalator&lt;br /&gt;is&lt;br /&gt;      stopped and not operating.?&lt;br /&gt;&lt;br /&gt; &lt;br /&gt; Iv. In A four team foot-ball tournament , all the teams played each&lt;br /&gt;other&lt;br /&gt;       in three rounds of matches are shown in the table -A . some of&lt;br /&gt;the  &lt;br /&gt;       results of the tournament are shown in the table _B . Using the&lt;br /&gt;clues&lt;br /&gt;       given below, please fill -in the blank columns in the result&lt;br /&gt;table-B&lt;br /&gt;       (goals for &amp;amp; goals for against ? ).&lt;br /&gt;    &lt;br /&gt;     Note : two points for win , one point for draw &amp;amp; zero points for&lt;br /&gt;defeat&lt;br /&gt;            are awarded .&lt;br /&gt;     &lt;br /&gt;      CLUES :&lt;br /&gt;      1. EAST ZONE WON THE TOURNAMENT DESPITE SCORING ONE LESS GOAL THAN&lt;br /&gt;THE&lt;br /&gt;      RUNNERS -UP.&lt;br /&gt;      2. NORTH ZONE SCORED AN ODD NUMBER OF GOALS IN THEIR FIRST ROUND&lt;br /&gt;GAME&lt;br /&gt;      3. SOUTH ZONE , WHO FAILED TO SCORE IN THEIR FINAL MATCH , WERE&lt;br /&gt;BEATEN&lt;br /&gt;      BY A TWO -GOAL MARGIN IN THE FIRST ROUND.&lt;br /&gt;      4. EAST ZONE LOST THEIR MATCH AGINST WEST ZONE.&lt;br /&gt;      5. ALL FOUR TEAMS SCORED GOALS IN THE SECOND ROUND MATCHES&lt;br /&gt;      6. WEST ZONE SCORED THE SAME NUMBER OF GOALS AGAINST EAST ZONE AS&lt;br /&gt;NORTH&lt;br /&gt;      ZONE SCORED AGINST THEM.&lt;br /&gt;      7. EAST ZONE SCORED FOUR GOALS IN ROUND TWO MATCH.&lt;br /&gt;                        Table - A ( MATCHES PLAYED )&lt;br /&gt;                ~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;      Round 1       North zone vs south zone&lt;br /&gt;                    west zone vs east zone.&lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;       round 2      south zone vs west zone&lt;br /&gt;                    east zone vs north zone&lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;      Round 3       south zone vs east zone&lt;br /&gt;                    west zone  vs north zone&lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;     &lt;br /&gt;                           Table - B   ( RESULTS )&lt;br /&gt;                          ~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;                    Played     won   draw   lost   golas for  goals  &lt;br /&gt;points&lt;br /&gt;                                                              against&lt;br /&gt;&lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  &lt;br /&gt;$&lt;br /&gt;      East Zone      3         -      -      -        ?         3       &lt;br /&gt;4&lt;br /&gt;&lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;      North zone     3         -      -      -        ?          ?      &lt;br /&gt;4&lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;      West zone      3         -      -      -        4          3      &lt;br /&gt;3&lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt;      South Zone     3         -      -      -        2           5     &lt;br /&gt;1&lt;br /&gt;        &lt;br /&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;br /&gt; &lt;br /&gt;   &lt;br /&gt;    V.&lt;br /&gt;       At six o'clock the wall clock struck 6 times . Checking with my&lt;br /&gt;watch&lt;br /&gt;       , I noticed that the time between the first &amp;amp; last strokes was 30&lt;br /&gt;seconds&lt;br /&gt;       . How long will the clock take to stike 12 at mid night (Ans 66&lt;br /&gt;seconds)&lt;br /&gt;   &lt;br /&gt;     &lt;br /&gt;   VI. PLESE ANSWER THE Q'S WITH REFERENCE TO THE FACTS GIVEN BELOW ( 8&lt;br /&gt;MARKS )&lt;br /&gt;   &lt;br /&gt;   &lt;br /&gt;      1 . THE MEMBERS OF A CERTAIN TRIBE ARE DIVIDED INTO THREE CASTES&lt;br /&gt;-ABHOR&lt;br /&gt;&lt;br /&gt;  , DRAVID , &amp;amp; MAGAR CASTES&lt;br /&gt;                   &lt;br /&gt;      2. AN ABHOR WOMEN CANNOT MARRY A DRAVID MAN &lt;br /&gt;      3. A MAGAR MAN CANNOT MARRY A DRAVID WOMEN.&lt;br /&gt;      4. A SON TAKES THE CASTE OF HIS FATHER , A DAUGHTER TAKES THE&lt;br /&gt;CASTE OF&lt;br /&gt;      HER MOTHER .&lt;br /&gt;     5. ALL MARRIAGES EXCEPT THOSE MENTIOND ABOVE ARE PERMITTED&lt;br /&gt;      6 . THERE ARE NO CHILDREN BORN OUT OF WED LOCK.&lt;br /&gt;&lt;br /&gt;     &lt;br /&gt;    Q'S&lt;br /&gt;&lt;br /&gt;    1)  AN ABHOR WOMAN MARRIES  AND HAS THREE CHILDREN , ALL OF WHOM ARE&lt;br /&gt;MEMBER$&lt;br /&gt;                                                             &lt;br /&gt;                I) IF SHE WERE TO BE MARRIED to a magar MAN , THEY HAVE&lt;br /&gt;NO MALE$&lt;br /&gt;             II) IF SHE WERE TO BE MARRIED TO ABHOR MAN ,THEY MAY HAVE&lt;br /&gt;MALE CHI$&lt;br /&gt;&lt;br /&gt;               III) IF HER FOURTH CHILD IS MALE , HE MAY BE A DRAVID,.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;       A) I ONLY&lt;br /&gt;       B) III ONLY&lt;br /&gt;        &lt;br /&gt;       C) I &amp;amp; II ONLY&lt;br /&gt;     &lt;br /&gt;          D) II &amp;amp; III ONLY&lt;br /&gt;       E) I , II &amp;amp; III&lt;br /&gt;      &lt;br /&gt;    2) IT CAN BE INFERRED FROM THE CONDITION DESCRIBED THAT A MAGAR MAN&lt;br /&gt;     &lt;br /&gt;&lt;br /&gt;    I)  CANNOT HAVE A DRAVID GRAND MOTHER&lt;br /&gt;    II) CANNOT HAVE A DRAVID MOTHER&lt;br /&gt;    III ) MAY HAVE A ABHOR MOTHER&lt;br /&gt;              A) I ONLY&lt;br /&gt;                 B) II ONLY&lt;br /&gt;                C) I  &amp;amp;  II ONLY&lt;br /&gt;                D) II &amp;amp; III ONLY&lt;br /&gt;                E) I , II &amp;amp; III&lt;br /&gt;     &lt;br /&gt;    &lt;br /&gt;   3) WHEN TWO DRAVIDS ARE MARRIED , IT IS IMPOSSIBLE FOR THEM TO HAVE&lt;br /&gt;&lt;br /&gt;      I) ANY FEMALE ABHOR DESCENDENTS&lt;br /&gt;      II ) AN ABHOR GREAT_ GRAND  DAUGHTER \&lt;br /&gt;       III) AN ABHOR GRAND DAUGHTER&lt;br /&gt;   &lt;br /&gt;        A)I ONLY&lt;br /&gt;        B) III ONLY&lt;br /&gt;        C) I &amp;amp; II ONLY&lt;br /&gt;          D) II &amp;amp; III ONLY&lt;br /&gt;       E) I , II &amp;amp; III&lt;br /&gt;                       &lt;br /&gt;   4) AN ABHOR WOMAN&lt;br /&gt;      &lt;br /&gt;    A)CAN HAVE A DRAVID GRANDSON&lt;br /&gt;    B) CANNOT HAVE A MAGAR SON&lt;br /&gt;    C) CANNOT HAVE A MAGAR GRAND SON&lt;br /&gt;    D) CAN HAVE A MAGAR DAUGHTER -IN - LAW&lt;br /&gt;    E) CANNOT HAVE A DRAVID GRAND DAUGHTER&lt;br /&gt; &lt;br /&gt;                       &lt;br /&gt;       VII.      ( 2 MARKS )&lt;br /&gt;               IN CERTAIN COMMUNITY THERE ARE THOUSAND MARRIED COUPLES .&lt;br /&gt;TWO&lt;br /&gt;      THIRDS OF THE HUSBANDS WHO ARE TALLER THAN THEIR WIVES ARE ALSO&lt;br /&gt;HEAVIER&lt;br /&gt;     AND THREE QUARTERS OF THE HUSBANDS WHO ARE HEAVIER THAN THEIR WIVES&lt;br /&gt;ARE&lt;br /&gt;     ALSO TALLER . IF THERE ARE 120 WIVES WHO ARE TALLER AND HEAVIER&lt;br /&gt;THAN THEIR&lt;br /&gt;     HUSBANDS, HOW MANY HUSBANDS ARE TALLER AND HEAVIER THAN THEIR WIVES&lt;br /&gt;?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;   VIII. BOTH THE GUPTHAS AND SINHAS HAVE TWO YOUNG SONS WHOSE AGES ARE&lt;br /&gt;        ANDER ELEVEN. THE NAMES OF THE BOYS, WHOSE AGES ROUNDEDOFF TO&lt;br /&gt;THE&lt;br /&gt;        NEAREST YEAR ARE ALL DIFFERENT, ARE RAJESH, PRAVEEN,LALITH AND&lt;br /&gt;PRATHAP.&lt;br /&gt;        TAKING THE AGES OF THE AGES OF THE BOYS ONLY TO THE NEAREST&lt;br /&gt;YEAR, THE&lt;br /&gt;        FOLLOWING STATEMENTS ARE TRUE.&lt;br /&gt;                RAJESH IS THREE YEARS YOUNGER THAN HIS BROTHER IS.&lt;br /&gt;                PRAVEEN IS THE OLDEST.&lt;br /&gt;                PRATHAP IS 5YRS OLDER THAN THE YOUNGER SINHAS BOY.&lt;br /&gt;                LALITH IS HALF AS OLD AS ONE OF THE GUPTHA BOYS.&lt;br /&gt;                THE TOTAL AGES OF THE BOYS IN EACH FAMILY DIFFER BY THE&lt;br /&gt;SAME&lt;br /&gt;                  AMOUNT TODAY AS THEY DID FIVE YEARS AGO.  --- 7 MARKS.&lt;br /&gt;       &lt;br /&gt;    &lt;br /&gt;   IX. A LONG DIVISION SUM                 ---- 7 MARKS&lt;br /&gt;                xx)xxxxxxxxx(xxxxxxx&lt;br /&gt;                    xx&lt;br /&gt;                   ---&lt;br /&gt;                     xxx&lt;br /&gt;                      xx&lt;br /&gt;                    ----&lt;br /&gt;                      xxx&lt;br /&gt;                       xx&lt;br /&gt;                      ---&lt;br /&gt;                        xx&lt;br /&gt;                        xx&lt;br /&gt;                        --&lt;br /&gt;                         xxx&lt;br /&gt;                         xxx&lt;br /&gt;                        -----&lt;br /&gt;&lt;br /&gt;                in the complete solution there are four 5's.&lt;br /&gt;                Find the missing digits.&lt;br /&gt;   &lt;br /&gt; &lt;br /&gt;&lt;br /&gt;  X. FOLLOWING SERVICES ARE OPERATED BY ASIAN AIRLINES BETWEEN THE TWO  &lt;br /&gt;&lt;br /&gt;           ARE LOCATED IN DIFFERENT COUNTRIES WITH DIFFERENT TIME ZONES.&lt;br /&gt;AS IS&lt;br /&gt;           IS NORMALLY DONE, THE TIME SHOWN IS THE LOCAL TIME-  VIZ IST&lt;br /&gt;&amp;amp; TST.&lt;br /&gt;                                        REGULAR FLIGHT  SUPERSONICFLIG&lt;br /&gt;                ARRIVE ALEXANDRIA       17:10 TST          15:40 TST&lt;br /&gt;                DEPART  "    "          20:50 TST           22:50 TST&lt;br /&gt;                ARRIVE RAMPUR           23:40 IST              ?   IST&lt;br /&gt;               &lt;br /&gt;                THE ARRIVAL TIME OF SUPERSONIC FLIGHT INTO RAMPUR FROM&lt;br /&gt;ALEXANDR$&lt;br /&gt;                ARRIVAL TIME OF THE SUPER SONIC FLIGHT ASSUMING EACH&lt;br /&gt;SERVICE-&lt;br /&gt;                REGULAR AND SUPERSONIC MAINTAINS ITS OWN CONSTANT SPEED&lt;br /&gt;OF FLIG$&lt;br /&gt;               &lt;br /&gt;   &lt;br /&gt;                DON'T ANSWER FOR  ALL THE PUZZULES. 8 ARE THE SAFE ONE.&lt;br /&gt;&lt;br /&gt;===============================================================================&lt;br /&gt;&lt;br /&gt;                       INFOSYS 1997.           TIME 1hr.&lt;br /&gt;               PAPER CODE Q/A.        marks 50. &lt;br /&gt;--------------------------------------------------------------&lt;br /&gt;1)At 6'o clock clock ticks 6 times.  The time between first and&lt;br /&gt;last ticks was 30sec.  How much time it takes at 12'o clock.&lt;br /&gt;&lt;br /&gt;Ans. 66 sec.                 2 marks.&lt;br /&gt;&lt;br /&gt;2)Three friends divided some bullets equally.  After all of them&lt;br /&gt;shot 4 bullets the total no.of remaining bullets is equal to that of&lt;br /&gt;one has after division.  Find the original number divided.&lt;br /&gt;&lt;br /&gt;Ans. x          x       x&lt;br /&gt;     x-4        x-4     x-4&lt;br /&gt;     3x-12 = x&lt;br /&gt;     x= 6&lt;br /&gt;     ans is 18                  2 marks&lt;br /&gt;&lt;br /&gt;3)A ship went on a voyage after 180 miles a plane statrted with 10 times&lt;br /&gt;speed  that of the ship.  Find the distance when they meet from&lt;br /&gt;starting point.&lt;br /&gt;&lt;br /&gt;Ans.  180 + (x/10) = x&lt;br /&gt;       x = 20&lt;br /&gt;       ans is 180+20=200miles.             2 marks&lt;br /&gt;&lt;br /&gt;4) Fill the empty slots.&lt;br /&gt;  Three FOOTBALL teams are there. Given below the list of maches.&lt;br /&gt;&lt;br /&gt;        played  won     lost    draw    Goals for       Goals against&lt;br /&gt;A        2       2       *0      *0        *7                 1         &lt;br /&gt;B        2       *0      *1       1        2                    4&lt;br /&gt;C        2       *0      *1      *1        3                   7&lt;br /&gt;&lt;br /&gt;the slots with stars are answers.   4 marks&lt;br /&gt;BC drew with  2-2&lt;br /&gt;A won on B by 2-0&lt;br /&gt;a won on C by 5-1&lt;br /&gt;( YOU HAVE TO FILL THE BLANKS AT APPROPRIATE STAR SYMBOLS.)`&lt;br /&gt;5) There are 3 societies a,b,c.  a lent tractors to b and c as many&lt;br /&gt;as they had.  After some time b gave as many tractors to a and c&lt;br /&gt;as many as they have.  After sometime c did the same thing.&lt;br /&gt;At the end of this transaction each one of them had 24.&lt;br /&gt;Find the tractors each orginally had.&lt;br /&gt;&lt;br /&gt;Ans a had 39, b had 21, c had 12,          4 marks&lt;br /&gt;&lt;br /&gt;6) There N stations on a railroad. After adding x stations 46 additional&lt;br /&gt;tickets have to be printed. Find N and X.&lt;br /&gt;&lt;br /&gt;Ans. let N(N-1) = t;&lt;br /&gt;         (N+x)(N+x-1)       = t+46;&lt;br /&gt;         trail and error method   x=2  and N=11   4 marks&lt;br /&gt;&lt;br /&gt;7)Given that April 1 is tuesday. a,b,c are 3 persons told that their&lt;br /&gt;farewell party was on&lt;br /&gt;  a  - may 8, thursday&lt;br /&gt;  b  - may 10,tuesday&lt;br /&gt;  c  - june 8, friday&lt;br /&gt;Out of a,b,c one is only correct one of the regarding month,day,date.&lt;br /&gt;Other told two correct and the third person told all wrong.What is&lt;br /&gt;correct date,month,day.                 5 marks&lt;br /&gt;(ans may be MAY 10 SUNDAY. check once again)&lt;br /&gt;8)There are 4 parties.  df,gs,dl(depositloss),ew ran for a contest.&lt;br /&gt;Anup,Sujit,John made the following statements regarding results.&lt;br /&gt;Anup said either df or ew will definitely win&lt;br /&gt;sujit said he is confident that df will not win&lt;br /&gt;John said he is confident that neither ew nor dl will win&lt;br /&gt;the result has come.  only one of the above three has made a correct&lt;br /&gt;statement.  Who has made the correct statement and who has won&lt;br /&gt;the contest.                    5 marks.&lt;br /&gt;  (ans DL )&lt;br /&gt;9)Five people a,b,c,d,e are related to each other.  Four of them make&lt;br /&gt;one true statement each as follows.&lt;br /&gt;i) b is my father's brother.       (ans. d said this)&lt;br /&gt;ii)e is my mother-in-law.          (     b          )&lt;br /&gt;iii)c is my son-in-law's brother.  (     e          )&lt;br /&gt;iv)a is my brother's wife.         (     c          )&lt;br /&gt;who said each statement.&lt;br /&gt;                                        10 marks.&lt;br /&gt;&lt;br /&gt;10) All members of d are also members of a&lt;br /&gt;    All   ''       e   ''                d&lt;br /&gt;    all    ''      c    ''               both a and b&lt;br /&gt;    not all ''     a  are members of d&lt;br /&gt;    not all ''     d   ''            e&lt;br /&gt;Some questions on these conditions.(5questions 5 marks)&lt;br /&gt;11)&lt;br /&gt;   boys are allowed to watch football at c.v.Raman auditorium subjected to&lt;br /&gt;   conditions.&lt;br /&gt;   i)the boy over age 16 can wear overcoat&lt;br /&gt;   ii)no boy over age  15 can wear cap&lt;br /&gt;   iii)to watch the football either he has to wear overcoat or cap&lt;br /&gt;        or both&lt;br /&gt;   iv) a boy with an umberella or above 16 or both cannot wear sweater.&lt;br /&gt;   v)  boys must either not watch football or wear sweater.&lt;br /&gt;&lt;br /&gt;   What is the appearence of the boy who is watching football.&lt;br /&gt;   Try to solve this question................&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;                      ok!!!!!&lt;br /&gt;                 infosys paper 1997.&lt;br /&gt;         paper code q/c.&lt;br /&gt;------------------------------------------------------------------------&lt;br /&gt;&lt;br /&gt;1)   be * be = acb&lt;br /&gt;     a,b,c,e   are non zero numbers find b,e.&lt;br /&gt;     ans)  b=1   e=9&lt;br /&gt;2)   a,b,c,d,e  are having numerical values. there are some conditions&lt;br /&gt;given&lt;br /&gt;   a)  a=c  &lt;===&gt;  b!=e&lt;br /&gt;   b) difference between a and c as same as difference between c and b &lt;br /&gt;      as same as difference between a and d&lt;br /&gt;   c)  c&lt;a&gt;d &lt;br /&gt;  then find a,b,c,d,e&lt;br /&gt;3)   there are six cards in which it has two king cards. all cards are &lt;br /&gt;turned down and two cards are opened&lt;br /&gt;  a) what is the possobility to get at least one king.&lt;br /&gt;  b) what is the possibility to get two kings.&lt;br /&gt;4) a person went to a shop and asked for change for 1.15paise.&lt;br /&gt;but he said that he could not only give change for one rupee.&lt;br /&gt;but also for 50p,25p,10p  and 5p. what were the coins he had &lt;br /&gt;   ans) 1--&gt;50   4---&gt;10p      1---&gt;25p&lt;br /&gt;5)  there are 3 nurses and they work altogether only once in a week.&lt;br /&gt;no nurse  is called to work for 3 consecutive days.&lt;br /&gt;  nurse 1 is off on tueseday,thursday and sunday.&lt;br /&gt;  nurse 2 is off on saturday.&lt;br /&gt;  nurse 3 is off on thursday,sunday.&lt;br /&gt;  no two nurses are off more than once a week.&lt;br /&gt;    find the day on which all the 3 nurses were on work.&lt;br /&gt;6)  there are 5 persons a,b,c,d,e and each is wearing a block or white &lt;br /&gt;cap on his head. a person can see the caps of the remaining 4 but can't&lt;br /&gt;see his own cap. a person wearing white says true and who wears block &lt;br /&gt;says false.&lt;br /&gt;  i) a says i see 3 whites and 1 block&lt;br /&gt; ii) b says i see 4 blocks&lt;br /&gt;iii) e says i see 4 whites&lt;br /&gt;iiii) c says i see 3 blocks and 1 white.&lt;br /&gt;    now find the caps weared by a,b,c,d and e&lt;br /&gt;7) there are two women, kavitha and shamili and two males shyam, aravind&lt;br /&gt;who are musicians. out of these four one is a pianist, one flutist,&lt;br /&gt;violinist and drummer.&lt;br /&gt;    i) across aravind beats pianist&lt;br /&gt;   ii) across shyam is not a flutist&lt;br /&gt;  iii) kavitha's left is a pianist&lt;br /&gt; iiii) shamili's left is not a drummer&lt;br /&gt; v)    flutist and drummer are married.&lt;br /&gt;8) 1/3 ed of the contents of a container evaporated on the 1 st day.&lt;br /&gt;   3/4 th of the remaining contents of the container evaporated the &lt;br /&gt;   second day. what part of the contents of the container are left at&lt;br /&gt;   the end of the second day.&lt;br /&gt;9) a man covered 28 steps in 30 seconds but he decided to move fast and &lt;br /&gt;covered 34 steps in 18 seconds. how many steps are there on the &lt;br /&gt;escalator when stationary.&lt;br /&gt;10)  all fair skinned, rich, handsome, muscular, lean and employed are &lt;br /&gt;tall men&lt;br /&gt; 1) all lean men are muscular.&lt;br /&gt; 2) no fairskinned person who is not rich is handsome.&lt;br /&gt; 3) some muscular men are handsome.&lt;br /&gt; 4) all handsome are fairskinned.&lt;br /&gt; 5) no person who is neither fair skinned nor muscular is enplyed.&lt;br /&gt; 6)  we unable to recall this condition  and  question also incomplete.&lt;br /&gt;------------------------------------------------------------------------&lt;br /&gt;&lt;br /&gt;===============================================================================&lt;br /&gt;&lt;br /&gt;&gt; Here are the infosys questions which appeared here.,  They had&lt;br /&gt;&gt; five sets of que. papers A,B,C,D,E.  This is set A.  Others&lt;br /&gt;&gt; are also of similar type with few modifications.&lt;br /&gt;&gt; love-kalya.&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; INFOSYS&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; &gt; There are 5 q papers  with them. this is only one of those.&lt;br /&gt;&gt; &gt; All otheres were more or less of the same model but different&lt;br /&gt;&gt; &gt; qs. This is just to give U an idea.&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;     The questions are not in order.&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 1)A,B,C,D,E related.4 of them made these statements each.&lt;br /&gt;&gt; &gt;   i)C is my son in law's brother.&lt;br /&gt;&gt; &gt;  ii)B is my father's brother.&lt;br /&gt;&gt; &gt;  iii)E is my mother in law.&lt;br /&gt;&gt; &gt;  iv)A is my brother's wife.&lt;br /&gt;&gt; &gt;  who made these statements?(person mentioned is one of A,B,C,D,E)(10 marks).&lt;br /&gt;&gt; &gt; 2)e means belong.&lt;br /&gt;&gt; &gt;   All members of E e D.&lt;br /&gt;&gt; &gt;   All members of D e A.&lt;br /&gt;&gt; &gt;   Not all members of D e E.&lt;br /&gt;&gt; &gt;   Not all members of A e D.&lt;br /&gt;&gt; &gt;   All members of C e both A and B.some questions are asked about relatio&lt;br /&gt;         n.use venn diagram.(5 marks). &lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 3)complete the table.&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;     Played  won lost draw  goals  goals&lt;br /&gt;&gt; &gt;                for    against&lt;br /&gt;&gt; &gt; A    2       2                     1&lt;br /&gt;&gt; &gt; B    2                 1    2      4&lt;br /&gt;&gt; &gt; C    2                      3      7&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;   A,B,C are 3 hockey teams.(2 marks).&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 4) A says Party was held on :Thursday ,May 8th.&lt;br /&gt;&gt; &gt;    B says Party was held on :Tuesday,May 10th.&lt;br /&gt;&gt; &gt;    C says party was held on :Friday ,June 8th.  &lt;br /&gt;&gt; &gt;    Given April 1 st was Tuesday.one of A,B,C says 1 correct.one says 1&lt;br /&gt;&gt; &gt;    wrong.and one was completely wrong of date,Month and day. Find the&lt;br /&gt;&gt; &gt;    Day  the party held.   (5marks).&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 5) A ship is away from the shore by 180 miles.A plane is travelling at&lt;br /&gt;&gt; &gt;    10 times speed of the ship.How long from the shore will they meet?&lt;br /&gt;&gt; &gt;    (2marks)&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 6) Every station in N railroad issue everyother station's ticket.&lt;br /&gt;&gt; &gt;    some stations are added.Now they have to issue 46 more tickets.&lt;br /&gt;&gt; &gt;    say the No.of stations after and before added.(5 marks).&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 7) 3 persons say these statements.&lt;br /&gt;&gt; &gt;    A says either Democratic or liberal wins the elections.&lt;br /&gt;&gt; &gt;    B says Democratic wins.C says neither democratic nor liberal wins&lt;br /&gt;&gt; &gt;    the election.of these only one is wrong.who wins the election?&lt;br /&gt;&gt; &gt;    (5 marks).&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 8) A clock showing 6 o'clock takes 30 secs to strike 6 times.How long&lt;br /&gt;&gt; &gt;    will it take to strike 12 at midnight?Ans.66 secs.(2marks)&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 9) Only boys aged &gt; 16 wear coats.&lt;br /&gt;&gt; &gt;    Boys aged &gt; 15 go to watch football.some more statements are given.&lt;br /&gt;&gt; &gt;    What can be said about those who are watching football ? (age and&lt;br /&gt;&gt; &gt;    costume)&lt;br /&gt;&gt; &gt;    (5 marks).&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; 10) There are 3 societies A,B,C having some tractors each.&lt;br /&gt;&gt; &gt;     A Gives  B and C as many tractors as they already have.&lt;br /&gt;&gt; &gt;     After some days B gives A and C as many tractors as they have.&lt;br /&gt;&gt; &gt;     After some days C gives A and B as many tractors as they have.&lt;br /&gt;&gt; &gt;     Finally each has 24 tractors.what is the original No.of&lt;br /&gt;&gt; &gt;     tractors each had in the beginning?&lt;br /&gt;&gt; &gt;         Ans.A -39.&lt;br /&gt;&gt; &gt;             B- 21.&lt;br /&gt;&gt; &gt;             C- 12.(7 marks).&lt;br /&gt;&gt; &gt; 11) 4,5 statements.From that find the answer.(7 marks).&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt; Reference books&lt;br /&gt;&gt; &gt; 1.) Puzzles and teasers by summer's&lt;br /&gt;&gt; &gt; 2.) Shakuntala Devi. (puzzles).&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;=================================================================================================&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;                 INFOSYS PAPER 1997.&lt;br /&gt;                 PAPER CODE Q/C.&lt;br /&gt;------------------------------------------------------------------------&lt;br /&gt;1a)   BE * BE = ACB&lt;br /&gt;     A,B,C,E   ARE NON ZERO NUMBERS FIND B,E.&lt;br /&gt;     ANS)  B=1   E=9&lt;br /&gt;2)   A,B,C,D,E  ARE HAVING NUMERICAL VALUES. THERE ARE SOME CONDITIONS&lt;br /&gt;GIVEN&lt;br /&gt;   a)  A=C  &lt;===&gt;  B!=E&lt;br /&gt;   b) DIFFERENCE BETWEEN A AND C AS SAME AS DIFFERENCE BETWEEN C AND B&lt;br /&gt;      AS SAME AS DIFFERENCE BETWEEN A AND D&lt;br /&gt;   c)  C&lt;a&gt;D&lt;br /&gt;  THEN FIND A,B,C,D,E&lt;br /&gt;3)   THERE ARE SIX CARDS IN WHICH IT HAS TWO KING CARDS. ALL CARDS ARE&lt;br /&gt;TURNED DOWN AND TWO CARDS ARE OPENED&lt;br /&gt;  a) WHAT IS THE POSSOBILITY TO GET AT LEAST ONE KING.&lt;br /&gt;  b) WHAT IS THE POSSIBILITY TO GET TWO KINGS.&lt;br /&gt;4) A PERSON WENT TO A SHOP AND ASKED FOR CHANGE FOR 1.15PAISE.&lt;br /&gt;BUT HE SAID THAT HE COULD NOT ONLY GIVE CHANGE FOR ONE RUPEE.&lt;br /&gt;BUT ALSO FOR 50P,25P,10P  AND 5P. WHAT WERE THE COINS HE HAD&lt;br /&gt;   ans) 1--&gt;50   4---&gt;10P      1---&gt;25P&lt;br /&gt;5)  THERE ARE 3 NURSES AND THEY WORK ALTOGETHER ONLY ONCE IN A WEEK.&lt;br /&gt;NO NURSE  IS CALLED TO WORK FOR 3 CONSECUTIVE DAYS.&lt;br /&gt;  NURSE 1 IS OFF ON TUESEDAY,THURSDAY AND SUNDAY.&lt;br /&gt;  NURSE 2 IS OFF ON SATURDAY.&lt;br /&gt;  NURSE 3 IS OFF ON THURSDAY,SUNDAY.&lt;br /&gt;  NO TWO NURSES ARE OFF MORE THAN ONCE A WEEK.&lt;br /&gt;    FIND THE DAY ON WHICH ALL THE 3 NURSES WERE ON WORK.&lt;br /&gt;6)  THERE ARE 5 PERSONS A,B,C,D,E AND EACH IS WEARING A BLOCK OR WHITE&lt;br /&gt;CAP ON HIS HEAD. A PERSON CAN SEE THE CAPS OF THE REMAINING 4 BUT CAN'T&lt;br /&gt;SEE HIS OWN CAP. A PERSON WEARING WHITE SAYS TRUE AND WHO WEARS BLOCK&lt;br /&gt;SAYS FALSE.&lt;br /&gt;  i) A SAYS I SEE 3 WHITES AND 1 BLOCK&lt;br /&gt; ii) B SAYS I SEE 4 BLOCKS&lt;br /&gt;iii) E SAYS I SEE 4 WHITES&lt;br /&gt;iiii) C SAYS I SEE 3 BLOCKS AND 1 WHITE.&lt;br /&gt;    NOW FIND THE CAPS WEARED BY A,B,C,D AND E&lt;br /&gt;7) THERE ARE TWO WOMEN, KAVITHA AND SHAMILI AND TWO MALES SHYAM, ARAVIND&lt;br /&gt;WHO ARE MUSICIANS. OUT OF THESE FOUR ONE IS A PIANIST, ONE FLUTIST,&lt;br /&gt;VIOLINIST AND DRUMMER.&lt;br /&gt;    i) ACROSS ARAVIND BEATS PIANIST&lt;br /&gt;   ii) ACROSS SHYAM IS NOT A FLUTIST&lt;br /&gt;  iii) KAVITHA'S LEFT IS A PIANIST&lt;br /&gt; iiii) SHAMILI'S LEFT IS NOT A DRUMMER&lt;br /&gt; V)    FLUTIST AND DRUMMER ARE MARRIED.&lt;br /&gt;8) 1/3 ED OF THE CONTENTS OF A CONTAINER EVAPORATED ON THE 1 ST DAY.&lt;br /&gt;   3/4 TH OF THE REMAINING CONTENTS OF THE CONTAINER EVAPORATED THE&lt;br /&gt;   SECOND DAY. WHAT PART OF THE CONTENTS OF THE CONTAINER ARE LEFT AT&lt;br /&gt;   THE END OF THE SECOND DAY.&lt;br /&gt;9) A MAN COVERED 28 STEPS IN 30 SECONDS BUT HE DECIDED TO MOVE FAST AND&lt;br /&gt;COVERED 34 STEPS IN 18 SECONDS. HOW MANY STEPS ARE THERE ON THE&lt;br /&gt;ESCALATOR WHEN STATIONARY.&lt;br /&gt;10)  ALL FAIR SKINNED, RICH, HANDSOME, MUSCULAR, LEAN AND EMPLOYED ARE&lt;br /&gt;TALL MEN&lt;br /&gt; 1) ALL LEAN MEN ARE MUSCULAR.&lt;br /&gt; 2) NO FAIRSKINNED PERSON WHO IS NOT RICH IS HANDSOME.&lt;br /&gt; 3) SOME MUSCULAR MEN ARE HANDSOME.&lt;br /&gt; 4) ALL HANDSOME ARE FAIRSKINNED.&lt;br /&gt; 5) NO PERSON WHO IS NEITHER FAIR SKINNED NOR MUSCULAR IS ENPLYED.&lt;br /&gt; 6)  we unable to recall this condition  and  question also incomplete.&lt;br /&gt;------------------------------------------------------------------------&lt;br /&gt;&lt;br /&gt;=====================================================================================&lt;br /&gt;&lt;br /&gt;This paper contains INFOSYS interview questions.&lt;br /&gt;=================================================&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1&gt; S: Satch         J:Jute           ANS&lt;br /&gt;    person1: Most of us are satch     J&lt;br /&gt;    person2:Most of us are jute       S&lt;br /&gt;    person3:Two of us are satch       J&lt;br /&gt;    person4: Three of us are jute     J&lt;br /&gt;    person5:I am satch                J&lt;br /&gt;    we have to find who is satch and who is jute.&lt;br /&gt;2&gt; Four perssons are three to cross a bridge they have one torch light&lt;br /&gt;    one person  A can cross in  1 min&lt;br /&gt;        person  B  can cross in 2 min&lt;br /&gt;        person  C can  cross in 5 min&lt;br /&gt;         person D can cross in 10 min&lt;br /&gt; They have to cross bridge with in 17 min.&lt;br /&gt;  At a time only two persons can cross&lt;br /&gt;  I    A&amp;amp;B    ----&gt; 2&lt;br /&gt;  II   A      &lt;----1&lt;br /&gt;  III  C&amp;amp;D     ----&gt;10&lt;br /&gt;  IV   B       &lt;----2&lt;br /&gt;  V    A&amp;amp;B      ----&gt;2&lt;br /&gt;               __________&lt;br /&gt;      Total         17&lt;br /&gt;&lt;br /&gt;3&gt;7 hours  hour glass&lt;br /&gt;  11 hours hour glass&lt;br /&gt;  you have to boil one egg for 15 hours&lt;br /&gt;4&gt; What is the maximum number of slices can you obtain by cutting&lt;br /&gt;  a cake with only 4 cuts?&lt;br /&gt;  ANS: 16&lt;br /&gt;5&gt; Three are three boxes ,&lt;br /&gt;   In one box Two white balls&lt;br /&gt;   In two box 2 black balls&lt;br /&gt;   In three box 1 white &amp;amp;1 black&lt;br /&gt;   The lables on the boxes are not correct.Then you have to open&lt;br /&gt;   one box and to find the colour of the balls in all boxes.&lt;br /&gt;&lt;br /&gt;Solution: Open the box labled black&amp;amp; white&lt;br /&gt;If white balls are there then the box labled with white balls&lt;br /&gt;contain black balls and labled with black balls contain one&lt;br /&gt;black and one white ball and viceversa if two black ballsare there.&lt;br /&gt;&lt;br /&gt;========================================================================================&lt;br /&gt;&lt;br /&gt;PART 1&lt;br /&gt;&lt;br /&gt;1). A beggr collects cigarette stubs and makes one ful  cigarette&lt;br /&gt;with every 7 stubs.  Once he gets 49 stubs .  How many cigarettes&lt;br /&gt;can he smoke totally.&lt;br /&gt;                     Ans. 8&lt;br /&gt;2). A soldiar looses his way in a thick jungle  at  random  walks&lt;br /&gt;from his camp but  mathematically  in  an  interestingg  fashion. &lt;br /&gt;First he walks one mile east then half mile to north.   Then  1/4&lt;br /&gt;mile to west, then 1/8 mile to south and so  on  making  a  loop.&lt;br /&gt;Finally hoe far he is from his camp and in which direction.&lt;br /&gt;ans: in north and south directions&lt;br /&gt;   1/2 - 1/8 + 1/32 - 1/128 + 1/512 - and so on&lt;br /&gt;                = 1/2/((1-(-1/4))&lt;br /&gt;      similarly in east and west directions&lt;br /&gt;     1- 1/4 + 1/16 - 1/64 + 1/256 -    and so on&lt;br /&gt;             = 1/(( 1- ( - 1/4))&lt;br /&gt;   add both the answers&lt;br /&gt;3). hoe 1000000000 can be written as a  product  of  two  factors&lt;br /&gt;neither of them containing zeros&lt;br /&gt;   Ans   2 power 9 x 5 ppower 9  ( check the answer )&lt;br /&gt;4). Conversation between two mathematcians:&lt;br /&gt;  first : I have three childern. Thew pproduct of their ages is  36&lt;br /&gt;.  If you sum their ages . it is exactly same as  my  neighbour's&lt;br /&gt;door number on my left. The  sacond  mathematiciaan  verfies  the&lt;br /&gt;door number and says that the not sufficient  .  Then  the  first&lt;br /&gt;says " o.k one more clue is that my youngest is the youngest"&lt;br /&gt;Immmediately the second mathematician answers . Can  you  aanswer&lt;br /&gt;the questoion asked by the first mathematician?&lt;br /&gt;What are the childeren ages?        ans  2 and 3 and 6&lt;br /&gt;5). Light glows for every 13 seconds .  How  many  times  did  it&lt;br /&gt;between 1:57:58 and 3:20:47 am&lt;br /&gt;      ans : 383 + 1 = 384&lt;br /&gt;&lt;br /&gt;6). 500 men are  arranged in an array of 10 rows and 50 columns .&lt;br /&gt;ALL tallest among each row aare asked to   fall  out  .  And  the&lt;br /&gt;shortest among THEM is A. Similarly after resuming that to  their&lt;br /&gt;originaal podsitions that the shorteest  among  each  column  are&lt;br /&gt;asked to fall out. And the longest among them is B . Now  who  is&lt;br /&gt;taller among A and B ?&lt;br /&gt;      ans  A &lt;br /&gt;7). A person spending out 1/3 for cloths , 1/5 of the  remsaining&lt;br /&gt;for food and 1/4 of the remaining for travelles  is left with&lt;br /&gt;Rs 100/- . How he had in the begining ?&lt;br /&gt;    ans    RS 250/-&lt;br /&gt;8). there are six  boxes containing 5 , 7 , 14 , 16  ,  18  ,  29&lt;br /&gt;balls of either red or blue in colour. Some  boxes  contain  only&lt;br /&gt;&lt;br /&gt;red balls and others contain only blue . One sales man  sold  one&lt;br /&gt;box out of them and then he says " I have the same number of  red&lt;br /&gt;balls left out  as that of blue ". Which box is the one he  solds&lt;br /&gt;out  ?&lt;br /&gt;Ans : total no of balls  =  89 and (89-29 /2 = 60/2 = 30&lt;br /&gt;     and also 14 + 16 = 5 + 7 + 18 = 30&lt;br /&gt;9).  A  chain  is  broken  into  three  pieces  of  equal  lenths&lt;br /&gt;conttaining 3 links each. It is taken to a backsmith to join into&lt;br /&gt;a single continuous one . How many links are to  tobe  opened  to&lt;br /&gt;make it ?&lt;br /&gt;    Ans : 2.&lt;br /&gt;10). Grass in lawn grows equally thickand in a uniform  rate.  It&lt;br /&gt;takes 24 days for 70 cows and 60 for 30 cows . How many cows  can&lt;br /&gt;eat away the same in 96 days.?&lt;br /&gt;    Ans : 18 or 19&lt;br /&gt;11). There is a certain four digit number whose fourth  digit  is&lt;br /&gt;twise the first digit.&lt;br /&gt;   Third digit is three more than second digit.&lt;br /&gt;   Sum of the first and fourth digits twise the third number.&lt;br /&gt;   What was that  number ?&lt;br /&gt;  Ans : 2034 and 4368&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;    If you qualify in the first part then you have to appear for&lt;br /&gt;the second i.e the following part.&lt;br /&gt;&lt;br /&gt;Part 2.&lt;br /&gt;&lt;br /&gt;1.  From  a  vessel  on  the  first  day,  1/3rd  of  the  liquid&lt;br /&gt;evaporates.  On the second day  3/4th  of  the  remaining  liquid&lt;br /&gt;evaporates.  what fraction of the volume is present at the end of&lt;br /&gt;the II day. &lt;br /&gt;2. an orange galss has orange juice.  and white glass  has  apple&lt;br /&gt;juice.  Bothe equal volume 50ml of the orange juice is taken  and&lt;br /&gt;poured into the apple juice.  50ml from the white glass is poured&lt;br /&gt;into the orange glass.  Of the  two  quantities,  the  amount  of&lt;br /&gt;apple juice in the orange glass and the amount of orange juice in&lt;br /&gt;the white glass, which one is greater and by how much?&lt;br /&gt;3.  there is a 4 inch cube painted on all  sides.   this  is  cut&lt;br /&gt;into no of 1 inch cubes.  what is the no of cubes which  have  no&lt;br /&gt;pointed sides.&lt;br /&gt;4. sam and mala have a conversation.  sam says i am vertainly not&lt;br /&gt;over 40.  mala says i am 38 and you are  atleast  5  years  older&lt;br /&gt;than me.  Now sam says you are atleast 39.  all the sattements by&lt;br /&gt;the two are false.  How hold are they realy.&lt;br /&gt;5.  ram singh goes to his office in the city, every day from  his&lt;br /&gt;suburbun house.  his driver mangaram drops  him  at  the  railway&lt;br /&gt;station in the morning and picks him up in  the  evening.   Every&lt;br /&gt;evening ram singh reaches the station  at  5  o'clock.   mangaram&lt;br /&gt;also reaches at the same time.  one day  ramsingh  started  early&lt;br /&gt;from his office and came  to  the  station  at  4  o'clock.   not&lt;br /&gt;wanting to wait for the car he  starts  walking  home.   Mangaram&lt;br /&gt;starts at normal time, picks him up on the way and takes him back&lt;br /&gt;house, half an hour early.  how much time did ram singh walk.&lt;br /&gt;6.  in a railway station, there are tow trains going.  One in the&lt;br /&gt;harbour line and one in the main line, each having a frequency of&lt;br /&gt;10 minutes.  the main line service  starts  at  5  o'clock.   the&lt;br /&gt;harbour line starts at 5.02a.m. a man goes to the  station  every&lt;br /&gt;day to catch the first train.  what is  the  probability  of  man&lt;br /&gt;catchinhg the first train&lt;br /&gt;7.  some people went for vaction.  unfortunately it rained for 13&lt;br /&gt;days when they  were  there.   but  whenever  it  rained  in  the&lt;br /&gt;morning, they had clean afternood and vice versa.   In  all  they&lt;br /&gt;enjoyed  11 morning and 12 afternoons.  how many  days  did  they&lt;br /&gt;stay there totally&lt;br /&gt;8. exalator problem repeat&lt;br /&gt;9. a survey was taken among 100 people to firn  their  preference&lt;br /&gt;of watching t.v. programmes.  there are 3 channels.  given no  of&lt;br /&gt;&lt;br /&gt;people who watch&lt;br /&gt;at least channel   1&lt;br /&gt;"         "        2&lt;br /&gt;"         "        3&lt;br /&gt;no channels at all&lt;br /&gt;atleast channels  1and 3&lt;br /&gt;"     "           1 and 2&lt;br /&gt;"        "        2 and 3&lt;br /&gt;find the no of people who watched all three.&lt;br /&gt;10. albert and fernandes they have two leg swimming  race.   both&lt;br /&gt;start from opposite and of the pool.  On the first leg, the  boys&lt;br /&gt;pass each other at 18 mt from the deep end of the  pool.   during&lt;br /&gt;the II leg they pass at 10 mt from the shallow end of  the  pool. &lt;br /&gt;Both go at const speed.  but one of them  is  faster.   each  boy&lt;br /&gt;rests for 4 sec to see at the end of the  i  leg.   what  is  the&lt;br /&gt;length of the pool.&lt;br /&gt;11. T  H  I  S                Each  alphabet  stands  for  one&lt;br /&gt;        I S                   digit, what is the maximum value T&lt;br /&gt;--------------                can take&lt;br /&gt;    X F X X&lt;br /&gt;  X X U X&lt;br /&gt;--------------&lt;br /&gt;  X X N X X&lt;br /&gt;--------------&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1. an escalator is descending at constant speed.   A  walks  down&lt;br /&gt;and takes 50 steps to reach the bottom.  B runs down and takes 90&lt;br /&gt;steps in the same time as A takes 10 steps.  how many  steps  are&lt;br /&gt;visible when the escalator is not operating.&lt;br /&gt;2. evvery day a cyclist meets a train at a  particular  crossing. &lt;br /&gt;the road is straignt before the crossing and both are  travelling&lt;br /&gt;in the same direction.  cyclist travels with a speed of 10  Kmph. &lt;br /&gt;One day the cyclist comes late by 25 min. and meets the train 5km&lt;br /&gt;before the crossing.  what is the seppd of the train.&lt;br /&gt;3. five persons muckerjee, misra, iyer,  patil  and  sharma,  all&lt;br /&gt;take then first or middle names in the full names.  There  are  4&lt;br /&gt;persons having I or middle name of kumar, 3 persons with mohan, 2&lt;br /&gt;persons withdev and 1 anil.&lt;br /&gt;--Either mukherjee and patil have a I or middle name  of  dev  or&lt;br /&gt;misra and iyer have their  I or middle name of dev&lt;br /&gt;--of mukherkjee and misre, either both of them have  a  first  or&lt;br /&gt;middle name of mohan or neither have a first or  middle  name  of&lt;br /&gt;mohan&lt;br /&gt;--either iyer of sharma has a I or middle name of kumar  hut  not&lt;br /&gt;both.&lt;br /&gt;who has the I or middle name of anil&lt;br /&gt;4. reading conprehension&lt;br /&gt;5. a bird keeper has got Ppigeon, M mynas and  S  sparrows.   the&lt;br /&gt;keeper goes for lunch leaving his assistant to watch the birds.&lt;br /&gt;a. suppose p=10, m=5, s=8 when the bird keeper  comes  back,  the&lt;br /&gt;assistant informs the x birds  have  escaped.   the  bird  keeper&lt;br /&gt;exclaims oh no! all my sparrows are gone.  how  many  birds  flew&lt;br /&gt;away.&lt;br /&gt;   b. when the bird keeper come back, the assistand told him that&lt;br /&gt;x birds have escaped.  the keeper realised that atleast2 sparrows&lt;br /&gt;have escaped.  what is minimum no of birds that can escape.&lt;br /&gt;6.  select from the five alternatives A,B,C,D,E&lt;br /&gt;   AT THE end of each question ,two  conditions  will  be  given. &lt;br /&gt;the choices are to filled at follows.&lt;br /&gt;a. if a definete conclusion can be drawn from condition 1&lt;br /&gt;b. if a definete conclusion can be drawn from condition 2&lt;br /&gt;c. if a definete conclusion can be drawn from condition 1 and 2&lt;br /&gt;d. if a definete conclusion can be drawn from condition 1 or 2&lt;br /&gt;e. no conclusion can be drawn using both conditions&lt;br /&gt;1. person 1 says N&lt;5&lt;br /&gt;   person says n&gt;5&lt;br /&gt;     person 3 says 3N&gt;20&lt;br /&gt;     person 4 says 3n&gt;10&lt;br /&gt;    person 5 says N&lt;8&lt;br /&gt;whaT IS value of N&lt;br /&gt;a) 1. no of persons who speak false being less than no of persons&lt;br /&gt;who tells the truth.&lt;br /&gt;   2. person 2 is telling the truth.&lt;br /&gt;b)  1. no of persong telling the truth  is  greater  than  no  of&lt;br /&gt;persons telling lies&lt;br /&gt;    2. person 5 is telling the truth.&lt;br /&gt;7.  there are N coins on a table.  there  are  two  players  A&amp;amp;B. &lt;br /&gt;you can take 1 or 2 coins at a time.  the person  who  takes  the&lt;br /&gt;last coin is the loser.  a always starts first&lt;br /&gt;--1. if N=7&lt;br /&gt;     a) A can always win by taking two coins in his first chanse&lt;br /&gt;     b) B can win only if A takes two coins in his first chance.&lt;br /&gt;     c) B can always win by proper play&lt;br /&gt;     d) none of the above&lt;br /&gt;--2.  A can win by proper play if N is equal to&lt;br /&gt; a) 13   b) 37  c) 22 d) 34   e) 48      ans. e.&lt;br /&gt;--3. B can win by proper play if N is equal to&lt;br /&gt;a) 25   b)26    c) 32   d) 41  e) none&lt;br /&gt;--4. if N&lt;4, can A win by proper play always&lt;br /&gt;&lt;br /&gt;8. Two turns have vertain pecular charcteristics.   One  of  them&lt;br /&gt;always lies on Monday, Wednesday, Friday.  \the other always lies&lt;br /&gt;on Tuesdays, thursdays and saturdays.  On the other days they tel&lt;br /&gt;the truth.  You are given a conversation.&lt;br /&gt;    person A--  today is sunday my name is anil&lt;br /&gt;    person B-- today is tuesday, my name is bill&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;answers for selected questions&lt;br /&gt;2.  equal                          1. 150&lt;br /&gt;3.  8                              2. 60 kmph&lt;br /&gt;4. 37(M),41(S)                     3.  Mukherjee&lt;br /&gt;5.   45 min.                       8. today is tuesday&lt;br /&gt;6.   0.8&lt;br /&gt;7.   18&lt;br /&gt;11.   T max value = 4&lt;br /&gt;&lt;br /&gt;===========================================================================================&lt;br /&gt;&lt;br /&gt;INFOSYS&lt;br /&gt;&lt;br /&gt;Release : '97&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;ROBERT,RAJ AND RASUL ALWAYS TAKE DINNER TOGETHER.AFTER DINNER EACH&lt;br /&gt;&lt;br /&gt;WILL ORDER EITHER COFFEE OR TEA.&lt;br /&gt;&lt;br /&gt;(a) IF RASUL ORDERS COFFEE, RAJ ORDERS WHAT ROBERT ORDERS&lt;br /&gt;&lt;br /&gt;(b)IF RAJ ORDERS COFFEE,RASUL ORDERS WHAT ROBERT DOESN'T ORDER&lt;br /&gt;&lt;br /&gt;(c) IF ROBERT ORDERS TEA,RASUL ORDERS WHAT RAJ ORDERS&lt;br /&gt;&lt;br /&gt;WHO ORDERS ALWAYS SAME DRINK?&lt;br /&gt;&lt;br /&gt;(i) RASUL (ii)RASUL&amp;amp;ROBERT&lt;br /&gt;&lt;br /&gt;(iii)ALL (iv)------- ans:ALL&lt;br /&gt;&lt;br /&gt;(2) MAJOR RAJ ,HIS BROTHER ,DAUGHTERS AND SON ARE ALWAYS PLAYING&lt;br /&gt;&lt;br /&gt;MIXED DOUBLES&lt;br /&gt;&lt;br /&gt;(i)MJ.RAJ BROTHER IS PLAYING ACROSS THE NET DIRECTLY OPPOSITE TO&lt;br /&gt;&lt;br /&gt;MJ.RAJ'S DAUGHTER.&lt;br /&gt;&lt;br /&gt;(ii) MJ.RAJ'S SON IS PLAYING DIAGONALLY ACROSS THE NET OF THE WORSTE&lt;br /&gt;&lt;br /&gt;PLAYERS SIBLING.&lt;br /&gt;&lt;br /&gt;(iii)EACH TEAM MADE UP OF BEST AND WORSTE PLAYER OF THE SLOT.&lt;br /&gt;&lt;br /&gt;WHO IS THE BEST PLAYER?&lt;br /&gt;&lt;br /&gt;(a)MJ.RAJ (b)HIS BROTHER (c)HIS SON (d) INSUFFICIENT DATA.&lt;br /&gt;&lt;br /&gt;(1) THERE ARE NINE CARDS NUMBERING 1-9.&lt;br /&gt;&lt;br /&gt;(i)CARD 5 IS IN CENTRE&lt;br /&gt;&lt;br /&gt;(ii) CARD 9 IS IN ROW IN WHICH CARD 5 IS THERE&lt;br /&gt;&lt;br /&gt;(iii) CARD FOUR IS IN SAME COLUMN IN WHICH CARD 9 IS PRESENT.&lt;br /&gt;&lt;br /&gt;(iv)CARD 1&amp;amp;CARD 6 ADJACENT TO EACH OTHER&lt;br /&gt;&lt;br /&gt;(v) SUM OF EACH ROW IS EQUAL TO 15 ONLY&lt;br /&gt;&lt;br /&gt;(vi) SUM OF EACH DIAGONAL IS ALSO 15&lt;br /&gt;&lt;br /&gt;(A) NINE CARDS ARE ARRANGED IN THREE ROWS AND THREE COLUMNS;&lt;br /&gt;&lt;br /&gt;(*) ADJACENT MEANS TWO CARDS ARE IN SAME ROW OR SAME COLUMN NEXT&lt;br /&gt;&lt;br /&gt;TO EACH OTHER.&lt;br /&gt;&lt;br /&gt;*) NOTE: CONSIDER ALL POSSIBLE ANSWERS&lt;br /&gt;&lt;br /&gt;(i)WHICH OF THE FOLLOWING ARE DEFINETLY ADJACENT TO EACH OTHER&lt;br /&gt;&lt;br /&gt;ANS;(1&amp;amp;5)&lt;br /&gt;&lt;br /&gt;(ii) WHICH TWO CARDS DO NOT COME DEFINETLY ADJACENT TO EACH OTHER a)5&amp;amp;4&lt;br /&gt;&lt;br /&gt;(iii) THE MIDDLE COLUMN ADDS UP TO (a) 15&lt;br /&gt;&lt;br /&gt;iv) THE MIDDLE COLUMN CAN'T BE ADDED UP TO&lt;br /&gt;&lt;br /&gt;X)10 Y)15 Z) 20 a) X&amp;amp;Z&lt;br /&gt;&lt;br /&gt;POSSIBLE TABLES&lt;br /&gt;&lt;br /&gt;2 7 6 6 2 7&lt;br /&gt;&lt;br /&gt;9 5 1 1 5 9&lt;br /&gt;&lt;br /&gt;4 3 8 3 8 4&lt;br /&gt;&lt;br /&gt;THERE ARE '3' CASTERS IN A VILLAGE MARGONS, ADHORS,AND DRAVIDS&lt;br /&gt;&lt;br /&gt;i)ADHOR WOMEN CAN'T MARRY DRAVID MEN&lt;br /&gt;&lt;br /&gt;ii) MARGON MAN CAN'T MARRY DRAVID WOMAN&lt;br /&gt;&lt;br /&gt;iii) IN A FAMILY SON WILL GET HIS FATHER'S CASTE AND DAUGHTER WILL&lt;br /&gt;&lt;br /&gt;GET HIS MOTHER'S CASTE&lt;br /&gt;&lt;br /&gt;iv) THERE IS NO CHILDREN OUT OF WEDLOCK&lt;br /&gt;&lt;br /&gt;v) OTHER COMBINATION'S ARE ALLOWED EXCEPT ABOVE COMBINATIONS&lt;br /&gt;&lt;br /&gt;QUESTIONS:&lt;br /&gt;&lt;br /&gt;i)ADHOR FEMALES CAN HEVE&lt;br /&gt;&lt;br /&gt;i) ii) iii)&lt;br /&gt;&lt;br /&gt;4) IF TWO DRAVIDAN'S GOT MARRIAGE IT IS IMPOSSIBLE FOR&lt;br /&gt;&lt;br /&gt;i)TO HAVE&lt;br /&gt;&lt;br /&gt;I5) ADHOR WOMAN HAVE&lt;br /&gt;&lt;br /&gt;A) B) C) D) E)&lt;br /&gt;&lt;br /&gt;------------------------- THE END ----------------------------------&lt;br /&gt;==========================================================================&lt;br /&gt;&lt;br /&gt;&gt; &gt;                         INFOSYS - 99 AT JU,CALCUTTA.&lt;br /&gt;&gt;         (1). ALL TALLMEN ARE HANDSOME,FAIRSKIMED,LEAN,NULSCULAR RICH&lt;br /&gt;&gt; &gt;AND EMPLOYED.&lt;br /&gt;&gt; &gt;                 1. ALL HANDSOME PERSONS ARE FAIRSKIMED&lt;br /&gt;&gt; &gt;                 2. SOME NULSCULAR ARE FAIR SKIMED AND HANDSOME&lt;br /&gt;&gt; &gt;                 3. ALL LEANS ARE NUSCULAR PERSONS&lt;br /&gt;&gt; &gt;                 4. NO FAIRSKIMED PERSON WHO IS NOT HANDSOME IS RICH&lt;br /&gt;&gt; &gt;                 5. NEITHER FAIRSKIMED NOR NULSCULAR ARE EMPLOYED.&lt;br /&gt;&gt; &gt;        THERE ARE 4 QUESTIONS BASED ON THE ABOVE.                8 MARKS&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;        (2). PROBLEM&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;         (3). A SOFTWARE ENGINEER STARTS FROM HOME AT 3PM FOR EVENING&lt;br /&gt;&gt;WALK.&lt;br /&gt;&gt; &gt;              HE WALKSPEED OF 4KMPH ON LEVEL GROUND AND THEN AT A SPEED&lt;br /&gt;&gt;OF&lt;br /&gt;&gt; &gt;              3KMPH ON THE UPHILL AND THEN DOWN THE HILL AT A SPEED OF&lt;br /&gt;&gt; &gt;6KMPH&lt;br /&gt;&gt; &gt;              TO THE LEVEL GROUND AND THEN AT A SPEED OF 4KMPH TO THE&lt;br /&gt;&gt;HOME&lt;br /&gt;&gt; &gt;AT 9PM&lt;br /&gt;&gt; &gt;              WHAT IS THE DISTENCE ON ONE WAY?           4MARKS&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;         (4). A BAG CONTAINS CERTAIN NUMBER OF FILES. EACH FILE IS&lt;br /&gt;&gt;NUMBERED&lt;br /&gt;&gt; &gt;WITH ONE&lt;br /&gt;&gt; &gt;              DIGIT OF 0 TO 9. SUPPOSE THE PERSON WANT TO GET THE NUMBER&lt;br /&gt;&gt; &gt;BETWEEN 1 TO&lt;br /&gt;&gt; &gt;              2000 (OR 7000 CHECK ) . HOW MANY MIN NO. OF FILES SHOULD BE&lt;br /&gt;&gt; &gt;PRESENT IN&lt;br /&gt;&gt; &gt;              THE BAG.                                           3MARKS&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;         (5). A + B + C +D = D + E + F + G = G + H + I =17.&lt;br /&gt;&gt; &gt;                 IF A = 4 WHAT ARE THE VALUES OF D AND G. EACH LETTER&lt;br /&gt;&gt;TAKEN&lt;br /&gt;&gt; &gt;              ONLY ONE OF THE DIGIT FROM 1 TO 9.                 8MARKS&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;              ANS        : A = 4 ,B = 2, C =6, D = 5, E = 3, F = 8, G =&lt;br /&gt;&gt;1,&lt;br /&gt;&gt; &gt;                           H = 7, I = 9.&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;         (6). SIX PERSONS A,B,C,D,E &amp;amp;F WENT TO SOLIDER CINIMA. THERE ARE&lt;br /&gt;&gt; &gt;SIX&lt;br /&gt;&gt; &gt;              CONSEUTIVE SEATS. A SITS IN THE FIRST SEAT FOLLOWED BY B ,&lt;br /&gt;&gt; &gt;              FOLLOWED BY C AND SOON. IF A TAKEN ONE OF THE SIX SEATS ,&lt;br /&gt;&gt; &gt;THEN&lt;br /&gt;&gt; &gt;              B SHOULD SIT ADJACENT TO A. C SHOULD SIT ADJACENT A OR B. D&lt;br /&gt;&gt; &gt;              SHOULD SIT ADJACENT TO A, B,ORC AND SOON. HOW MANY&lt;br /&gt;&gt; &gt;              POSSIBILITIES ARE THERE?&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;         (7). SUPPOSE THERE ARE 4 GRADES A, B, C, D. (A IS THE BEST AND D&lt;br /&gt;&gt; &gt;              IS THE WORST) 4 PERSONS JACK,JEAN,POUL,LUCY WROTE THEN&lt;br /&gt;&gt;FINAL&lt;br /&gt;&gt; &gt;EXAM&lt;br /&gt;&gt; &gt;              AND MADE THE STATEMENTS LIKE THIS.&lt;br /&gt;&gt; &gt;              1.  JACK: IF I WILL GET A THEN LUCY WILL GET D&lt;br /&gt;&gt; &gt;              2.  LUCY: IF I WILL GET C THEN JACK WILL GET D . JACK GRADE&lt;br /&gt;&gt; &gt;                        IN BETTER THEN POUL GRADE.&lt;br /&gt;&gt; &gt;              3.  JEAN: IF JEAN DOESNOT GET A , THEN LACK WILL NOT GET A.&lt;br /&gt;&gt; &gt;              4.  POUL: IF JACK GET A , THEN JEAN WILL NOT GET B.&lt;br /&gt;&gt; &gt;                        LUCY WILL GET C, I WON'T EITHER A OR B.&lt;br /&gt;&gt; &gt;              IF ALL THE ABOVE STATEMENTS ARE TRUE, THEN WHICH PERSON&lt;br /&gt;&gt; &gt;              WILLGET WHICH GRADES.&lt;br /&gt;&gt; &gt;         (8). EACH MAN DANCES WITH 3 WOMEN. EACH WOMEN DANCES WITH 3&lt;br /&gt;&gt;MENS.&lt;br /&gt;&gt; &gt;              AMONG EACHG PAIR OF MEN THEY HAVE EXACTLY TWO WOMEN IN&lt;br /&gt;&gt; &gt;COMMAN.&lt;br /&gt;&gt; &gt;              FIND THE NO. OF MEN &amp;amp; WOMEN.&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;              ESSAY.&lt;br /&gt;&gt; &gt;&lt;br /&gt;&gt; &gt;              1. INTERNET REVOLUTION&lt;br /&gt;&gt; &gt;              2. ROLE OF MEDIA FOR YOUNG.&lt;br /&gt;&lt;br /&gt;=====================================================================================&lt;br /&gt;&lt;br /&gt;THIS IS INFOSYS 99 FROM IITBHU&lt;br /&gt;&lt;br /&gt; q1) a frog jumps 3 ft comres bacfk 2ft in a day in how many day&lt;br /&gt;it will come out of 30ft deep well (2 marks   )ns ans 28 day&lt;br /&gt; q2) A-B=C &lt;br /&gt;D/E=F&lt;br /&gt; G+H=I&lt;br /&gt; C.F=I ANS A=9,B=5,C=4,F=2.&lt;br /&gt; Q3)when the asctual time pass 1hr wall clock ic 10 min behind it&lt;br /&gt;when 1 hr is shown by wall clock, table clock shows 10 min ahead&lt;br /&gt;of ih when table clock shows 1 hr the alarm closck goes 5minbehind&lt;br /&gt;it,when alarm clock goes 1 hr wrist watch is 5 min ahead of it assuming&lt;br /&gt;that all clocks are correcrt with actual tinme at 12 noon what will&lt;br /&gt;be time shown by wrist watch after 6 hr&lt;br /&gt; ans---5:47:32.5&lt;br /&gt; (n X 60 )50/60 X 70/60 X 55/60 X 65/60&lt;br /&gt;&lt;br /&gt; q4)a soft. engr just returned from US has eaten too mucgh fat &amp;amp;put&lt;br /&gt;a lot of weight everfy sunday he starts walking 4 km/hr on level&lt;br /&gt;ground then up at 3 km\hr then back down hill at 6km\hr then again&lt;br /&gt;on level grounggd at 4km\hr till he reaches his destination if he&lt;br /&gt;returned home at 9 p.m. e what distance did he covered&lt;br /&gt; ans  24km&lt;br /&gt; it is aslo asked in some parer fingd half of the dist covered by hoim&lt;br /&gt; in this case ans 12km&lt;br /&gt; q5 ] one family promblem was there in wich varios relations are givern&lt;br /&gt; ans is correcr  3 families&lt;br /&gt;&lt;br /&gt; q6] one question was about cards 4 persons play yjthe game of cards&lt;br /&gt;cards are of 4 colour yellow green blue,red fing colours of playing cards which are hiddeden&lt;br /&gt; ans remember the sequence yellow,yellow,green,blue,probably frob gre&lt;br /&gt; q7]  one very easy question just solve two eq, about ages&lt;br /&gt; i rembemer only ans boy=10yrsor girl =4 yrs or vice versa pl. check it&lt;br /&gt; q8]ans the questions from facts 8marks&lt;br /&gt; the member of certain tribe are divided into 3 casts abhor,dravid magar&lt;br /&gt; 1 an abhor woman cant marry dravid nman&lt;br /&gt; 2 a nmagar man cant marry a dravid man&lt;br /&gt; 3 a son takes the caste of his father a dauhgter takes cast of her mother&lt;br /&gt; 5 all marriages except those mentioned are not permirtted 6 there&lt;br /&gt;are no childern born out of a wedlock&lt;br /&gt; 1]aliertnates cant rem u acan previos question&lt;br /&gt; some ans are&lt;br /&gt; q]an abhor woman a alternative istrue ---can have a dra, grandson&lt;br /&gt; q]it can be refer from conditions described that a magar man&lt;br /&gt; ans is 2 &amp;amp;3 only iii choice --cant have a dravid mother &amp;amp; may have abhou mother&lt;br /&gt; q]an abhor marreies &amp;amp; has thr e childern allof are members aof&lt;br /&gt;abhor acsateeich of ffoloewing is true&lt;br /&gt; ans i &amp;amp; ii only --i is if c were married toa magar man&lt;br /&gt; ii-- if c were married to abhor man,they have a male child&lt;br /&gt;&lt;br /&gt;&lt;br /&gt; eassy topic&lt;br /&gt; role of media ohun youth&lt;br /&gt; &amp;amp; interner revolution&lt;br /&gt; one more q] there are 2 scales of temp A &amp;amp; B it was given  A 14 to 133&lt;br /&gt; B 36 to 87&lt;br /&gt; find the temp when temp oa a is eq. to temp of b&lt;br /&gt; ans 52.5&lt;br /&gt; let t=mx+c c=-70 m=51/119&lt;br /&gt; a=a.51/119 -70  ; a=52.5&lt;br /&gt; this all came in bhu&lt;br /&gt; TCS paper was repeat of 98 some crirtical reasoning new ques 50 t0 70&lt;br /&gt; some new quants question i rembember some bnew q&lt;br /&gt; ans are 1]some rect question %of widthis inscresaded like that ans is 4%&lt;br /&gt; 2]some packet of 1 a and 2  ---- ans 114 etCn&lt;br /&gt;&lt;br /&gt;===============================================================================&lt;br /&gt;&gt;                        INFOSYS(1998-99)&lt;br /&gt;&gt;                  &lt;br /&gt;&gt;                     LOGICAL THINKING TEST&lt;br /&gt;&gt;                     ---------------------&lt;br /&gt;&gt;                                                       Time : 1 hour&lt;br /&gt;&gt;                                                       Total Marks : 50&lt;br /&gt;&gt;&lt;br /&gt;&gt;           ALL QUESTIONS DO NOT CARRY EQUAL MARKS&lt;br /&gt;&gt;&lt;br /&gt;&gt; INSTRUCTIONS&lt;br /&gt;&gt;      &lt;br /&gt;&gt;       *  Please do not write/mark on the question paper.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       *  Read the questions carefully. Work your answers carefully&lt;br /&gt;&gt; and as&lt;br /&gt;&lt;br /&gt;&gt;          rapidly as possible.&lt;br /&gt;&gt;      &lt;br /&gt;&gt;       *  Do not spend too much time on questions that seem difficult&lt;br /&gt;&gt; for you.&lt;br /&gt;&lt;br /&gt;&gt;          If time permits go back to the ones you have left out.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       *  Answer sheets have been provided to you separately. Use the&lt;br /&gt;&gt; first&lt;br /&gt;&lt;br /&gt;&gt;          sheet to write your answers and use the remaining sheets for&lt;br /&gt;&gt; the&lt;br /&gt;&lt;br /&gt;&gt;          detailed working.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       *  On completion, please put your answer sheet along with the&lt;br /&gt;&gt; working&lt;br /&gt;&lt;br /&gt;&gt;          sheets, inside the question paper and hand them back to us.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       *  There is no negative marking.  &lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;       Do not turn the page until you are told to do so.&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;  1.                                                           2 Marks&lt;br /&gt;&gt;      &lt;br /&gt;&gt;       When Arthur is as old as his father Hailey is now, he shall be&lt;br /&gt;&gt; 5 times&lt;br /&gt;&lt;br /&gt;&gt;       as old as his son Clarke is now. By then, Clarke will be 8&lt;br /&gt;&gt; times older&lt;br /&gt;&lt;br /&gt;&gt;       than Arthur is now. The combined ages of Hailey and Arthur are&lt;br /&gt;&gt; 100&lt;br /&gt;&lt;br /&gt;&gt;       years. How old is Clarke ?&lt;br /&gt;&gt;&lt;br /&gt;&gt; 2.                                                            3 Marks&lt;br /&gt;&gt;        &lt;br /&gt;&gt;       I have one orange glass containing orange juice and another&lt;br /&gt;&gt; white glass&lt;br /&gt;&lt;br /&gt;&gt;       containing apple juice. I take 50 ml of orange juice and mix it&lt;br /&gt;&gt; with&lt;br /&gt;&lt;br /&gt;&gt;       apple juice in the white glass. Then I take 50 ml of this&lt;br /&gt;&gt; mixture and&lt;br /&gt;&lt;br /&gt;&gt;       transfer it back to the orange glass.&lt;br /&gt;&gt;       If you compare amount of apple juice in the orange glass and&lt;br /&gt;&gt; amount of&lt;br /&gt;&lt;br /&gt;&gt;       orange juice in the white glass which is more than the other&lt;br /&gt;&gt; and by&lt;br /&gt;&lt;br /&gt;&gt;       how much ?&lt;br /&gt;&gt;&lt;br /&gt;&gt; 3.                                                            3 Marks&lt;br /&gt;&gt;      &lt;br /&gt;&gt;         The seven digits in this subtraction problem are&lt;br /&gt;&gt; 0,1,2,3,4,5 and 6.&lt;br /&gt;&lt;br /&gt;&gt;       Each letter represents the same digit whenever it occurs.&lt;br /&gt;&gt;      &lt;br /&gt;&gt;&lt;br /&gt;&gt;                    D   A   D   C   B&lt;br /&gt;&gt;                      &lt;br /&gt;&gt;                        E   B   E   G&lt;br /&gt;&gt;&lt;br /&gt;&gt;                   --------------------&lt;br /&gt;&gt;                  &lt;br /&gt;&gt;                          B   F   E   G&lt;br /&gt;&gt;                       &lt;br /&gt;&gt;                     --------------------&lt;br /&gt;&gt;                       &lt;br /&gt;&gt;       What digit is represented by each letter ?&lt;br /&gt;&gt;      &lt;br /&gt;&gt;&lt;br /&gt;&gt; 4.                                                            4 Marks&lt;br /&gt;&gt;&lt;br /&gt;&gt;         The Jones have named their four boys after favourite&lt;br /&gt;&gt; relatives; their&lt;br /&gt;&lt;br /&gt;&gt;       friends, the Smiths, have done the same thing with their three&lt;br /&gt;&gt; boys.&lt;br /&gt;&lt;br /&gt;&gt;       One of the families has twin boys. From the following clues,&lt;br /&gt;&gt; can you&lt;br /&gt;&lt;br /&gt;&gt;       determine the families of all seven children and their ages ?&lt;br /&gt;&gt;&lt;br /&gt;&gt;       i)   Valantine is 4 years older than his twin brothers.&lt;br /&gt;&gt;       ii)  Winston, who is 8, and Benedict are not brothers. They are&lt;br /&gt;&gt; each&lt;br /&gt;&lt;br /&gt;&gt;            named after a grandfather.&lt;br /&gt;&gt;       iii) Briscoe is two years younger than his brother Hamilton,&lt;br /&gt;&gt; But three&lt;br /&gt;&lt;br /&gt;&gt;            years older than Dewey.&lt;br /&gt;&gt;       iv)  Decatur is 10 years old.&lt;br /&gt;&gt;       v)   Benedict is 3 years younger than Valentine; they are not&lt;br /&gt;&gt; related.&lt;br /&gt;&lt;br /&gt;&gt;       vi)  The twins are named for uncles.&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; 5.                                                            4 Marks&lt;br /&gt;&gt;&lt;br /&gt;&gt;         Motorboat A leaves shore P as B leaves Q; they move&lt;br /&gt;&gt; across the lake at&lt;br /&gt;&lt;br /&gt;&gt;       a constant speed. They meet first time 600 yards from P. Each&lt;br /&gt;&gt; returns&lt;br /&gt;&lt;br /&gt;&gt;       from the opposite shore without halting, and they meet 200&lt;br /&gt;&gt; yards from&lt;br /&gt;&lt;br /&gt;&gt;       Q. How long is the lake ?&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; 6.                                                            4 Marks&lt;br /&gt;&gt;&lt;br /&gt;&gt;         Fanta, Pepsi and Citra often eat dinner out.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       a) Each orders either coffee or tea after dinner.&lt;br /&gt;&gt;       b) If Fanta orders coffee, then Pepsi orders the drink that&lt;br /&gt;&gt; Citra orders.&lt;br /&gt;&lt;br /&gt;&gt;       c) If Pepsi orders coffee, then Fanta orders the drink that&lt;br /&gt;&gt; Citra does&lt;br /&gt;&lt;br /&gt;&gt;          not order.&lt;br /&gt;&gt;       d) If Citra orders tea, then Fanta orders the drink that Pepsi&lt;br /&gt;&gt; orders.&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;       Which person/persons always orders the same drink after dinner&lt;br /&gt;&gt; ?&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; 7.                                                            6 Marks&lt;br /&gt;&gt;&lt;br /&gt;&gt;         On the Island of imperfection there is a special road,&lt;br /&gt;&gt; Logic Lane, on&lt;br /&gt;&lt;br /&gt;&gt;       which the houses are usually reserved for the more mathematical&lt;br /&gt;&gt; inhabitants.&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;       Add, Divide and Even live in three different houses on this&lt;br /&gt;&gt; road (which&lt;br /&gt;&lt;br /&gt;&gt;       has houses numbered from 1-50). One of them is a member of the&lt;br /&gt;&gt; Pukka Tribe,&lt;br /&gt;&lt;br /&gt;&gt;       who always tell the truth; another is a member of the Wotta&lt;br /&gt;&gt; Tribe, who never&lt;br /&gt;&lt;br /&gt;&gt;       tell the truth; and the third is a member of the Shalla Tribe,&lt;br /&gt;&gt; who make&lt;br /&gt;&lt;br /&gt;&gt;       statements which are alternately true and false, or false and&lt;br /&gt;&gt; true.&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;       They make statements as follows :&lt;br /&gt;&gt;&lt;br /&gt;&gt;       ADD     : 1 The number of my house is greater than that of&lt;br /&gt;&gt; Divide's.&lt;br /&gt;&lt;br /&gt;&gt;                 2 My number is divisible by 4.&lt;br /&gt;&gt;                 3 Even's number differs by 13 from that of one of the&lt;br /&gt;&gt; others.&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;       DIVIDE  : 1 Add's number is divisible by 12.&lt;br /&gt;&gt;                 2 My number is 37.&lt;br /&gt;&gt;                 3 Even's number is even.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       EVEN    : 1 No one's number is divisible by 10.  &lt;br /&gt;&gt;                   2 My number is 30.&lt;br /&gt;&gt;                 3 Add's number is divisible by 3.&lt;br /&gt;&gt;              &lt;br /&gt;&gt;       Find to which tribe each of them belongs, and the number of&lt;br /&gt;&gt; each of&lt;br /&gt;&lt;br /&gt;&gt;       their houses.&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; 8.                                                            8 Marks&lt;br /&gt;&gt;&lt;br /&gt;&gt;      &lt;br /&gt;&gt;         The names of the inhabitants of Walkie Talkie Land sound&lt;br /&gt;&gt; strange to the&lt;br /&gt;&lt;br /&gt;&gt;       visitors, and they find it difficult to pronounce them, due to&lt;br /&gt;&gt; their&lt;br /&gt;&lt;br /&gt;&gt;       length and a few vowel sounds they contain. The Walkie Talkie&lt;br /&gt;&gt; guide is&lt;br /&gt;&lt;br /&gt;&gt;       discussing the names of four inhabitants - A,B,C and D. Their&lt;br /&gt;&gt; names&lt;br /&gt;&lt;br /&gt;&gt;       each contain upto eight syllables, although none of the four&lt;br /&gt;&gt; names&lt;br /&gt;&lt;br /&gt;&gt;       contain the same number. Two of the names contain no vowel&lt;br /&gt;&gt; sounds; one&lt;br /&gt;&lt;br /&gt;&gt;       contains one vowel sound; and one contains two vowel sounds.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       From the Guide's statements below, determine the number of&lt;br /&gt;&gt; syllables&lt;br /&gt;&lt;br /&gt;&gt;       and vowel sounds in each of the four Walkie Talkie names :&lt;br /&gt;&gt;&lt;br /&gt;&gt;       i)  The one whose name contains two vowel sounds is not A.&lt;br /&gt;&gt;       ii) C's name does not contain more than one vowel sound or&lt;br /&gt;&gt; fewer than&lt;br /&gt;&lt;br /&gt;&gt;           seven syllables.&lt;br /&gt;&gt;       iii)The name with seven syllables does not contain exactly one&lt;br /&gt;&gt; vowel&lt;br /&gt;&lt;br /&gt;&gt;           sound.&lt;br /&gt;&gt;       iv) B and C do not have names with the same number of vowel&lt;br /&gt;&gt; sounds.&lt;br /&gt;&lt;br /&gt;&gt;       v)  Neither the name with five syllables nor the name with&lt;br /&gt;&gt; seven syllables&lt;br /&gt;&lt;br /&gt;&gt;           contains more than one vowel sound.&lt;br /&gt;&gt;       vi) Neither the name with six syllables, nor the B's name,&lt;br /&gt;&gt; contains&lt;br /&gt;&lt;br /&gt;&gt;           two vowel sounds.&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt; 9.                                                            8 Marks&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;         Multiplication and Division.&lt;br /&gt;&gt;&lt;br /&gt;&gt;      &lt;br /&gt;&gt;       (1)    - - - - -  (a)          (2)          _ _ . _    (d)&lt;br /&gt;&gt;                                           ________________&lt;br /&gt;&gt;                  - - -  (b)          - - -) - - - - - . 0&lt;br /&gt;&gt;             _________________                 - - -&lt;br /&gt;&gt;                                            _______________&lt;br /&gt;&gt;              - - - - -                       &lt;br /&gt;&gt;          - - - - -                            - - - -&lt;br /&gt;&gt;         ___________________                     - - -&lt;br /&gt;&gt;                                              _______________&lt;br /&gt;&gt;          - - - - - - -  (c)                      &lt;br /&gt;&gt;                                                   - - -&lt;br /&gt;&gt;                                                   - - -&lt;br /&gt;&gt;                                                  ========&lt;br /&gt;&gt;&lt;br /&gt;&gt;         (1) and (2) represent the same two numbers - in one case&lt;br /&gt;&gt; multiplied&lt;br /&gt;&lt;br /&gt;&gt;  9together, in other case divided.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       Given that there are no 3's. find the missing digits indicated&lt;br /&gt;&gt; as&lt;br /&gt;&lt;br /&gt;&gt;       (a), (b), (c) and (d) above ?&lt;br /&gt;&gt;&lt;br /&gt;&gt; 10.                                                           8 Marks&lt;br /&gt;&gt;&lt;br /&gt;&gt;         Two identical twins have a very unusual characteristic.&lt;br /&gt;&gt; One tells&lt;br /&gt;&lt;br /&gt;&gt;       nothing but lies on Mondays, Wednesdays and Fridays, and tells&lt;br /&gt;&gt; nothing&lt;br /&gt;&lt;br /&gt;&gt;       but the truth all other days. The other tells nothing but lies&lt;br /&gt;&gt; on&lt;br /&gt;&lt;br /&gt;&gt;       Tuesdays, Thursdays and Saturdays, and tells nothing but the&lt;br /&gt;&gt; truth&lt;br /&gt;&lt;br /&gt;&gt;       all other days. On Sundays both children speak the truth.&lt;br /&gt;&gt;&lt;br /&gt;&gt;       1) Which of the following statements can be deduced from the&lt;br /&gt;&gt;          information presented ?&lt;br /&gt;&gt;        &lt;br /&gt;&gt;          i)  If it is Sunday, the twins will both say so.&lt;br /&gt;&gt;          ii) If it is not Sunday,one twin will give the correct day&lt;br /&gt;&gt; and&lt;br /&gt;&lt;br /&gt;&gt;              the other will lie about everything.&lt;br /&gt;&gt;          iii)On any given day, only one twin will give his correct&lt;br /&gt;&gt; name.&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;                a)  i only&lt;br /&gt;&gt;                b)  i and ii only&lt;br /&gt;&gt;                c)  i and iii only&lt;br /&gt;&gt;                d)  ii and iii only&lt;br /&gt;&gt;                e)  i,ii and iii&lt;br /&gt;&gt;&lt;br /&gt;&gt;       2) According to the information presented, which of the&lt;br /&gt;&gt; following&lt;br /&gt;&lt;br /&gt;&gt;          conversations will be impossible.&lt;br /&gt;&gt;&lt;br /&gt;&gt;          a) Twin A  :  "Today you are a lier"&lt;br /&gt;&gt;             Twin B  :  "You are telling the truth"&lt;br /&gt;&gt;&lt;br /&gt;&gt;          b) Twin A  :  "Today you are a lier"&lt;br /&gt;&gt;             Twin B  :  "Today I am a truthteller"&lt;br /&gt;&gt;            &lt;br /&gt;&gt;          c) Twin A  :  "Tommorow I shall be a lier"&lt;br /&gt;&gt;             Twin B  :  "That's correct"&lt;br /&gt;&gt;            &lt;br /&gt;&gt;          d) Twin A  :  "Tommorow you will be a lier"&lt;br /&gt;&gt;             Twin B  :  "Today you are a truthteller"&lt;br /&gt;&gt;            &lt;br /&gt;&gt;          e) Twin A  :  "Yesterday we were both truthtellers"&lt;br /&gt;&gt;             Twin B  :  "You are lying"&lt;br /&gt;&gt;&lt;br /&gt;&gt;        3) Assume that the twins followed a different set of&lt;br /&gt;&gt; rules, so that on&lt;br /&gt;&lt;br /&gt;&gt;           a given day both told only the truth while next day&lt;br /&gt;&gt; both only lied,&lt;br /&gt;&lt;br /&gt;&gt;         alternating days of truth telling and lying. Under these&lt;br /&gt;&gt; rules,which&lt;br /&gt;&lt;br /&gt;&gt;         of the following conversations would be possible ?&lt;br /&gt;&gt;&lt;br /&gt;&gt;          a) Twin A  :  "Today you are a lier"&lt;br /&gt;&gt;             Twin B  :  "That is correct"&lt;br /&gt;&gt;            &lt;br /&gt;&gt;          b) Twin A  :  "Today you are a lier"&lt;br /&gt;&gt;             Twin B  :  "That is not so"&lt;br /&gt;&gt;            &lt;br /&gt;&gt;          c) Twin A  :  "Tommorow we will be liers"&lt;br /&gt;&gt;             Twin B  :  "Yesterday we were truthtellers"&lt;br /&gt;&gt;            &lt;br /&gt;&gt;          d) Twin A  :  "Tommorow we will be liers"&lt;br /&gt;&gt;             Twin B  :  "You are 1 year older than I am"&lt;br /&gt;&gt;            &lt;br /&gt;&gt;          e) Twin A  :  "We always tell the truth"&lt;br /&gt;&gt;             Twin B  :  "We some times tell the truth"&lt;br /&gt;&gt;&lt;br /&gt;&gt;       4) If the twins are heard saying the following on the same&lt;br /&gt;&gt; day, which&lt;br /&gt;&lt;br /&gt;&gt;          choice presents a correct statement ?&lt;br /&gt;&gt;&lt;br /&gt;&gt;         Twin A   :   "It is Sunday Today"&lt;br /&gt;&gt;         Twin B   :   "Yesterday was Sunday"&lt;br /&gt;&gt;         Twin A   :   "it is summer season now"&lt;br /&gt;&gt;&lt;br /&gt;&gt;           a) it is a summer sunday&lt;br /&gt;&gt;         b) it is a summer monday&lt;br /&gt;&gt;         c) it is Monday but not summer&lt;br /&gt;&gt;         d) it is Sunday but not summer&lt;br /&gt;&gt;         e) it is impossible to determine whether it is Sunday or&lt;br /&gt;&gt; Monday.&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;                  -----------------*-----------------&lt;br /&gt;&lt;br /&gt;==================================================================================&lt;br /&gt;   Here is theinfosys paper It may not contain full&lt;br /&gt;&gt; questions and ans&lt;br /&gt;&lt;br /&gt;&gt;         Understand it&lt;br /&gt;&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;         1. There are 4 married couples out ofwhich 3 a group&lt;br /&gt;&gt; isneeded . Butther&lt;br /&gt;&lt;br /&gt;&gt;         should not be his of her spouse .How nmany groups are&lt;br /&gt;&gt; possible ?&lt;br /&gt;&lt;br /&gt;&gt;         Ans 32&lt;br /&gt;&gt;&lt;br /&gt;&gt;         2.In the 4 digits 1,2,3,4 how many 4 digited numbers are&lt;br /&gt;&gt; possible&lt;br /&gt;&lt;br /&gt;&gt;          which are divisable by 4? Repetetions are allowed&lt;br /&gt;&gt;         Ans 64&lt;br /&gt;&gt;&lt;br /&gt;&gt;         3. Twow men are goingalong a trackf rail in the opposite&lt;br /&gt;&gt; direction.&lt;br /&gt;&lt;br /&gt;&gt;             One goods train crossed the first person in 20 sec.&lt;br /&gt;&gt; After 10 min&lt;br /&gt;&lt;br /&gt;&gt;         the train crossed the other person who is commingin&lt;br /&gt;&gt; opposite direction&lt;br /&gt;&lt;br /&gt;&gt;          in 18 sec .Afterthe train haspassed, when thetwo&lt;br /&gt;&gt; persons will meet?&lt;br /&gt;&lt;br /&gt;&gt;&lt;br /&gt;&gt;         Approx  72min check it once.&lt;br /&gt;&gt;&lt;br /&gt;&gt;         4. Theno. of children &gt; adults . Theno .of adults &gt; the&lt;br /&gt;&gt; no .of boys .&lt;br /&gt;&lt;br /&gt;&gt;           The no.of boys &gt; no. of girls .The no.of girls &gt; no.of&lt;br /&gt;&gt; familyi                 conditions&lt;br /&gt;&lt;br /&gt;&gt;               1.No family is without a child&lt;br /&gt;&gt;       2 Every girl has atleast one brotherand sister .&lt;br /&gt;&gt;&lt;br /&gt;&gt;       Ans c&gt;a&gt;b&gt;g&gt;f; 9 6 5 4 3 .&lt;br /&gt;&gt;&lt;br /&gt;&gt;       6.There are4 boys Anand ,Anandya ,Madan and Murali with&lt;br /&gt;&gt; nicmnames perich       ,zomie ,drummy and madeena not in thesame&lt;br /&gt;&gt; order&lt;br /&gt;&lt;br /&gt;&gt;        Some com=nditons&lt;br /&gt;&gt;&lt;br /&gt;&gt;       Ans Anand : Perich&lt;br /&gt;&gt;               Anandya: drummy&lt;br /&gt;&gt;               Madan : Zombie&lt;br /&gt;&gt;               murali: Madeena&lt;br /&gt;&gt;       7.Thereare2diomans ,1 spadeand1 club and 1ace and also 1king ,1&lt;br /&gt;&gt; jack&lt;br /&gt;&lt;br /&gt;&gt;       and 1 aceare arranged in a straight line&lt;br /&gt;&gt;      &lt;br /&gt;&gt;       1.The king is at third place&lt;br /&gt;&gt;       2.Theleft of jack is a heart and itsright is king&lt;br /&gt;&gt;       3. No two red colours arein consequtive.&lt;br /&gt;&gt;       4.The queensareseperated by two cards. Write the orderor which&lt;br /&gt;&gt; suits&lt;br /&gt;&lt;br /&gt;&gt;       (hearts ,clubs )and names(jacks queensetc.)&lt;br /&gt;&gt;       8. Writeeach statementis true or false 8M&lt;br /&gt;&gt;       1.The sum of the1st three statements and the2nd false statement&lt;br /&gt;&gt; givesthe true statement.&lt;br /&gt;&lt;br /&gt;&gt;       2.The no.oftrue statements &gt;falsestatement&lt;br /&gt;&gt;       3. The sum of2nd true statement and 1st falsestatement gives&lt;br /&gt;&gt; the first&lt;br /&gt;&lt;br /&gt;&gt;       true statement.&lt;br /&gt;&gt;       4. Thereareatmost 3 falsestatements&lt;br /&gt;&gt;       5.There is no two consequtive true statements&lt;br /&gt;&gt;       6.If this containsonly 1-5 statements ,theanswer of this is&lt;br /&gt;&gt; sameasthe an  answer of the following question&lt;br /&gt;&lt;br /&gt;&gt;      &lt;br /&gt;&gt;       9.Question on Venn diagram.&lt;br /&gt;&gt;       All handsome are also fair skinned&lt;br /&gt;&gt;       Sme musularsare fair skinned&lt;br /&gt;&gt;       Some musculars are also handsome&lt;br /&gt;&gt;       All lean are  also muscular&lt;br /&gt;&gt;       Some lean are also fair skinned.&lt;br /&gt;&gt;       All rich  man inot fair skinned but all rich manare handsome&lt;br /&gt;&gt;&lt;br /&gt;&gt;       Some questions follows.&lt;br /&gt;&gt;       10&lt;br /&gt;&gt;       There are 3 pileseach containe 10 15 20 stones. There are&lt;br /&gt;&gt; A,B,C,D,F,G&lt;br /&gt;&lt;br /&gt;&gt;       and h persons .One man can catch upto four stones from any&lt;br /&gt;&gt; pile.&lt;br /&gt;&lt;br /&gt;&gt;       The last manwho takeswill  win. If first A starts next B. and&lt;br /&gt;&gt; so on&lt;br /&gt;&lt;br /&gt;&gt;       who will win?&lt;br /&gt;&gt;       Ans May be F&lt;br /&gt;&gt;      &lt;br /&gt;&gt;       Essay writing&lt;br /&gt;&gt;       1 Intrnet revolution&lt;br /&gt;&gt;       2.Media for youth&lt;br /&gt;&lt;br /&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-192313353295625070?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/192313353295625070/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=192313353295625070' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/192313353295625070'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/192313353295625070'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/10/infosys-puzzles.html' title='INFOSYS  Puzzles'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-772877686461275922</id><published>2008-08-04T18:43:00.000-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>cakes</title><content type='html'>n errand boy was collecting boxes of cakes for the Summer Fair. He collected boxes from various people in his local village and each box was labelled in Roman Numerals with the number of cakes in the box. By the time the errand boy had collected the last box, he was quite hungry, and really needed to eat at least one cake. Luckily the last box was marked with an underlined XI, meaning there were 11 cakes in it. He had the brain wave of turning the box around and underlining the number again to give the impression there were IX, that is 9 cakes. However, after eating the 2 cakes, he was still hungry. How can he change the number shown on the box again and eat more cakes?&lt;br /&gt;&lt;br /&gt;&lt;h3 class="smller"&gt;answer&lt;/h3&gt; &lt;div class="para"&gt; He could add an S to the IX to give the impression there were supposed to be 6 cakes in the box, thus eating 5 cakes in total. &lt;/div&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-772877686461275922?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/772877686461275922/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=772877686461275922' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/772877686461275922'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/772877686461275922'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/cakes.html' title='cakes'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-1566285804771309106</id><published>2008-08-04T18:32:00.001-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Crossing the Bridge</title><content type='html'>&lt;h3 class="smller"&gt;&lt;br /&gt;&lt;/h3&gt; &lt;div class="para"&gt; &lt;b&gt;Four people are on one side of a bridge that they need to cross. It is dark, there are holes in the bridge surface, and they only have one flashlight which must travel with them as they cross. To add to their troubles, the bridge will only support two people at a time.&lt;br /&gt;&lt;br /&gt;* Now person #1 can walk cross in 1 minute.&lt;br /&gt;* Person #2 has a limp and can cross in 2 minutes.&lt;br /&gt;* Person #3 has a sprained ankle can cross in 5 minutes.&lt;br /&gt;* Person #4 has a cast on his leg and takes 10 minutes to cross.&lt;br /&gt;&lt;br /&gt;No matter how fast a person can cross, he must wait for his companion. If #1 goes with #2, it takes 2 minutes for them to cross.If number #2 goes with #4,it takes them 10 minutes.The flashlight cannot be thrown, instead it must be carried and delivered from person to person. The 4 people are trying to make it across in the fastest possible time to make an important appointment in the next town.&lt;br /&gt;&lt;br /&gt;What is the fastest time in minutes for them to cross the bridge?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Answers:&lt;br /&gt;&lt;br /&gt;&lt;/b&gt;Trip 1: Person 1 &amp;amp; Person 2 across = 2 min&lt;br /&gt;Trip 2: Person 2 back = 2 min&lt;br /&gt;Trip 3: Person 10 &amp;amp;Person 5 across = 10 min&lt;br /&gt;Trip 4: Person 1 back = 1 min&lt;br /&gt;Trip 5: Person 1 w/ Person 2 = 2 min.&lt;br /&gt;Total 17 minutes.&lt;br /&gt;&lt;/div&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-1566285804771309106?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/1566285804771309106/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=1566285804771309106' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1566285804771309106'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1566285804771309106'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/crossing-bridge.html' title='Crossing the Bridge'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-2752390270153685762</id><published>2008-08-04T18:30:00.000-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Lockers Puzzle</title><content type='html'>&lt;b&gt;A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:&lt;br /&gt;&lt;br /&gt;There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it.The third goes to every third locker and, if it is closed, he opens it, and if it is open,he closes it.The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?&lt;br /&gt;&lt;br /&gt;Answer;&lt;br /&gt;&lt;br /&gt;&lt;/b&gt;open lockers are perfect squares 1, 4, 9, 16, 25........ they are the only numbers divisible by an odd number of whole numbers.&lt;br /&gt;every factor other than the number's square root is paired up with another.&lt;br /&gt;so these lockers will be "changed" an odd number of times.&lt;br /&gt;which means they will be left open.&lt;br /&gt;All the other numbers are divisible by an even number of factors and will consequently end up closed.&lt;br /&gt;&lt;br /&gt;the number of open lockers = the number of perfect squares less than or equal to 1000.&lt;br /&gt;numbers are 1 squared, 2 squared, 3 squared, 4 squared, and so on, up to 31 squared. 32 squared&gt;1000&lt;br /&gt;out of range.&lt;br /&gt;ans is 31&lt;br /&gt;&lt;br /&gt;so answer is 31&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-2752390270153685762?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/2752390270153685762/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=2752390270153685762' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2752390270153685762'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2752390270153685762'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/lockers-puzzle.html' title='Lockers Puzzle'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-2742756330302409522</id><published>2008-08-04T18:28:00.001-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Microsoft Puzzle</title><content type='html'>"U2" has a concert that starts in 17 minutes and they must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them.&lt;br /&gt;&lt;br /&gt;The flashlight must be walked back and forth, it cannot be thrown, etc.&lt;br /&gt;Each band member walks at a different speed. A pair must walk together at the rate of the slower man's pace:&lt;br /&gt;&lt;br /&gt;* Bono: - 1 minute to cross&lt;br /&gt;* Edge: - 2 minutes to cross&lt;br /&gt;* Adam: - 5 minutes to cross&lt;br /&gt;* Larry: - 10 minutes to cross&lt;br /&gt;&lt;br /&gt;For example: if Bono and Larry walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Larry then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Notes: There is no trick behind this. It is the simple movement of resources in the appropriate order. There are two known answers to this problem. This is based on a question that Microsoft gives to all prospective employees.&lt;br /&gt;&lt;br /&gt;Note: Microsoft expects you to answer this question in under 5 minutes!&lt;br /&gt;&lt;br /&gt;Answer:&lt;br /&gt;&lt;br /&gt;* Bono: - 1 minute to cross&lt;br /&gt;* Edge: - 2 minutes to cross&lt;br /&gt;* Adam: - 5 minutes to cross&lt;br /&gt;* Larry: - 10 minutes to cross&lt;br /&gt;&lt;br /&gt;Bono and Edge Bono comes back ---(2+1)&lt;br /&gt;Adam and Larry Edge comes back----(10+2)&lt;br /&gt;Edge and Bono over---------------(2)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;(2+1)+(10+2)+2= 17 mins&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Microsoft, hire me because it took me less then 3 MIns to solve :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-2742756330302409522?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/2742756330302409522/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=2742756330302409522' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2742756330302409522'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2742756330302409522'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/microsoft-puzzle.html' title='Microsoft Puzzle'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-1084248432159356190</id><published>2008-08-04T18:27:00.000-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Prisoners puzzle</title><content type='html'>&lt;b&gt;A prisoner is faced with a decision where he must open one of two doors.&lt;br /&gt;Behind each door is either a lady or a tiger. They may be both tigers,both&lt;br /&gt;ladies or one of each.&lt;br /&gt;&lt;br /&gt;If the prisoner opens a door to find a lady he will marry her and if he opens&lt;br /&gt;a door to find a tiger he will be eaten alive.Of course,the prisoner would be&lt;br /&gt;prefer to be married than eaten alive.&lt;br /&gt;&lt;br /&gt;Each of the doors has a sign bearing a statement that may be either&lt;br /&gt;true or false.The statement on door one says,"In this room there is a&lt;br /&gt;lady,and in the other room there is a tiger."&lt;br /&gt;&lt;br /&gt;The statement on door two says,"In one of these rooms there is lady,&lt;br /&gt;and in one of these rooms there is a tiger."&lt;br /&gt;&lt;br /&gt;The prisoner is informed that one of the statements is true and one is&lt;br /&gt;false.&lt;br /&gt;&lt;br /&gt;Which door should the Prisoner open?&lt;br /&gt;&lt;br /&gt;&lt;/b&gt;&lt;br /&gt;&lt;b&gt;Solution&lt;br /&gt;there are two possibilities&lt;br /&gt;&lt;br /&gt;possibility#1&lt;br /&gt;&gt;&gt;statement on door 1 is true and statement on door 2 is false.&lt;br /&gt;but this possibility is not feasible because if it is confirm that door 1 has lady and door 2 has tiger then the statement on door 2 can't be wrong because it states that " In one of these rooms there is lady,&lt;br /&gt;and in one of these rooms there is a tiger."&lt;br /&gt;&lt;br /&gt;So this possibilty is not feasible.&lt;br /&gt;&lt;br /&gt;possibility#2&lt;br /&gt;&gt;&gt;statement on door 1 is false and statement on door 2 is true.&lt;br /&gt;this possibilty is feasible beacase the true statement says "In one of these rooms there is lady,&lt;br /&gt;and in one of these rooms there is a tiger."&lt;br /&gt;and false one thats on door 1 says "In this room there is a&lt;br /&gt;lady,and in the other room there is a tiger."&lt;br /&gt;now correcting this false statement we get our answer i.e. there is tiger behind door 1 and lady is behind door 2.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Prisoner must open door 2..... but Pray that LADY MUST BE BEAUTIFUL&lt;/b&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-1084248432159356190?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/1084248432159356190/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=1084248432159356190' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1084248432159356190'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1084248432159356190'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/prisoners-puzzle.html' title='Prisoners puzzle'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-5622592483671538641</id><published>2008-08-04T18:24:00.000-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>There r some cows &amp; hens in field.Total no of their legs(legs of cows &amp; hens)  is equal to twice their head + 14. how many cows r their in fields</title><content type='html'>&lt;h3 class="smller" style="font-weight: normal;"&gt; Anonymous &lt;/h3&gt; &lt;div class="para"&gt; say x cows and y hens&lt;br /&gt;&lt;br /&gt;cows have 4 legs, hens have 2&lt;br /&gt;all have 1 head (unless you have beheaded some already &lt;img src="http://www.orkut.co.in/img/smiley/i_funny.gif" alt="" border="0" /&gt;)&lt;br /&gt;&lt;br /&gt;so:&lt;br /&gt;&lt;br /&gt;4x + 2y = 2*(x + y) +14&lt;br /&gt;&lt;br /&gt;4x + 2y = 2x + 2y + 14&lt;br /&gt;&lt;br /&gt;4x = 2x + 14&lt;br /&gt;&lt;br /&gt;2x = 14&lt;br /&gt;x = 7&lt;br /&gt;7 cows. &lt;/div&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-5622592483671538641?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/5622592483671538641/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=5622592483671538641' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/5622592483671538641'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/5622592483671538641'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/there-r-some-cows-hens-in-fieldtotal-no.html' title='There r some cows &amp; hens in field.Total no of their legs(legs of cows &amp; hens)  is equal to twice their head + 14. how many cows r their in fields'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-1877979733503940206</id><published>2008-08-04T18:22:00.000-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>How many points are there on the globe where, by walking one mile south, one mile east,and one mile north,you reach the place where you started?</title><content type='html'>&lt;b&gt;Answer:&lt;br /&gt;&lt;br /&gt;&lt;/b&gt;If you are on north pole, any direction you take is south.&lt;br /&gt;So you walk south, then east back to north makeing equilateral trangle of 1 mile side. Then yes north pole is one point.&lt;br /&gt;&lt;br /&gt;Now assume your are 1.159 miles from south pole.&lt;br /&gt;You walk 1 mile towards south. You are 0.159 mile from south pole. by walking 1 mile east (directions on earth are cirular) you would complete full circle 2*Pi*r (r =0.159) = 1 mile and be back at same spot where you started your east walk. Then 1 mile to north. will take you to the same point you started. (1.159, just example, quick thinking, further calculating, you can get, more accurate).&lt;br /&gt;&lt;br /&gt;so this is 2nd point.&lt;br /&gt;&lt;br /&gt;now 1 + (0.159)/2) from south pole. In this case after walking south for 1 mile, and then east, you wil complete 2 rounds and be at same place where you started your eastward walk. then 1 mile toward north will take you to same point where you sarted.&lt;br /&gt;&lt;br /&gt;so this is 3rd pont.&lt;br /&gt;&lt;br /&gt;Like this&lt;br /&gt;1 + (0.159)/3 from south pole.. 4th point&lt;br /&gt;1 + (0.159)/4) frrom south pole ..5th point..&lt;br /&gt;.&lt;br /&gt;.&lt;br /&gt;.&lt;br /&gt;.&lt;br /&gt;&lt;br /&gt;infinity..&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-1877979733503940206?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/1877979733503940206/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=1877979733503940206' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1877979733503940206'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/1877979733503940206'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/how-many-points-are-there-on-globe.html' title='How many points are there on the globe where, by walking one mile south, one mile east,and one mile north,you reach the place where you started?'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-2450386283391597974</id><published>2008-08-01T07:26:00.001-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Finding a ball less in weight than other 6 balls outof 7 balls.</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questions:&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;There are 7 balls which looks similar in shape. however 6 balls are of same weight and one ball is some what less in weight than other 6 balls.&lt;br /&gt;you have a weighing scale.&lt;br /&gt;&lt;br /&gt;how you will find ball with less weight with two  scaling attempts ?&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;answers.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;1 scaling: -- weight 3 balls against 3 balls&lt;br /&gt;   if it weigh's equal it means remaing ball is the smaller one.&lt;br /&gt;  if one of the side lower than other then do the second scaling with those 3 balls&lt;br /&gt;&lt;br /&gt;2 scalling:&lt;br /&gt;  1 against 1.&lt;br /&gt;  if equal remaining will be smaller else one of the balls currently weighing is smaller.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-2450386283391597974?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/2450386283391597974/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=2450386283391597974' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2450386283391597974'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/2450386283391597974'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/finding-ball-less-in-weight-than-other.html' title='Finding a ball less in weight than other 6 balls outof 7 balls.'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-6160281525717344985</id><published>2008-08-01T07:01:00.000-07:00</published><updated>2008-12-23T21:42:41.354-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>25 horses puzzle - find minimum number of races.</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questions....&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Given 25 Horses, each of them runs at different speeds.&lt;br /&gt;one can race max 5 horses at a time.&lt;br /&gt;there is no way to calculate the timings.&lt;br /&gt;What is the minimum number of races needed to find out fastest 3 horses.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Answer.....&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1. Divide 25 horses into 5 groups of 5 horses each. Race all groups. we will get 3 fastest from each of the groups.&lt;br /&gt;&lt;br /&gt;2. now race fastest 5 horses from each race. this will give 1 fastest horse from all horses. elimiate the last two horses.&lt;br /&gt;&lt;br /&gt;3. take 2nd and 3rd horse from last race, 2 and 3 from last from the group to which fastes horse belongs( it may possible that 3 fastes horses may be from same group). take the 2nd fasted horse from the group two which 2nd fastest horse belogs from race in last step.&lt;br /&gt;&lt;br /&gt;4. do the race for horses selected in step 3, this will give 2nd and 3rd fastes horse.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;IT will take minimum 7 races to find fastest 3 horses.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/2267127033347614408-6160281525717344985?l=puzzleszone.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://puzzleszone.blogspot.com/feeds/6160281525717344985/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://www.blogger.com/comment.g?blogID=2267127033347614408&amp;postID=6160281525717344985' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/6160281525717344985'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/2267127033347614408/posts/default/6160281525717344985'/><link rel='alternate' type='text/html' href='http://puzzleszone.blogspot.com/2008/08/25-horses-puzzle-find-minimum-number-of.html' title='25 horses puzzle - find minimum number of races.'/><author><name>Siebel Expert</name><uri>http://www.blogger.com/profile/11533458660230230361</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-2267127033347614408.post-2340076827471052104</id><published>2008-07-23T08:41:00.000-07:00</published><updated>2008-12-23T21:42:41.355-08:00</updated><category scheme='http://www.blogger.com/atom/ns#' term='Programming Puzzles'/><title type='text'>Puzzles</title><content type='html'>Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;18&lt;br /&gt;&lt;br /&gt;Assume that initial there were 3*X bullets.&lt;br /&gt;&lt;br /&gt;So they got X bullets each after division.&lt;br /&gt;&lt;br /&gt;All of them shot 4 bullets. So now they have (X - 4) bullets each.&lt;br /&gt;&lt;br /&gt;But it is given that,after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X&lt;br /&gt;&lt;br /&gt;Therefore, the equation is&lt;br /&gt;3 * (X - 4) = X&lt;br /&gt;3 * X - 12 = X&lt;br /&gt;2 * X = 12&lt;br /&gt;X = 6&lt;br /&gt;&lt;br /&gt;Therefore the total bullets before division is = 3 * X = 18&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Find sum of digits of D.&lt;br /&gt;&lt;br /&gt;Let&lt;br /&gt;A= 19991999&lt;br /&gt;B = sum of digits of A&lt;br /&gt;C = sum of digits of B&lt;br /&gt;D = sum of digits of C&lt;br /&gt;&lt;br /&gt;(HINT : A = B = C = D (mod 9))&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The sum of the digits od D is 1.&lt;br /&gt;&lt;br /&gt;Let E = sum of digits of D.&lt;br /&gt;&lt;br /&gt;It follows from the hint that A = E (mod 9)&lt;br /&gt;Consider,&lt;br /&gt;&lt;br /&gt;        A = 19991999&lt;br /&gt;&lt;br /&gt;          &lt; 20002000&lt;br /&gt;&lt;br /&gt;          = 22000 * 10002000&lt;br /&gt;&lt;br /&gt;          = 1024200 * 106000&lt;br /&gt;&lt;br /&gt;          &lt; 10800 * 106000&lt;br /&gt;&lt;br /&gt;          = 106800&lt;br /&gt;&lt;br /&gt;       &lt;br /&gt;&lt;br /&gt;        i.e. A &lt; 106800&lt;br /&gt;&lt;br /&gt;        i.e. B &lt; 6800 * 9 = 61200&lt;br /&gt;&lt;br /&gt;        i.e. C &lt; 5 * 9 = 45&lt;br /&gt;&lt;br /&gt;        i.e. D &lt; 2 * 9 = 18&lt;br /&gt;&lt;br /&gt;        i.e. E &lt;= 9&lt;br /&gt;&lt;br /&gt;        i.e. E is a single digit number.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;        Also,&lt;br /&gt;&lt;br /&gt;        1999 = 1 (mod 9)&lt;br /&gt;&lt;br /&gt;        so 19991999 = 1 (mod 9)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Therefore we conclude that E=1.&lt;br /&gt;There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.&lt;br /&gt;&lt;br /&gt;In the mean time the whole platoon has moved ahead by 50m.&lt;br /&gt;&lt;br /&gt;The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The last person covered 120.71 meters.&lt;br /&gt;&lt;br /&gt;It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.&lt;br /&gt;&lt;br /&gt;Let's assume that when the last person reached the first person, the platoon moved X meters forward.&lt;br /&gt;&lt;br /&gt;Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.&lt;br /&gt;&lt;br /&gt;Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.&lt;br /&gt;&lt;br /&gt;Now, as the ratios are equal,&lt;br /&gt;(50+X)/X = X/(50-X)&lt;br /&gt;(50+X)*(50-X) = X*X&lt;br /&gt;&lt;br /&gt;Solving, X=35.355 meters&lt;br /&gt;&lt;br /&gt;Thus, total distance covered by the last person&lt;br /&gt;= (50+X) + X&lt;br /&gt;= 2*X + 50&lt;br /&gt;= 2*(35.355) + 50&lt;br /&gt;= 120.71 meters&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.&lt;br /&gt;&lt;br /&gt;If you take a marker &amp;amp; start from a corner on a cube, what is the maximum number of edges you can trace across if you never trace across the same edge twice, never remove the marker from the cube, &amp;amp; never trace anywhere on the cube, except for the corners &amp;amp; edges?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;9&lt;br /&gt;&lt;br /&gt;To verify this, you can make a drawing of a cube, &amp;amp; number each of its 12 edges. Then, always starting from 1 corner &amp;amp; 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube.&lt;br /&gt;&lt;br /&gt;There is no need to start from other corners or edges of the cube, as you will only be repeating the same combinations. The process is a little more involved than this, but is useful for solving many types of spatial puzzles.&lt;br /&gt;One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and another is a Doctor.&lt;br /&gt;    If the Doctor is a male, then the Engineer is a male.&lt;br /&gt;    If the Engineer is younger than the Doctor, then the Engineer and the Doctor are not blood relatives.&lt;br /&gt;    If the Engineer is a female, then she and the Doctor are blood relatives.&lt;br /&gt;Can you tell who is the Doctor and the Engineer?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.&lt;br /&gt;&lt;br /&gt;Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.&lt;br /&gt;&lt;br /&gt;Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr. Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.&lt;br /&gt;&lt;br /&gt;Now from 2, Mr. Bajaj's mother can not be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further.&lt;br /&gt;&lt;br /&gt;Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not necessarily in the same order.&lt;br /&gt;&lt;br /&gt;Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge. No wife partners her husband and Cam does not play bridge.&lt;br /&gt;&lt;br /&gt;Who is married to Cam?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Carrie is married to Cam.&lt;br /&gt;&lt;br /&gt;"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."&lt;br /&gt;&lt;br /&gt;It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina.&lt;br /&gt;&lt;br /&gt;As Cam does not play bridge, Billy's husband must be Laurie.&lt;br /&gt;&lt;br /&gt;Hence, Carrie is married to Cam.&lt;br /&gt;&lt;br /&gt;There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.&lt;br /&gt;&lt;br /&gt;Find the tractors each originally had?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.&lt;br /&gt;&lt;br /&gt;It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X &amp;amp; Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)&lt;br /&gt;&lt;br /&gt;Similarly, before Y gave tractors to X &amp;amp; Z, they had 6 &amp;amp; 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)&lt;br /&gt;&lt;br /&gt;Again, before X gave tractors to Y &amp;amp; Z, they had 21 &amp;amp; 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)&lt;br /&gt;&lt;br /&gt;Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.&lt;br /&gt;A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The sign-maker will need 192 zeroes.&lt;br /&gt;&lt;br /&gt;Divide 1000 building numbers into groups of 100 each as follow:&lt;br /&gt;(1..100), (101..200), (201..300), ....... (901..1000)&lt;br /&gt;&lt;br /&gt;For the first group, sign-maker will need 11 zeroes.&lt;br /&gt;For group numbers 2 to 9, he will require 20 zeroes each.&lt;br /&gt;And for group number 10, he will require 21 zeroes.&lt;br /&gt;&lt;br /&gt;The total numbers of zeroes required are&lt;br /&gt;= 11 + 8*20 + 21&lt;br /&gt;= 11 + 160 + 21&lt;br /&gt;= 192&lt;br /&gt;&lt;br /&gt;There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.&lt;br /&gt;1.    Take 8 coins and weigh 4 against 4.&lt;br /&gt;o    If both are not equal, goto step 2&lt;br /&gt;o    If both are equal, goto step 3&lt;br /&gt;&lt;br /&gt;2.    One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.&lt;br /&gt;o    If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.&lt;br /&gt;    If both are equal, L4 is the odd coin and is lighter.&lt;br /&gt;    If L2 is light, L2 is the odd coin and is lighter.&lt;br /&gt;    If L3 is light, L3 is the odd coin and is lighter.&lt;br /&gt;&lt;br /&gt;o    If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2&lt;br /&gt;    If both are equal, there is some error.&lt;br /&gt;    If H1 is heavy, H1 is the odd coin and is heavier.&lt;br /&gt;    If H2 is heavy, H2 is the odd coin and is heavier.&lt;br /&gt;&lt;br /&gt;o    If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4&lt;br /&gt;    If both are equal, L1 is the odd coin and is lighter.&lt;br /&gt;    If H3 is heavy, H3 is the odd coin and is heavier.&lt;br /&gt;    If H4 is heavy, H4 is the odd coin and is heavier.&lt;br /&gt;&lt;br /&gt;3.    The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.&lt;br /&gt;o    If both are equal, there is some error.&lt;br /&gt;o    If X is heavy, X is the odd coin and is heavier.&lt;br /&gt;o    If X is light, X is the odd coin and is lighter.&lt;br /&gt;In a sports contest there were m medals awarded on n successive days (n &gt; 1).&lt;br /&gt;1.    On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.&lt;br /&gt;2.    On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.&lt;br /&gt;3.    On the nth and last day, the remaining n medals were awarded.&lt;br /&gt;How many days did the contest last, and how many medals were awarded altogether?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Total 36 medals were awarded and the contest was for 6 days.&lt;br /&gt;&lt;br /&gt;On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals&lt;br /&gt;On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals&lt;br /&gt;On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals&lt;br /&gt;On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals&lt;br /&gt;On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals&lt;br /&gt;On day 6: Medals awarded 6&lt;br /&gt;&lt;br /&gt;I got this answer by writing small program. If anyone know any other simpler method, do submit it.&lt;br /&gt;A number of 9 digits has the following properties:&lt;br /&gt;    The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2&lt;=n&lt;=9.&lt;br /&gt;    Each digit in the number is different i.e. no digits are repeated.&lt;br /&gt;    The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.&lt;br /&gt;Find the number.&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The answer is 381654729&lt;br /&gt;&lt;br /&gt;One way to solve it is Trial-&amp;amp;-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.&lt;br /&gt;&lt;br /&gt;The other way to solve this problem is by writing a computer program that systematically tries all possibilities&lt;br /&gt;&lt;br /&gt;1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day.&lt;br /&gt;&lt;br /&gt;What part of the contents of the container is left at the end of the second day?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Assume that contents of the container is X&lt;br /&gt;&lt;br /&gt;On the first day 1/3rd is evaporated.&lt;br /&gt;(1 - 1/3) of X is remaining i.e. (2/3)X&lt;br /&gt;&lt;br /&gt;On the Second day 3/4th is evaporated. Hence,&lt;br /&gt;(1- 3/4) of (2/3)X is remaining&lt;br /&gt;i.e. (1/4)(2/3)X = (1/6) X&lt;br /&gt;&lt;br /&gt;Hence 1/6th of the contents of the container is remaining&lt;br /&gt;&lt;br /&gt;Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.&lt;br /&gt;&lt;br /&gt;For how long did Vipul study in candle light?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Vipul studied for 3 hours in candle light.&lt;br /&gt;&lt;br /&gt;Assume that the initial lenght of both the candle was L and Vipul studied for X hours.&lt;br /&gt;&lt;br /&gt;In X hours, total thick candle burnt = XL/6&lt;br /&gt;In X hours, total thin candle burnt = XL/4&lt;br /&gt;&lt;br /&gt;After X hours, total thick candle remaining = L - XL/6&lt;br /&gt;After X hours, total thin candle remaining = L - XL/4&lt;br /&gt;&lt;br /&gt;Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.&lt;br /&gt;(L - XL/6) = 2(L - XL/4)&lt;br /&gt;(6 - X)/6 = (4 - X)/2&lt;br /&gt;(6 - X) = 3*(4 - X)&lt;br /&gt;6 - X = 12 - 3X&lt;br /&gt;2X = 6&lt;br /&gt;X = 3&lt;br /&gt;&lt;br /&gt;Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.&lt;br /&gt;&lt;br /&gt;If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, &amp;amp; so on.&lt;br /&gt;&lt;br /&gt;How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Rs.333,062,500&lt;br /&gt;&lt;br /&gt;To begin with, you want to know the total number of days: 365 x 50 = 18250.&lt;br /&gt;&lt;br /&gt;By experimentation, the following formula can be discovered, &amp;amp; used to determine the amount earned for any particular day: 1 + 2(x-1), with x being the number of the day. Take half of the 18250 days, &amp;amp; pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, &amp;amp; so on, &amp;amp; you will see that if you add these pairs together, they always equal Rs.36500.&lt;br /&gt;&lt;br /&gt;Multiply this number by the total number of pairs (9125), &amp;amp; you have the amount you would have earned in 50 years.&lt;br /&gt;&lt;br /&gt;Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11.....upto 18250 terms)&lt;br /&gt;A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, &amp;amp; now his salary is Rs. 22702.68&lt;br /&gt;&lt;br /&gt;What was his salary to begin with?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Rs.22176&lt;br /&gt;&lt;br /&gt;Assume his salary was Rs. X&lt;br /&gt;&lt;br /&gt;He earns 5% raise. So his salary is (105*X)/100&lt;br /&gt;&lt;br /&gt;A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100) which is Rs. 22702.68&lt;br /&gt;&lt;br /&gt;Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68&lt;br /&gt;X = 22176&lt;br /&gt;At 6'o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12'o.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;66 seconds&lt;br /&gt;&lt;br /&gt;It is given that the time between first and last ticks at 6'o is 30 seconds.&lt;br /&gt;Total time gaps between first and last ticks at 6'o = 5&lt;br /&gt;(i.e. between 1 &amp;amp; 2, 2 &amp;amp; 3, 3 &amp;amp; 4, 4 &amp;amp; 5 and 5 &amp;amp; 6)&lt;br /&gt;&lt;br /&gt;So time gap between two ticks = 30/5 = 6 seconds.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Now, total time gaps between first and last ticks at 12'o = 11&lt;br /&gt;Therefore time taken for 12 ticks = 11 * 6 = 66 seconds (and not 60 seconds)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;500 men are arranged in an array of 10 rows and 50 columns according to their heights.&lt;br /&gt;&lt;br /&gt;Tallest among each row of all are asked to come out. And the shortest among them is A.&lt;br /&gt;&lt;br /&gt;Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B.&lt;br /&gt;&lt;br /&gt;Now who is taller A or B ?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;No one is taller, both are same as A and B are the same person.&lt;br /&gt;&lt;br /&gt;As it is mentioned that 500 men are arranged in an array of 10 rows and 50 columns according to their heights. Let's assume that position numbers represent their heights. Hence, the shortest among the 50, 100, 150, ... 450, 500 is person with height 50 i.e. A. Similarly the tallest among 1, 2, 3, 4, 5, ..... 48, 48, 50 is person with height 50 i.e. B&lt;br /&gt;&lt;br /&gt;Now, both A and B are the person with height 50. Hence both are same.&lt;br /&gt;&lt;br /&gt;In Mr. Mehta's family, there are one grandfather, one grandmother, two fathers, two mothers, one father-in-law, one mother-in-law, four children, three grandchildren, one brother, two sisters, two sons, two daughters and one daughter-in-law.&lt;br /&gt;&lt;br /&gt;How many members are there in Mr. Mehta's family? Give minimal possible answer.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 7 members in Mr. Mehta's family. Mother &amp;amp; Father of Mr. Mehta, Mr. &amp;amp; Mrs. Mehta, his son and two daughters.&lt;br /&gt;&lt;br /&gt;Mother &amp;amp; Father of Mr. Mehta&lt;br /&gt;|&lt;br /&gt;|&lt;br /&gt;Mr. &amp;amp; Mrs. Mehta&lt;br /&gt;|&lt;br /&gt;|&lt;br /&gt;One Son &amp;amp; Two Daughters&lt;br /&gt;When Alexander the Great attacked the forces of Porus, an Indian soldier was captured by the Greeks. He had displayed such bravery in battle, however, that the enemy offered to let him choose how he wanted to be killed. They told him, "If you tell a lie, you will put to the sword, and if you tell the truth you will be hanged."&lt;br /&gt;&lt;br /&gt;The soldier could make only one statement. He made that statement and went free. What did he say?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The soldier said, "You will put me to the sword."&lt;br /&gt;&lt;br /&gt;The soldier has to say a Paradox to save himself. If his statement is true, he will be hanged, which is not the sword and hence false. If his statement is false, he will be put to the sword, which will make it true. A Paradox !!!&lt;br /&gt;&lt;br /&gt;A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither the person nor the cashier noticed that.&lt;br /&gt;&lt;br /&gt;After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.&lt;br /&gt;&lt;br /&gt;Find X and Y. (1 Rupee = 100 Paise)&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;As given, the person wanted to withdraw 100X + Y paise.&lt;br /&gt;&lt;br /&gt;But he got 100Y + X paise.&lt;br /&gt;&lt;br /&gt;After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is&lt;br /&gt;&lt;br /&gt;        2 * (100X + Y) = 100Y + X - 20&lt;br /&gt;&lt;br /&gt;        200X + 2Y = 100Y +X - 20&lt;br /&gt;&lt;br /&gt;        199X - 98Y = -20&lt;br /&gt;&lt;br /&gt;        98Y - 199X = 20&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X &amp;amp; Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1&lt;br /&gt;&lt;br /&gt;Case I : Y=2X&lt;br /&gt;Solving two equations simultaneously&lt;br /&gt;98Y - 199X = 20&lt;br /&gt;Y - 2X = 0&lt;br /&gt;We get X = - 20/3 &amp;amp; Y = - 40/2&lt;br /&gt;&lt;br /&gt;Case II : Y=2X+1&lt;br /&gt;Solving two equations simultaneously&lt;br /&gt;98Y - 199X = 20&lt;br /&gt;Y - 2X = 1&lt;br /&gt;We get X = 26 &amp;amp; Y = 53&lt;br /&gt;&lt;br /&gt;Now, its obvious that he wanted to withdraw Rs. 26.53&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Submit&lt;br /&gt;Answer         Users&lt;br /&gt;Answer (2)         BrainVista&lt;br /&gt;Answer    &lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;The game of Tic-Tac-Toe is being played between two players. Only the last mark to be placed in the game as shown.&lt;br /&gt;&lt;br /&gt;Who will win the game, O or X? Can you tell which was the sixth mark and at which position? Do explain your answer.&lt;br /&gt;At the Party:&lt;br /&gt;1.    There were 9 men and children.&lt;br /&gt;2.    There were 2 more women than children.&lt;br /&gt;3.    The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.&lt;br /&gt;&lt;br /&gt;Also, of the three groups - men, women and children - at the party:&lt;br /&gt;4.    There were 4 of one group.&lt;br /&gt;5.    There were 6 of one group.&lt;br /&gt;6.    There were 8 of one group.&lt;br /&gt;Exactly one of the above 6 statements is false.&lt;br /&gt;&lt;br /&gt;Can you tell which one is false? Also, how many men, women and children are there at the party&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Assume that both the players are intelligent enough.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;O will win the game. The sixth mark was X in square 9.&lt;br /&gt;&lt;br /&gt;The 7th mark must be placed in square 5 which is the win situation for both X and O. Hence, the 6th mark must be placed in a line already containing two of the opponents marks. There are two such possibilities - the 6th mark would have been either O in square 7 or X in square 9.&lt;br /&gt;&lt;br /&gt;As we know both the players are intelligent enough, the 6th mark could not be O in square 7. Instead, he would have placed O in square 5 and would have won.&lt;br /&gt;&lt;br /&gt;Hence, the sixth mark must be X placed in square 9. And the seventh mark will be O. Thus O will win the game.&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Statement (4) is false. There are 3 men, 8 women and 6 children.&lt;br /&gt;&lt;br /&gt;Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly one statement is false.&lt;br /&gt;&lt;br /&gt;So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true.&lt;br /&gt;&lt;br /&gt;From (1) and (2), there are 11 men and women. Then from (3), there are 2 possible cases - either there are 8 men and 3 women or there are 3 men and 8 women.&lt;br /&gt;&lt;br /&gt;If there are 8 men and 3 women, then there is 1 child. Then Statements (4) and (5) both are false, which is not possible.&lt;br /&gt;&lt;br /&gt;Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.&lt;br /&gt;&lt;br /&gt;There is a shortage of tubelights, bulbs and fans in a village - Kharghar. It is found that&lt;br /&gt;    All houses do not have either tubelight or bulb or fan.&lt;br /&gt;    exactly 19% of houses do not have just one of these.&lt;br /&gt;    atleast 67% of houses do not have tubelights.&lt;br /&gt;    atleast 83% of houses do not have bulbs.&lt;br /&gt;    atleast 73% of houses do not have fans.&lt;br /&gt;What percentage of houses do not have tubelight, bulb and fan?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;42% houses do not have tubelight, bulb and fan.&lt;br /&gt;&lt;br /&gt;Let's assume that there are 100 houses. Hence, there should be total 300 items i.e. 100 tubelights, 100 bulbs and 100 fans.&lt;br /&gt;&lt;br /&gt;From the given data, we know that there is shortage of atleast (67+83+73) 223 items in every 100 houses.&lt;br /&gt;&lt;br /&gt;Also, exactly 19 houses do not have just one item. It means that remaining 81 houses should account for the shortage of remaining (223-19) 204 items. If those remaining 81 houses do not have 2 items each, there would be a shortage of 162 items. But total of 204 items are short. Hence, atleast (204-162) 42 houses do not have all 3 items - tubelight, bulb and fan.&lt;br /&gt;&lt;br /&gt;Thus, 42% houses do not have tubelight, bulb and fan.&lt;br /&gt;Mr. Subramaniam rents a private car for Andheri-Colaba-Andheri trip. It costs him Rs. 300 everyday.&lt;br /&gt;&lt;br /&gt;One day the car driver informed Mr. Subramaniam that there were two students from Bandra who wished to go from Bandra to Colaba and back to Bandra. Bandra is halfway between Andheri and Colaba. Mr. Subramaniam asked the driver to let the students travel with him.&lt;br /&gt;&lt;br /&gt;On the first day when they came, Mr. Subramaniam said, "If you tell me the mathematically correct price you should pay individually for your portion of the trip, I will let you travel for free."&lt;br /&gt;&lt;br /&gt;How much should the individual student pay for their journey?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The individual student should pay Rs. 50 for their journey.&lt;br /&gt;&lt;br /&gt;Note that 3 persons are travelling between Bandra and Colaba.&lt;br /&gt;&lt;br /&gt;The entire trip costs Rs. 300 to Mr. Subramanian. Hence, half of the trip costs Rs. 150.&lt;br /&gt;&lt;br /&gt;For Andheri-Bandra-Andheri, only one person i.e. Mr. Subramaniam is travelling. Hence, he would pay Rs. 150.&lt;br /&gt;&lt;br /&gt;For Bandra-Colaba-Bandra, three persons i.e Mr. Subramaniam and two students, are travelling. Hence, each student would pay Rs. 50.&lt;br /&gt;&lt;br /&gt;Substitute digits for the letters to make the following Division true&lt;br /&gt;                   O U T&lt;br /&gt;&lt;br /&gt;           -------------&lt;br /&gt;&lt;br /&gt;   S T E M | D E M I S E&lt;br /&gt;&lt;br /&gt;           | D M O C&lt;br /&gt;&lt;br /&gt;           -------------&lt;br /&gt;&lt;br /&gt;               T U I S&lt;br /&gt;&lt;br /&gt;               S T E M&lt;br /&gt;&lt;br /&gt;              ----------&lt;br /&gt;&lt;br /&gt;                 Z Z Z E&lt;br /&gt;&lt;br /&gt;                 Z U M M&lt;br /&gt;&lt;br /&gt;                --------&lt;br /&gt;&lt;br /&gt;                   I S T&lt;br /&gt;Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;C=0, U=1, S=2, T=3, O=4, M=5, I=6, Z=7, E=8, D=9&lt;br /&gt;&lt;br /&gt;It is obvious that U=1 (as U*STEM=STEM) and C=0 (as I-C=I).&lt;br /&gt;&lt;br /&gt;S*O is a single digit and also S*T is a single digit. Hence, their values (O, S, T) must be 2, 3 or 4 (as they can not be 0 or 1 or greater than 4).&lt;br /&gt;&lt;br /&gt;Consider, STEM*O=DMOC, where C=0. It means that M must be 5. Now, its simple. O=4, S=2, T=3, E=8, Z=7, I=6 and D=9.&lt;br /&gt;                   O U T                        4 1 3&lt;br /&gt;&lt;br /&gt;           -------------                -------------&lt;br /&gt;&lt;br /&gt;   S T E M | D E M I S E        2 3 8 5 | 9 8 5 6 2 8&lt;br /&gt;&lt;br /&gt;           | D M O C                    | 9 5 4 0&lt;br /&gt;&lt;br /&gt;           -------------                -------------&lt;br /&gt;&lt;br /&gt;               T U I S                      3 1 6 2&lt;br /&gt;&lt;br /&gt;               S T E M                      2 3 8 5&lt;br /&gt;&lt;br /&gt;              ----------                   ----------&lt;br /&gt;&lt;br /&gt;                 Z Z Z E                      7 7 7 8&lt;br /&gt;&lt;br /&gt;                 Z U M M                      7 1 5 5&lt;br /&gt;&lt;br /&gt;                --------                     --------&lt;br /&gt;&lt;br /&gt;                   I S T                        6 2 3&lt;br /&gt;Also, when arranged from 0 to 9, it spells CUSTOMIZED.&lt;br /&gt;At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with the hour hand?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;4:21:49.5&lt;br /&gt;&lt;br /&gt;Assume that X minutes after 4.00 PM minute hand exactly aligns with and hour hand.&lt;br /&gt;&lt;br /&gt;For every minute, minute hand travels 6 degrees.&lt;br /&gt;Hence, for X minutes it will travel 6 * X degrees.&lt;br /&gt;&lt;br /&gt;For every minute, hour hand travels 1/2 degrees.&lt;br /&gt;Hence, for X minutes it will travel X/2 degrees.&lt;br /&gt;&lt;br /&gt;At 4.00 PM, the angle between minute hand and hour hand is 120 degrees. Also, after X minutes, minute hand and hour hand are exactly aligned. So the angle with respect to 12 i.e. Vertical Plane will be same. Therefore,&lt;br /&gt;&lt;br /&gt;6 * X = 120 + X/2&lt;br /&gt;12 * X = 240 + X&lt;br /&gt;11 * X = 240&lt;br /&gt;X = 21.8182&lt;br /&gt;X = 21 minutes 49.5 seconds&lt;br /&gt;&lt;br /&gt;Hence, at 4:21:49.5 minute hand is exactly aligned with the hour hand.&lt;br /&gt;&lt;br /&gt;A soldier looses his way in a thick jungle. At random he walks from his camp but mathematically in an interesting fashion.&lt;br /&gt;&lt;br /&gt;First he walks one mile East then half mile to North. Then 1/4 mile to West, then 1/8 mile to South and so on making a loop.&lt;br /&gt;&lt;br /&gt;Finally how far he is from his camp and in which direction?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The soldier is 0.8944 miles away from his camp towards East-North.&lt;br /&gt;&lt;br /&gt;It is obvious that he is in East-North direction.&lt;br /&gt;&lt;br /&gt;Distance travelled in North and South directions&lt;br /&gt;= 1/2 - 1/8 + 1/32 - 1/128 + 1/512 - 1/2048 + and so on... (a geometric series with r = (-1/4) )&lt;br /&gt;&lt;br /&gt;   (1/2) * ( 1 - (-1/4)n )&lt;br /&gt;= ---------------------------&lt;br /&gt;         ( 1 - (-1/4) )&lt;br /&gt;&lt;br /&gt;= 1 / ( 2 * ( 1 - (-1/4) ) )&lt;br /&gt;= 2/5&lt;br /&gt;&lt;br /&gt;Similarly in East and West directions&lt;br /&gt;= 1 - 1/4 + 1/16 - 1/64 + 1/256 - and so on... (a geometric series with r = (-1/4) )&lt;br /&gt;&lt;br /&gt;   (1) * ( 1 - (-1/4)n )&lt;br /&gt;= ---------------------------&lt;br /&gt;         ( 1 - (-1/4) )&lt;br /&gt;&lt;br /&gt;= 1 / ( ( 1- (-1/4) )&lt;br /&gt;= 4/5&lt;br /&gt;&lt;br /&gt;So the soldier is 4/5 miles away towards East and 2/5 miles away towards North. So using right angled triangle, soldier is 0.8944 miles away from his camp.&lt;br /&gt;&lt;br /&gt;Raj has a jewel chest containing Rings, Pins and Ear-rings. The chest contains 26 pieces. Raj has 2 1/2 times as many rings as pins, and the number of pairs of earrings is 4 less than the number of rings.&lt;br /&gt;&lt;br /&gt;How many earrings does Raj have?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;12 earrings&lt;br /&gt;&lt;br /&gt;Assume that there are R rings, P pins and E pair of ear-rings.&lt;br /&gt;&lt;br /&gt;It is given that, he has 2 1/2 times as many rings as pins.&lt;br /&gt;R = (5/2) * P or P = (2*R)/5&lt;br /&gt;&lt;br /&gt;And, the number of pairs of earrings is 4 less than the number of rings.&lt;br /&gt;E = R - 4 or R = E + 4&lt;br /&gt;&lt;br /&gt;Also, there are total 26 pieces.&lt;br /&gt;R + P + 2*E = 26&lt;br /&gt;R + (2*R)/5 + 2*E = 26&lt;br /&gt;5*R + 2*R + 10*E = 130&lt;br /&gt;7*R + 10*E = 130&lt;br /&gt;7*(E + 4) + 10*E = 130&lt;br /&gt;7*E + 28 + 10*E = 130&lt;br /&gt;17*E = 102&lt;br /&gt;E = 6&lt;br /&gt;&lt;br /&gt;Hence, there are 6 pairs of Ear-rings i.e. total 12 Ear-rings&lt;br /&gt;How many ways are there of arranging the sixteen black or white pieces of a standard international chess set on the first two rows of the board?&lt;br /&gt;&lt;br /&gt;Given that each pawn is identical and each rook, knight and bishop is identical to its pair.&lt;br /&gt;Submitted&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;6,48,64,800 ways&lt;br /&gt;&lt;br /&gt;There are total 16 pieces which can be arranged on 16 places in 16P16 = 16! ways.&lt;br /&gt;(16! = 16 * 15 * 14 * 13 * 12 * ..... * 3 * 2 * 1)&lt;br /&gt;&lt;br /&gt;But, there are some duplicate combinations because of identical pieces.&lt;br /&gt;    There are 8 identical pawn, which can be arranged in 8P8 = 8! ways.&lt;br /&gt;    Similarly there are 2 identical rooks, 2 identical knights and 2 identical bishops. Each can be arranged in 2P2 = 2! ways.&lt;br /&gt;Hence, the require answer is&lt;br /&gt;= (16!) / (8! * 2! * 2! * 2!)&lt;br /&gt;= 6,48,64,800&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;A person with some money spends 1/3 for cloths, 1/5 of the remaining for food and 1/4 of the remaining for travel. He is left with Rs 100/-&lt;br /&gt;&lt;br /&gt;How much did he have with him in the begining?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Rs. 250/-&lt;br /&gt;&lt;br /&gt;Assume that initially he had Rs. X&lt;br /&gt;He spent 1/3 for cloths =. (1/3) * X&lt;br /&gt;Remaining money = (2/3) * X&lt;br /&gt;&lt;br /&gt;He spent 1/5 of remaining money for food = (1/5) * (2/3) * X = (2/15) * X&lt;br /&gt;Remaining money = (2/3) * X - (2/15) * X = (8/15) * X&lt;br /&gt;&lt;br /&gt;Again, he spent 1/4 of remaining maoney for travel = (1/4) * (8/15) * X = (2/15) * X&lt;br /&gt;Remaining money = (8/15) * X - (2/15) * X = (6/15) * X&lt;br /&gt;&lt;br /&gt;But after spending for travel he is left with Rs. 100/- So&lt;br /&gt;(6/15) * X = 100&lt;br /&gt;X = 250&lt;br /&gt;&lt;br /&gt;Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass.&lt;br /&gt;&lt;br /&gt;How many cows are needed to eat the grass in 96 days?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;20 cows&lt;br /&gt;&lt;br /&gt;g - grass at the beginning&lt;br /&gt;r - rate at which grass grows, per day&lt;br /&gt;y - rate at which one cow eats grass, per day&lt;br /&gt;n - no of cows to eat the grass in 96 days&lt;br /&gt;&lt;br /&gt;From given data,&lt;br /&gt;g + 24*r = 70 * 24 * y ---------- A&lt;br /&gt;g + 60*r = 30 * 60 * y ---------- B&lt;br /&gt;g + 96*r = n * 96 * y ---------- C&lt;br /&gt;&lt;br /&gt;Solving for (B-A),&lt;br /&gt;(60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y)&lt;br /&gt;36 * r = 120 * y ---------- D&lt;br /&gt;&lt;br /&gt;Solving for (C-B),&lt;br /&gt;(96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y)&lt;br /&gt;36 * r = (n * 96 - 30 * 60) * y&lt;br /&gt;120 * y = (n * 96 - 30 * 60) * y [From D]&lt;br /&gt;120 = (n * 96 - 1800)&lt;br /&gt;n = 20&lt;br /&gt;&lt;br /&gt;Hence, 20 cows are needed to eat the grass in 96 days.&lt;br /&gt;There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than the second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11.&lt;br /&gt;&lt;br /&gt;Find the number.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;65292&lt;br /&gt;&lt;br /&gt;As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)&lt;br /&gt;&lt;br /&gt;It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9)&lt;br /&gt;&lt;br /&gt;Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.&lt;br /&gt;Four friends - Arjan, Bhuvan, Guran and Lakha were comparing the number of sheep that they owned.&lt;br /&gt;&lt;br /&gt;It was found that Guran had ten more sheep than Lakha.&lt;br /&gt;&lt;br /&gt;If Arjan gave one-third to Bhuvan, and Bhuvan gave a quarter of what he then held to Guran, who then passed on a fifth of his holding to Lakha, they would all have an equal number of sheep.&lt;br /&gt;&lt;br /&gt;How many sheep did each of them possess? Give the minimal possible answer&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Arjan, Bhuvan, Guran and Lakha had 90, 50, 55 and 45 sheep respectively.&lt;br /&gt;&lt;br /&gt;Assume that Arjan, Bhuvan, Guran and Lakha had A, B, G and L sheep respectively. As it is given that at the end each would have an equal number of sheep, comparing the final numbers from the above table.&lt;br /&gt;&lt;br /&gt;Arjan's sheep = Bhuvan's sheep&lt;br /&gt;2A/3 = A/4 + 3B/4&lt;br /&gt;8A = 3A + 9B&lt;br /&gt;5A = 9B&lt;br /&gt;&lt;br /&gt;Arjan's sheep = Guran's sheep&lt;br /&gt;2A/3 = A/15 + B/5 + 4G/5&lt;br /&gt;2A/3 = A/15 + A/9 + 4G/5 (as B=5A/9)&lt;br /&gt;30A = 3A + 5A + 36G&lt;br /&gt;22A = 36G&lt;br /&gt;11A = 18G&lt;br /&gt;&lt;br /&gt;Arjan's sheep = Lakha's sheep&lt;br /&gt;2A/3 = A/60 + B/20 + G/5 + L&lt;br /&gt;2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18)&lt;br /&gt;2A/3 = A/6 + L&lt;br /&gt;A/2 = L&lt;br /&gt;A = 2L&lt;br /&gt;&lt;br /&gt;Also, it is given that Guran had ten more sheep than Lakha.&lt;br /&gt;G = L + 10&lt;br /&gt;11A/18 = A/2 + 10&lt;br /&gt;A/9 = 10&lt;br /&gt;A = 90 sheep&lt;br /&gt;&lt;br /&gt;Thus, Arjan had 90 sheep, Bhuvan had 5A/9 i.e. 50 sheep, Guran had 11A/18 i.e. 55 sheep and Lakha had A/2 i.e. 45 sheep.&lt;br /&gt;Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5.&lt;br /&gt;&lt;br /&gt;How many such 3-digit numbers are there?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 45 different 3-digit numbers.&lt;br /&gt;&lt;br /&gt;The last digit can not be 0.&lt;br /&gt;&lt;br /&gt;If the last digit is 1, the only possible number is 101. (Note that 011 is not a 3-digit number)&lt;br /&gt;&lt;br /&gt;If the last digit is 2, the possible numbers are 202 and 112.&lt;br /&gt;&lt;br /&gt;If the last digit is 3, the possible numbers are 303, 213 and 123.&lt;br /&gt;&lt;br /&gt;If the last digit is 4, the possible numbers are 404, 314, 224 and 134.&lt;br /&gt;&lt;br /&gt;If the last digit is 5, the possible numbers are 505, 415, 325, 235 and 145.&lt;br /&gt;&lt;br /&gt;Note the pattern here - If the last digit is 1, there is only one number. If the last digit is 2, there are two numbers. If the last digit is 3, there are three numbers. If the last digit is 4, there are four numbers. If the last digit is 5, there are five numbers. And so on.....&lt;br /&gt;&lt;br /&gt;Thus, total numbers are&lt;br /&gt;1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45&lt;br /&gt;&lt;br /&gt;Altogether then, there are 45 different 3-digit numbers, where last digit is the sum of first two digits.&lt;br /&gt;&lt;br /&gt;Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The answer is 285714.&lt;br /&gt;&lt;br /&gt;If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714.&lt;br /&gt;&lt;br /&gt;The simplest way is to write a small program. And the other way is trial and error !!!&lt;br /&gt;Two identical pack of cards A and B are shuffled throughly. One card is picked from A and shuffled with B. The top card from pack A is turned up. If this is the Queen of Hearts, what are the chances that the top card in B will be the King of Hearts?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;52 / 2703&lt;br /&gt;&lt;br /&gt;There are two cases to be considered.&lt;br /&gt;&lt;br /&gt;CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B&lt;br /&gt;&lt;br /&gt;Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)&lt;br /&gt;Probability of having King of Hearts on the top of the Pack B = 2/53&lt;br /&gt;&lt;br /&gt;So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)&lt;br /&gt;&lt;br /&gt;CASE 2 : King of Hearts is not drawn from Pack A&lt;br /&gt;&lt;br /&gt;Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)&lt;br /&gt;Probability of having King of Hearts on the top of the Pack B = 1/53&lt;br /&gt;&lt;br /&gt;So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)&lt;br /&gt;&lt;br /&gt;Now adding both the probability, the required probability is&lt;br /&gt;= 2 / (51 * 53) + 50 / (51 * 53)&lt;br /&gt;= 52 / (51 * 53)&lt;br /&gt;= 52 / 2703&lt;br /&gt;= 0.0192378&lt;br /&gt;There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner.&lt;br /&gt;&lt;br /&gt;What is the probability that they don't collide?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Let's mark the corners of the triangle as A,B,C. There are total 8 ways in which ants can move.&lt;br /&gt;1.    A-&gt;B, B-&gt;C, C-&gt;A&lt;br /&gt;2.    A-&gt;B, B-&gt;C, C-&gt;B&lt;br /&gt;3.    A-&gt;B, B-&gt;A, C-&gt;A&lt;br /&gt;4.    A-&gt;B, B-&gt;A, C-&gt;B&lt;br /&gt;5.    A-&gt;C, C-&gt;B, B-&gt;A&lt;br /&gt;6.    A-&gt;C, C-&gt;B, B-&gt;C&lt;br /&gt;7.    A-&gt;C, C-&gt;A, B-&gt;A&lt;br /&gt;8.    A-&gt;C, C-&gt;A, B-&gt;C&lt;br /&gt;&lt;br /&gt;Out of which, there are only two cases under which the ants won't collide :&lt;br /&gt;    A-&gt;B, B-&gt;C, C-&gt;A&lt;br /&gt;    A-&gt;C, C-&gt;B, B-&gt;A&lt;br /&gt;&lt;br /&gt;Find all sets of consecutive integers that add up to 1000.&lt;br /&gt;Submitted by : James Barberousse&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are total 8 such series:&lt;br /&gt;1.    Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to 1000.&lt;br /&gt;(-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000&lt;br /&gt;2.    Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202.&lt;br /&gt;(-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000&lt;br /&gt;3.    Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70.&lt;br /&gt;(-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000&lt;br /&gt;4.    Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52.&lt;br /&gt;(-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000&lt;br /&gt;5.    Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52.&lt;br /&gt;28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000&lt;br /&gt;6.    Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70.&lt;br /&gt;55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000&lt;br /&gt;7.    Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202.&lt;br /&gt;198 + 199 + 200 +201 + 202 = 1000&lt;br /&gt;8.    Sum of 1 number starting from 1000.&lt;br /&gt;1000 = 1000&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;There is a 4-character code, with 2 of them being letters and the other 2 being numbers.&lt;br /&gt;&lt;br /&gt;How many maximum attempts would be necessary to find the correct code? Note that the code is case-sensitive.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The maximum number of attempts required are 16,22,400&lt;br /&gt;&lt;br /&gt;There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 4C2 i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).&lt;br /&gt;&lt;br /&gt;Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@&lt;br /&gt;&lt;br /&gt;Hence, the required answer is&lt;br /&gt;= 52*52*10*10*6&lt;br /&gt;= 16,22,400 attempts&lt;br /&gt;= 1.6 million approx.&lt;br /&gt;&lt;br /&gt;Thanks to Tim Sanders for opening BrainVista's brain !!!&lt;br /&gt;How many possible combinations are there in a 3x3x3 rubics cube?&lt;br /&gt;&lt;br /&gt;In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)?&lt;br /&gt;&lt;br /&gt;How many for a 4x4x4 cube?&lt;br /&gt;Submitted&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Let's consider 3x3x3 Rubics first.&lt;br /&gt;&lt;br /&gt;There are 8 corner cubes, which can be arranged in 8! ways.&lt;br /&gt;Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)&lt;br /&gt;&lt;br /&gt;Similarly, 12 edge cubes can be arranged in 12! ways.&lt;br /&gt;Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)&lt;br /&gt;&lt;br /&gt;Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.&lt;br /&gt;&lt;br /&gt;Total different possible combinations are&lt;br /&gt;= [(8!) * (3^7)] * [(12!) * (2^11)] / 2&lt;br /&gt;= (8!) * (3^7) * (12!) * (2^10)&lt;br /&gt;= 4.3252 * 10^19&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Similarly, for 4x4x4 Rubics total different possible combinations are&lt;br /&gt;= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24&lt;br /&gt;= 7.4011968 * 10^45&lt;br /&gt;&lt;br /&gt;Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.&lt;br /&gt;Substitute digits for the letters to make the following relation true.&lt;br /&gt;               N  E  V  E  R&lt;br /&gt;&lt;br /&gt;               L  E  A  V  E&lt;br /&gt;&lt;br /&gt;            +           M  E&lt;br /&gt;&lt;br /&gt;            -----------------&lt;br /&gt;&lt;br /&gt;               A  L  O  N  E&lt;br /&gt;Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A tough one!!!&lt;br /&gt;&lt;br /&gt;Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to&lt;br /&gt;&lt;br /&gt;solve it. Now use trial-n-error method.&lt;br /&gt;&lt;br /&gt;     N  E  V  E  R               2  1  4  1  9&lt;br /&gt;&lt;br /&gt;     L  E  A  V  E               3  1  5  4  1&lt;br /&gt;&lt;br /&gt;  +           M  E            +           6  1&lt;br /&gt;&lt;br /&gt;  -----------------           -----------------&lt;br /&gt;&lt;br /&gt;     A  L  O  N  E               5  3  0  2  1&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;One of the four people - Mr. Clinton, his wife Monika, their son Mandy and their daughter Cindy - is a singer and another is a dancer. Mr. Clinton is older than his wife and Mady is older than his sister.&lt;br /&gt;1.    If the singer and the dancer are the same sex, then the dancer is older than the singer.&lt;br /&gt;2.    If neither the singer nor the dancer is the parent of the other, then the singer is older than the dancer.&lt;br /&gt;3.    If the singer is a man, then the singer and the dancer are the same age.&lt;br /&gt;4.    If the singer and the dancer are of opposite sex then the man is older than the woman.&lt;br /&gt;5.    If the dancer is a woman, then the dancer is older than the singer.&lt;br /&gt;Whose occupation do you know? And what is his/her occupation?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Cindy is the Singer. Mr. Clinton or Monika is the Dancer.&lt;br /&gt;&lt;br /&gt;From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman.&lt;br /&gt;&lt;br /&gt;CASE I : Singer is a woman and Dancer is also a woman&lt;br /&gt;Then, the dancer is Monika and the singer is Cindy.&lt;br /&gt;&lt;br /&gt;CASE II : Singer is a woman and Dancer is also a man&lt;br /&gt;Then, the dancer is Mr. Clinton and the singer is Cindy.&lt;br /&gt;&lt;br /&gt;In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.&lt;br /&gt;There are 20 people in your applicant pool, including 5 pairs of identical twins.&lt;br /&gt;&lt;br /&gt;If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins? (Needless to say, this could cause trouble ;))&lt;br /&gt;Submitted&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The probability to hire 5 people with at least 1 pair of identical twins is 25.28%&lt;br /&gt;&lt;br /&gt;5 people from the 20 people can be hired in 20C5 = 15504 ways.&lt;br /&gt;&lt;br /&gt;Now, divide 20 people into two groups of 10 people each :&lt;br /&gt;G1 - with all twins&lt;br /&gt;G2 - with all people other than twins&lt;br /&gt;&lt;br /&gt;Let's find out all possible ways to hire 5 people without a single pair of indentical twins.&lt;br /&gt;People from G1    People from G2    No of ways to hire G1 without a single pair of indentical twins    No of ways to hire G2    Total ways&lt;br /&gt;0    5    10C0    10C5    252&lt;br /&gt;1    4    10C1    10C4    2100&lt;br /&gt;2    3    10C2 * 8/9    10C3    4800&lt;br /&gt;3    2    10C3 * 8/9 * 6/8    10C2    3600&lt;br /&gt;4    1    10C4 * 8/9 * 6/8 * 4/7    10C1    800&lt;br /&gt;5    0    10C5 * 8/9 * 6/8 * 4/7 * 2/6    10C0    32&lt;br /&gt;Total    11584&lt;br /&gt;&lt;br /&gt;Thus, total possible ways to hire 5 people without a single pair of indentical twins = 11584 ways&lt;br /&gt;&lt;br /&gt;So, total possible ways to hire 5 people with at least a single pair of indentical twins = 15504 - 11584 = 3920 ways&lt;br /&gt;&lt;br /&gt;Hence, the probability to hire 5 people with at least a single pair of indentical twins&lt;br /&gt;= 3920/15504&lt;br /&gt;= 245/969&lt;br /&gt;= 0.2528&lt;br /&gt;= 25.28%&lt;br /&gt;In a hotel, rooms are numbered from 101 to 550. A room is chosen at random. What is the probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are total 450 rooms.&lt;br /&gt;&lt;br /&gt;Out of which 299 room number starts with either 1, 2 or 3. (as room number 100 is not there) Now out of those 299 rooms only 90 room numbers end with 4, 5 or 6&lt;br /&gt;&lt;br /&gt;So the probability is 90/450 i.e. 1/5 or 0.20&lt;br /&gt;Draw 9 dots on a page, in the shape of three rows of three dots to form a square. Now place your pen on the page, draw 4 straight lines and try and cover all the dots.&lt;br /&gt;&lt;br /&gt;You're not allowed to lift your pen.&lt;br /&gt;&lt;br /&gt;Note: Don't be confined by the dimensions of the square.&lt;br /&gt;Submitted&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.&lt;br /&gt;&lt;br /&gt;Find the tractors each originally had?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.&lt;br /&gt;&lt;br /&gt;It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X &amp;amp; Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)&lt;br /&gt;&lt;br /&gt;Similarly, before Y gave tractors to X &amp;amp; Z, they had 6 &amp;amp; 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)&lt;br /&gt;&lt;br /&gt;Again, before X gave tractors to Y &amp;amp; Z, they had 21 &amp;amp; 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)&lt;br /&gt;&lt;br /&gt;Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.&lt;br /&gt;&lt;br /&gt;There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.&lt;br /&gt;&lt;br /&gt;In the mean time the whole platoon has moved ahead by 50m.&lt;br /&gt;&lt;br /&gt;The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.&lt;br /&gt;Submitted&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The last person covered 120.71 meters.&lt;br /&gt;&lt;br /&gt;It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.&lt;br /&gt;&lt;br /&gt;Let's assume that when the last person reached the first person, the platoon moved X meters forward.&lt;br /&gt;&lt;br /&gt;Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.&lt;br /&gt;&lt;br /&gt;Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.&lt;br /&gt;&lt;br /&gt;Now, as the ratios are equal,&lt;br /&gt;(50+X)/X = X/(50-X)&lt;br /&gt;(50+X)*(50-X) = X*X&lt;br /&gt;&lt;br /&gt;Solving, X=35.355 meters&lt;br /&gt;&lt;br /&gt;Thus, total distance covered by the last person&lt;br /&gt;= (50+X) + X&lt;br /&gt;= 2*X + 50&lt;br /&gt;= 2*(35.355) + 50&lt;br /&gt;= 120.71 meters&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.&lt;br /&gt;Assume that you have enough coins of 1, 5, 10, 25 and 50 cents.&lt;br /&gt;&lt;br /&gt;How many ways are there to make change for a dollar? Do explain your answer.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;There are 292 ways to make change for a dollar using coins of 1, 5, 10, 25 and 50 cents.&lt;br /&gt;&lt;br /&gt;Let's generalised the teaser and make a table as shown above.&lt;br /&gt;&lt;br /&gt;If you wish to make change for 75 cents using only 1, 5, 10 and 25 cent coins, go to the .25 row and the 75 column to obtain 121 ways to do this.&lt;br /&gt;&lt;br /&gt;The table can be created from left-to-right and top-to-bottom. Start with the top left i.e. 1 cent row. There is exactly one way to make change for every amount. Then calculate the 5 cents row by adding the number of ways to make change for the amount using 1 cent coins plus the number of ways to make change for 5 cents less using 1 and 5 cent coins.&lt;br /&gt;&lt;br /&gt;Let's take an example:&lt;br /&gt;To get change for 50 cents using 1, 5 and 10 cent coins.&lt;br /&gt;* 50 cents change using 1 and 5 cent coins = 11 ways&lt;br /&gt;* (50-10) 40 cents change using 1, 5 and 10 cent coins = 25 ways&lt;br /&gt;* 50 cents change using 1, 5 and 10 cent coins = 11+25 = 36 ways&lt;br /&gt;&lt;br /&gt;Let's take another example:&lt;br /&gt;To get change for 75 cents using all coins up to 50 cent i.e. 1, 5, 10, 25 and 50 cents coins.&lt;br /&gt;* 75 cents change using coins upto 25 cent = 121 ways&lt;br /&gt;* (75-50) 25 cents change using coins upto 50 cent = 13 ways&lt;br /&gt;* 75 cents change using coins upto 50 cent = 121+13 = 134 ways&lt;br /&gt;&lt;br /&gt;For people who don't want to tease their brain and love to do computer programming, there is a simple way. Write a small multi-loop program to solve the equation: A + 5B + 10C + 25D + 50E = 100&lt;br /&gt;where,&lt;br /&gt;A = 0 to 100&lt;br /&gt;B = 0 to 20&lt;br /&gt;C = 0 to 10&lt;br /&gt;D = 0 to 4&lt;br /&gt;E = 0 to 2&lt;br /&gt;&lt;br /&gt;The program should output all the possible values of A, B, C, D and E for which the equation is satisfied.&lt;br /&gt;&lt;br /&gt;In a Road Race, one of the three bikers was doing 15km less than the first and 3km more than the third. He also finished the race 12 minutes after the first and 3 minutes before the third.&lt;br /&gt;&lt;br /&gt;Can you find out the speed of each biker, the time taken by each biker to finish the race and the length of the course?&lt;br /&gt;&lt;br /&gt;Assume that there were no stops in the race and also they were driving with constant speeds through out the&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Let us assume that&lt;br /&gt;Speed of First biker = V1 km/min&lt;br /&gt;Speed of Second biker = V2 km/min&lt;br /&gt;Speed of Third biker = V3 km/min&lt;br /&gt;Total time take by first biker = T1 min&lt;br /&gt;Total distance = S km&lt;br /&gt;&lt;br /&gt;Now as per the data given in the teaser, at a time T min&lt;br /&gt;   X1 = V1 * T         ----&gt; 1&lt;br /&gt;&lt;br /&gt;   X1 - 15 = V2 * T    ----&gt; 2&lt;br /&gt;&lt;br /&gt;   X1 - 18 = V3 * T    ----&gt; 3&lt;br /&gt;&lt;br /&gt;At a Distance S Km.&lt;br /&gt;   S = V1 * T1         ----&gt; 4&lt;br /&gt;&lt;br /&gt;   S = V2 * (T1 + 12)  ----&gt; 5&lt;br /&gt;&lt;br /&gt;   S = V3 * (T1 + 15)  ----&gt; 6&lt;br /&gt;&lt;br /&gt;Thus there are 6 equations and 7 unknown data that means it has infinite number of solutions.&lt;br /&gt;&lt;br /&gt;By solving above 6 equations we get,&lt;br /&gt;Time taken by first biker, T1 = 60 Min.&lt;br /&gt;Time taken by Second biker, T2 = 72 Min.&lt;br /&gt;Time taken by first biker, T3 = 75 Min.&lt;br /&gt;&lt;br /&gt;Also, we get&lt;br /&gt;Speed of first biker, V1 = 90/T km/min&lt;br /&gt;Speed of second biker, V2 = (5/6)V1 = 75/T km/min&lt;br /&gt;Speed of third biker, V3 = (4/5)V1 = 72/T km/min&lt;br /&gt;&lt;br /&gt;Also, the length of the course, S = 5400/T km&lt;br /&gt;&lt;br /&gt;Thus, for the data given, only the time taken by each biker can be found i.e. 60, 72 and 75 minutes. For other quantities, one more independent datum is required i.e. either T or V1 or V2 or V3&lt;br /&gt;&lt;br /&gt;Thanks to Theertham Srinivas for the answer !!!&lt;br /&gt;What is the four-digit number in which the first digit is 1/3 of the second, the third is the sum of the first and second, and the last is three times the second?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The 4 digit number is 1349.&lt;br /&gt;&lt;br /&gt;It is given that the first digit is 1/3 of the second. There are 3 such possibilities.&lt;br /&gt;1.    1 and 3&lt;br /&gt;2.    2 and 6&lt;br /&gt;3.    3 and 9&lt;br /&gt;Now, the third digit is the sum of the first and second digits.&lt;br /&gt;1.    1 + 3 = 4&lt;br /&gt;2.    2 + 6 = 8&lt;br /&gt;3.    3 + 9 = 12&lt;br /&gt;It is clear that option 3 is not possible. So we are left with only two options. Also, the last digit is three times the second, which rules out the second option. Hence, the answer is 1349.&lt;br /&gt;&lt;br /&gt;Difference between Bholu's and Molu's age is 2 years and the difference between Molu's and Kolu's age is 5 years.&lt;br /&gt;&lt;br /&gt;What is the maximum possible value of the sum of the difference in their ages, taken two at a time?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The maximum possible value of the sum of the difference in their ages - taken two at a time - is 14 years.&lt;br /&gt;&lt;br /&gt;It is given that -&lt;br /&gt;"Difference between Bholu's and Molu's age is 2 years"&lt;br /&gt;"Difference between Molu's and Kolu's age is 5 years"&lt;br /&gt;&lt;br /&gt;Now, to get the maximum possible value, the difference between Bholu's and Kolu's age should be maximum i.e. Molu's age should be in between Bholu's and Kolu's age. Then, the difference between Bholu's and Kolu's age is 7 years.&lt;br /&gt;&lt;br /&gt;Hence, the maximum possible value of the sum of the difference in their ages - taken two at a time - is (2 + 5 + 7) 14 years.&lt;br /&gt;If it is given that:&lt;br /&gt;25 - 2 = 3&lt;br /&gt;100 x 2 = 20&lt;br /&gt;36 / 3 = 2&lt;br /&gt;&lt;br /&gt;What is 144 - 3 = ?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 3 possible answers to it.&lt;br /&gt;&lt;br /&gt;Answer 1 : 9&lt;br /&gt;Simply replace the first number by its square root.&lt;br /&gt;(25) 5 - 2 = 3&lt;br /&gt;(100) 10 x 2 = 20&lt;br /&gt;(36) 6 / 3 = 2&lt;br /&gt;(144) 12 - 3 = 9&lt;br /&gt;&lt;br /&gt;Answer 2 : 11&lt;br /&gt;Drop the digit in the tens position from the first number.&lt;br /&gt;(2) 5 - 2 = 3&lt;br /&gt;1 (0) 0 x 2 = 20&lt;br /&gt;(3) 6 / 3 = 2&lt;br /&gt;1 (4) 4 - 3 = 11&lt;br /&gt;&lt;br /&gt;You will get the same answer on removing left and right digit alternatively from the first number i.e remove left digit from first (2), right digit from second (0), left digit from third (3) and right digit from forth (4).&lt;br /&gt;(2) 5 - 2 = 3&lt;br /&gt;10 (0) x 2 = 20&lt;br /&gt;(3) 6 / 3 = 2&lt;br /&gt;14 (4) - 3 = 11&lt;br /&gt;&lt;br /&gt;Answer 3 : 14&lt;br /&gt;Drop left and right digit alternatively from the actual answer.&lt;br /&gt;25 - 2 = (2) 3 (drop left digit i.e. 2)&lt;br /&gt;100 * 2 = 20 (0) (drop right digit i.e. 0)&lt;br /&gt;36 / 3 = (1) 2 (drop left digit i.e. 1)&lt;br /&gt;144 - 3 = 14 (1) (drop right digit i.e. 1)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;A 3 digit number is such that it's unit digit is equal to the product of the other two digits which are prime. Also, the difference between it's reverse and itself is 396.&lt;br /&gt;&lt;br /&gt;What is the sum of the three digits?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The required number is 236 and the sum is 11.&lt;br /&gt;&lt;br /&gt;It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e. 22, 23, 32 or 33 which means that there are four possible numbers - 224, 236, 326 and 339.&lt;br /&gt;&lt;br /&gt;Now, it is also given that - the difference between it's reverse and itself is 396. Only 236 satisfies this condition. Hence, the sum of the three digits is 11.&lt;br /&gt;&lt;br /&gt;There are 4 mugs placed upturned on the table. Each mug have the same number of marbles and a statement about the number of marbles in it. The statements are: Two or Three, One or Four, Three or One, One or Two.&lt;br /&gt;&lt;br /&gt;Only one of the statement is correct. How many marbles are there under each mug?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A simple one.&lt;br /&gt;&lt;br /&gt;As it is given that only one of the four statement is correct, the correct number can not appear in more than one statement. If it appears in more than one statement, then more than one statement will be correct.&lt;br /&gt;&lt;br /&gt;Hence, there are 4 marbles under each mug.&lt;br /&gt;At University of Probability, there are 375 freshmen, 293 sophomores, 187 juniors, &amp;amp; 126 seniors. One student will randomly be chosen to receive an award.&lt;br /&gt;&lt;br /&gt;What percent chance is there that it will be a junior? Round to the nearest whole percent&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;19%&lt;br /&gt;&lt;br /&gt;This puzzle is easy. Divide the number of juniors (187) by the total number of students (981), &amp;amp; then multiply the number by 100 to convert to a percentage.&lt;br /&gt;&lt;br /&gt;Hence the answer is (187/981)*100 = 19%&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;If you were to dial any 7 digits on a telephone in random order, what is the probability that you will dial your own phone number?&lt;br /&gt;&lt;br /&gt;Assume that your telephone number is 7-digits.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;1 in 10,000,000&lt;br /&gt;&lt;br /&gt;There are 10 digits i.e. 0-9. First digit can be dialed in 10 ways. Second digit can be dialed in 10 ways. Third digit can be dialed in 10 ways. And so on.....&lt;br /&gt;&lt;br /&gt;Thus, 7-digit can be dialed in 10*10*10*10*10*10*10 (=10,000,000) ways. And, you have just one telephone number. Hence, the possibility that you will dial your own number is 1 in 10,000,000.&lt;br /&gt;&lt;br /&gt;Note that 0123456 may not be a valid 7-digit telephone number. But while dialing in random order, that is one of the possible 7-digit number which you may dial.&lt;br /&gt;An anthropologist discovers an isolated tribe whose written alphabet contains only six letters (call the letters A, B, C, D, E and F). The tribe has a taboo against using the same letter twice in the same word. It's never done.&lt;br /&gt;&lt;br /&gt;If each different sequence of letters constitues a different word in the language, what is the maximum number of six-letter words that the language can employ?&lt;br /&gt;Submitted&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The language can employ maximum of 720 six-letter words.&lt;br /&gt;&lt;br /&gt;It is a simple permutation problem of arranging 6 letters to get different six-letter words. And it can be done in in 6! ways i.e. 720 ways.&lt;br /&gt;&lt;br /&gt;In otherwords, the first letter can be any of the given 6 letters (A through F). Then, whatever the first letter is, the second letter will always be from the remaining 5 letters (as same letter can not be used twice), and the third letter always be from the remaining 4 letters, and so on. Thus, the different possible six-letter words are 6*5*4*3*2*1 = 720&lt;br /&gt;&lt;br /&gt;Kate, Demi, Madona, Sharon, Britney and Nicole decided to lunch together in a restaurant. The waiter led them to a round table with six chairs.&lt;br /&gt;&lt;br /&gt;How many different ways can they seat?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 120 different possible seating arrangments.&lt;br /&gt;&lt;br /&gt;Note that on a round table ABCDEF and BCDEFA is the same.&lt;br /&gt;&lt;br /&gt;The first person can sit on any one of the seats. Now, for the second person there are 5 options, for the third person there are 4 options, for the forth person there are 3 options, for the fifth person there are 2 options and for the last person there is just one option.&lt;br /&gt;&lt;br /&gt;Thus, total different possible seating arrangements are&lt;br /&gt;= 5 * 4 * 3 * 2 * 1&lt;br /&gt;= 120&lt;br /&gt;3 blocks are chosen randomly on a chessboard. What is the probability that they are in the same diagonal?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are total of 64 blocks on a chessboard. So 3 blocks can be chosen out of 64 in 64C3 ways.&lt;br /&gt;So the sample space is = 41664&lt;br /&gt;&lt;br /&gt;There are 2 diagonal on chessboard each one having 8 blocks. Consider one of them.&lt;br /&gt;3 blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.&lt;br /&gt;But there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 = 112&lt;br /&gt;&lt;br /&gt;The require probability is&lt;br /&gt;= 112 / 41664&lt;br /&gt;= 1 / 372&lt;br /&gt;= 0.002688&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;What is the area of the triangle ABC with A(e,p) B(2e,3p) and C(3e,5p)?&lt;br /&gt;&lt;br /&gt;where p = PI (3.141592654)&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A tricky ONE.&lt;br /&gt;&lt;br /&gt;Given 3 points are colinear. Hence, it is a straight line.&lt;br /&gt;&lt;br /&gt;Hence area of triangle is 0.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Silu and Meenu were walking on the road.&lt;br /&gt;&lt;br /&gt;Silu said, "I weigh 51 Kgs. How much do you weigh?"&lt;br /&gt;&lt;br /&gt;Meenu replied that she wouldn't reveal her weight directly as she is overweight. But she said, "I weigh 29 Kgs plus half of my weight."&lt;br /&gt;&lt;br /&gt;How much does Meenu weigh?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Meenu weighs 58 Kgs.&lt;br /&gt;&lt;br /&gt;It is given that Meenu weighs 29 Kgs plus half of her own weight. It means that 29 Kgs is the other half. So she weighs 58 Kgs.&lt;br /&gt;&lt;br /&gt;Solving mathematically, let's assume that her weight is X Kgs.&lt;br /&gt;X = 29 + X/2&lt;br /&gt;2*X = 58 + X&lt;br /&gt;X = 58 Kgs&lt;br /&gt;Consider the sum: ABC + DEF + GHI = JJJ&lt;br /&gt;&lt;br /&gt;If different letters represent different digits, and there are no leading zeros, what does J represent?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The value of J must be 9.&lt;br /&gt;&lt;br /&gt;Since there are no leading zeros, J must be 7, 8, or 9. (JJJ = ABC + DEF + GHI = 14? + 25? + 36? = 7??)&lt;br /&gt;&lt;br /&gt;Now, the remainder left after dividing any number by 9 is the same as the remainder left after dividing the sum of the digits of that number by 9. Also, note that 0 + 1 + ... + 9 has a remainder of 0 after dividing by 9 and JJJ has a remainder of 0, 3, or 6.&lt;br /&gt;&lt;br /&gt;The number 9 is the only number from 7, 8 and 9 that leaves a remainder of 0, 3, or 6 if you remove it from the sum 0 + 1 + ... + 9. Hence, it follows that J must be 9.&lt;br /&gt;A man has Ten Horses and nine stables as shown here.&lt;br /&gt;  [] [] [] [] [] [] [] [] []&lt;br /&gt;The man wants to fit Ten Horses into nine stables. How can he fit Ten horses into nine stables?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The answer is simple. It says the man wants to fit "Ten Horses" into nine stables. There are nine letters in the phrase "Ten Horses". So you can put one letter each in all nine stables.&lt;br /&gt;  [T] [E] [N] [H] [O] [R] [S] [E] [S]&lt;br /&gt;&lt;br /&gt;A man is at a river with a 9 gallon bucket and a 4 gallon bucket. He needs exactly 6 gallons of water.&lt;br /&gt;&lt;br /&gt;How can he use both buckets to get exactly 6 gallons of water?&lt;br /&gt;&lt;br /&gt;Note that he cannot estimate by dumping some of the water out of the 9 gallon bucket or the 4 gallon bucket&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;For the sack of explanation, let's identify 4 gallon bucket as Bucket P and 9 gallon bucket as Bucket Q.&lt;br /&gt;&lt;br /&gt;Operation    4 gallon bucket&lt;br /&gt;(Bucket P)    9 gallon bucket&lt;br /&gt;(Bucket Q)&lt;br /&gt;Initially    0    0&lt;br /&gt;Fill the bucket Q with 9 gallon water    0    9&lt;br /&gt;Pour 4 gallon water from bucket Q to bucket P    4    5&lt;br /&gt;Empty bucket P    0    5&lt;br /&gt;Pour 4 gallon water from bucket Q to bucket P    4    1&lt;br /&gt;Empty bucket P    0    1&lt;br /&gt;Pour 1 gallon water from bucket Q to bucket P    1    0&lt;br /&gt;Fill the bucket Q with 9 gallon water    1    9&lt;br /&gt;Pour 3 gallon water from bucket Q to bucket P    4    6&lt;br /&gt;&lt;br /&gt;9 gallon bucket contains 6 gallon of water, as required.&lt;br /&gt;&lt;br /&gt;Each of the five characters in the word BRAIN has a different value between 0 and 9. Using the given grid, can you find out the value of each character?&lt;br /&gt;        B   R   A   I   N   31&lt;br /&gt;&lt;br /&gt;        B   B   R   B   A   31&lt;br /&gt;&lt;br /&gt;        N   I   A   B   B   32&lt;br /&gt;&lt;br /&gt;        N   I   B   A   I   30&lt;br /&gt;&lt;br /&gt;        I   R   A   A   A   23&lt;br /&gt;&lt;br /&gt;       37  29  25  27  29&lt;br /&gt;The numbers on the extreme right represent the sum of the values represented by the characters in that row. Also, the numbers on the last raw represent the sum of the values represented by the characters in that column. e.g. B + R + A + I + N = 31 (from first row)&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;B=7, R=6, A=4, I=5 and N=9&lt;br /&gt;&lt;br /&gt;Make total 10 equations - 5 for rows and 5 for columns - and sovle them.&lt;br /&gt;&lt;br /&gt;From Row3 and Row4,&lt;br /&gt;N + I + A + B + B = N + I + B + A + I + 2&lt;br /&gt;B = I + 2&lt;br /&gt;&lt;br /&gt;From Row1 and Row3,&lt;br /&gt;B + R + A + I + N = N + I + A + B + B - 1&lt;br /&gt;R = B - 1&lt;br /&gt;&lt;br /&gt;From Column2,&lt;br /&gt;R + B + I + I + R = 29&lt;br /&gt;B + 2R + 2I = 29&lt;br /&gt;B + 2(B - 1) + 2I = 29&lt;br /&gt;3B + 2I = 31&lt;br /&gt;3(I + 2) + 2I = 31&lt;br /&gt;5I = 25&lt;br /&gt;I = 5&lt;br /&gt;&lt;br /&gt;Hence, B=7 and R=6&lt;br /&gt;&lt;br /&gt;From Row2,&lt;br /&gt;B + B + R + B + A = 31&lt;br /&gt;3B + R + A = 31&lt;br /&gt;3(7) + 6 + A = 31&lt;br /&gt;A = 4&lt;br /&gt;&lt;br /&gt;From Row1,&lt;br /&gt;B + R + A + I + N = 31&lt;br /&gt;7 + 6 + 4 + 5 + N = 31&lt;br /&gt;N = 9&lt;br /&gt;&lt;br /&gt;Thus, B=7, R=6, A=4, I=5 and N=9&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Submit&lt;br /&gt;Answer         Users&lt;br /&gt;Answer (24)         BrainV&lt;br /&gt;&lt;br /&gt;There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.&lt;br /&gt;1.    Take 8 coins and weigh 4 against 4.&lt;br /&gt;o    If both are not equal, goto step 2&lt;br /&gt;o    If both are equal, goto step 3&lt;br /&gt;&lt;br /&gt;2.    One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.&lt;br /&gt;o    If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.&lt;br /&gt;    If both are equal, L4 is the odd coin and is lighter.&lt;br /&gt;    If L2 is light, L2 is the odd coin and is lighter.&lt;br /&gt;    If L3 is light, L3 is the odd coin and is lighter.&lt;br /&gt;&lt;br /&gt;o    If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2&lt;br /&gt;    If both are equal, there is some error.&lt;br /&gt;    If H1 is heavy, H1 is the odd coin and is heavier.&lt;br /&gt;    If H2 is heavy, H2 is the odd coin and is heavier.&lt;br /&gt;&lt;br /&gt;o    If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4&lt;br /&gt;    If both are equal, L1 is the odd coin and is lighter.&lt;br /&gt;    If H3 is heavy, H3 is the odd coin and is heavier.&lt;br /&gt;    If H4 is heavy, H4 is the odd coin and is heavier.&lt;br /&gt;&lt;br /&gt;3.    The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.&lt;br /&gt;o    If both are equal, there is some error.&lt;br /&gt;o    If X is heavy, X is the odd coin and is heavier.&lt;br /&gt;o    If X is light, X is the odd coin and is lighter.&lt;br /&gt;In a sports contest there were m medals awarded on n successive days (n &gt; 1).&lt;br /&gt;1.    On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.&lt;br /&gt;2.    On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.&lt;br /&gt;3.    On the nth and last day, the remaining n medals were awarded.&lt;br /&gt;How many days did the contest last, and how many medals were awarded altogether?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Total 36 medals were awarded and the contest was for 6 days.&lt;br /&gt;&lt;br /&gt;On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals&lt;br /&gt;On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals&lt;br /&gt;On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals&lt;br /&gt;On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals&lt;br /&gt;On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals&lt;br /&gt;On day 6: Medals awarded 6&lt;br /&gt;&lt;br /&gt;I got this answer by writing small program. If anyone know any other simpler method, do submit it.&lt;br /&gt;&lt;br /&gt;A number of 9 digits has the following properties:&lt;br /&gt;    The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2&lt;=n&lt;=9.&lt;br /&gt;    Each digit in the number is different i.e. no digits are repeated.&lt;br /&gt;    The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.&lt;br /&gt;Find the number.&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The answer is 381654729&lt;br /&gt;&lt;br /&gt;One way to solve it is Trial-&amp;amp;-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.&lt;br /&gt;&lt;br /&gt;The other way to solve this problem is by writing a computer program that systematically tries all possibilities.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day.&lt;br /&gt;&lt;br /&gt;What part of the contents of the container is left at the end of the second day?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Assume that contents of the container is X&lt;br /&gt;&lt;br /&gt;On the first day 1/3rd is evaporated.&lt;br /&gt;(1 - 1/3) of X is remaining i.e. (2/3)X&lt;br /&gt;&lt;br /&gt;On the Second day 3/4th is evaporated. Hence,&lt;br /&gt;(1- 3/4) of (2/3)X is remaining&lt;br /&gt;i.e. (1/4)(2/3)X = (1/6) X&lt;br /&gt;&lt;br /&gt;Hence 1/6th of the contents of the container is remaining&lt;br /&gt;&lt;br /&gt;There are four people in a room (not including you). Exactly two of these four always tell the truth. The other two always lie.&lt;br /&gt;&lt;br /&gt;You have to figure out who is who IN ONLY 2 QUESTIONS. Your questions have to be YES or NO questions and can only be answered by one person. (If you ask the same question to two different people then that counts as two questions). Keep in mind that all four know each other's characteristics whether they lie or not.&lt;br /&gt;&lt;br /&gt;What questions would you ask to figure out who is who? Remember that you can ask only 2 questions.&lt;br /&gt;Submitted&lt;br /&gt;You have 3 baskets, &amp;amp; each one contains exactly 4 balls, each of which is of the same size. Each ball is either red, black, white, or purple, &amp;amp; there is one of each color in each basket.&lt;br /&gt;&lt;br /&gt;If you were blindfolded, &amp;amp; lightly shook each basket so that the balls would be randomly distributed, &amp;amp; then took 1 ball from each basket, what chance is there that you would have exactly 2 red balls?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 64 different possible outcomes, &amp;amp; in 9 of these, exactly 2 of the balls will be red. There is thus a slightly better than 14% chance [(9/64)*100] that exactly 2 balls will be red.&lt;br /&gt;&lt;br /&gt;A much faster way to solve the problem is to look at it this way. There are 3 scenarios where exactly 3 balls are red:&lt;br /&gt;&lt;br /&gt;  1   2   3&lt;br /&gt;&lt;br /&gt; -----------&lt;br /&gt;&lt;br /&gt;  R   R   X&lt;br /&gt;&lt;br /&gt;  R   X   R&lt;br /&gt;&lt;br /&gt;  X   R   R&lt;br /&gt;&lt;br /&gt;X is any ball that is not red.&lt;br /&gt;There is a 4.6875% chance that each of these situations will occur.&lt;br /&gt;&lt;br /&gt;Take the first one, for example: 25% chance the first ball is red, multiplied by a 25% chance the second ball is red, multiplied by a 75% chance the third ball is not red.&lt;br /&gt;&lt;br /&gt;Because there are 3 scenarios where this outcome occurs, you multiply the 4.6875% chance of any one occurring by 3, &amp;amp; you get 14.0625%&lt;br /&gt;&lt;br /&gt;Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetiton allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if atleast 7 of your numbers are there in the 11 chosen by the Lottery Commission.&lt;br /&gt;&lt;br /&gt;What is the probablity of winning the lottery?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The probability of winning the lottery is two in one billion i.e. only two person can win from one billion !!!&lt;br /&gt;&lt;br /&gt;Let's find out sample space first. The Lottery Commission chooses 11 numbers from the 80. Hence, the 11 numbers from the 80 can be selected in 80C11 ways which is very very high and is equal to 1.04776 * 1013&lt;br /&gt;&lt;br /&gt;Now, you have to select 8 numbers from 80 which can be selected in 80C8 ways. But we are interested in only those numbers which are in 11 numbers selected by the Lottery Commision. There are 2 cases.&lt;br /&gt;    You might select 8 numbers which all are there in 11 numbers choosen by the Lottery Commission. So there are 11C8 ways.&lt;br /&gt;    Another case is you might select 7 lucky numbers and 1 non-lucky number from the remaining 69 numbers. There are ( 11C7 ) * ( 69C1 ) ways to do that.&lt;br /&gt;So total lucky ways are&lt;br /&gt;= ( 11C8 ) + ( 11C7 ) * ( 69C1 )&lt;br /&gt;= (165) + (330) * (69)&lt;br /&gt;= 165 + 22770&lt;br /&gt;= 22935&lt;br /&gt;&lt;br /&gt;Hence, the probability of the winning lottery is&lt;br /&gt;= (Total lucky ways) / (Total Sample space)&lt;br /&gt;= (22935) / ( 1.04776 * 1013)&lt;br /&gt;= 2.1889 * 10-9&lt;br /&gt;i.e. 2 in a billion.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Submit&lt;br /&gt;Answer         Users&lt;br /&gt;Answer (4)         BrainVista&lt;br /&gt;Answer         Puzzle&lt;br /&gt;To move a Safe, two cylindrical steel bars 7 inches in diameter are used as rollers.&lt;br /&gt;&lt;br /&gt;How far will the safe have moved forward when the rollers have made one revolution?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The safe must have moved 22 inches forward.&lt;br /&gt;&lt;br /&gt;If the rollers make one revolution, the safe will move the distance equal to the circumference of the roller. Hence, the distance covered by the safe is&lt;br /&gt;= PI * Diameter (or 2 * PI * Radius)&lt;br /&gt;= PI * 7&lt;br /&gt;= 3.14159265 * 7&lt;br /&gt;= 21.99115&lt;br /&gt;= 22 inches approx.&lt;br /&gt;SubmittIf a rook and a bishop of a standard chess set are randomly placed on a chessboard, what is the probability that one is attacking the other?&lt;br /&gt;&lt;br /&gt;Note that both are different colored pieces.&lt;br /&gt;SubmAnswer&lt;br /&gt;&lt;br /&gt;The probability of either the Rook or the Bishop attacking the other is 0.3611&lt;br /&gt;&lt;br /&gt;A Rook and a Bishop on a standard chess-board can be arranged in 64P2 = 64*63 = 4032 ways&lt;br /&gt;&lt;br /&gt;Now, there are 2 cases - Rook attacking Bishop and Bishop attacking Rook. Note that the Rook and the Bishop never attack each other simultaneously. Let's consider both the cases one by one.&lt;br /&gt;&lt;br /&gt;Case I - Rook attacking Bishop&lt;br /&gt;The Rook can be placed in any of the given 64 positions and it always attacks 14 positions. Hence, total possible ways of the Rook attacking the Bishop = 64*14 = 896 ways&lt;br /&gt;&lt;br /&gt;Case II - Bishop attacking Rook&lt;br /&gt;View the chess-board as a 4 co-centric hollow squares with the outermost square with side 8 units and the innermost square with side 2 units.&lt;br /&gt;&lt;br /&gt;If the bishop is in one of the outer 28 squares, then it can attack 7 positions. If the bishop is in one of the 20 squares at next inner-level, then it can attack 9 positions. Similarly if the bishop is in one of the 12 squares at next inner-level, then it can attack 11 positions. And if the bishop is in one of the 4 squares at next inner-level (the innermost level), then it can attack 13 positions.&lt;br /&gt;&lt;br /&gt;Hence, total possible ways of the Bishop attacking the Rook&lt;br /&gt;= 28*7 + 20*9 + 12*11 + 4*13&lt;br /&gt;= 560 ways&lt;br /&gt;&lt;br /&gt;Thus, the required probability is&lt;br /&gt;= (896 + 560) / 4032&lt;br /&gt;= 13/36&lt;br /&gt;= 0.3611&lt;br /&gt;itted ed&lt;br /&gt;Here in England McDonald's has just launched a new advertising campaign. The poster shows 8 McDonald's products and underneath claims there are 40312 combinations of the above items.&lt;br /&gt;&lt;br /&gt;Given that the maximum number of items allowed is 8, and you are allowed to have less than 8 items, and that the order of purchase does not matter (i.e. buying a burger and fries is the same as buying fries and a burger)&lt;br /&gt;&lt;br /&gt;How many possible combinations are there? Are McDonald's correct in claiming there are 40312 combinations?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Total possible combinations are 12869.&lt;br /&gt;&lt;br /&gt;It is given that you can order maximum of 8 items and you are allowed to have less than 8 items. Also, the order of purchase does not matter. Let's create a table for ordering total N items using X products.&lt;br /&gt;Items&lt;br /&gt;Ordered&lt;br /&gt;(N)    Products Used (X)&lt;br /&gt;    1    2    3    4    5    6    7    8&lt;br /&gt;1    1    -    -    -    -    -    -    -&lt;br /&gt;2    1    1    -    -    -    -    -    -&lt;br /&gt;3    1    2    1    -    -    -    -    -&lt;br /&gt;4    1    3    3    1    -    -    -    -&lt;br /&gt;5    1    4    6    4    1    -    -    -&lt;br /&gt;6    1    5    10    10    5    1    -    -&lt;br /&gt;7    1    6    15    20    15    6    1    -&lt;br /&gt;8    1    7    21    35    35    21    7    1&lt;br /&gt;Total (T)    8    28    56    70    56    28    8    1&lt;br /&gt;Ways to choose&lt;br /&gt;X products from&lt;br /&gt;8 products (W)    8C1    8C2    8C3    8C4    8C5    8C6    8C7    8C8&lt;br /&gt;Total combinations&lt;br /&gt;(T*W)    64    784    3136    4900    3136    784    64    1&lt;br /&gt;&lt;br /&gt;Thus, total possible combinations are&lt;br /&gt;= 64 + 784 + 3136 + 4900 + 3136 + 784 + 64 + 1&lt;br /&gt;= 12869&lt;br /&gt;What are the chances that at least two out of a group of fifty people share the same birthday?&lt;br /&gt;SubmittedAnswer&lt;br /&gt;&lt;br /&gt;The probability of atleast two out of a group of 50 people share the same birthday is 97%&lt;br /&gt;&lt;br /&gt;Probability of atleast two share the same birthday = 1 - probability of all 50 have different birthdays&lt;br /&gt;&lt;br /&gt;Probability of all 50 have different birthday&lt;br /&gt;= 365/365 * 364/365 * 363/365 * ... * 317/365 * 316/365&lt;br /&gt;= (365 * 364 * 363 * 362 * ... * 317 * 316)/36550&lt;br /&gt;= 0.0296264&lt;br /&gt;&lt;br /&gt;Probability of atleast two share the same birthday&lt;br /&gt;= 1 - 0.0296264&lt;br /&gt;= 0.9703735&lt;br /&gt;= 97% approx.&lt;br /&gt;&lt;br /&gt;Thus, the probability of atleast two out of a group of 50 people share the same birthday is 97%&lt;br /&gt;&lt;br /&gt;This explains why in a school/college with classrooms of 50 students, there are at least two students with a birthday on the same day of the year. Also, if there are 23 people in the room, then there are 50% chances that atleast two of them have a birthday on the same day of the year!!!&lt;br /&gt;&lt;br /&gt;A tank can be filled by pipe A in 30 minutes and by pipe B in 24 minutes. Outlet pipe C can empty the full tank in X minutes.&lt;br /&gt;&lt;br /&gt;If the tank is empty initially and if all the three pipes A, B and C are opened simultaneously, the tank will NEVER be full. Give the maximal possible value of X.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The maximum possible value of X is 13 minutes 20 seconds.&lt;br /&gt;&lt;br /&gt;In one minute,&lt;br /&gt;pipe A can fill 1/30 part of the tank.&lt;br /&gt;pipe B can fill 1/24 part of the tank.&lt;br /&gt;&lt;br /&gt;Thus, the net water level increase in one minute is&lt;br /&gt;= 1/30 + 1/24&lt;br /&gt;= 3/40 part of the tank&lt;br /&gt;&lt;br /&gt;In order to keep the tank always empty, outlet pipe C should empty at least 3/40 part of the tank in one minute. Thus, pipe C can empty the full tank in 40/3 i.e. 13 minutes 20 seconds.&lt;br /&gt;A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, &amp;amp; now his salary is Rs. 22702.68&lt;br /&gt;&lt;br /&gt;What was his salary to begin with?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Rs.22176&lt;br /&gt;&lt;br /&gt;Assume his salary was Rs. X&lt;br /&gt;&lt;br /&gt;He earns 5% raise. So his salary is (105*X)/100&lt;br /&gt;&lt;br /&gt;A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100) which is Rs. 22702.68&lt;br /&gt;&lt;br /&gt;Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68&lt;br /&gt;X = 22176&lt;br /&gt;&lt;br /&gt;500 men are arranged in an array of 10 rows and 50 columns according to their heights.&lt;br /&gt;&lt;br /&gt;Tallest among each row of all are asked to come out. And the shortest among them is A.&lt;br /&gt;&lt;br /&gt;Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B.&lt;br /&gt;&lt;br /&gt;Now who is taller A or B ?&lt;br /&gt;A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither the person nor the cashier noticed that.&lt;br /&gt;&lt;br /&gt;After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.&lt;br /&gt;&lt;br /&gt;Find X and Y. (1 Rupee = 100 Paise)&lt;br /&gt;&lt;br /&gt;As given, the person wanted to withdraw 100X + Y paise.&lt;br /&gt;&lt;br /&gt;But he got 100Y + X paise.&lt;br /&gt;&lt;br /&gt;After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is&lt;br /&gt;&lt;br /&gt;        2 * (100X + Y) = 100Y + X - 20&lt;br /&gt;&lt;br /&gt;        200X + 2Y = 100Y +X - 20&lt;br /&gt;&lt;br /&gt;        199X - 98Y = -20&lt;br /&gt;&lt;br /&gt;        98Y - 199X = 20&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X &amp;amp; Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1&lt;br /&gt;&lt;br /&gt;Case I : Y=2X&lt;br /&gt;Solving two equations simultaneously&lt;br /&gt;98Y - 199X = 20&lt;br /&gt;Y - 2X = 0&lt;br /&gt;We get X = - 20/3 &amp;amp; Y = - 40/2&lt;br /&gt;&lt;br /&gt;Case II : Y=2X+1&lt;br /&gt;Solving two equations simultaneously&lt;br /&gt;98Y - 199X = 20&lt;br /&gt;Y - 2X = 1&lt;br /&gt;We get X = 26 &amp;amp; Y = 53&lt;br /&gt;&lt;br /&gt;Now, its obvious that he wanted to withdraw Rs. 26.53&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Submit&lt;br /&gt;Answer         Users&lt;br /&gt;Answer (2)         BrainVista&lt;br /&gt;Answer         Puzzle&lt;br /&gt;&lt;br /&gt;At the Party:&lt;br /&gt;1.    There were 9 men and children.&lt;br /&gt;2.    There were 2 more women than children.&lt;br /&gt;3.    The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.&lt;br /&gt;&lt;br /&gt;Also, of the three groups - men, women and children - at the party:&lt;br /&gt;4.    There were 4 of one group.&lt;br /&gt;5.    There were 6 of one group.&lt;br /&gt;6.    There were 8 of one group.&lt;br /&gt;Exactly one of the above 6 statements is false.&lt;br /&gt;&lt;br /&gt;Can you tell which one is false? Also, how many men, women and children are there at the party?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Statement (4) is false. There are 3 men, 8 women and 6 children.&lt;br /&gt;&lt;br /&gt;Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly one statement is false.&lt;br /&gt;&lt;br /&gt;So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true.&lt;br /&gt;&lt;br /&gt;From (1) and (2), there are 11 men and women. Then from (3), there are 2 possible cases - either there are 8 men and 3 women or there are 3 men and 8 women.&lt;br /&gt;&lt;br /&gt;If there are 8 men and 3 women, then there is 1 child. Then Statements (4) and (5) both are false, which is not possible.&lt;br /&gt;&lt;br /&gt;Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.&lt;br /&gt;    Brain Teaser No : 00242&lt;br /&gt;&lt;br /&gt;There is a shortage of tubelights, bulbs and fans in a village - Kharghar. It is found that&lt;br /&gt;    All houses do not have either tubelight or bulb or fan.&lt;br /&gt;    exactly 19% of houses do not have just one of these.&lt;br /&gt;    atleast 67% of houses do not have tubelights.&lt;br /&gt;    atleast 83% of houses do not have bulbs.&lt;br /&gt;    atleast 73% of houses do not have fans.&lt;br /&gt;What percentage of houses do not have tubelight, bulb and fan?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;42% houses do not have tubelight, bulb and fan.&lt;br /&gt;&lt;br /&gt;Let's assume that there are 100 houses. Hence, there should be total 300 items i.e. 100 tubelights, 100 bulbs and 100 fans.&lt;br /&gt;&lt;br /&gt;From the given data, we know that there is shortage of atleast (67+83+73) 223 items in every 100 houses.&lt;br /&gt;&lt;br /&gt;Also, exactly 19 houses do not have just one item. It means that remaining 81 houses should account for the shortage of remaining (223-19) 204 items. If those remaining 81 houses do not have 2 items each, there would be a shortage of 162 items. But total of 204 items are short. Hence, atleast (204-162) 42 houses do not have all 3 items - tubelight, bulb and fan.&lt;br /&gt;&lt;br /&gt;Thus, 42% houses do not have tubelight, bulb and fan.&lt;br /&gt;What is the remainder left after dividing 1! + 2! + 3! + … + 100! By 7?&lt;br /&gt;&lt;br /&gt;Think carefully !!!&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A tricky one.&lt;br /&gt;&lt;br /&gt;7! onwards all terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms i.e. remainder left after dividing 7! + 8! + 9! + ... + 100! is 0.&lt;br /&gt;&lt;br /&gt;The only part to be consider is&lt;br /&gt;= 1! + 2! + 3! + 4! + 5! + 6!&lt;br /&gt;= 1 + 2 + 6 + 24 + 120 + 720&lt;br /&gt;= 873&lt;br /&gt;&lt;br /&gt;The remainder left after dividing 873 by 7 is 5&lt;br /&gt;&lt;br /&gt;Hence, the remainder is 5.&lt;br /&gt;&lt;br /&gt;Imagine that you have 26 constants, labelled A through Z. Each constant is assigned a value in the following way: A = 1; the rest of the values equal their position in the alphabet (B corresponds to the second position so it equals 2, C = 3, etc.) raised to the power of the preceeding constant value. So, B = 2 ^ (A's value), or B = 2^1 = 2. C = 3^2 = 9. D = 4^9, etc.&lt;br /&gt;&lt;br /&gt;Find the exact numerical value to the following equation: (X - A) * (X - B) * (X - C) * ... * (X - Y) * (X - Z)&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;(X - A) * (X - B) * (X - C) * ... * (X - Y) * (X - Z) equals 0 since (X - X) is zero&lt;br /&gt;If three babies are born every second of the day, then how many babies will be born in the year 2001?&lt;br /&gt;SuAnswer&lt;br /&gt;&lt;br /&gt;9,46,08,000 babies&lt;br /&gt;&lt;br /&gt;The total seconds in year 2001&lt;br /&gt;= 365 days/year * 24 hours/day * 60 minutes/hours * 60 seconds/minute&lt;br /&gt;= 365 * 24 * 60 * 60 seconds&lt;br /&gt;= 3,15,36,000 seconds&lt;br /&gt;&lt;br /&gt;Thus, there are 3,15,36,000 seconds in the year 2001. Also, three babies born are every second. Hence, total babies born&lt;br /&gt;= 3 * 3,15,36,000 seconds&lt;br /&gt;= 9,46,08,000bmitted&lt;br /&gt;&lt;br /&gt;Replace the letters with the correct numbers.&lt;br /&gt;     T W O&lt;br /&gt;&lt;br /&gt; X   T W O&lt;br /&gt;&lt;br /&gt; ---------&lt;br /&gt;&lt;br /&gt; T H R E E&lt;br /&gt;&lt;br /&gt;Submitted by : Timmy Chan&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;T=1, W=3, O=8, H=9, R=2, E=4&lt;br /&gt;      1 3 8&lt;br /&gt;&lt;br /&gt; x    1 3 8&lt;br /&gt;&lt;br /&gt;------------&lt;br /&gt;&lt;br /&gt;  1 9 0 4 4&lt;br /&gt;You can reduce the number of trials. T must be 1 as there is multiplication of T with T in hundred's position. Also, O can not be 0 or 1. Now, you have to find three digit number whose square satisfies above conditions and square of that has same last two digits. Hence, it must be between 102 and 139.&lt;br /&gt;    Brain Teaser No : 00052&lt;br /&gt;&lt;br /&gt;Four words add up to a fifth word numerically:&lt;br /&gt;&lt;br /&gt;     mars&lt;br /&gt;&lt;br /&gt;    venus&lt;br /&gt;&lt;br /&gt;   uranus&lt;br /&gt;&lt;br /&gt;   saturn&lt;br /&gt;&lt;br /&gt; -------- +&lt;br /&gt;&lt;br /&gt;  neptune&lt;br /&gt;&lt;br /&gt;Each of the ten letters (m, a, r, s, v, e, n, u, t, and p) represent a unique number from the range 0 .. 9.&lt;br /&gt;&lt;br /&gt;Furthermore, numbers 1 and 6 are being used most frequently.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The easiest way to solve this problem is by writing a computer program that systematically tries all possible mappings from the numbers onto the letters. This will give you only one solution which meets the condition that numbers 1 and 6 are most frequently used.&lt;br /&gt;&lt;br /&gt;     mars         m = 4&lt;br /&gt;&lt;br /&gt;    venus         a = 5&lt;br /&gt;&lt;br /&gt;   uranus         r = 9&lt;br /&gt;&lt;br /&gt;   saturn         s = 3&lt;br /&gt;&lt;br /&gt; -------- +       v = 2            4593&lt;br /&gt;&lt;br /&gt;  neptune         e = 0           20163&lt;br /&gt;&lt;br /&gt;                  n = 1          695163&lt;br /&gt;&lt;br /&gt;                  u = 6          358691&lt;br /&gt;&lt;br /&gt;                  t = 8        -------- +&lt;br /&gt;&lt;br /&gt;                  p = 7         1078610   &lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;There are 4 army men. They have been captured by a rebel group and have been held at ransom. An army intelligent officer orders them to be burried deep in dirt up to their necks. The format of their burrial are as shown in the figure.&lt;br /&gt;&lt;br /&gt;Conditions&lt;br /&gt;    They each have hats on their heads. either black(b) or white (w) look at diagram above. There are total 2 white hats and 2 black hats.&lt;br /&gt;    They only look in front of them not behind. They are not allowed to communicate by talking.&lt;br /&gt;    Between army man 1 and 2, there is a wall.&lt;br /&gt;    Captive man 4 can see the colour of hats on 2 and 3&lt;br /&gt;    3 can only see 2's hat&lt;br /&gt;    2 can only see a wall and 1 can see a wall too, but is on the other side&lt;br /&gt;The officer speaks up, "If one of you can correctly tell me the colour of your hat, you will all go scott free back to your contries. If you are wrong, you will all be killed.&lt;br /&gt;&lt;br /&gt;How can one of them be certain about the hat they are wearing and not risk the lives of their fellow souldiers by taking a 50/50 guess!&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Either soldier 3 or soldier 4 can save the life as soldier 1 and soldier 2 can not see colour of any hat, even not their own.. In our case soldier 3 will tell the colour of his hat.&lt;br /&gt;&lt;br /&gt;Soldier 4 can see the hat on soldier 2 and soldier 3. If both are white, then he can be sure about colour of his hat which will be black and vice-versa. But if one of them is white and one is black, then soldier 4 can not say anything as he can have either of them. So he will keep mum.&lt;br /&gt;&lt;br /&gt;If soldier 4 won't say anyhing for a while, then soldier 3 will know that soldier 4 is not in position to tell the colour of hat on his hat. It means that colour of soldier 3's hat is opposite of colour of soldier 2's hat. So soldier 3 can tell correctly the colour of hat on his head which is Black.&lt;br /&gt;&lt;br /&gt;Here, we are assuming that all the soldiers are intelligent enough. Also, this solution will work for any combination of 2 Black hats and 2 White hats.&lt;br /&gt;One side of the bottom layer of a triangular pyramid has 12 balls. How many are there in the whole pyramid?&lt;br /&gt;&lt;br /&gt;Note that the pyramid is equilateral and solid.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are total 364 balls.&lt;br /&gt;&lt;br /&gt;As there are 12 balls along one side, it means that there are 12 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55, 66 and 78 balls in the remaining layers.&lt;br /&gt;&lt;br /&gt;Hence, the total number of balls are&lt;br /&gt;= 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78&lt;br /&gt;= 364 balls&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;A blindfolded man is asked to sit in the front of a carrom board. The holes of the board are shut with lids in random order, i.e. any number of all the four holes can be shut or open.&lt;br /&gt;&lt;br /&gt;Now the man is supposed to touch any two holes at a time and can do the following.&lt;br /&gt;    Open the closed hole.&lt;br /&gt;    Close the open hole.&lt;br /&gt;    Let the hole be as it is.&lt;br /&gt;After he has done it, the carrom board is rotated and again brought to some position. The man is again not aware of what are the holes which are open or closed.&lt;br /&gt;&lt;br /&gt;How many minimum number of turns does the blindfolded man require to either open all the holes or close all the holes?&lt;br /&gt;&lt;br /&gt;Note that whenever all the holes are either open or close, there will be an alarm so that the blindfolded man will know that he has won.&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The blindfolded man requires 5 turns.&lt;br /&gt;1.    Open two adjacent holes.&lt;br /&gt;2.    Open two diagonal holes. Now atleast 3 holes are open. If 4th hole is also open, then you are done. If not, the 4th hole is close.&lt;br /&gt;3.    Check two diagonal holes.&lt;br /&gt;o    If one is close, open it and all the holes are open.&lt;br /&gt;o    If both are close, open any one hole. Now, two holes are open and two are close. The diagonal holes are in the opposite status i.e. in both the diagonals, one hole is open and one is close.&lt;br /&gt;&lt;br /&gt;4.    Check any two adjacent holes.&lt;br /&gt;o    If both are open, close both of them. Now, all holes are close.&lt;br /&gt;o    If both are close, open both of them. Now, all holes are open.&lt;br /&gt;o    If one is open and one is close, invert them i.e. close the open hole and open the close hole. Now, the diagonal holes are in the same status i.e. two holes in one diagonal are open and in other are close.&lt;br /&gt;&lt;br /&gt;5.    Check any two diagonal holes.&lt;br /&gt;o    If both are open, close both of them. Now, all holes are close.&lt;br /&gt;o    If both are close, open both of them. Now, all holes are open.&lt;br /&gt;&lt;br /&gt;In the middle of the confounded desert, there is the lost city of "Ash". To reach it, I will have to travel overland by foot from the coast. On a trek like this, each person can only carry enough rations for five days and the farthest we can travel in one day is 30 miles. Also, the city is 120 miles from the starting point.&lt;br /&gt;&lt;br /&gt;What I am trying to figure out is the fewest number of persons, including myself, that I will need in our Group so that I can reach the city, stay overnight, and then return to the coast without running out of supplies.&lt;br /&gt;&lt;br /&gt;How many persons (including myself) will I need to accomplish this mission?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Total 4 persons (including you) required.&lt;br /&gt;&lt;br /&gt;It is given that each person can only carry enough rations for five days. And there are 4 persons. Hence, total of 20 days rations is available.&lt;br /&gt;1.    First Day : 4 days of rations are used up. One person goes back using one day of rations for the return trip. The rations remaining for the further trek is for 15 days.&lt;br /&gt;2.    Second Day : The remaining three people use up 3 days of rations. One person goes back using 2 days of rations for the return trip. The rations remaining for the further trek is for 10 days.&lt;br /&gt;3.    Third Day : The remaining two people use up 2 days of rations. One person goes back using 3 days of rations for the return trip. The rations remaining for the further trek is for 5 days.&lt;br /&gt;4.    Fourth Day : The remaining person uses up one day of rations. He stays overnight. The next day he returns to the coast using 4 days of rations.&lt;br /&gt;&lt;br /&gt;Thus, total 4 persons, including you are required.&lt;br /&gt;At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with the hour hand?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;4:21:49.5&lt;br /&gt;&lt;br /&gt;Assume that X minutes after 4.00 PM minute hand exactly aligns with and hour hand.&lt;br /&gt;&lt;br /&gt;For every minute, minute hand travels 6 degrees.&lt;br /&gt;Hence, for X minutes it will travel 6 * X degrees.&lt;br /&gt;&lt;br /&gt;For every minute, hour hand travels 1/2 degrees.&lt;br /&gt;Hence, for X minutes it will travel X/2 degrees.&lt;br /&gt;&lt;br /&gt;At 4.00 PM, the angle between minute hand and hour hand is 120 degrees. Also, after X minutes, minute hand and hour hand are exactly aligned. So the angle with respect to 12 i.e. Vertical Plane will be same. Therefore,&lt;br /&gt;&lt;br /&gt;6 * X = 120 + X/2&lt;br /&gt;12 * X = 240 + X&lt;br /&gt;11 * X = 240&lt;br /&gt;X = 21.8182&lt;br /&gt;X = 21 minutes 49.5 seconds&lt;br /&gt;&lt;br /&gt;Hence, at 4:21:49.5 minute hand is exactly aligned with the hour hand.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Substitute digits for the letters to make the following Division true&lt;br /&gt;                   O U T&lt;br /&gt;&lt;br /&gt;           -------------&lt;br /&gt;&lt;br /&gt;   S T E M | D E M I S E&lt;br /&gt;&lt;br /&gt;           | D M O C&lt;br /&gt;&lt;br /&gt;           -------------&lt;br /&gt;&lt;br /&gt;               T U I S&lt;br /&gt;&lt;br /&gt;               S T E M&lt;br /&gt;&lt;br /&gt;              ----------&lt;br /&gt;&lt;br /&gt;                 Z Z Z E&lt;br /&gt;&lt;br /&gt;                 Z U M M&lt;br /&gt;&lt;br /&gt;                --------&lt;br /&gt;&lt;br /&gt;                   I S T&lt;br /&gt;Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.&lt;br /&gt;Submitted by : Calon&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;C=0, U=1, S=2, T=3, O=4, M=5, I=6, Z=7, E=8, D=9&lt;br /&gt;&lt;br /&gt;It is obvious that U=1 (as U*STEM=STEM) and C=0 (as I-C=I).&lt;br /&gt;&lt;br /&gt;S*O is a single digit and also S*T is a single digit. Hence, their values (O, S, T) must be 2, 3 or 4 (as they can not be 0 or 1 or greater than 4).&lt;br /&gt;&lt;br /&gt;Consider, STEM*O=DMOC, where C=0. It means that M must be 5. Now, its simple. O=4, S=2, T=3, E=8, Z=7, I=6 and D=9.&lt;br /&gt;                   O U T                        4 1 3&lt;br /&gt;&lt;br /&gt;           -------------                -------------&lt;br /&gt;&lt;br /&gt;   S T E M | D E M I S E        2 3 8 5 | 9 8 5 6 2 8&lt;br /&gt;&lt;br /&gt;           | D M O C                    | 9 5 4 0&lt;br /&gt;&lt;br /&gt;           -------------                -------------&lt;br /&gt;&lt;br /&gt;               T U I S                      3 1 6 2&lt;br /&gt;&lt;br /&gt;               S T E M                      2 3 8 5&lt;br /&gt;&lt;br /&gt;              ----------                   ----------&lt;br /&gt;&lt;br /&gt;                 Z Z Z E                      7 7 7 8&lt;br /&gt;&lt;br /&gt;                 Z U M M                      7 1 5 5&lt;br /&gt;&lt;br /&gt;                --------                     --------&lt;br /&gt;&lt;br /&gt;                   I S T                        6 2 3&lt;br /&gt;Also, when arranged from 0 to 9, it spells CUSTOMIZED.&lt;br /&gt;    Brain Teaser No : 00015&lt;br /&gt;&lt;br /&gt;In the town called Alibaug, the following facts are true:&lt;br /&gt;    No two inhabitants have exactly the same number of hairs.&lt;br /&gt;    No inhabitants has exactly 2025 hairs.&lt;br /&gt;    There are more inhabitants than there are hairs on the head of any one inhabitants.&lt;br /&gt;What is the largest possible number of the inhabitants of Alibaug?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;2025&lt;br /&gt;&lt;br /&gt;It is given that no inhabitants have exactly 2025 hairs. Hence there are 2025 inhabitants with 0 to 2024 hairs in the head.&lt;br /&gt;&lt;br /&gt;Suppose there are more than 2025 inhabitants. But these will violate the condition that "There are more inhabitants than there are hairs on the head of any one inhabitants." As for any number more than 2025, there will be same number of inhabitants as the maximum number of hairs on the head of any inhabitant.&lt;br /&gt;There are four groups of Mangoes, Apples and Bananas as follows:&lt;br /&gt;Group I : 1 Mango, 1 Apples and 1 Banana&lt;br /&gt;Group II : 1 Mango, 5 Apples and 7 Bananas&lt;br /&gt;Group III : 1 Mango, 7 Apples and 10 Bananas&lt;br /&gt;Group IV : 9 Mango, 23 Apples and 30 Bananas&lt;br /&gt;&lt;br /&gt;Group II costs Rs 300 and Group III costs Rs 390.&lt;br /&gt;&lt;br /&gt;Can you tell how much does Group I and Group IV cost?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Group I costs Rs 120 and Group IV costs Rs 1710&lt;br /&gt;&lt;br /&gt;Assume that the values of one mango, one apple and one banana are M, A and B respectively.&lt;br /&gt;&lt;br /&gt;From Group II : M + 5A + 7B = 300&lt;br /&gt;From Group III : M + 7A + 10B = 390&lt;br /&gt;&lt;br /&gt;Subtracting above to equations : 2A + 3B = 90&lt;br /&gt;&lt;br /&gt;For Group I :&lt;br /&gt;= M + A + B&lt;br /&gt;= (M + 5A + 7B) - (4A + 6B)&lt;br /&gt;= (M + 5A + 7B) - 2(2A + 3B)&lt;br /&gt;= 300 - 2(90)&lt;br /&gt;= 300 - 180&lt;br /&gt;= 120&lt;br /&gt;&lt;br /&gt;Similarly, for Group IV :&lt;br /&gt;= 9M + 23A + 30B&lt;br /&gt;= 9(M + 5A + 7B) - (22A + 33B)&lt;br /&gt;= 9(M + 5A + 7B) - 11(2A + 3B)&lt;br /&gt;= 9(300) - 11(90)&lt;br /&gt;= 2700 - 990&lt;br /&gt;= 1710&lt;br /&gt;&lt;br /&gt;Thus, Group I costs Rs 120 and Group IV costs Rs 1710.&lt;br /&gt;Tic-Tac-Toe is being played. One 'X' has been placed in one of the corners. No 'O' has been placed yet.&lt;br /&gt;&lt;br /&gt;Where does the player that is playing 'O' has to put his first 'O' so that 'X' doesn't win?&lt;br /&gt;&lt;br /&gt;Assume that both players are very intelligent. Explain your answer&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;"O" should be placed in the center.&lt;br /&gt;&lt;br /&gt;Let's number the positions as:&lt;br /&gt;                  1 | 2 | 3&lt;br /&gt;&lt;br /&gt;                  ---------&lt;br /&gt;&lt;br /&gt;                  4 | 5 | 6&lt;br /&gt;&lt;br /&gt;                  ---------&lt;br /&gt;&lt;br /&gt;                  7 | 8 | 9&lt;br /&gt;It is given that "X" is placed in one of the corner position. Let's assume that its at position 1.&lt;br /&gt;&lt;br /&gt;Now, let's take each position one by one.&lt;br /&gt;    If "O" is placed in position 2, "X" can always win by choosing position 4, 5 or 7.&lt;br /&gt;    If "O" is placed in position 3, "X" can always win by choosing position 4, 7 or 9.&lt;br /&gt;    If "O" is placed in position 4, "X" can always win by choosing position 2, 3 or 5.&lt;br /&gt;    If "O" is placed in position 6, "X" can always win by choosing position 3, 5 or 7.&lt;br /&gt;    If "O" is placed in position 7, "X" can always win by choosing position 2, 3 or 9.&lt;br /&gt;    If "O" is placed in position 8, "X" can always win by choosing position 3, 5 or 7.&lt;br /&gt;    If "O" is placed in position 9, "X" can always win by choosing position 3, or 7.&lt;br /&gt;If "O" is placed in position 5 i.e. center position, "X" can't win unless "O" does something foolish ;))&lt;br /&gt;&lt;br /&gt;Hence, "O" should be placed in the center.&lt;br /&gt;&lt;br /&gt;Amit, Bhavin, Himanshu and Rakesh are sitting around a table.&lt;br /&gt;    The Electonics Engineer is sitting to the left of the Mechanical Engineer.&lt;br /&gt;    Amit is sitting opposite to Computer Engineer.&lt;br /&gt;    Himanshu likes to play Computer Games.&lt;br /&gt;    Bhavin is sitting to the right of the Chemical Engineer.&lt;br /&gt;Can you figure out everyone's profession?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Amit is the Mechanical Engineer. Bhavin is the Computer Engineer. Himanshu and Rakesh are either Chemical Engineer or Elecronics Engineer.&lt;br /&gt;&lt;br /&gt;Amit and Bhavin are sitting opposite to each other. Whereas Chemical Engineer and Elecronics Engineer are sitting opposite to each other.&lt;br /&gt;&lt;br /&gt;We cannot find out who is Chemical Engineer and Elecronics Engineer as data provided is not sufficient&lt;br /&gt;&lt;br /&gt;Five friends with surname Batliwala, Pocketwala, Talawala, Chunawala and Natakwala have their first name and middle name as follow.&lt;br /&gt;1.    Four of them have a first and middle name of Paresh.&lt;br /&gt;2.    Three of them have a first and middle name of Kamlesh.&lt;br /&gt;3.    Two of them have a first and middle name of Naresh.&lt;br /&gt;4.    One of them have a first and middle name of Elesh.&lt;br /&gt;5.    Pocketwala and Talawala, either both are named Kamlesh or neither is named Kamlesh.&lt;br /&gt;6.    Either Batliwala and Pocketwala both are named Naresh or Talawala and Chunawala both are named Naresh.&lt;br /&gt;7.    Chunawala and Natakwala are not both named Paresh.&lt;br /&gt;Who is named Elesh?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Pocketwala is named Elesh.&lt;br /&gt;&lt;br /&gt;From (1) and (7), it is clear that Batliwala, Pocketwala and Talawala are named Paresh.&lt;br /&gt;&lt;br /&gt;From (6) and (5), if Pocketwala or Talawala both are named Kamlesh, then either of them will have three names i.e. Paresh, Kamlesh and Naresh. Hence, Pocketwala and Talawala both are not named Kamlesh. It means that Batliwala, Chunawala and Natakwala are named Kamlesh.&lt;br /&gt;&lt;br /&gt;Now it is clear that Talawala and Chunawala are named Naresh. Also, Pocketwala is named Elesh.&lt;br /&gt;&lt;br /&gt;Mr. Wagle goes to work by a bus. One day he falls asleep when the bus still has twice as far to go as it has already gone.&lt;br /&gt;&lt;br /&gt;Halfway through the trip he wakes up as the bus bounces over some bad potholes. When he finally falls asleep again, the bus still has half the distance to go that it has already travelled. Fortunately, Mr. Wagle wakes up at the end of his trip.&lt;br /&gt;&lt;br /&gt;What portion of the total trip did Mr. Wagle sleep?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Mr. wagle slept through half his trip.&lt;br /&gt;&lt;br /&gt;Let's draw a timeline. Picture the bus route on a line showen below:&lt;br /&gt;&lt;br /&gt;    ---------------- ________ -------- ________________&lt;br /&gt;&lt;br /&gt;  Start            1/3      1/2      2/3             End&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt; ----- shows time for which Mr. Wagle was not sleeping&lt;br /&gt;&lt;br /&gt; _____ shows time for which Mr. Wagle was sleeping&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;When Mr. Wagle fell asleep the first time, the bus sill had twice as far to go as it had already gone, that marks the first third of his trip.&lt;br /&gt;&lt;br /&gt;He wake up halfway through the trip i.e slept from 1/3 mark to the 1/2 mark. He fell sleep again when the bus still had half the distance to go that it had already traveled i.e 2/3 mark.&lt;br /&gt;&lt;br /&gt;Adding up, all sleeping times,&lt;br /&gt;= (1/2 - 1/3) + (1 - 2/3)&lt;br /&gt;= 1/6 + 1/3&lt;br /&gt;= 1/2&lt;br /&gt;&lt;br /&gt;Hence, Mr. wagle slept through half his trip.&lt;br /&gt;    Brain Teaser No : 00068&lt;br /&gt;&lt;br /&gt;In your sock drawer, you have a ratio of 5 pairs of blue socks, 4 pairs of brown socks, and 6 pairs of black socks.&lt;br /&gt;&lt;br /&gt;In complete darkness, how many socks would you need to pull out to get a matching pair of the same color?&lt;br /&gt;4 If you don't agree, try it yourself!&lt;br /&gt;You have a bucket of jelly beans. Some are red, some are blue, and some green. With your eyes closed, pick out 2 of a like color.&lt;br /&gt;&lt;br /&gt;How many do you have to grab to be sure you have 2 of the same?&lt;br /&gt;&lt;br /&gt;You have a bucket of jelly beans. Some are red, some are blue, and some green. With your eyes closed, pick out 2 of a like color.&lt;br /&gt;&lt;br /&gt;How many do you have to grab to be sure you have 2 of the same?&lt;br /&gt;If you select 4 Jelly beans you are guarenteed that you will have 2 that are the same color.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;There are 70 employees working with BrainVista of which 30 are females. Also,&lt;br /&gt;    30 employees are married&lt;br /&gt;    24 employees are above 25 years of age&lt;br /&gt;    19 married employees are above 25 years, of which 7 are males&lt;br /&gt;    12 males are above 25 years of age&lt;br /&gt;    15 males are married.&lt;br /&gt;How many unmarried females are there and how many of them are above 25?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;15 unmarried females &amp;amp; none are above 25 years of age.&lt;br /&gt;&lt;br /&gt;Simply put all given information into the table structure and you will get the answer.&lt;br /&gt;     Married    Unmarried&lt;br /&gt;    Below 25    Above 25    Below 25    Above 25&lt;br /&gt;Female    3    12    15    0&lt;br /&gt;Male    8    7    20    5&lt;br /&gt;&lt;br /&gt;There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than the second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11.&lt;br /&gt;&lt;br /&gt;Find the number.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;65292&lt;br /&gt;&lt;br /&gt;As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)&lt;br /&gt;&lt;br /&gt;It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9)&lt;br /&gt;&lt;br /&gt;Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.&lt;br /&gt;My friend collects antique stamps. She purchased two, but found that she needed to raise money urgently. So she sold them for Rs. 8000 each. On one she made 20% and on the other she lost 20%.&lt;br /&gt;&lt;br /&gt;How much did she gain or lose in the entire transaction?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;She lost Rs 666.67&lt;br /&gt;&lt;br /&gt;Consider the first stamp. She mades 20% on it after selling it for Rs 8000.&lt;br /&gt;&lt;br /&gt;So the original price of first stamp is&lt;br /&gt;= (8000 * 100) / 80&lt;br /&gt;= Rs 6666.67&lt;br /&gt;&lt;br /&gt;Similarly, consider second stamp. She lost 20% on it after selling it for Rs 8000&lt;br /&gt;&lt;br /&gt;So the original price of second stamp is&lt;br /&gt;= (8000 * 100) / 80&lt;br /&gt;= Rs 10000&lt;br /&gt;&lt;br /&gt;Total buying price of two stamps&lt;br /&gt;= Rs 6666.67 + Rs 10000&lt;br /&gt;= Rs 16666.67&lt;br /&gt;&lt;br /&gt;Total selling price of two stamps&lt;br /&gt;= Rs 8000 + Rs 8000&lt;br /&gt;= Rs 16000&lt;br /&gt;&lt;br /&gt;Hence, she lost Rs 666.67&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Assume for a moment that the earth is a perfectly uniform sphere of radius 6400 km. Suppose a thread equal to the length of the circumference of the earth was placed along the equator, and drawn to a tight fit.&lt;br /&gt;&lt;br /&gt;Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions.&lt;br /&gt;&lt;br /&gt;By how many cm. will the thread be separated from the earth's surface?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The cicumference of the earth is&lt;br /&gt;= 2 * PI * r&lt;br /&gt;= 2 * PI * 6400 km&lt;br /&gt;= 2 * PI * 6400 * 1000 m&lt;br /&gt;= 2 * PI * 6400 * 1000 * 100 cm&lt;br /&gt;= 1280000000 * PI cm&lt;br /&gt;&lt;br /&gt;where r = radius of the earth, PI = 3.141592654&lt;br /&gt;&lt;br /&gt;Hence, the length of the thread is = 1280000000 * PI cm&lt;br /&gt;&lt;br /&gt;Now length of the thread is increasd by 12 cm. So the new length is = (1280000000 * PI) + 12 cm&lt;br /&gt;&lt;br /&gt;This thread will make one concentric circle with the earth which is slightly away from the earth. The circumfernce of that circle is nothing but (1280000000 * PI) + 12 cm&lt;br /&gt;&lt;br /&gt;Assume that radius of the outer circle is R cm&lt;br /&gt;Therefore,&lt;br /&gt;2 * PI * R = (1280000000 * PI) + 12 cm&lt;br /&gt;&lt;br /&gt;Solving above equation, R = 640000001.908 cm&lt;br /&gt;Radius of the earth is r = 640000000 cm&lt;br /&gt;&lt;br /&gt;Hence, the thread will be separatedfrom the earth by&lt;br /&gt;= R - r cm&lt;br /&gt;= 640000001.908 - 640000000&lt;br /&gt;= 1.908 cm&lt;br /&gt;Scientist decided to do a study on the population growth of rabbits. Inside a controlled environment, 1000 rabbits were placed.&lt;br /&gt;&lt;br /&gt;Six months later, there were 1000Z rabbits. At the beginning of the 3rd year, there were roughly 2828Z rabbits, which was 4 times what the scientists placed in there at the beginning of the 1st year.&lt;br /&gt;&lt;br /&gt;If Z is a positive variable, how many rabbits would be there at the beginning of the 11th year?&lt;br /&gt;SubmiAnswer&lt;br /&gt;&lt;br /&gt;At the beginning of the 11th year, there would be 1,024,000 rabbits.&lt;br /&gt;&lt;br /&gt;At the beginning, there were 1000 rabbits. Also, there were 4000 rabbits at the beginning of third year which is equal to 2828Z. Thus, Z = 4000/2828 i.e. 1.414 (the square root of 2)&lt;br /&gt;&lt;br /&gt;Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further simplified as 1000*Z*Z*Z*Z&lt;br /&gt;&lt;br /&gt;Also, it is given that at the end of 6 months, there were 1000Z rabbits.&lt;br /&gt;&lt;br /&gt;It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(Z^(2N)) i.e. 1000*(2^N)&lt;br /&gt;&lt;br /&gt;Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 rabbits.&lt;br /&gt;tted&lt;br /&gt;A class of 100 students. 24 of them are girls and 32 are not. Which base am I using?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Let the base be X.&lt;br /&gt;&lt;br /&gt;Therefore&lt;br /&gt;(X*X + X*0 + 0) = (2*X +4) + (3*X + 2)&lt;br /&gt;X*X = 5*X + 6&lt;br /&gt;X*X - 5*X -6 = 0&lt;br /&gt;(X-6)(X+1) = 0&lt;br /&gt;&lt;br /&gt;Therefore base is 6&lt;br /&gt;&lt;br /&gt;A man is stranded on a desert island. All he has to drink is a 20oz bottle of sprite.&lt;br /&gt;&lt;br /&gt;To conserve his drink he decides that on the first day he will drink one oz and the refill the bottle back up with water. On the 2nd day he will drink 2oz and refill the bottle. On the 3rd day he will drink 3oz and so on...&lt;br /&gt;&lt;br /&gt;By the time all the sprite is gone, how much water has he drunk?&lt;br /&gt;SubmittAnswer&lt;br /&gt;&lt;br /&gt;The man drunk 190oz of water.&lt;br /&gt;&lt;br /&gt;It is given that the man has 20oz bottle of sprite. Also, he will drink 1oz on the first day and refill the bottle with water, will drink 2oz on the second day and refill the bottle, will drink 3oz on the third day and refill the bottle, and so on till 20th day. Thus at the end of 20 days, he must have drunk (1 + 2 + 3 + 4 + ..... +18 + 19 + 20) = 210oz of liquid.&lt;br /&gt;&lt;br /&gt;Out of that 210oz, 20oz is the sprite which he had initially. Hence, he must have drunk 190oz of water.ed&lt;br /&gt;You have four 9's and you may use any of the (+, -, /, *) as many times as you like. I want to see a mathematical expression which uses the four 9's to = 100&lt;br /&gt;&lt;br /&gt;How many such expressions can you make?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 5 such expressions.&lt;br /&gt;&lt;br /&gt;99 + (9/9) = 100&lt;br /&gt;&lt;br /&gt;(99/.99) = 100&lt;br /&gt;&lt;br /&gt;(9/.9) * (9/.9) = 100&lt;br /&gt;&lt;br /&gt;((9*9) + 9)/.9 = 100&lt;br /&gt;&lt;br /&gt;(99-9)/.9 = 100&lt;br /&gt;Two planes take off at the same exact moment. They are flying across the Atlantic. One leaves New York and is flying to Paris at 500 miles per hour. The other leaves Paris and is flying to New York at only 450 miles per hour ( because of a strong head wind ).&lt;br /&gt;&lt;br /&gt;Which one will be closer to Paris when they meet?&lt;br /&gt;They will both be the same distance from Paris when they meet!!!&lt;br /&gt;&lt;br /&gt;12 members were present at a board meeting. Each member shook hands with all of the other members before &amp;amp; after the meeting.&lt;br /&gt;&lt;br /&gt;How many hand shakes were there?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;132&lt;br /&gt;&lt;br /&gt;Think of it this way: the first person shakes hands with 11 people, the second person also shakes hands with 11 people, but you only count 10, because the hand shake with the first person was already counted. Then add 9 for the third person, 8 for the fourth, &amp;amp; so on.&lt;br /&gt;&lt;br /&gt;66 hand shakes took place before &amp;amp; 66 after the meeting, for a total of 132.&lt;br /&gt;Arrange five planets such that 4 of them add up to 5th planet numerically. Each of the letters of the planet should represent a unique number from the range 0 - 9. You have to use all ten digits.&lt;br /&gt;&lt;br /&gt;There is an amazing mathematical relationship exists among the names of the planet.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The tought process is initially to find planets such that the total number of alphabets in them is 10.&lt;br /&gt;&lt;br /&gt;The only possible combination of planets is Saturn, Uranus, Venus, Mars and Neptune because for other combinations there will be more than 10 alphabets. Among these five, Neptune is the lenghtiest, so it must be the sum of the other four.&lt;br /&gt;&lt;br /&gt;        S A T U R N&lt;br /&gt;&lt;br /&gt;        U R A N U S&lt;br /&gt;&lt;br /&gt;          V E N U S&lt;br /&gt;&lt;br /&gt;     +      M A R S&lt;br /&gt;&lt;br /&gt;      --------------&lt;br /&gt;&lt;br /&gt;      N E P T U N E&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Now the only possible value for N is 1. By finding the value for S, we can reach the result:&lt;br /&gt;&lt;br /&gt;        3 5 8 6 9 1&lt;br /&gt;&lt;br /&gt;        6 9 5 1 6 3&lt;br /&gt;&lt;br /&gt;          2 0 1 6 3&lt;br /&gt;&lt;br /&gt;     +      4 5 9 3&lt;br /&gt;&lt;br /&gt;      --------------&lt;br /&gt;&lt;br /&gt;      1 0 7 8 6 1 0&lt;br /&gt;&lt;br /&gt;You have 14 apples. Your Friend Marge takes away 3 and gives you 2. You drop 7 but pick up 4. Bret takes 4 and gives 5. You take one from Marge and give it to Bret in exchange for 3 more. You give those 3 to Marge and she gives you an apple and an orange. Frank comes and takes the apple Marge gave you and gives you a pear. You give the pear to Bret in exchange for an apple. Frank then takes an apple from Marge, gives it to Bret for an orange, gives you the orange for an apple.&lt;br /&gt;&lt;br /&gt;How many pears do you have?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;None&lt;br /&gt;&lt;br /&gt;Frank gave you a pear in exchange of the apple which Marge gave you. And you gave that pear to Bret in exchange for an apple. All the others exchanges involved apples and/or organges.&lt;br /&gt;Four couples are going to the movie. Each row holds eight seats. Betty and Jim don't want to sit next to Alice and Tom. Alice and Tom don't want to sit next to Gertrude and Bill. On the otherhand, Sally and Bob don't want to sit next to Betty and Jim.&lt;br /&gt;&lt;br /&gt;How can the couples arrange themselves in a row so that they all sit where they would like?&lt;br /&gt;Submitted by : Tara Smith&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;From the given data, it can be inferred that:&lt;br /&gt;(Sally &amp;amp; Bob) NOT (Betty &amp;amp; Jim) NOT (Alice &amp;amp; Tom) NOT (Gertrude &amp;amp; Bill)&lt;br /&gt;&lt;br /&gt;(A) NOT (B) means A and B can not seat next to each other.&lt;br /&gt;&lt;br /&gt;Now, it is obvious that (Betty &amp;amp; Jim) and (Alice &amp;amp; Tom) will occupy the corner seats as both of them can have only one neighbour. Therefore,&lt;br /&gt;(Gertrude &amp;amp; Bill) will seat next to (Betty &amp;amp; Jim)&lt;br /&gt;(Sally &amp;amp; Bob) will seat next to (Gertrude &amp;amp; Bill)&lt;br /&gt;(Alice &amp;amp; Tom) will seat next to (Sally &amp;amp; Bob)&lt;br /&gt;&lt;br /&gt;Thus, there are two possible arrangements - a mirror images of each other.&lt;br /&gt;&lt;br /&gt;1. (Betty &amp;amp; Jim) - (Gertrude &amp;amp; Bill) - (Sally &amp;amp; Bob) - (Alice &amp;amp; Tom)&lt;br /&gt;2. (Alice &amp;amp; Tom) - (Sally &amp;amp; Bob) - (Gertrude &amp;amp; Bill) - (Betty &amp;amp; Jim)&lt;br /&gt;Substitute digits for the letters to make the following addition problem true.&lt;br /&gt;               W  H  O  S  E&lt;br /&gt;&lt;br /&gt;               T  E  E  T  H&lt;br /&gt;&lt;br /&gt;                     A  R  E&lt;br /&gt;&lt;br /&gt;          +             A  S&lt;br /&gt;&lt;br /&gt;          -------------------&lt;br /&gt;&lt;br /&gt;            S  W  O  R  D  S&lt;br /&gt;Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter H, no other letter can be 3 and all other H in the puzzle must be 3.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;It is obvious that S=1 and T=9.&lt;br /&gt;&lt;br /&gt;Also, (H + E) should be greater than 10 and hence, (E + H + E) must 20. Thus, there are 3 possible values for (E, H) pair: (6, 8) or (7, 6) or (8, 4). Use trial-n-error and everything will fit-in.&lt;br /&gt;&lt;br /&gt;       W  H  O  S  E                2  8  5  1  6&lt;br /&gt;&lt;br /&gt;       T  E  E  T  H                9  6  6  9  8&lt;br /&gt;&lt;br /&gt;             A  R  E                      4  7  6&lt;br /&gt;&lt;br /&gt;  +             A  S           +             4  1&lt;br /&gt;&lt;br /&gt;  -------------------          -------------------&lt;br /&gt;&lt;br /&gt;    S  W  O  R  D  S             1  2  5  7  3  1&lt;br /&gt;&lt;br /&gt;When Socrates was imprisoned for being a disturbing influence, he was held in high esteem by his guards. All four of them hoped that something would occur that would facilitate his escape. One evening, the guard who was on duty intentionally left the cell door open so that Socrates could leave for distant parts.&lt;br /&gt;&lt;br /&gt;Socrates did not attempt to escape, as it was his philosophy that if you accept society's rules, you must also accept it's punishments. However, the open door was considered by the authorities to be a serious matter. It was not clear which guard was on that evening. The four guards make the following statements in their defense:&lt;br /&gt;&lt;br /&gt;Aaron:&lt;br /&gt;A) I did not leave the door open.&lt;br /&gt;B) Clement was the one who did it.&lt;br /&gt;&lt;br /&gt;Bob:&lt;br /&gt;A) I was not the one who was on duty that evening.&lt;br /&gt;B) Aaron was on duty.&lt;br /&gt;&lt;br /&gt;Clement:&lt;br /&gt;A) Bob was the one on duty that evening.&lt;br /&gt;B) I hoped Socrates would escape.&lt;br /&gt;&lt;br /&gt;David:&lt;br /&gt;A) I did not leave the door open.&lt;br /&gt;B) I was not surprised that Socrates did not attempt to escape.&lt;br /&gt;&lt;br /&gt;Considering that, in total, three statements are true, and five statements are false, which guard is guiltyAnswer&lt;br /&gt;&lt;br /&gt;David is the guilty.&lt;br /&gt;&lt;br /&gt;Note that "All four of them hoped that something would occur that would facilitate his escape". It makes Clement's statement B True and David's statement B False.&lt;br /&gt;&lt;br /&gt;Now consider each of them as a guilty, one at a time.&lt;br /&gt;     Aaron    Bob    Clement    David    True&lt;br /&gt;Stmts&lt;br /&gt;     A    B    A    B    A    B    A    B   &lt;br /&gt;If Aaron is guilty    False    False    True    True    False    True    True    False    4&lt;br /&gt;If Bob is guilty    True    False    False    False    True    True    True    False    4&lt;br /&gt;If Clement is guilty    True    True    True    False    False    True    True    False    5&lt;br /&gt;If David is guilty    True    False    True    False    False    True    False    False    3&lt;br /&gt;&lt;br /&gt;Since in total, three statements are true and five statements are false. It is clear from the above table that David is?&lt;br /&gt;    Brain Teaser No : 00474&lt;br /&gt;&lt;br /&gt;Given any whole number take the sum of the digits, and the product of the digits, and multiply these together to get a new whole number.&lt;br /&gt;&lt;br /&gt;For example, starting with 6712, the sum of the digits is (6+7+1+2) = 16, and the product of the digits is (6*7*1*2) = 84. The answer in this case is then 84 x 16 = 1344.&lt;br /&gt;&lt;br /&gt;If we do this again starting from 1344, we get (1+3+4+4) * (1*3*4*4) = 576&lt;br /&gt;&lt;br /&gt;And yet again (5+7+6) * (5*7*6) = 3780&lt;br /&gt;&lt;br /&gt;At this stage we know what the next answer will be (without working it out) because, as one digit is 0, the product of the digits will be 0, and hence the answer will also be 0.&lt;br /&gt;&lt;br /&gt;Can you find any numbers to which when we apply the above mentioned rule repeatedly, we never end up at 0?&lt;br /&gt;    Brain Teaser No : 00474&lt;br /&gt;&lt;br /&gt;Given any whole number take the sum of the digits, and the product of the digits, and multiply these together to get a new whole number.&lt;br /&gt;&lt;br /&gt;For example, starting with 6712, the sum of the digits is (6+7+1+2) = 16, and the product of the digits is (6*7*1*2) = 84. The answer in this case is then 84 x 16 = 1344.&lt;br /&gt;&lt;br /&gt;If we do this again starting from 1344, we get (1+3+4+4) * (1*3*4*4) = 576&lt;br /&gt;&lt;br /&gt;And yet again (5+7+6) * (5*7*6) = 3780&lt;br /&gt;&lt;br /&gt;At this stage we know what the next answer will be (without working it out) because, as one digit is 0, the product of the digits will be 0, and hence the answer will also be 0.&lt;br /&gt;&lt;br /&gt;Can you find any numbers to which when we apply the above mentioned rule repeatedly, we never end up at 0?&lt;br /&gt;&lt;br /&gt;There were N stations on a railroad. After adding X stations 46 additional tickets have to be printed.&lt;br /&gt;&lt;br /&gt;Find N and X.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Let before adding X stations, total number of tickets&lt;br /&gt;t = N(N-1)&lt;br /&gt;&lt;br /&gt;After adding X stations total number of tickets are&lt;br /&gt;t + 46 = (N+X)(N+X-1)&lt;br /&gt;&lt;br /&gt;Subtracting 1st from 2nd&lt;br /&gt;46 = (N+X)(N+X-1) - N(N-1)&lt;br /&gt;46 = N2 + NX - N + NX + X2 - X - N2 + N&lt;br /&gt;46 = 2NX + X2 - X&lt;br /&gt;46 = (2N - 1)X + X2&lt;br /&gt;X2 + (2N - 1)X - 46 = 0&lt;br /&gt;&lt;br /&gt;Now there are only two possible factors of 46. They are (46,1) and (23,2)&lt;br /&gt;&lt;br /&gt;Case I: (46,1)&lt;br /&gt;2N - 1 = 45&lt;br /&gt;2N = 46&lt;br /&gt;N = 23&lt;br /&gt;And X = 1&lt;br /&gt;&lt;br /&gt;Case II: (23,2)&lt;br /&gt;2N - 1 = 21&lt;br /&gt;2N = 22&lt;br /&gt;N = 11&lt;br /&gt;And X = 2&lt;br /&gt;&lt;br /&gt;Hence, there are 2 possible answers.&lt;br /&gt;&lt;br /&gt;An emergency vehicle travels 10 miles at a speed of 50 miles per hour.&lt;br /&gt;&lt;br /&gt;How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;75 miles per hour&lt;br /&gt;&lt;br /&gt;While going to the destination, the vehicle travels 10 mils at the speed of 50 miles per hour. So the time taken to travel 10 miles is&lt;br /&gt;= (60 * 10) / 50&lt;br /&gt;= 12 minutes&lt;br /&gt;&lt;br /&gt;Now it's given that round-trip travel time is 20 minutes. So the vehicle should complete its return trip of 10 miles in 8 minutes. So the speed of the vehicle must&lt;br /&gt;= (60 * 10) / 8&lt;br /&gt;= 75 miles per hour&lt;br /&gt;All of the students at a college are majoring in psychology, business, or both. 73% of the students are psychology majors, &amp;amp; 62% are business majors.&lt;br /&gt;&lt;br /&gt;If there are 200 students, how many of them are majoring in both psychology &amp;amp; business?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;70 students are majoring in both, psychology &amp;amp; business&lt;br /&gt;&lt;br /&gt;If 73% of the students are psychology majors, we know that 27% are not psychology majors. By the same reasoning, 38% are not business majors, because 62% of the students do major in business. So: 27 + 38 = 65&lt;br /&gt;&lt;br /&gt;65% of the students are not majoring in both psychology &amp;amp; business, so 35% are double majors, a total of 70 students.&lt;br /&gt;Two trains starting at same time, one from Bangalore to Mysore and other in opposite direction arrive at their destination 1hr and 4hrs respectively after passing each other.&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;The speed of Bangalore-Mysore train is TWICE the speed of Mysore-Bangalore train.&lt;br /&gt;&lt;br /&gt;Let the distance between Bangalore and Mysore is D kms.&lt;br /&gt;Also, let speed of the train from Bangalore to Mysore is X km/hr and speed of the tain from Mysore to Bangalore is Y km/hr.&lt;br /&gt;&lt;br /&gt;Now, assume that both the trains met each other at T kms from the Bangalore (point P in figure)&lt;br /&gt;Time taken by Bangalore-Mysore train to reach P = Time taken by Mysore-Bangalore train to reach P&lt;br /&gt;( T / X ) = ( D - T ) / Y -----equ(I)&lt;br /&gt;&lt;br /&gt;Also, Bangalore-Mysore train and Mysore-Bangalore train arrive destination 1 hr and 4 hrs respectively after passing each other. It means that Bangalore-Mysore train travels (D - T) kms in 1 hr at X km/hr and Mysore-Bangalore train travels T kms in 4 hrs at Y km/hr. Hence,&lt;br /&gt;( D - T ) = X and&lt;br /&gt;T = 4 * Y&lt;br /&gt;&lt;br /&gt;Substituting these values in equation I, we get&lt;br /&gt;( 4 * Y ) / X = X / Y&lt;br /&gt;4 * Y * Y = X * X&lt;br /&gt;2 * Y = X&lt;br /&gt;&lt;br /&gt;Hence, the speed of Bangalore-Mysore train is TWICE the speed of Mysore-Bangalore train.How much faster is one train from other?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;49 times&lt;br /&gt;&lt;br /&gt;Let's assume that everyone clinked their mug with friend to his left only. It means that there are total 49 clinks. Now the right clink of each person is left clink of the person on right which is already happened. Hence, there are only 49 clinks.&lt;br /&gt;&lt;br /&gt;Mrs. Watsherface had a garage sale. A custmer named Gina bought an old lamp and a rug. She paid a total of $5.25 for everything. The rug cost 25 cents more than the lamp.&lt;br /&gt;&lt;br /&gt;How much did each cost?&lt;br /&gt;Submitted by : Kimi&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The lamp cost $ 2.50 and the rug cost $ 2.75&lt;br /&gt;&lt;br /&gt;A simple one.&lt;br /&gt;&lt;br /&gt;Assume that the lamp cost $ L.&lt;br /&gt;Hence the rug must have cost $ (L + 0.25)&lt;br /&gt;Also, total cost is $ 5.25, Hence the equation :&lt;br /&gt;L + L + 0.25 = 5.25&lt;br /&gt;2 * L = 5&lt;br /&gt;L = 2.50&lt;br /&gt;&lt;br /&gt;Hence, the lamp cost $ 2.50 and the rug cost $ 2.75&lt;br /&gt;&lt;br /&gt;Brain Teaser No : 00518&lt;br /&gt;&lt;br /&gt;Write 1111......(243 times) i.e. a 243 digit number with all 1s.&lt;br /&gt;&lt;br /&gt;Prove that it is divisible by 243.&lt;br /&gt;SubmittAnswer&lt;br /&gt;&lt;br /&gt;Prove it using the mathematical induction.&lt;br /&gt;&lt;br /&gt;First here are a couple of things to note:&lt;br /&gt;&lt;br /&gt;[1] A number whose digits add up to a multiple of three is divisable by 3.&lt;br /&gt;e.g. 369: 3+6+9=18: 1+8=9 which is a multiple of 3 hence 369 is divisable by 3.&lt;br /&gt;&lt;br /&gt;[2] Whenever a number (X) is multiplied with another number (Y) then the product (X*Y) will have all the factors of X as well as all the factors of Y in its set of factors.&lt;br /&gt;e.g. if X has factors of (1,P,Q,X) and Y has factors of (1,Q,R,Y) then X*Y has factors of (1,P,Q,Q,R,X,Y).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Let&lt;br /&gt;N = any series of digits (e.g. N=369)&lt;br /&gt;D = the number of digits in N (e.g. if N=369 then D=3)&lt;br /&gt;P = is a number constructed in the following way : a 1, followed by (D-1) 0s, followed by another 1, followed by (D-1) 0s, followed by another 1. (e.g. if N=369 then D=3 and P would be 1001001) Note that P will always be divisible by 3.&lt;br /&gt;&lt;br /&gt;Also, if we multiply N with P we are essentially repeating N for (D-1) times.&lt;br /&gt;e.g. if N=369 then D=3, P=1001001 and N*P=369369369&lt;br /&gt;&lt;br /&gt;Let's start with N=111. It is clear that N is divisible by 3. (From [1])&lt;br /&gt;Also, D=3 and P=1001001&lt;br /&gt;N*P=111111111 (9 times)&lt;br /&gt;The resulting number 111111111 must be divisible by 9 as N and P both are divisible by 3.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Now, let's start with N=111111111. It is clear that N is divisible by 9.&lt;br /&gt;Also, D=9 and P=1000000001000000001&lt;br /&gt;N*P=111111111... (27 times)&lt;br /&gt;The resulting number 1111111... (27 times) must be divisible by 27 as N is divisible by 9 and P is divisible by 3.&lt;br /&gt;&lt;br /&gt;Repeat the same procedure for N=1111111... (27 times) The resulting number 1111111... (81 times) must be divisible by 81 as N is divisible by 27 and P is divisible by 3.&lt;br /&gt;&lt;br /&gt;Similarly, for N=1111111... (81 times) The resulting number 1111111... (243 times) must be divisible by 243 as N is divisible by 81 and P is divisible by 3.&lt;br /&gt;&lt;br /&gt;Thus, 1111111... (243 times) is divisible by 243.&lt;br /&gt;&lt;br /&gt;Thanks to Ryan Hutcherson for solution !!!&lt;br /&gt;edKaran bought a little box of midget matches, each one inch in length. He found that he could arrange them all in the form of a triangle whose area was just as many square inches as there were matches.&lt;br /&gt;&lt;br /&gt;He then used up six of the matches, and found that with the remainder he could again construct another triangle whose area was just as many square inches as there were matches.&lt;br /&gt;&lt;br /&gt;And using another six matches he could again do precisely the same.&lt;br /&gt;&lt;br /&gt;How many matches were there in the box originally?&lt;br /&gt;&lt;br /&gt;Note that the match-box can hold maximum of 50 matches.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Initially, there were 42 or 36 matches in the match-box.&lt;br /&gt;&lt;br /&gt;There are 42 matches in the box with which he could form a triangle 20, 15, 7, with an area of 42 square inches. After 6 matches had been used, the remaining 36 matches would form a triangle 17, 10, 9, with an area of 36 square inches. After using another 6 matches, the remaining 30 matches would form a triangle 13, 12, 5, with an area of 30 square inches. After using another 6, the 24 remaining would form a triangle 10, 8, 6, with an area of 24 square inches.&lt;br /&gt;&lt;br /&gt;Thus, there are two possible answers. There were either 42 or 36 matches in the match-box.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Also it is interesting to know that there are just 5 such triangles for which the perimeter and the area is the same (assuming all sides are integers) and they are :&lt;br /&gt;1.    24 (10, 8, 6)&lt;br /&gt;2.    30 (13, 12, 5)&lt;br /&gt;3.    36 (17, 10, 9)&lt;br /&gt;4.    42 (20, 15, 7)&lt;br /&gt;5.    60 (29, 25, 6)&lt;br /&gt;&lt;br /&gt;Find the values of each of the alphabets.&lt;br /&gt;&lt;br /&gt;            N O O N&lt;br /&gt;&lt;br /&gt;            S O O N&lt;br /&gt;&lt;br /&gt;         +  M O O N&lt;br /&gt;&lt;br /&gt;         ----------&lt;br /&gt;&lt;br /&gt;            J U N E&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Using trial and error. There are 2 solutions to it and may be more.&lt;br /&gt;&lt;br /&gt;            2 4 4 2&lt;br /&gt;&lt;br /&gt;            1 4 4 2&lt;br /&gt;&lt;br /&gt;         +  5 4 4 2&lt;br /&gt;&lt;br /&gt;         ----------&lt;br /&gt;&lt;br /&gt;            9 3 2 6&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;            4 1 1 4&lt;br /&gt;&lt;br /&gt;            5 1 1 4&lt;br /&gt;&lt;br /&gt;         +  0 1 1 4&lt;br /&gt;&lt;br /&gt;         ----------&lt;br /&gt;&lt;br /&gt;            9 3 4 2&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;We have to fill number from 1 to 12 at the intersection point of two or more lines. We have to construct a star using two triangle. The sum of all number lying in straight lines should be same. This can be easilty understood by the fig. and hence solved.&lt;br /&gt;Submitted by : Vaibhav Gupta&lt;br /&gt;Submit&lt;br /&gt;Answer         Users&lt;br /&gt;Answer (3)         BrainVista&lt;br /&gt;Answer         Puzzle A&lt;br /&gt;Friend         Add to&lt;br /&gt;Favourite&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;We have one answer where sum of all the numbers lying in straight lines is 26.&lt;br /&gt;&lt;br /&gt;If you have others, do submit them.&lt;br /&gt;    Brain Teaser No : 00355&lt;br /&gt;&lt;br /&gt;Montu, Bantu, Chantu and Pintu have pets.&lt;br /&gt;&lt;br /&gt;Montu says, "If Pintu and I each have a dog, then exactly one of Bantu and Chantu has a dog."&lt;br /&gt;&lt;br /&gt;Bantu says, "If Chantu and I each have a cat, then exactly one of Montu and Pintu has a dog."&lt;br /&gt;&lt;br /&gt;Chantu says, "If Montu and I each have a dog, then exactly one of Bantu and Pintu has a cat."&lt;br /&gt;&lt;br /&gt;Pintu says, "If Bantu and I each have a cat, then exactly one of Bantu and I has a dog."&lt;br /&gt;&lt;br /&gt;Only one of the four is telling the truth. Who is telling the truth?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Bantu is telling the truth.&lt;br /&gt;&lt;br /&gt;For a IF-THEN statement to be false, IF part has to be true and THEN part has to be false.&lt;br /&gt;&lt;br /&gt;Since only one statement is true and remaining three are false, IF part of three statements are true &amp;amp; THEN part of one statement is true. Let's put the given information in table. The pet-name in the normal text represents the IF part and the pet-name in round brackets represents the THEN part.&lt;br /&gt;     Montu    Bantu    Chantu    Pintu&lt;br /&gt;Montu says    Dog    (Dog)    (Dog)    Dog&lt;br /&gt;Bantu says    (Dog)    Cat    Cat    (Dog)&lt;br /&gt;Chantu says    Dog    (Cat)    Dog    (Cat)&lt;br /&gt;Pintu says         Cat&lt;br /&gt;(Dog)         Cat&lt;br /&gt;(Dog)&lt;br /&gt;&lt;br /&gt;It is clear that the IF part of the statements made by Montu, Chantu and Pintu are true as they do not contradict each other. And the IF part of the statement made by Bantu is false.&lt;br /&gt;&lt;br /&gt;Thus, Bantu is telling the truth.&lt;br /&gt;&lt;br /&gt;Montu have a Dog and may or may not have a Cat.&lt;br /&gt;Bantu have a Cat.&lt;br /&gt;Chantu have a Dog.&lt;br /&gt;Pintu have a Dog and a Cat.&lt;br /&gt;    Brain Teaser No : 00520&lt;br /&gt;&lt;br /&gt;Somebody marked the six faces of a die with the numbers 1, 2 and 3 - each number twice. The die was put on a table. Four people - Abu, Babu, Calu and Dabu - sat around the table so that each one was able to see only three sides of the die at a glance.&lt;br /&gt;    Abu sees the number 1 and two even numbers.&lt;br /&gt;    Babu and Calu can see three different numbers each.&lt;br /&gt;    Dabu sees number 2 twice and he can't remember the third number.&lt;br /&gt;What number is face down on the table?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Number 3 is face down on the table.&lt;br /&gt;&lt;br /&gt;If Abu can see two even numbers i.e. number 2 twice, and if Dabu can see number 2 twice, then number 2 must be facing up.&lt;br /&gt;&lt;br /&gt;Now everything else is simple. (see the following diagram)&lt;br /&gt;&lt;br /&gt;       Dabu            Abu&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;                1&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;            3   2   2&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;                1&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;       Calu            Babu&lt;br /&gt;Thus, the number hidden from the view is number 3 and hence the answer.&lt;br /&gt;Two identical pack of cards A and B are shuffled throughly. One card is picked from A and shuffled with B. The top card from pack A is turned up. If this is the Queen of Hearts, what are the chances that the top card in B will be the King of Hearts?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;52 / 2703&lt;br /&gt;&lt;br /&gt;There are two cases to be considered.&lt;br /&gt;&lt;br /&gt;CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B&lt;br /&gt;&lt;br /&gt;Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)&lt;br /&gt;Probability of having King of Hearts on the top of the Pack B = 2/53&lt;br /&gt;&lt;br /&gt;So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)&lt;br /&gt;&lt;br /&gt;CASE 2 : King of Hearts is not drawn from Pack A&lt;br /&gt;&lt;br /&gt;Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)&lt;br /&gt;Probability of having King of Hearts on the top of the Pack B = 1/53&lt;br /&gt;&lt;br /&gt;So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)&lt;br /&gt;&lt;br /&gt;Now adding both the probability, the required probability is&lt;br /&gt;= 2 / (51 * 53) + 50 / (51 * 53)&lt;br /&gt;= 52 / (51 * 53)&lt;br /&gt;= 52 / 2703&lt;br /&gt;= 0.0192378&lt;br /&gt;How many possible combinations are there in a 3x3x3 rubics cube?&lt;br /&gt;&lt;br /&gt;In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)?&lt;br /&gt;&lt;br /&gt;How many for a 4x4x4 cube?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Let's consider 3x3x3 Rubics first.&lt;br /&gt;&lt;br /&gt;There are 8 corner cubes, which can be arranged in 8! ways.&lt;br /&gt;Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)&lt;br /&gt;&lt;br /&gt;Similarly, 12 edge cubes can be arranged in 12! ways.&lt;br /&gt;Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)&lt;br /&gt;&lt;br /&gt;Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.&lt;br /&gt;&lt;br /&gt;Total different possible combinations are&lt;br /&gt;= [(8!) * (3^7)] * [(12!) * (2^11)] / 2&lt;br /&gt;= (8!) * (3^7) * (12!) * (2^10)&lt;br /&gt;= 4.3252 * 10^19&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Similarly, for 4x4x4 Rubics total different possible combinations are&lt;br /&gt;= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24&lt;br /&gt;= 7.4011968 * 10^45&lt;br /&gt;&lt;br /&gt;Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.&lt;br /&gt;&lt;br /&gt;    Brain Teaser No : 00528&lt;br /&gt;&lt;br /&gt;Substitute digits for the letters to make the following relation true.&lt;br /&gt;               N  E  V  E  R&lt;br /&gt;&lt;br /&gt;               L  E  A  V  E&lt;br /&gt;&lt;br /&gt;            +           M  E&lt;br /&gt;&lt;br /&gt;            -----------------&lt;br /&gt;&lt;br /&gt;               A  L  O  N  E&lt;br /&gt;Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A tough one!!!&lt;br /&gt;&lt;br /&gt;Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to&lt;br /&gt;&lt;br /&gt;solve it. Now use trial-n-error method.&lt;br /&gt;&lt;br /&gt;     N  E  V  E  R               2  1  4  1  9&lt;br /&gt;&lt;br /&gt;     L  E  A  V  E               3  1  5  4  1&lt;br /&gt;&lt;br /&gt;  +           M  E            +           6  1&lt;br /&gt;&lt;br /&gt;  -----------------           -----------------&lt;br /&gt;&lt;br /&gt;     A  L  O  N  E               5  3  0  2  1&lt;br /&gt;There are 20 people in your applicant pool, including 5 pairs of identical twins.&lt;br /&gt;&lt;br /&gt;If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins? (Needless to say, this could cause trouble ;))&lt;br /&gt;SubmAnswer&lt;br /&gt;&lt;br /&gt;The probability to hire 5 people with at least 1 pair of identical twins is 25.28%&lt;br /&gt;&lt;br /&gt;5 people from the 20 people can be hired in 20C5 = 15504 ways.&lt;br /&gt;&lt;br /&gt;Now, divide 20 people into two groups of 10 people each :&lt;br /&gt;G1 - with all twins&lt;br /&gt;G2 - with all people other than twins&lt;br /&gt;&lt;br /&gt;Let's find out all possible ways to hire 5 people without a single pair of indentical twins.&lt;br /&gt;People from G1    People from G2    No of ways to hire G1 without a single pair of indentical twins    No of ways to hire G2    Total ways&lt;br /&gt;0    5    10C0    10C5    252&lt;br /&gt;1    4    10C1    10C4    2100&lt;br /&gt;2    3    10C2 * 8/9    10C3    4800&lt;br /&gt;3    2    10C3 * 8/9 * 6/8    10C2    3600&lt;br /&gt;4    1    10C4 * 8/9 * 6/8 * 4/7    10C1    800&lt;br /&gt;5    0    10C5 * 8/9 * 6/8 * 4/7 * 2/6    10C0    32&lt;br /&gt;Total    11584&lt;br /&gt;&lt;br /&gt;Thus, total possible ways to hire 5 people without a single pair of indentical twins = 11584 ways&lt;br /&gt;&lt;br /&gt;So, total possible ways to hire 5 people with at least a single pair of indentical twins = 15504 - 11584 = 3920 ways&lt;br /&gt;&lt;br /&gt;Hence, the probability to hire 5 people with at least a single pair of indentical twins&lt;br /&gt;= 3920/15504&lt;br /&gt;= 245/969&lt;br /&gt;= 0.2528&lt;br /&gt;= 25.28%itted&lt;br /&gt;Veeru says to Jay, "Can you figure out how many Eggs I have in my bucket?" He gives 3 clues to Jay: If the number of Eggs I have&lt;br /&gt;1.    is a multiple of 5, it is a number between 1 and 19&lt;br /&gt;2.    is not a multiple of 8, it is a number between 20 and 29&lt;br /&gt;3.    is not a multiple of 10, it is a number between 30 and 39&lt;br /&gt;How many Eggs does Veeru have in his bucket?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;32 eggs&lt;br /&gt;&lt;br /&gt;Let's apply all 3 condition separately and put all possible numbers together.&lt;br /&gt;&lt;br /&gt;First condition says that if multiple of 5, then the number is between 1 and 19. Hence, the possible numbers are (5, 10, 15, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39)&lt;br /&gt;&lt;br /&gt;Second condition says that if not a multiple of 8, then the number is between 20 and 29. Hence, the possible numbers are (8, 16, 20, 21, 22, 23, 25, 26, 27, 28, 29, 32)&lt;br /&gt;&lt;br /&gt;Third condition says that if not a multiple of 10, then the number is between 30 and 39. Hence, the possible numbers are (10, 20, 31, 32, 33, 34, 35, 36, 37, 38, 39)&lt;br /&gt;&lt;br /&gt;Only number 32 is there in all 3 result sets. That means that only number 32 satisfies all three conditions. Hence, Veeru have 32 eggs in his bucket.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Submit&lt;br /&gt;Answer         Users&lt;br /&gt;Answer (14)         BrainV&lt;br /&gt;&lt;br /&gt;Mr. Black, Mr. White and Mr. Grey were chatting in the Yahoo conference. They were wearing a black suit, a white suit and a grey suit, not necessarily in the same order.&lt;br /&gt;&lt;br /&gt;Mr. Grey sent message, "We all are wearing suit that are of the same color as our names but none of us is wearing a suit that is the same color as his name."&lt;br /&gt;&lt;br /&gt;On that a person wearing the white suit replied, "What difference does that make?"&lt;br /&gt;&lt;br /&gt;Can you tell what color suit each of the three persons had on?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Mr. Grey is wearing Black suit.&lt;br /&gt;Mr. White is wearing Grey suit.&lt;br /&gt;Mr. Black is wearing White suit.&lt;br /&gt;&lt;br /&gt;Mr. Grey must not be wearing grey suit as that is the same colour as his name. Also, he was not wearing white suit as the person wearing white suit responded to his comment. So Mr Grey must be wearing a black suit.&lt;br /&gt;&lt;br /&gt;Similarly, Mr. White must be wearing either black suit or grey suit. But Mr. Grey is wearing a black suit. Hence, Mr. White must be wearing a grey suit.&lt;br /&gt;&lt;br /&gt;And, Mr. Black must be wearing white suit.&lt;br /&gt;Substitute numbers for the letters so that the following mathematical expressions are correct.&lt;br /&gt;   ABC           DEF           GHI&lt;br /&gt;&lt;br /&gt;   --- = IE      --- = IE      --- = IE&lt;br /&gt;&lt;br /&gt;    3             6             9&lt;br /&gt;Note that the same number must be used for the same letter whenever it appears.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7&lt;br /&gt;&lt;br /&gt;Let's start with GHI = 9 * IE. Note that I appears on both the side. Also, after multiplying IE by 9 the answer should have I at the unit's place. The possible values of IE are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (as all alphabet should represent different digits)&lt;br /&gt;&lt;br /&gt;Now, consider DEF = 6 * IE. Out of three short-listed values, only 73 satisfies the equation. Also, ABC = 3 * IE is satisfied by 73.&lt;br /&gt;&lt;br /&gt;Hence, A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7&lt;br /&gt;     219           438           657&lt;br /&gt;&lt;br /&gt;     --- = 73      --- = 73      --- = 73&lt;br /&gt;&lt;br /&gt;      3             6             9&lt;br /&gt;&lt;br /&gt;Brain Teaser No : 00374&lt;br /&gt;&lt;br /&gt;A, B, C and D are related to each other.&lt;br /&gt;    One of the four is the opposite sex from each of the other three.&lt;br /&gt;    D is A's brother or only daughter.&lt;br /&gt;    A or B is C's only son.&lt;br /&gt;    B or C is D's sister.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A, B &amp;amp; D are males; C is female. B is C's only son. A &amp;amp; D are C's brothers.&lt;br /&gt;       A(male) --- C(female) --- D(male)&lt;br /&gt;&lt;br /&gt;                       |&lt;br /&gt;&lt;br /&gt;                       |&lt;br /&gt;&lt;br /&gt;                    B(male)&lt;br /&gt;Work out which relation can hold and discard the contradictory options.&lt;br /&gt;&lt;br /&gt;From (2) and (4), D can not be a only daughter and have a sister (B or C). Hence, D is A's brother i.e. D is a Male.&lt;br /&gt;&lt;br /&gt;From (4), let's say that B is D's sister i.e. B is Female.&lt;br /&gt;From (3), A is C's only son i.e. A is Male.&lt;br /&gt;But D is A's brother which means that A is not C's only son. Hence, our assumption was wrong.&lt;br /&gt;&lt;br /&gt;Thus, C is D's sister i.e. C is Female. And B must be C's only son.&lt;br /&gt;&lt;br /&gt;Now it is clear that D &amp;amp; B are Males and C is Female. A must be a Male as only one of them is of opposite sex from each of the other three. And he is C &amp;amp; D's brother.How are they related to each other?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Dr. DoLittle always goes walking to the clinic and takes the same time while going and while coming back. One day he noticed something.&lt;br /&gt;&lt;br /&gt;When he left the home, the hour hand and the minute hand were exactly opposite to each other and when he reached the clinic, they were together.&lt;br /&gt;&lt;br /&gt;Similarly, when he left the clinic, the hour hand and the minute hand were together and when he reached the home, they were exactly opposite to each other.&lt;br /&gt;&lt;br /&gt;How much time does Dr. DoLittle take to reach home from the clinic? Give the minimal possible answer.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;32 minutes 43.6 seconds&lt;br /&gt;&lt;br /&gt;In twelve hours, the minute hand and the hour hand are together for 11 times. It means that after every 12/11 hours, both the hands are together.&lt;br /&gt;&lt;br /&gt;Similarly in twelve hours, the minute hand and the hour hand are exactly opposite to each other for 11 times. It means that after every 12/11 hours, both the hands are opposite.&lt;br /&gt;&lt;br /&gt;Now, let's take an example. We know that at 12 both the hands are together and at 6 both the hands are exactly opposite to each other.&lt;br /&gt;&lt;br /&gt;After 6, both the hands are in opposition at [6+(12/11)] hours, [6+2*(12/11)] hours, [6+3*(12/11)] hours and so on. The sixth such time is [6+6*(12/11)] hours which is the first time after 12. Thus after 12, both the hands are opposite to each other at 12:32:43.6&lt;br /&gt;&lt;br /&gt;Hence, Dr. DoLittle takes 32 minutes and 43.6 seconds to reach home from the clinic.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;SlowRun Express runs between Bangalore and Mumbai, For the up as well as the down journey, the train leaves the starting station at 10:00 PM everyday and reaches the destination at 11:30 PM after three days.&lt;br /&gt;&lt;br /&gt;Mr. Haani once travelled by SlowRun Express from Mumbai to Bangalore. How many SlowRun Express did he cross during his journey?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Mr. Haani crossed 7 SlowRun Expresses during his journey.&lt;br /&gt;&lt;br /&gt;Let's say that Mr. Haani travelled by SlowRun Express on Wednesday 10:00PM from Mumbai. The first train he would have crossed is the one scheduled to arrive at Mumbai at 11:30 PM the same day i.e. the one that left Bangalore at 10:00 PM on last Sunday.&lt;br /&gt;&lt;br /&gt;Also, he would have crossed the last train just before reaching Bangalore on Saturday.&lt;br /&gt;&lt;br /&gt;Thus, Mr. Haani must have crossed 7 SlowRun Expresses during his journey.&lt;br /&gt;Six cabins numbered 1-6 consecutively, are arranged in a row and are separated by thin dividers. These cabins must be assigned to six staff members based on following facts.&lt;br /&gt;1.    Miss Shalaka's work requires her to speak on the phone frequently throughout the day.&lt;br /&gt;2.    Miss Shudha prefers cabin number 5 as 5 is her lucky number.&lt;br /&gt;3.    Mr. Shaan and Mr. Sharma often talk to each other during their work and prefers to have adjacent cabins.&lt;br /&gt;4.    Mr. Sinha, Mr. Shaan and Mr. Solanki all smoke. Miss Shudha is allergic to smoke and must have non-smokers adjacent to her.&lt;br /&gt;5.    Mr. Solanki needs silence during work.&lt;br /&gt;Can you tell the cabin numbers of each of them?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The cabins from left to right (1-6) are of Mr. Solanki, Mr. Sinha, Mr. Shaan, Mr. Sharma, Miss Shudha and Miss Shalaka.&lt;br /&gt;&lt;br /&gt;From (2), cabin number 5 is assigned to Miss Shudha.&lt;br /&gt;&lt;br /&gt;As Miss Shudha is allergic to smoke and Mr. Sinha, Mr. Shaan &amp;amp; Mr. Solanki all smoke, they must be in cabin numbers 1, 2 and 3 not necessarily in the same order. Also, Miss Shalaka and Mr. Sharma must be in cabin 4 and 6.&lt;br /&gt;&lt;br /&gt;From (3), Mr. Shaan must be in cabin 3 and Mr. Sharma must be in cabin 4. Thus, Miss Shalaka is in cabin 6.&lt;br /&gt;&lt;br /&gt;As Mr. Solanki needs silence during work and Mr. Shaan is in cabin 3 who often talks to Mr. Sharma during work, Mr. Solanki must be in cabin 1. Hence, Mr. Sinha is in cabin 2.&lt;br /&gt;&lt;br /&gt;Thus, the cabins numbers are&lt;br /&gt;1# Mr. Solanki,&lt;br /&gt;2# Mr. Sinha,&lt;br /&gt;3# Mr. Shaan,&lt;br /&gt;4# Mr. Sharma,&lt;br /&gt;5# Miss Shudha,&lt;br /&gt;6# Miss Shalaka&lt;br /&gt;SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.&lt;br /&gt;    When it snows, morning service on line E is delayed.&lt;br /&gt;    When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.&lt;br /&gt;    When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.&lt;br /&gt;    When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.&lt;br /&gt;    When service on line A is delayed or cancelled, service on line I is also delayed.&lt;br /&gt;    When service on line Z is delayed or cancelled, service on line E is also delayed.&lt;br /&gt;On February 10, it snows all day with the temperature at 18C. On how many lines service will be delayed or cancelled, including both morning and afternoon?&lt;br /&gt;SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.&lt;br /&gt;    When it snows, morning service on line E is delayed.&lt;br /&gt;    When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.&lt;br /&gt;    When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.&lt;br /&gt;    When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.&lt;br /&gt;    When service on line A is delayed or cancelled, service on line I is also delayed.&lt;br /&gt;    When service on line Z is delayed or cancelled, service on line E is also delayed.&lt;br /&gt;On February 10, it snows all day with the temperature at 18C. On how many lines service will be delayed or cancelled, including both morning and afternoon?&lt;br /&gt;In a certain game, if 2 wixsomes are worth 3 changs, and 4 changs are worth 1 plut, then 6 plutes are worth how many wixsomes?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;It is given that&lt;br /&gt;2 wixsomes = 3 changs&lt;br /&gt;8 wixsomes = 12 changs ----- (I)&lt;br /&gt;&lt;br /&gt;Also, given that&lt;br /&gt;4 changs = 1 plut&lt;br /&gt;12 changs = 3 plutes&lt;br /&gt;8 wixsomes = 3 plutes ----- From (I)&lt;br /&gt;&lt;br /&gt;Therefore,&lt;br /&gt;6 plutes = 16 wixsomes&lt;br /&gt;&lt;br /&gt;In a certain year, the number of girls who graduated from City High School was twice the number of boys. If 3/4 of the girls and 5/6 of the boys went to college immediately after graduation, what fraction of the graduates that year went to college immediately after graduation?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Assume that number of boys graduated from City High School = B&lt;br /&gt;Therefore, number of girls graduated from City High School = 2*B&lt;br /&gt;&lt;br /&gt;It is given that 3/4 of the girls and 5/6 of the boys went to college immediately after graduation.&lt;br /&gt;Hence, total students went to college&lt;br /&gt;= (3/4)(2*B) + (5/6)(B)&lt;br /&gt;= B * (3/2 + 5/6)&lt;br /&gt;= (7/3)B&lt;br /&gt;&lt;br /&gt;Fraction of the graduates that year went to college immediately after graduation&lt;br /&gt;= [(7/3)B] / [3*B]&lt;br /&gt;= 7/9&lt;br /&gt;&lt;br /&gt;Therefore, the answer is 7/9&lt;br /&gt;&lt;br /&gt;A mule and a donkey were carrying full sacks on their backs.&lt;br /&gt;&lt;br /&gt;The mule started complaining that his load was too heavy. The donkey said to him "Why are you complaining? If you gave me one of your sacks I'd have double what you have and if I give you one of my sacks we'd have an even amount."&lt;br /&gt;&lt;br /&gt;How many sacks were each of them carrying? Give the minimal possible answer.&lt;br /&gt;SubmittAnswer&lt;br /&gt;&lt;br /&gt;The mule was carrying 5 sacks and the donkey was carrying 7 sacks.&lt;br /&gt;&lt;br /&gt;Let's assume that the mule was carrying M sacks and the donkey was carrying D sacks.&lt;br /&gt;&lt;br /&gt;As the donkey told the mule, "If you gave me one of your sacks I'd have double what you have."&lt;br /&gt;D + 1 = 2 * (M-1)&lt;br /&gt;D + 1 = 2M - 2&lt;br /&gt;D = 2M - 3&lt;br /&gt;&lt;br /&gt;The donkey also said, "If I give you one of my sacks we'd have an even amount."&lt;br /&gt;D - 1 = M + 1&lt;br /&gt;D = M + 2&lt;br /&gt;&lt;br /&gt;Comparing both the equations,&lt;br /&gt;2M - 3 = M + 2&lt;br /&gt;M = 5&lt;br /&gt;&lt;br /&gt;Substituting M=5 in any of above equation, we get D=7&lt;br /&gt;&lt;br /&gt;Hence, the mule was carrying 5 sacks and the donkey was carrying 7 sacks.&lt;br /&gt;edTwo people enter a race in whick you run to a point and back. Person A runs 20 mph to and from the point. Person B runs to the point going 10 mph and 30 mph going back.&lt;br /&gt;&lt;br /&gt;Who came in first?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Person A came in first.&lt;br /&gt;&lt;br /&gt;Let's assume that the distance between start and the point is D miles.&lt;br /&gt;&lt;br /&gt;Total time taken by Person A to finish&lt;br /&gt;= (D/20) + (D/20)&lt;br /&gt;= D/10&lt;br /&gt;= 0.1D&lt;br /&gt;&lt;br /&gt;Total time taken by Person B to finish&lt;br /&gt;= (D/10) + (D/30)&lt;br /&gt;= 2D/15&lt;br /&gt;= 0.1333D&lt;br /&gt;&lt;br /&gt;Thus, Person A is the Winner.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Alternatively (if you don't like mathematics ;)), analyse the situation as follow:&lt;br /&gt;&lt;br /&gt;Note that initially speed of Person A (20 mph) was twice the speed of Person B (10 mph). Hence, when Person A (20 mph forward) reached the point, Person B (10 mph forward) was halfway. When Person A (20 mph back) finished, Person B (still 10 mph forward) reached the point.&lt;br /&gt;&lt;br /&gt;Thus, Person A wins the race and by that time Person B covers only half the distance, no matter how far the point is!!!&lt;br /&gt;Mark ate half of a pizza on Monday. He ate half of what was left on Tuesday and so on. He followed this pattern for one week.&lt;br /&gt;&lt;br /&gt;How much of the pizza would he have eaten during the week?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Mark would have ate 127/128 (99.22%) of the pizza during the week.&lt;br /&gt;&lt;br /&gt;Mark ate half the pizza on Monday. On Tuesday, he would have ate half of the remaining pizza i.e. 1/4 of the original pizza. Similarly, he would have ate 1/8 of the original pizza on Wednesday and so on for the seven days.&lt;br /&gt;&lt;br /&gt;Total pizza Mark ate during the week is&lt;br /&gt;= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128&lt;br /&gt;= 127/128&lt;br /&gt;= 99.22% of the original pizza&lt;br /&gt;In the General meeting of "Friends Club", Sameer said, "The repairs to the Club will come to a total of Rs 3120 and I propose that this amount should be met by the members, each paying an equal amount."&lt;br /&gt;&lt;br /&gt;The proposal was immediately agreed. However, four members of the Club chose to resign, leaving the remaining members to pay an extra Rs 26 each.&lt;br /&gt;&lt;br /&gt;How many members did the Club originally have?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The Club originally had 24 members.&lt;br /&gt;&lt;br /&gt;Assume that there were initially N members.&lt;br /&gt;&lt;br /&gt;As 4 members resigned and remaining members paid Rs 26 each, it means that total amount of 4 members is equal to Rs 26 each from remaining (N-4) members. Thus,&lt;br /&gt;&lt;br /&gt;4 * (3120 / N) = 26 * (N - 4)&lt;br /&gt;12480 = 26N2 - 104N&lt;br /&gt;26N2 - 104N - 12480 = 0&lt;br /&gt;&lt;br /&gt;Solving the quadratic equation we get N=24.&lt;br /&gt;&lt;br /&gt;Hence, the Club originally had 24 members.&lt;br /&gt;    Brain Teaser No : 00206&lt;br /&gt;&lt;br /&gt;A tank can be filled by pipe A in 30 minutes and by pipe B in 24 minutes. Outlet pipe C can empty the full tank in one hour and twenty minutes.&lt;br /&gt;&lt;br /&gt;If the tank is empty initially and if all the three pipes A, B and C are opened simultaneously, in how much time will the tank be full?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The tank will be full in 16 minutes.&lt;br /&gt;&lt;br /&gt;In one minute,&lt;br /&gt;pipe A can fill 1/30 part of the tank.&lt;br /&gt;pipe B can fill 1/24 part of the tank.&lt;br /&gt;pipe C can empty 1/80 part of the tank.&lt;br /&gt;&lt;br /&gt;Thus, the net water level in one minute is&lt;br /&gt;= 1/30 + 1/24 - 1/80&lt;br /&gt;= 15/240 part of the tank&lt;br /&gt;&lt;br /&gt;Hence, the tank will be full in 240/15 i.e. 16 minutes.&lt;br /&gt;&lt;br /&gt;A rich old Arab has three sons. When he died, he willed his 17 camels to the sons, to be divided as follows:&lt;br /&gt;&lt;br /&gt;First Son to get 1/2 of the camels Second Son to get 1/3rd of the camels Third Son to get 1/9th of the camels.&lt;br /&gt;&lt;br /&gt;The sons are sitting there trying to figure out how this can possibly be done, when a very old wise man goes riding by. They stop him and ask him to help them solve their problem. Without hesitation he divides the camels properly and continues riding on his way.&lt;br /&gt;&lt;br /&gt;How did he do it?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The old man temporarily added his camel to the 17, making a total of 18 camels.&lt;br /&gt;&lt;br /&gt;First son got 1/2 of it = 9&lt;br /&gt;&lt;br /&gt;Second son got 1/3 of it = 6&lt;br /&gt;&lt;br /&gt;Third son got 1/9 of it = 2&lt;br /&gt;&lt;br /&gt;For a total of 17. He then takes his camel back and rides away......&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;There were two men standing on a street. The one says to the other, "I have 3 daughters, the product of their ages is 36. What is the age of the OLDEST daughter?"&lt;br /&gt;&lt;br /&gt;The second guy says, "I need more information." So, the first guy says, "The sum of their ages is equal to the address of the house across the street."&lt;br /&gt;&lt;br /&gt;The second guy looks at the address and says, "I still need more information." So, the first guy says, "My oldest daughter wears a red dress."&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The answer is 9 years.&lt;br /&gt;&lt;br /&gt;First you need to find all the possible sets of three numbers that when multiplied equals 36:&lt;br /&gt;&lt;br /&gt;1 1 36&lt;br /&gt;1 2 18&lt;br /&gt;1 3 12&lt;br /&gt;1 4 9&lt;br /&gt;1 6 6&lt;br /&gt;2 2 9&lt;br /&gt;2 3 6&lt;br /&gt;3 3 4&lt;br /&gt;&lt;br /&gt;Then you add the numbers together to find the sum&lt;br /&gt;1 1 36 = 38&lt;br /&gt;1 2 18 = 21&lt;br /&gt;1 3 12 = 16&lt;br /&gt;1 4 9 = 14&lt;br /&gt;1 6 6 = 13&lt;br /&gt;2 2 9 = 13&lt;br /&gt;2 3 6 = 11&lt;br /&gt;3 3 4 = 10&lt;br /&gt;&lt;br /&gt;Even though we don't know the address the guy knows it. For him to need more information that means that at least two of the sets of numbers has the same sum. Two of them do, 1 6 6 and 2 2 9.&lt;br /&gt;&lt;br /&gt;When the first guy said that his OLDEST daugher wears a red dress that meant that there had to be the oldest. So 1 6 6 can't possibly be the answer. So the possible possiblity is 2 2 9 and the OLDEST daughter is 9 years old.&lt;br /&gt;&lt;br /&gt;Therefore, the answer is 9.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Submit&lt;br /&gt;Answer         User&lt;br /&gt;There are 3 colored boxes - Red, Green and Blue. Each box contains 2 envelopes. Each envelope contains money - two of them contain Rs. 25000 each, two of them contain Rs. 15000 each and remaining two contain Rs. 10000 each.&lt;br /&gt;&lt;br /&gt;There is one statement written on the cover of each box.&lt;br /&gt;* Red Box: Both, a red box and a blue box contain Rs. 10000 each.&lt;br /&gt;* Green Box: Both, a green box and a red box contain Rs. 25000 each.&lt;br /&gt;* Blue Box: Both, a blue box and a green box contain Rs. 15000 each.&lt;br /&gt;&lt;br /&gt;Only one of the above 3 statements is true and the corresponding box contains the maximum amount.&lt;br /&gt;&lt;br /&gt;Can you tell which box contains the maximum amount and how much?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Blue box contains the maximum amount Rs. 40000&lt;br /&gt;&lt;br /&gt;As it is given that only one of the given 3 statements is true; assume in turn, each statement to be true &amp;amp; the other 2 false and check whether the corresponding box contains the maximum amount.&lt;br /&gt;&lt;br /&gt;Let's assume that the statement on the Blue box is true. Thus, the given 3 statements can be interpreted as&lt;br /&gt;* Atmost one, a red box or a blue box contains Rs. 10000.&lt;br /&gt;* Atmost one, a green box or a red box contains Rs. 25000.&lt;br /&gt;* Both, a blue box and a green box contain Rs. 15000 each.&lt;br /&gt;&lt;br /&gt;Going through all possible combinations, we can conclude that&lt;br /&gt;Red Box : Rs. 10000 + Rs. 25000 = Rs. 35000&lt;br /&gt;Green Box : Rs. 10000 + Rs. 15000 = Rs. 25000&lt;br /&gt;Blue Box : Rs. 15000 + Rs. 25000 = Rs. 40000&lt;br /&gt;&lt;br /&gt;You can test out for other two statements i.e. assuming Red box statement true and then Green box statement true. In both the cases, other statements will contradict the true statement.&lt;br /&gt;&lt;br /&gt;Sachin, Dravid and Ganguly played in a Cricket match between India and England.&lt;br /&gt;    None of them scored more than 99 runs.&lt;br /&gt;    If you add the digits of the runs scored by Sachin to his own score, you will get the runs scored by Dravid.&lt;br /&gt;    If you reverse the digits of the runs scored by Dravid, you will get the runs scored by Ganguly.&lt;br /&gt;    The total runs scored by them is 240.&lt;br /&gt;Can you figure out their individual scores?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Sachin, Dravid and Ganguly scored 75, 87 and 78 respectively.&lt;br /&gt;&lt;br /&gt;Sachin's score must be less than 86, otherwise Dravid's score would be more than 99. Also, he must have scored atleast 42 - incase Dravid and Ganguly scored 99 each.&lt;br /&gt;&lt;br /&gt;Also, as none of them scored more than 99 and the total runs scored by them is 240; their individual scores must be around 80.&lt;br /&gt;&lt;br /&gt;Now, use trial-n-error method to solve the teaser.&lt;br /&gt;Three men, including Gianni and three woman, including Sachi are in line at the BrentWood post office. Each has two different pieces of business to conduct.&lt;br /&gt;1.    The first person is a woman.&lt;br /&gt;2.    Carlos wants to send an overnight package.&lt;br /&gt;3.    Lau is just ahead of Pimentelli who is the same sex as Lau.&lt;br /&gt;4.    Gianni is two places ahead of the person who wants to buy stamps.&lt;br /&gt;5.    Knutson - who is the opposite sex than Rendler - isn't the person who wanted to complain about a mail carrier.&lt;br /&gt;6.    The six people, not necessarily in the same order are - Anthony, Donna, the person who wants to fill out a change-of-address form, the one who wants to buy a money order, the one who wants to send Airmail to Tibet and the second person in the line.&lt;br /&gt;7.    The four tasks of the last two people in line, not necessarily in the same order are - sending books fourth class, buying a money order, picking up a package and complaining about a mail carrier.&lt;br /&gt;8.    The person who wants to send books fourth class is just behind a person of the same sex.&lt;br /&gt;9.    Mary is just behind a person who wants to send an insured package.&lt;br /&gt;10.    The person who wants to send Airmail to Tibet is either two places ahead of or two places behind the one who wants to add postage to his or her meter.&lt;br /&gt;11.    Anthony isn't two places behind the who wants to pickup a registered letter.&lt;br /&gt;12.    Toriseza is two places ahead of the person who wants to pick up a package.&lt;br /&gt;13.    Knutson isn't just ahead of the person who wants to send an item parcel post.&lt;br /&gt;Can you figure out where each customer is in the line, his or her full name (one surname is Loti) and the two things he or she wants to accomplish? Provide your answer is POSITION - FIRST NAME - LAST NAME - BUSINESS format.&lt;br /&gt;Submit&lt;br /&gt;Answer         Users&lt;br /&gt;Answer (8)    &lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A very TOUGH puzzle !!!&lt;br /&gt;POS    FIRST NAME     LAST NAME     BUSINESS&lt;br /&gt;1    Sachi    Loti    • Fill Out a Change-of-Address Form&lt;br /&gt;• Add Postage to Meter&lt;br /&gt;2    Gianni    Lau    • Pick Up a Registered Letter&lt;br /&gt;• Send an Item Parcel Post&lt;br /&gt;3    Carlos    Pimentelli    • Overnight Package&lt;br /&gt;• Send Airmail to Tibet&lt;br /&gt;4    Donna    Toriseza    • Buy Stamps&lt;br /&gt;• Send an Insured Package&lt;br /&gt;5    Mary    Knutson    • Buy a Money Order&lt;br /&gt;• Send Books fourth Class&lt;br /&gt;6    Anthony    Rendler    • Complain About a Mail Carrier&lt;br /&gt;• Pick Up a Package&lt;br /&gt;&lt;br /&gt;    Brain Teaser No : 00164&lt;br /&gt;&lt;br /&gt;Substitute digits for the letters to make the following relation true.&lt;br /&gt;                  W O R L D&lt;br /&gt;&lt;br /&gt;               +  T R A D E&lt;br /&gt;&lt;br /&gt;               -------------&lt;br /&gt;&lt;br /&gt;                C E N T E R&lt;br /&gt;Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter W, no other letter can be 3 and all other W in the puzzle must be 3.&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A tough one.&lt;br /&gt;&lt;br /&gt;It is obvious that C=1. Also, the maximum possible value of E is 7. Now, start putting possible values of D, E and R as they occure frequently and use trial-n-error.&lt;br /&gt;&lt;br /&gt;     W O R L D          5 3 6 8 4&lt;br /&gt;&lt;br /&gt;  +  T R A D E       +  7 6 0 4 2&lt;br /&gt;&lt;br /&gt;  ------------       ------------&lt;br /&gt;&lt;br /&gt;   C E N T E R        1 2 9 7 2 6&lt;br /&gt;    Brain Teaser No : 00107&lt;br /&gt;&lt;br /&gt;If you look at a clock and the time is 3:15.&lt;br /&gt;&lt;br /&gt;What is the angle between the hour and the minute hands? ( The answer to this is not zero!)&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;7.5 degrees&lt;br /&gt;&lt;br /&gt;At 3:15 minute hand will be perfactly horizontal pointing towards 3. Whereas hour hand will be towards 4. Also, hour hand must have covered 1/4 of angle between 3 and 4.&lt;br /&gt;&lt;br /&gt;The angle between two adjacent digits is 360/12 = 30 degrees.&lt;br /&gt;&lt;br /&gt;Hence 1/4 of it is 7.5 degrees.&lt;br /&gt;An apple vendor has 1000 apples and 10 empty boxes. He asks his son to place all the 1000 apples in all the 10 boxes in such a manner that if he asks for any number of apples from 1 to 1000, his son should be able to pick them in terms of boxes.&lt;br /&gt;&lt;br /&gt;How did the son place all the apples among the 10 boxes, given that any number of apples can be put in one box.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;1, 2, 4, 8, 16, 32, 64, 128, 256, 489&lt;br /&gt;&lt;br /&gt;Let's start from scratch.&lt;br /&gt;    The apple vandor can ask for only 1 apple, so one box must contain 1 apple.&lt;br /&gt;    He can ask for 2 apples, so one box must contain 2 apples.&lt;br /&gt;He can ask for 3 apples, in that case box one and box two will add up to 3.&lt;br /&gt;    He can ask for 4 apples, so one box i.e. third box must contain 4 apples.&lt;br /&gt;    Now using box number one, two and three containing 1, 2 and 4 apples respectively, his son can give upto 7 apples. Hence, forth box must contain 8 apples.&lt;br /&gt;    Similarly, using first four boxes containing 1, 2, 4 and 8 apples, his son can give upto 15 apples. Hence fifth box must contain 16 apples.&lt;br /&gt;    You must have noticed one thing till now that each box till now contains power of 2 apples. Hence the answer is 1, 2, 4, 8, 16, 32, 64, 128, 256, 489. This is true for any number of apples, here in our case only upto 1000.&lt;br /&gt;&lt;br /&gt;    Brain Teaser No : 00261&lt;br /&gt;&lt;br /&gt;The letters P, Q, R, S, T, U and V, not necessarily in that order represents seven consecutive integers from 22 to 33.&lt;br /&gt;    U is as much less than Q as R is greater than S.&lt;br /&gt;    V is greater than U.&lt;br /&gt;    Q is the middle term.&lt;br /&gt;    P is 3 greater than S.&lt;br /&gt;Can you find the sequence of letters from the lowest value to the highest value?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The sequence of letters from the lowest value to the highest value is TUSQRPV.&lt;br /&gt;&lt;br /&gt;From (3), Q is the middle term.&lt;br /&gt;___ ___ ___ _Q_ ___ ___ ___&lt;br /&gt;&lt;br /&gt;From (4), there must be exactly 2 numbers between P and S which gives two possible positions.&lt;br /&gt;&lt;br /&gt;[1] ___ _S_ ___ _Q_ _P_ ___ ___&lt;br /&gt;&lt;br /&gt;[2] ___ ___ _S_ _Q_ ___ _P_ ___&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;From (1), the number of letters between U and Q must be same as the number of letters between S and R. Also, the number of letters between them can be 1, 2 or 3.&lt;br /&gt;&lt;br /&gt;Using trial and error, it can be found that there must be 2 letters between them. Also, it is possible only in option [2] above.&lt;br /&gt;&lt;br /&gt;[2] ___ _U_ _S_ _Q_ _R_ _P_ ___&lt;br /&gt;&lt;br /&gt;From (2) V must be the highest and the remaining T must be the lowest number.&lt;br /&gt;&lt;br /&gt;_T_ _U_ _S_ _Q_ _R_ _P_ _V_&lt;br /&gt;&lt;br /&gt;Thus, the sequence of letters from the lowest value to the highest value is TUSQRPV.&lt;br /&gt;A contractor had employed 100 labourers for a flyover construction task. He did not allow any woman to work without her husband. Also, atleast half the men working came with their wives.&lt;br /&gt;&lt;br /&gt;He paid five rupees per day to each man, four ruppes to each woman and one rupee to each child. He gave out 200 rupees every evening.&lt;br /&gt;&lt;br /&gt;How many men, women and children were working with the constructor?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;16 men, 12 women and 72 children were working with the constructor.&lt;br /&gt;&lt;br /&gt;Let's assume that there were X men, Y women and Z children working with the constructor. Hence,&lt;br /&gt;&lt;br /&gt;X + Y + Z = 100&lt;br /&gt;5X + 4Y + Z = 200&lt;br /&gt;&lt;br /&gt;Eliminating X and Y in turn from these equations, we get&lt;br /&gt;X = 3Z - 200&lt;br /&gt;Y = 300 - 4Z&lt;br /&gt;&lt;br /&gt;As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get&lt;br /&gt;&lt;br /&gt;if Y = X,&lt;br /&gt;300 - 4Z = 3Z - 200&lt;br /&gt;7Z = 500&lt;br /&gt;Z = 500/7 i.e. 71.428&lt;br /&gt;&lt;br /&gt;if Y = X/2,&lt;br /&gt;300 - 4Z = (3Z - 200)/2&lt;br /&gt;600 - 8Z = 3Z - 200&lt;br /&gt;11Z = 800&lt;br /&gt;Z = 800/11 i.e. 72.727&lt;br /&gt;&lt;br /&gt;But Z must be an integer, hence Z=72. Also, X=16 and Y=12&lt;br /&gt;&lt;br /&gt;There were 16 men, 12 women and 72 children working with the constructor.&lt;br /&gt;&lt;br /&gt;Because cigars cannot be entirely smoked, a Bobo who collects cigar butts can make a cigar to smoke out of every 3 butts that he finds.&lt;br /&gt;&lt;br /&gt;Today, he has collected 27 cigar butts. How many cigars will he be able to smoke?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;13 not 12&lt;br /&gt;&lt;br /&gt;He makes 9 originals from the 27 butts he found, and after he smokes them he has 9 butts left for another 3 cigars. And then he has 3 butts for another cigar.&lt;br /&gt;&lt;br /&gt;So 9+3+1=13&lt;br /&gt;In a small town, there are three temples in a row and a well in front of each temple. A pilgrim came to the town with certain number of flowers.&lt;br /&gt;&lt;br /&gt;Before entering the first temple, he washed all the flowers he had with the water of well. To his surprise, flowers doubled. He offered few flowers to the God in the first temple and moved to the second temple. Here also, before entering the temple he washed the remaining flowers with the water of well. And again his flowers doubled. He offered few flowers to the God in second temple and moved to the third temple. Here also, his flowers doubled after washing them with water. He offered few flowers to the God in third temple.&lt;br /&gt;&lt;br /&gt;There were no flowers left when pilgrim came out of third temple and he offered same number of flowers to the God in all three temples.&lt;br /&gt;&lt;br /&gt;What is the minimum number of flowers the pilgrim had initially? How many flower did he offer to each God?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;The pilgrim had 7 flowers, initially and he offered 8 flowers to each God.&lt;br /&gt;&lt;br /&gt;Assume that the pilgrim had X flowers initially and he offered Y flowers to each God.&lt;br /&gt;&lt;br /&gt;From the above figure, there are (8X - 7Y) flowers when the pilgrim came out of the third temple. But it is given that there were no flowers left when he came out of third temple. It means that&lt;br /&gt;(8X - 7Y) = 0&lt;br /&gt;8X = 7Y&lt;br /&gt;&lt;br /&gt;The minimum values of X and Y are 7 and 8 respectively to satisfy above equation. Hence, the pilgrim had 7 flowers and he offered 8 flowers to each God.&lt;br /&gt;&lt;br /&gt;In general, the pilgrim had 7N flowers initially and he offered 8N flowers to each God, where N = 1, 2, 3, 4, .....&lt;br /&gt;&lt;br /&gt;    Brain Teaser No : 00432&lt;br /&gt;&lt;br /&gt;Tanya wants to go on a date and prefers her date to be tall, dark and handsome.&lt;br /&gt;1.    Of the preferred traits - tall, dark and handsome - no two of Adam, Bond, Cruz and Dumbo have the same number.&lt;br /&gt;2.    Only Adam or Dumbo is tall and fair.&lt;br /&gt;3.    Only Bond or Cruz is short and handsome.&lt;br /&gt;4.    Adam and Cruz are either both tall or both short.&lt;br /&gt;5.    Bond and Dumbo are either both dark or both fair.&lt;br /&gt;Who is Tanya's date?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Cruz is Tanya's date.&lt;br /&gt;&lt;br /&gt;As no two of them have the same number of preferred traits - from (1), exactly one of them has none of the preferred traits and exactly one of them has all the preferred traits.&lt;br /&gt;&lt;br /&gt;From (4) and (5), there are only two possibilities:&lt;br /&gt;* Adam &amp;amp; Cruz both are tall and Bond &amp;amp; Dumbo both are fair.&lt;br /&gt;* Adam &amp;amp; Cruz both are short and Bond &amp;amp; Dumbo both are dark.&lt;br /&gt;&lt;br /&gt;But from (2), second possibility is impossible. So the first one is the correct possibility i.e. Adam &amp;amp; Cruz both are tall and Bond &amp;amp; Dumbo both are fair.&lt;br /&gt;&lt;br /&gt;Then from (3), Bond is short and handsome.&lt;br /&gt;&lt;br /&gt;Also, from (1) and (2), Adam is tall and fair. Also, Dumbo is the person without any preferred traits. Cruz is Dark. Adam and Cruz are handsome. Thus, following are the individual preferred traits:&lt;br /&gt;&lt;br /&gt;Cruz - Tall, Dark and Handsome&lt;br /&gt;Adam - Tall and Handsome&lt;br /&gt;Bond - Handsome&lt;br /&gt;Dumbo - None :-(&lt;br /&gt;&lt;br /&gt;Hence, Cruz is Tanya's date.&lt;br /&gt;Consider a game of Tower of Hanoi (like the one that you can play on BrainVista).&lt;br /&gt;&lt;br /&gt;If the tower has 2 discs, the least possible moves with which you can move the entire tower to another peg is 3.&lt;br /&gt;&lt;br /&gt;If the tower has 3 discs, the least possible moves with which you can move the entire tower to another peg is 7.&lt;br /&gt;&lt;br /&gt;What is the least possible moves with which you can move the entire tower to another peg if the tower has N discs?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are number of ways to find the answer.&lt;br /&gt;&lt;br /&gt;To move the largest disc (at level N) from one tower to the other, it requires 2(N-1) moves. Thus, to move N discs from one tower to the other, the number of moves required is&lt;br /&gt;= 2(N-1) + 2(N-2) + 2(N-3) + ..... + 22 + 21 + 20&lt;br /&gt;= 2N - 1&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;For N discs, the number of moves is one more than two times the number of moves for N-1 discs. Thus, the recursive function is&lt;br /&gt;F(1) = 1&lt;br /&gt;F(N) = 2*[F(N-1)] + 1&lt;br /&gt;where N is the total number of discs&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Also, one can arrive at the answer by finding the number of moves for smaller number of discs and then derive the pattern.&lt;br /&gt;For 1 disc, number of moves = 1&lt;br /&gt;For 2 discs, number of moves = 3&lt;br /&gt;For 3 discs, number of moves = 7&lt;br /&gt;For 4 discs, number of moves = 15&lt;br /&gt;For 5 discs, number of moves = 31&lt;br /&gt;&lt;br /&gt;Thus, the pattern is 2N – 1&lt;br /&gt;A boy found that he had a 48 inch strip of paper. He could cut an inch off every second.&lt;br /&gt;&lt;br /&gt;How long would it take for him to cut 48 pieces? He can not fold the strip and also, can not stack two or more strips and cut them together.&lt;br /&gt;SubmiAnswer&lt;br /&gt;&lt;br /&gt;47 seconds.&lt;br /&gt;&lt;br /&gt;To get 48 pieces, the boy have to put only 47 cuts. i.e. he can cut 46 pieces in 46 seconds. After getting 46 pieces, he will have a 2 inches long piece. He can cut it into two with just a one cut in 1 second. Hence, total of 47 seconds.tted by : Kimi&lt;br /&gt;&lt;br /&gt;The cricket match between India and Pakistan was over.&lt;br /&gt;    Harbhajan scored more runs than Ganguly.&lt;br /&gt;    Sachin scored more runs than Laxman but less than Dravid&lt;br /&gt;    Badani scored as much runs as Agarkar but less than Dravid and more than Sachin.&lt;br /&gt;    Ganguly scored more runs than either Agarkar or Dravid.&lt;br /&gt;Each batsman scored 10 runs more than his immediate batsman. The lowest score was 10 runs. How much did each one of them score&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;A simple one. Use the given facts and put down all the players in order. The order is as follow with Harbhajan, the highest scorer and Laxman, the lowest scorer.&lt;br /&gt;1.    Harbhajan&lt;br /&gt;2.    Ganguly&lt;br /&gt;3.    Dravid&lt;br /&gt;4.    Badani, Agarkar&lt;br /&gt;5.    Sachin&lt;br /&gt;6.    Laxman&lt;br /&gt;Also, as the lowest score was 10 runs. Laxman must have scored 10, Sachin 20, Badani &amp;amp; Agarkar 30 and so on.&lt;br /&gt;1.    Harbhajan - 60 runs&lt;br /&gt;2.    Ganguly - 50 runs&lt;br /&gt;3.    Dravid - 40 runs&lt;br /&gt;4.    Badani, Agarkar - 30 runs each&lt;br /&gt;5.    Sachin - 20 runs&lt;br /&gt;6.    Laxman - 10 runs&lt;br /&gt;There are 10 statements written on a piece of paper:&lt;br /&gt;1.    At least one of statements 9 and 10 is true.&lt;br /&gt;2.    This either is the first true or the first false statement.&lt;br /&gt;3.    There are three consecutive statements, which are false.&lt;br /&gt;4.    The difference between the numbers of the last true and the first true statement divides the number, that is to be found.&lt;br /&gt;5.    The sum of the numbers of the true statements is the number, that is to be found.&lt;br /&gt;6.    This is not the last true statement.&lt;br /&gt;7.    The number of each true statement divides the number, that is to be found.&lt;br /&gt;8.    The number that is to be found is the percentage of true statements.&lt;br /&gt;9.    The number of divisors of the number, that is to be found, (apart from 1 and itself) is greater than the sum of the numbers of the true statements.&lt;br /&gt;10.    There are no three consecutive true statements.&lt;br /&gt;Find the minimal possible number?&lt;br /&gt;Submitted&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The numebr is 420.&lt;br /&gt;&lt;br /&gt;If statement 6 is false, it creates a paradox. Hence, Statement 6 must be true.&lt;br /&gt;&lt;br /&gt;Consider Statement 2:&lt;br /&gt;    If it is true, it must be the first true statement. Otherwise, it creates a paradox.&lt;br /&gt;    If it is false, it must be the second false statement. Otherwise, it creates a paradox.&lt;br /&gt;In both the cases, Statement 1 is false.&lt;br /&gt;&lt;br /&gt;As Statement 1 is false, Statement 9 and Statement 10 both are false i.e. there are three consecutive true statements.&lt;br /&gt;1    2    3    4    5    6    7    8    9    10&lt;br /&gt;False    -    -    -    -    True    -    -    False    False&lt;br /&gt;&lt;br /&gt;Let\'s assume that Statement 3 is false i.e. there are no three consecutive false statements. It means that Statement 2 and Statement 8 must be true, else there will be three consecutive false statements.&lt;br /&gt;1    2    3    4    5    6    7    8    9    10&lt;br /&gt;False    True    False    -    -    True    -    True    False    False&lt;br /&gt;&lt;br /&gt;Also, atleast two of Statements 4, 5 and 7 must be true as there are three consecutive true statements.&lt;br /&gt;&lt;br /&gt;According to Statement 8, the number that is to be found is the percentage of true statements. Hence, number is either 50 or 60. Now if Statement 7 is true, then the number of each true statement divides the number, that is to be found. But 7 and 8 do not divide either 50 or 60. Hence, Statement 7 is false which means that Statement 4 and 5 are true. But Statement 5 contradicts the Statement 8. Hence, our assumption that Statement 3 is false is wrong and Statement 3 is true i.e. there are 3 consecutive false statements which means that Statement 8 is false as there is no other possibilities of 3 consecutive false statements.&lt;br /&gt;&lt;br /&gt;Also, Statement 7 is true as Statement 6 is not the last true statement.&lt;br /&gt;1    2    3    4    5    6    7    8    9    10&lt;br /&gt;False    -    True    -    -    True    True    False    False    False&lt;br /&gt;&lt;br /&gt;According to Statement 7, the number of each true statement divides the number, that is to be found. And according to Statement 5, the sum of the numbers of the true statements is the number, that is to be found. For all possible combinations Statement 5 is false.&lt;br /&gt;&lt;br /&gt;There 3 consecutive true statements. Hence, Statement 2 and Statement 4 are true.&lt;br /&gt;1    2    3    4    5    6    7    8    9    10&lt;br /&gt;False    True    True    True    False    True    True    False    False    False&lt;br /&gt;&lt;br /&gt;Now, the conditions for the number to be found are:&lt;br /&gt;1.    The numebr is divisible by 5 (Statement 4)&lt;br /&gt;2.    The numebr is divisible by 2, 3, 4, 6, 7 (Statement 7)&lt;br /&gt;3.    The number of divisors of the number, that is to be found, (apart from 1 and itself) is not greater than the sum of the numbers of the true statements. (Statement 9)&lt;br /&gt;The minimum possible number is 420.&lt;br /&gt;&lt;br /&gt;The divisors of 420, apart from 1 and itself are 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210. There are total of 22 divisors. Also, the sum of the numbers of the true statements is 22 (2+3+4+6+7=22), which satisfies the third condition.&lt;br /&gt;Ankit and Tejas divided a bag of Apples between them.&lt;br /&gt;&lt;br /&gt;Tejas said, "It's not fair! You have 3 times as many Apples I have." Ankit said, "OK, I will give you one Apple for each year of your age." Tejas replied, "Still not fair. Now, you have twice as many Apples as I have." "Dear, that's fair enough as I am twice older than you.", said Ankit.&lt;br /&gt;&lt;br /&gt;Ankit went to Kitchen to drink water. While Ankit was in Kitchen, Tejas took apples from Ankit's pile equal to Ankit's age.&lt;br /&gt;&lt;br /&gt;Who have more apples now?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;At the end, Ankit and Tejas, both have the same number of apples.&lt;br /&gt;&lt;br /&gt;Let's assume that initially Tejas got N apples and his age is T years. Hence, initially Ankit got 3N apples and his age is 2T years.&lt;br /&gt;Operation    Ankit's Apples    Tejas's Apples&lt;br /&gt;Initially    3N    N&lt;br /&gt;Ankit gave T apples to Tejas&lt;br /&gt;(equals age of Tejas)    3N - T    N + T&lt;br /&gt;Tejas took 2T apples from Ankit's pile&lt;br /&gt;(equals age of Ankit)    3N - 3T    N + 3T&lt;br /&gt;&lt;br /&gt;It is given that after Ankit gave T apples to Tejas, Ankit had twice as many apples as Tejas had.&lt;br /&gt;3N - T = 2*(N + T)&lt;br /&gt;3N - T = 2N + 2T&lt;br /&gt;N = 3T&lt;br /&gt;&lt;br /&gt;From the table, at the end Ankit have (3N - 3T) apples and Tejas have (N + 3T) apples. Substituting N = 3T, we get&lt;br /&gt;Ankit's apples = 3N - 3T = 9T - 3T = 6T&lt;br /&gt;Tejas's apples = N + 3T = 3T + 3T = 6T&lt;br /&gt;&lt;br /&gt;Thus, at the end Ankit and Tejas, both have the same number of apples.&lt;br /&gt;&lt;br /&gt;On evey Sunday Amar, Akbar and Anthony lunch together at Preetam-Da-Dhaba where they order lassi based on following facts.&lt;br /&gt;1.    Unless neither Amar nor Akbar have lassi, Anthony must have it.&lt;br /&gt;2.    If Amar does not have lassi, either Akbar or Anthony or both have it.&lt;br /&gt;3.    Anthony has lassi only if either Amar or Akbar or both have it.&lt;br /&gt;4.    Akbar and Anthony never have lassi together.&lt;br /&gt;Who order(s) lassi?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Amar and Anthony both have lassi whereas Akbar never does.&lt;br /&gt;&lt;br /&gt;Fact (2) can be alternatively stated that "either Amar or Akbar or Anthony must have lassi".&lt;br /&gt;&lt;br /&gt;From Fact (3), it can be infered that either Amar or Akbar must have lassi.&lt;br /&gt;&lt;br /&gt;Now, from Fact (1), it is apparent that Anthony too must have lassi. But according to Fact (4), Akbar cannot have lassi when Anthony does.&lt;br /&gt;    Brain Teaser No : 00191&lt;br /&gt;&lt;br /&gt;Decipher this sentence.&lt;br /&gt;&lt;br /&gt;         B R W Q H L F K W   H J K   Q I B W K  &lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;         Q I C E D W   Z B G   W K K   M I K E&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;         Z B G   Q H S K   Z B G J   K Z K W&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;         B U U   Z B G J   D B H F W.&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Start with ZBG and ZBGJ. It should be either "the/then" or "you/your" combination as they appear more.&lt;br /&gt;&lt;br /&gt;         B R W Q H L F K W   H J K   Q I B W K  &lt;br /&gt;&lt;br /&gt;         o b s t a c l e s   a r e   t h o s e&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;         Q I C E D W   Z B G   W K K   M I K E&lt;br /&gt;&lt;br /&gt;         t h i n g s   y o u   s e e   w h e n&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;         Z B G   Q H S K   Z B G J   K Z K W&lt;br /&gt;&lt;br /&gt;         y o u   t a k e   y o u r   e y e s&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;         B U U   Z B G J   D B H F W.&lt;br /&gt;&lt;br /&gt;         o f f   y o u r   g o a l s.&lt;br /&gt;    Brain Teaser No : 00001&lt;br /&gt;&lt;br /&gt;At what time immediately prior to Six O'clock the hands of the clock are exactly opposite to each other. Give the exact time in hours, minutes and seconds.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;It is obvious that between 5 O'clock and 6 O'clock the hands will not be exactly opposite to each other. It is also obvious that the hands will be opposite to each other just before 5 O'clock. Now to find exact time:&lt;br /&gt;&lt;br /&gt;The hour hand moves 1 degree for every 12 degrees that the minute hand moves. Let the hour hand be X degree away from 5 O'clock. Therefore the minute hand is 12X degree away from 12 O'clock.&lt;br /&gt;&lt;br /&gt;Therefore solving for X&lt;br /&gt;&lt;br /&gt;Angle between minute hand and 12 O'clock + Angle between 12 O'clock and 4 O'clock + Angle between 4 O'clock and hour hand = 180&lt;br /&gt;12X + 120 + (30-X) = 180&lt;br /&gt;11X = 30&lt;br /&gt;Hence X = 30/11 degrees&lt;br /&gt;(hour hand is X degree away from 5 O'clock)&lt;br /&gt;&lt;br /&gt;Now each degree the hour hand moves is 2 minutes.&lt;br /&gt;&lt;br /&gt;Therefore minutes are&lt;br /&gt;= 2 * 30/11&lt;br /&gt;= 60/11&lt;br /&gt;= 5.45 (means 5 minutes 27.16 seconds)&lt;br /&gt;&lt;br /&gt;Therefore the exact time at which the hands are opposite to each other is&lt;br /&gt;= 4 hrs. 54 min. 32.74 seconds&lt;br /&gt;&lt;br /&gt;Ali Baba had four sons, to whom he bequeathed his 39 camels, with the proviso that the legacy be divided in the following way :&lt;br /&gt;&lt;br /&gt;The oldest son was to receive one half the property, the next a quarter, the third an eighth and the youngest one tenth. The four brothers were at a loss as how to divide the inheritance among themselves without cutting up a camel, until a stranger appeared upon the scene.&lt;br /&gt;Dismounting from his camel, he asked if he might help, for he knew just what to do. The brothers gratefully accepted his offer.&lt;br /&gt;&lt;br /&gt;Adding his own camel to Ali Baba's 39, he divided the 40 as per the will. The oldest son received 20, the next 10, the third 5 and the youngest 4. One camel remained : this was his, which he mounted and rode away.&lt;br /&gt;&lt;br /&gt;Scratching their heads in amazement, they started calculating. The oldest thought : is not 20 greater than the half of 39? Someone must have received less than his proper share ! But each brother discovered that he had received more than his due. How is it possible?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;They took their percentages from 40 and not from 39, so they got more than their share.&lt;br /&gt;&lt;br /&gt;The oldest son got 1/2 of 40 = 20 which is 0.5 more&lt;br /&gt;The second son got 1/4 of 40 = 10 which is 0.25 more&lt;br /&gt;The third son got 1/8 of 40 = 5 which is 0.125 more&lt;br /&gt;The youngest son got 1/10 of 40 = 4 which is 0.1 more&lt;br /&gt;&lt;br /&gt;And the stranger got 1/40 of 40 = 1 which is 0.025 more (As he is not supposed to get anything)&lt;br /&gt;&lt;br /&gt;All these fractions add to = 0.5 + 0.25 + 0.125 + 0.1 + 0.025 = 1 which stranger took away.&lt;br /&gt;There is a family party consisting of two fathers, two mothers, two sons, one father-in-law, one mother-in-law, one daughter-in-law, one grandfather, one grandmother and one grandson.&lt;br /&gt;&lt;br /&gt;What is the minimum number of persons required so that this is possible?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;There are total 2 couples and a son. Grandfather and Grand mother, their son and his wife and again their son. So total 5 people.&lt;br /&gt;&lt;br /&gt;Grandfather, Grandmother&lt;br /&gt;|&lt;br /&gt;|&lt;br /&gt;Son, wife&lt;br /&gt;|&lt;br /&gt;|&lt;br /&gt;Son&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;A man went into a fast food restaurant and ate a meal costing Rs. 105, giving the accountant a Rs. 500 note. He kept the change, came back a few minutes later and had some food packed for his girl friend. He gave the accountant a Rs. 100 note and received Rs. 20 in change. Later the bank told the accountant that both the Rs. 500 and the Rs. 100 notes were counterfeit.&lt;br /&gt;&lt;br /&gt;How much money did the restaurant lose? Ignore the profit of the food restaurant.&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;He lost Rs.600&lt;br /&gt;&lt;br /&gt;First time restaurant has given food worth Rs.105 and Rs. 395 change. Similarly second time, food worth Rs.80 and Rs.20 change. Here, we are not considering food restaurant profits.&lt;br /&gt;&lt;br /&gt;               S L I D E&lt;br /&gt;&lt;br /&gt;             -   D E A N&lt;br /&gt;&lt;br /&gt;               ---------&lt;br /&gt;&lt;br /&gt;                 3 6 5 1&lt;br /&gt;Each of seven digits from 0-9 are represented by a different letter above such that the subtraction is true.&lt;br /&gt;&lt;br /&gt;What word represents 3651?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;3651 represents LENS.&lt;br /&gt;&lt;br /&gt;Let's assign possible values to each letter and then use trial-n-error.&lt;br /&gt;&lt;br /&gt;S must be 1.&lt;br /&gt;&lt;br /&gt;Then D (under L) must be greater than 5. If D is 6, then L is 0. But then A must be 0 or 1 which is impossible. Hence, the possible values of D are 7, 8 or 9.&lt;br /&gt;&lt;br /&gt;N must be E + 1. Also, D must be A + 5 as the possible values of D are 7, 8 or 9, D can not be (10+A) + 5.&lt;br /&gt;&lt;br /&gt;Now using trial-n-error, we get S=1, I=2, L=3, A=4, N=5, E=6 and D=9&lt;br /&gt;&lt;br /&gt;         S  L  I  D  E             1  3  2  9  6&lt;br /&gt;&lt;br /&gt;       -    D  E  A  N           -    9  6  4  5&lt;br /&gt;&lt;br /&gt;         --------------            --------------&lt;br /&gt;&lt;br /&gt;            3  6  5  1                L  E  N  S&lt;br /&gt;Hence, 3651 represents LENS.&lt;br /&gt;Adam, Burzin, Clark and Edmund each live in an apartment. Their apartments are arranged in a row numbered 1 to 4 from left to right. Also, one of them is the landlord.&lt;br /&gt;1.    If Clark's apartment is not next to Burzin's apartment, then the landlord is Adam and lives in apartment 1.&lt;br /&gt;2.    If Adam's apartment is right of Clark's apartment, then the landlord is Edmund and lives in apartment 4.&lt;br /&gt;3.    If Burzin's apartment is not next to Edmund's apartment, then the landlord is Clark and lives in apartment 3.&lt;br /&gt;4.    If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin and lives in apartment 2.&lt;br /&gt;Who is the landlord?&lt;br /&gt;&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;Clark is the landlord.&lt;br /&gt;&lt;br /&gt;Assume each statement true, one at a time and see that no other statement is contradicted.&lt;br /&gt;&lt;br /&gt;Let's assume that Statement (1) is true. Then, Adam is the landlord and lives in apartment 1. Also, other three's apartments will be on the right of his apartment - which contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Adam is not the landlord.&lt;br /&gt;&lt;br /&gt;Let's assume that Statement (2) is true. Then, Edmund is the landlord and lives in apartment 4. Also, other three's apartments will be on the left of his apartment - which again contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Edmund is not the landlord either.&lt;br /&gt;&lt;br /&gt;Let's assume that Statement (3) is true. Then, Clark is the landlord and lives in apartment 3. It satisfies all the statements for&lt;br /&gt;(1) Adam - (2) Edmund - (3) Clark - (4) Burzin&lt;br /&gt;&lt;br /&gt;Hence, Clark is the landlord.&lt;br /&gt;&lt;br /&gt;Similarly, you can assume Statement (4) true and find out that it also contradicts.&lt;br /&gt;    Brain Teaser No : 00456&lt;br /&gt;&lt;br /&gt;B, J and P are related to each other.&lt;br /&gt;1.    Among the three are B's legal spouse, J's sibling and P's sister-in-law.&lt;br /&gt;2.    B's legal spouse and J's sibling are of the same sex.&lt;br /&gt;Who is the married man?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;J is the married man.&lt;br /&gt;&lt;br /&gt;Note that a person's sister-in-law may be the wife of that person's brother or the sister of that person's spouse.&lt;br /&gt;&lt;br /&gt;There are 2 cases:&lt;br /&gt;1.    If B's legal spouse is J, then J's sibling must be P and P's sister-in-law must be B.&lt;br /&gt;2.    If B's legal spouse is P, then P's sister-in-law must be J and J's sibling must be B.&lt;br /&gt;It is given that B's legal spouse and J's sibling are of the same sex. Also, it is obvious that P's sister-in-law is female. Then, B's legal spouse and J's sibling both must be males.&lt;br /&gt;&lt;br /&gt;          B's spouse   J's sibling   P's sister-in-law&lt;br /&gt;&lt;br /&gt;           (male)        (male)          (female)&lt;br /&gt;&lt;br /&gt;------------------------------------------------------&lt;br /&gt;&lt;br /&gt;Case I       J             P                B&lt;br /&gt;&lt;br /&gt;Case II      P             B                J&lt;br /&gt;&lt;br /&gt;Case II is not possible as B &amp;amp; P are married to each other and both are male. Hence, J is the married man.&lt;br /&gt;    Brain Teaser No : 00041&lt;br /&gt;&lt;br /&gt;A polygon has 1325 diagonals. How many vertices does it have?&lt;br /&gt;Answer&lt;br /&gt;&lt;br /&gt;The formula to find number of diagonals (D) given total number of vertices or sides (N) is&lt;br /&gt;&lt;br /&gt;                 N * (N - 3)   &lt;br /&gt;&lt;br /&gt;           D  =  -----------      &lt;br /&gt;&lt;br /&gt;                     2&lt;br /&gt;&lt;br /&gt;Using the formula, we get&lt;br /&gt;1325 * 2 = N * (N - 3)&lt;br /&gt;N2 - 3N - 2650 = 0&lt;br /&gt;&lt;br /&gt;Solving the quadratic equation, we get N = 53 or -50&lt;br /&gt;&lt;br /&gt;It is obvious that answer is 5
