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.
Abu sees the number 1 and two even numbers.
Babu and Calu can see three different numbers each.
Dabu sees number 2 twice and he can't remember the third number.
What number is face down on the table?
Answer
Number 3 is face down on the table.
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.
Now everything else is simple. (see the following diagram)
Dabu Abu
1
3 2 2
1
Calu Babu
Thus, the number hidden from the view is number 3 and hence the answer.
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?
Answer
52 / 2703
There are two cases to be considered.
CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B
Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 2/53
So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)
CASE 2 : King of Hearts is not drawn from Pack A
Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 1/53
So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)
Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378
How many possible combinations are there in a 3x3x3 rubics cube?
In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)?
How many for a 4x4x4 cube?
Submitted
Answer
There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.
Let's consider 3x3x3 Rubics first.
There are 8 corner cubes, which can be arranged in 8! ways.
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)
Similarly, 12 edge cubes can be arranged in 12! ways.
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)
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.
Total different possible combinations are
= [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19
Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45
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.
Brain Teaser No : 00528
Substitute digits for the letters to make the following relation true.
N E V E R
L E A V E
+ M E
-----------------
A L O N E
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.
Answer
A tough one!!!
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
solve it. Now use trial-n-error method.
N E V E R 2 1 4 1 9
L E A V E 3 1 5 4 1
+ M E + 6 1
----------------- -----------------
A L O N E 5 3 0 2 1
There are 20 people in your applicant pool, including 5 pairs of identical twins.
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 ;))
SubmAnswer
The probability to hire 5 people with at least 1 pair of identical twins is 25.28%
5 people from the 20 people can be hired in 20C5 = 15504 ways.
Now, divide 20 people into two groups of 10 people each :
G1 - with all twins
G2 - with all people other than twins
Let's find out all possible ways to hire 5 people without a single pair of indentical twins.
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
0 5 10C0 10C5 252
1 4 10C1 10C4 2100
2 3 10C2 * 8/9 10C3 4800
3 2 10C3 * 8/9 * 6/8 10C2 3600
4 1 10C4 * 8/9 * 6/8 * 4/7 10C1 800
5 0 10C5 * 8/9 * 6/8 * 4/7 * 2/6 10C0 32
Total 11584
Thus, total possible ways to hire 5 people without a single pair of indentical twins = 11584 ways
So, total possible ways to hire 5 people with at least a single pair of indentical twins = 15504 - 11584 = 3920 ways
Hence, the probability to hire 5 people with at least a single pair of indentical twins
= 3920/15504
= 245/969
= 0.2528
= 25.28%itted
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
1. is a multiple of 5, it is a number between 1 and 19
2. is not a multiple of 8, it is a number between 20 and 29
3. is not a multiple of 10, it is a number between 30 and 39
How many Eggs does Veeru have in his bucket?
Answer
32 eggs
Let's apply all 3 condition separately and put all possible numbers together.
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)
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)
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)
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.
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Answer (14) BrainV
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.
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."
On that a person wearing the white suit replied, "What difference does that make?"
Can you tell what color suit each of the three persons had on?
Answer
Mr. Grey is wearing Black suit.
Mr. White is wearing Grey suit.
Mr. Black is wearing White suit.
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.
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.
And, Mr. Black must be wearing white suit.
Substitute numbers for the letters so that the following mathematical expressions are correct.
ABC DEF GHI
--- = IE --- = IE --- = IE
3 6 9
Note that the same number must be used for the same letter whenever it appears.
Answer
A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
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)
Now, consider DEF = 6 * IE. Out of three short-listed values, only 73 satisfies the equation. Also, ABC = 3 * IE is satisfied by 73.
Hence, A=2, B=1, C=9, D=4, E=3, F=8, G=6, H=5, I=7
219 438 657
--- = 73 --- = 73 --- = 73
3 6 9
Brain Teaser No : 00374
A, B, C and D are related to each other.
One of the four is the opposite sex from each of the other three.
D is A's brother or only daughter.
A or B is C's only son.
B or C is D's sister.
Answer
A, B & D are males; C is female. B is C's only son. A & D are C's brothers.
A(male) --- C(female) --- D(male)
|
|
B(male)
Work out which relation can hold and discard the contradictory options.
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.
From (4), let's say that B is D's sister i.e. B is Female.
From (3), A is C's only son i.e. A is Male.
But D is A's brother which means that A is not C's only son. Hence, our assumption was wrong.
Thus, C is D's sister i.e. C is Female. And B must be C's only son.
Now it is clear that D & 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 & D's brother.How are they related to each other?
Dr. DoLittle always goes walking to the clinic and takes the same time while going and while coming back. One day he noticed something.
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.
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.
How much time does Dr. DoLittle take to reach home from the clinic? Give the minimal possible answer.
Answer
32 minutes 43.6 seconds
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.
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.
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.
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
Hence, Dr. DoLittle takes 32 minutes and 43.6 seconds to reach home from the clinic.
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.
Mr. Haani once travelled by SlowRun Express from Mumbai to Bangalore. How many SlowRun Express did he cross during his journey?
Answer
Mr. Haani crossed 7 SlowRun Expresses during his journey.
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.
Also, he would have crossed the last train just before reaching Bangalore on Saturday.
Thus, Mr. Haani must have crossed 7 SlowRun Expresses during his journey.
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.
1. Miss Shalaka's work requires her to speak on the phone frequently throughout the day.
2. Miss Shudha prefers cabin number 5 as 5 is her lucky number.
3. Mr. Shaan and Mr. Sharma often talk to each other during their work and prefers to have adjacent cabins.
4. Mr. Sinha, Mr. Shaan and Mr. Solanki all smoke. Miss Shudha is allergic to smoke and must have non-smokers adjacent to her.
5. Mr. Solanki needs silence during work.
Can you tell the cabin numbers of each of them?
Answer
The cabins from left to right (1-6) are of Mr. Solanki, Mr. Sinha, Mr. Shaan, Mr. Sharma, Miss Shudha and Miss Shalaka.
From (2), cabin number 5 is assigned to Miss Shudha.
As Miss Shudha is allergic to smoke and Mr. Sinha, Mr. Shaan & 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.
From (3), Mr. Shaan must be in cabin 3 and Mr. Sharma must be in cabin 4. Thus, Miss Shalaka is in cabin 6.
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.
Thus, the cabins numbers are
1# Mr. Solanki,
2# Mr. Sinha,
3# Mr. Shaan,
4# Mr. Sharma,
5# Miss Shudha,
6# Miss Shalaka
SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.
When it snows, morning service on line E is delayed.
When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.
When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.
When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.
When service on line A is delayed or cancelled, service on line I is also delayed.
When service on line Z is delayed or cancelled, service on line E is also delayed.
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?
SkyFi city is served by 6 subway lines - A, E, I, O, U and Z.
When it snows, morning service on line E is delayed.
When it rains or snows, service on the lines A, U and Z is delayed both morning and afternoon.
When the temperature drops below 20 C, afternoon service is cancelled on either line A or line O, but not both.
When the temperature rises above 40 C, afternoon service is cancelled on either line I or line Z, but not both.
When service on line A is delayed or cancelled, service on line I is also delayed.
When service on line Z is delayed or cancelled, service on line E is also delayed.
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?
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?
Answer
It is given that
2 wixsomes = 3 changs
8 wixsomes = 12 changs ----- (I)
Also, given that
4 changs = 1 plut
12 changs = 3 plutes
8 wixsomes = 3 plutes ----- From (I)
Therefore,
6 plutes = 16 wixsomes
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?
Answer
Assume that number of boys graduated from City High School = B
Therefore, number of girls graduated from City High School = 2*B
It is given that 3/4 of the girls and 5/6 of the boys went to college immediately after graduation.
Hence, total students went to college
= (3/4)(2*B) + (5/6)(B)
= B * (3/2 + 5/6)
= (7/3)B
Fraction of the graduates that year went to college immediately after graduation
= [(7/3)B] / [3*B]
= 7/9
Therefore, the answer is 7/9
A mule and a donkey were carrying full sacks on their backs.
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."
How many sacks were each of them carrying? Give the minimal possible answer.
SubmittAnswer
The mule was carrying 5 sacks and the donkey was carrying 7 sacks.
Let's assume that the mule was carrying M sacks and the donkey was carrying D sacks.
As the donkey told the mule, "If you gave me one of your sacks I'd have double what you have."
D + 1 = 2 * (M-1)
D + 1 = 2M - 2
D = 2M - 3
The donkey also said, "If I give you one of my sacks we'd have an even amount."
D - 1 = M + 1
D = M + 2
Comparing both the equations,
2M - 3 = M + 2
M = 5
Substituting M=5 in any of above equation, we get D=7
Hence, the mule was carrying 5 sacks and the donkey was carrying 7 sacks.
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.
Who came in first?
Submitted
Answer
Person A came in first.
Let's assume that the distance between start and the point is D miles.
Total time taken by Person A to finish
= (D/20) + (D/20)
= D/10
= 0.1D
Total time taken by Person B to finish
= (D/10) + (D/30)
= 2D/15
= 0.1333D
Thus, Person A is the Winner.
Alternatively (if you don't like mathematics ;)), analyse the situation as follow:
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.
Thus, Person A wins the race and by that time Person B covers only half the distance, no matter how far the point is!!!
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.
How much of the pizza would he have eaten during the week?
Submitted
Answer
Mark would have ate 127/128 (99.22%) of the pizza during the week.
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.
Total pizza Mark ate during the week is
= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
= 127/128
= 99.22% of the original pizza
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."
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.
How many members did the Club originally have?
Answer
The Club originally had 24 members.
Assume that there were initially N members.
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,
4 * (3120 / N) = 26 * (N - 4)
12480 = 26N2 - 104N
26N2 - 104N - 12480 = 0
Solving the quadratic equation we get N=24.
Hence, the Club originally had 24 members.
Brain Teaser No : 00206
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.
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?
Answer
The tank will be full in 16 minutes.
In one minute,
pipe A can fill 1/30 part of the tank.
pipe B can fill 1/24 part of the tank.
pipe C can empty 1/80 part of the tank.
Thus, the net water level in one minute is
= 1/30 + 1/24 - 1/80
= 15/240 part of the tank
Hence, the tank will be full in 240/15 i.e. 16 minutes.
A rich old Arab has three sons. When he died, he willed his 17 camels to the sons, to be divided as follows:
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.
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.
How did he do it?
Answer
The old man temporarily added his camel to the 17, making a total of 18 camels.
First son got 1/2 of it = 9
Second son got 1/3 of it = 6
Third son got 1/9 of it = 2
For a total of 17. He then takes his camel back and rides away......
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?"
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."
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."
Answer
The answer is 9 years.
First you need to find all the possible sets of three numbers that when multiplied equals 36:
1 1 36
1 2 18
1 3 12
1 4 9
1 6 6
2 2 9
2 3 6
3 3 4
Then you add the numbers together to find the sum
1 1 36 = 38
1 2 18 = 21
1 3 12 = 16
1 4 9 = 14
1 6 6 = 13
2 2 9 = 13
2 3 6 = 11
3 3 4 = 10
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.
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.
Therefore, the answer is 9.
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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.
There is one statement written on the cover of each box.
* Red Box: Both, a red box and a blue box contain Rs. 10000 each.
* Green Box: Both, a green box and a red box contain Rs. 25000 each.
* Blue Box: Both, a blue box and a green box contain Rs. 15000 each.
Only one of the above 3 statements is true and the corresponding box contains the maximum amount.
Can you tell which box contains the maximum amount and how much?
Answer
Blue box contains the maximum amount Rs. 40000
As it is given that only one of the given 3 statements is true; assume in turn, each statement to be true & the other 2 false and check whether the corresponding box contains the maximum amount.
Let's assume that the statement on the Blue box is true. Thus, the given 3 statements can be interpreted as
* Atmost one, a red box or a blue box contains Rs. 10000.
* Atmost one, a green box or a red box contains Rs. 25000.
* Both, a blue box and a green box contain Rs. 15000 each.
Going through all possible combinations, we can conclude that
Red Box : Rs. 10000 + Rs. 25000 = Rs. 35000
Green Box : Rs. 10000 + Rs. 15000 = Rs. 25000
Blue Box : Rs. 15000 + Rs. 25000 = Rs. 40000
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.
Sachin, Dravid and Ganguly played in a Cricket match between India and England.
None of them scored more than 99 runs.
If you add the digits of the runs scored by Sachin to his own score, you will get the runs scored by Dravid.
If you reverse the digits of the runs scored by Dravid, you will get the runs scored by Ganguly.
The total runs scored by them is 240.
Can you figure out their individual scores?
Answer
Sachin, Dravid and Ganguly scored 75, 87 and 78 respectively.
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.
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.
Now, use trial-n-error method to solve the teaser.
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.
1. The first person is a woman.
2. Carlos wants to send an overnight package.
3. Lau is just ahead of Pimentelli who is the same sex as Lau.
4. Gianni is two places ahead of the person who wants to buy stamps.
5. Knutson - who is the opposite sex than Rendler - isn't the person who wanted to complain about a mail carrier.
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.
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.
8. The person who wants to send books fourth class is just behind a person of the same sex.
9. Mary is just behind a person who wants to send an insured package.
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.
11. Anthony isn't two places behind the who wants to pickup a registered letter.
12. Toriseza is two places ahead of the person who wants to pick up a package.
13. Knutson isn't just ahead of the person who wants to send an item parcel post.
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.
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Answer (8)
Answer
A very TOUGH puzzle !!!
POS FIRST NAME LAST NAME BUSINESS
1 Sachi Loti • Fill Out a Change-of-Address Form
• Add Postage to Meter
2 Gianni Lau • Pick Up a Registered Letter
• Send an Item Parcel Post
3 Carlos Pimentelli • Overnight Package
• Send Airmail to Tibet
4 Donna Toriseza • Buy Stamps
• Send an Insured Package
5 Mary Knutson • Buy a Money Order
• Send Books fourth Class
6 Anthony Rendler • Complain About a Mail Carrier
• Pick Up a Package
Brain Teaser No : 00164
Substitute digits for the letters to make the following relation true.
W O R L D
+ T R A D E
-------------
C E N T E R
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.
Answer
A tough one.
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.
W O R L D 5 3 6 8 4
+ T R A D E + 7 6 0 4 2
------------ ------------
C E N T E R 1 2 9 7 2 6
Brain Teaser No : 00107
If you look at a clock and the time is 3:15.
What is the angle between the hour and the minute hands? ( The answer to this is not zero!)
Answer
7.5 degrees
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.
The angle between two adjacent digits is 360/12 = 30 degrees.
Hence 1/4 of it is 7.5 degrees.
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.
How did the son place all the apples among the 10 boxes, given that any number of apples can be put in one box.
Answer
1, 2, 4, 8, 16, 32, 64, 128, 256, 489
Let's start from scratch.
The apple vandor can ask for only 1 apple, so one box must contain 1 apple.
He can ask for 2 apples, so one box must contain 2 apples.
He can ask for 3 apples, in that case box one and box two will add up to 3.
He can ask for 4 apples, so one box i.e. third box must contain 4 apples.
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.
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.
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.
Brain Teaser No : 00261
The letters P, Q, R, S, T, U and V, not necessarily in that order represents seven consecutive integers from 22 to 33.
U is as much less than Q as R is greater than S.
V is greater than U.
Q is the middle term.
P is 3 greater than S.
Can you find the sequence of letters from the lowest value to the highest value?
Answer
The sequence of letters from the lowest value to the highest value is TUSQRPV.
From (3), Q is the middle term.
___ ___ ___ _Q_ ___ ___ ___
From (4), there must be exactly 2 numbers between P and S which gives two possible positions.
[1] ___ _S_ ___ _Q_ _P_ ___ ___
[2] ___ ___ _S_ _Q_ ___ _P_ ___
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.
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.
[2] ___ _U_ _S_ _Q_ _R_ _P_ ___
From (2) V must be the highest and the remaining T must be the lowest number.
_T_ _U_ _S_ _Q_ _R_ _P_ _V_
Thus, the sequence of letters from the lowest value to the highest value is TUSQRPV.
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