### Puzzles Set I

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.

Now the man is supposed to touch any two holes at a time and can do the following.
 Open the closed hole.
 Close the open hole.
 Let the hole be as it is.
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.

How many minimum number of turns does the blindfolded man require to either open all the holes or close all the holes?

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.
Submitted

The blindfolded man requires 5 turns.
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.
3. Check two diagonal holes.
o If one is close, open it and all the holes are open.
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.

4. Check any two adjacent holes.
o If both are open, close both of them. Now, all holes are close.
o If both are close, open both of them. Now, all holes are open.
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.

5. Check any two diagonal holes.
o If both are open, close both of them. Now, all holes are close.
o If both are close, open both of them. Now, all holes are open.

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.

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.

How many persons (including myself) will I need to accomplish this mission?

Total 4 persons (including you) required.

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.
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.
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.
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.
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.

Thus, total 4 persons, including you are required.
At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with the hour hand?

4:21:49.5

Assume that X minutes after 4.00 PM minute hand exactly aligns with and hour hand.

For every minute, minute hand travels 6 degrees.
Hence, for X minutes it will travel 6 * X degrees.

For every minute, hour hand travels 1/2 degrees.
Hence, for X minutes it will travel X/2 degrees.

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,

6 * X = 120 + X/2
12 * X = 240 + X
11 * X = 240
X = 21.8182
X = 21 minutes 49.5 seconds

Hence, at 4:21:49.5 minute hand is exactly aligned with the hour hand.

Substitute digits for the letters to make the following Division true
O U T

-------------

S T E M | D E M I S E

| D M O C

-------------

T U I S

S T E M

----------

Z Z Z E

Z U M M

--------

I S T
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.
Submitted by : Calon

C=0, U=1, S=2, T=3, O=4, M=5, I=6, Z=7, E=8, D=9

It is obvious that U=1 (as U*STEM=STEM) and C=0 (as I-C=I).

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).

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.
O U T 4 1 3

------------- -------------

S T E M | D E M I S E 2 3 8 5 | 9 8 5 6 2 8

| D M O C | 9 5 4 0

------------- -------------

T U I S 3 1 6 2

S T E M 2 3 8 5

---------- ----------

Z Z Z E 7 7 7 8

Z U M M 7 1 5 5

-------- --------

I S T 6 2 3
Also, when arranged from 0 to 9, it spells CUSTOMIZED.
Brain Teaser No : 00015

In the town called Alibaug, the following facts are true:
 No two inhabitants have exactly the same number of hairs.
 No inhabitants has exactly 2025 hairs.
 There are more inhabitants than there are hairs on the head of any one inhabitants.
What is the largest possible number of the inhabitants of Alibaug?

2025

It is given that no inhabitants have exactly 2025 hairs. Hence there are 2025 inhabitants with 0 to 2024 hairs in the head.

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.
There are four groups of Mangoes, Apples and Bananas as follows:
Group I : 1 Mango, 1 Apples and 1 Banana
Group II : 1 Mango, 5 Apples and 7 Bananas
Group III : 1 Mango, 7 Apples and 10 Bananas
Group IV : 9 Mango, 23 Apples and 30 Bananas

Group II costs Rs 300 and Group III costs Rs 390.

Can you tell how much does Group I and Group IV cost?

Group I costs Rs 120 and Group IV costs Rs 1710

Assume that the values of one mango, one apple and one banana are M, A and B respectively.

From Group II : M + 5A + 7B = 300
From Group III : M + 7A + 10B = 390

Subtracting above to equations : 2A + 3B = 90

For Group I :
= M + A + B
= (M + 5A + 7B) - (4A + 6B)
= (M + 5A + 7B) - 2(2A + 3B)
= 300 - 2(90)
= 300 - 180
= 120

Similarly, for Group IV :
= 9M + 23A + 30B
= 9(M + 5A + 7B) - (22A + 33B)
= 9(M + 5A + 7B) - 11(2A + 3B)
= 9(300) - 11(90)
= 2700 - 990
= 1710

Thus, Group I costs Rs 120 and Group IV costs Rs 1710.
Tic-Tac-Toe is being played. One 'X' has been placed in one of the corners. No 'O' has been placed yet.

Where does the player that is playing 'O' has to put his first 'O' so that 'X' doesn't win?

"O" should be placed in the center.

Let's number the positions as:
1 | 2 | 3

---------

4 | 5 | 6

---------

7 | 8 | 9
It is given that "X" is placed in one of the corner position. Let's assume that its at position 1.

Now, let's take each position one by one.
 If "O" is placed in position 2, "X" can always win by choosing position 4, 5 or 7.
 If "O" is placed in position 3, "X" can always win by choosing position 4, 7 or 9.
 If "O" is placed in position 4, "X" can always win by choosing position 2, 3 or 5.
 If "O" is placed in position 6, "X" can always win by choosing position 3, 5 or 7.
 If "O" is placed in position 7, "X" can always win by choosing position 2, 3 or 9.
 If "O" is placed in position 8, "X" can always win by choosing position 3, 5 or 7.
 If "O" is placed in position 9, "X" can always win by choosing position 3, or 7.
If "O" is placed in position 5 i.e. center position, "X" can't win unless "O" does something foolish ;))

Hence, "O" should be placed in the center.

Amit, Bhavin, Himanshu and Rakesh are sitting around a table.
 The Electonics Engineer is sitting to the left of the Mechanical Engineer.
 Amit is sitting opposite to Computer Engineer.
 Himanshu likes to play Computer Games.
 Bhavin is sitting to the right of the Chemical Engineer.
Can you figure out everyone's profession?

Amit is the Mechanical Engineer. Bhavin is the Computer Engineer. Himanshu and Rakesh are either Chemical Engineer or Elecronics Engineer.

Amit and Bhavin are sitting opposite to each other. Whereas Chemical Engineer and Elecronics Engineer are sitting opposite to each other.

We cannot find out who is Chemical Engineer and Elecronics Engineer as data provided is not sufficient

Five friends with surname Batliwala, Pocketwala, Talawala, Chunawala and Natakwala have their first name and middle name as follow.
1. Four of them have a first and middle name of Paresh.
2. Three of them have a first and middle name of Kamlesh.
3. Two of them have a first and middle name of Naresh.
4. One of them have a first and middle name of Elesh.
5. Pocketwala and Talawala, either both are named Kamlesh or neither is named Kamlesh.
6. Either Batliwala and Pocketwala both are named Naresh or Talawala and Chunawala both are named Naresh.
7. Chunawala and Natakwala are not both named Paresh.
Who is named Elesh?

Pocketwala is named Elesh.

From (1) and (7), it is clear that Batliwala, Pocketwala and Talawala are named Paresh.

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.

Now it is clear that Talawala and Chunawala are named Naresh. Also, Pocketwala is named Elesh.

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.

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.

What portion of the total trip did Mr. Wagle sleep?

Mr. wagle slept through half his trip.

Let's draw a timeline. Picture the bus route on a line showen below:

---------------- ________ -------- ________________

Start 1/3 1/2 2/3 End

----- shows time for which Mr. Wagle was not sleeping

_____ shows time for which Mr. Wagle was sleeping

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.

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.

= (1/2 - 1/3) + (1 - 2/3)
= 1/6 + 1/3
= 1/2

Hence, Mr. wagle slept through half his trip.
Brain Teaser No : 00068

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.

In complete darkness, how many socks would you need to pull out to get a matching pair of the same color?
4 If you don't agree, try it yourself!
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.

How many do you have to grab to be sure you have 2 of the same?

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.

How many do you have to grab to be sure you have 2 of the same?
If you select 4 Jelly beans you are guarenteed that you will have 2 that are the same color.

There are 70 employees working with BrainVista of which 30 are females. Also,
 30 employees are married
 24 employees are above 25 years of age
 19 married employees are above 25 years, of which 7 are males
 12 males are above 25 years of age
 15 males are married.
How many unmarried females are there and how many of them are above 25?

15 unmarried females & none are above 25 years of age.

Simply put all given information into the table structure and you will get the answer.
Married Unmarried
Below 25 Above 25 Below 25 Above 25
Female 3 12 15 0
Male 8 7 20 5

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.

Find the number.

65292

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)

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)

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.
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%.

How much did she gain or lose in the entire transaction?

She lost Rs 666.67

Consider the first stamp. She mades 20% on it after selling it for Rs 8000.

So the original price of first stamp is
= (8000 * 100) / 80
= Rs 6666.67

Similarly, consider second stamp. She lost 20% on it after selling it for Rs 8000

So the original price of second stamp is
= (8000 * 100) / 80
= Rs 10000

Total buying price of two stamps
= Rs 6666.67 + Rs 10000
= Rs 16666.67

Total selling price of two stamps
= Rs 8000 + Rs 8000
= Rs 16000

Hence, she lost Rs 666.67

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.

Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions.

By how many cm. will the thread be separated from the earth's surface?

The cicumference of the earth is
= 2 * PI * r
= 2 * PI * 6400 km
= 2 * PI * 6400 * 1000 m
= 2 * PI * 6400 * 1000 * 100 cm
= 1280000000 * PI cm

where r = radius of the earth, PI = 3.141592654

Hence, the length of the thread is = 1280000000 * PI cm

Now length of the thread is increasd by 12 cm. So the new length is = (1280000000 * PI) + 12 cm

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

Assume that radius of the outer circle is R cm
Therefore,
2 * PI * R = (1280000000 * PI) + 12 cm

Solving above equation, R = 640000001.908 cm
Radius of the earth is r = 640000000 cm

Hence, the thread will be separatedfrom the earth by
= R - r cm
= 640000001.908 - 640000000
= 1.908 cm
Scientist decided to do a study on the population growth of rabbits. Inside a controlled environment, 1000 rabbits were placed.

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.

If Z is a positive variable, how many rabbits would be there at the beginning of the 11th year?

At the beginning of the 11th year, there would be 1,024,000 rabbits.

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)

Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further simplified as 1000*Z*Z*Z*Z

Also, it is given that at the end of 6 months, there were 1000Z rabbits.

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)

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.
tted
A class of 100 students. 24 of them are girls and 32 are not. Which base am I using?

Let the base be X.

Therefore
(X*X + X*0 + 0) = (2*X +4) + (3*X + 2)
X*X = 5*X + 6
X*X - 5*X -6 = 0
(X-6)(X+1) = 0

Therefore base is 6

A man is stranded on a desert island. All he has to drink is a 20oz bottle of sprite.

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...

By the time all the sprite is gone, how much water has he drunk?

The man drunk 190oz of water.

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.

Out of that 210oz, 20oz is the sprite which he had initially. Hence, he must have drunk 190oz of water.ed
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

How many such expressions can you make?
Submitted

There are 5 such expressions.

99 + (9/9) = 100

(99/.99) = 100

(9/.9) * (9/.9) = 100

((9*9) + 9)/.9 = 100

(99-9)/.9 = 100
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 ).

Which one will be closer to Paris when they meet?
They will both be the same distance from Paris when they meet!!!

12 members were present at a board meeting. Each member shook hands with all of the other members before & after the meeting.

How many hand shakes were there?

132

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, & so on.

66 hand shakes took place before & 66 after the meeting, for a total of 132.
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.

There is an amazing mathematical relationship exists among the names of the planet.

The tought process is initially to find planets such that the total number of alphabets in them is 10.

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.

S A T U R N

U R A N U S

V E N U S

+ M A R S

--------------

N E P T U N E

Now the only possible value for N is 1. By finding the value for S, we can reach the result:

3 5 8 6 9 1

6 9 5 1 6 3

2 0 1 6 3

+ 4 5 9 3

--------------

1 0 7 8 6 1 0

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.

How many pears do you have?

None

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.

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