1. 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

2. 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.

Answer :  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.

3. 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/-    How much did he have with him in the begining?

Answer :   Rs. 250/-

Assume that initially he had Rs. X
He spent 1/3 for cloths =. (1/3) * X
Remaining money = (2/3) * X

He spent 1/5 of remaining money for food = (1/5) * (2/3) * X = (2/15) * X
Remaining money = (2/3) * X - (2/15) * X = (8/15) * X

Again, he spent 1/4 of remaining maoney for travel = (1/4) * (8/15) * X = (2/15) * X
Remaining money = (8/15) * X - (2/15) * X = (6/15) * X

But after spending for travel he is left with Rs. 100/- So
(6/15) * X = 100
X = 250

4. 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 ?

Answer :

There are total 450 rooms.

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

So the probability is 90/450 i.e. 1/5 or 0.20

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.

You're not allowed to lift your pen.

Note: Don't be confined by the dimensions of the square.

5. If you take a marker & 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, & never trace anywhere on the cube, except for the corners & edges ?

Answer :  9

To verify this, you can make a drawing of a cube, & number each of its 12 edges. Then, always starting from 1 corner & 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube.

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.

6. 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 

7. 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. Find the tractors each originally had?

Answer :

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.

It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & 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)

Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)

Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)

Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.

8. 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?

Given that each pawn is identical and each rook, knight and bishop is identical to its pair.

Answer :    6,48,64,800 ways

There are total 16 pieces which can be arranged on 16 places in 16P16 = 16! ways.
(16! = 16 * 15 * 14 * 13 * 12 * ..... * 3 * 2 * 1)

But, there are some duplicate combinations because of identical pieces.

There are 8 identical pawn, which can be arranged in 8P8 = 8! ways.

Similarly there are 2 identical rooks, 2 identical knights and 2 identical bishops. Each can be arranged in 2P2 = 2! ways.

Hence, the require answer is
= (16!) / (8! * 2! * 2! * 2!)
= 6,48,64,800

9. 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 ?

Answer :

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.

Take 8 coins and weigh 4 against 4.

If both are not equal, goto step 2

If both are equal, goto step 3

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.

If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.

If both are equal, L4 is the odd coin and is lighter.

If L2 is light, L2 is the odd coin and is lighter.

If L3 is light, L3 is the odd coin and is lighter.

If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2

If both are equal, there is some error.

If H1 is heavy, H1 is the odd coin and is heavier.

If H2 is heavy, H2 is the odd coin and is heavier.

If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4

If both are equal, L1 is the odd coin and is lighter.

If H3 is heavy, H3 is the odd coin and is heavier.

If H4 is heavy, H4 is the odd coin and is heavier.

The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.

If both are equal, there is some error.

If X is heavy, X is the odd coin and is heavier.

If X is light, X is the odd coin and is lighter.

10. 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?

Answer :

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