Mind blowing puzzles

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.

Answer

18

Assume that initial there were 3*X bullets.

So they got X bullets each after division.

All of them shot 4 bullets. So now they have (X - 4) bullets each.

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

Therefore, the equation is
3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X=6

Therefore the total bullets before division is = 3 * X = 18

Find sum of digits of D.

Let
A= 19991999
B = sum of digits of A
C = sum of digits of B
D = sum of digits of C

(HINT: A = B = C = D (mod 9))

Answer

the sum of the digits od D is 1.

Let E = sum of digits of D.

It follows from the hint that A = E (mod 9)

Consider,

A = 19991999

< 20002000

= 22000 * 10002000

= 1024200 * 106000

< 10800 * 106000

= 106800

i.e. A < 106800

i.e. B < 6800 * 9 = 61200

i.e. C < 5 * 9 = 45

i.e. D < 2 * 9 = 18

i.e. E <= 9

i.e. E is a single digit number.

Also,

1999 = 1 (mod 9)

so 19991999 = 1 (mod 9)

Therefore we conclude that E=1.
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.

In the mean time the whole platoon has moved ahead by 50m.

The question is how much distance did the last person cover in that time. Assuming that

he ran the whole distance with uniform speed.
Submitted
Answer

The last person covered 120.71 meters.

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.

Let's assume that when the last person reached the first person, the platoon
moved X meters forward.

Thus, while moving forward the last person moved (50+X) meters whereas the
platoon moved X meters.

Similarly, while moving back the last person moved [50-(50-X)] X meters
whereas the platoon moved (50-X) meters.

Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X

Solving, X=35.355 meters

Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters

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.

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.

One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and
another is a Doctor.

• If the Doctor is a male, then the Engineer is a male.
• If the Engineer is younger than the Doctor, then the Engineer and the
Doctor are not blood relatives.
• If the Engineer is a female, then she and the Doctor are blood relatives.

Can you tell who is the Doctor and the Engineer?
Answer

Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.

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.

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.

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.

Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not
necessarily in the same order.

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.

Who is married to Cam?

Answer

Carrie is married to Cam.

"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."

It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to
Tina.

As Cam does not play bridge, Billy's husband must be Laurie.

Hence, Carrie is married to Cam.

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.

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?

Answer

The sign-maker will need 192 zeroes.

Divide 1000 building numbers into groups of 100 each as follow:
(1..100), (101..200), (201..300), ....... (901..1000)

For the first group, sign-maker will need 11 zeroes.
For group numbers 2 to 9, he will require 20 zeroes each.
And for group number 10, he will require 21 zeroes.

The total numbers of zeroes required are
= 11 + 8*20 + 21
= 11 + 160 + 21
= 192

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.

1. Take 8 coins and weigh 4 against 4.
o If both are not equal, goto step 2
o If both are equal, goto step 3

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.

o 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.

o 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.

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

3. The remaining coin X is the odd one. Weigh X against the anyone coin
used in initial weighing.
o If both are equal, there is some error.
o If X is heavy, X is the odd coin and is heavier.
o If X is light, X is the odd coin and is lighter.

In a sports contest there were m medals awarded on n successive days (n > 1).

1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were
awarded.
2. On the second day 2 medals and 1/7 of the now remaining medals was
awarded; and so on.
3. On the nth and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded
altogether?
Answer

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler
method, do submit it.

A number of 9 digits has the following properties:

• 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<=n<=9.
• Each digit in the number is different i.e. no digits are repeated.

• The digit 0 does not occur in the number i.e. it is comprised only of the
digits 1-9 in some order.

Find the number.

Answer

The answer is 381654729

One way to solve it is Trial-&-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.

The other way to solve this problem is by writing a computer program that
systematically tries all possibilities

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.

What part of the contents of the container is left at the end of the second day?

Answer

Assume that contents of the container is X

On the first day 1/3rd is evaporated.
(1 - 1/3) of X is remaining i.e. (2/3)X

On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X

Hence 1/6th of the contents of the container is remaining

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.

For how long did Vipul study in candle light?

Answer

Vipul studied for 3 hours in candle light.

Assume that the initial lenght of both the candle was L and Vipul studied for X
hours.

In X hours, total thick candle burnt = XL/6
In X hours, total thin candle burnt = XL/4

After X hours, total thick candle remaining = L - XL/6
After X hours, total thin candle remaining = L - XL/4

Also, it is given that the thick candle was twice as long as the thin one when he
finally went to sleep.
(L - XL/6) = 2(L - XL/4)
(6 - X)/6 = (4 - X)/2
(6 - X) = 3*(4 - X)
6 - X = 12 - 3X
2X = 6
X=3

Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.

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, & so on.

How much would you have earned with this business after 50 years (assuming
there are exactly 365 days in every year)?

Answer

Rs.333,062,500

To begin with, you want to know the total number of days: 365 x 50 = 18250.

By experimentation, the following formula can be discovered, & 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, & pair them up with the other
half in the following way: day 1 with day 18250, day 2 with day 18249, & so on, &
you will see that if you add these pairs together, they always equal Rs.36500.

Multiply this number by the total number of pairs (9125), & you have the amount
you would have earned in 50 years.

Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11.....upto 18250
terms)
A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, &
now his salary is Rs. 22702.68

What was his salary to begin with?
Answer

Rs.22176

Assume his salary was Rs. X

He earns 5% raise. So his salary is (105*X)/100

A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100)
which is Rs. 22702.68

Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68
X = 22176
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.

Answer

66 seconds

It is given that the time between first and last ticks at 6'o is 30 seconds.
Total time gaps between first and last ticks at 6'o = 5
(i.e. between 1 & 2, 2 & 3, 3 & 4, 4 & 5 and 5 & 6)

So time gap between two ticks = 30/5 = 6 seconds.

Now, total time gaps between first and last ticks at 12'o = 11
Therefore time taken for 12 ticks = 11 * 6 = 66 seconds (and not 60 seconds)

500 men are arranged in an array of 10 rows and 50 columns according to their
heights.

Tallest among each row of all are asked to come out. And the shortest among
them is A.

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.

Now who is taller A or B ?
Answer

No one is taller, both are same as A and B are the same person.

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

Now, both A and B are the person with height 50. Hence both are same.

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.

How many members are there in Mr. Mehta's family? Give minimal possible
answer.
Answer

There are 7 members in Mr. Mehta's family. Mother & Father of Mr. Mehta, Mr. &
Mrs. Mehta, his son and two daughters.

Mother & Father of Mr. Mehta
|
|
Mr. & Mrs. Mehta
|
|
One Son & Two Daughters

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."

The soldier could make only one statement. He made that statement and went
free. What did he say?

Answer

The soldier said, "You will put me to the sword."

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 !!!

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.

After spending 20 paise, the person counts the money. And to his surprise, he
has double the amount he wanted to withdraw.

Find X and Y. (1 Rupee = 100 Paise)

Answer

As given, the person wanted to withdraw 100X + Y paise.

But he got 100Y + X paise.

After spending 20 paise, he has double the amount he wanted to withdraw.
Hence, the equation is

2 * (100X + Y) = 100Y + X - 20

200X + 2Y = 100Y +X - 20

199X - 98Y = -20

98Y - 199X = 20

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 & 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

Case I : Y=2X
Solving two equations simultaneously
98Y - 199X = 20
Y - 2X = 0
We get X = - 20/3 & Y = - 40/2

Case II : Y=2X+1
98Y - 199X = 20
Y - 2X = 1
We get X = 26 & Y = 53

Now, its obvious that he wanted to withdraw Rs. 26.53

The game of Tic-Tac-Toe is being played between two players. Only the last
mark to be placed in the game as shown.

Who will win the game, O or X? Can you tell which was the sixth mark and at
which position? Do explain your answer.

Assume that both the players are intelligent enough.

Answer

O will win the game. The sixth mark was X in square 9.

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.

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.

Hence, the sixth mark must be X placed in square 9. And the seventh mark will
be O. Thus O will win the game.

At the Party:

1. There were 9 men and children.
2. There were 2 more women than children.
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.

Also, of the three groups - men, women and children - at the party:
4. There were 4 of one group.

Exactly one of the above 6 statements is false.

Can you tell which one is false? Also, how many men, women and children are
there at the party?

Answer

Statement (4) is false. There are 3 men, 8 women and 6 children.

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.

So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1),
(2) and (3) all are true.

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.

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.

Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.

There is a shortage of tubelights, bulbs and fans in a village - Kharghar. It is
found that

• All houses do not have either tubelight or bulb or fan.
• exactly 19% of houses do not have just one of these.
• atleast 67% of houses do not have tubelights.
• atleast 83% of houses do not have bulbs.
• atleast 73% of houses do not have fans.

What percentage of houses do not have tubelight, bulb and fan?
Answer

42% houses do not have tubelight, bulb and fan.

Let's assume that there are 100 houses. Hence, there should be total 300 items
i.e. 100 tubelights, 100 bulbs and 100 fans.

From the given data, we know that there is shortage of atleast (67+83+73) 223
items in every 100 houses.

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.

Thus, 42% houses do not have tubelight, bulb and fan.
Mr. Subramaniam rents a private car for Andheri-Colaba-Andheri trip. It costs him
Rs. 300 everyday.

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.

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."

How much should the individual student pay for their journey?
Answer

The individual student should pay Rs. 50 for their journey.

Note that 3 persons are travelling between Bandra and Colaba.

The entire trip costs Rs. 300 to Mr. Subramanian. Hence, half of the trip costs
Rs. 150.

For Andheri-Bandra-Andheri, only one person i.e. Mr. Subramaniam is travelling.
Hence, he would pay Rs. 150.

For Bandra-Colaba-Bandra, three persons i.e Mr. Subramaniam and two
students, are travelling. Hence, each student would pay Rs. 50.

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
Answer

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.
At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with
the hour hand?

Answer

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.

A soldier looses his way in a thick jungle. At random he walks from his camp but
mathematically in an interesting fashion.

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.

Finally how far he is from his camp and in which direction?

Answer

The soldier is 0.8944 miles away from his camp towards East-North.

It is obvious that he is in East-North direction.

Distance travelled in North and South directions
= 1/2 - 1/8 + 1/32 - 1/128 + 1/512 - 1/2048 + and so on... (a geometric series with
r = (-1/4) )

(1/2) * ( 1 - (-1/4)n )
= ---------------------------
( 1 - (-1/4) )

= 1 / ( 2 * ( 1 - (-1/4) ) )
= 2/5

Similarly in East and West directions
= 1 - 1/4 + 1/16 - 1/64 + 1/256 - and so on... (a geometric series with r = (-1/4) )

(1) * ( 1 - (-1/4)n )
= ---------------------------
( 1 - (-1/4) )

= 1 / ( ( 1- (-1/4) )
= 4/5

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.

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.

How many earrings does Raj have?

Answer

12 earrings

Assume that there are R rings, P pins and E pair of ear-rings.

It is given that, he has 2 1/2 times as many rings as pins.
R = (5/2) * P or P = (2*R)/5

And, the number of pairs of earrings is 4 less than the number of rings.
E = R - 4 or R = E + 4

Also, there are total 26 pieces.
R + P + 2*E = 26
R + (2*R)/5 + 2*E = 26
5*R + 2*R + 10*E = 130
7*R + 10*E = 130
7*(E + 4) + 10*E = 130
7*E + 28 + 10*E = 130
17*E = 102
E=6

Hence, there are 6 pairs of Ear-rings i.e. total 12 Ear-rings
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.
Submitted

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

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

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.

How many cows are needed to eat the grass in 96 days?
Answer

20 cows

g - grass at the beginning
r - rate at which grass grows, per day
y - rate at which one cow eats grass, per day
n - no of cows to eat the grass in 96 days

From given data,
g + 24*r = 70 * 24 * y ---------- A
g + 60*r = 30 * 60 * y ---------- B
g + 96*r = n * 96 * y ---------- C

Solving for (B-A),
(60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y)
36 * r = 120 * y ---------- D

Solving for (C-B),
(96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y)
36 * r = (n * 96 - 30 * 60) * y
120 * y = (n * 96 - 30 * 60) * y [From D]
120 = (n * 96 - 1800)
n = 20

Hence, 20 cows are needed to eat the grass in 96 days.
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.
Four friends - Arjan, Bhuvan, Guran and Lakha were comparing the number of
sheep that they owned.

It was found that Guran had ten more sheep than Lakha.

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.

How many sheep did each of them possess? Give the minimal possible answer

Answer

Arjan, Bhuvan, Guran and Lakha had 90, 50, 55 and 45 sheep respectively.

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.

Arjan's sheep = Bhuvan's sheep
2A/3 = A/4 + 3B/4
8A = 3A + 9B
5A = 9B

Arjan's sheep = Guran's sheep
2A/3 = A/15 + B/5 + 4G/5
2A/3 = A/15 + A/9 + 4G/5 (as B=5A/9)
30A = 3A + 5A + 36G
22A = 36G
11A = 18G

Arjan's sheep = Lakha's sheep
2A/3 = A/60 + B/20 + G/5 + L
2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18)
2A/3 = A/6 + L
A/2 = L
A = 2L

Also, it is given that Guran had ten more sheep than Lakha.
G = L + 10
11A/18 = A/2 + 10
A/9 = 10
A = 90 sheep

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.

Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 =
5.

How many such 3-digit numbers are there?

Answer

There are 45 different 3-digit numbers.

The last digit can not be 0.

If the last digit is 1, the only possible number is 101. (Note that 011 is not a 3-
digit number)

If the last digit is 2, the possible numbers are 202 and 112.

If the last digit is 3, the possible numbers are 303, 213 and 123.

If the last digit is 4, the possible numbers are 404, 314, 224 and 134.

If the last digit is 5, the possible numbers are 505, 415, 325, 235 and 145.

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.....

Thus, total numbers are
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Altogether then, there are 45 different 3-digit numbers, where last digit is the sum
of first two digits.

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.
Answer

The answer is 285714.

If its rightmost digit is placed at its left end, then new number is 428571 which is
50% larger than the original number 285714.

The simplest way is to write a small program. And the other way is trial and error
!!!
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)

Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378
There are 3 ants at 3 corners of a triangle, they randomly start moving towards
another corner.

What is the probability that they don't collide?

Answer

Let's mark the corners of the triangle as A,B,C. There are total 8 ways in which
ants can move.

1. A->B, B->C, C->A
2. A->B, B->C, C->B
3. A->B, B->A, C->A
4. A->B, B->A, C->B
5. A->C, C->B, B->A
6. A->C, C->B, B->C
7. A->C, C->A, B->A

8. A->C, C->A, B->C

Out of which, there are only two cases under which the ants won't collide :

• A->B, B->C, C->A
• A->C, C->B, B->A

Answer

There are total 8 such series:

1. Sum of 2000 numbers starting from -999 i.e. summation of numbers
from -999 to 1000.
(-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999
+ 1000 = 1000
2. Sum of 400 numbers starting from -197 i.e. summation of numbers
from -197 to 202.
(-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201
+ 202 = 1000
3. Sum of 125 numbers starting from -54 i.e. summation of numbers from
-54 to 70.
(-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000
4. Sum of 80 numbers starting from -27 i.e. summation of numbers from -
27 to 52.
(-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000
5. Sum of 25 numbers starting from 28 i.e. summation of numbers from 28
to 52.
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
6. Sum of 16 numbers starting from 55 i.e. summation of numbers from 55
to 70.
55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 +
69 + 70 = 1000
7. Sum of 5 numbers starting from 198 i.e. summation of numbers from
198 to 202.
198 + 199 + 200 +201 + 202 = 1000
8. Sum of 1 number starting from 1000.
1000 = 1000

There is a 4-character code, with 2 of them being letters and the other 2 being
numbers.

How many maximum attempts would be necessary to find the correct code? Note
that the code is case-sensitive.

Answer

The maximum number of attempts required are 16,22,400

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).

Consider an example : Let's assume that @ represents letter and # represents
number. the 6 possible ways of arranging them are : @@##, @#@#, @##@,
#@@#, #@#@, ##@@

Hence, the required answer is
= 52*52*10*10*6
= 16,22,400 attempts
= 1.6 million approx.

Thanks to Tim Sanders for opening BrainVista's brain !!!
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.

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.
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
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

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.

1. If the singer and the dancer are the same sex, then the dancer is older
than the singer.
2. If neither the singer nor the dancer is the parent of the other, then the
singer is older than the dancer.
3. If the singer is a man, then the singer and the dancer are the same age.
4. If the singer and the dancer are of opposite sex then the man is older than
the woman.
5. If the dancer is a woman, then the dancer is older than the singer.

Whose occupation do you know? And what is his/her occupation?
Answer

Cindy is the Singer. Mr. Clinton or Monika is the Dancer.

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.

CASE I : Singer is a woman and Dancer is also a woman
Then, the dancer is Monika and the singer is Cindy.

CASE II : Singer is a woman and Dancer is also a man
Then, the dancer is Mr. Clinton and the singer is Cindy.

In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or
Monika is the Dancer.
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 ;))
Submitted

Answer

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 People No of ways to hire G1 without No of ways Total
from G1 from G2 a single pair of indentical to hire G2 ways
twins
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%
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.
Submitted

Assume that you have enough coins of 1, 5, 10, 25 and 50 cents.

How many ways are there to make change for a dollar? Do explain your answer.

There are 292 ways to make change for a dollar using coins of 1, 5, 10, 25
and 50 cents.

Let's generalised the teaser and make a table as shown above.

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.

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.

Let's take an example:

To get change for 50 cents using 1, 5 and 10 cent coins.
* 50 cents change using 1 and 5 cent coins = 11 ways
* (50-10) 40 cents change using 1, 5 and 10 cent coins = 25 ways
* 50 cents change using 1, 5 and 10 cent coins = 11+25 = 36 ways

Let's take another example:
To get change for 75 cents using all coins up to 50 cent i.e. 1, 5, 10, 25 and 50
cents coins.
* 75 cents change using coins upto 25 cent = 121 ways
* (75-50) 25 cents change using coins upto 50 cent = 13 ways
* 75 cents change using coins upto 50 cent = 121+13 = 134 ways

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
where,
A = 0 to 100
B = 0 to 20
C = 0 to 10
D = 0 to 4
E = 0 to 2

The program should output all the possible values of A, B, C, D and E for which
the equation is satisfied.

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.

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?

Assume that there were no stops in the race and also they were driving with
constant speeds through out the

Answer

Let us assume that
Speed of First biker = V1 km/min
Speed of Second biker = V2 km/min
Speed of Third biker = V3 km/min
Total time take by first biker = T1 min
Total distance = S km

Now as per the data given in the teaser, at a time T min
X1 = V1 * T ----> 1

X1 - 15 = V2 * T ----> 2

X1 - 18 = V3 * T ----> 3

At a Distance S Km.
S = V1 * T1 ----> 4

S = V2 * (T1 + 12) ----> 5

S = V3 * (T1 + 15) ----> 6

Thus there are 6 equations and 7 unknown data that means it has infinite number
of solutions.

By solving above 6 equations we get,
Time taken by first biker, T1 = 60 Min.
Time taken by Second biker, T2 = 72 Min.
Time taken by first biker, T3 = 75 Min.

Also, we get
Speed of first biker, V1 = 90/T km/min

Speed of second biker, V2 = (5/6)V1 = 75/T km/min
Speed of third biker, V3 = (4/5)V1 = 72/T km/min

Also, the length of the course, S = 5400/T km

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

Thanks to Theertham Srinivas for the answer !!!
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?

Answer

The 4 digit number is 1349.

It is given that the first digit is 1/3 of the second. There are 3 such possibilities.

1. 1 and 3
2. 2 and 6
3. 3 and 9

Now, the third digit is the sum of the first and second digits.

1. 1 + 3 = 4
2. 2 + 6 = 8
3. 3 + 9 = 12

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.

Difference between Bholu's and Molu's age is 2 years and the difference
between Molu's and Kolu's age is 5 years.

What is the maximum possible value of the sum of the difference in their ages,
taken two at a time?
Answer

The maximum possible value of the sum of the difference in their ages -
taken two at a time - is 14 years.

It is given that -
"Difference between Bholu's and Molu's age is 2 years"

"Difference between Molu's and Kolu's age is 5 years"

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.

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.

If it is given that:
25 - 2 = 3
100 x 2 = 20
36 / 3 = 2

What is 144 - 3 = ?
Submitted
Answer

There are 3 possible answers to it.

Answer 1 : 9
Simply replace the first number by its square root.
(25) 5 - 2 = 3
(100) 10 x 2 = 20
(36) 6 / 3 = 2
(144) 12 - 3 = 9

Answer 2 : 11
Drop the digit in the tens position from the first number.
(2) 5 - 2 = 3
1 (0) 0 x 2 = 20
(3) 6 / 3 = 2
1 (4) 4 - 3 = 11

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).
(2) 5 - 2 = 3
10 (0) x 2 = 20
(3) 6 / 3 = 2
14 (4) - 3 = 11

Answer 3 : 14
Drop left and right digit alternatively from the actual answer.
25 - 2 = (2) 3 (drop left digit i.e. 2)
100 * 2 = 20 (0) (drop right digit i.e. 0)

36 / 3 = (1) 2 (drop left digit i.e. 1)
144 - 3 = 14 (1) (drop right digit i.e. 1)

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.

What is the sum of the three digits?

Answer

The required number is 236 and the sum is 11.

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.

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.

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.

Only one of the statement is correct. How many marbles are there under each
mug?

Answer

A simple one.

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.

Hence, there are 4 marbles under each mug.
At University of Probability, there are 375 freshmen, 293 sophomores, 187
juniors, & 126 seniors. One student will randomly be chosen to receive an award.

What percent chance is there that it will be a junior? Round to the nearest whole
percent
Answer

19%

This puzzle is easy. Divide the number of juniors (187) by the total number of
students (981), & then multiply the number by 100 to convert to a percentage.

Hence the answer is (187/981)*100 = 19%

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?

Assume that your telephone number is 7-digits.
Answer

1 in 10,000,000

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.....

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.

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.
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.

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?
Submitted

Answer

The language can employ maximum of 720 six-letter words.

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.

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

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.

How many different ways can they seat?

Answer

There are 120 different possible seating arrangments.

Note that on a round table ABCDEF and BCDEFA is the same.

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.

Thus, total different possible seating arrangements are
=5*4*3*2*1
= 120
3 blocks are chosen randomly on a chessboard. What is the probability that they
are in the same diagonal?

Answer

There are total of 64 blocks on a chessboard. So 3 blocks can be chosen out
of 64 in 64C3 ways.
So the sample space is = 41664

There are 2 diagonal on chessboard each one having 8 blocks. Consider one
of them.
3 blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.
But there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 = 112

The require probability is
= 112 / 41664
= 1 / 372
= 0.002688

What is the area of the triangle ABC with A(e,p) B(2e,3p) and C(3e,5p)?

where p = PI (3.141592654)

Answer

A tricky ONE.

Given 3 points are colinear. Hence, it is a straight line.

Hence area of triangle is 0.

Silu and Meenu were walking on the road.

Silu said, "I weigh 51 Kgs. How much do you weigh?"

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."

How much does Meenu weigh?

Answer

Meenu weighs 58 Kgs.

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.

Solving mathematically, let's assume that her weight is X Kgs.
X = 29 + X/2
2*X = 58 + X
X = 58 Kgs

Consider the sum: ABC + DEF + GHI = JJJ

If different letters represent different digits, and there are no leading zeros, what
does J represent?

Answer

The value of J must be 9.

Since there are no leading zeros, J must be 7, 8, or 9. (JJJ = ABC + DEF + GHI
= 14? + 25? + 36? = 7??)

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.

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.
A man has Ten Horses and nine stables as shown here.
[] [] [] [] [] [] [] [] []
The man wants to fit Ten Horses into nine stables. How can he fit Ten horses
into nine stables?
Submitted
Answer

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.
[T] [E] [N] [H] [O] [R] [S] [E] [S]

A man is at a river with a 9 gallon bucket and a 4 gallon bucket. He needs exactly
6 gallons of water.

How can he use both buckets to get exactly 6 gallons of water?

Note that he cannot estimate by dumping some of the water out of the 9 gallon
bucket or the 4 gallon bucket
Answer

For the sack of explanation, let's identify 4 gallon bucket as Bucket P and 9
gallon bucket as Bucket Q.
4 gallon 9 gallon
bucket bucket
Operation
(Bucket P) (Bucket Q)
Initially 0 0
Fill the bucket Q with 9 gallon water 0 9
Pour 4 gallon water from bucket Q to bucket
4 5
P
Empty bucket P 0 5
4 1
P
Empty bucket P 0 1

1 0
P
Fill the bucket Q with 9 gallon water 1 9
4 6
P

9 gallon bucket contains 6 gallon of water, as required.

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?
B R A I N 31

B B R B A 31

N I A B B 32

N I B A I 30

I R A A A 23

37 29 25 27 29
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)

Answer

B=7, R=6, A=4, I=5 and N=9

Make total 10 equations - 5 for rows and 5 for columns - and sovle them.

From Row3 and Row4,
N+I+A+B+B=N+I+B+A+I+2
B=I+2

From Row1 and Row3,
B+R+A+I+N=N+I+A+B+B-1
R=B-1

From Column2,
R + B + I + I + R = 29
B + 2R + 2I = 29

B + 2(B - 1) + 2I = 29
3B + 2I = 31
3(I + 2) + 2I = 31
5I = 25
I=5

Hence, B=7 and R=6

From Row2,
B + B + R + B + A = 31
3B + R + A = 31
3(7) + 6 + A = 31
A=4

From Row1,
B + R + A + I + N = 31
7 + 6 + 4 + 5 + N = 31
N=9

Thus, B=7, R=6, A=4, I=5 and N=9

Submit Users
BrainV
Answer Answer (24)

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.

1. Take 8 coins and weigh 4 against 4.

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.





3. The remaining coin X is the odd one. Weigh X against the anyone coin
used in initial weighing.
o If both are equal, there is some error.
o If X is heavy, X is the odd coin and is heavier.
o If X is light, X is the odd coin and is lighter.

In a sports contest there were m medals awarded on n successive days (n > 1).

1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were
awarded.
2. On the second day 2 medals and 1/7 of the now remaining medals was
awarded; and so on.
3. On the nth and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded
altogether?
Answer

Total 36 medals were awarded and the contest was for 6 days.


On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler
method, do submit it.

A number of 9 digits has the following properties:

• 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<=n<=9.
• Each digit in the number is different i.e. no digits are repeated.
• The digit 0 does not occur in the number i.e. it is comprised only of the
digits 1-9 in some order.

Find the number.

Answer

The answer is 381654729

One way to solve it is Trial-&-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.

The other way to solve this problem is by writing a computer program that
systematically tries all possibilities.

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.

What part of the contents of the container is left at the end of the second day?
Answer

Assume that contents of the container is X

On the first day 1/3rd is evaporated.
(1 - 1/3) of X is remaining i.e. (2/3)X

On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X

Hence 1/6th of the contents of the container is remaining

There are four people in a room (not including you). Exactly two of these four
always tell the truth. The other two always lie.

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.

What questions would you ask to figure out who is who? Remember that you can
ask only 2 questions.
Submitted
You have 3 baskets, & each one contains exactly 4 balls, each of which is of the
same size. Each ball is either red, black, white, or purple, & there is one of each
color in each basket.

If you were blindfolded, & lightly shook each basket so that the balls would be
randomly distributed, & then took 1 ball from each basket, what chance is there
that you would have exactly 2 red balls?

Answer

There are 64 different possible outcomes, & 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.

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:

1 2 3

-----------

R R X

R X R

X R R

X is any ball that is not red.
There is a 4.6875% chance that each of these situations will occur.

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.

Because there are 3 scenarios where this outcome occurs, you multiply the
4.6875% chance of any one occurring by 3, & you get 14.0625%

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.

What is the probablity of winning the lottery?

Answer

The probability of winning the lottery is two in one billion i.e. only two person
can win from one billion !!!

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
80
C11 ways which is very very high and is equal to 1.04776 * 1013

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.

• You might select 8 numbers which all are there in 11 numbers choosen
by the Lottery Commission. So there are 11C8 ways.
• 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.

So total lucky ways are
= ( 11C8 ) + ( 11C7 ) * ( 69C1 )
= (165) + (330) * (69)
= 165 + 22770
= 22935

Hence, the probability of the winning lottery is
= (Total lucky ways) / (Total Sample space)
= (22935) / ( 1.04776 * 1013)
= 2.1889 * 10-9
i.e. 2 in a billion.

Submit Users BrainVista
Puzzle
Answer Answer (4) Answer
To move a Safe, two cylindrical steel bars 7 inches in diameter are used as
rollers.

How far will the safe have moved forward when the rollers have made one
revolution?
Answer

The safe must have moved 22 inches forward.

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
= PI * Diameter (or 2 * PI * Radius)
= PI * 7
= 3.14159265 * 7
= 21.99115
= 22 inches approx.
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?

Note that both are different colored pieces.
SubmAnswer

The probability of either the Rook or the Bishop attacking the other is
0.3611

A Rook and a Bishop on a standard chess-board can be arranged in 64P2 = 64*63
= 4032 ways

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.

Case I - Rook attacking Bishop
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

Case II - Bishop attacking Rook
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.

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.

Hence, total possible ways of the Bishop attacking the Rook

= 28*7 + 20*9 + 12*11 + 4*13
= 560 ways

Thus, the required probability is
= (896 + 560) / 4032
= 13/36
= 0.3611
itted ed
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.

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)

How many possible combinations are there? Are McDonald's correct in claiming
there are 40312 combinations?
Answer

Total possible combinations are 12869.

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.

Items Products Used (X)
Ordered
(N) 1 2 3 4 5 6 7 8

1 1 - - - - - - -
2 1 1 - - - - - -
3 1 2 1 - - - - -
4 1 3 3 1 - - - -
5 1 4 6 4 1 - - -
6 1 5 10 10 5 1 - -
7 1 6 15 20 15 6 1 -
8 1 7 21 35 35 21 7 1
Total (T) 8 28 56 70 56 28 8 1
Ways to choose 8C1 8C2 8C3 8C4 8C5 8C6 8C7 8C8

X products from
8 products (W)
Total combinations 64 784 3136 4900 3136 784 64 1
(T*W)

Thus, total possible combinations are
= 64 + 784 + 3136 + 4900 + 3136 + 784 + 64 + 1
= 12869
What are the chances that at least two out of a group of fifty people share the
same birthday?
SubmittedAnswer

The probability of atleast two out of a group of 50 people share the same
birthday is 97%

Probability of atleast two share the same birthday = 1 - probability of all 50 have
different birthdays

Probability of all 50 have different birthday
= 365/365 * 364/365 * 363/365 * ... * 317/365 * 316/365
= (365 * 364 * 363 * 362 * ... * 317 * 316)/36550
= 0.0296264

Probability of atleast two share the same birthday
= 1 - 0.0296264
= 0.9703735
= 97% approx.

Thus, the probability of atleast two out of a group of 50 people share the same
birthday is 97%

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!!!

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.

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.
Answer

The maximum possible value of X is 13 minutes 20 seconds.

In one minute,
pipe A can fill 1/30 part of the tank.
pipe B can fill 1/24 part of the tank.

Thus, the net water level increase in one minute is
= 1/30 + 1/24
= 3/40 part of the tank

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.
A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, &
now his salary is Rs. 22702.68

What was his salary to begin with?
Answer

Rs.22176

Assume his salary was Rs. X

He earns 5% raise. So his salary is (105*X)/100

A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100)
which is Rs. 22702.68

Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68
X = 22176

500 men are arranged in an array of 10 rows and 50 columns according to their
heights.

Tallest among each row of all are asked to come out. And the shortest among
them is A.

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.

Now who is taller A or B ?
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.

After spending 20 paise, the person counts the money. And to his surprise, he
has double the amount he wanted to withdraw.

Find X and Y. (1 Rupee = 100 Paise)

As given, the person wanted to withdraw 100X + Y paise.

But he got 100Y + X paise.

After spending 20 paise, he has double the amount he wanted to withdraw.
Hence, the equation is

2 * (100X + Y) = 100Y + X - 20

200X + 2Y = 100Y +X - 20

199X - 98Y = -20

98Y - 199X = 20

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 & 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

Case I : Y=2X
98Y - 199X = 20
Y - 2X = 0
We get X = - 20/3 & Y = - 40/2

Case II : Y=2X+1
98Y - 199X = 20
Y - 2X = 1
We get X = 26 & Y = 53

Now, its obvious that he wanted to withdraw Rs. 26.53

Puzzle

At the Party:

1. There were 9 men and children.
2. There were 2 more women than children.

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.

Also, of the three groups - men, women and children - at the party:

Exactly one of the above 6 statements is false.

Can you tell which one is false? Also, how many men, women and children are
there at the party?
Answer

Statement (4) is false. There are 3 men, 8 women and 6 children.

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.

So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1),
(2) and (3) all are true.

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.

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.

Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.
Brain Teaser No : 00242

There is a shortage of tubelights, bulbs and fans in a village - Kharghar. It is
found that

• All houses do not have either tubelight or bulb or fan.
• exactly 19% of houses do not have just one of these.
• atleast 67% of houses do not have tubelights.
• atleast 83% of houses do not have bulbs.
• atleast 73% of houses do not have fans.

What percentage of houses do not have tubelight, bulb and fan?

Answer

42% houses do not have tubelight, bulb and fan.

Let's assume that there are 100 houses. Hence, there should be total 300 items
i.e. 100 tubelights, 100 bulbs and 100 fans.

From the given data, we know that there is shortage of atleast (67+83+73) 223
items in every 100 houses.

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.

Thus, 42% houses do not have tubelight, bulb and fan.
What is the remainder left after dividing 1! + 2! + 3! + … + 100! By 7?

Think carefully !!!
Answer

A tricky one.

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.

The only part to be consider is
= 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873

The remainder left after dividing 873 by 7 is 5

Hence, the remainder is 5.

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.

Find the exact numerical value to the following equation: (X - A) * (X - B) * (X - C)
* ... * (X - Y) * (X - Z)

Answer

(X - A) * (X - B) * (X - C) * ... * (X - Y) * (X - Z) equals 0 since (X - X) is zero
If three babies are born every second of the day, then how many babies will be
born in the year 2001?
SuAnswer

9,46,08,000 babies

The total seconds in year 2001
= 365 days/year * 24 hours/day * 60 minutes/hours * 60 seconds/minute
= 365 * 24 * 60 * 60 seconds
= 3,15,36,000 seconds

Thus, there are 3,15,36,000 seconds in the year 2001. Also, three babies born
are every second. Hence, total babies born
= 3 * 3,15,36,000 seconds
= 9,46,08,000bmitted

Replace the letters with the correct numbers.
T W O

X T W O

---------

T H R E E

Submitted by : Timmy Chan
Answer

T=1, W=3, O=8, H=9, R=2, E=4
1 3 8

x 1 3 8

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

1 9 0 4 4
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.

Four words add up to a fifth word numerically:

mars

venus

uranus

saturn

-------- +

neptune

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.

Furthermore, numbers 1 and 6 are being used most frequently.
Answer

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.

mars m = 4

venus a = 5

uranus r = 9

saturn s = 3

-------- + v = 2 4593

neptune e = 0 20163

n = 1 695163

u = 6 358691

t = 8 -------- +

p = 7 1078610

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.

Conditions

• 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.
• They only look in front of them not behind. They are not allowed to
communicate by talking.
• Between army man 1 and 2, there is a wall.
• Captive man 4 can see the colour of hats on 2 and 3
• 3 can only see 2's hat
• 2 can only see a wall and 1 can see a wall too, but is on the other side

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.

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!
Submitted
Answer

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.

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.

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.

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.
One side of the bottom layer of a triangular pyramid has 12 balls. How many are
there in the whole pyramid?

Note that the pyramid is equilateral and solid.
Answer

There are total 364 balls.

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.

Hence, the total number of balls are
= 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78
= 364 balls

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
Answer

The blindfolded man requires 5 turns.

1. Open two adjacent holes.

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?
Answer

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?
Answer

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.

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
Submitted by : Calon
Answer

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.

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?

Answer

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?
Answer

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?

Assume that both players are very intelligent. Explain your answer
Answer

"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.
or 9.
or 5.
or 7.
or 9.
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?
Answer

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?
Answer

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?

Answer

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.

Adding up, all sleeping times,
= (1/2 - 1/3) + (1 - 2/3)
= 1/6 + 1/3
= 1/2

Hence, Mr. wagle slept through half his trip.

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?
Answer

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.
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.
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?
Answer

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?
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
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?
SubmiAnswer

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?

Answer

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?
SubmittAnswer

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
Answer

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?
Answer

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.
Answer

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?
Submitted
Answer

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.
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.

How can the couples arrange themselves in a row so that they all sit where they
would like?
Submitted by : Tara Smith

Answer

From the given data, it can be inferred that:
(Sally & Bob) NOT (Betty & Jim) NOT (Alice & Tom) NOT (Gertrude & Bill)

(A) NOT (B) means A and B can not seat next to each other.

Now, it is obvious that (Betty & Jim) and (Alice & Tom) will occupy the corner
seats as both of them can have only one neighbour. Therefore,
(Gertrude & Bill) will seat next to (Betty & Jim)
(Sally & Bob) will seat next to (Gertrude & Bill)
(Alice & Tom) will seat next to (Sally & Bob)

Thus, there are two possible arrangements - a mirror images of each other.

1. (Betty & Jim) - (Gertrude & Bill) - (Sally & Bob) - (Alice & Tom)
2. (Alice & Tom) - (Sally & Bob) - (Gertrude & Bill) - (Betty & Jim)
Substitute digits for the letters to make the following addition problem true.

W H O S E

T E E T H

A R E

+ A S

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

S W O R D S
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.
Answer

It is obvious that S=1 and T=9.

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.

W H O S E 2 8 5 1 6

T E E T H 9 6 6 9 8

A R E 4 7 6

+ A S + 4 1

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

S W O R D S 1 2 5 7 3 1

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.

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:

Aaron:
A) I did not leave the door open.

B) Clement was the one who did it.

Bob:
A) I was not the one who was on duty that evening.
B) Aaron was on duty.

Clement:
A) Bob was the one on duty that evening.
B) I hoped Socrates would escape.

David:
A) I did not leave the door open.
B) I was not surprised that Socrates did not attempt to escape.

Considering that, in total, three statements are true, and five statements are
false, which guard is guiltyAnswer

David is the guilty.

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.

Now consider each of them as a guilty, one at a time.
Aaron Bob Clement David True
Stmts
A B A B A B A B
If Aaron is guilty False False True True False True True False 4
If Bob is guilty True False False False True True True False 4
If Clement is
True True True False False True True False 5
guilty
If David is guilty True False True False False True False False 3

Since in total, three statements are true and five statements are false. It is clear
from the above table that David is?

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.

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.

If we do this again starting from 1344, we get (1+3+4+4) * (1*3*4*4) = 576

And yet again (5+7+6) * (5*7*6) = 3780

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.

Can you find any numbers to which when we apply the above mentioned rule
repeatedly, we never end up at 0?


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.

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.

If we do this again starting from 1344, we get (1+3+4+4)
* (1*3*4*4) = 576

And yet again (5+7+6) * (5*7*6) = 3780

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.

Can you find any numbers to which when we apply the above
mentioned rule repeatedly, we never end up at 0?

There were N stations on a railroad. After adding X stations 46 additional tickets
have to be printed.

Find N and X.
Answer

Let before adding X stations, total number of tickets
t = N(N-1)

After adding X stations total number of tickets are
t + 46 = (N+X)(N+X-1)

Subtracting 1st from 2nd
46 = (N+X)(N+X-1) - N(N-1)
46 = N2 + NX - N + NX + X2 - X - N2 + N
46 = 2NX + X2 - X
46 = (2N - 1)X + X2
X2 + (2N - 1)X - 46 = 0

Now there are only two possible factors of 46. They are (46,1) and (23,2)

Case I: (46,1)
2N - 1 = 45
2N = 46
N = 23
And X = 1

Case II: (23,2)
2N - 1 = 21
2N = 22
N = 11
And X = 2

Hence, there are 2 possible answers.

An emergency vehicle travels 10 miles at a speed of 50 miles per hour.

How fast must the vehicle travel on the return trip if the round-trip travel time is to
be 20 minutes?
Answer

75 miles per hour

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
= (60 * 10) / 50
= 12 minutes

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
= (60 * 10) / 8
= 75 miles per hour
All of the students at a college are majoring in psychology, business, or both.
73% of the students are psychology majors, & 62% are business majors.

If there are 200 students, how many of them are majoring in both psychology &
business?
Answer

70 students are majoring in both, psychology & business

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

65% of the students are not majoring in both psychology & business, so 35% are
double majors, a total of 70 students.
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.

Answer

The speed of Bangalore-Mysore train is TWICE the speed of Mysore-
Bangalore train.

Let the distance between Bangalore and Mysore is D kms.
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.

Now, assume that both the trains met each other at T kms from the Bangalore
(point P in figure)
Time taken by Bangalore-Mysore train to reach P = Time taken by Mysore-
Bangalore train to reach P
( T / X ) = ( D - T ) / Y -----equ(I)

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,
( D - T ) = X and
T=4*Y

Substituting these values in equation I, we get
(4*Y)/X=X/Y
4*Y*Y=X*X

2*Y=X

Hence, the speed of Bangalore-Mysore train is TWICE the speed of Mysore-
Bangalore train.How much faster is one train from other?

Answer

49 times

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.

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.

How much did each cost?
Submitted by : Kimi
Answer

The lamp cost $ 2.50 and the rug cost $ 2.75

A simple one.

Assume that the lamp cost $ L.
Hence the rug must have cost $ (L + 0.25)
Also, total cost is $ 5.25, Hence the equation :
L + L + 0.25 = 5.25
2*L=5
L = 2.50

Hence, the lamp cost $ 2.50 and the rug cost $ 2.75


Write 1111......(243 times) i.e. a 243 digit number with all 1s.

Prove that it is divisible by 243.
SubmittAnswer

Prove it using the mathematical induction.

First here are a couple of things to note:

[1] A number whose digits add up to a multiple of three is divisable by 3.

e.g. 369: 3+6+9=18: 1+8=9 which is a multiple of 3 hence 369 is divisable by 3.

[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.
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).

Let
N = any series of digits (e.g. N=369)
D = the number of digits in N (e.g. if N=369 then D=3)
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.

Also, if we multiply N with P we are essentially repeating N for (D-1) times.
e.g. if N=369 then D=3, P=1001001 and N*P=369369369

Let's start with N=111. It is clear that N is divisible by 3. (From [1])
Also, D=3 and P=1001001
N*P=111111111 (9 times)
The resulting number 111111111 must be divisible by 9 as N and P both are
divisible by 3.

Now, let's start with N=111111111. It is clear that N is divisible by 9.
Also, D=9 and P=1000000001000000001
N*P=111111111... (27 times)
The resulting number 1111111... (27 times) must be divisible by 27 as N is
divisible by 9 and P is divisible by 3.

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.

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.

Thus, 1111111... (243 times) is divisible by 243.

Thanks to Ryan Hutcherson for solution !!!
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.

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.

And using another six matches he could again do precisely the same.

How many matches were there in the box originally?

Note that the match-box can hold maximum of 50 matches.
Answer

Initially, there were 42 or 36 matches in the match-box.

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.

Thus, there are two possible answers. There were either 42 or 36 matches in the
match-box.

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 :

1. 24 (10, 8, 6)
2. 30 (13, 12, 5)
3. 36 (17, 10, 9)
4. 42 (20, 15, 7)
5. 60 (29, 25, 6)

Find the values of each of the alphabets.

N O O N

S O O N

+ M O O N

----------

J U N E
Answer

Using trial and error. There are 2 solutions to it and may be more.

2 4 4 2

1 4 4 2

+ 5 4 4 2

----------

9 3 2 6

4 1 1 4

5 1 1 4

+ 0 1 1 4

----------

9 3 4 2

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.
Submitted by : Vaibhav Gupta

We have one answer where sum of all the numbers lying in straight lines is 26.

If you have others, do submit them.

Montu, Bantu, Chantu and Pintu have pets.

Montu says, "If Pintu and I each have a dog, then exactly one of Bantu and
Chantu has a dog."

Bantu says, "If Chantu and I each have a cat, then exactly one of Montu and
Pintu has a dog."

Chantu says, "If Montu and I each have a dog, then exactly one of Bantu and
Pintu has a cat."

Pintu says, "If Bantu and I each have a cat, then exactly one of Bantu and I
has a dog."

Only one of the four is telling the truth. Who is telling the truth?

Answer

Bantu is telling the truth.

For a IF-THEN statement to be false, IF part has to be true and THEN part
has to be false.

Since only one statement is true and remaining three are false, IF part of three
statements are true & 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.

Montu Bantu Chantu Pintu
Montu says Dog (Dog) (Dog) Dog
Bantu says (Dog) Cat Cat (Dog)
Chantu says Dog (Cat) Dog (Cat)
Pintu says Cat Cat
(Dog) (Dog)

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.

Thus, Bantu is telling the truth.

Montu have a Dog and may or may not have a Cat.
Bantu have a Cat.
Chantu have a Dog.
Pintu have a Dog and a Cat.

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)


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)


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.
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.
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.


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 People No of ways to hire G1 without No of ways Total
from G1 from G2 a single pair of indentical to hire G2 ways
twins
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.

Submit Users
BrainV
Answer Answer (14)

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


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.

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
149

166
229
236
334

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.

Submit
User
Answer
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.
Submit Users
Answer 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


W O R L D

+ T R A D E

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

C E N T E R
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

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.


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.
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.

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.

How many men, women and children were working with the constructor?
Answer

16 men, 12 women and 72 children were working with the constructor.

Let's assume that there were X men, Y women and Z children working with the
constructor. Hence,

X + Y + Z = 100
5X + 4Y + Z = 200

Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z

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

if Y = X,
300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428

if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727

But Z must be an integer, hence Z=72. Also, X=16 and Y=12

There were 16 men, 12 women and 72 children working with the constructor.

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.

Today, he has collected 27 cigar butts. How many cigars will he be able to
smoke?
Answer

13 not 12

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.

So 9+3+1=13
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.

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.

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.

What is the minimum number of flowers the pilgrim had initially? How many
flower did he offer to each God?
Answer

The pilgrim had 7 flowers, initially and he offered 8 flowers to each God.

Assume that the pilgrim had X flowers initially and he offered Y flowers to each
God.

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
(8X - 7Y) = 0
8X = 7Y

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.

In general, the pilgrim had 7N flowers initially and he offered 8N flowers to each
God, where N = 1, 2, 3, 4, .....


Tanya wants to go on a date and prefers her date to be tall, dark and
handsome.

1. Of the preferred traits - tall, dark and handsome - no two of Adam,
Bond, Cruz and Dumbo have the same number.
2. Only Adam or Dumbo is tall and fair.
3. Only Bond or Cruz is short and handsome.
4. Adam and Cruz are either both tall or both short.
5. Bond and Dumbo are either both dark or both fair.

Who is Tanya's date?

Answer

Cruz is Tanya's date.

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.

From (4) and (5), there are only two possibilities:
* Adam & Cruz both are tall and Bond & Dumbo both are fair.
* Adam & Cruz both are short and Bond & Dumbo both are dark.

But from (2), second possibility is impossible. So the first one is the correct
possibility i.e. Adam & Cruz both are tall and Bond & Dumbo both are fair.

Then from (3), Bond is short and handsome.

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:

Cruz - Tall, Dark and Handsome
Adam - Tall and Handsome
Bond - Handsome
Dumbo - None :-(

Hence, Cruz is Tanya's date.
Consider a game of Tower of Hanoi (like the one that you can play on
BrainVista).

If the tower has 2 discs, the least possible moves with which you can move the
entire tower to another peg is 3.

If the tower has 3 discs, the least possible moves with which you can move the
entire tower to another peg is 7.

What is the least possible moves with which you can move the entire tower to
another peg if the tower has N discs?
Submitted
Answer

There are number of ways to find the answer.

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
= 2(N-1) + 2(N-2) + 2(N-3) + ..... + 22 + 21 + 20
= 2N - 1

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
F(1) = 1
F(N) = 2*[F(N-1)] + 1
where N is the total number of discs

Also, one can arrive at the answer by finding the number of moves for smaller
number of discs and then derive the pattern.
For 1 disc, number of moves = 1
For 2 discs, number of moves = 3

Thus, the pattern is 2N – 1
A boy found that he had a 48 inch strip of paper. He could cut an inch off every
second.

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.

SubmiAnswer

47 seconds.

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

The cricket match between India and Pakistan was over.

• Harbhajan scored more runs than Ganguly.
• Sachin scored more runs than Laxman but less than Dravid
• Badani scored as much runs as Agarkar but less than Dravid and more
than Sachin.
• Ganguly scored more runs than either Agarkar or Dravid.

Each batsman scored 10 runs more than his immediate batsman. The lowest
score was 10 runs. How much did each one of them score
Answer

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.

1. Harbhajan
2. Ganguly
3. Dravid
4. Badani, Agarkar
5. Sachin
6. Laxman

Also, as the lowest score was 10 runs. Laxman must have scored 10, Sachin 20,
Badani & Agarkar 30 and so on.

1. Harbhajan - 60 runs
2. Ganguly - 50 runs
3. Dravid - 40 runs
4. Badani, Agarkar - 30 runs each
5. Sachin - 20 runs
6. Laxman - 10 runs

There are 10 statements written on a piece of paper:

1. At least one of statements 9 and 10 is true.
2. This either is the first true or the first false statement.

3. There are three consecutive statements, which are false.
4. The difference between the numbers of the last true and the first true
statement divides the number, that is to be found.
5. The sum of the numbers of the true statements is the number, that is to be
found.
6. This is not the last true statement.
7. The number of each true statement divides the number, that is to be
found.
8. The number that is to be found is the percentage of true statements.
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.
10. There are no three consecutive true statements.

Find the minimal possible number?
Submitted
Answer

The numebr is 420.

If statement 6 is false, it creates a paradox. Hence, Statement 6 must be true.

Consider Statement 2:

• If it is true, it must be the first true statement. Otherwise, it creates a
paradox.
• If it is false, it must be the second false statement. Otherwise, it creates a
paradox.

In both the cases, Statement 1 is false.

As Statement 1 is false, Statement 9 and Statement 10 both are false i.e. there
are three consecutive true statements.

1 2 3 4 5 6 7 8 9 10
False - - - - True - - False False

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.

1 2 3 4 5 6 7 8 9 10
False True False - - True - True False False

Also, atleast two of Statements 4, 5 and 7 must be true as there are three
consecutive true statements.

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.

Also, Statement 7 is true as Statement 6 is not the last true statement.

1 2 3 4 5 6 7 8 9 10
False - True - - True True False False False

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.

There 3 consecutive true statements. Hence, Statement 2 and Statement 4 are
true.

1 2 3 4 5 6 7 8 9 10
False True True True False True True False False False

Now, the conditions for the number to be found are:

1. The numebr is divisible by 5 (Statement 4)
2. The numebr is divisible by 2, 3, 4, 6, 7 (Statement 7)
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)

The minimum possible number is 420.

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.
Ankit and Tejas divided a bag of Apples between them.

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.

Ankit went to Kitchen to drink water. While Ankit was in Kitchen, Tejas took
apples from Ankit's pile equal to Ankit's age.

Who have more apples now?
Answer

At the end, Ankit and Tejas, both have the same number of apples.

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.

Operation Ankit's Apples Tejas's Apples

Initially 3N N

Ankit gave T apples to Tejas
3N - T N+T
(equals age of Tejas)

Tejas took 2T apples from Ankit's pile
3N - 3T N + 3T
(equals age of Ankit)

It is given that after Ankit gave T apples to Tejas, Ankit had twice as many apples
as Tejas had.
3N - T = 2*(N + T)
3N - T = 2N + 2T
N = 3T

From the table, at the end Ankit have (3N - 3T) apples and Tejas have (N + 3T)
apples. Substituting N = 3T, we get
Ankit's apples = 3N - 3T = 9T - 3T = 6T
Tejas's apples = N + 3T = 3T + 3T = 6T

Thus, at the end Ankit and Tejas, both have the same number of apples.

On evey Sunday Amar, Akbar and Anthony lunch together at Preetam-Da-Dhaba
where they order lassi based on following facts.

1. Unless neither Amar nor Akbar have lassi, Anthony must have it.
2. If Amar does not have lassi, either Akbar or Anthony or both have it.
3. Anthony has lassi only if either Amar or Akbar or both have it.
4. Akbar and Anthony never have lassi together.

Who order(s) lassi?
Answer

Amar and Anthony both have lassi whereas Akbar never does.

Fact (2) can be alternatively stated that "either Amar or Akbar or Anthony must
have lassi".

From Fact (3), it can be infered that either Amar or Akbar must have lassi.

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.

Decipher this sentence.

B R W Q H L F K W H J K Q I B W K

Q I C E D W Z B G W K K M I K E

Z B G Q H S K Z B G J K Z K W

B U U Z B G J D B H F W.

Answer

Start with ZBG and ZBGJ. It should be either "the/then" or "you/your"
combination as they appear more.

B R W Q H L F K W H J K Q I B W K

o b s t a c l e s a r e t h o s e

Q I C E D W Z B G W K K M I K E

t h i n g s y o u s e e w h e n

Z B G Q H S K Z B G J K Z K W

y o u t a k e y o u r e y e s

B U U Z B G J D B H F W.

o f f y o u r g o a l s.

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.

Answer

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:

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.

Therefore solving for X

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
12X + 120 + (30-X) = 180
11X = 30
Hence X = 30/11 degrees
(hour hand is X degree away from 5 O'clock)

Now each degree the hour hand moves is 2 minutes.

Therefore minutes are
= 2 * 30/11
= 60/11
= 5.45 (means 5 minutes 27.16 seconds)

Therefore the exact time at which the hands are opposite to each other is
= 4 hrs. 54 min. 32.74 seconds

Ali Baba had four sons, to whom he bequeathed his 39 camels, with the proviso
that the legacy be divided in the following way :

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.
Dismounting from his camel, he asked if he might help, for he knew just what to
do. The brothers gratefully accepted his offer.

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.

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?
Answer

They took their percentages from 40 and not from 39, so they got more than their
share.

The oldest son got 1/2 of 40 = 20 which is 0.5 more
The second son got 1/4 of 40 = 10 which is 0.25 more
The third son got 1/8 of 40 = 5 which is 0.125 more
The youngest son got 1/10 of 40 = 4 which is 0.1 more

And the stranger got 1/40 of 40 = 1 which is 0.025 more (As he is not supposed
to get anything)

All these fractions add to = 0.5 + 0.25 + 0.125 + 0.1 + 0.025 = 1 which stranger
took away.
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.

What is the minimum number of persons required so that this is possible?
Answer

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.

Grandfather, Grandmother
|
|
Son, wife
|
|
Son

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.

How much money did the restaurant lose? Ignore the profit of the food
restaurant.
Answer

He lost Rs.600

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.

S L I D E

- D E A N

---------

3 6 5 1
Each of seven digits from 0-9 are represented by a different letter above such
that the subtraction is true.

What word represents 3651?
Answer

3651 represents LENS.

Let's assign possible values to each letter and then use trial-n-error.

S must be 1.

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.

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.

Now using trial-n-error, we get S=1, I=2, L=3, A=4, N=5, E=6 and D=9

S L I D E 1 3 2 9 6

- D E A N - 9 6 4 5

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

3 6 5 1 L E N S
Hence, 3651 represents LENS.
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.

1. If Clark's apartment is not next to Burzin's apartment, then the landlord is
Adam and lives in apartment 1.
2. If Adam's apartment is right of Clark's apartment, then the landlord is
Edmund and lives in apartment 4.
3. If Burzin's apartment is not next to Edmund's apartment, then the landlord
is Clark and lives in apartment 3.
4. If Edmund's apartment is right of Adam's apartment, then the landlord is
Burzin and lives in apartment 2.

Who is the landlord?

Answer

Clark is the landlord.

Assume each statement true, one at a time and see that no other statement is
contradicted.

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.

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.

Let's assume that Statement (3) is true. Then, Clark is the landlord and lives in
apartment 3. It satisfies all the statements for
(1) Adam - (2) Edmund - (3) Clark - (4) Burzin

Hence, Clark is the landlord.

Similarly, you can assume Statement (4) true and find out that it also contradicts.

B, J and P are related to each other.

1. Among the three are B's legal spouse, J's sibling and P's sister-in-law.
2. B's legal spouse and J's sibling are of the same sex.

Who is the married man?

Answer

J is the married man.

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.

There are 2 cases:

1. If B's legal spouse is J, then J's sibling must be P and P's sister-in-law
must be B.
2. If B's legal spouse is P, then P's sister-in-law must be J and J's sibling
must be B.

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.

B's spouse J's sibling P's sister-in-law

(male) (male) (female)

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

Case I J P B

Case II P B J

Case II is not possible as B & P are married to each other and both are male.
Hence, J is the married man.

A polygon has 1325 diagonals. How many vertices does it have?
Answer

The formula to find number of diagonals (D) given total number of vertices or
sides (N) is

N * (N - 3)

D = -----------

2

Using the formula, we get
1325 * 2 = N * (N - 3)
N2 - 3N - 2650 = 0

Solving the quadratic equation, we get N = 53 or -50

It is obvious that answer is 53 as number of vertices can not be negative.

Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel
has 2, pentagon has 5, hexagon has 9 and so on......

Hence the series is 0, 0, 0, 2, 5, 9, 14, ........ (as diagram with 1,2 or 3 vertices
will have 0 diagonals).

Using the series one can arrive to the formula given above.

A cube is made of a white material, but the exterior is painted black.

If the cube is cut into 125 smaller cubes of exactly the same size, how many of
the cubes will have atleast 2 of their sides painted black?

Answer

44

36 of the cubes have EXACTLY 2 of their sides painted black, but because a
cube with 3 of its sides painted black has 2 of its sides painted black, you must
also include the corner cubes. This was a trick question, but hopefully the title of
the puzzle tipped you off to this.

Imagine a triangle of coins on a table so that the first row has one coin in it
and the second row has two coins in it and so on. If you can only move one
coin at a time, how many moves does it take to make the triangle point the
other way?

For a triangle with two row it is one, for a triangle with three rows it is two, for a
triangle with four rows it is three.

For a traingle with five rows is it four?
Submitted
Answer

It takes 5 moves to make the triangle with 5 rows point the other way.

0 = a coin that has not been moved.
X = the old position of the moved coin
8 = the new position of the moved coin.

________X
_______X X
____8 0 0 0 8
_____0 0 0 0
____X 0 0 0 X
_______8 8
________8

For traingle of any number of rows, the optimal number of moves can be
achieved by moving the vertically symmetrical coins i.e. by moving same number
of coins from bottom left and right, and remaining coins from the top.

For a triangle with an odd number of rows, the total moves require are :
(N2/4) - (N-4) Where N = 4, 6, 8, 10, ...

For a triangle with even number of rows, the total moves require are :
((N2-1)/4) - (N-4) Where N = 5, 7, 9, 11, ...

Thanks to Alex Crosse for submitting above formulas.

A man is going to an Antique Car auction. All purchases must be paid for in
cash. He goes to the bank and draws out $25,000.

Since the man does not want to be seen carrying that much money, he places
it in 15 evelopes numbered 1 through 15. Each envelope contains the least
number of bills possible of any available US currency (i.e. no two tens in place
of a twenty).

At the auction he makes a successful bid of $8322 for a car. He hands the
auctioneer envelopes number(s) 2, 8, and 14. After opening the envelopes the
auctioneer finds exactly the right amount.

How many ones did the auctioneer find in the envelopes?

Answer

Each envelope contains the money equal to the 2 raised to the envelope number
minus 1. The sentence "Each envelope contains the least number of bills
possible of any available US currency" is only to misguide you. This is always
possible for any amount !!!

One more thing to notice here is that the man must have placed money in
envelopes in such a way that if he bids for any amount less than $25000, he
should be able to pick them in terms of envelopes.

First envelope contains, 20 = $1
Second envelope contains, 21 = $2
Third envelope contains, 22 = $4
Fourth envelope contains, 23 = $8 and so on...

Hence the amount in envelopes are $1, $2, $4, $8, $16, $32, $64, $128, $256,
$512, $1024, $2048, $4096, $8192, $8617

Last envelope (No. 15) contains only $8617 as total amount is only $25000.

Now as he bids for $8322 and gives envelope number 2, 8 and 14 which

contains $2, $128 and $8192 respectively.

Envelope No 2 conrains one $2 bill
Envelope No 8 conrains one $100 bill, one $20 bill, one $5 bill, one $2 bill and
one $1 bill
Envelope No 14 conrains eighty-one $100 bill, one $50 bill, four $10 bill and one
$2 bill

Hence the auctioneer will find one $1 bill in the envelopes.

The minute and the hour hand of a watch meet every 65 minutes.

How much does the watch lose or gain time and by how much?

Answer

The minute and the hour hand meet 11 times in 12 hours in normal watch i.e.
they meet after every
= (12 * 60) / 11 minutes
= 65.45 minutes
= 65 minutes 27.16 seconds

But in our case they meet after every 65 minutes means the watch is gaining
27.16 seconds.

There is a number that is 5 times the sum of its digits. What is this number?
Answer is not 0.

Answer

The number is 45, simply because
45 = 5 * (4 + 5)
How does one find this number?

Let T be the digit in the tens place and U be the digit in the units place. Then, the
number is 10*T + U, and the sum of its digits is T + U.

The following equation can be readily written:
10*T + U = 5*(T + U) or
10*T + U = 5*T + 5*U or
5*T = 4*U

Thus, T / U = 4 / 5

Since T and U are digits, T must be 4 and U must be 5.
There are six boxes containing 5, 7, 14, 16, 18, 29 balls of either red or blue in
colour. Some boxes contain only red balls and others contain only blue.

One sales man sold one box out of them and then he says, "I have the same
number of red balls left out as that of blue."

Which box is the one he solds out?
Answer

Total no of balls = 5 + 7 + 14 + 16 + 18 + 29 = 89

Total number of balls are odd. Also, same number of red balls and blue balls are
left out after selling one box. So it is obvious that the box with odd number of
balls in it is sold out i.e. 5, 7 or 29.

Now using trial and error method,
(89-29) /2 = 60/2 = 30 and
14 + 16 = 5 + 7 + 18 = 30

So box with 29 balls is sold out.

Ekta got chocolates to give her friends on her Birthday. If she gives 3
chocolates to each friend, one friend will get only 2 chocolates. Also, if she
gives 2 chocolates to each friends, she will left with 15 chocolates.

How many chocolates Ekta got on her Birthday? and how many friends are
there?

Answer

47 Chocolates and 16 Friends

Let's assume that there are total C chocolates and F friends.

According to first case, if she gives 3 chocolates to each friend, one friend will get
only 2 chocolates.
3*(F - 1) + 2 = C

Similarly, if she gives 2 chocolates to each friends, she will left with 15
chocolates.
2*F + 15 = C

Solving above 2 equations, F = 16 and C = 47. Hence, Ekta got 47 chocolates
and 16 friends
Pooja and Esha met each other after long time. In the course of their
conversation, Pooja asked Esha her age. Esha replied, "If you reverse my age,
you will get my husbund's age. He is of course older than me. Also, the
difference between our age is 1/11th of the sum of our age."

Can you help out Pooja in finding Esha's age?
Answer

Esha's age is 45 years.

Assume that Esha's age is 10X+Y years. Hence, her hunsbands age is (10Y + X)
years.

It is given that difference between their age is 1/11th of the sum of their age.
Hence,
[(10Y + X) - (10X + Y)] = (1/11)[(10Y + X) + (10X + Y)]
(9Y - 9X) = (1/11)(11X + 11Y)
9Y - 9X = X + Y
8Y = 10X
4Y = 5X

Hence, the possible values are X=4, Y=5 and Esha's age is 45 years.
A fish had a tail as long as its head plus a quarter the lenght of its body. Its body
was three-quarters of its total length. Its head was 4 inches long.

What was the length of the fish?
Submitted

The fish is 128 inches long.

It is obvious that the lenght of the fish is the summation of lenghts of the head,
the body and the tail. Hence,
Fish (F) = Head (H) + Body (B) + Tail (T)

But it is given that the lenght of the head is 4 inches i.e. H = 4. The body is three-
quarters of its total length i.e. B = (3/4)*F. And the tail is its head plus a quarter
the lenght of its body i.e. T = H + B/4. Thus, the equation is
F=H+B+T
F = 4 + (3/4)*F + H + B/4
F = 4 + (3/4)*F + 4 + (1/4)*(3/4)*F
F = 8 + (15/16)*F
(1/16)*F = 8
F = 128 inches

Thus, the fish is 128 inches long.

Assume that you have just heard of a scandal and you are the first one to know.
You pass it on to four person in a matter of 30 minutes. Each of these four in turn
passes it to four other persons in the next 30 minutes and so on.

How long it will take for everybody in the World to get to know the scandal?

Assume that nobody hears it more than once and the population of the World is
approximately 5.6 billions.
Answer

Everybody in the World will get to know the scandal in 8 hours.

You came to know of a scandal and you passed it on to 4 persons in 30 minutes.
So total (1+4) 5 persons would know about it in 30 minutes.

By the end of one hour, 16 more persons would know about it. So total of
(1+4+16) 21 persons would know about it in one hour.

Similarly, the other (1+4+16+64) persons would have know about it in one and a
half hours. (1+4+16+64+256) persons would have know about it in two hours and
so on...

It can be deduced that the terms of the above series are the power of 4 i.e. 4^0,
4^1, 4^2, 4^3 and so on upto (2N+1) terms. Also, the last term would be 4^2N
where N is the number of hours.

Sum of the above mentioned series = [4^(2N+1)-1]/3

The sum of the series must be 5.6 billions. Hence, equating the sum of the series
with 5.6 billions, we get N=8 hours.

Scandals travel FAST !!!
A B C

D

E F G

H

I
Each of the digits from 1 to 9 is represented by a different letter above. Also, A +
B + C = C + D + E = E + F + G = G + H + I = 13

Which digit does E represent?
Answer

E represents 4.

Find out all possible groups of three different numbers that add up to 13 and
arrange them according to given condition.

If one number is 9, it must go with 1 and 3.
If one number is 8, it must go with either 1 and 4 or 2 and 3.

It is clear that 9 must go with 1 and 3. Also, no digit may be used in more than
two sums. Hence, there are 2 cases:
Case I: If 8 goes with 1 and 4, then 7 goes with 2 and 4, then 6 goes with 2 and
5.
Case II: If 8 goes with 2 and 3, then 7 goes with 2 and 4, then 6 goes with 3 and
4.

But in case II, 3 is used in three sums. Hence, Case I is correct. And the possible
arrangements are:

9 3 1 5 6 2

8 7

4 7 2 4 8 1

6 3

5 9
Thus, E must be 4.
A, B and C are three points on a straight line, not necessarily equidistant with B
being between A and C. Three semicircles are drawn on the same side of the
line with AB, BC and AC as the diameters. BD is perpendicular to the line ABC,
and D lies on the semicircle AC.

If the funny shaped diagram between the three semicircles has an area of 1000
square cms, find the length of BD.

Answer

The length of BD is 35.68 cms

There are 3 right-angled triangles - ABD, CBD and ADC.

From ABD, AB^2 + BD^2 = AD^2 ------ I
From CBD, CB^2 + BD^2 = CD^2 ------ II
From ADC, AD^2 + CD^2 = AC^2 ------ III

Adding I and II,
AB^2 + BC^2 + 2*BD^2 = AD^2 + CD^2 ------ IV

FROM III and IV
AB^2 + BC^2 + 2*BD^2 = AC^2
AB^2 + BC^2 + 2*BD^2 = (AB+CB)^2
2*BD^2 = 2*AB*CB
BD^2 = AB*CB
BD = SQRT(AB*CB)

Given that funny shaped diagram beween three semicircles has an area of
1000 square cms.
[PI/2 * (AC/2)^2] - [PI/2 * (AB/2)^2] - [PI/2 * (BC/2)^2] = 1000
PI/8 * [AC^2 - AB^2 - BC^2] = 1000
PI * [(AB+BC)^2 - AB^2 - BC^2] = 8000
PI * [2*AB*BC] = 8000
AB * BC = 4000/PI

Hence BD = SQRT(4000/PI) = 35.68 cms
where PI = 3.141592654

Hence, the length of BD is 35.68 cms.

P


Gomzi has 3 timepieces in his house - a wall clock, an alarm clock and a
wristwatch. The wristwatch is always accurate, whereas the wall clock gains 2
minutes everyday and the alarm clock loses 2 minutes everyday.

At exactly midnight last night, all three watches were showing the same time.

If today is 25 July 2003, then on which date all three clocks will show the same
time again?

Answer

All three clocks will show the same time again on midnight between 19 July
2004 and 20 July 2004.

A clock finishes on round in 12*60 i.e. 720 minutes.

If a clock gains 2 minutes everyday, then it would be 720 minutes ahead after
360 days. Thus, after 360 days, it will show the same time again.

Similary, if a clock loses 2 minutes everyday, then it would be 720 minutes
behind after 360 days. Thus, after 360 days, it will show the same time again.

Thus, after 360 days all three clocks will show the same time again i.e. midnight
between 19 July 2004 and 20 July 2004.
You have 9 marbles. 8 marbles weigh 1 ounce each, & one marble weighs 1.5
ounces. You are unable to determine which is the heavier marble by looking at
them. You have a weighing scale that consists of 2 pans, but the scale is only
good for 2 total weighings.

How can you determine which marble is the heaviest one using the scale & in 2
weighings?
Answer

Divide 9 marbles into 3 groups of 3 marbles each.

Take any 2 groups and place them on each pan. If they balance, remove the
marbles from the pans, & place any 2 of the marbles from the remaining
unweighed group on the pans, 1 on each pan.

If one is heavier, it is the heavier marble, but if they balance, the remaining
unweighed marble is the heavier one.

If your first weighing does not balance, remove the marbles from the lighter pan,

& place 1 marble on each pan from the heavier pan. The heavier 1 is the 1.5
ounce marble, but if they balance, then the marble from the heavy pan from the
first weighing that was not weighed in the second weighing is the heavy 1.
Once a week a wagon driver leaves his hut and drives his wagon to the river
dock to pick up supplies for his town. At 4:05 PM, one-fifth of the way to the dock,
he passes the Temple. At 4:15 PM, one-third of the way, he passes the Preetam-
Da-Dhabaa.

At what time does he reached the dock?

Answer

5:05 PM

At 4:05 PM, the wagon driver passes the temple, one-fifth of the way to the dock.
Also, at 4:15 PM, he passes the Preetam-Da-Dhabaa, one-third of the way.
Thus, he travels 2/15 (1/3 - 1/5) of the distance in 10 minutes.

At 4:15 PM, he has already travelled 1/3 of the distance. Thus 2/3 of the way is
remaining, which can be travelled in
= ( (2/3) * 10 ) / (2/15)
= 50 minutes

At 4:15, he was at Preetam-Da-Dhabaa.and remaining way will take 50 more
minutes. Hence, the driver will reach at 5:05 PM to the dock.

Four prisoners escape from a prison.
The prisoners, Mr. East, Mr. West, Mr. South, Mr. North head towards different
directions after escaping.
The following information of their escape was supplied:

• The escape routes were North Road, South Road, East Road and West
Road
• None of the prisoners took the road which was their namesake
• Mr. East did not take the South Road
• Mr.West did not the South Road
• The West Road was not taken by Mr. East

What road did each of the prisoners take to make their escape
Answer

Put all the given information into the table structure as follow:

North Road South Road East Road West Road

Mr. North No
Mr. South No
Mr. East No No No
Mr. West No No

Now from table, two things are obvious and they are:

• Mr.North took the South Road
• Mr.East took the North Road

Put this information into the table, Also keep in mind that the prisoners head
towards different directions after escaping.

North Road South Road East Road West Road
Mr. North No YES No No
Mr. South No No
Mr. East YES No No No
Mr. West No No No

Now from the table:

• Mr.West took the East Road
• Mr.South took the West Road

So the answer is:

• Mr.North took the South Road
• Mr.South took the West Road
• Mr.East took the North Road
• Mr.West took the East Road

Shahrukh speaks truth only in the morning and lies in the afternoon, whereas
Salman speaks truth only in the afternoon and lies in the morning.

A says that B is Shahrukh.

Is it morning or afternoon and who is A - Shahrukh or Salman?
Answer

It is Afternoon and A can be Salman or Shahrukh. If A is Salman, he is speaking
truth. If A is Shahrukh, he is lying.

Want to confirm it? Consider following 4 possible answers and check for its
truthness individually.

1. It is Morning and A is Shahrukh
2. It is Morning and A is Salman
3. It is Afternoon and A is Shahrukh
4. It is Afternoon and A is Salman

A rich man died. In his will, he has divided his gold coins among his 5 sons, 5
daughters and a manager.

According to his will: First give one coin to manager. 1/5th of the remaining to the
elder son. Now give one coin to the manager and 1/5th of the remaining to
second son and so on..... After giving coins to 5th son, divided the remaining
coins among five daughters equally.

All should get full coins. Find the minimum number of coins he has?
Answer

We tried to find out some simple mathematical method and finally we wrote small
C program to find out the answer. The answer is 3121 coins.

Here is the breakup:
First son = 624 coins
Second son = 499 coins
Third son = 399 coins
Forth son = 319 coins
Fifth son = 255 coins
Daughters = 204 each
Manager = 5 coins
There is a grid of 20 squares by 10 squares. How many different rectangles are
possible?

Note that square is a rectangle.
Answer

11550

The Generic solution to this is:
Total number of rectangles = (Summation of row numbers) * (Summation of
column numbers)

Here there are 20 rows and 10 columns or vice versa. Hence, total possible
rectangles
= ( 20 + 19 + 18 + 17 + 16 + .... + 3 + 2 + 1 ) * ( 10 + 9 +8 + 7 + .... + 3 + 2 + 1)
= ( 210 ) * (55)
= 11550

Hence, total 11,550 different rectangles are possible.

If you don't believe it, try formula on some smaller grids like 4x2, 3x2, 3x3 etc...
If A+B=C, D-C=A and E-B=C, then what does D+F stands for? Provide your
answer in letter terms as well as in number terms.
Submitted by : David
Answer

J or 10

A simple one.

Assume that each character represents the number equivalent to the position in
the alphabet i.e. A = 1, B = 2, C = 3, D = 4 and so on. Now let's check our
assumption.

A + B = C i.e. 1 + 2 = 3
D - C = A i.e. 4 - 3 = 1
E - B = C i.e. 5 - 2 = 3

Thus, our assumption was Correct. Hence, D + F = J i.e. 4 + 6 = 10
A woman took a certain number of eggs to the market and sold some of them.

The next day, through the industry of her hens, the number left over had been
doubled, and she sold the same number as the previous day.

On the third day the new remainder was tripled, and she sold the same number
as before.

On the fourth day the remainder was quadrupled, and her sales the same as
before.

On the fifth day what had been left over were quintupled, yet she sold exactly the
same as on all the previous occasions and so disposed of her entire stock.

What is the smallest number of eggs she could have taken to market the first
day, and how many did she sell daily? Note that the answer is not zero.
Submitted
Answer

She took 103 eggs to market on the first day and sold 60 eggs everyday.

Let's assume that she had N eggs on the first day and she sold X eggs everyday.
Putting down the given information in the table as follow.

Days Eggs at the start of the day Eggs Sold Eggs Remaining
Day 1 N X N-X
Day 2 2N-2X X 2N-3X
Day 3 6N-9X X 6N-10X
Day 4 24N-40X X 24N-41X
Day 5 120N-205X X 120N-206X

It is given that she disposed of her entire stock on the fifth day. But from the table
above, the number of eggs remaining are (120N-206X). Hence,
120N - 206X = 0
120N = 206X
60N = 103X

The smallest value of N and X must be 103 and 60 respectively. Hence, she took
103 eggs to market on the first day and sold 60 eggs everyday.
John lives in "Friends Society" where all the houses are in a row and are
numbered sequentially starting from 1. His house number is 109.

Jessy lives in the same society. All the house numbers on the left side of Jessy's
house add up exactly the same as all the house numbers on the right side of her
house.

What is the number of Jessy's house? Find the minimal possible answer.
Answer

There are 288 houses and Jessy's house number is 204.

Let's assume that in the "Friends Society" there are total N houses numbered
from 1 to N and Jessy's house number is X.

Now it is given that all the house numbers on the left side of Jessy's house add
up exactly the same as all the house numbers on the right side of her house.
Hence,
1 + 2 + 3 + ..... + (X-1) = (X+1) + (X+2) + (X+3) + ..... + N

Both the sides of the above equations are in A.P. Hence, using A.P. summation
formaula,

[(X-1)/2][2*(1) + (X-1-1)] = [(N-X)/2][2*(X+1) + (N-X-1)]
[X-1][(2) + (X-2)] = [N-X][(2X+2) + (N-X-1)]
(X-1)(X) = (N-X)(N+X+1)
X2 - X = N2 + NX + N - NX - X2 - X
X2 = N2 + N - X2
2X2 = N2 + N
X2 = (N2 + N)/2
X2 = N(N+1)/2

Now, using Trial and Error method to find values of N and X such that above
equation is satisfied, we get

1. N = 8, X = 6
2. N = 49, X = 35
3. N = 288, X = 204
4. N = 1681, X = 1189
5. N = 9800, X = 6930

But we require minimal possible answer and it is given that John's house number
is 109. It means that there are atleast 109 houses. Hence, first two are not
possible. And the answer is : there are 288 houses and Jessy's house number is
204.
Makayla had $1.19 in change. None of the coins was a dollar.

Nicole ask her for change for a dollar, but Makayla could not make change.

What coins did she have?
Submitted
Answer

As it is given that Makayla had $1.19, it means she would have four pennies.
Now, the remaining $1.15 in coins must not add up for exactly a dollar. Therefore
she would not have 4 quarters or 2 quarters and 5 dimes. But she would have
either 1 quarter or 3 quarters. Hence, there are 2 solutions.

Solution I
1 Quarter, 9 Dimes, 4 Pennies (0.25 + 0.90 + 0.04 = $1.19)

Solution II
3 Quarters, 4 Dimes, 4 Pennies (0.75 + 0.40 + 0.04 = $1.19)
A group of friends went on a holiday to a hill station. It rained for 13 days. But
when it rained in the morning, the afternoon was lovely. And when it rained in the
afternoon, the day was preceded by clear morning.

Altogether there were 11 very nice mornings and 12 very nice afternoons. How
many days did their holiday last?

Answer

The holiday last for 18 days.

Let's assume the number of days as follows:
Rain in the morning and lovely afternoon = X days
Clear morning and rain in the afternoon = Y days
No rain in the morning and in the afternoon = Z days

Number of days with rain = X + Y = 13 days
Number of days with clear mornings = Y + Z = 11 days
Number of days with clear afternoons = X + Z = 12 days

Solving above 3 equations, we get X = 7, Y = 6 and Z = 5

Hence, total number of days on holiday = 18 days

Y F Y

-----------

A Y | N E L L Y

| N L Y

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

P P L

P N H

----------

N L Y

N L Y

----------

0 0 0
letter N, no other letter can be 3 and all other N in the puzzle must be 3.
Submitted by : Calon

Answer

See the pattern of the Y. AY * Y = NLY i.e. Y is multiplied by Y and the last digit
of the answer is also Y. Thus, the value of Y would be 5 or 6.

Also,
H=0 as L - H = L
P = 2N as P - N = N
L - Y = P = 2N
E-L=p

Let's find out the minimum possible values. If N=1, then P=2, Y=5, L=7 and E=9.
Note that the value of Y can not be 6 as it makes L=8 and E=10 which is not
possible. Hence, Y=5, N=1, P=2, L=7, E=9, H=0

Now, using trial-n-error or rather solving F*AY=PNH, we get F=6 and A=3.

5 6 5 Y F Y

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

3 5 | 1 9 7 7 5 A Y | N E L L Y

| 1 7 5 | N L Y

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

2 2 7 P P L

2 1 0 P N H

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

1 7 5 N L Y

1 7 5 N L Y

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

0 0 0 0 0 0


Here is the family tree of Mr. RAHUL
RAHUL

|

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

| | | |

RATISH YASH OM TRILOK

| | ?

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

| | | | |

AMAR AMIT RAM HARSH ASHOK

| |

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

| | | | |

How many children does Mr. TRILOK have?

Answer

TRILOK have 5 children.

Name of the person and number of his children are related by some pattern.

Assign each vowel following values.
A=0 E=1 I=2 O=3 U=4

The number of children to any person is the sum of the values represented by
vowels in his name.
RATISH = 0(A) + 2(I) = 2
OM = 3(O) = 3
AMIT = 0(A) + 2(I) = 2
ASHOK = 0(A) + 3(O) = 3
TRILOK = 2(I) + 3(O) = 5

Hence, TRILOK have 5 children.

Major Jasbir is forming five-person Special Task Group. The group must contain

one leader, two bomb-experts and two soldiers.

P, Q and R are possible bomb-experts. R, S and T are possible leaders. U, V
and W are possible soldiers. Also, P and R prefers to work with each other in the
same team. T prefers to work only if V works.

How many different possible Groups, Major Jasbir can make?
Answer

Major Jasbir can make 8 different possible groups.

As 2 bomb-experts to be selected from the given 3 and also P & R prefers to
work together, PR must be there in all the possible Groups. Also, T prefers to
work only if V works. It doesn't mean that V won't work without T.

Hence, possible groups are:
PR - S - UV
PR - S - VW
PR - S - WU

PR - T - UV
PR - T - VW

PQ - R - UV
PQ - R - VW
PQ - R - WU

Hence, there 8 different groups are possible.
The secret agent X emailed some code to his head office. They are "RADAR,
LEVEL, ROTOR, REDIVIDER, MOTOR". But four of these five words have
something in common and one is fake.

Can you tell which one is fake? Ignore the fact that four of the code-words are of
the same length.
Answer

The fake code-word is MOTOR.

All the code-words except MOTOR are Palindromes.

In the village called TALAJA, only three TV channels are available - Moon
Plus, Mony and Mee TV.

Out of 4000 TV viewers in the village, 1500 watch Moon TV, 2000 watch Mony

and 2500 watch Mee TV.

Amongst these, 500 viewers watch Moon Plus and Mony, 800 watch Moon
Plus and Mee TV, and 1000 watch Mony and Mee TV.

How many viewers watch all three channels?

Answer

300 viewers watch all three channels.

Let's assume that total X viewers watch all three channels.

total viewers who watch only Moon Plus and Mony = 500-X
total viewers who watch only Moon Plus and Mee TV = 800-X
total viewers who watch only Mony and Mee TV = 1000-X

total viewers who watch only Moon Plus
= 1500 - (500-X) - (800-X) - X
= 200 + X

total viewers who watch only Mony
= 2000 - (500-X) - (1000-X) - X
= 500 + X

total viewers who watch only Mee TV
= 2500 - (1000-X) - (800-X) - X
= 700 + X

We know that total viewers are 4000. Summing up all 7 values,
X + (500 - X) + (800 - X) + (1000 - X) + (200 + X) + (500 + X) + (700 + X) = 4000
X + 3700 = 4000
X = 300

Hence, total 300 viewers watch all three channels.
A man was looking at a portrait. Someone asked him, "Whose picture are you
looking at?"

He replied, pointing at the portrait: "Brothers and sisters have I none, but this
man's son is my father's son."

Now whose picture is the man looking at?
Answer

The man is looking at his FATHER's portrait.

"my father's son" is the man himself as he do not have any brothers and sisters.
So the statement reduces to "this man's son is myself." Now it is clear that the
portrait is of his father.
Given the following facts:

1. Dinesh is younger than Farukh and older than Gurmit.
2. Jatin is younger than Chandu and older than Eshrat.
3. Amit is younger than Irfan and older than Chandu.
4. Farukh is younger than Bhavin and older than Hemant.
5. Irfan is younger than Gurmit and older than Jatin.
6. Hemant is older than Gurmit.

Who is the Youngest?
Answer

Eshrat is the youngest.

Discard whoever are older than someone.

From (1) Gurmit is younger than Dinesh and Farukh.
From (5) Jatin is younger than Irfan and Gurmit.
From (2) Eshrat is younger than Jatin and Chandu.

From above 3 deductions, Eshrat is younger than Dinesh, Farukh, Irfan, Gurmit,
Jatin and Chandu.

Also,
From (3) Chandu is younger than Amit and Irfan.
From (4) Hemant is younger than Farukh and Bhavin.
From (6) Gurmit is younger than Hemant.

From above 3 deductions, Gurmit is younger than Farukh, Bhavin and Hemant.
Also, Chandu is younger than Amit and Irfan. But as seen earlier, Eshrat is
younger than Gurmit and Chandu.

Hence, Eshrat is the youngest.
Last Saturday Milan went for the late night show and came late. In the morning
family members asked him which movie did he see. He gave different answers to
everyone.

• He told to his father that he had gone to see MONEY.
• According to his mom, he saw either JOHNY or BABLU.
• His elder brother came to know that he saw BHABI.
• To his sister, he told ROBOT.
• And his grandpa heard that he saw BUNNY.

Thus, Milan gave six movie names, all five letter words. But he saw some other
movie with five letter word. Moreover, each of the six movie names mentioned
above has exactly two letters common with the movie he saw. (with the same
positions)

Can you tell which movie did Milan see?
Answer

Milan saw BOBBY.

The six movie names are - MONEY, JOHNY, BABLU, BHABI, ROBOT and
BUNNY.

Compare MONEY and JOHNY. They have O common at the second place and Y
common at the fifth place. Also, they can't have two different letters each,
common with the required movie as the letters in remaining three places are all
different. Thus, the required movie must have either O at the second place or Y
at the fifth place or both.

Similarly, comparing JOHNY and BUNNY - the required movie must have either
N at the fourth place or Y at the fifth place or both. Also, comparing MONEY and
BUNNY - the required movie must have either N at the third place or Y at the fifth
place or both.

From the above 3 deduction, either Y is at fifth place or O is at the second place
and N is at the third & fourth place. The later combination is not possible as
BABLU, BHABI & ROBOT will need at least 3 other letters which makes the
required movie 6 letter long. Hence, the required movie must have Y at the fifth
place.

Now Y is not there in BABLU and BHABI at the fifth place and they have only B
common at the first place. Hence, B must be the first letter.

As B is at the first place and Y is at the fifth place and every movie has exactly 2
letters common with the required movie. From BUNNY, the required movie do
not have U at the second place and N at the third and fourth place. Now looking
at JOHNY and MONEY, they must have O common at the second place.

Using the same kind of arguments for BABLU, BHABI and ROBOT, we can
conclude that Milan saw BOBBY.

Jim lies a lot. He tells the truth on only one day in a week.

One day he said: "I lie on Mondays and Tuesdays."
The next day he said: "Today is either Sunday, Saturday or Thursday."
The next day he said: "I lie on Fridays and Wednesdays."

On which day of the week does Jim tell the truth?
Answer

Jim tells the truth on Tuesday.

As Jim tells truth only on one day in a week, his statement on day 1 and day 3
both can not be false. Otherwise he tells truth on more than one days in a week.
Also, all three statements are mad on three consecutive days, statement made
on day 1 and day 3 both can not be true. Thus, either the statement made on day
1 or day 3 is true and other is false. Also, the statement made on day 2 must be
false i.e. day 1 is not Saturday, Friday or Wednesday.

Let's assume that the statement 1 is true. Then from the statement 3, day 1 must
be either Friday or Wednesday. But it is already deduced that day 1 is not
Saturday, Friday or Wednesday.

Hence, the statement made on day 1 is false and the last statement is true. then
from the statement 1, day 3 must be either Monday or Tuesday. But it is already
deduced that day 1 can not be Saturday i.e. day 3 can't be Monday. Hence, Jim
tells the truth on Tuesday.

4 men can dig 4 holes in 4 days.

How many hours does it take for 1 man to dig half a hole?
Submitted
Answer

There is nothing like "HALF HOLE".
Consider a chessboard with a single Rook. A Rook can move any number of
square sideways/forward, but not diagonally.

What is the minimum number of moves the Rook needs to make, in order to pass
over all the squares on the chessboard and return to the original position?
Answer

16 moves

As a Rook can move any number of square sideways/forward, but not diagonally
and there are 8 rows and 8 columns on the chessboard; the Rook needs
minimum 16 moves to pass over all the squares and return to the original
position.
A farmer needs 8 gallons of water. He has only three unmared buckets, two 6
gallon and one 11 gallon bucket.

How can he collect 8 gallons of water using three unmarked buckets? Provide
solution with minimal water wastage.
Answer

Here is the solution with 10 gallon water wastage.
OPERATIONS 6 6 11
Fill 6 gallon bucket with water 6 0 0
Empty 6 gallon bucket into 11 gallon bucket 0 0 6
Fill 6 gallon bucket with water 6 0 6
Fill 11 gallon bucket to full using filled 6 gallon bucket. This will leave 1
1 0 11
gallon water in 6 gallon bucket
Empty 11 gallon bucket into second 6 gallon bucket. 1 6 5
Empty 11 gallon bucket - wastage of 5 gallon water 1 6 0
Empty second 6 gallon bucket into 11 gallon bucket 1 0 6
Fill seccond 6 gallon bucket with water 1 6 6
Fill 11 gallon bucket to full using filled second 6 gallon bucket. This will
1 1 11
leave 1 gallon water in second 6 gallon bucket
Fill first 6 gallon bucket with 1 gallon water which is in second 6 gallon
2 0 11
bucket
Empty 11 gallon bucket into second 6 gallon bucket. 2 6 5
Empty 11 gallon bucket - wastage of 5 gallon water 2 6 0
Fill 11 gallon bucket with water in both the 6 gallon buckets 0 0 11
I bought a car with a peculiar 5 digit numbered licence plate which on reversing
could still be read. On reversing value is increased by 78633.

Whats the original number if all digits are different?
Answer

Only 0 1 6 8 and 9 can be read upside down. So on rearranging these digits we
get the answer as 10968.
Jack and Jill are playing cards for a stake of $1 a game. At the end of the
evening, Jack has won 3 games and Jill has won $3. How many games did they
play?
Submitted by : Nathalie Drouin
Answer

They played total of 9 games. Jack won 3 games and Jill won 6 games.

If Jack has won three games and Jill has won $3, she lost a dollar for each loss,

therefore she has won 6 and lost 3 to make $3 and he won the other 3 that she
lost!

Sam and Mala have a conversation.

• Sam says I am certainly not over 40
• Mala says I am 38 and you are atleast 5 years older than me
• Now Sam says you are atleast 39

All the statements by the two are false. How old are they really?
Answer

Sam is 41 and Mala is 37.

Let's invert the teaser and read it like this :

• Sam says I am certainly over 40
• Mala says I am not 38 and you are atmost 4 years older than me
• Now Sam says you are atmost 38

From first statement it is clear that Sam is over 40. Also, from next 2 statements
it is clear that Mala is less then 38. Hence the possibilities are :
Sam = 41, 42, 43, 44, 45, ......
Mala = 37, 36, 35, 34, 33, ......

It also says that the difference between their age is maximum 4 years. Hence,
there is only one possible pair i.e. 41 and 37, all other combination have
differences more then 4.

Hence the answer - Sam is 41 and Mala is 37.
A person travels on a cycle from home to church on a straight road with wind
against him. He took 4 hours to reach there.

On the way back to the home, he took 3 hours to reach as wind was in the same
direction.

If there is no wind, how much time does he take to travel from home to church?
Answer

Let distance between home and church is D.

A person took 4 hours to reach church. So speed while travelling towards church
is D/4.

Similarly, he took 3 hours to reach home. So speed while coming back is D/3.

There is a speed difference of 7*D/12, which is the wind helping person in 1
direction, & slowing him in the other direction. Average the 2 speeds, & you have
the speed that person can travel in no wind, which is 7*D/24.

Hence, person will take D / (7*D/24) hours to travel distance D which is 24/7
hours.

Answer is 3 hours 25 minutes 42 seconds
There are N secret agents each know a different piece of secret information.
They can telephone each other and exchange all the information they know. After
the telephone call, they both know anything that either of them knew before the
call.

What are the minimum number of telephone calls needed so that all of the them
know everything?
Answer

(2N - 3) telephone calls, for N = 2,3
(2N - 4) telephone calls, for N > 3

Divide the N secret agents into two groups. If N is odd, one group will contain
one extra agent.

Consider first group: agent 1 will call up agent 2, agent 2 will call up agent 3 and
so on. Similarly in second group, agent 1 will call up agent 2, agent 2 will call up
agent 3 and so on. After (N - 2) calls, two agents in each the group will know
anything that anyone knew in his group, say they are Y1 & Y2 from group 1 and
Z1 & Z2 from group 2.

Now, Y1 will call up Z1 and Y2 will call up Z2. Hence, in next two calls total of 4
agents will know everything.

Now (N - 4) telephone calls are reqiured for remaining (N - 4) secret agents.

Total telephone calls require are
= (N - 2) + 2 + (N - 4)
= 2N - 4

Let's take an example. Say there are 4 secret agents W, X, Y & Z. Divide them
into two groups of 2 each i.e. (W, X) and (Y, Z). Here, 4 telephone calls are
required.

1. W will call up X.
2. Y will call up Z.
3. W, who knows WX will call up Y, who knows YZ.
4. X, who knows WX will call up Z, who knows YZ.

Take an another example. Say there are 5 secret agents J, K, L, M & N. Divide
them into two groups i.e. (J, K) and (L, M, N). Here, 6 telephone calls are
required.

1. J will call up K.
2. L will call up M.
3. M will call up N. Now M and N know LMN.
4. J, who knows JK will call up M, who knows LMN.
5. K, who knows JK will call up N, who knows LMN.
6. L will call up to anyone of four.

Mrs. F has invited several wives of delegates to the United Nations for an
informal luncheon. She plans to seat her 9 guests ina row such that each lady
will be able to converse with the person directly to her left and right. She has
prepared the following list.

Mrs. F speaks English only.
Mrs. G speaks English and French.
Mrs. H speaks English and Russian.
Mrs. J speaks Russian only.
Mrs. K speaks English only.
Mrs. L speaks French only.
Mrs. M speaks French and German.
Mrs. N speaks English and German.
Mrs. O speaks English only.

How many distinct seating arrangements are possible? Give all possible seating
arrangements.

Note that ABCD and DCBA are the same.
Answer

126 distinct seating arrangements are possible.

Mrs. J and Mrs. H must be together and Mrs. J must be at the end as Mrs. J
speaks only Russian and Mrs. H is the only other Russian speaker.

Mrs. L speaks only French and there are two others - Mrs. G and Mrs. M - who
speak French. Here there are 2 cases.

• CASE A : Mrs. L is at the other end
If Mrs. L is at the other end, either Mrs. G or Mrs. M must seat next to her.

o CASE AA : Mrs. G seats next to Mrs. L
Then, Mrs. M must seat next to Mrs. G and Mrs. N must seat next
to Mrs. M. This is because Mrs. M speaks French and German, and
Mrs. N is the only other German speaker. Thus, the possible

seating arrangement is JHxxxNMGL, where x is the English
speakers. Mrs. F, Mrs. K and Mrs. O can be arranged in remaining
3 positions in 3! different ways i.e. 6 ways.

o CASE AB : Mrs. M seats next to Mrs. L
If so, then either Mrs. N or Mrs. G must seat next to Mrs. M

CASE ABA : Mrs. N seats next to Mrs. M
Thus, the possible seating arrangement is JHxxxxNML,
where x is the English speakers. Mrs. F, Mrs. G, Mrs. K and
Mrs. O can be arranged in remaining 4 positions in 4!
different ways i.e. 24 ways.

CASE ABB : Mrs. G seats next to Mrs. M
Thus, the possible seating arrangement is JHxxxxGML,
where x is the English speakers. Mrs. F, Mrs. K, Mrs. N and

• CASE B : Mrs. L does not seat at the end
It means that Mrs. G, Mrs. L and Mrs. M must seat together. Also, Mrs. L
must seat between Mrs. G and Mrs. M.

o CASE BA : Mrs. G seats left and Mrs. M seats right to Mrs. L i.e.
GLM

CASE BAA : GLM is at the other end
Thus, the possible seating arrangement is JHxxxxGLM,
where x is the English speakers. Mrs. F, Mrs. K, Mrs. N and

CASE BAB : GLM is not at the other end
Then Mrs. N must seat next to Mrs. M. Now, we have a
group of four GLMN where Mrs. G and Mrs. N speak
English. Thus, the possible seating arrangement is JHxxxX,
where x is the individual English speakers and X is the group
of four females with English speakers at the both ends.
Thus, there are 4! different ways i.e. 24 ways.

o CASE BB : Mrs. M seats left and Mrs. G seats right to Mrs. L i.e.
MLG
Then, Mrs. N must seat next to Mrs. M. Now, we have a group of
four NMLG where Mrs. G and Mrs. N speak English. Thus, the
possible seating arrangement is JHxxxX, where x is the individual
English speakers and X is the group of four females with English
speakers at the both ends. Thus, there are 4! different ways i.e. 24
ways.

Thus, total different possible seating arrangements are :
= 6 (case AA) + 24 (case ABA) + 24 (case ABB) + 24 (case BAA) + 24 (case
BAB) + 24 (case BB)
= 126 seating arrangements

Thus, 126 distinct seating arrangements are poosible.
What is the smallest number which when divided by 10 leaves a remainder of 9,
when divided by 9 leaves a remainder of 8, when divided by 8 leaves a
remainder of 7, when divided by 7 leaves a remainder of 6 and so on until when
divided by 2 leaves a remainder of 1?
Answer

The smallest such number is 2519.

The easiest way is to find the Least Common Multiple (LCM) of 2, 3, 4, 5, 6, 7, 8
and 9. And subtract 1 from it.

The LCM of 2, 3, 4, 5, 6, 7, 8 and 9 is given by 2520. Hence, the required
number is 2519

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.
Answer

18

Assume that initial there were 3*X bullets.

So they got X bullets each after division.

All of them shot 4 bullets. So now they have (X - 4) bullets each.

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

Therefore, the equation is

3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X=6

Therefore the total bullets before division is = 3 * X = 18

Everyday in his business a merchant had to weigh amounts from 1 kg to 121
kgs, to the nearest kg. What are the minimum number of different weights
required and how heavy should they be?

The minimum number is 5 and they should weigh 1, 3, 9, 27 and 81 kgs
Replace each letter by a digit. Each letter must be represented by the same digit
and no beginning letter of a word can be 0.

O N E

O N E

O N E

+ O N E

-------

T E N
Answer

Use trial and error. 0 =1, N = 8 ,E = 2, T = 7

1 8 2

1 8 2

1 8 2

+ 1 8 2

------

7 2 8
A man is on a search for Atlantis and comes upon an island where all the
inhabitants know whether Atlantis is still around or not.

However, all of the inhabitants are either Fairies or Trolls and they all use a spell
to appear humanoid so you cannot tell which is which. And the Faries always tell
the truth and the Trolls always lie, but there is a slight complication, some of the
Fairies have gone insane and always lie and some of the Trolls have also gone
insane and always tell the truth.

So here is your task: you must ask the first inhabitant that you come to ONE
question and from that ONE question you must determine wether Atlantis is still
around or not.

What is the question that you must ask?
Answer

There are 2 answers to it:

Answer I"Is the statement that you are reliable equivalent to the statement that
Atlantis is still around?"

Answer II"Do you believe that the Statement that you are a Fairy is equivalent to
the statement that Atlantis is still around?"

A frog starts climbing 15 feet wall. Each hour he climbs 3 feet and rests for 30
minutes. During rest, he slips back 2 feet.

How many hours does the frog take to reach the top?

Answer

19 hours

A frog climbs 1 foot per 1 1/2 hours as during 30 minutes rest he slips back 2
feet. This way he will climb 12 feet in 18 hours. In next hour he will climb 3 more
feet i.e. he will complete 15 feet in 19 hours and will reach the top of the wall.
If a bear eats 65 pounds in fish every day EXCEPT every 6th day which it only
eats 45 pounds of fish.

If the bear continues this, how many pounds of fish will it eat in 200 days?
Submitted by : David
Answer

The bear will eat 12,340 pounds of fish in 200 days.

It is given that on every 6th day beareats 45 pounds of fish i.e. on day number 6,
12, 18, 24, .... 192, 198 the bear eats 45 pounds of fish.

Total number of 6th days = 200/6 = 33 (the bear eats 45 pounds)
Hence, the normal days are = 200 - 33 = 167 (the bear eats 65 pounds)

Thus, in 200 days, the bear will eat
= (167) * (65) + (33) * (45)
= 10855 + 1485
= 12,340 pounds
You have 3 points labelled A, B and C. You then have another 3 points labelled
1, 2 and 3. The aim of the puzzle is to connect point A with point 1, 2 and 3. Point
B with point 1, 2 and 3 and point C with point 1, 2 and 3.

Now while connecting the points you have to follow one rule - the lines cannot
cross over each other.
A B C

1 2 3
PS : You can arrange the points in order as long as the lines DO NOT cross over
each other.
Answer

There is no solution to it, if you consider 2 dimensions. It is impossible to join
each of points A, B and C with points 1, 2 and 3 without lines crossing each
other.

There is solution, if you consider 3 dimensions. Consider a circular base and a
line perpendicular to it passing from the center. Now take any 3 points along the
perimeter of the circular base as points 1, 2 and 3. Similarly take any 3 points
along the perpendicular line as points A, B and C. Now it is quite simple to join
each of points A, B and C with points 1, 2 and 3 without any of the lines crossing
each other.

The other possible 3D structure is Pyramid. Take points 1, 2 and 3 as a vertices
of the triangular base and points A, B and C along the height of the Pyramid
which is perpendicular to the triangular base and passing through the apex.

Suppose five bales of hay are weighed two at a time in all possible ways. The
weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121.

How much does each bale weigh?
Submitted by : Travis Lara

Answer

They weigh 54, 56, 58, 59, 62 pounds.

Let's assume that the weight of five bales are B1, B2, B3, B4 and B5 pounds
respectively. Also, B1 <= B2 <= B3 <= B4 <= B5

It is given that five bales of hay are weighed two at a time in all possible ways. It
means that each of the bale is weighted four times.
Thus,
4*(B1 + B2 + B3 + B4 + B5) = (110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 +
120 + 121)
4*(B1 + B2 + B3 + B4 + B5) = 1156
(B1 + B2 + B3 + B4 + B5) = 289 pounds

Now, B1 and B2 must add to 110 as they are the lightest one.
B1 + B2 = 110

Similarly, B4 and B5 must add to 121 as they are the heaviest one.
B4 + B5 = 121

From above three equation, we get B3 = 58 pounds

Also, it is obvious that B1 and B3 will add to 112 - the next possible higher value.
Similarly, B3 and B5 will add to 120 - the next possible lower value.
B1 + B3 = 112
B3 + B5 = 120

Substituting B3 = 58, we get B1 = 54 and B5 = 62
From 2 & 3 equations, we get B2 = 56 and B4 = 59

Hence, the weight of five bales are 54, 56, 58, 59 and 62 pounds.

Pinto says, "The horse is not Black."
Sandy says, "The horse is either Brown or Grey."
Andy says, "The horse is Brown."

At least one is telling truth and at least one is lying.

Can you tell the color of the horse?
Answer

The color of the horse can be any color other than Black and Brown.

If the color of the horse is Black - all are lying.

If the color of the horse is Brown - all are telling truth.

Thus, the horse is neither Black nor Brown.

If the color of the horse is Grey - Pinto and Sandy are telling truth whereas Andy
is lying.

If the color of the horse is other than Black, Brown and Grey - Pinto is telling truth
whereas Sandy and Andy are lying.

You must have noticed that for the given conditions, Pinto is always telling truth
whereas Andy is always lying

Three convicts are brought into the warden's office. He says he can parole one of them and to
decide which one he will parole he takes out 5 hats (3 red and 2 white). He stands behind them
and places a hat on each one of their heads and puts the other two remaining hats in a drawer.

He tells the prisioners they can look at the others hats and if they can tell which hat they have
on they will be the one who is paroled.

The first man looks at the other two and says, "I don't know."

The second man looks at the others hats and says, "I don't know."

The third man who is blind says, "Even though I have not the gift of sight I can tell by what the
others have said that the color of my hat is..."

What color is the blind mans hat and how does he know?
Submitted

Answer Hello sajeesh murali
• my Answers
The color of blind man's hat is Red. • my Favourites
• Modify Personal
It is sure that the first man saw either both Red hats or one White hat Info
and one Red hat. There are 6 such possibilities: • Subscribe
1) R R R • Logout
2) R R W

3) R W R

4) W R R

5) W R W

6) W W R
In all above possibilities, the first man won't be sure of the color of
his hat.

Now, the second man knows that the first man saw either both Red
hats or one White hat and one Red hat. And, he also knows that its
one of the above 6 possibilities. (like we know ;)) But he says, "I don't • Jigsaw Puzzle
know". That means that (2) and (5) are not the possibilities as in • Join the Dots
either case he would be sure of the color of his hat (Red) by just • Marbles Game
looking at the third man's color of hat (White). • Balls Game
• Towers of Hanoi
Now, the blind man knows that there are just 4 possibilities - (1), (3),
(4), (6) - and in all, the color of his hat is Red.
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Three Gold (G) coins, three Silver (S) coins and three Copper (C) coins are
arranged in a single row as follow:
G S C G S C G S C

• Only 2 adjacent unlike coins can be moved at any one time.
• The moved coins must be in contact with at least one other coin in line. i.e.
no pair of coins is to be moved and placed away from the remaining ones.
• No coin pairs can be reversed i.e. a S-C combination must remain in that
order in its new positionwhen it is moved.

What is the minimum number of moves required to get all the coins in following
order?
C C C S S S G G G
Show all moves.
Answer

Minimum number of moves are 8.

Move Order of Coins

0 G S C G S C G S C

1 G S G S C G S C C

2 G S C G S C C S G

3 G S C G S C C S G

4 G S C C S G S C G

5 G S C C S C S G G

6 G S C C C S S G G

7 C C C S S G G S G

8 C C C S S S G G G
A fly is flying between two trains, each travelling towards each other on the same
track at 60 km/h. The fly reaches one engine, reverses itself immediately, and
flies back to the other engine, repeating the process each time.

The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far
apart were the trains initially?
Answer

Initially, the trains were 240 km apart.

The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies
for 2 hours after trains started.

It's obvious that trains met 2 hours after they started travelling towards each
other. Also, trains were travelling at the speed of 60 km/h. So, each train traveled
120 km before they met.

Hence, the trains were 240 km apart initially.

What is the minimum number of numbers needed to form every number from 1 to
7,000?

Example: To form 4884, you would need 2 4s & 2 8s. 4822 requires a 4, a 8, & 2
2s, but you would not count the numbers again that you had already counted
from making 4884.
Answer

36

You will need 3 of numbers 0, 7, 8 & 9, & 4 of numbers 1-6.

A drinks machine offers three selections - Tea, Coffee or Random (Either tea or
Coffee) but the machine has been wired up wrongly so that each button does not
give what it claims.

If each drink costs 50p, how much minimum money do you have to put into the
machine to work out which button gives which selection?
Submitted
Answer

You have to put just 50p.

Put 50p and push the button for Random. There are only 2 possibilities. It will
give either Tea or Coffee.

• If it gives Tea, then the button named Random is for Tea. The button
named Coffee is for Random selection. And the button named Tea is for
Coffee.
• If it gives Coffee, then the button named Random is for Coffee. The button
named Tea is for Random selection. And the button named Coffee is for
Tea.

Thus, you can make out which button is for what by putting just 50p and pressing
Random selection first.
You have 13 balls which all look identical. All the balls are the same weight
except for one. Using only a balance scale, can find the odd one out with only 3
weighings?

Is it possible to always tell if the odd one out is heavier or lighter than the other
balls?
Submitted by : Brett Hurrell
Answer

It is always possible to find odd ball in 3 weighings and in most of the cases it is
possible to tell whether the odd ball is heavier or lighter. Only in one case, it is
not possible to tell the odd ball is whether heavier or lighter.

1. Take 8 balls and weigh 4 against 4.

2. One of these 8 balls is the odd one. Name the balls on heavier side of the
scale as H1, H2, H3 and H4. Similarly, name the balls 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 ball from
the remaining 5 balls in intial weighing.

If both are equal, L4 is the odd ball and is lighter.
If L2 is light, L2 is the odd ball and is lighter.
If L3 is light, L3 is the odd ball and is lighter.

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

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

3. One of the remaining 5 balls is the odd one. Name the balls as C1, C2,
C3, C4, C5. Weight (C1, C2, C3) against (X1, X2, X3) where X1, X2, X3
are any three balls from the first weighing of 8 balls.

o If both are equal, one of remaining 2 balls is the odd i.e. either C4
or C5. Weigh C4 with X1
If both are equal, C5 is the odd ball. But you can not tell
whether it is heavier or lighter.
If C4 is heavy, C4 is the odd ball and is heavier.
If C4 is light, C4 is the odd ball and is lighter.

o If (C1, C2, C3) is heavier side, one of C1, C2, C3 is the odd ball
and is heavier. Weigh C1 and C2.
If both are equal, C3 is the odd ball and is heavier.

o If (C1, C2, C3) is lighter side, one of C1, C2, C3 is the odd ball and
is lighter. Weigh C1 and C2.
If both are equal, C3 is the odd ball and is heavier.

How many squares are there in a 5 inch by 5 inch square grid? Note that the grid
is made up of one inch by one inch squares.
Submitted by : Kristin Monroe
Answer

There are 55 squares in a 5 by 5 grid.

There are 25 squares of one grid.
There are 16 squares of four grids i.e. 2 by 2
There are 9 squares of nine grids i.e. 3 by 3
There are 4 squares of sixteen grids i.e. 4 by 4
There is 1 square of twenty-five girds i.e. 5 by 5

Hence, there are total 25 + 16 + 9 + 4 + 1 = 55 squares.

You must have noticed one thing that total number squares possible of each size
is always a perfact square i.e. 25, 16, 9, 4, 1

For a grid of N by N, the possible number of squares are
= N2 + (N - 1)2 + (N - 2)2 + (N - 3)2 + ......... + 32 + 22 + 12

For 1 by 1 grid, total squares = 12 = 1
For 2 by 2 grid, total squares = 22 + 12 = 5
For 3 by 3 grid, total squares = 32 + 22 + 12 = 14
For 4 by 4 grid, total squares = 42 + 32 + 22 + 12 = 30
For 5 by 5 grid, total squares = 52 + 42 + 32 + 22 + 12 = 55
Five horses ran in the race.

• There were no ties.
• Sikandar did not come first.
• Star was neither first nor last.
• Mughal Glory came in one place after Sikandar.
• Zozo was not second.
• Rangila was two place below Zozo.

In what order did the horses finish?
Answer

It's simple.

Let's find the possible places horses can finish. Possibilities are:
Sikandar - 2,3,4 (not 5th as Mughal Glory came one place after him)
Star - 2,3,4
Mughal Glory - 3,4,5
Zozo - 1,3 (not 4th & 5th as Rangila is two place after him)
Rangila - 3,5

So the result is:
1 Zozo
2 Star
3 Rangila
4 Sikandar
5 Mughal Glory

If you added together the number of 2's in each of the following sets of numbers,
which set would contain the most 2's: 1-333, 334-666, or 667-999?
Answer

1-333

The reason why is because 200-299 each begins with a 2!

If one person sends the e-mail to two friends, asking each of them to copy the
mail and send it to two of their friends, those in turn send it to two of their friends
and so on.

How many e-mails would have been sent by the time it did 30 sets?
Answer

2147483646

First person sent the mail to 2 persons. Those 2 sent the mail to 2 persons each,
total 4 persons. Now, those 4 person sent mail to total 8 persons, then 8 to 16
persons, 16 to 32 persons and so on.... Hence, it a series of 2, 4, 8, 16, 32 upto
30 numbers

It is a Geometric series with common ratio 2 and first number is also 2.
Summation of such series is given by A * (Rn - 1) / (R - 1) where
A = First term
R = Common Ratio
n = total numbers

So total number of times mail sent by the time it did 30 sets
= 2 * (230 - 1) / (2 - 1)
= 2 * (1073741824 - 1)

= 2 * 1073741823
= 2147483646

At the entrance to a members club stands a stranger seeking admission. A
friend told him that it's easy to get in. You just have to answer a question
corrcetly! Answering wrong, however, will result in being shot!

To live a little longer, the man waits in a back alley near the entrance for
people to go in. After a while a man comes to the entrance. The door warden
asks him: "Twelve?" to which he replies "Six!" and goes in.

"That's easy." our friend thinks, but he waits a little longer.

Another man comes to the door. "Six?" the door warden asks, to which he
replies "Three!" and goes in.

"That's too good to be true" our friend thinks, and he was right. Because, when
asked "Four?", he answered "Two!" and was found dead in the alley.

What was the correct answer?
Submitted by : Milind Gadagkar

Answer

The correct answer was "Four".

The answer is the number of letters in the word spoken by the door warden.

"Twelve" contains "Six" letters i.e. T, W, E, L, V, E
"Six" contains "Three" letters i.e. S, I, X
Similarly, "Four" contains "Four" letters i.e. F, O, U, R
There is a perfect sphere of diameter 40 cms. resting up against a perfectly
straight wall and a perfectly straight floor i.e. the wall and the floor make a perfect
right angle.

Can a perfect sphere of diameter 7 cms. pass through the space between the big
sphere, the wall and the floor? Support your answer with valid arguments. Don't
submit just
"Yes" or
"No". Submit Users BrainVista Puzzle A Add to
Answer Answer (23) Answer Friend Favourite

For the sake of simplicity, consider two-dimension i.e view sphere as a two
dimensional circle with diameter 40 cms.

From Figure I, (40 cms diameter sphere)
OC2 = OD2 + CD2
OC2 = 202 + 202
OC = 28.28427 cms

Also, X is the closest point to origin O on the sphere.
CX = 20 cms (radius)
OX = OC - CX
OX = 28.28427 - 20
OX = 8.28427 cms

From Figure II, (7 cms diameter sphere)
OP2 = OQ2 + PQ2
OP2 = (3.5)2 + (3.5)2
OP = 4.94974 cms

Also, Y is the farthest point to origin O on the sphere.
PY = 3.5 cms (radius)
OY = OP + PY
OY = 4.94974 + 3.5
OY = 8.44974 cms

Now, as OY > OX i.e. smaller sphere requires more space than the space
available. Hence, smaller sphere of 7 cms diameter can not pass through the
space between the big sphere, the wall and the floor.

The puzzle can be solved by another method.
Draw a line tangent to the big sphere at the point X such that X is the closest
point to the origin O on sphere. The tanget will cut X and Y axes at A and B
respectively such that OA=OB. [See Fig III] From above, OX=8.28427 cms.

From the right angle triangle OAB, we can deduct that
OA = OB = 11.71572 cms
AB = 16.56854 cms

Now, the diameter of the inscribed circle of right angle triangle is given by d =
a + b - c where a <= b < c

The maximum possible diameter of the circle which can pass through the
space between the big sphere, the wall and the floor is
= OA + OB - AB
= 11.71572 + 11.71572 - 16.56854
= 6.86291 cms

Hence, the sphere with 7 cms diameter can not pass through the space
between the big sphere, the wall and the floor.

Submit Users BrainVista Puzzle A
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Answer Answer (23) Answer Friend
Sarika multiplied 414 by certain number and obtained 69958 as the answer. But
she found that there is some error in the answer - both the 9s in the answer are
wrong and all the other digits are correct.

Can you find the correct answer?
Answer

The correct answer is 60858.

If you divide 69958 by 414, you will get 168.98. Hence, assume some three digit
number and multiply it by 414 and use 6**58 as the answer.

Assume three digit number such that

* * *

4 1 4

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

* * *

* * * 0

* * * 0 0

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

6 * * 5 8
It is obvious that the last digit of the assumed number must be 7.
* * 7

4 1 4

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

* * 8

* * 7 0

* * 8 0 0

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

6 * * 5 8
Now, the second last digit of the assumed number must be 4 or 9. Also, the first
digit of the assumed number must be 1 as the first digit of the answer is 6. Using
trial and error for above two conditions, the answer is

1 4 7

4 1 4

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

5 8 8

1 4 7 0

5 8 8 0 0

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

6 0 8 5 8
Find the least number which when divided by 35, leaves remainder 25; when
divided by 45, leaves remainder 35 and when divided by 55, leaves remainder
45.

Answer

3455

The answer is LCM of (35, 45, 55) minus 10.
LCM of (35, 45, 55) is 3465.
Hence, the answer is 3455.
The ratio of Boys to Girls is 6:4. 60% of the boys and 40% of the girls take lunch
in the canteen. What % of class takes lunch in canteen?
Answer

Assume there are 6X boys and 4X Girls

Total Students taking lunch in canteen
= (6X)(60/100) + (4X)(40/100)
= 36(X/10) + 16(X/10)
= 52(X/10)

Total students are = 6X + 4X = 10X

% of class taking lunch in canteen
= ((52X/10) * 100 ) / 10X
= 52 %
In the following multiplication, certain digits have been replaced with asterisks (*).
Replace all the asterisks such that the problem holds the result.

* * 7

X 3 * *

----------

* 0 * 3

* 1 *

* 5 *

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

* 7 * * 3
Answer

A simple one.
1 1 7

X 3 1 9

----------

1 0 5 3

1 1 7

3 5 1

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

3 7 3 2 3
How long would it take you to count 1 billion orally if you could count 200 every
minute and were given a day off every four years?

Assume that you start counting on 1 January 2001.
SubmitteAnswer

9 Years, 187 Days, 5 Hours, 20 minutes

As you can count 200 per minute, to count 1 billion you require
= 1,000,000,000/200 minutes
= 5,000,000 minutes
= 83,333.3333 hours
= 3,472.2222 days
= 9.512937 years

= 9 Years, 187 Days, 5 Hours, 20 minutes

Note that a day off every four year will be a Leap day. Hence, no need to
consider leap year.
dFive students - Akash, Chintan, Jignesh, Mukund and Venky - appeared for an
exam. There were total five questions - two multiple choice (a, b or c) and three
true/false questions. They answered five questions each and answered as follow.
I II III IV V

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

Chintan c b True True False

Akash c c True True True

Jignesh a c False True True

Mukund b a True True False

Venky b b True False True

--------------------------------------------------
Also, no two students got the same number of correct answers.

Can you tell which are the correct answers? What are their individual score?
Answer

The correct answers are b, a, True, False and False. Also, the scores are
Jignesh (0), Akash (1), Chintan (2), Venky (3) and Mukund (4).

As no two students got the same number of correct answers, the total number of
correct answers must be either 15 (1+2+3+4+5) or 10 (0+1+2+3+4).

Let's find out the maximum number of correct answers possible from the answers
given by them.
For Question I = 2 (b or c)
For Question II = 2 (b or c)
For Question III = 4 (True)
For Question IV = 4 (True)
For Question V = 3 (True)

Thus, the maximum number of correct answers possible are 15 (2+2+4+4+3)
which means that Akash would have given all correct answers as only he
answered True for questions III, IV and V. But then Chintan and Jignesh would
have exactly 3 correct answers. And also, Mukund and Venky would have 2
correct answers. So no one got all five correct. One can also arrive at this
conclusion by trial-and-error, but that would be bit lengthy.

Now, it is clear that total number of correct answers are 10 (0+1+2+3+4).
Questions III and IV both can not be False. If so, total number of correct answers
would not be 10. So the student who got all wrong can not be Chintan, Akash
and Mukund.

If Venky got all wrong, then Chintan, Jignesh and Mukund each would have
atleast 2 correct answers. It means that Akash would have to be the student with
only one correct answer and the correct answers for questions I and II would be
a and a respectively. But then the total number of correct answers would be 1 (a)
+ 1 (a) + 1 (False) + 4 (True) + 2 (Flase) = 9.

Thus, Jignesh is the student with all wrong answers. The correct answers are b,
a, True, False and False. Also, the scores are Jignesh (0), Akash (1), Chintan
(2), Venky (3) and Mukund (4).

Eleven boys and girls wait to take their seats in the same row in a movie theater.
There are exactly 11 seats in the row.

They decided that after the first person sits down, the next person has to sit next
to the first. The third sits next to one of the first two and so on until all eleven are
seated. In other words, no person can take a seat that separates him/her from at
least one other person.

How many different ways can this be accomplished? Note that the first person
can choose any of the 11 seats.
Answer

There are 1024 different ways.

This is the type of Brain Teaser that can be solved using the method of induction.

If there is just a one person and one seat, that person has only one option.

If there are two persons and two seats, it can be accomplished in 2 different
ways.

If there are three persons and three seats, it can be accomplished in 4 different
ways. Remember that no person can take a seat that separates him/her from at
least one other person.

Similarly, four persons and four seats produce 8 different ways. And five persons
with five seats produce 16 different ways.

It can be seen that with each additional person and seat, the different ways
increase by the power of two. For six persons with six seats, there are 32

different ways.

For any number N, the different possible ways are 2(N-1)

Thus, for 11 persons and 11 seats, total different ways are 210 i.e. 1024
The secret agent X emailed a code word to his head office. They are "AIM DUE
OAT TIE MOD". But four of these five words are fake and only one contains the
information.

The agent X also mailed a sentence as a clue - if I tell you any one character of
the code word, you would be able to tell the number of vowels in the code word.

Can you tell which is the code word?

Answer

The code word is TIE.

If you were told any one character of MOD, then you would not be able to
determine whether the number of vowels are one or two. e.g. if you were told M,
there are two words with M - AIM with 2 vowels and MOD with 1 vowel. So you
would not be able to say the number of vowels. Same arguments can be given
for characters O and D.

Hence, the word with any one of M, O or D is not a code word i.e. AIM, DUE,
OAT and MOD are not the code word. Thus, TIE is the code word.
T : two words - TIE and OAT, both with 2 vowels
I : two words - TIE and AIM, both with 2 vowels
E : two words - TIE and DUE, both with 2 vowels.

Four men - Abraham, Bobby, Clinton and Denial - are standing in a straight
line.

1. One man is fair, handsome and unscarred.
2. Two men who are not fair, are each standing next to Abraham.
3. Bobby is the only man standing next to exactly one handsome man.
4. Clinton is the only man not standing next to exactly one scarred man.

Who is fair, handsome and unscarred?

Answer

Clinton is fair, handsome and unscarred.

From (2), both the men standing next to Abraham are not fair. Also, exactly one
man is fair, handsom and unscarred. Hence, there are two cases:

Case 1 :: ? (N, ?, ?) : Abraham (Y, Y, N) : ? (N, ?, ?) : ? (?, ?, ?)
Case 2 :: ? (N, ?, ?) : Abraham (?, ?, ?) : ? (N, ?, ?) : ? (Y, Y, N)

Note the representation - Name (Fair, Handsome, Scarred). "Y" stands for Yes
and "N" stabds for No. Abraham (Y, Y, N) means Abraham is Fair, Handsome
and Unscarred.

It is clear that either Abraham or the man at the extreme right is fair, handsome
and unscarred.

From (4), it is deduced that Clinton is standing next to unscarred man and each
of the other men standing next to exactly one scarred man.

Case 1 :: Clinton (N, ?, N) : Abraham (Y, Y, N) : ? (N, ?, Y) : ? (?, ?, Y)
Case 2 :: ? (N, ?, Y) : Abraham (?, ?, Y) : ? (N, ?, N) : Clinton (Y, Y, N)

From (3), Bobby is the only man standing next to exactly one handsome man.
But in Case 1, Clinton is standing next to exactly one handsome man. Hence,
Case 1 is not possible and Case 2 is the correct one.

Case 2 :: ? (N, ?, Y) : Abraham (?, ?, Y) : ? (N, ?, N) : Clinton (Y, Y, N)

Again from (3) and (4), there are 2 possibilities as shown below.

Case 2a :: Denial (N, N, Y) : Abraham (?, N, Y) : Bobby (N, N, N) : Clinton (Y, Y,
N)
Case 2b :: Bobby (N, N, Y) : Abraham (?, Y, Y) : Denial (N, N, N) : Clinton (Y, Y,
N)

Thus, Clinton is fair, handsome and unscarred. Also, Abraham may be either fair
or not fair.
An orange colored glass has Orange juice and white colored glass has Apple
juice both of equal volumes. 50ml of the orange juice is taken and poured into the
white glass. After that similarly, 50ml from the white glass is poured into the
orange glass.

Of the two quantities, the amount of apple juice in the orange glass and the
amount of orange juice in the white glass, which one is greater and by how
much?
Answer

The two quantities are equal.

Solve it by taking example. Let's assume that both glasses contain 450 ml of
juice each.

Now, 50ml of the orange juice is taken and poured into the White glass. Hence,
orange colored glass contains 400 ml of Orange juice and white glass contains
450 ml of Apple juice and 50 ml of Orange juice i.e. total of 500 ml from white
glass contains 450 ml of Apple juice and 50 ml of Orange juice. It means that
every 50 ml from white glass contains 45 ml of Apple juice and 5 ml of
Orange juice.

Similary, 50 ml of juice from white glass is poured into orange glass. Now this 50
ml is not a pure apple juice. It contains 45 ml of Apple juice and 5 ml of Orange
juice.

Hence, Orange glass contains 405 ml of Orange juice and 45 ml of Apple juice.
Similary, white glass contains 405 ml of Apple juice and 45 ml of Orange juice.

Orange Glass White Glass
Orange Apple Orange Apple
Juice Juice Juice Juice
Initially 450 ml 0 ml 0 ml 450 ml
50 ml from Orange Glass is
400 ml 0 ml 50 ml 450 ml
poured into White Glass
50 ml from White Glass is poured
405 ml 45 ml 45 ml 405 ml
into Orange Glass

Now it is clear that the amount of apple juice in the orange glass and the amount
of orange juice in the white glass are the same.

P.S. Here we assumed 450 ml as initial quantity in both the glasses just for
simplicity. You can try the same by assuming any other number. But the answer
is the same.

Annie, Bunnie, Candy and Dina visited Edy on 14th February.

1. The time of each visit was as follows:
- Annie at 8:00
- Bunnie at 9:00
- Candy at 10:00
- Dina at 11:00
Each time mentioned above may be either AM or PM.

2. Candy did not visit Edy between Bunnie and Dina.
3. At least one female visited Edy between Annie and Bunnie.
4. Annie did not visit Edy before both Candy and Dina.

Can you tell at what time did they individually visit Edy?

Answer

Bunnie (9:00AM) - Dina (11:00AM) - Annie (8:00PM) - Candy (10:00PM)

From the given data, it is clear that at least one female visited Edy in the morning
and at least one female visited Edy in the evening. Also, from (4), Annie did not
visit Edy first. It means that Annie visited Edy at 8:00 PM

From (3), Bunnie must have visited Edy at 9:00 AM. Also, either Candy or Dina
or both visited Edy in the morning.

But from (2), only Dina must have visited Edy in the morning at 11:00 AM and
hence, Candy visited Edy at 10:00 PM.

The order of visits must be:
Bunnie (9:00AM) - Dina (11:00AM) - Annie (8:00PM) - Candy (10:00PM)

In training for a competition, you find that swimming downstream (with the
current) in a river, you can swim 2 miles in 40 minutes, & upstream (against the
current), you can swim 2 miles in 60 minutes.

How long would it take you to swim a mile in still water?
Answer

You are able to swim downstream at 3 miles an hour, & upstream at 2 miles an
hour. There is a difference of 1 mile an hour, which is the river helping you in 1
direction, & slowing you in the other direction.

Average the 2 rates, & you have the rate that you can swim in still water, which is
2.5 miles an hour.

You can thus swim a mile in still water in 24 minutes.
Father's age is three years more than three times the son's age. After three
years, father's age will be ten years more than twice the son's age.

What is the father's present age?

Answer

Let son's present age is X years.
Hence, father's present age is (3X + 3) years.

After 3 years, son's age will be (X + 3) years.
and father's age will be (3X + 6) years.

But given that after 3 years father's age will be ten years more than twice the
son's age.
(3X + 6) = 2 * (X + 3) + 10
3X + 6 = 2X + 16
X = 10

Therefore, father's present age is 33 years.

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Answer Answer (17)


• A is the father of two children - B and D who are of different sexes.
• C is B's spouse.
• E is the same sex as D.
• B and C have the two children - F who is the same sex as B and G
who is the same sex as C.
• E's mother, H who is married to L, is the sister of D's mother, M.
• E and E's spouse, I have two children - J and K who are the same
sex as I.

Note that no persons have married more than once. Also, there are more
number of females than males. Can you tell how many females are there?

Answer

There are 7 females and 6 males.

Assume that there are four sexes - male, female, X and Y. Prepare the following
tree based on the data given :

sister

L(m) - H(f) -------------------- M(f) - A(m)

| |

| |

E(x) - I(y) D(x) B(y) -
C(x)

| |

| |

J(y) K(y) F(y)
G(x)

It is clear that there are altogether 13 persons - 2 males, 2 females, 4 Xs and 5
Ys.

It is given that there are more number of females than male. Hence, all Y must
represent female. Thus, there are 7 females and 6 males.
A positive integer that, when added to 1000 gives a sum which is greater than
when multiplied by 1000.

Find the positive integer.
Answer

The positive integer is 1.

Sum of 1 and 1000 = 1 + 1000 = 1001
Multiplication of 1 and 1000 = 1 * 1000 = 1000

Thus, sum of 1 and 1000 is greater than the multiplication of 1 and 1000.
Mr. D'souza has bought four cars - Merc, Honda, Ford, Zen - as presents for his
sons' birthdays, all of which are next week. Given the following information, what
will each son get?

Alan will not get the Honda unless Barry gets the Merc and Denzil gets the Ford.
Barry will not get the Ford unless Carl gets the Zen and Alan gets the Merc.
Denzil will not get the Zen unless Alan gets the Honda and Barry gets the Merc.
Alan will not get the Merc unless Carl gets the Zen and Denzil gets the Ford.
Barry will not get the Merc unless Alan gets the Zen and Denzil gets the Ford.
Alan will not get the Zen unless Barry gets the Honda and Carl gets the Merc.
Carl will not get the Zen unless Barry gets the Honda and Alan gets the Ford.

Alan will not get the Ford unless Barry gets the Zen and Denzil gets the Honda.
Carl will not get the Merc unless Denzil gets the Honda.
Answer

Let's put given 9 information in a table. The person in Bold Font will not get the
corresponding car unless the persons in Normal Font get the corresponding cars.
Also, the person will Italics will get the remaining car.

Merc Honda Ford Zen
1 Barry Alan Denzil Carl
2 Alan Denzil Barry Carl
3 Barry Alan Carl Denzil
4 Alan Barry Denzil Carl
5 Barry Carl Denzil Alan
6 Carl Barry Denzil Alan
7 Denzil Barry Alan Carl
8 Carl Denzil Alan Barry
9 Carl Denzil ? ?

Now, let's assume that Alan gets the Merc. Then from (4), Barry gets the Honda,
Denzil gets the Ford and Carl gets the Zen. But from (7), Carl will not get the Zen
unless Barry gets the Honda and Alan gets the Ford. Thus, it contradicts the
original assumption. Hence, Alan will not get the Merc.

Let's assume that Alan gets the Honda. Then from (1), Barry gets the Merc,
Denzil gets the Ford and Carl gets the Zen. But from (5) or from (7), it contradicts
the original assumption. Hence, Alan will not get the Honda.

Let's assume that Alan gets the Ford. Then from (8), Carl gets the Merc, Denzil
gets the Ford and Barry gets the Zen - which does not contradict any of the
statement.

Similaly, you can assume that Alan gets the Zen. (which is contradictory to (9))

Hence, Alan gets the Ford, Barry gets the Zen, Carl gets the Merc and Denzil
gets the Honda.
Yesterday in a party, I asked Mr. Shah his birthday. With a mischievous glint in
his eyes he replied. "The day before yesterday I was 83 years old and next year I
will be 86."

Can you figure out what is the Date of Birth of Mr. Shah? Assume that the current
year is 2000.
Answer

Mr. Shah's date of birth is 31 December, 1915

Today is 1 January, 2000. The day before yesterday was 30 December, 1999
and Mr. Shah was 83 on that day. Today i.e. 1 January, 2000 - he is 84. On 31
December 2000, he will be 85 and next year i.e. 31 December, 2001 - he will be
86. Hence, the date of birth is 31 December, 1915.

Many people do think of Leap year and date of birth as 29th February as 2000 is
the Leap year and there is difference of 3 years in Mr. Shah's age. But that is not
the answer.


There are 4 mathematicians - Brahma, Sachin, Prashant and Nakul - having
lunch in a hotel. Suddenly, Brahma thinks of 2 integer numbers greater than 1
and says, "The sum of the numbers is..." and he whispers the sum to Sachin.
Then he says, "The product of the numbers is..." and he whispers the product
to Prashant. After that following conversation takes place :

Sachin : Prashant, I don't think that we know the numbers.
Prashant : Aha!, now I know the numbers.
Sachin : Oh, now I also know the numbers.
Nakul : Now, I also know the numbers.

What are the numbers? Explain your answer.
Submitted
Answer

The numbers are 4 and 13.

As Sachin is initially confident that they (i.e. he and Prashant) don't know the
numbers, we can conclude that -
1) The sum must not be expressible as sum of two primes, otherwise Sachin
could not have been sure in advance that Prashant did not know the numbers.
2) The product cannot be less than 12, otherwise there would only be one choice
and Prashant would have figured that out also.

Such possible sum are - 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67,
71, 77, 79, 83, 87, 89, 93, 95, 97, 101, 107, 113, 117, 119, 121, 123, 125, 127,
131, 135, 137, 143, 145, 147, 149, 155, 157, 161, 163, 167, 171, 173, 177, 179,
185, 187, 189, 191, 197, ....

Let's examine them one by one.

If the sum of two numbers is 11, Sachin will think that the numbers would be
(2,9), (3,8), (4,7) or (5,6).

Sachin : "As 11 is not expressible as sum of two primes, Prashant can't know the
numbers."

Here, the product would be 18(2*9), 24(3*8), 28(4*7) or 30(5*6). In all the cases
except for product 30, Prashant would know the numbers.

- if product of two numbers is 18:
Prashant : "Since the product is 18, the sum could be either 11(2,9) or 9(3,6). But
if the sum was 9, Sachin would have deduced that I might know the numbers as
(2,7) is the possible prime numbers pair. Hence, the numbers must be 2 and 9."
(OR in otherwords, 9 is not in the Possible Sum List)

Prashant : "Since the product is 24, the sum could be either 14(2,12), 11(3,8) or
10(4,6). But 14 and 10 are not in the Possible Sum List. Hence, the numbers
must be 3 and 8."

Prashant : "Since the product is 28, the sum could be either 16(2,14) or 11(4,7).
But 16 is not in the Possible Sum List. Hence, the numbers must be 4 and 7."

Prashant : "Since the product is 30, the sum could be either 17(2,15), 13(3,10) or
11(5,6). But 13 is not in the Possible Sum List. Hence, the numbers must be
either (2,15) or (5,6)." Here, Prashant won't be sure of the numbers.

Hence, Prashant will be sure of the numbers if product is either 18, 24 or 28.

Sachin : "Since Prashant knows the numbers, they must be either (3,8), (4,7) or
(5,6)." But he won't be sure. Hence, the sum is not 11.

Summerising data for sum 11:
Possible Sum PRODUCT Possible Sum
2+9 18 2+9=11 (possible)
3+6=9
3+8 24 2+12=14
3+8=11 (possible)
4+6=10
4+7 28 2+12=14
3+8=11 (possible)

4+6=10
5+6 30 2+15=17 (possible)
3+10=13
5+6=11 (possible)

Following the same procedure for 17:
Possible Sum PRODUCT Possible Sum
2+15 30 2+15=17 (possible)
3+10= 13
5+6=11 (possible)
3+14 42 2+21=23 (possible)
3+14=17 (possible)
6+7=13
4+13 52 2+26=28
4+13=17 (possible)
5+12 60 2+30=32
3+20=23 (possible)
4+15=19
5+12=17 (possible)
6+10=16
6+11 66 2+33=35 (possible)
3+22=25
6+11=17 (possible)
7+10 70 2+35=37 (possible)
5+14=19
7+10=17 (possible)
8+9 72 2+36=38
3+24=27 (possible)
4+18=22
6+12=18
8+9=17 (possible)

Here, Prashant will be sure of the numbers if the product is 52.

Sachin : "Since Prashant knows the numbers, they must be (4,13)."

For all other numbers in the Possible Sum List, Prashant might be sure of the
numbers but Sachin won't.

Here is the step by step explaination:

Sachin : "As the sum is 17, two numbers can be either (2,15), (3,14), (4,13),
(5,12), (6,11), (7,10) or (8,9). Also, as none of them is a prime numbers pair,
Prashant won't be knowing numbers either."

Prashant : "Since Sachin is sure that both of us don't know the numbers, the sum
must be one of the Possible Sum List. Further, as the product is 52, two numbers
can be either (2,26) or (4,13). But if they were (2,26), Sachin would not have
been sure in advance that I don't know the numbers as 28 (2+26) is not in the
Possible Sum List. Hence, two numbers are 4 and 13."

Sachin : "As Prashant now knows both the numbers, out of all possible products -
30(2,15), 42(3,14), 52(4,13), 60(5,12), 66(6,11), 70(7,10), 72(8,9) - there is one
product for which list of all possible sum contains ONLY ONE sum from the
Possible Sum List. And also, no such two lists exist. [see table above for 17]
Hence, two numbers are 4 and 13."

Nakul figured out both the numbers just as we did by observing the conversation
between Sachin and Prashant.

It is interesting to note that there are no other such two numbers. We checked all
the possible sums till 500 !!!
Substitute digits for the letters to make the following subtraction problem true.
S A N T A

- C L A U S

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

X M A S
Answer

One of the simplest brain teaser as there are total 26 possible answers.

It is obvious that S=C+1. Since A-S=S, it is clear that A=2*S or 2*s-10. Also, L
and X are interchangeable.
SANTA - CLAUS = XMAS
24034 - 16492 = 7542
24034 - 17492 = 6542
24074 - 15432 = 8642

24074 - 18432 = 5642
24534 - 16492 = 8042
24534 - 18492 = 6042
24794 - 16452 = 8342
24794 - 18452 = 6342
24804 - 15462 = 9342
24804 - 19462 = 5342
24974 - 16432 = 8542
24974 - 18432 = 6542
36806 - 27643 = 9163
36806 - 29643 = 7163
36156 - 27693 = 8463
36156 - 28693 = 7463
62132 - 54206 = 7926
62132 - 57206 = 4926
62172 - 53246 = 8926
62172 - 58246 = 3926
62402 - 53276 = 9126
62402 - 59276 = 3126
62712 - 53286 = 9426
62712 - 59286 = 3426
62932 - 58206 = 4726
62932 - 54206 = 8726

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