1.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.
18

2.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))

3.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.

4.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? 9

5.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?
Doc is

6.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?
Cam carrie

7.There are 3 persons X, Y and Z. On some day, X lent tractors to
Y and Z as many as they had.
After a month Y gave as many tractors to X and Z as many as they
have. After a month Z did
the same thing. At the end of this transaction each one of them
had 24.
Find the tractors each originally had?

8.A certain street has 1000 buildings. A signmaker is contracted
to number the houses from 1
to 1000. How many zeroes will he need?

9.There are 9 coins. Out of which one is odd one i.e weight is
less or more. How many iterations
of weighing are required to find odd coin?
10.In a sports contest there were m medals awarded on n successive
days (n > 1).
On the first day 1 medal and 1/7 of the remaining m  1 medals
were awarded.
On the second day 2 medals and 1/7 of the now remaining medals was
awarded; and so on.
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?

11.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 19 in some order.
Find the number.

12. 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?
(x1/3x)((x1/3x)3/4)=2/3x1/2x= 1/6x

13.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? 3 hrs

14.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)?
Math gurus may use series formula to solve it.(series:
1,3,5,7,9,11.....upto 18250 terms)

15.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?

16.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.

17. 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&b same person

18.In Mr. Mehta's family, there are one grandfather, one
grandmother, two fathers, two
mothers, one fatherinlaw, one motherinlaw, four children,
three grandchildren, one brother,
two sisters, two sons, two daughters and one daughterinlaw. How
many members are there in
Mr. Mehta's family? Give minimal possible answer. 7

19.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?
20.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)
21.The game of TicTacToe 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.
22At the Party:
There were 9 men and children.
There were 2 more women than children.
The number of different manwoman couples possible was 24. Note
that if there were 7 men
and 5 women, then there would have been 35 manwoman couples
possible.
Also, of the three groups  men, women and children  at the
party:
There were 4 of one group.
There were 6 of one group.
There were 8 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
23There is a shortage of tube lights , 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?
23.Mr. Subramaniam rents a private car for AndheriColabaAndheri
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?
ANSWERS
1. 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
2. The sum of the digits of 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.
3.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(50X)] X
meters whereas the platoon
moved (50X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50X)
(50+X)*(50X) = 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 length 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.
4. Ans.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.
5.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.
6.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.
7.Answer
One way to solve it is by making 3 equations and solve them
simultaneously. But there is rather
easier way to solve it using Backtracing.
It's given that at the end, each had 24 tractors (24, 24, 24) i.e.
after Z gave tractors to X & Y
as many as they had. It means that after getting tractors from Z
their tractors got doubled. So
before Z gave them tractors, they had 12 tractors each and Z had
48 tractors. (12, 12, 48)
Similarly, before Y gave tractors to X & Z, they had 6 &
24 tractors respectively and Y had 42
tractors i.e. (6, 42, 24)
Again, before X gave tractors to Y & Z, they had 21 & 12
tractors respectively and X had 39
tractors i.e. (39, 21, 12)
Hence, initially X had 39 tractors, Y had 21 tractors and Z
had 12 tractors.
8.Answer
The signmaker 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, signmaker 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
9.ANSWER
It is always possible to find odd coin in 3 weighings and to tell
whether the odd coin is heavier
or lighter.
Take 8 coins and weigh 4 against 4.
If both are not equal, goto step 2
If both are equal, goto step 3
One of these 8 coins is the odd one. Name the coins on heavier
side of the scale as H1, H2, H3
and H4. Similarly, name the coins on the lighter side of the scale
as L1, L2, L3 and L4. Either one
of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1)
against (H3, H4, X) where X is one
coin remaining in intial weighing.
If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against
L3.
If both are equal, L4 is the odd coin and is lighter.
If L2 is light, L2 is the odd coin and is lighter.
If L3 is light, L3 is the odd coin and is lighter.
If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is
heavier. Weight H1 against H2
If both are equal, there is some error.
If H1 is heavy, H1 is the odd coin and is heavier.
If H2 is heavy, H2 is the odd coin and is heavier.
If (H3, H4, X) is heavier side on the scale, either H3 or H4 is
heavier or L1 is lighter. Weight
H3 against H4
If both are equal, L1 is the odd coin and is lighter.
If H3 is heavy, H3 is the odd coin and is heavier.
If H4 is heavy, H4 is the odd coin and is heavier.
The remaining coin X is the odd one. Weigh X against the anyone
coin used in initial weighing.
If both are equal, there is some error.
If X is heavy, X is the odd coin and is heavier.
If X is light, X is the odd coin and is lighter.
10.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 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 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.
11.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
12.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
13.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.
14.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(x1), 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.
15.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
16.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)
17.
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.
18.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
19.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 !!!
20.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
Solving two equations simultaneously
98Y  199X = 20
Y  2X = 1
We get X = 26 & Y = 53
Now, its obvious that he wanted to withdraw Rs. 26.53
21.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.
22.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.
23.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 AndheriBandraAndheri, only one person i.e. Mr. Subramaniam
is travelling. Hence, he would
pay Rs. 150.
For BandraColabaBandra, three persons i.e Mr. Subramaniam and
two students, are travelling.
Hence, each student would pay Rs. 50.
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 (22319) 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 (204162) 42 houses do not have all 3 items 
tubelight, bulb and fan.
Thus, 42% houses do not have tubelight, bulb and fan.
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