21.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.
22At the Party:
There were 9 men and children.
There were 2 more women than children.
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:
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
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?
24.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.
25.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.
25.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.
26.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
27.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
28.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
29.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.
30.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.
31.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.
32.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.
33.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.
34.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)
Probability of having King of Hearts on the top of the Pack B = 1/53
So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)
Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378
35.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.
A->B, B->C, C->A
A->B, B->C, C->B
A->B, B->A, C->A
A->B, B->A, C->B
A->C, C->B, B->A
A->C, C->B, B->C
A->C, C->A, B->A
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
|
36.Find
all sets of consecutive integers that add up to 1000. |
|
|
37.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 !!!
38.How many possible combinations are there in a
3x3x3 rubics cube?
In other words, if you wanted to solve the rubics cube by trying different
combinations, how many might it take you (worst case senerio)?
How many for a 4x4x4 cube?
Submitted
Answer
There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 *
10^45 possible combinations for 4x4x4 Rubics.
Let's consider 3x3x3 Rubics first.
There are 8 corner cubes, which can be arranged in 8! ways.
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8
orientations altogether. But if you get all but one of the corner cube into
chosen positions and orientations, only one of 3 orientations of the final
corner cube is possible. Thus, total ways corner cubes can be placed = (8!) *
(3^8)/8 = (8!) * (3^7)
Similarly, 12 edge cubes can be arranged in 12! ways.
Each of these 12 cubes can be turned in 2 different directions, so there are
2^12 orientations altogether. But if you get all but one of the edge cube into
chosen positions and orientations, only one of 2 orientations of the final edge
cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2
= (12!) * (2^11)
Here, we have essentially pulled the cubes apart and stuck cubes back in place
wherever we please. In reality, we can only move cubes around by turning the
faces of the cubes. It turns out that you can't turn the faces in such a way as
to switch the positions of two cubes while returning all the others to their
original positions. Thus if you get all but two cubes in place, there is only
one attainable choice for them (not 2!). Hence, we must divide by 2.
Total different possible combinations are
= [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19
Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45
Note that there are 24 edge cubes, which you can not turn in 2 orientations
(hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) /
(4!^6). You can switch 2 cubes without affecting the rest of the combination as
4*4*4 has even dimensions (hence no division by 2). But pattern on one side is
rotated in 4 directions over 6 faces, hence divide by 24.
38.Substitute digits for the letters to make the
following relation true.
N E
V E R
L E
A V E
+ M
E
-----------------
A L
O N E
Note that the leftmost letter can't be zero in any
word. Also, there must be a one-to-one mapping between digits and letters. e.g.
if you substitute 3 for the letter M, no other letter can be 3 and all other M
in the puzzle must be 3.
Answer
A tough one!!!
Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is
equal to 0 or 5. This is the only entry point to
solve it. Now use trial-n-error method.
N
E V E
R 2 1
4 1 9
L
E A V
E 3 1
5 4 1
+ M
E + 6
1
----------------- -----------------
A
L O N
E 5 3
0 2 1
39.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.
If the singer and the dancer are the same sex, then
the dancer is older than the singer.
If neither the singer nor the dancer is the parent
of the other, then the singer is older than the dancer.
If the singer is a man, then the singer and the
dancer are the same age.
If the singer and the dancer are of opposite sex
then the man is older than the woman.
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.
40.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 from G1 |
People from G2 |
No of ways to hire G1 without a
single pair of indentical twins |
No of ways to hire G2 |
Total ways |
|
0 |
5 |
10C0 |
10C5 |
252 |
|
1 |
4 |
10C1 |
10C4 |
2100 |
|
2 |
3 |
10C2 * 8/9 |
10C3 |
4800 |
|
3 |
2 |
10C3 * 8/9 * 6/8 |
10C2 |
3600 |
|
4 |
1 |
10C4 * 8/9 * 6/8 * 4/7 |
10C1 |
800 |
|
5 |
0 |
10C5 * 8/9 * 6/8 * 4/7 * 2/6 |
10C0 |
32 |
|
Total |
11584 |
|||
Thus, total possible ways to hire 5 people without a single pair of indentical
twins = 11584 ways
So, total possible ways to hire 5 people with at least a single pair of
indentical twins = 15504 - 11584 = 3920 ways
Hence, the probability to hire 5 people with at least a single pair of
indentical twins
= 3920/15504
= 245/969
= 0.2528
= 25.28%
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