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Placement Papers with Solutions

(#M40001390) CTS QUESTION Keep an EYE

Rs 50000 is divided into two parts One part is given to a person with 10% interest and another part is given to a person with 20 % interest. At the end of first year he gets profit 7000 Find money given by 10%?

Asked In CTS Anil (14 years ago)

Solved Abhishek Divakar Read Solution (6)

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(#M40001389) CTS QUESTION Keep an EYE

2 power 300,3 power 200. Which is greater?

Asked In CTS Anil (14 years ago)

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(#M40001388) CTS QUESTION Keep an EYE

In how many ways the letters in the word MANAGEMENT can be arranged such that A’s does not come together.

Solution

Methods to arrange all the letters leaving A = 8*7*6*5*4*3*2*1/(2*1)*(2*1)*(2*1)= 5040
Now between these 8 letters , there are 9 places so methods to place A and A in these positions = 9*8/2*1 = 36
thus total methods are 5040*36 = 181440

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Asked In CTS Anil (14 years ago)

Solved Read Solution (3)

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(#M40001387) CTS QUESTION Keep an EYE

Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are
opened simultaneously, after how much time should B be closed so that the tank is full in 18
minutes?

Asked In CTS Anil (14 years ago)

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(#M40001386) CTS QUESTION Keep an EYE

Bhatia, Ashtikar and Singh begin to play with Rs 70 each. At the end the ratio of the amounts left
with Ashtikar and Singh is 1 : 2 and of those with Singh and Bhatia is 4 : 1. What is Singh's gain
(in Rs)?

Asked In CTS Anil (14 years ago)

Solved Abhishek Divakar Read Solution (5)

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(#M40001385) CTS QUESTION Keep an EYE

Rs. 4536 is divided among 4 men, 5 women and 2 boys. The ratio of share of a man, a woman and a boy is 7: 4: 3. What is the share of a woman?

Solution

Ratio of the share of a man, a woman and a boy is 7:4:3
No of men, women and boys are 4, 5, 2
Thus effective ratio of men, women and boys is 7*4 : 4*5 : 3*2 = 28 : 20 : 6
so part of women = (20/54)* 4536 = 1680
so part of 1 woman = 1680/5 =336

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Asked In CTS Anil (14 years ago)

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(#M40001384) CTS QUESTION Keep an EYE

Find out the wrong number in the following series:
5, 15, 30, 80, 180, 540, 1080

Asked In CTS Anil (14 years ago)

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(#M40001383) CTS QUESTION Keep an EYE

A car takes 3hrs less than a truck to cover the distance frm Bombay to pune(192 km). The speed of the truck is 16km/hr slower than car.
Find the speed of the car?

Solution

Let the speed of car = x km/hrs
then that of truck = (x-16) km/hrs
According to the given condition
[192/(x-16)]- [192/x] = 3
on solving the equation we have x = 8[(sq-root of 17) + 1]


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Asked In CTS Anil (14 years ago)

Solved Read Solution (2)

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(#M40001382) CTS QUESTION Keep an EYE

how many zeros are there from 1 to 10000?

Solution

for solving this problem first we would break the whole range in 5 sections
1) from 1 to 9
Total number of zero in this range = 0
2) from 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)
3) from 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180
4) from 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

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Asked In CTS Anil (14 years ago)

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(#M40001090) TCS QUESTION Keep an EYE

A and B play a game between them. The dice consist of colors on their faces(instead of number). When the dice are thrown, A wins if both show the same color, otherwise B wins. One die has 3 red faces and 3 blue faces. How many red and blue faces should the other die have if the both players have if the both players have the same chances of winning?
a) 5 red and 1 blue faces.
b) 1 red and 5 blue faces.
c) 3 red and 3 blue faces.

Asked In TCS Anil (14 years ago)

Unsolved Read Solution (2)

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