Quantitative Aptitude

Question 1

Find the number of trailing zeroes in 155!

• A. 30
• B. 38
• C. 42
• D. 44

Answer: (B)

Multiplication of 2x5 results into 10. So number of zeroes depend on the number of pairs of 2 and 5. In any factorial, number of 5’s is lesser than the number of 2’s. So, we need to count the maximum power of 5 in 155! [155/5] + [155/5²] + [155/5³] =31 + 6 +1 =38 Hence, number of zeroes is 38.

Question 2

Find the maximum value of n such that 671! is perfectly divisible by 45n.

• A. 163
• B. 164
• C. 165
• D. 166

Answer: (C)

Prime Factor of 45= 3²x5 
We will count the number of 3^2 and 5 in 671!, and which one is lesser in number would be the answer. 
No of 3’s= 671/3 + 671/9 + 671/27 + 671/81 + 671/243 
= 223 + 74 + 24 + 8 + 2 
= 331 
No of 3²= 331/2 = 165
No of 5= 671/5 + 671/25 + 671/125 + 671/625 
= 134 + 26 + 5 + 1 = 166
165 will be the answer because 3² is lower in number than 5.

Question 3

The LCM of two numbers is 15 times of HCF. The sum of HCF and LCM is 480. If both number are smaller than LCM. Find both the numbers.

• A. 60 and 90
• B. 90 and 120
• C. 120 and 150
• D. 90 and 150

Answer: (D)

LCM = 15 * HCF 
We know that LCM + HCF = 480 
16 * HCF = 480 
HCF = 30 
Then LCM = 450 
we know that LCM*HCF = x*y.
So, by calculating we get the numbers
3 * 30 = 90 and 5 * 30 = 150

Question 4

Find the least perfect square number which when divided by 4, 6, 7, 9 gives remainder zero.

• A. 42
• B. 40
• C. 36
• D. 32

Answer: (A)

Find the LCM for 4, 6, 7, 9 
LCM= 2² * 3² * 7 = 252 
To become perfect square all factors should be in power of 2. 
So, multiply it by 7 LCM = 2² * 3² * 7² = 1764 
And it is perfect square of 42.

Question 5

A, B and C can do a piece of work in 10, 12 and 15 days respectively.They all start the work together but A leaves after the 2 days of work and B leaves 3 days before the work is completed.Find the number of days the work completed.

• A. 4 days
• B. 7 days
• C. 6 days
• D. 9 days

Answer: (B)

Total work done is LCM(10, 12, 15)=60 unit 
A’s efficiency = 60/10= 6 
B’s efficiency = 60/12= 5 
C’s efficiency = 60/15= 4 
First two days all work together So, the work completed in first two days= 15 x 2 = 30 unit 
Remaining work= 60 - 30 = 30 unit 
If B completes 3 day work also = 3 x 5 = 15 unit 
Total work remaining= 30 + 15 = 45 unit 
Number of days B and C works= 45/9=5 
Total number of days to complete the work = 2 + 5 = 7 days.

Question 6

In a factory same number of women and children are present. Women works for 6 hours in a day and children work 4 hours in a day.In festival season workload increases by 60% and government does not allow children to work more than 6 hours per day.If their efficiency are equal and remain work is done by women then how many extra hours/day increased by women?

• A. 2 hours/day
• B. 3 hours/day
• C. 4 hours/day
• D. 5 hours/day

Answer: (C)

Shortcut Let they earn 1 Rs/hr.

Woman     Child      Earns
  6    +    4     =   10
  |         |         |60%
 __        max 6   =  16

Workload increases by 60% from 10 to 16. Children can work maximum 6 hours Then women work per day 16 - 6 = 10 So, it increases by 4 hours/day extra.

Question 7

A alone would take 64 hours more to complete a work then A + B work together. B take 4 hours more to complete a work alone than A and B work together.Find in how much time A alone complete the work.

• A. 16 hours
• B. 80 hours
• C. 72 hours
• D. 48 hours

Answer: (B)

Let A and B take x hours to complete a work together. 
A alone would take (x + 64) and B alone would take (x + 4)hours to complete the work. 
A( x + 64) = x (A + B) 
64A =x B …………(1) 
B(x + 4)= x(A + B) 
4B = x A……………(2) 
from (1)and (2)
64A = x * x A/4 
x² = 256 
x = 16 
A alone = 16 + 64 = 80 hours

Question 8

A leak can empty a completely filled tank in 10 hours. If a tap is opened in completely filled tank which admits 4 liters of water per minute, then leak takes 15 hours to empty the tank. How many litres of water does the tank hold?

• A. 2400 litres
• B. 4800 litres
• C. 7200 litres
• D. 9600 litres

Answer: (C)

Take LCM (10, 15) = 30 
Let leak pipe is A and A’s efficiency = 30/10 = 3 
Let inlet pipe B and B’s efficiency= 30/15 = 2 
Pipe A is emptying at 3 units/hour and Pipe B is filling using then emptying rate down to 2 units/hour. 
So, filling pipe efficiency is 3 - 2 = 1 unit/ hour. 
Pipe B will fill tank in 30/1=30 hours 
Filling rate = 4 litre/minute 
It will fill 4 x 60 = 240 litre/hour. 
Total capacity= 240 x 30 = 7200 litres

Question 9

Two pipes A and B independently can fill a tank in 20 hours and 25 hours. Both are opened together for 5 hours after which the second pipe is turned off. What is the time taken by first pipe alone to fill the remaining portion of the tank?

• A. 11 hours
• B. 13 hours
• C. 15 hours
• D. 17 hours

Answer: (A)

Total unit water = LCM(20, 25) = 100 unit 
A’s efficiency = 100/20 = 5 unit/hour 
B’s efficiency =100/25 = 4 unit/hour 
After 5 hour the water filled by A and B together = 5 x 9 =45 unit 
Remaining unit = 100 - 45 = 55 unit 
Time taken by A alone = 55/5 = 11 hours

Question 10

If an employee walks at speed of 10 km at 3 km/hr, he will be late by 20 minutes. If he walks at 4 km/hr, how early from the fixed time he will reach ?

• A. 18 minutes
• B. 24 minutes
• C. 30 minutes
• D. 36 minutes

Answer: (C)

Time taken at 3 km/hr = Distance/speed = 10/3 
Actual time is obtained by subtracting the late time 
So, Actual time = 10/3 - 1/3 = 9/3 = 3 hour 
Time taken at 4 km/hr = 10/4 hr 
Time difference = Actual time - time taken at 4 km/hr = 3 - 10/4 = 1/2 hour 
Hence, he will be early by 30 minutes.