# Quantitative Aptitude

 Question 1

Find the number of trailing zeroes in 155! 30 38 42 44

Question 1-Explanation:

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. 163 164 165 166

Question 2-Explanation:

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. 60 and 90 90 and 120 120 and 150 90 and 150

Question 3-Explanation:

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. 42 40 36 32

Question 4-Explanation:

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. 4 days 7 days 6 days 9 days

Question 5-Explanation:

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? 2 hours/day 3 hours/day 4 hours/day 5 hours/day

Question 6-Explanation:
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. 16 hours 80 hours 72 hours 48 hours

Question 7-Explanation:

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? 2400 litres 4800 litres 7200 litres 9600 litres

Question 8-Explanation:

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? 11 hours 13 hours 15 hours 17 hours

Question 9-Explanation:

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 ? 18 minutes 24 minutes 30 minutes 36 minutes

Question 10-Explanation:

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.

There are 47 questions to complete.

• Last Updated : 27 Sep, 2023