# Quantitative Aptitude

Please wait while the activity loads.

If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

Question 1 |

Find the number of zeroes in 155!

30 | |

38 | |

42 | |

44 |

**Quantitative Aptitude**

**Discuss it**

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] + [155/5

^{2}] + [155/5^{3}] =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 |

**Quantitative Aptitude**

**Discuss it**

Question 2 Explanation:

Prime Factor of 45= 3

^{2}x5 We will count the number of 32 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 3^{2}= 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 3^{2}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 |

**Quantitative Aptitude**

**Discuss it**

Question 3 Explanation:

LCM = 15 * HCF
We know that
LCM + HCF = 480
16 * HCF = 480
HCF = 30
Then LCM = 450
LCM = 15 HCF
30 * x * y = 15 * 30
x * y = 15
Factors are {1 x 15} and { 3 x 5}
Both numbers less than LCM so take {3 x 5}
Hence numbers are 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 |

**Quantitative Aptitude**

**Discuss it**

Question 4 Explanation:

Find the LCM for 4, 6, 7, 9
LCM= 2

^{2}* 3^{2}* 7 = 252 To become perfect square all factors should be in power of 2. So, multiply it by 7 LCM = 2^{2}* 3^{2}* 7^{2}= 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 |

**Quantitative Aptitude**

**Discuss it**

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 |

**Quantitative Aptitude**

**Discuss it**

Question 6 Explanation:

**Shortcut**Let they earn 1 Rs/hr.

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

**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 | |

60 hours | |

72 hours | |

48 hours |

**Quantitative Aptitude**

**Discuss it**

Question 7 Explanation:

**First method**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 x

^{2}= 256 x = 16 A alone = 16 + 64 =

**60 hours**

**Shortcut method -**x

^{2}= more of A * more of B x

^{2}= 64 * 4 x

^{2}= 256 x= 16

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 |

**Quantitative Aptitude**

**Discuss it**

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 |

**Quantitative Aptitude**

**Discuss it**

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 |

**Quantitative Aptitude**

**Discuss it**

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.