# Quantitative Aptitude

 Question 1
Find the number of zeroes in 155!
 A 30 B 38 C 42 D 44
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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.
 A 163 B 164 C 165 D 166
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Question 2 Explanation:
Prime Factor of 45= 32x5 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 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
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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.
 A 42 B 40 C 36 D 32
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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.
 A 4 days B 7 days C 6 days D 9 days
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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?
 A 2 hours/day B 3 hours/day C 4 hours/day D 5 hours/day
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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.
 A 16 hours B 60 hours C 72 hours D 48 hours
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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 x2 = 256 x = 16 A alone = 16 + 64 = 60 hours Shortcut method - x2 = more of A * more of B x2= 64 * 4 x2= 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?
 A 2400 litres B 4800 litres C 7200 litres D 9600 litres
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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?
 A 11 hours B 13 hours C 15 hours D 17 hours
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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 ?
 A 18 minutes B 24 minutes C 30 minutes D 36 minutes
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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.

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