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  • SBI SO Exam Quantitative Aptitude Study Material

Digitization help student to explore and study their academic courses online, as this gives them flexibility and scheduling their learning at their convenience. Kidsfront has prepared unique course material of Quantitative Aptitude Pipes and Cisterns for SBI SO Exam student. This free online Quantitative Aptitude study material for SBI SO Exam will help students in learning and doing practice on Pipes and Cisterns topic of SBI SO Exam Quantitative Aptitude . The study material on Pipes and Cisterns, help SBI SO Exam Quantitative Aptitude students to learn every aspect of Pipes and Cisterns and prepare themselves for exams by doing online test exercise for Pipes and Cisterns, as their study progresses in class. Kidsfront provide unique pattern of learning Quantitative Aptitude with free online comprehensive study material and loads of SBI SO Exam Quantitative Aptitude Pipes and Cisterns exercise prepared by the highly professionals team. Students can understand Pipes and Cisterns concept easily and consolidate their learning by doing practice test on Pipes and Cisterns regularly till they excel in Quantitative Aptitude Pipes and Cisterns.


Pipes and Cisterns
A pipe can fill a tank in x hours and another can empty it in y hours. They can together fill it in (y > x )

a) X-y
b) Y-x
c) Xy/x-y
d) Xy/y-x



Answer
Solution
Correct Answer Is : Xy/y-x
Solution Is : Work done by A in one hour =1/x
work done by B in one hour=1/y
Both A and B together work in one hour=(1/x)-(1/y) =(y-x)/xy.
Both A and B fill tank in xy/(y-x) hours.
A tap can empty a tank in 30 minutes. A second tap can empty it in 45 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank ?

a) 18 minutes
b) 14 minutes
c) 15 minutes
d) 30 minutes



Answer
Solution
Correct Answer Is : 18 minutes
Solution Is : Work done by 1st tap in one minute =1/30
work done by 2nd tap in one minute =1/45
Both taps one minute work =(1/30)+(1/45)
=(45+30)/1350
=75/1350 =1/18.
Both tap will empty the tank in 18 minutes.
Water is flowing at the rate of 3 km/hr through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10m and depth 2m. In how much time will the cistern be filled ?

a) 1 hour
b) 1 hour 40 minutes
c) 1 hour 20 minutes
d) 2 hours 40 minutes



Answer
Solution
Correct Answer Is : 1 hour 40 minutes
Solution Is :
An empty pool being filled with water at a constant rate takes 8 hours to fill 3/5 th of its capacity. How much more time will it take to finish filling the pool ?

a) 4 hours 50 minutes
b) 5 hours 30 minutes
c) 5 hours 20 minutes
d) 4 hours 48 minutes



Answer
Solution
Correct Answer Is : 5 hours 20 minutes
Solution Is :
A pipe can fill a cistern in 9 hours. Due to leak in its bottom, the cistern fills up in 10 hours. If cistern is full, in how much time will it be emptied by the leak?

a) 70 hours
b) 80 hours
c) 90 hours
d) 100 hours



Answer
Solution
Correct Answer Is : 90 hours
Solution Is : Part filled without leak in 1 hour = 1/9
Part filled with leak in 1 hour = 1/10
Work done by leak in 1 hour =1/9?1/10=1/90
We used subtraction as it is getting empty.
So total time to empty the cistern is 90 hours
Pipe A and B can fill a tank in 10 hours and 8 hours respectively. After certain time, pipe A was closed. It took a total of 6 hours to fill the tank completely. For how many hours did pipe A work?

a) 4 1/4 hours
b) 2 1/2 hours
c) 3 1/3 hours
d) 5 1/3 hours



Answer
Solution
Correct Answer Is : 2 1/2 hours
Solution Is : (2) Let pipe A remained open for x hours. According to the question, (x/10)+(6/8)=1 =>x/10=1-3/4=1/4 => x=10/4 = 2 1/2 hours
A water tank can be filled by a tap in 10 minutes and another tap can be filled it in 60 minutes. If both the taps are kept open for 5 minutes and then the first tap is closed, how long will it take for the tank to be full?

a) 45 minutes
b) 30 minutes
c) 25 minutes
d) 20 minutes



Answer
Solution
Correct Answer Is : 45 minutes
Solution Is :
Two pipes can fill a cistern in 6 mins and 7 minutes respectively. Both the pipe are opened alternately for 1 minute each. In what time will they fill the cistern?

a) 5 minutes
b) 17/3minutes
c) 45/7 minutes
d) 5/4 minutes



Answer
Solution
Correct Answer Is : 45/7 minutes
Solution Is : It is an alternate men`s problem A` s 1 min work=(1/60) B` s 1 min work=(1/7) (A+B)`s 2 min work=(1/6)+(1/7)=13/42 (A+B)`s 6 min work=(3*13/42)=13/14 Remaining work =1-(13/14)=1/14 It is A` turn and 1/14 part to be filled Time taken to fill 1/14 part =(1/14)*6=(3/7)mins The time taken to fill the take=6mins+(3/7)mins =45/7 mins
Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. The water pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes C is opened. Total time (in minutes) in which the tank will be filled up is :

a) 30
b) 12
c) 36
d) 24



Answer
Solution
Correct Answer Is : 24
Solution Is : Part of tank filled by pipes A and B in 1 minute = (1/30)-(1/45)=1/18 part.
Therefore,part of tank filled in 12 minutes=(12/15)=2/3 part
Remaning part=1-(2/3)=1/3 part
when pipe Cis opened,part of tank filled by all 3 pipes=(1/30)(1/45)-(1/36)=5/180=1/36.
therefore,Time taken in filling 1/3 part = (1/3)*36=12 minutes,
therefore,Total time=12+12 =24 minutes.
Pipe A can fill a tank in 4 hours and pipe B can fill it in 6 hours. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?

a) 4 1/2
b) 3 1/2
c) 3 1/4
d) 4 2/3



Answer
Solution
Correct Answer Is : 4 2/3
Solution Is : Part of tank filled in first two hours = 1/4 + 1/6 = (3+2)/12 = 5/12. Part of tank filled I first 4 hours =10/12 =5/6. Remaining part 1 - 5/6 =1/6. This remaining part will be filled by pipe A. Time taken by Pipe A = (1/6) *4 = 2/3 hour. Therefore total time = 4+(2/3) = 4 2/3
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