Aptitude - Pipes & Cisterns Online Quiz



Following quiz provides Multiple Choice Questions (MCQs) related to Pipes & Cisterns. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Questions and Answers

Q 1 - In 1 minute 3/7 of a basin is filled. Whatever remains of the container can be filled in:

A - 2 min

B - 4/3 min

C - 7/3 min

D - none of these

Answer : B

Explanation

Part filled in 1 min. = 3/7. remaining part = (1- 3/7)= 4/7
Let the required time be x min.
More part, more time taken. (Direct)
3/7: 4/7:: 1: x ⇒ 3x/7 = (4/7*1) ⇒ x= 4/3 min.

Q 2 - A channels can fill a tank in x hours and another funnel can exhaust it in y (y>x) hours. In the event that both the funnels are open, in how long will the tank be filled?

A - (x-y) hours

B - (y-x) hours

C - xy/(x-y) hours

D - xy/(y-x) hours

Answer : D

Explanation

Work done by filling pipe in 1 hr = 1/x
Work done by emptying pipe in 1 hr = 1/y
Net filling work done by both in 1 hr = (1/x- 1/y) = (y-x)/xy
∴The tank will be filled in xy/(y-x) hrs.

Q 3 - A storage has two funnels. One can fill it with water in 8hours and the other can exhaust it in 5 hours. In how long will the reservoir be purged if both the channels are opened together when 3/4 of the storage is now loaded with water?

A - 10/3 hours

B - 6 hours

C - 10 hours

D - 40/3 hours

Answer : C

Explanation

Net part emptied by both in 1 hr = (1/5-1/8)= 3/40
3/40 part is emptied in 1 hr.
3/4 part will be emptied in (40/3*3/4) hrs = 10 hrs.

Q 4 - Two channels can fill a tank in 15 hours and 12 hours separately and a third pipe can purge it in 4 hours. In the event that the channels are opened all together at 8 am, 9 am and 11am separately, the tank will be exhausted at

A - 11.40 am

B - 12.40 pm

C - 1.40 pm

D - 2.40 pm

Answer : D

Explanation

Let the tank be emptied in x hrs after 8 am.
Work done by A in x hrs, by B in (x-1) hrs and C in (x-3) hrs = 0
⇒x/15+ (x-1)/12- (x-3)/4 = 0 ⇒ 4x+5(x-1) - 15(x-3) = 0
⇒6x= 40 ⇒x= 20/3 hrs.
⇒x= 6 hrs. 40 min after 8 am
Hence the tank will be emptied at 14 hrs 40 min, i.e., 2:40 pm

Q 5 - A reservoir has three channels A, B and C. A and B can fill it in 3 hrs and 4 hrs. individually while C can exhaust the totally filled reservoir in 1 hours. On the off chance that the funnels are opened all together at 3 pm, 4 pm and 5 pm individually, at what the truth will surface eventually reservoir void?

A - 6.15 pm

B - 7.12 pm

C - 8.12 pm

D - 8.35 pm

Answer : B

Explanation

Let the cistern be emptied in x hrs after 3 pm
Work done by A in x hrs, by B in(x-1) hrs and by C in (x-2) hrs= 0
⇒x/3 +x-1/4 ? (x-2) =0 ⇒ 4x+3(x-1)-12(x-2) = 0
⇒5x=21 ⇒x= 4 hrs 12 min.
Required time is 7.12 pm.

Q 6 - Three funnels A, B; C can fill a tank in 6 hours. In the wake of working at it together for 2 hours, C is shut and A and B can fill the remaining part in 7 hours. The quantity of hours taken by C alone to fill the tank is:

A - 10 hr.

B - 12 hr.

C - 14 hr.

D - 16 hr.

Answer : C

Explanation

Part filled by (A+B+c) in 2 hours= (1/6*2)=1/3
2/3 part is filled by (A+B) in 7 hours.
Whole is filled by (A+B) in (7*3/2) hr=21/2hrs.
Part filled by C in 1 hour = (1/6-2/21) = 3/42 = 1/14
∴C alone can fill it in 14 hours.

Q 7 - A storage has a hole which would exhaust it in 8 hours. A tap is transformed on which concedes 6 liters a moment into the reservoir and it is currently purged in 12 hours. What number of liters does the reservoir hold?

A - 7580 ltr.

B - 7960 ltr.

C - 8290 ltr.

D - 8640 ltr.

Answer : D

Explanation

Part filled in 1 hour = (1/8- 1/12)= 1/24
Time taken to fill the cistern= 24 hours
Water moved in it 24 hours = (6*60*24) = 8640 liters.
Capacity of the cistern = 8640 liters.

Q 8 - Two pipes A and B can ill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

A - 10 hours

B - 15 hours

C - 18 hours

D - 20 hours

Answer : D

Explanation

T = xy/(x+y)
= (36*45)/(36+45)
= 1620/80
= 20 hours

Or,

Part filled by A in 1 hour = 1/36
Part filled by B in 1 hour = 1/45
Part filled by (A+B) in 1 hour = (1/36 + 1/45) = 1/20

∴ Both the pipes can fill the tank in 20 hours.

Q 9 - Two pipes A and B can fill a water tank in 20 and 24 min respectively. A third pipe C can empty at the rate of 3 gallons per minute. If A, B and C opened together fill the tank in 15 min, the capacity of the tank (in gallons) is:

A - 180

B - 150

C - 120

D - 60

Answer : C

Explanation

Let the capacity of the tank = x gallons
Quantity of the water filled in the tank in 1 min when all the pipes A, B and C are opened simultaneously= x/20 + x/24 - 3
According to question,
x/20 + x/24 - 3 = x/15
or, x/20 + x/24 - x/15 = 3
or, (6x + 5x  - 8x)/120 = 3
or, 3x/120 = 3
or, x = 120 gallons

Q 10 - Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for 1 hour each alternatively, the tank will be full in -

A - 5 hr

B - 7.5 hr

C - 6 hr

D - 7 hr

Answer : D

Explanation

Tank filled by (A+B) in 1 hour = (1/12 + 1/15) = 3/20
Tank filled by (A+C) in 1 hr = (1/12 + 1/20) = 2/15

∴ Part of tank filled in 2 hr = (3/20 + 2/15) = 17/60
∴ Part of tank filled in 6 hr = (3*17/60) = 17/20

Remaining part = 1 - 17/20 = 3/20

Now it is the turn of A and B and 3/20 is filled by A and B in 1 hour.

∴ Total time taken to fill the tank = (6+1) = 7 hr
aptitude_pipes_cisterns.htm
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