Pipes and Cistern- Aptitude Questions and Answers

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A tank can be filled by one tap in 10 minutes and by another in 30 minutes. Both the taps are kept open for 5 minutes and then the first one is shut off. In how many minutes more is the tank completely filled?

A)
5

B)
7.5

C)
10

D)

12



Correct Answer :

10


Explanation :


Work done by first tap in 1 min is 1/10
Work done by second tap in 1 min is 1/30

In 5 minutes work done by both pipes is = 5*(1/10 + 1/30)
=5*(4/30)
=2/3

After 5 min first tap is closed,
Now, remaining work is = 1 - 2/3 = 1/3

So, remaining work done by second tap is = 30*(1/3) = 10
After closing first tap, the second tap will be filled the tank completely in 10 minutes.

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Three pipes A, B and C together can fill a cistern in 6 hours. After working together for 2 hours, C is closed and A and B fill the remaining cistern in 8 hours. Then the time in which the cistern can be filled by pipe C alone will be:

A)
11 hours

B)
9 hours

C)
12 hours

D)
10 hours


Correct Answer :

12 hours


Explanation :

Three Pipes A,B and C together filled a tank in 6 hours
So 1 hour work of A,B and C =1/6
C closed after 2 hour's, Part filled in 2 hours working together(A,B and C) = 2/6 = 1/3
Remaining part = (1-1/3) = 2/3

A and B can fill remaining in 8 hours, (A + B)'s 8 hour's work = 2/3

So, (A+B)'s 1 hour work = 2/24=1/12

Therefore C's 1 hour's work
= [(A+B+C)'s 1 hour]-[(A+B)'s 1 hr work]
=(1/6 - 1/12) = 1/12

C alone can fill the tank in 12 hours

Two pipes A and B can fill a cistern in 4 and 6 minutes respectively. If both the pipes kept turned on alternatively for one minute each starting from A, how long will it take to fill the cistern?

A)
3(3/7) minutes

B)
6 minutes

C)
4(2/3) minutes

D)
6(6/7) minutes


Correct Answer :

4(2/3) minutes


Explanation :

As the pipes are operating alternatively for one minute, thus (A+B)'s 2 minutes job is =(1/4 +1/6) =5/12
in next, 2 minutes both pipe can fill another 5/12 part.
So in 4 minutes A and B are operating alternatively will fill 5/12 + 5/12 = 5/6

Remaining part of cistern= 1 - 5/6 =1/6

Pipe A can fill 1/4 part of the cistern in 1 minute,and can fill 1/6 part of the cistern in = 4*(1/6) =2/3 minutes

Total time taken to fill the Cistern = time taken to fill (5/6 part + 1/6 part)
=> 4 minutes + 2/3 minutes = 4(2/3) minutes Or, 4 minutes 40 seconds

Pipe A can fill a tank in 45 hrs and pipe B can fill it in 36 hrs. If both the pipes are opened in the empty tank. In how many hours will it be full?

A)
10 hr

B)
15 hr

C)
20 hr

D)
28 hr


Correct Answer :

20 hr


Explanation :

If a pipe can fill a tank in x hours, then : part filled in 1 hour = 1/x

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

So, The tank will be full in 20 hr.

If a pipe fills a tank in 6 h, then what part of the tank will the pipe fill in 1 h?

A)
1/3

B)
1/6

C)
1/4

D)
1/5


Correct Answer :

1/6


Explanation :

Let the pipe can fill a tank in x hours, then part filled in 1 hour = 1/x

A pipe fills a tank in 6 h, then the part of tank filled in 1 h = 1/6
Hence, required part of the tank to be filled in 1h = 1/6 part

A tap can fill a tank in 16 minutes and another can empty it in 8 minutes. If the tank is already half full and both the taps are opened together, the tank will be ?

A)
Filled in 12 min

B)
Emptied in 12 min

C)
Filled in 8 min

D)
Emptied in 8 min


Correct Answer :

Emptied in 8 min.


Explanation :

Let a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), when opening both the pipes, the net part emptied in 1 hour = (1/y)-(1/x)
=> (1/8 - 1/16) = 1/16

Time taken to empty the full tank = 16 min.
Hence, time taken to empty the half tank = 8 min

One tap can fill a cistern in 2 hours and another can empty the cistern in 3 hours. How long will they take to fill the cistern if both the taps are opened ?

A)
5 hours

B)
6 hours

C)
7 hours

D)
8 hours


Correct Answer :

6 hours


Explanation :

Let a pipe can fill a tank in x hours, and another pipe can empty the full tank in y hours (where y> x), then on opening both the pipes, the net part filled in 1 hour = (1/x)-(1/y)
=> (1/2 - 1/3) = 1/6

Hence, time taken to fill the cistern = 6 hours

44 pipes can fill a large water tank in 27 hours. How many hours it take for 66 pipes to fill four such tanks ?

A)
72

B)
54

C)
63

D)
84


Correct Answer :

72


Explanation :

44 pipes => 27 hours
66 pipes=> ?

Amount of work done is same for filling a pipe, So 44*27 = 66*X
X = 44*27 / 66
X = 18

66 pipes can fill one tanks in 18 hours.
Hence, they will fill 4 tanks in 18*4 = 72 hours

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