45+ Solved Pipes and Cisterns Problems With Solutions

Q:

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

A) 10 min. 20 sec.	B) 11 min. 45 sec.
C) 12 min. 30 sec.	D) 14 min. 40 sec.

Answer & Explanation Answer: D) 14 min. 40 sec.

Explanation:

Part filled in 4 minutes =4(1/15+1/20) = 7/15

Remaining part =(1-7/15) = 8/15

Part filled by B in 1 minute =1/20 : 8/15 :: 1:x

x = (8/15*1*20) = $10 \frac{2}{3} m i n = 10 m i n 40 s e c$

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec

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Q:

A water tank is two-fifth full.Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes.If both the pipes are open,how long will it take to empty or fill the tank completely?

A) 6 min.to empty	B) 6 min.to fill
C) 9 min.to empty	D) 9 min.to fill

Answer & Explanation Answer: A) 6 min.to empty

Explanation:

Clearly,pipe B is faster than pipe A and so,the tank will be emptied.

part to be emptied = 2/5

part emptied by (A+B) in 1 minute= $(\frac{1}{6} - \frac{1}{10}) = \frac{1}{15}$

$\frac{1}{15} : \frac{2}{5} : : 1 : x$

$\frac{2}{5} * 15 = 6 m i n s$

so, the tank will be emptied in 6 min

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Q:

A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

A) 3 hrs 15 min	B) 3 hrs 45 min
C) 4 hrs 15 min	D) 4 hrs 1

Answer & Explanation Answer: B) 3 hrs 45 min

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =4*1/6 =2/3

Remaining part = $(1 - \frac{1}{2}) = \frac{1}{2}$

$\frac{2}{3} : \frac{1}{2} : : 1 : x$

=> x = $(\frac{1}{2} * 1 * \frac{3}{2}) = \frac{3}{4}$

So, total time taken = 3 hrs. 45 mins.

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Q:

A pump can fill a tank with water in 2 hours. Because of a leak, it took $2 \frac{1}{3}$ hours to fill the tank. The leak can drain all the water of the tank in:

A) 7 hours	B) 8 hours
C) 12 hours	D) 14 hours

Answer & Explanation Answer: D) 14 hours

Explanation:

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part

The pumop take1 hour to fill 1/2 portion

Due to lickage

The pumop take 7/3 hrs to fill 1 part

The pumop take1 hour to fill 3/7 portion

Now the difference of area = (1/2-3/7)=1/14

This 1/14 part of water drains in 1 hour

Total area=1 part of water drains in (1x14/1)hours= 14 hours

So the leak can drain all the water of the tank in 14 hours.

Leak will empty the tank in 14 hrs

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Q:

A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in ?

A) 20 hrs	B) 28 hrs
C) 36 hrs	D) 40 hrs

Answer & Explanation Answer: A) 20 hrs

Explanation:

1/8 - 1/x = 1/10

=> x = 40 hrs

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Q:

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

A) 10	B) 12
C) 14	D) 16

Answer & Explanation Answer: C) 14

Explanation:

Part filled in 2 hours = 2/6=1/3

Remaining part = $(1 - \frac{1}{3}) = \frac{2}{3}$

(A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

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77 22808

Q:

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 ,C are open for one hour each alternatively, the tank will be full in:

A) 6 hrs	B) 20/3 hrs
C) 7 hrs	D) 15/2 hrs

Answer & Explanation Answer: C) 7 hrs

Explanation:

$(A + B)' s 1 h o u r w o r k = (\frac{1}{12} + \frac{1}{15}) = \frac{9}{60} = \frac{3}{20}$

$(A + C)' s 1 h o u r w o r k = (\frac{1}{12} + \frac{1}{20}) = \frac{8}{60} = \frac{2}{15}$

$p a r t f i l l e d i n 2 h r s = (\frac{3}{20} + \frac{2}{15}) = \frac{17}{60}$

$p a r t f i l l e d i n 6 h r s = (3 * \frac{17}{60}) = \frac{17}{20}$

$r e m a i n i n g p a r t = (1 - \frac{17}{20}) = \frac{3}{20}$

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

Therefore, Total time taken to fill the tank =(6+1)hrs= 7 hrs

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Q:

A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 $m^{3}$ . The emptying capacity of the tank is 10 $m^{3}$ per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

A) 50 m^3/min	B) 60 m^3/min
C) 72 m^3/min	D) None of these

Answer & Explanation Answer: A) 50 m^3/min

Explanation:

Let the filling capacity of the pump be x $m^{3}$ /min.

Then, emptying capacity of the pump=(x+10) $m^{3}$ /min.

so, $\frac{2400}{x} - \frac{2400}{x + 10} = 8 \Leftrightarrow x^{2} + 10 x - 3000 = 0$

$\Rightarrow (x - 50) + (x + 60) = 0 \Leftrightarrow x = 50$

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