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Here you can download CMAT 2022 – important Pipes and Cisterns Questions PDF by Cracku. Very Important Pipes and Cisterns Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.

Question 1:Â Pipe ‘P’ can fill a tank in 36 hours and pipe ‘Q’ can fill this tank in 45 hours. If both the pipes are opened simultaneously, then how much time will be taken to fill this tank ?

a)Â 20 hours

b)Â $40 \frac{1}{2}$ hours

c)Â 9 hours

d)Â 42 hours

Question 2:Â A tank has three pipes. The first one fills up the tank in 30 min and the second one can fill it up in 45 min. The third one drains out the tank. With all the pipes open it takes 27 min to fill up the tank. Find out how long it takes the third pipe alone to empty a full tank?

a)Â 56 min

b)Â 50 min

c)Â 52 min

d)Â 54 min

Question 3:Â Pipes A, B and C together can fill a tank in 4 hours. Pipes A and B together can fill the tank in 9 hours. If pipes A, B and C are opened simultaneously, and after 3 hours, pipes A and B are closed, then how many hours will C alone require to fill the tank?

a)Â 2.25

b)Â 2

c)Â 1.5

d)Â 1.8

Question 4:Â A water tank is filled in 5 hours by three pipes X, Y and Z the pipe Z is thrice as fast as Y and Y is twice as fast as X. How much time will pipe X alone take to fill the water tank?

a)Â 60 hrs

b)Â 45 hrs

c)Â 40 hrs

d)Â 35 hrs

Question 5:Â Pipe A is an inlet pipe that can fill an empty cistern in 69 hours. Pipe B can drain the filled cistern in 46 hours. When the cistern was filled the two pipes are opened one at a time for an hour each, starting with Pipe B. how long will it take for the cistern to be empty?

a)Â 11 days 10 hours

b)Â 11 days 7 hours

c)Â 11 days 12 hours

d)Â 1 days 13 hours

Question 6:Â Pipes A and B, when working together, can fill an empty tank in 8 hours. If B is stoppedÂ after 2 hours, it will take a total of 11 hoursto fill the tank. How long would it take A to fill the empty tank alone?

a)Â 15 hours

b)Â 12 hours

c)Â 13 hours

d)Â 16 hours

Question 7:Â Pipe A could fill an empty cistern in 18 hours while Pipe B can drain filled cistern in 30 hours. When the cistern is empty. Pipe A is turned on for an hour and then turned off. Now Pipe B is allowed to drain out water from the cistern for an hour and then turned off. The pipes were alternately left open for an hour each time till the cistern was full. How much time did it take for the cistern to be full?

a)Â 90 hours

b)Â 86 hours 40 minutes

c)Â 45 hours

d)Â 86 hours 48 minutes

Question 8:Â Two pipes P and Q can fill a cistern in 18 and 24 minutes respectively. Both pipes are opened, but after some time, the slower pipe is closed. The cistern gets filled in 12 minutes from the start. When is the slower pipe closed?

a)Â After 9 minutes

b)Â After 8 minutes

c)Â After 10 minutes

d)Â After 6 minutes

Question 9:Â Two taps can separately fill a cistern in 20 minutes and 25 minutes. Both taps are open for 10 minutes after which the slower one is closed. How long will it take to fill the remaining portion by the other tap alone?

a)Â 2 minutes

b)Â 4 minutes

c)Â 10 minutes

d)Â 5 minutes

Question 10:Â Two pipes can fill a tank in 5 and 6 hours respectively. A third pipe will empty it in 12 hours. If all the pipes areÂ opened, then find the time taken to either fill or empty the tank.

a)Â Filled in 30/17 hours

b)Â Filled in 60/17 hours

c)Â Filled in 30/11 hours

d)Â Emptied in 9/2 hours

= $\frac{1}{36} + \frac{1}{45} = \frac{81}{36 \times 45} = \frac{1}{20}$

If both the pipes are opened simultaneously, then 20 hrs will be taken to fill this tank.

So, the answer would be option a)20 hours

1st and 2nd pipes fill the tank so addition occurs but 3rd can drain out,

$\ \frac{\ 1}{30}+\ \frac{\ 1}{45}-\ \frac{\ 1}{x}=\ \frac{\ 1}{27}$

by solving this we get the value of x=54

Hence option ‘D’ is correct.

A and B works for 3 hours , so remaining work =36-3$\times\ 9$=9

so C alon can fill the tank inÂ $\ \frac{\ 9}{5}=1.8\ hours$,

hence option ‘D’ is correct.

A water tank is filled in 5 hours by three pipes X, Y and Z the pipe Z is thrice as fast as Y and Y is twice as fast as X.

$\frac{Z}{Y}=\frac{3}{1}Â andÂ \frac{Y}{X}=\frac{2}{1}$

Now equate both the ratios by making Y equal in both,

$\frac{Z}{Y}=\frac{6}{2} and \frac{Y}{X}=\frac{2}{1}$

So, Z : Y : X = 6 : 2 : 1 (this is the ratio of efficiencies of the pipes)

If three pipes fill the tank in 5 hours, there fore the total work done is

$=Time \times TotalÂ EfficiencyÂ ofÂ pipes$

$=5\times (6+2+1) = 45$

So X will take time = $\frac{Work}{X’sÂ efficiency}$

$=\frac{45}{1} = 45 hours$

Option B is correct.

Pipe A can fill cistern in 69 hours.

PipeÂ B can empty cistern in 46 hours.

Let’s assume total work = 138

Efficiency of A = 2

Efficiency of B = -3 (Subtraction sign is due to outlet pipe.)

For empty theÂ cistern, Alternatively pipe B and A respectively work. So in every 2 hours 1 part of tank is empty out of total 138.

Then we can say that for emptying 135 parts, we need 270Â hours and last 1 hour B will empty 3 parts.

So 270 + 1 = 271 hours

If we convertÂ 271 hours in days by dividing 24. Then got 11 days 7 hours.

A can fill the empty tank alone in ( 12/1 Ã— 1 ) = 12 hours.

Step-by-step explanation:

let, ( A + B ) means A and B work together.

( A + B) can fill an empty tank in 8 hours and after 2 hours B was stopped. That means first 2 hours A and B worked together.

( A + B) can fill a 2/8 = 1/4 part in 2 hours.

after 2 two hours, A fill the remaining part of tank which is ( 1 – 1/4 ) = 3/4 inÂ  Â  ( 11- 2) = 9 hours

It means A can fill ( 3/4 Ã— 1/9 ) = 1/12 part in 1 hour.

Therefore, A can fill the empty tank alone in ( 12/1 Ã— 1 ) = 12 hours.

A can fill the cistern in 18 hours and B can empty in 30 hours
LCM of 18 and 30 is 90 units
In one hour A can fill 5 units
In one hour B can empty 3 units
So in 2 hours 2 units of work is done
For 86 hours 86 units of work is done
Now that 4 units is filled by A in 4/5 hours
So total time 86 hours 48 minutes

P fill the cistern = 18 min

Q fill the cistern = 24 min

Total work done = LCM (18,24)= 72

P’s efficiency = $\frac{72}{18}$ = 4

Q’s efficiency = $\frac{72}{24}$ = 3

Tank will be fiiled= $\frac{72}{7}$hours

Tap P is open for 12 minÂ  Â and taps Q closed after some time

so

as we know that ,

efficiency multiplied by time = work

$4\times 12 + 3\times T$= 72

T = 8 min

$work done = efficiency \times time$

Two taps P1 and P2Â  can separately fill a cistern in 20 minutes and 25 minutes.

work done = LCM( 20,25 ) =100

efficiency of P1 = $\frac{100}{20} = 5$

efficiency of P2 = $\frac{100}{25} = 4$

Both taps are open for 10 minutes after which the slower one is closed.

$work done = ( efficiency of P1 + efficiency of P2 ) \times 10 + efficiency of P1 \times x$

where x = time required by P1 to fill the remaining portion by the other tap alone

substituting

$100 = ( 5 + 4) \times 10 + 5x$

solving x = 2 minutes

$work = efficiency \times time$

P1,P2 and P3 be the pipes

P1 fill the tank in 5 hrs

P2 fill the tank in 6 hrs

P3 empty the tank in 12 hrs

work done = LCM of P1,P2 and P3 = LCM (5,6,12) = 60

efficiency of P1 = $\frac{60}{5} = 12$

efficiency of P2 = $\frac{60}{6} = 10$

efficiency of P3 = $\frac{60}{12} = 5$

efficiency of P1 + P2 + P3 = 12 + 10 – 5 = 17

$60 = 17 \times time$

$time = \frac{60}{17} hrs$