__Pipes and Cisterns Practice Questions – Set 1__**1. Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in?**

a) 7 min

b) 13 min

c) 23 min

d) 9 min

**2. Two pipes can fill a tank in 10 and 14 minutes respectively and a waste pipe can empty 4 gallons per minute. If all the pipes working together can fill the tank in 6 minutes, what is the capacity of the tank?**

a) 120 gallons

b) 240 gallons

c) 450 gallons

d) 840 gallons

e) 540 gallons

**3. 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) Can’t be determined

e) None of these

**4. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:**

a) 6 hours

b) 10 hours

c) 15 hours

d) 25 hours

e) 30 hours

**5. Two pipes A and B can fill a tank in 15 minutes and 40 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 10 sec

b) 25 min 20 sec

c) 29 min 20 sec

d) 20 min 10 sec

e) 22 min 20 sec

**6. Two pipes A and B can separately fill a tank in 12 and 15 minutes respectively. A third pipe C can drain off 45 liters of water per minute. If all the pipes are opened, the tank can be filled in 15 minutes. What is the capacity of the tank?**

a) 480liters

b) 540 litres

c) 600 litres

d) 675 litres

**7. Two pipes fills a cistern in 15 hrs and 20 hrs respectively. The pipes are opened simultaneously and it is observed that it took 26 min more to fill the cistern because of leakage at the bottom. If the cistern is full, then in what time will the leak empty it?**

a) 35 hrs

b) 70 hrs

c) 180 hrs

d) 300 hrs

e) 240 hrs

**8. Two pipes A and B can separately fill a cistern in 10 and 15 minutes respectively. A person opens both the pipes together when the cistern should have been was full he finds the waste pipe open. He then closes the waste pipe and in another 4 minutes the cistern was full. In what time can the waste pipe empty the cistern when fill?**

a) 7 min

b) 8 min

c) 9 min

d) 10 min

e) 11 min

**9. If two pipes function simultaneously, the reservoir will be filled in 24 hrs. One pipe fills the reservoir 20 hours faster than the other. How many hours does it take for the second pipe to fill the reservoir?**

a) 12 hrs

b) 30 hrs

c) 44 hrs

d) 60 hrs

e) 64 hrs

**10. It is observed that the pipe A can fill the tank in 15 hrs and the same tank is filled by pipe B in 20 hrs. The third pipe C can vacant the tank in 25 hrs. If all the pipes get opened initially and after 10 hrs, the pipe C is closed, then how long will it take to fill the tank?**

a) 3 hrs

b) 7 hrs

c) 12 hrs

d) 20 hrs

e) 22 hrs

**Solutions:**

**1. D)**1/10 + 1/15 - 1/x = 1/18

x = 9

**2. D)**Work done by waste pipe in 1 min = (Part filled by total pipes together) -(part filled by first pipe + part filled by second pipe)

= 1/6 – [(1/10) + (1/14)] = 1/6 – (24/140) = -1/210

Here, negative (-) sign indicates emptying of tank.

To find the capacity, we need to determine the
volume of 1/210 part.

Therefore, volume of 1/210 part = 4 gallons
---------(given condition)

Hence, the capacity of tank = volume of whole = 4 x
210 = 840 gallons.

**3. B)**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 – 1/2) = 1/2

Therefore 2/3 : 1/2 :: 1 : x => x = (1/2 × 1 ×
3/2) = 3/4

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

**4. C)**Suppose, first pipe alone takes x hours to fill the tank

Then, second and third pipes will take (x – 5) and
(x – 9) hours respectively to fill the tank.

=> 1/x + 1/(x – 5) = 1/(x – 9) => [(x – 5 +
x)/(x{x – 5})] = 1/(x – 9) => (2x – 5)(x – 9) = x (x – 5) => x^2 – 18x +
45 = 0 => (x – 15)(x – 3) = 0

X = 15 [neglecting x = 3]

**5. C)**Part filled by pipe A in 1 min = 1/15; Part filled by pipe B in 1 min = 1/40

Part filled by pipe A and pipe B together in 1
minute = 1/15 + 1/40 = 11/120

pipe A and pipe B were open for 4 minutes. Part
filled by pipe A and pipe B together in these 4 minutes = 4 × 11/120 = 11/30

Remaining part to be filled = 1 – 11/30 = 19/30

Time taken by pipe B to fill this remaining part =
(19/30)/(1/40) = (19 × 4)/3

= 76/3 = 25 1/3min = 25 min 20 sec

Therefore, total time required = 4 minutes + (25
minutes 20 seconds) = 29 minutes 20 seconds.

**6. B)**1/12 + 1/15 - 1/x = 1/15

x = 12

12 * 45 = 540

**7. C)**Consider a pipe fills the tank in 'x' hrs. If there is a leakage in the bottom, the tank is filled in 'y' hrs.

The time taken by the leak to empty the full tank = xy/(y – x) hrs

But, direct values of x & y are not given. So,
we need to find the work done by the two pipes in 1 hr = (1/15) + (1/20) = 7/60

Hence, the time taken by these pipes to fill the
tank = 60/7 hrs = 8 .57 hrs = 8 hrs 34 min --------( by multiplying '0.57' hrs
x 60 = 34 minutes)

Due to leakage, time taken = 8 hrs 34 min + 26 min =
8 hrs 60 min = 9 hrs ----------( because 60 min = 1 hr)

Thus, work done by (two pipes + leak) in 1 hr = 1/9

Hence, work done by leak in 1 hr = work done by two
pipes – 1/9

= 7 /60 – 1/9 = 3/ 540 = 1/180

Therefore, leak will empty the full cistern in 180
hours.

**8. B)**1/10 + 1/15 = 1/6 * 4 = 2/3

1 - 2/3 = 1/3

1/10 + 1/15 - 1/x = 1/3

x = 8

**9. D)**Assume that the reservoir is filled by first pipe in 'x' hours.

So, the reservoir is filled by second pipe in 'x +
20' hours.

Now, from these above conditions, we can form the
equations as,

1/x + 1/ (x + 20) = 1/24

[x + 20 + x] / [x(x + 20)] = 1/24

x2 – 28x – 480 = 0

By solving this quadratic equation , we get the
factors (x – 40) (x+12) = 0

Hence, we get two values : x = 40 & x = -6

Since filling of reservoir is positive work , we can
neglect the negative value of 'x'.

Thus, x = 40

This means that the second pipe will take (x+ 20)
hrs = 40 + 20 = 60 hrs to fill the reservoir.

**10. C)**Assume that the tank will be full in '10 + x' hrs.

Part filled by pipe A in 1 hr = 1/15

Part filled by pipe B in 1 hr = 1/20

Part emptied by pipe C in 1 hr = 1/ 25

(A+ B)'s part filled in 1 hr = 1/15 + 1/20 = 7 /60

As Pipe C is closed after 10 hours, let us find the
part of tank filled in 10 hrs.

Tank filled in 10 hrs = 10 (part filled by A in 1 hr
+ part filled by B in 1 hr– part emptied by C in 1 hr) = 10 [1/15 + 1/20 –
1/25] = 23 /30

Remaining part = 1 – part filled in 10 hours = 1 –
23/30 = 7/30

We know that,

Part filled by (A + B) in 1 hr/Remaining part ::
Total part filled/Time taken

So, we get the ratio as ,

7/60 : 7/30 :: 1 : x

So, the value of x = 2/30 x 1 x 60/7 = 2 hrs.

Hence, the tank will be full in x + 2 = 10 + 2 hrs =
12 hrs