Pipes & Cisterns: Concepts and Tricks

The approach to solve time and work questions is the same as the questions on time and work. In this case, the filling rate or capacity of the pipes and given and then question asks about the time it will take to fill a tank or about the capacity of the tank.  There are two ways to deal with the questions on time and work, let us discuss those one by one.

If a pipe fills a tank in A minutes and another pipe fills the same tank in B minutes. Then in how much time the tank will be filled completely?

(1/A) + (1/B) = (1/C), where C is the time in which tank will be filled, (in minutes). These have been added because both the pipes are filling the tank.

If a pipe fills a tank in A minutes and another pipe empties the same full tank in B minutes. Then in how much time the tank will be filled completely, if both the pipes are opened simultaneously?

(1/A) - (1/B) = (1/C), where C is the time in which tank will be filled (in minutes). It is subtracted because the second pipe is a drain pipe and it reduces the work done by the first pipe.

Illustration 1: If a pipe fills a tank in 20 minutes and a pipe empties the same tank in 60 minutes. Then in how much time the tank will be filled completely if both the pipes are opened together?

Sol: Let the tank be filled in X minutes.

By unitary method,

(1/20) - (1/60) = (1/X).

Solving the equation, we get, x = 30 minutes. So, the tank will be filled in 30 minutes.

Let the capacity of the tank = 60 units (or say 60 litres),

LCM = (20,60) = 60.

Rate of the filling pipe = 60/20

⇒ 3 litres per minute

Rate of the draining pipe = 60/60

⇒ 1 litre per minute

In one minute (3 – 1) = 2 litres of water is collected in the tank

Thus, 60 litres of water will be filled in 60/2 = 30 minutes.

Illustration 2:  Two pipes fill a tank in 20 minutes and 60 minutes. A third pipe empties the same tank in 40 minutes. In how much time will the tank be filled if all three pipes are opened together?

Sol: Let the tank is filled in X minutes. By unitary method,

(1/20)+(1/60)-(1/40)=(1/X).

Solving, we get x = 24 minutes. Hence, the tank will be filled in 24 minutes.

Let the capacity of the tank = 120 units (or say 120 litres),

LCM = (20, 60, 40) = 120

Rate of the first filling pipe = 120/20

⇒ 6 litres per minute

Rate of the second filling pipe = 120/60

⇒ 2 litre per minute

Rate of the draining pipe = 120/40

⇒3 litre per minute      

Now in a minute, (6 + 2 – 3) = 5 litre of water is filled in the tank

Hence, 120 litres of water will be collected in 120/5 = 24 minutes.

A pipe can fill a cistern in 12 minutes and another can fill it in 15 minutes, but a third pipe can empty it in 6 minutes. The first two are kept open for 6 minutes in the beginning and then the third pipe is also opened, in what time will the cistern be emptied?

Let the capacity of the tank = LCM of (12,15, 6) = 60 litres.

Rate of work done by the first pipe = 60/12

⇒ 5 L/min

Rate of work done by the second pipe = 60/15

⇒ 4L/min

Rate of work done by the draining pipe = 60/6

⇒10L/min

If all the pipes are opened simultaneously, (5+4-10) litres will be filled in a minute.

Since this comes out to be negative, no water will be filled at any point of time.

Rate of both filling pipes = 9L/min, (5+4).

So in 6 minutes, 9 × 6 = 54 litres water is filled.

When the third pipe is opened, all three pipes will work simultaneously, at a rate of (5+4 – 10) = - 1 ltr per min. This implies that in a minute, 1 litre water will be drained.

So, 54 litres water will be drained in 54 minutes. So the tank will be emptied in 54 minutes.