Let the time taken to fill the tank with the smaller pipe be T. This pipe fills at the rate of 20 liters/ minute.

So the capacity of the tank is 20T.

The bigger pipe fills at 25 liters/ minute and takes 1 minute less.

So we have 20T /...

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Let the time taken to fill the tank with the smaller pipe be T. This pipe fills at the rate of 20 liters/ minute.

So the capacity of the tank is 20T.

The bigger pipe fills at 25 liters/ minute and takes 1 minute less.

So we have 20T / 25 = T - 1

=> 20T = 25T - 25

=> 5T = 25

=> T = 5

Therefore the smaller pipe takes **5 minutes** to fill the tank.

Let the volume of the tank be V ( in litres)

The rate of filling the tank with a smaller pipe is 20 L / m

Then, the time will be taker to fill the tank is T1 = V/20 .......(1)

Now with the bigger pipe, the rate is 25 L / m

Then the time will be taken to fill the tank is T2 = V/25

But the time taken is one minute less than the time required to fill with smaller pipe.

==> T2 = T1 - 1

==> T1 -1 = v/25

==> T1 = v/25 + 1............(2)

Now from (1) and (2) we conclude that:

v/25 + 1 = v/ 20

==> (v+25) / 25 = v/20

==> (v+25) /5 = v/ 4

We will multiply by 20.

==> 4(v+25) = 5v

==> 4v + 100 = 5v

==> V = 100 litre

==> T1 = v/20 = 100/20 = 5 minutes.

**Then, the time required to fill the tank with the smaller pipe is 5 minutes.**