A tank can be fully filled with water using a pipe that fills 20 liters in a minute. A bigger pipe that can fill 25 liters in a minute ..will take one minute less to fill the same tank. How...
A tank can be fully filled with water using a pipe that fills 20 liters in a minute. A bigger pipe that can fill 25 liters in a minute ..
will take one minute less to fill the same tank. How many minutes does the smaller pipe take 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.
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 V be the volume of the tank in litres.
Then at the rate of 20 litre per minute the smaller pipe takes V/20 minutes.
Then the bigger pipe takes V/20-1 minutes by data which should also be equal to V/ 25 minutes.
Therefore V/20 -1 = V/25. Multiplying by 20*25 we get:
25V - 20*25 = 20 V.
So 25V- 20 V = 20*25.
5V = 20* 25.
V = 20*25/ 5 = 100 litre.
So the smaller pipe takes 100/20 = 5 minutes.