Pipes A and B are fixed to a tank. A is the filling pipe and B can be used for filling or emptying at the same rate. When B is used for filling. It takes time 't' along with A to fill the tank. If it is used for emptying when A is filling the tank, the time taken for the tank to fill up would be '5t'. Find the ratio of the rate of A and B.

Correct option is C

Given:

Pipe A is a filling pipe.

Pipe B can either fill or empty at the same rate.

Time to fill the tank when both A and B are filling = $t$.

Time to fill the tank when A fills and B empties = $5t$.

Formula used:

Let the rate of pipe A be 'a' and the rate of pipe B be 'b'.

When both pipes are filling:

(a + b) × t = 1 (tank full)

When pipe A is filling and pipe B is emptying:

(a - b) × 5t = 1 (tank full)

Solution:

(a + b) × t = 1

=> a + b = $$\frac 1t$$​​

(a - b) × 5t = 1

=> a - b = $$\frac1{(5t)}$$​​

Adding the two equations:

(a + b) + (a - b) = $$\frac1t + \frac1{(5t)}$$​​

=> 2a = $$\frac1t + \frac1{(5t)}$$​​

=> 2a = $$\frac 6{(5t)}$$​​

=> a = $$\frac3{(5t)}$$​​

Subtracting the two equations:

(a + b) - (a - b) = $$\frac1t - \frac1{(5t)}$$​​

=> 2b = $$\frac1t - \frac1{(5t)}$$​​

=> 2b = $$\frac4{(5t)}$$​​

=> b = $$\frac2{(5t)}$$​​

Ratio of the rates of A and B:

a : b = $$\frac3{(5t)} : \frac2{(5t)}$$​​

=> a : b = 3 : 2