A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m³. The emptying capacity of the tank is 10 m³ per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

Given,

Capacity of tank = 2400 m³

Let filling capacity of pump = x m³/min,

Time required to fill the tank = $\dfrac{2400}{x}$

Emptying capacity = (x + 10) m³/min

Time required to empty the tank = $\dfrac{2400}{x + 10}$

Given,

The pump needs 8 minutes lesser to empty the tank than it needs to fill it.

$\Rightarrow \dfrac{2400}{x} - \dfrac{2400}{x + 10} = 8 \\[1em] \Rightarrow \dfrac{2400(x + 10) - 2400x}{x(x + 10)} = 8 \\[1em] \Rightarrow \dfrac{2400x + 24000 - 2400x}{x^2 + 10x} = 8 \\[1em] \Rightarrow \dfrac{24000}{x^2 + 10x} = 8 \\[1em] \Rightarrow 24000 = 8(x^2 + 10x) \\[1em] \Rightarrow 8x^2 + 80x - 24000 = 0 \\[1em] \Rightarrow 8(x^2 + 10x - 3000) = 0 \\[1em] \Rightarrow x^2 + 10x - 3000 = 0 \\[1em] \Rightarrow x^2 + 60x - 50x - 3000 = 0 \\[1em] \Rightarrow x(x + 60) - 50(x + 60) = 0 \\[1em] \Rightarrow (x - 50)(x + 60) = 0 \\[1em] \Rightarrow (x - 50) = 0 \text{ or }(x + 60) = 0 \\[1em] \Rightarrow x = 50 \text{ or } x = -60.$

Since, rate must be positive.

∴ x = 50.

Hence, filling capacity of pump is 50 m³/min.