Question

A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is $$2400 \text{ m}^3$$. The emptying capacity of the tank is $$10 \text{ m}^3$$ 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?
[Hint : Let the filling capacity of the pump be $$x \text{ m}^3/\text{min}$$.
Then, emptying capacity of the pump $$= (x + 10) \text{ m}^3/\text{min}$$.
$$\therefore \frac{2400}{x} - \frac{2400}{(x + 10)} = 8. \text{ Solve for } x.$$]

Answer 1

Let the filling capacity of the pump be $$x\text{ m}^3/\text{min}$$.

Then, the emptying capacity is $$(x + 10)\text{ m}^3/\text{min}$$. 

Time to fill the tank $$= \frac{2400}{x}\text{ min}$$. 

Time to empty the tank $$= \frac{2400}{x + 10}\text{ min}$$. 

According to the problem: $$\frac{2400}{x} - \frac{2400}{x + 10} = 8$$ 

Dividing throughout by 8: $$\frac{300}{x} - \frac{300}{x + 10} = 1$$ 

$$300\left(\frac{x + 10 - x}{x(x + 10)}\right) = 1$$

$$\frac{3000}{x^2 + 10x} = 1$$

$$x^2 + 10x - 3000 = 0$$ 

Factoring: $$(x + 60)(x - 50) = 0$$

$$x = -60 \quad \text{or} \quad x = 50$$ 

Since capacity cannot be negative, reject $$x = -60$$. 

Hence, the filling capacity of the pump is $$50\text{ m}^3/\text{min}$$.