Question 21

# A pump can be operated both for filling a tank and for emptying it. The capacity of tank is 2400 $$m^3$$ .The emptying capacity of the pump is 10 $$m^3$$ per min higher than its filling capacity. Consequently, the pump needs 8 min less to empty the tank than to fill it. Find the filling capacity of the pump.

Solution

Let the filling capacity of the pump be x $$m^3$$/min

Then the emptying capacity of the pump = (x+10) $$m^3$$/min

The time required for filling the tank = $$\ \frac{\ 2400}{x}$$

The time required for emptying the tank = $$\ \frac{\ 2400}{x+10}$$

The pump needs 8 min less to empty the tank than to fill it.

$$\ \frac{\ 2400}{x}$$ - $$\ \frac{\ 2400}{x+10}$$ = 8

$$\ \frac{\ 300}{x}$$ - $$\ \frac{\ 300}{x+10}$$ = 1

300(x+10) - 300(x) = x(x+10)

$$x^2+10x-3000$$ = 0

(x+60)(x-50) = 0

x = 50 or -60

Since x cannot be negative, x = 50

Filling capacity of the pump = 50 $$m^3$$/min

• All Quant Formulas and shortcuts PDF
• 170+ previous papers with solutions PDF