A booster pump can be used for filling as wel | vedclass

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Question

(A) Let the filling capacity of the pump be $x m^3/min$.The emptying capacity of the pump is $(x + 10) m^3/min$.The time taken to fill the tank is $T_

Vedclass · Accepted Answer

$50 m^3/min$

Vedclass · Answer

(A) Let the filling capacity of the pump be $x m^3/min$.The emptying capacity of the pump is $(x + 10) m^3/min$.The time taken to fill the tank is $T_f = \frac{2400}{x}$ minutes.The time taken to empty the tank is $T_e = \frac{2400}{x+10}$ minutes.According to the problem,$T_f - T_e = 8$.$\frac{2400}{x} - \frac{2400}{x+10} = 8$Divide by $8$: $\frac{300}{x} - \frac{300}{x+10} = 1$$300(x + 10) - 300x = x(x + 10)$$300x + 3000 - 300x = x^2 + 10x$$x^2 + 10x - 3000 = 0$$(x + 60)(x - 50) = 0$Since capacity cannot be negative,$x = 50$.Thus,the filling capacity of the pump is $50 m^3/min$.

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