Question

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

• A. 45 \(m^3\)/min
• B. 30 \(m^3\)/min
• C. 50 \(m^3\)/min
• D. 55 \(m^3\)/min

Hint:

Write filling and emptying times as 2400/x and 2400/(x+10), set their difference to 8 minutes, and solve the resulting quadratic.

Answer 1

The Correct Option is C

Step 1: Set up expressions for the filling and emptying times.
Let the filling capacity of the pump be \(x \, m^3\)/min. Then the emptying capacity is \((x + 10) \, m^3\)/min. Time to fill the 2400 \(m^3\) tank: \(\dfrac{2400}{x}\) minutes. Time to empty it: \(\dfrac{2400}{x + 10}\) minutes.

Step 2: Use the condition on the time difference.
The pump needs 8 minutes less to empty than to fill: \[ \frac{2400}{x} - \frac{2400}{x + 10} = 8 \] Dividing both sides by 8: \[ \frac{300}{x} - \frac{300}{x + 10} = 1 \]

Step 3: Clear denominators and solve the quadratic. \[ 300(x + 10) - 300x = x(x + 10) \] \[ 3000 = x^2 + 10x \] \[ x^2 + 10x - 3000 = 0 \] Factoring: \((x - 50)(x + 60) = 0\), so \(x = 50\) or \(x = -60\). Since capacity cannot be negative, \(x = 50\).

Final Answer:
The filling capacity of the pump is 50 \(m^3\)/min. \[ \boxed{x = 50 \, m^3/\text{min}} \]