Question

A pump can be operated both for filling a tank and for emptying it. The capacity of tank is 2400 m³ .The emptying capacity of the pump is 10 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.

• A. 45 m³ / min
• B. 30 m³ / min
• C. 50 m³ / min
• D. 55 m³ / min

Answer 1

The Correct Option is C

Let's denote the filling capacity of the pump as \( x \) m³ per minute. Therefore, the emptying capacity of the pump will be \( (x + 10) \) m³ per minute. The tank has a capacity of 2400 m³.

If it takes \( t \) minutes to fill the tank, then:

\[ x \times t = 2400 \]

\[ t = \frac{2400}{x} \]

For emptying the tank, it takes \( t - 8 \) minutes, so:

\[ (x + 10) \times (t - 8) = 2400 \]

Substitute \( t = \frac{2400}{x} \) into the emptying equation:

\[ (x + 10) \left(\frac{2400}{x} - 8\right) = 2400 \]

Expand and simplify:

\[ (x + 10) \left(\frac{2400 - 8x}{x}\right) = 2400 \]

\[ 2400(x + 10) - 8x(x + 10) = 2400x \]

Simplify further:

\[ 2400x + 24000 - 8x² - 80x = 2400x \]

Cancel out \( 2400x \) from both sides:

\[ 24000 - 80x - 8x² = 0 \]

Rearrange terms:

\[ 8x² + 80x - 24000 = 0 \]

Divide the entire equation by 8 for simplicity:

\[ x² + 10x - 3000 = 0 \]

We solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 10, c = -3000 \):

\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times (-3000)}}{2 \times 1} \]

\[ x = \frac{-10 \pm \sqrt{100 + 12000}}{2} \]

\[ x = \frac{-10 \pm \sqrt{12100}}{2} \]

\[ x = \frac{-10 \pm 110}{2} \]

Solving the quadratic formula gives us two potential solutions:

• \( x = \frac{100}{2} = 50 \)
• \( x = \frac{-120}{2} = -60 \)

Since a negative capacity doesn't make sense in this context, the filling capacity must be:

50 m³ / min