Question:medium

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.

Updated On: May 6, 2026
  • \(50\) m\(^3\)/min
  • \(56\) m\(^3\)/min
  • \(55\) m\(^3\)/min
  • \(52\) m\(^3\)/min
  • \(54\) m\(^3\)/min
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The problem requires us to determine the filling capacity of a pump in cubic meters per minute.
We know the total volume capacity of the tank is 2400 m$^3$.
The pump's emptying capacity is stated to be 10 m$^3$/min faster than its filling capacity.
Due to this difference in speeds, emptying the entire tank takes 8 minutes less than filling it completely.
Step 2: Key Formula or Approach:
The relationship between volume, rate, and time is given by $\text{Time} = \frac{\text{Volume}}{\text{Rate}}$.
Let the filling rate of the pump be $x$ m$^3$/min.
Consequently, the emptying rate will be $(x + 10)$ m$^3$/min.
We will set up an algebraic equation based on the difference in the time required for both processes.
Step 3: Detailed Explanation:

Let the filling capacity of the pump be represented by the variable $x$ m$^3$/min.

According to the problem, the emptying capacity is $x + 10$ m$^3$/min.

The time taken to fill the tank is determined by dividing the total volume by the filling rate, which is $\frac{2400}{x}$ minutes.

Similarly, the time taken to empty the tank is $\frac{2400}{x + 10}$ minutes.

We are given that the difference between the filling time and the emptying time is exactly 8 minutes.

Therefore, we can write the equation: \[ \frac{2400}{x} - \frac{2400}{x + 10} = 8 \]

To simplify the arithmetic, we divide the entire equation by 8.

\[ \frac{300}{x} - \frac{300}{x + 10} = 1 \]

Next, we factor out 300 to solve the rational equation more easily.

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

We find a common denominator for the terms inside the parentheses.

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

This simplifies down to: \[ 300 \left( \frac{10}{x^2 + 10x} \right) = 1 \]

By cross-multiplying, we obtain a standard quadratic equation.

\[ 3000 = x^2 + 10x \]

Rearranging the terms into the standard $ax^2 + bx + c = 0$ format gives:

\[ x^2 + 10x - 3000 = 0 \]

We now factor the quadratic equation by finding two numbers that multiply to -3000 and add to 10.

These two numbers are 60 and -50.

\[ (x + 60)(x - 50) = 0 \]

This yields two possible values for $x$: $x = -60$ or $x = 50$.

Since a physical pumping capacity cannot be a negative value, we discard -60.

Therefore, the filling capacity $x$ is 50 m$^3$/min.

Step 4: Final Answer:
The filling capacity of the pump is 50 m$^3$/min.
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