Question

Pipes P and Q can fill a storage tank in full with water in 10 and 6 minutes, respectively. Pipe R draws the water out from the storage tank at a rate of 34 litres per minute. P, Q and R operate at a constant rate.
If it takes one hour to completely empty a full storage tank with all the pipes operating simultaneously, what is the capacity of the storage tank (in litres)?

• A. 26.8
• B. 60.0
• C. 120.0
• D. 127.5

Hint:

When dealing with problems involving multiple rates of change (such as filling and emptying), always express the net rate of change and use the total time to set up an equation to solve for the unknown quantity.

Answer 1

The Correct Option is C

Let the capacity of the storage tank be \( x \) litres.
- Pipe P fills the tank in 10 minutes, so it fills \( \frac{x}{10} \) litres per minute.
- Pipe Q fills the tank in 6 minutes, so it fills \( \frac{x}{6} \) litres per minute.
- Pipe R draws out water at a rate of 34 litres per minute. When all pipes are operating simultaneously, the net rate of change in the tank's water level is: \[ \text{Net rate} = \left( \frac{x}{10} + \frac{x}{6} - 34 \right) \text{ litres per minute}. \] We are told that it takes 1 hour (or 60 minutes) to empty the tank. Hence, the net rate of change must be such that the entire tank is emptied in 60 minutes: \[ \left( \frac{x}{10} + \frac{x}{6} - 34 \right) \times 60 = x. \]
Step 1: Solve for \( x \).
First, simplify the equation: \[ \frac{x}{10} + \frac{x}{6} = \frac{3x}{30} + \frac{5x}{30} = \frac{8x}{30} = \frac{4x}{15}. \] Thus, the equation becomes: \[ \left( \frac{4x}{15} - 34 \right) \times 60 = x. \] Distribute the 60: \[ \frac{240x}{15} - 2040 = x. \] Simplify the first term: \[ 16x - 2040 = x. \] Move all terms involving \( x \) to one side: \[ 16x - x = 2040, \] \[ 15x = 2040. \] Now, solve for \( x \): \[ x = \frac{2040}{15} = 120. \] Therefore, the capacity of the tank is \( \boxed{120} \) litres.