Question

A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am, pump A is started. At what time will the swimming pool be filled if pump B is started at 10 am?

• A. 11:20 a.m.
• B. 11:05 a.m.
• C. 11:10 a.m.
• D. 10:50 a.m.

Hint:

Find how much of the pool pipe A fills alone in the first hour, then use the combined rate for the remaining part.

Answer 1

The Correct Option is A

This can also be solved using the LCM-of-capacity (unit work) method instead of fractions of the pool.
Let the pool hold 6 units of water (LCM of 3 and 6).
Pipe A's rate = $\frac{6}{3} = 2$ units per hour. Pipe B's rate = $\frac{6}{6} = 1$ unit per hour.
From 9 am to 10 am, only pipe A runs for 1 hour, filling $2 \times 1 = 2$ units.
Remaining water to be filled = $6 - 2 = 4$ units.
From 10 am, both pipes run together at a combined rate of $2 + 1 = 3$ units per hour.
Time to fill the remaining 4 units = $\frac{4}{3}$ hours = 1 hour 20 minutes.
Adding this to 10:00 am gives the pool full at 11:20 am.\[\boxed{11{:}20 \text{ a.m.}}\]