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Working at constant rate, pump X pumped out . . . . .

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Working at constant rate, pump X pumped out . . . . .

by VJesus12 » Sat Jan 06, 2018 4:08 am
Working at a constant rate, pump X pumped out half of the water in a flooded basement in 4 hours. The pump Y was started and the two pumps, working independently at their respective constant rates, pumped out rest of the water in 3 hours. How many hours would it have taken pump Y, operating alone at its own constant rate, to pump out all of the water that was pumped out of the basement?

a. 10
b. 12
c. 14
d. 18
e. 24

The OA is the option E.

I am really confused here. Can any expert clarify this for me? Please.
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by elias.latour.apex » Sat Jan 06, 2018 10:14 am
The easiest solution is to assume that at the time both pumps started working on the basement, there were 12 units of water in the basement. This is the least common multiple of 4 and 3. Accordingly, the basement had 24 units of water in it when it was full.

Since the pump X evacuated 12 units of water in 4 hours, it must be pumping 3 units of water per hour. Together the pumps finished the rest of the job in 3 hours. That means that they must be pumping at 4 units per hour -- 3 units per hour from pump X and 1 unit per hour from pump Y.

Therefore, it would have taken the pump Y 24 hours to pump out the 24 units of water. This is answer choice E.
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