AAPL wrote:GMAT Prep
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?
A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3
OA E
We are given that the rate of 1 pump is 1.5 times faster than the rate of the other pump. Since 1 pool is being filled and rate = work/time, the rate of the faster pump is 1/x, for which x = the time it takes for the faster pump to fill the pool, and the rate of the slower pump = 1/(1.5x) = 1/(3x/2) = 2/(3x).
Since, when the pumps work together, they take 4 hours to fill 1 pool, we can create the following equation:
work of faster pump + work of slower pump = 1
(1/x)(4) + (2/(3x))(4) = 1
4/x + 8/(3x) = 1
Multiplying the entire equation by 3x, we have:
12 + 8 = 3x
20 = 3x
20/3 = x
Alternate Solution:
If the rate of the faster pump is 1.5 times that of the slower pump, then it will take the slower pump 1.5 times the amount of time to fill the entire pool compared to the faster pump. Let t denote the time (in hours) it takes for the faster pump to fill the pool alone. Then, it will take 1.5t hours for the slower pump to fill the pool alone.
Notice that in one hour, 1/t of the pool is filled by the faster pump and 1/(1.5t) of the pool is filled by the slower pump. We are also given that together they fill the entire pool in 4 hours; therefore, 1/4 of the pool is filled in one hour when both pumps are running. We can create the following equation:
1/t + 1/(1.5t) = 1/4
Let's multiply each side of this equation by 12t:
12 + 8 = 3t
20 = 3t
t = 20/3
Answer: E
