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29 Jan 2024, 01:02
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D
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95% (01:06) correct
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Ed's own pump, operating alone at its constant rate, can empty his swimming pool in 15 hours. The pump Ed rented, operating alone at its constant rate, can empty his swimming pool in 5 hours. How many hours will it take the two pumps, operating simultaneously and independently at their respective constant rates, to empty Ed's pool?
A. \(3\)
B. \(3 \frac{1}{4}\)
C. \(3 \frac{1}{2}\)
D. \(3 \frac{3}{4}\)
E. \(4\)
2024-01-29_11-58-03.png [ 51.9 KiB | Viewed 5459 times ]
A. \(3\)
B. \(3 \frac{1}{4}\)
C. \(3 \frac{1}{2}\)
D. \(3 \frac{3}{4}\)
E. \(4\)
Attachment:
2024-01-29_11-58-03.png [ 51.9 KiB | Viewed 5459 times ]
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29 Jan 2024, 01:10
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guddo
Assuming the rate of Ed's own pump is x pool/hour and the rate of the rented pump is y pool/hour, we'd have:
- x = 1/15
y = 1/5
Summing these up to get the combined rate of the two pumps gives:
- x + y = 1/15 + 1/5 = 4/15.
Since rate is the reciprocal of time, the two pumps will take \(\frac{15}{4 }= 3 \frac{3}{4}\) hours to empty the pool.
Answer: D.
Alternatively, notice that the rented pump, being thrice as fast as Ed's pump, does the job equal to three of Ed's pumps. Thus, when the two pumps are working together, it's like 1 + 3 = 4 of Ed's pumps working together. Consequently, 4 of Ed's pumps would need \(\frac{15}{4 }= 3 \frac{3}{4}\) hours to empty the pool.
Answer: D.
25 Oct 2024, 10:55
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Standard combined work problem:
