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16 Sep 2014, 00:36
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Two pumps, A and B, are used to fill a swimming pool. Pump A, when working alone, can fill the pool in 4 hours, whereas Pump B, when working alone, can fill it in 6 hours. If Pump B starts filling the pool by itself for 1 hour before Pump A joins in to work together with Pump B, in total, how long will it take to fill the pool completely?
A. 2 hours
B. 2.4 hours
C. 2.5 hours
D. 3 hours
E. 3.5 hours
A. 2 hours
B. 2.4 hours
C. 2.5 hours
D. 3 hours
E. 3.5 hours
16 Sep 2014, 00:37
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Official Solution:
Two pumps, A and B, are used to fill a swimming pool. Pump A, when working alone, can fill the pool in 4 hours, whereas Pump B, when working alone, can fill it in 6 hours. If Pump B starts filling the pool by itself for 1 hour before Pump A joins in to work together with Pump B, in total, how long will it take to fill the pool completely?
A. 2 hours
B. 2.4 hours
C. 2.5 hours
D. 3 hours
E. 3.5 hours
In the first hour that Pump B works alone, it can fill 1/6 of the pool, leaving 5/6 of the pool to be filled by Pump A and Pump B working together. The combined rate of the two pumps is the sum of their individual rates, which is \(\frac{1}{4} + \frac{1}{6} = \frac{5}{12}\) pool per hour. Therefore, the remaining 5/6 of the pool will be filled in: \(time = \frac{job}{rate} = \frac{\frac{5}{6}}{\frac{5}{12}} = \frac{5}{6} * \frac{12}{5}\) = 2 hours.
So, both pumps work together for 2 hours to complete the remaining work. Since Pump B started an hour before Pump A, the total time taken to fill the pool is 2 hours (both working together) + 1 hour (Pump B working alone) = 3 hours.
Answer: D
Two pumps, A and B, are used to fill a swimming pool. Pump A, when working alone, can fill the pool in 4 hours, whereas Pump B, when working alone, can fill it in 6 hours. If Pump B starts filling the pool by itself for 1 hour before Pump A joins in to work together with Pump B, in total, how long will it take to fill the pool completely?
A. 2 hours
B. 2.4 hours
C. 2.5 hours
D. 3 hours
E. 3.5 hours
In the first hour that Pump B works alone, it can fill 1/6 of the pool, leaving 5/6 of the pool to be filled by Pump A and Pump B working together. The combined rate of the two pumps is the sum of their individual rates, which is \(\frac{1}{4} + \frac{1}{6} = \frac{5}{12}\) pool per hour. Therefore, the remaining 5/6 of the pool will be filled in: \(time = \frac{job}{rate} = \frac{\frac{5}{6}}{\frac{5}{12}} = \frac{5}{6} * \frac{12}{5}\) = 2 hours.
So, both pumps work together for 2 hours to complete the remaining work. Since Pump B started an hour before Pump A, the total time taken to fill the pool is 2 hours (both working together) + 1 hour (Pump B working alone) = 3 hours.
Answer: D
General Discussion
23 Apr 2023, 02:39
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
