AAPL wrote:e-GMAT

Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?

A. 4:30 PM
B. 6:00 PM
C. 6:30 PM
D. 8:30 PM
E. 9:30 PM

OA C

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Since pipe P is faster than pipe Q, we will have pipe P working (i.e., kept open) as much as possible. That is, we will use pipe Q only when pipe P needs to take a break (i.e., is kept closed). Using this strategy, let's keep track of the times and the amount of the pool that is filled at those times.

At 11 AM, the pool has been filled by pipe P for one hour and also drained by pipe R for one hour, so the portion of the pool that is filled is 1/3 - 1/5 = 2/15.

At 11:15 AM, the pool has been filled by pipe Q for one quarter hour and also drained by pipe R for one quarter hour, so the cumulative total portion of the pool that is filled is 2/15 + (1/4 - 1/5) x 1/4 = 2/15 + 1/80 = 32/240 + 3/240 = 35/240 = 7/48.

At this point, we can see that for every 1 hour and 15 minutes, or 5/4 hours, 7/48 of the pool is filled. If we repeat this process, we see that 6 x 7/48 = 7/8 of the pool will be filled in 6 x 5/4 = 15/2 or 7½ hours. Therefore, for the remaining 1/8 of the pool, we fill it using pipe P, which will take approximately (but no longer than) 1 hour to fill since 2/15 is slightly greater than 1/8 or 2/16.

To summarize, it takes approximately 7½ + 1 = 8½ hours to fill the pool. Since we begin at 10 AM, then at about 10 AM + 8½ hours = 6:30 PM, the entire pool will be filled.

Answer: C