EgmatQuantExpert wrote:
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. \(9:30 PM\)
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate \(= \frac{60}{3} = 20\) liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate \(= \frac{60}{4} = 15\) liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate \(= \frac{60}{-5} = -12\) liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
To MINIMIZE the time required to fill the tank, we must MAXIMIZE the time for P, since P's input rate is greater than Q's input rate.
Since P must close for \(\frac{1}{4}\) hour after each hour of work, P's time will be maximized if -- over a 5-hour period -- P works for 4 hours and takes 4 quarter-hour breaks, with Q working for \(\frac{1}{4}\) hour during each break.
Thus, the input rate for each 5-hour period = (4 hours of work for P) + (1 hour of work for Q) - (5 hours of work for R) \(= (4*20) + (1*15) - (5*12) = 35\) liters for every 5 hours of work.
Since the volume increases by 35 liters every 5 hours, the hourly rate \(≈ \frac{35}{5} ≈ 7\) liters per hour.
(The hourly rate is an approximation because it increases when P works but decreases when Q works.)
Since the hourly rate is about 7 liters per hour, the approximate time to fill the 60-liter tank \(= \frac{60}{7} ≈ 8.5\) hours.
10am + about 8.5 hours ≈ 6:30pm.
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