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Three pipes P, Q, and R are attached to a tank. P and Q indi

Expert replies
by BTGmoderatorDC » Fri Feb 01, 2019 5:44 am

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

OA C

Source: e-GMAT
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Source: — Problem Solving |

by GMATGuruNY » Fri Feb 01, 2019 5:57 am
BTGmoderatorDC 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. 4:30pm
B. 6:00pm
C. 6:30pm
D. 8:30pm
E. 9:30pm
Let the tank = 60 liters.

Since P takes 3 hours to fill the 60-liter tank, P's rate = 60/3 = 20 liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate = 60/4 = 15 liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate = -(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 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 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 ≈ 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 ≈ 60/7 ≈ 8.5 hours.
10am + about 8.5 hours ≈ 6:30pm.

The correct answer is C.
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by swerve » Fri Feb 01, 2019 7:54 am
Let total capacity of the tank be = 60 liters.

P's one hour work = 60/3 = +20 litres/hour
Q's one hour work = 60/4 = +15 litres/hour
R's one hour work = 60/5 = -12 litres/hour

We would require P to work for a maximum time as P+R work = 8 l/hour which is higher than Q+R = 3 l/hour.
P would work for 1 hour remain closed for 15 minutes when Q will work and the P will resume. Q+R 15 minute work = 3/4 =0.75 litres/hour

tank filled in 1 hour 15 minutes (75 minutes) = 8.75 litres.
Number of such cycles required = 60/8.75 = 6000/875
Minutes required = 6000*75/875
Hours required = 6000*75/(875*60) = 8.5 hours approx

Thus 10:00 AM + 8.5 hours = 6:30 PM
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