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

Question

e-GMAT          Question of the Week #7

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

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chetan2u · Accepted Answer

To get the minimum time, the faster that is P should function the most  ......
so the functioning should be P for 1 hr and then Q for 15 min and then P for 1 hour and so on

two cases:-
1) when P and R function for 1 hour.
 \frac{1}{3}-\frac{1}{5}=\frac{2}{15} ...
2) when Q and R function for 15 min or 1/4 hr
 \frac{1}{4}-\frac{1}{5}=\frac{1}{20}  in one hour, so  \frac{1}{20}*\frac{1}{4} = \frac{1}{80}  in 15 min

so 1 hr 15 min or 5/4 hr work =  \frac{2}{15}+\frac{1}{80}=\frac{35}{240}  and thus 1 hr work =  \frac{35}{240}*\frac{4}{5}=\frac{7}{60} 
and the work will get completed in  \frac{60}{7}  = 8 hr 35 min approx

time wise it will finish at 10:00 + 8 hr 35 min = 18:35 or 06:35 pm ~ 06:30 pm

C

GMATGuruNY · Answer

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.
 C

EgmatQuantExpert · Answer

Hey everyone,

We will post the solution very soon. Till then, try it one more time and post your analysis.

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sidsst · Answer

Let total capacity of tank be =  60 litres .

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 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

Answer = option C

Mo2men · Answer

Dear  GMATGuruNY

I believe the highlighted should be modified as 35 liter per 5 hrs. Correct?

GMATGuruNY · Answer

Good catch.
I've corrected the typo.

slayer1983 · Answer

It is understood that for least time to fill up the tank, P should be opened for the maximum time.
=> P should be opened for 1 hour after 10 am
R is open at all times
In this one hour, the proportion of tank which is has been filled is (1/3 - 1/5) = 2/15
After the 1 hour, Q is opened for a mandatory 15 mins while P is closed
In these 15 mins, the proportion of tank which has been filled up is (1/16 - 1/20) = 1/80
So, for a cycle of 1 hours 15 mins, tank filled is (2/15) + (1/80) = 35/240
The above cycle has to be repeated till the tank is filled.
It can be seen that this will run for 6 complete cycles, and the tank will be filled in the 7th cycle when P is opened with R (Since 35/240*6 is 210/40, and 35*7/240>1)
6 cycles of 1 hours 15 mins yields 7 hours 30 mins. => Time after 6 cycles is 17:30 hours

Tank remaining empty is 1 - 35*6/240 = 1/8

1/8 tank will be filled by P and R open
(1/8)/(2/15) = approximately 1 hour.
Time at the time of finishing 18:30.

I think assuming hourly rates of a cycle works only when rates of P and Q are comparable and the size of the cycle is not too big. Otherwise, the approximation may result into wrong answer.

EgmatQuantExpert · Answer

Solution

Given:

• Three pipes P, Q, and R are attached to a tank
 • P and Q can fill the tank in 3 hours and 4 hours respectively. 
 • R can empty the tank in 5 hours.
 • P is opened at 10 am and Q is opened at 11 am, while R is open all the time.
 • 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.

To find:

• Time when the tank will be full.

Approach and Working:

Let us assume that the capacity of tank is 60 Litres. Hence:
  • In 1 hour, P can fill= 20 L
 • In 1 hour, Q can fill= 15 L
 • In 1 hour, R can empty= 12 L

Since, we are asked the earliest time to fill the tank, P should be opened for as much time as it can.

Now, from 10 AM to 11 AM:
  • P and Rare open.
 • Hence, amount of water in the tank= 20-12= 8 L

From 11 AM to 11:15AM:
  • Q and R are open.
 • Hence, amount of water in the tank now= 8+(\frac{15}{4}-\frac{12}{4})=8+\frac{3}{4}

From 11:15 AM to 12:15 PM:
  • P and R are open.
 • Hence, amount of water in the tank=  (8+\frac{3}{4})+8

From 12:15 PM to 12:30 PM:
  • Q and R are open.
 • Hence, amount of water in the tank now=  (8+\frac{3}{4})+8+\frac{3}{4}= (8+\frac{3}{4})+(8+\frac{3}{4})= 2(8+\frac{3}{4})

From 12:30 PM to 1:30 PM
  • The amount of water in the tank=  2(8+\frac{3}{4}) +8

From 1:30 PM to 1:45 PM
  • The amount of water in the tank= 3(8+\frac{3}{4})

So, if you can understand then:

• In every 1 hour 15 min or  \frac{5}{4}  hour,  8+\frac{3}{4} or \frac{35}{4} l  water is getting filled.
• Hence, in 1 hour=  \frac{35}{4}* \frac{4}{5}= 7  litres.

• Thus, time taken to fill 60 litres=  \frac{60}{7}≈ 8.5  hours
• Hence, Time= 10AM+8h 30 min= 6:30 PM

Hence, the correct answer is option C.

Answer: C

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durgamadhab · Answer

We have to decide ...which combination will us maximum rate of filling.
combinations are 
1. P alone with 15 minutes breaks
2. P, then 15 mins break, then again P....continue
3. P for 1 hour then 15 mins break, during 15 mins break run Q.....stop Q in 15 mins and then again run P.

Combination three will give maximum filling rate. solution is as given by Chetan sir

ScottTargetTestPrep · Answer

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

serenayong · Answer

hi is this way ok?
10-11am: work done = P-R = 1/3-1/5 = 2/15
11-12pm: work done = 3/4 x 2/15 (P-R part) + 1/4 x (1/4-1/5) (Q-R part) = 1/10+1/80 = 9/80
12-1pm: same as above.... 
total work = 1 = 2/15 + 9/80 (how many rounds of 1 hour)
rounds of 1 hours = 13/15 X 80/9 = 7.7hours
total time = 7.7 hours + 1 hour = 8.7 hours -> which is close to 8.5 hours given in answer C. 
I cant figure out why there's about 0.2 hours of gap. Can someone please help..? :-/ 
thanksss

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

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