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Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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eGMAT Question of the Week #7Three 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. \(9:30 PM\) To access all the questions: Question of the Week: Consolidated List
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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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13 Jul 2018, 09:09
EgmatQuantExpert wrote: eGMAT Question of the Week #7Three 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. \(9:30 PM\) To access all the questions: Question of the Week: Consolidated List 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
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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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Updated on: 18 Jul 2018, 13:09
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 60liter tank, P's rate \(= \frac{60}{3} = 20\) liters per hour. Since Q takes 4 hours to fill the 60liter tank, Q's rate \(= \frac{60}{4} = 15\) liters per hour. Since R takes 5 hours to empty the 60liter 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 5hour period  P works for 4 hours and takes 4 quarterhour breaks, with Q working for \(\frac{1}{4}\) hour during each break. Thus, the input rate for each 5hour 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 60liter tank \(= \frac{60}{7} ≈ 8.5\) hours. 10am + about 8.5 hours ≈ 6:30pm.
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Originally posted by GMATGuruNY on 18 Jul 2018, 04:43.
Last edited by GMATGuruNY on 18 Jul 2018, 13:09, edited 1 time in total.




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Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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Hey everyone, We will post the solution very soon. Till then, try it one more time and post your analysis.
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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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18 Jul 2018, 04:01
Let total capacity of tank be = 60 litres. P's one hour work = 60/3 = +20 litres/hourQ's one hour work = 60/4 = +15 litres/hourR's one hour work = 60/5 = 12 litres/hourWe 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 PMAnswer = option C
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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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18 Jul 2018, 13:05
GMATGuruNY wrote: Let the tank = 60 liters. Since P takes 3 hours to fill the 60liter tank, P's rate \(= \frac{60}{3} = 20\) liters per hour. Since Q takes 4 hours to fill the 60liter tank, Q's rate \(= \frac{60}{4} = 15\) liters per hour. Since R takes 5 hours to empty the 60liter 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 5hour period  P works for 4 hours and takes 4 quarterhour breaks, with Q working for \(\frac{1}{4}\) hour during each break. Thus, the rate for each 5hour 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 per hour. 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 60liter tank \(= \frac{60}{7} ≈ 8.5\) hours. 10am + about 8.5 hours ≈ 6:30pm. Dear GMATGuruNYI believe the highlighted should be modified as 35 liter per 5 hrs. Correct?



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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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18 Jul 2018, 13:17
Mo2men wrote: I believe the highlighted should be modified as 35 liter per 5 hrs. Correct? Good catch. I've corrected the typo.
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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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18 Jul 2018, 23:58
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.



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Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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Updated on: 13 Aug 2018, 00:37
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= 2012= 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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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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25 Jul 2018, 06:36
EgmatQuantExpert wrote: 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= 2012= 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: CWe 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



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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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22 Mar 2019, 09:32
EgmatQuantExpert wrote: eGMAT Question of the Week #7Three 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. \(9:30 PM\) 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
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Re: Question of the Week 7 ( Three pipes P, Q, and R are attached to a..)
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