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Two pipes can separately fill a tank in 20 hours and 30 hours
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23 Mar 2019, 21:48
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Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank? A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours
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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 04:22
DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours Let the tank = 180 gallons. Since the first pipe takes 20 hours to fill the 180gallon tank, the rate for the first pipe\(= \frac{work}{time}= \frac{180}{20} = 9\) gallons per hour. Since the second pipe takes 30 hours to fill the 180gallon tank, the rate for the second pipe \(= \frac{work}{time} = \frac{180}{30} = 6\) gallons per hour. Combined rate for the two pipes = 9+6 = 15 gallons per hour. \(\frac{1}{3}\) of the 180gallon tank \(= \frac{1}{3}*180 = 60\) gallons. Since the combined rate for the two pipes = 15 gallons per hour, the time for the two pipes to pump in 60 gallons \(= \frac{work}{rate} = \frac{60}{15} = 4\) hours. Remaining volume = 18060 = 120 gallons. Since the leak reduces the rate by 1/3, the resulting rate \(= \frac{2}{3}*15 = 10\) gallons per hour. Since the new rate = 10 gallons per hour, the time for the remaining 120 gallons \(= \frac{work}{rate} = \frac{120}{10} = 12\) hours. Total time to fill the tank = (4 hours for the first 1/3 of the tank) + (12 hours for the remaining volume) = 4+12 = 16 hours.
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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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23 Mar 2019, 22:35
DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours Let's assume the size of the tank to be LCM(20,30) = 60 units. The individual rates of the two pipes filling are 3 units & 2 units respectively. Together the tanks fill 5 units in an hour. When the tank is \(\frac{1}{3}\)rd full(20 units have been filled). The leak develops causing the tank to fill at \(\frac{2}{3}\)rds of its usual rate. The time taken to fill the first 20 units is \(\frac{20}{5} = 4\) hours. For the remaining 40 units, the tanks will fill \(\frac{10}{3}(5*\frac{2}{3})\) units in an hour. Because of the leak. the pipe takes \(\frac{40}{\frac{10}{3}} = \frac{120}{10} = 12\) hours to fill the remaining tank. Therefore, the total time taken for the two pipes to fill the tank is 4 + 12 = 16 hours(Option C)
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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 00:00
Pipe A  can fill the tank in 20 hrs. Hence, in 1 hr it can fill 1/20th of the tank. Pipe B  can fill the tank in 30 hrs. Hence, in 1 hr it can fill 1/30th of the tank.
Therefore, when both pipes are opened, (1/20 + 1/30 =) 1/12th of the tank is filled in an hr. It'll take both the pipes 12 hrs to fill the tank completely.
When 1/3rd of the tank is filled, both the pipes would have been opened for (1/3 = 4/12 = 4*(1/12)) 4 hrs. Now, 2/3rd (= 8/12) of the tank remains to be filled.
However, at this stage due to the leak in the tank, 1/3rd of the water supplied by the pipes leaks. So, now only (2/3 * 1/12 = ) 1/18th of the tank is filled in an hour instead of 1/12th earlier. At this rate it would have taken 18 hrs to fill the tank if it were completely empty. However, since only 8/12th of the tank needs to be filled, it'd take another (8/12 * 18 =) 12 hrs to completely fill the tank.
Therefore, the tank would be completely filled in (4 + 12 =) 16 hrs.



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 05:53
GMATGuruNY wrote: DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours Let the tank = 180 gallons. Since the first pipe takes 20 hours to fill the 180gallon tank, the rate for the first pipe\(= \frac{work}{time}= \frac{180}{20} = 9\) gallons per hour. Since the second pipe takes 30 hours to fill the 180gallon tank, the rate for the second pipe \(= \frac{work}{time} = \frac{180}{30} = 6\) gallons per hour. Combined rate for the two pipes = 9+6 = 15 gallons per hour. \(\frac{1}{3}\) of the 180gallon tank \(= \frac{1}{3}*180 = 60\) gallons. Since the combined rate for the two pipes = 15 gallons per hour, the time for the two pipes to pump in 60 gallons \(= \frac{work}{rate} = \frac{60}{15} = 4\) hours. Remaining volume = 18060 = 120 gallons. Since the leak reduces the rate by 1/3, the resulting rate \(= \frac{2}{3}*15 = 10\) gallons per hour. Since the new rate = 10 gallons per hour, the time for the remaining 120 gallons \(= \frac{work}{rate} = \frac{120}{10} = 12\) hours. Total time to fill the tank = (4 hours for the first 1/3 of the tank) + (12 hours for the remaining volume) = 4+12 = 16 hours. Dear GMATGuruNYI do not understand the reasoning behind the leak in the tank and the be 2/3 of the combines rate of both pumps? does not the leak produce 1/3 of the water supplied to lost, lowering the water to 40 (1/3 * 60 =20 so remaining water to be 40)? Can you please elaborate more? I'm confused. Thanks



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 08:09
GMATGuruNY wrote: DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours Let the tank = 180 gallons. Since the first pipe takes 20 hours to fill the 180gallon tank, the rate for the first pipe\(= \frac{work}{time}= \frac{180}{20} = 9\) gallons per hour. Since the second pipe takes 30 hours to fill the 180gallon tank, the rate for the second pipe \(= \frac{work}{time} = \frac{180}{30} = 6\) gallons per hour. Combined rate for the two pipes = 9+6 = 15 gallons per hour. \(\frac{1}{3}\) of the 180gallon tank \(= \frac{1}{3}*180 = 60\) gallons. Since the combined rate for the two pipes = 15 gallons per hour, the time for the two pipes to pump in 60 gallons \(= \frac{work}{rate} = \frac{60}{15} = 4\) hours. Remaining volume = 18060 = 120 gallons. Since the leak reduces the rate by 1/3, the resulting rate \(= \frac{2}{3}*15 = 10\) gallons per hour. Since the new rate = 10 gallons per hour, the time for the remaining 120 gallons \(= \frac{work}{rate} = \frac{120}{10} = 12\) hours. Total time to fill the tank = (4 hours for the first 1/3 of the tank) + (12 hours for the remaining volume) = 4+12 = 16 hours. Is there any specic reason to take tak capacity as 180 gallons?(LCM formula?)



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 08:47
Mo2men wrote: Dear GMATGuruNYI do not understand the reasoning behind the leak in the tank and the be 2/3 of the combines rate of both pumps? does not the leak produce 1/3 of the water supplied to lost, lowering the water to 40 (1/3 * 60 =20 so remaining water to be 40)? Can you please elaborate more? I'm confused. Thanks Prompt: When the tank is 1/3 full, a leak develops through which 1/3 of the water supplied by both the pipes leaks out.This wording is intended to convey the following: Once the tank is 1/3 full, a leak develops. This leak decreases the input rate, as follows: For every 3 gallons supplied by the two pipes, the leak removes 1/3 of the 3 gallons. In other words, the leak removes 1 gallon for every 3 gallons supplied by the two pipes, reducing the input rate by 1/3. The resulting input rate is thus 2/3 of the original input rate.
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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 09:00
mangamma wrote: Is there any specic reason to take tak capacity as 180 gallons?(LCM formula?)
The capacity must be divided by the two given times (20 hours and 30 hours) and then by 3 (since the rate is reduced by 1/3). To get a good value for the capacity, I multiplied the LCM of 20 and 30 by 3: LCM (20,30) * 3 = 60*3 = 180
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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 09:17
mangamma wrote: GMATGuruNY wrote: DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours Let the tank = 180 gallons. Since the first pipe takes 20 hours to fill the 180gallon tank, the rate for the first pipe\(= \frac{work}{time}= \frac{180}{20} = 9\) gallons per hour. Since the second pipe takes 30 hours to fill the 180gallon tank, the rate for the second pipe \(= \frac{work}{time} = \frac{180}{30} = 6\) gallons per hour. Combined rate for the two pipes = 9+6 = 15 gallons per hour. \(\frac{1}{3}\) of the 180gallon tank \(= \frac{1}{3}*180 = 60\) gallons. Since the combined rate for the two pipes = 15 gallons per hour, the time for the two pipes to pump in 60 gallons \(= \frac{work}{rate} = \frac{60}{15} = 4\) hours. Remaining volume = 18060 = 120 gallons. Since the leak reduces the rate by 1/3, the resulting rate \(= \frac{2}{3}*15 = 10\) gallons per hour. Since the new rate = 10 gallons per hour, the time for the remaining 120 gallons \(= \frac{work}{rate} = \frac{120}{10} = 12\) hours. Total time to fill the tank = (4 hours for the first 1/3 of the tank) + (12 hours for the remaining volume) = 4+12 = 16 hours. Is there any specic reason to take tak capacity as 180 gallons?(LCM formula?) Yes, this is the only reason this value has been chosen. You could also take 60 (LCM of 20 & 30), but that would become a fraction after initial division. So, choose LCM of 20 and 30 and multiply it by 3. Means 60 * 3 = 180 This way you can get rid of fractions and can increase you calculation speed.
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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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24 Mar 2019, 20:29
Work rate formula: AB/(A+B)=T(for both to work together) 20x30/(20+30)=12 While working simultaneously, it takes the two pipes 12 hours to fill the tank. Given 1/3 of the water filled by the pipes are lost, meaning 12x(1/3) of time is wasted, =4 hours. So, it takes an additional 4 hours to fill the tank: 12+4=16hours.



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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25 Mar 2019, 06:11
DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours Hi DisciplinedPrep I hope you are doing well in GMAT prep as your nickname says. What is the source of the question? it is really good.



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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30 Mar 2019, 10:06
DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours let total capacity of tank = 60 ltrs so rate of A = 60/20 ; 3 and rate of B = 60/3 ; 2 together ; 5 given 1/3 of tank is filled ; i.e 20 ltrs ; so time taken would have been 20/5 ; 4 hrs and later leak happens and the rate decreases by 1/3 or original value or say the rate becomes 2/3 so new combined flow rate 5*2/3 ; 10/3 so for 40 ltrs balance vol the time it will take 40/10 /3 ; 12 hrs total time taken 12 + 4 ; 16 hrs IMO C



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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18 Apr 2019, 05:20
First post gmatclub. I guess this is a positive sign. ^^ Well, I did it in this way: Rate of pipe A: 1/20 Rate of pipe B: 1/30 Both pipes are opened to fill the tank; so combined rate is 5/60 or 1/12 Combined rate is 1/12 so tank will be full after 12h. When the tank is 1/3 full a leak develops in the tank through which 1/3 of the water leak out Translated: When tank is at 4/12 all the water leaks out. (This is after 4 Hours) So there are 4 extra hours to add because of the leak. 12h+4h = 16h Please correct me if I made a mistake.



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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04 Jun 2020, 22:41
DM Me to know the simple method. Let us assume capacity of the tank is 600 Liters. Efficiency of pipe 1  30 L/H Efficiency of Pipe 2  20 L/H Pipe 1 + Pipe 2 = 50 L/H which means Tank will take 12 H to fill it if there is no leak. now Opening C will leak 200 Liters from tank which means another 200/50 i.e 4 Hrs to fill the void hence total 12 + 4 = 16 HRS. Hope it helps!! Vikas Singla
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Two pipes can separately fill a tank in 20 hours and 30 hours
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03 Jul 2020, 05:25
Puyol wrote: First post gmatclub. I guess this is a positive sign. ^^ Well, I did it in this way: Rate of pipe A: 1/20 Rate of pipe B: 1/30 Both pipes are opened to fill the tank; so combined rate is 5/60 or 1/12 Combined rate is 1/12 so tank will be full after 12h. When the tank is 1/3 full a leak develops in the tank through which 1/3 of the water leak out Translated: When tank is at 4/12 all the water leaks out. (This is after 4 Hours) So there are 4 extra hours to add because of the leak. 12h+4h = 16h Please correct me if I made a mistake. I have also done this in the same way as you did but don't know everyone started calculating the rate of water loss due to leak in the tank. I assumed the leak got sealed after the tank empty due to leak.



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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03 Jul 2020, 07:22
DisciplinedPrep wrote: Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes leak out. What is the total time taken to fill the tank?
A. 15 hours B. 25 hours C. 16 hours D. 12 hours E. 10 hours combined rate=1/20+1/30=1/12 12/3=4 hours to fill 1/3 of tank before leak let t=time to fill tank after leak t*(1/12)(2/3)=2/3→ t=12 hours+4 original hours=16 total hours to fill tank C



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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13 Jul 2020, 00:56
A: 1/20. B: 1/30. A+B: 1/12 (they are filling the tank in 12 hours).
Suppose the tank is 180 liters, then they fill 15 liters each hour.
Now, they worked (12*1/3) 4 hours and filled (180*1/3) 60 liters  we are left with (18060) 120 liters to be filled.
From now on, their rate changes to (10*2/3) 10 liters per hour to fill 120 liters. It will take A&B (120/10) 12 hours to fill the rest of the tank.
C: 12+4 = 16.



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Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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13 Jul 2020, 15:14
Break into into parts A and B work to fill the tank 1/3 A and B work together against the leak to fill 2/3
A and B together 1/3 = (1/20+1/30)t t=4
AB against leak 2/3 = (1/20+1/30)t  1/3(1/20+1/30)t t=12
12+4=16




Re: Two pipes can separately fill a tank in 20 hours and 30 hours
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