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Two water pumps, working simultaneously at their respective constant
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Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate? A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3 Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!
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Originally posted by kmasonbx on 12 Jul 2013, 07:26.
Last edited by Bunuel on 23 Sep 2019, 04:09, edited 3 times in total.
Renamed the topic and edited the question.




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Re: Two water pumps, working simultaneously at their respective constant
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12 Jul 2013, 08:25
kmasonbx wrote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3
Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks! Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour. Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 > x=3/20 pool hour. The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone. Answer: E.
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Re: Two water pumps, working simultaneously at their respective constant
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15 Sep 2014, 06:40
AMITAGARWAL2 wrote: yes it does. Thanks... Let me elaborate the math so that it's absolutely clear: Let's calculate the combined rate first: Rate x Time = Work Rate x 4 = 1 [It takes 4 hours for both the pumps to fill up the pool] Rate = 1/4 [So, 1/4 is the rate for the pumps working together] Now, the let's assume the rate for the slower pump is x ; so the rate for the faster pump will be 1.5x According to our previous calculations, Slower pump + faster pump = 1/4 x + 1.5x = 1/4 2.5x = 1/4 x = 1/10 [slower pump's rate] so, the faster pump's rate is 1/10 x 1.5 = 3/20 Now let's calculate the time it will take for the faster pump Rate x Time = Work 3/20 x Time = 1 Time = 1 x 20/3 = 20/3 the answer




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Re: Two water pumps, working simultaneously at their respective constant
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12 Jul 2013, 10:01
i am little confused. If i take slower pump as X then 1/X + 1/1.5X = 1/4 which would result X = 20/3 and so faster pump as 10..... where i am going wrong...



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Re: Two water pumps, working simultaneously at their respective constant
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12 Jul 2013, 10:07
AMITAGARWAL2 wrote: i am little confused. If i take slower pump as X then 1/X + 1/1.5X = 1/4 which would result X = 20/3 and so faster pump as 10..... where i am going wrong... In my solution x is the rate in your solution x is the time. In your solution x is the time of the faster pump and 1.5x is the time of the slower pump (faster pump needs less time). Hope it's clear.
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Re: Two water pumps, working simultaneously at their respective constant
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08 Sep 2013, 12:25
kmasonbx wrote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3
Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks! sol: Rates be A and B (A+B) * Time = Work (A+B) * 4 = 1 >eq 1 A= 3B/2 >eq 2 substituting eq 2 in eq 1 B = 1/10 >eq 3 substituting eq 3 in eq 2 A= 3/20 Time= work/rate = 1/(3/20) =>20/3



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Re: Two water pumps, working simultaneously at their respective constant
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25 Sep 2013, 05:13
I did, A+B = 4 A = 1.5B
4B = 1.5B => B = 8/5
Why is this wrong?



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Re: Two water pumps, working simultaneously at their respective constant
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25 Sep 2013, 08:48
Skag55 wrote: I did, A+B = 4 A = 1.5B
4B = 1.5B => B = 8/5
Why is this wrong? We are told that "two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool". If one pump needs A hours to fill the pool (rate=1/A) and another need B hours to fill the same pool (rate=1/B), then 1/A + 1/B = 1/4. Solving 1/A + 1/B = 1/4 and A = 1.5B gives A=10 and B=20/3. Check the links provided here: twowaterpumpsworkingsimultaneouslyattheirrespective155865.html#p1245761 for more.
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Re: Two water pumps, working simultaneously at their respective constant
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14 Feb 2015, 14:38
kmasonbx wrote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3
Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks! fast pump takes x hour Slow pump takes 1.5x hour so 1/x+1/1.5x = 1/4 > (1.5+1)/1.5x = 1/4 > 2.5/1.5x = 1/4 > 1.5 x = 10 >x = 10/1.5 >x = 20/3 Somebody confirm whether this is a right approach to do this type of problem or not. Thanks



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Re: Two water pumps, working simultaneously at their respective constant
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14 Feb 2015, 21:23
Hi Salvetor, Yes, your approach is correct. In 'Work' questions, there are usually several different ways to organize the given information, but they all end up involving a ratio at some point. GMAT assassins aren't born, they're made, Rich
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Re: Two water pumps, working simultaneously at their respective constant
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06 Feb 2016, 10:15
How would you solve this problem by saying the faster pump 1.5x and slower x. I cant seem to figure that out.



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Re: Two water pumps, working simultaneously at their respective constant
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06 Feb 2016, 10:48
xLUCAJx wrote: How would you solve this problem by saying the faster pump 1.5x and slower x. I cant seem to figure that out. Plug in some values Attachment:
Plug in.PNG [ 2 KiB  Viewed 32969 times ]
Quote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. Capacity of the swimming pool is  5*4 = 20 Quote: how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate? The faster pump is pump B, so time required by fill the swimming pool alone will be = Total Capacity of the pool/Efficiency of Pipe B 20/3 Hence answer is (E)
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Re: Two water pumps, working simultaneously at their respective constant
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06 Mar 2017, 18:13
kmasonbx wrote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3 We are given that the rate of 1 pump is 1.5 times faster than the rate of the other pump. Since 1 pool is being filled and rate = work/time, the rate of the faster pump is 1/x, in which x = the time it takes for the faster pump to fill the pool, and the rate of the slower pump = 1/(1.5x) = 1/(3x/2) = 2/3x. Since when the pumps work together they take 4 hours to fill 1 pool, we can create the following equation: work of faster pump + work of slower pump = 1 (1/x)4 + (2/3x)4 = 1 4/x + 8/3x = 1 Multiplying the entire equation by 3x, we have: 12 + 8 = 3x 20 = 3x 20/3 = x Answer: E
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Re: Two water pumps, working simultaneously at their respective constant
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08 Jun 2017, 08:00
fast pump takes x hour Slow pump takes 1.5x hour 1/x + 1/1.5x = 1/4 1/x = 1/4  1/1.5x 1/x = (1.5x  4)/6x 6x = 1.5x^2  4x 10x = 1.5x^2 10 = 1.5x x = 10/1.5 > stop here during practice exam because it seems strange and you have a serious lack of confidence x = 20/3 could someone confirm please ??
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Re: Two water pumps, working simultaneously at their respective constant
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10 Jun 2017, 12:22
daviddaviddavid wrote: fast pump takes x hour Slow pump takes 1.5x hour 1/x + 1/1.5x = 1/4 1/x = 1/4  1/1.5x 1/x = (1.5x  4)/6x 6x = 1.5x^2  4x 10x = 1.5x^2 10 = 1.5x x = 10/1.5 > stop here during practice exam because it seems strange and you have a serious lack of confidence x = 20/3 could someone confirm please ?? Quote: EMPOWERgmatRichCQuote: Salvetor fast pump takes x hour Slow pump takes 1.5x hour so 1/x+1/1.5x = 1/4 > (1.5+1)/1.5x = 1/4 > 2.5/1.5x = 1/4 > 1.5 x = 10 >x = 10/1.5 >x = 20/3 Somebody confirm whether this is a right approach to do this type of problem or not. Thanks Hi Salvetor, Yes, your approach is correct. In 'Work' questions, there are usually several different ways to organize the given information, but they all end up involving a ratio at some point. daviddaviddavid , your approach is almost identical to Salvetor's. You just used slightly different algebra. I did the algebra your way one time and a second time using a fusion of your and Salvetor's approach. He added. You subtracted. You used a different LCM or this shortcut: \(\frac{a}{b} + \frac{c}{d} = \frac{(da + bc)}{bd}\). So I fused by adding, as Salvetor did, and by using your LCM. Both your method and fusion method gave me the correct answer. I think you should swagger a bit in your head. You took the way four people here asked about, which also seems to me to be the most direct way. Here's the math done in a way that fuses Salvetor's method and yours: \(\frac{1}{x} + \frac{1}{1.5x} = \frac{1}{4}\) \(\frac{(1.5x + x)}{1.5x^2} = \frac{1}{4}\) \(4 (1.5x + x) = 1.5 x^2\) \(4 (2.5x) = 1.5 x^2\) \(10 x = 1.5 x^2\), divide by x (we're allowed b/c we know x is not 0): \(10 = 1.5x\) Get rid of the decimal in the denominator. Multiply by sides by 10 → \(100=15x\) \(x=\frac{100}{15}=\frac{20}{3}\) Or convert 1.5 in the denominator to the fraction \(\frac{3}{2}\) Thus: \(x = \frac{10}{1.5}=\frac{10}{\frac{3}{2}}=\) \((10*\frac{2}{3})=\frac{20}{3}\) Decimals in denominators look strange to me, too. Make a little rule: if decimal in denominator isn't in answer choices, either multiply the fraction by 10/10 or 100/100 etc., or convert the decimal to a fraction. You just got stuck on the part where multiplying your answer by 2 would match one of the choices. Very frustrating. And common. Not to worry. :wink: Hope it helps.
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Re: Two water pumps, working simultaneously at their respective constant
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10 Jun 2017, 14:23
I prefer using numbers for rate problems such as these. So, here goes my method of solving the question : Since both pumps take exactly 4 hours to fill the tank, assume the tank has 200 units. Both the pumps fill the tank is 50 units/hour. It has been given that one of the pumps fill the tank at 1.5 times the rate of the other pump. If, Pump A(slower pump) fills at 20 units/hour, Pump B(faster pump  1.5*rate of slower pump) fills at 30 units/hour Therefore, if the faster pump alone worked, it would have filled the tank in \(\frac{200}{30}\) = \(\frac{20}{3}\) hours(Option E)
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Re: Two water pumps, working simultaneously at their respective constant
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07 May 2018, 05:38
Bunuel wrote: kmasonbx wrote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3
Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks! Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour. Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 > x=3/20 pool hour. The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone. Answer: E. I'm having trouble seeing how you got x= 3/20 from x+2x = 1/4. Do you mind showing step by step? thank you!!



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Re: Two water pumps, working simultaneously at their respective constant
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07 May 2018, 05:44
rnz wrote: Bunuel wrote: kmasonbx wrote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3
Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks! Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour. Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 > x=3/20 pool hour. The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone. Answer: E. I'm having trouble seeing how you got x= 3/20 from x+2x = 1/4. Do you mind showing step by step? thank you!! \(x+\frac{2x}{3}=\frac{1}{4}\); \(\frac{3x+2x}{3}=\frac{1}{4}\); \(\frac{5x}{3}=\frac{1}{4}\); \(20x=3\); \(x=\frac{3}{20}\).
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Re: Two water pumps, working simultaneously at their respective constant
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22 Dec 2018, 11:41
Hi All, This is an example of a Work Formula question. Any time you have two entities (people, machines, water pumps, etc.) working on a job together, you can use the following formula: (AB)/(A+B) = Total time to do the job together Here, we're told that the total time = 4 hours and that one machine's rate is 1.5 times the other machine's rate… If B = 1.5A then we have… (A)(1.5A)/(A + 1.5A) = 4 1.5(A^2)/2.5A = 4 (3/2)(A^2) = 10A A^2 = 20A/3 A = 20/3 hours to fill the pool alone Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Two water pumps, working simultaneously at their respective constant
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19 Jan 2019, 14:12
kmasonbx wrote: Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3 Drawing a chart: Since they are working together we add rates: One pump has a rate of 1 and the other 3/2, together their t=4 so combined 2/2+3/2 * 4 = 20/2 = 10 together they do 10 units of work To find the individual time we divide total work by individual rate so 10/(3/2) = 20/3, E Or... 1/1r + 1/(3/2r) = 1/4 1r+(3/2r) / (3/2)r^2 = 1/4 (3/2)r^2 / (2/2)r+(3/2r) = 4 (3/2)r / (5/2) = 4 (3/5)r = 4 r = 4*(5/3) = 20/3




Re: Two water pumps, working simultaneously at their respective constant
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