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Two water pumps, working simultaneously at their respective
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10 Jun 2017, 13:23
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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Two water pumps, working simultaneously at their respective
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21 Jul 2017, 09:47
Bunuel wrote: 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: http://gmatclub.com/forum/twowaterpum ... l#p1245761 for more. how can A = 1.5B be substituted here, as A stands for total time one pump needs to fill the pool and B stands for total time another pump needs to fill the pool. the ratio speaks about rates, so 1/A = 3/2B needs to be solved...



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Re: Two water pumps, working simultaneously at their respective
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07 Aug 2017, 08:36
Its always good to assume total work in such problems. The aim should be to find per hour work of each pump
Since total time is given to be 4 hours, we will assume total work to be a multiple of 4. Lets assume total work=40 units Per hour work of two pumps=40/4 = 10 units Ratio of rate(per hour work) of two pumps = 1.5/1 or 3/2 So out of a total of 10 units filled per hour, faster pump fills 6 units and slower pump fills 4 units.
If faster pump works alone Total work= 40 units Rate = 6 units/ hour Time = 40/6 = 20/3
Hence E.



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Re: Two water pumps, working simultaneously at their respective
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07 Aug 2017, 11:06
lets assume 2 pumps capacity 2.5 (as one is 1.5 more productive than another) then amount of work produced in 4 hours will be: 2.5 * 4 = 10 keeping the same amount of work and taking into account that only one pump will be working with the productivity rate of 1.5 how many hours is needed: 1.5*x = 10 x = 100/15 = 20/3 hours



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Re: Two water pumps, working simultaneously at their respective
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24 Apr 2018, 01:12
There’s a couple of steps we have to take in this question, but all hinge on the same formula. R*T = W Let’s look at what we have and what we need. The rate of two machines is ¼. Because R * 4 = 4. → R = ¼. We also know that the rate of anything is R = W/T. This means that 1/x + 1/x = ¼  because we have two machines. We are given that one machine is 1.5 times faster than the other. So we can adapt this formula to: 1/x + 1/x *3/2 = 1x+3/2x = ¼ → 5/2x = ¼ → x = 2/20 → x = 1/10. We know that the rate of one machine is 1/10. And one is faster 1.5x times faster. 1/10x * 3/2 = 3/20. The rate of the faster machine is 3/20. Let’s fill this in the formula to find out the time for the faster machine. 3/20 * T = 1 T = 20/3. This is answer choice E
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Re: Two water pumps, working simultaneously at their respective
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07 May 2018, 04:38
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
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07 May 2018, 04:44
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
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07 May 2018, 19:58
I substituted numbers and got to the answer choice E. pushpitkc  Most of the time for work/rate problems i use number substitution method. Though it is useful most of the time, in couple of occasions, i ended up doing little large calculations. Please advise the best strategy to follow especially for harder questions.
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Re: Two water pumps, working simultaneously at their respective
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07 May 2018, 22:44
Hey akadiyanTry another set of numbers if you feel the calculations are taking a lot of time. Spend no more than 30 seconds once you start the calculations. Only if you are close to your answer, go ahead! To choose the numbers correctly, you should start by using the LCM of the numbers. IMO, Choosing the right number is all about practice. With a lot of practice, you will be able to easily find the numbers. However, if you are unable to find the right number, go with the traditional method. If Tap A fills the tank is 2 hours, it fills 1/2 of the tank in an hour......
Hope this helps you!
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Re: Two water pumps, working simultaneously at their respective
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22 Dec 2018, 10:41
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
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19 Jan 2019, 13:12
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
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