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Official Solution:Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in \(\frac{1}{3}\) hours. How many minutes does it take pump Y, working alone, to fill the pool? \(rate*time=job\). We are told that \((x+y)*20=c\), where \(x\) is the rate of pump X in gallons per minute, \(y\) is the rate of pump Y in gallons per minute and \(c\) is the capacity of the pool in gallons. The question asks to find the value of \(y\). (1) The capacity of the pool is 900 gallons. \(c=900\), hence we have that \((x+y)*20=900\), so \(x+y=45\). Not sufficient. (2) The rate of Pump X is 30 gallons per minute. \(x=30\), hence we have that \((30+y)*20=c\). We have two unknowns and only one equation. Not sufficient. Consider this if \(c=1,000\), then \(y=20\) but if \(c=900\), then \(y=15\). (1)+(2) \(x=30\) and \(x+y=45\). Solving gives \(y=15\). Sufficient. Answer: C
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Re: M3010 [#permalink]
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18 Dec 2014, 13:07
BunuelI seem to really struggle on the rate/time/distance and rate/time/work problems because I have trouble deciphering when to state the rate as 1/x or just x. In this problem, I tried solving with 1/x + 1/y = 1/20 instead of x+y=20. I haven't figured out a rhyme or rhythm to them.. any advice?



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Bunuel wrote: codeblue wrote: BunuelI seem to really struggle on the rate/time/distance and rate/time/work problems because I have trouble deciphering when to state the rate as 1/x or just x. In this problem, I tried solving with 1/x + 1/y = 1/20 instead of x+y=20. I haven't figured out a rhyme or rhythm to them.. any advice? Time is reciprocal of rate, so if we denote rate as x (as done in my solution above), then the time would be 1/x. If we instead denoted the time by x, then the rate would be 1/x. For more check theory on work/rate problems: workwordproblemsmadeeasy87357.htmlHope it helps. hi Bunuel i am having the same concern as above. following your link i ve found an example identical to this quesiton. however there the author rather states the following formula: 1/a+1/b = 1/24 min. can you please help Example 3. Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A?
Solution: This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’
‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of 124 per minute.
‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of 160 per minute.
At this point, it should strike you that with just this much information, it is possible to calculate the rate at which B works: Rate at which B works = 124−160=140.
‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and A complete in one minute corresponds to 5 pages. So, let us calculate that difference. It will be 140−160=1120
‘How many pages does the task contain?’ If 1120 of the job consists of 5 pages, then the 1 job will consist of (5∗1)1120=600 pages.
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Re: M3010 [#permalink]
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02 Jan 2016, 07:04
can anyone please help with my question? i am really confused
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02 Jan 2016, 07:46
shasadou wrote: can anyone please help with my question? i am really confused Hi, we know that combined, x and y take 20 mins.. and we require to determine the time that y will take... how can we ansewr the Q.. 1) we know the time taken by x, we can find time taken by y.. 2) if we are given speed of any x or y, then we require the capacity of tank to determine the time.. now lets see the problem... Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in 1/3 hours. How many minutes does it take pump Y, working alone, to fill the pool? (1) The capacity of the pool is 900 gallons. the capacity is given, we know the combined time, so we know combined speed = 900/20=45 l/ min but nothing can be said about indl speed or time taken.. insuff (2) The rate of Pump X is 30 gallons per minute. we know only speed of pump x, but nothing on time taken... insuff combined, we know x speed is 30 g/min and speed combined is 45g/m.. so y's speed =15g/m.. time taken by y=900/15=60 min... suff C
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Re: M3010 [#permalink]
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02 Jan 2016, 08:55
chetan2u wrote: shasadou wrote: can anyone please help with my question? i am really confused Hi, we know that combined, x and y take 20 mins.. and we require to determine the time that y will take... how can we ansewr the Q.. 1) we know the time taken by x, we can find time taken by y.. 2) if we are given speed of any x or y, then we require the capacity of tank to determine the time.. now lets see the problem... Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in 1/3 hours. How many minutes does it take pump Y, working alone, to fill the pool? (1) The capacity of the pool is 900 gallons. the capacity is given, we know the combined time, so we know combined speed = 900/20=45 l/ min but nothing can be said about indl speed or time taken.. insuff (2) The rate of Pump X is 30 gallons per minute. we know only speed of pump x, but nothing on time taken... insuff combined, we know x speed is 30 g/min and speed combined is 45g/m.. so y's speed =15g/m.. time taken by y=900/15=60 min... suff C Hi chetan2u thank you for your response. i still have a concern: why we cannot follow the approach set out in here workwordproblemsmadeeasy87357.html example 3. it an identical problem to my eyes. kudos in advance!
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codeblue wrote: BunuelI seem to really struggle on the rate/time/distance and rate/time/work problems because I have trouble deciphering when to state the rate as 1/x or just x. In this problem, I tried solving with 1/x + 1/y = 1/20 instead of x+y=20. I haven't figured out a rhyme or rhythm to them.. any advice? i think i got it now: 1/x +1/y = 1/20 is absolutely correct. it denotes that working together the two pumps complete 1/20 of the job per 1 minute. in st 2 we are only given the productivity rate: just randomly how many gallons one of the machines pumps. BUT this is not yet the rate of completing the given job (jobrate). there can be a huge tank and a small tank, hence with the given productivity (productivityrate) the machine would have different rate (workrate)to fill in the tanks. and this is the latter rate (jobrate) in relation to job is the one we should use in the above formula. so. together 2 statements help us find the 1/x  that is the rate in relation to the given work (jobrate). 900/ 30 = 30 min to fulfill the job. hence only now we can insert 30 into 1/x > 1/30 + 1/y = 1/20 > y=60 hence y will fill the tank in 60 minutes. what confused me, a nonsophisticated math solver, is that the "correct" rate we derive is 1/30 just the same as one can get with staement 2 alone. should the capacity be some other amount this question would be more representative for teaching purposes and less confusing.
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Re: M3010 [#permalink]
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09 Apr 2016, 20:30
I have the same confusion of either to use x or 1/x but if you look at the solution above, y takes 15 mins to fill up the tank not 60 mins. Still not clear when to use x or 1/x though.



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Re: M3010 [#permalink]
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03 Jul 2016, 00:11
Bunuel wrote: Official Solution:
Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in \(\frac{1}{3}\) hours. How many minutes does it take pump Y, working alone, to fill the pool?
\(rate*time=job\). We are told that \((x+y)*20=c\), where \(x\) is the rate of pump X in gallons per minute, \(y\) is the rate of pump Y in gallons per minute and \(c\) is the capacity of the pool in gallons. The question asks to find the value of \(y\). (1) The capacity of the pool is 900 gallons. \(c=900\), hence we have that \((x+y)*20=900\), so \(x+y=45\). Not sufficient. (2) The rate of Pump X is 30 gallons per minute. \(x=30\), hence we have that \((30+y)*20=c\). We have two unknowns and only one equation. Not sufficient. Consider this if \(c=1,000\), then \(y=20\) but if \(c=900\), then \(y=15\). (1)+(2) \(x=30\) and \(x+y=45\). Solving gives \(y=15\). Sufficient.
Answer: C Please tell me, if below mentioned steps are right 1/x+1/y=20min from question stem statement one agree not sufficient statement two 1/201/30=1/60 thus y works at the rate 60 per minute Is this right? where am I going wrong? Please guide ,Thankyou!



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Re: M3010 [#permalink]
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03 Jul 2016, 03:41
89renegade wrote: Bunuel wrote: Official Solution:
Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in \(\frac{1}{3}\) hours. How many minutes does it take pump Y, working alone, to fill the pool?
\(rate*time=job\). We are told that \((x+y)*20=c\), where \(x\) is the rate of pump X in gallons per minute, \(y\) is the rate of pump Y in gallons per minute and \(c\) is the capacity of the pool in gallons. The question asks to find the value of \(y\). (1) The capacity of the pool is 900 gallons. \(c=900\), hence we have that \((x+y)*20=900\), so \(x+y=45\). Not sufficient. (2) The rate of Pump X is 30 gallons per minute. \(x=30\), hence we have that \((30+y)*20=c\). We have two unknowns and only one equation. Not sufficient. Consider this if \(c=1,000\), then \(y=20\) but if \(c=900\), then \(y=15\). (1)+(2) \(x=30\) and \(x+y=45\). Solving gives \(y=15\). Sufficient.
Answer: C Please tell me, if below mentioned steps are right 1/x+1/y=20min from question stem statement one agree not sufficient statement two 1/201/30=1/60 thus y works at the rate 60 per minute Is this right? where am I going wrong? Please guide ,Thankyou! You are not getting the correct answer therefore the solution is obviously not correct. Now, when you write a solution you should explain what each variable there represents and how you are constructing the equations. x and y in my solution are the rates. What are x and y in yours? What's the rationale behind the equations you've written? Also, it might be useful to read the whole discussion above.
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Re: M3010 [#permalink]
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12 Jul 2016, 13:04
i have doubt about the question stem. please clarify if i m wrong.
question asks to find the time to fill the pool when Y works alone. i think i hv to find out the value of (C/Y)? not the rate of Y. (gallon/min) even the choice C is right but the answer would be (900/15) = 60 min. As for DS the choice is right but for PS, the answer could be wrong.



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Re: M3010 [#permalink]
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13 Oct 2016, 13:50
here is why answer is C. pumps X and Y can fill an empty pool to capacity in 1/3 hours (20 mins) let the time taken by pump X be 'x' mins to fill pool to capacity let the time taken by pump Y be 'y' mins to fill pool to capacity let the pool capacity be 'c' => work Now, we know rate * time = work; rate = work/time; rate of pump X + rate of pump Y gives c/x + c/y = c/20; (eq. 1) simplified; 1/x + 1/y = 1/20. remember here that the rate is defined in terms of pool capacity and time required to fill the pool to capacity. x and y are times in mins taken by Pump X and Pump Y to fill the pool to capacity. 20 mins is time taken by Pump X and Pump Y working together to fill the pool to capacity. now statement 1 tell us the value of 'c' which is 900. this statement alone is clearly insuff. statement 2 tell us the rate of Pump X is 30 gallons per minute. 30 gallons per minute does not tell us how much time will be required to fill the pool to capacity because we dont know the pool capacity state 2 is not the same as c/x (rate to fill pool to capacity) or 1/x (rate to perform 1 unit of work which is to fill pool to capacity) statement 1 and 2 taken together gives us 'c' and 'x' => x = 900/30 = 30 mins from (eq. 1) (900/30) + (900/y) = (900/20) y=60



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Re: M3010 [#permalink]
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06 Feb 2017, 15:59
Bunuel wrote: Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in \(\frac{1}{3}\) hours. How many minutes does it take pump Y, working alone, to fill the pool?
(1) The capacity of the pool is 900 gallons. (2) The rate of Pump X is 30 gallons per minute. Hi Bunuel, I think need to practice more rate questions, but I want to get your thoughts on something. I got this question, and the one here ( https://gmatclub.com/forum/d01183490.html#p1781823) incorrect. I believe the root of my misunderstanding is a simple unit conversion issue. In some problems I have seen, when one must consider "one unit of work", r_1+r_2 = 1/ (t_1,2). For example, for this problem I set r_x+r_y=1/(1/3)=3, which was incorrect and I should have set it to 3W, which would give r_x+r_y= 2700 gal in one hour. When are you able to set W=1 (100%), as you did in the link I provided, and when do you have to set W equal to an actual number, not 100%? Additionally, if I wanted to consider 1 unit of work to consist of pumping 900 gallons, how would I have formulated the equations to figure this problem out? This would help me determine if my approach to rate questions is incorrect due to my inattention to units and their conversions (gallons/hour, units of work/hour, etc.).



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Re: M3010 [#permalink]
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06 Feb 2017, 16:08
brooklyndude wrote: Bunuel wrote: Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in \(\frac{1}{3}\) hours. How many minutes does it take pump Y, working alone, to fill the pool?
(1) The capacity of the pool is 900 gallons. (2) The rate of Pump X is 30 gallons per minute. Hi Bunuel, I think need to practice more rate questions, but I want to get your thoughts on something. I got this question, and the one here ( https://gmatclub.com/forum/d01183490.html#p1781823) incorrect. I believe the root of my misunderstanding is a simple unit conversion issue. In some problems I have seen, when one must consider "one unit of work", r_1+r_2 = 1/ (t_1,2). For example, for this problem I set r_x+r_y=1/(1/3)=3, which was incorrect and I should have set it to 3W, which would give r_x+r_y= 2700 gal in one hour. When are you able to set W=1 (100%), as you did in the link I provided, and when do you have to set W equal to an actual number, not 100%? Additionally, if I wanted to consider 1 unit of work to consist of pumping 900 gallons, how would I have formulated the equations to figure this problem out? This would help me determine if my approach to rate questions is incorrect due to my inattention to units and their conversions (gallons/hour, units of work/hour, etc.). I think I just answered my own question. If I set 1 work unit = 900 gal, then converting 3 units of work to gallons means r_x+r_y=2700 gal in one hour, or 2700/60=45 gal in one minute. Looks like I just have to pay closer attention to units when doing these work problems.



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Re: M3010 [#permalink]
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13 Jul 2017, 13:02
"pumps X and Y can fill an empty pool to capacity in 1/3 hours"  meaning time (X&Y) is 20 mins and the rate (X&Y) is 1/20 (2) The rate of Pump X is 30 gallons per minute I'm trying to understand why (2) by itself wouldn't be sufficient is this case... Is it because the fact that "The rate of Pump X is 30 gallons per minute" doesn't necessarily mean that (X) is 1/30? (given we don't know what the pool's capacity is) Hope my question is clear! Thank you



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Re: M3010 [#permalink]
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13 Jul 2017, 21:16
vicm730 wrote: "pumps X and Y can fill an empty pool to capacity in 1/3 hours"  meaning time (X&Y) is 20 mins and the rate (X&Y) is 1/20 (2) The rate of Pump X is 30 gallons per minute I'm trying to understand why (2) by itself wouldn't be sufficient is this case... Is it because the fact that "The rate of Pump X is 30 gallons per minute" doesn't necessarily mean that (X) is 1/30? (given we don't know what the pool's capacity is) Hope my question is clear! Thank you The reason as to why the second statement is not sufficient is given in the solution: (2) The rate of Pump X is 30 gallons per minute. \(x=30\), hence we have that \((30+y)*20=c\). We have two unknowns and only one equation. Not sufficient. Consider this if \(c=1,000\), then \(y=20\) but if \(c=900\), then \(y=15\).(\(x\) is the rate of pump X in gallons per minute, \(y\) is the rate of pump Y in gallons per minute and \(c\) is the capacity of the pool in gallons)
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Re: M3010 [#permalink]
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11 Sep 2017, 20:29
shasadou wrote: chetan2u wrote: shasadou wrote: can anyone please help with my question? i am really confused Hi, we know that combined, x and y take 20 mins.. and we require to determine the time that y will take... how can we ansewr the Q.. 1) we know the time taken by x, we can find time taken by y.. 2) if we are given speed of any x or y, then we require the capacity of tank to determine the time.. now lets see the problem... Working together at their constant rates, pumps X and Y can fill an empty pool to capacity in 1/3 hours. How many minutes does it take pump Y, working alone, to fill the pool? (1) The capacity of the pool is 900 gallons. the capacity is given, we know the combined time, so we know combined speed = 900/20=45 l/ min but nothing can be said about indl speed or time taken.. insuff (2) The rate of Pump X is 30 gallons per minute. we know only speed of pump x, but nothing on time taken... insuff combined, we know x speed is 30 g/min and speed combined is 45g/m.. so y's speed =15g/m.. time taken by y=900/15=60 min... suff C Hi chetan2u thank you for your response. i still have a concern: why we cannot follow the approach set out in here workwordproblemsmadeeasy87357.html example 3. it an identical problem to my eyes. kudos in advance! The reason you can't use the printer problem logic here is : We are not given the "Time or Rate of X to do the job alone by itself" Example – Tom and Jerry have to stuff and mail 1000 envelopes for a new marketing campaign. Jerry can do the job alone in 6 hours. If Tom helps, they can get the job done in 4 hours. How long would it take Tom to do the job by himself?
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