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Two musicians, Maria and Perry, work at independent constant [#permalink]
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Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments? A. 1 hr 20 min B. 1 hr 45 min C. 2 hr D. 2 hr 20 min E. 3 hr
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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\(rate*time=work\) Working together : \((P+M)*\frac{3}{4}=1\) If P=2M: \((2M+M)*\frac{1}{3}=1\) from this we get \(M=1\). We plug M=1 in the above equation: \(\frac{3}{4}P+\frac{3}{4}=1\) or \(\frac{3}{4}P=\frac{1}{4}\) \(P=\frac{1}{3}\). So it would take 3 hours to complete the job, \(\frac{1}{3}*3h=1job\)
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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Rock750 wrote: Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
A. 1 hr 20 min B. 1 hr 45 min C. 2 hr D. 2 hr 20 min E. 3 hr Let the whole work be 180 units. In this case combined rate of Maria and Perry is 180/45=4 units per minute > M+P=4, where M and P are individual rates of Maria and Perry. If Perry were to work at twice Maria’s rate, they would take only 20 minutes > in this case combined rate of Maria and Perry is 180/20=9 units per minute > M+2M=9 > M=3 units per minute. M+P=4 > 3+P=4 > P=1 unit per minute. Time to do 180 units = (job)/(rate) = 180/1 =180 minutes = 3 hours. Answer: E.
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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Rock750 wrote: Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
A. 1 hr 20 min B. 1 hr 45 min C. 2 hr D. 2 hr 20 min E. 3 hr Sol: Lets Perry Rate be P and Rate of Maria be M (rate)*(time)= Work or rate = work/time first equation=> P+M = 1/45 converting it to hrs P+M= 1/(45/60) => 1/(3/4) =>4/3 second equation => M+2M =>1/20 converting it to hrs 3M=1/(20/60) =>1/(1/3) =>3 therefore M= 1 and P=1/3 Rate of Perry = 1/3 time= work/rate (work = 1 job) Time= 3 hrs



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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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Hi Bunuel, I cant seem to understand what I am doing wrong by solving it like this.. please guide me .. given: let call maria =m perry = p working together we have \(\frac{1}{m} + \frac{1}{p} = \frac{1}{45}\) eq 1. now if perry works twice at maria's rate we get .... \(\frac{1}{m} + \frac{1}{2m} = \frac{1}{20}\) solving this we get... M = 30 now substituting it in the first eq 1. \(\frac{1}{30} + \frac{1}{p} = \frac{1}{45}\) obviously since I am solving inccorectly i get a negative value .. I understand the simple methods give in the above discussions .. i get it .. but what em i doing wrong.. i need to rectify my thought process otherwise I will keep approaching such problems in a similar fashion even in the future.. so comment will be greatly appreciated Bunuel .. Thanks a ton!



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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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rawjetraw wrote: Hi Bunuel, I cant seem to understand what I am doing wrong by solving it like this.. please guide me .. given: let call maria =m perry = p working together we have \(\frac{1}{m} + \frac{1}{p} = \frac{1}{45}\) eq 1. now if perry works twice at maria's rate we get .... \(\frac{1}{m} + \frac{1}{2m} = \frac{1}{20}\) solving this we get... M = 30 now substituting it in the first eq 1. \(\frac{1}{30} + \frac{1}{p} = \frac{1}{45}\) obviously since I am solving inccorectly i get a negative value .. I understand the simple methods give in the above discussions .. i get it .. but what em i doing wrong.. i need to rectify my thought process otherwise I will keep approaching such problems in a similar fashion even in the future.. so comment will be greatly appreciated Bunuel .. Thanks a ton! In your solution m and p are individual times, while 1/m and 1/p are respective rates. If Perry were to work at twice Maria’s rate, then his rate would be 2*(1/m), not 1/(2m). Therefore the correct equations are 1/m+1/p=1/45 and 1/m+2/m=1/20
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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04 Feb 2014, 04:49
Bunuel wrote: rawjetraw wrote: Hi Bunuel, I cant seem to understand what I am doing wrong by solving it like this.. please guide me .. given: let call maria =m perry = p working together we have \(\frac{1}{m} + \frac{1}{p} = \frac{1}{45}\) eq 1. now if perry works twice at maria's rate we get .... \(\frac{1}{m} + \frac{1}{2m} = \frac{1}{20}\) solving this we get... M = 30 now substituting it in the first eq 1. \(\frac{1}{30} + \frac{1}{p} = \frac{1}{45}\) obviously since I am solving inccorectly i get a negative value .. I understand the simple methods give in the above discussions .. i get it .. but what em i doing wrong.. i need to rectify my thought process otherwise I will keep approaching such problems in a similar fashion even in the future.. so comment will be greatly appreciated Bunuel .. Thanks a ton! In your solution m and p are individual times, while 1/m and 1/p are respective rates. If Perry were to work at twice Maria’s rate, then his rate would be 2*(1/m), not 1/(2m). Therefore the correct equations are 1/m+1/p=1/45 and 1/m+2/m=1/20 buneul by the equations 1/m+2/m =1/ 20 => m = 60. when i substituted in 1/m+1/p = 1/45. i got p = 18. How to proceed after that??



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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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04 Feb 2014, 07:15
nishanthadithya wrote: Bunuel wrote: rawjetraw wrote: Hi Bunuel, I cant seem to understand what I am doing wrong by solving it like this.. please guide me .. given: let call maria =m perry = p working together we have \(\frac{1}{m} + \frac{1}{p} = \frac{1}{45}\) eq 1. now if perry works twice at maria's rate we get .... \(\frac{1}{m} + \frac{1}{2m} = \frac{1}{20}\) solving this we get... M = 30 now substituting it in the first eq 1. \(\frac{1}{30} + \frac{1}{p} = \frac{1}{45}\) obviously since I am solving inccorectly i get a negative value .. I understand the simple methods give in the above discussions .. i get it .. but what em i doing wrong.. i need to rectify my thought process otherwise I will keep approaching such problems in a similar fashion even in the future.. so comment will be greatly appreciated Bunuel .. Thanks a ton! In your solution m and p are individual times, while 1/m and 1/p are respective rates. If Perry were to work at twice Maria’s rate, then his rate would be 2*(1/m), not 1/(2m). Therefore the correct equations are 1/m+1/p=1/45 and 1/m+2/m=1/20 buneul by the equations 1/m+2/m =1/ 20 => m = 60. when i substituted in 1/m+1/p = 1/45. i got p = 18. How to proceed after that?? From 1/60 + 1/p = 1/45 it follows that p = 180 minutes, not 18. p there denotes is time Perry needs to compete the job alone, and this is what we need to find, hence p = 180 minutes = 3 hours is the answer. Hope it's clear.
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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So we have the following 1/M + 1/P = 1/45 We also know that 1/M + 2/M = 1/20 Therefore we know that M=60 and by substituting in the first equation we have that 1/P = 1/180 Therefore P = 180 = 3 hours Hope this clarifies Cheers! J



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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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Hello Bunuel, If rates are 1/m & 1/p then pm/(p+m)=45. Now if 1/p=2/m, I substituted m=2p in pm/(p+m)=20, which resulted in p=30... where I am going wrong... Please help.
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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04 Aug 2014, 21:08
1/P + 1/M = 1/45
2/M + 1/M = 1/20
3/M = 1/20 => M=60
1/P + 1/60 = 1/45
P=180 mins, =3 hours



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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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14 Nov 2015, 15:10
shuvabrata88 wrote: Hello Bunuel, If rates are 1/m & 1/p then pm/(p+m)=45. Now if 1/p=2/m, I substituted m=2p in pm/(p+m)=20, which resulted in p=30... where I am going wrong... Please help. Made same mistake. if rate p = 2m then time p = m/2. So: [m/2 * m] / [ m/2 + m] = 20. Calculate and you will get m = 60.



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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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14 Nov 2015, 21:36
maria's rate=m if 3m=1/20, then m=1/60 1/451/60=1/180=perry's rate perry needs 180 minutes=3 hours to complete job alone



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Two musicians, Maria and Perry, work at independent constant [#permalink]
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28 Feb 2016, 11:12
Rock750 wrote: Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
A. 1 hr 20 min B. 1 hr 45 min C. 2 hr D. 2 hr 20 min E. 3 hr first of all. the question is ambiguous from the start. is 45 min the rate they work both together? or each independently? I suppose together... so \(\frac{1}{M}+\frac{1}{P}=\frac{1}{45}\) then we are told: \(\frac{2}{M}+\frac{1}{M}=\frac{1}{20}\) we can find the rate for M: \(\frac{1}{60}\) > so M needs 60 min to finish the job. now, we are given \(\frac{1}{M}\), we can find \(\frac{1}{P}\) \(\frac{1}{60}+\frac{1}{P}=\frac{1}{45}\) \(\frac{1}{P}=\frac{1}{180}\) So M needs 180 mins, or 3hours to complete the job alone.



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Re: Two musicians, Maria and Perry, work at independent constant rates to [#permalink]
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20 Jan 2017, 05:53
M's time to complete the job alone = M => time taken to complete 1 unit of work = 1/M P's time to complete the job alone = P => time taken to complete 1 unit of work = 1/P
Second case, P's rate = 2(M's rate) => to complete the job alone = M/2 => time taken to complete 1 unit of work = 2/M Now, 1/M + 2/M = 1/20 => M=60mins
First case, 1/M + 1/P = 45 => 1/60 + 1/P = 1/45 => P = 180mins = 3hrs



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Re: Two musicians, Maria and Perry, work at independent constant rates to [#permalink]
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20 Jan 2017, 06:52
robu wrote: Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
1 hr 20 min 1 hr 45 min 2 hr 2 hr 20 min  3 hr 
How we can solve it by considering time not by rate. Please explain. hi Ofcourse best method would be by rate only but since you want to know a method by time, one such will beM and P together do in 45 min.. Twice M + M or 3 Ms do in 20 min, so M will do in 20*3= 60 min. So if P joins M, it takes 45 min instead of 60 min.. But for 45 minutes M is also working, so work P does in 45 minutes saves 15 minutes of M work... This means P takes 45/15 times M's time, so 60*3=180 min..
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Two musicians, Maria and Perry, work at independent constant [#permalink]
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26 Apr 2017, 02:11
Rock750 wrote: Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
A. 1 hr 20 min B. 1 hr 45 min C. 2 hr D. 2 hr 20 min E. 3 hr 3m*20=(r+m)*45=1(work) 60m=45r+45m 15m=45p m=3p(substitute in above eq) 3m*20 3(3p)*20 9p*20 p*180 p takes 180 minutes to complete the work at its normal speed
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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29 Apr 2017, 08:56
Rock750 wrote: Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
A. 1 hr 20 min B. 1 hr 45 min C. 2 hr D. 2 hr 20 min E. 3 hr We can let the time it takes Perry to complete the job alone = p and the time it takes Maria to complete the job alone = m. Thus, Perry’s rate = 1/p and Maria’s rate = 1/m. Since they complete the job in 45 minutes, we use the formula work = rate x time to get: (1/m)45 + (1/p)45 = 1 45/m + 45/p = 1 Multiplying the entire equation by mp, we have: 45p + 45m = mp We are also given that if Perry were to work at twice Maria’s rate, they would take only 20 minutes. Since Maria’s rate is 1/m, Perry’s rate would be 2/m. We can create the following equation to determine p: (2/m)20 + (1/m)20 = 1 40/m + 20/m = 1 Multiplying the entire equation by m, we have: 40 + 20 = m m = 60 Recalling that 45p + 45m = mp, we can substitute m = 60 in the equation and solve for p: 45p + 45(60) = 60p 3p + 3(60) = 4p 180 = p Since Perry’s time is 180 minutes, and 60 minutes = 1 hour, it takes him 3 hours to complete the job at his normal rate. Answer: E
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink]
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17 Jan 2018, 13:56
Hi All, This question is a complex version of a WorkFormula question, but can still be solved using the Work Formula (you have to be careful to make sure that you're using the formula properly though... Work = (A)(B)/(A+B) where A and B are the individual rates of the two entities working on their own to complete a task. Here, we're told that Maria (M) and Perry (P) work on a task together. Working their standard rates, they will complete the job in 45 minutes. We can write this as.... (M)(P)/(M+P) = 45 Next, we're told that IF Perry worked TWICE Maria's rate, then they would take only 20 minutes to complete the task. This means that Perry works TWICE AS FAST as Maria  to write this mathematically, instead of writing P, we have to write (M/2)  since M represents the amount of time that Maria would take to complete the job, M/2 is the equivalent of TWICE Maria's rate... (M/2)(M)/(M/2 + M) = 20 From here, we have a 'system'  two variables and two unique equations, so we CAN solve for P... With the second equation, we have... (M^2)/2 = 10M + 20M (M^2)/2 = 30M M^2 = 60M M = 60 Plugging this value back into the first equation, we have... MP/(M+P) = 45 60P/(60 + P) = 45 60P = 60(45) + 45P 15P = 60(45) P = 60(3) P = 180 minutes Final Answer: GMAT assassins aren't born, they're made, Rich
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Two musicians, Maria and Perry, work at independent constant [#permalink]
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10 Apr 2018, 11:34
Bunuel chetan2uQuote: Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
A. 1 hr 20 min B. 1 hr 45 min C. 2 hr D. 2 hr 20 min E. 3 hr Let the whole work be 180 units.Is the highlighted text LCM of 45 and 20? If so, are not these mentioned as time units in main question stem? How can I use smart numbers for two different variables wiz units and time?
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