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Two musicians, Maria and Perry, work at independent constant

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Two musicians, Maria and Perry, work at independent constant [#permalink] New post 08 Jun 2013, 07:13
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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
[Reveal] Spoiler: OA

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Re: Two musicians, Maria and Perry, work at independent constant [#permalink] New post 08 Jun 2013, 07:22
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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] New post 08 Jun 2013, 07:30
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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] New post 08 Sep 2013, 11:09
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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] New post 08 Sep 2013, 15:55
Hi all

Can't this question be solved with the Work Formula?

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Re: Two musicians, Maria and Perry, work at independent constant [#permalink] New post 20 Jan 2014, 23:48
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! :idea:
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink] New post 21 Jan 2014, 02:02
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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! :idea:


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] New post 21 Jan 2014, 05:23
how dumb of me... :evil: :x ... Thank you so much Bunuel! .. I get it now... :D
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink] New post 04 Feb 2014, 03:49
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! :idea:


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] New post 04 Feb 2014, 06:15
Expert's post
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! :idea:


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] New post 14 Apr 2014, 14:05
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
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Re: Two musicians, Maria and Perry, work at independent constant [#permalink] New post 04 Aug 2014, 09:27
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] New post 04 Aug 2014, 20:08
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] New post 22 Aug 2015, 08:38
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Re: Two musicians, Maria and Perry, work at independent constant   [#permalink] 22 Aug 2015, 08:38
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