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# 3 machines have a productivity ratio of 1 to 2 to 5. All 3

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Director
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3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]

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04 Aug 2007, 08:05
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3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in 9 hours. What was the ratio of the work performed by the fastest machine as compared to the slowest???

Director
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04 Aug 2007, 10:07
jimmyjamesdonkey wrote:
3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in 9 hours. What was the ratio of the work performed by the fastest machine as compared to the slowest???

What is productivity ration..? Never came across this term earlier. Is this GMAT Q ?
Director
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Posts: 792
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04 Aug 2007, 11:18
Manhattan GMAT question, so it is a tougher one.
VP
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04 Aug 2007, 16:19
jimmyjamesdonkey wrote:
3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in 9 hours. What was the ratio of the work performed by the fastest machine as compared to the slowest???

Machine A, B, C
A <-- Slowest
C <-- Fastest
Working:
3 hours, A+B+C
3 hours, B+C
3 hours, A+B+C
We know that A worked 6 hours where C worked 9 hours
Productivity A:B:C = 1:2:5
This means...
Machine: time: Job
A: t : 1 => A: 6 : 6/t
B: t : 2
C: t : 5 => C: 9: (5*(9/t))

Ratio = (45/t) / (6/t) = 15/2
Director
Joined: 09 Aug 2006
Posts: 525
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04 Aug 2007, 21:15
bkk145 wrote:
jimmyjamesdonkey wrote:
3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in 9 hours. What was the ratio of the work performed by the fastest machine as compared to the slowest???

Machine A, B, C
A <-- Slowest
C <Fastest> A: 6 : 6/t
B: t : 2
C: t : 5 => C: 9: (5*(9/t))

Ratio = (45/t) / (6/t) = 15/2

This means...
Machine: time: Job
A: t : 1 => A: 6 : 6/t
B: t : 2
C: t : 5 => C: 9: (5*(9/t))

Ratio = (45/t) / (6/t) = 15/2

Could you please elaborate this part...
VP
Joined: 10 Jun 2007
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04 Aug 2007, 21:47
Amit05 wrote:
bkk145 wrote:
jimmyjamesdonkey wrote:
3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in 9 hours. What was the ratio of the work performed by the fastest machine as compared to the slowest???

Machine A, B, C
A <-- Slowest
C <Fastest> A: 6 : 6/t
B: t : 2
C: t : 5 => C: 9: (5*(9/t))

Ratio = (45/t) / (6/t) = 15/2

This means...
Machine: time: Job
A: t : 1 => A: 6 : 6/t
B: t : 2
C: t : 5 => C: 9: (5*(9/t))

Ratio = (45/t) / (6/t) = 15/2

Could you please elaborate this part...

It's the directly proportional rule for time and job. Say you do 1 job in 2 hours, you will spend 4 hours for 2 job. So (job / time) = (job / time)
Same logic here. A uses t hour to do 1 job. You want to find how many job A can do in 6 hours. You get 1/ t = job / 6; thus, job = 6/t.
Re: PS - Machines!   [#permalink] 04 Aug 2007, 21:47
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