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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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New post 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???

Answer to follow.
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Re: PS - Machines! [#permalink]

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New post 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???

Answer to follow.


What is productivity ration..? Never came across this term earlier. Is this GMAT Q ?
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 [#permalink]

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New post 04 Aug 2007, 11:18
Manhattan GMAT question, so it is a tougher one.
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Re: PS - Machines! [#permalink]

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New post 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???

Answer to follow.


Really not sure about the answer, but here we go.
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
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Re: PS - Machines! [#permalink]

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New post 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???

Answer to follow.


Really not sure about the answer, but here we go.
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...
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Re: PS - Machines! [#permalink]

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New post 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???

Answer to follow.


Really not sure about the answer, but here we go.
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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3 machines have a productivity ratio of 1 to 2 to 5. All 3

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