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soaringAlone
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Post subject: 4th Ed-SG# 4, Chapter 3, Problem 14 Posted: Sat Aug 06, 2011 9:46 am |
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Can someone explain how did we arrive at 107 minutes if partial loading were allowed ?
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mithunsam
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Post subject: Re: 4th Ed-SG# 4, Chapter 3, Problem 14 Posted: Tue Aug 09, 2011 9:57 pm |
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Assuming that you understand how we arrive at 80 mins & 64 mins for the two machines...
Machine A takes 80 mins for 1 unit of work Machine B takes 64 mins for 1 unit of work
How much time would the machines together take for 1 unit of work?
1/80 + 1/64 = 1/h 9/320 = 1/h 320/9 = h
How much time would the machines together take for 3 units of work?
h = (320/9) * 3 h = 320/3 ~ 107 mins
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mithunsam
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Post subject: Re: 4th Ed-SG# 4, Chapter 3, Problem 14 Posted: Tue Aug 09, 2011 10:05 pm |
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One more point...
Book says to divide the load between the machines as 4/9 and 5/9 respectively. However, we can avoid this computation.
We just have to remember that when two machines work together, the fastest they could finish a job is when none of the machines sit idle. That means, both the machines have to work for the same amount of time. In other words, the fastest completion time will depend on the average rate of the machines together. That is all I have done above.
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tim
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Post subject: Re: 4th Ed-SG# 4, Chapter 3, Problem 14 Posted: Fri Aug 12, 2011 11:28 pm |
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Posts: 2183 Location: Southwest Airlines, seat 21C
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thanks..
_________________ Tim Sanders Manhattan GMAT Instructor
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soaringAlone
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Post subject: Re: 4th Ed-SG# 4, Chapter 3, Problem 14 Posted: Wed Mar 14, 2012 1:38 am |
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Thanks mithunsam, You have explained it from a different angle and I get it now.
_________________ ~soaringAlone
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tim
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Post subject: Re: 4th Ed-SG# 4, Chapter 3, Problem 14 Posted: Sun Mar 18, 2012 1:40 am |
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Posts: 2183 Location: Southwest Airlines, seat 21C
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glad to hear it!
_________________ Tim Sanders Manhattan GMAT Instructor
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