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If 6 machines run at the same constant rate, they can

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If 6 machines run at the same constant rate, they can [#permalink] New post 05 Nov 2010, 12:47
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A
B
C
D
E

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73% (02:27) correct 27% (01:49) wrong based on 113 sessions
If 6 machines run at the same constant rate, they can complete a job in 8 hours. If only 5 of these machines run at this rate, how many more minutes will be required to complete the same job?

A. 38
B. 72
C. 80
D. 90
E. 96
[Reveal] Spoiler: OA
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Re: Hard Work Problem [#permalink] New post 05 Nov 2010, 13:17
My guess is E.

Six machines can do 1/8 of the job within an hour. => 1 machine can do 1/48 of the job within a hour. => 5 machines can do 5/48 of the job within a hour.

=> 5 machines can do a full job in 48/5 hours.

The difference between 48/5 hours and 5 hours is 8/5, which is 96 minutes (the answer to the question).
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Re: Hard Work Problem [#permalink] New post 06 Nov 2010, 04:44
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6 machines can do a job in 8 hrs.
5 machines con do a job in ?

Now think, will 5 machines take more than 8 hrs, or less? Of course more than 8 hrs since fewer machines are working.

So number of hours 5 machines will take = 8 * (\frac{6}{5})

Basically, you need hours so given hours term i.e. 8 * \frac{6 machines}{5 machines} to get 9.6 hours.

We multiply by 6/5 to increase 8 since more hours are required.
It is actually a Variation question (Inverse Variation here) but it is just easier to think in terms of more/less.

Another e.g. 10 people make 5 chairs in a day. How many people do we need to make 12 chairs.
Simply multiply 10 (the number that you want to change) by 12/5 because you want to increase the number of people to make more chairs: 10 * (\frac{12}{5}) = 24 people

Yet another e.g. 10 people need 4 hours to complete a job. How many hours do 18 people need to complete the same job?
4 * (\frac{10}{18}) = 2.2 hrs
If there are more people, they will need fewer hours so multiply by 10/18 (not 18/10).
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Re: Hard Work Problem [#permalink] New post 15 Jan 2011, 07:01
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a simpler way to tackling such questions is to convert the information into man-hours or, in this case, machine-hours.

6 machines in 8 hours = 6*8= 48 machine-hours
5 machines will take = 48/5= 9.6 hrs

so 1.6hrs more or \frac{8}{5} hrs = \frac{8*60}{5} mins = 96mins

HTH

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Re: Hard Work Problem [#permalink] New post 23 Feb 2012, 08:33
I agree with E

start with x/6=8 so x = 48

next divide 48/5 which =9.6

so it takes 9.6 hours but you need to convert that to minutes so .6=36 mins so it takes 1 hour and 36 mins more or 96 mins. so E
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Re: If 6 machines run at the same constant rate, they can [#permalink] New post 23 Feb 2012, 08:44
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rtaha2412 wrote:
If 6 machines run at the same constant rate, they can complete a job in 8 hours. If only 5 of these machines run at this rate, how many more minutes will be required to complete the same job?

A. 38
B. 72
C. 80
D. 90
E. 96


The question can be answered in 30 sec if you have a fundamental understanding of simple principle:
Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

time*speed=distance <--> time*rate=job \ done.

So, if you decrease the rate of work 5/6 times (from 6 to 5 machines) the time needed to complete the same job will increase 6/5 times.

Time needed for 6 machines = 80*60 = 480 minutes;
Time needed for 5 machines = 480*6/5 = 576 minutes;

Difference: 576-480=96 minutes.

Answer: E.

For more on Work/Rate Problems check this: two-consultants-can-type-up-a-report-126155.html#p1030079

Hope it helps.
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Re: If 6 machines run at the same constant rate, they can [#permalink] New post 26 Feb 2012, 07:56
rtaha2412 wrote:
If 6 machines run at the same constant rate, they can complete a job in 8 hours. If only 5 of these machines run at this rate, how many more minutes will be required to complete the same job?

A. 38
B. 72
C. 80
D. 90
E. 96


On the same lines as Bunuel.
The new rate is 5/6 of the earlier. The new total time will be 6/5 of original.
To get the extra time 6/5 - 1 = 1/5. It will take 1/5 (20%) of original time to complete the same task. 1/5 * 480 = 96mins
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Re: If 6 machines run at the same constant rate, they can [#permalink] New post 01 Jul 2012, 00:18
if 6 machines need 8 h, then 1 machine needs 6*8=48 hours

48/5 -8=9 3/5-8=8/5 h

8/5h*60=96
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Re: If 6 machines run at the same constant rate, they can [#permalink] New post 15 Nov 2012, 00:40
Setup the rate equation for 6 machines working for 8 hours to finish a job:
(\frac{1}{m}+\frac{1}{m}+\frac{1}{m}+\frac{1}{m}+\frac{1}{m}+\frac{1}{m})(8hours)=1==>m=48hours

Setup the rate equation for 5 machines to finish the same job:
\frac{5}{48}t=1==>t=\frac{48}{5}=9.6hours

ANswer: 9.6 - 8 = 1.6 hours or 96 minutes (E)
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Re: If 6 machines run at the same constant rate, they can   [#permalink] 15 Nov 2012, 00:40
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