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6 machines, all working at a constant rate, together can do

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6 machines, all working at a constant rate, together can do [#permalink] New post 10 Oct 2005, 06:49
6 machines, all working at a constant rate, together can do a job in 12 days. How many add'l machines, each working at the same constant rate, will be needed to finish the job in 8 days?
a. 2
b. 3
c. 4
d. 6
e. 8
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 [#permalink] New post 10 Oct 2005, 07:03
If 6 machines work at the same constant rate and all together do a job in 12 days, aech machine works at a constant rate of 1/(12*6) job per day. The new job must be done is 8 days and we will use n machines. Therefore, it must be true that

time=(amount of job)/rate of work

8=1 job / (n*1/72) ---> n=72/8 = 9

we have to add 2 machines.

B is the correct answer
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 [#permalink] New post 10 Oct 2005, 07:05
This question was discussed previously...

http://www.gmatclub.com/phpbb/viewtopic ... highlight=
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[#permalink] New post 10 Oct 2005, 07:23
B.3. There are many ways to go about getting this answer. The most confortable way for me, and it may not be the fastest, is to find what each machines individual rate is.

1(job)=rate*12(days)
rate=(x+x+x+x+x+x), where x=the rate of each individual machine.

Therefore, 1=(x+x+x+x+x+x)(12), deduce x=1/72.

Now that you have that, use:
1(job)=rate*8(days). You know that the rate=(x+x+x+x+x+x+ax), where a is the number of extra machines. Try values in the answer to see what works. I got 3, B. I would like to know how other s got this answer, I know there is a faster way, even though this took me about 1.5 minutes.
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 [#permalink] New post 10 Oct 2005, 07:51
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I did it similar to automan but slightly differently.

The job that needs to be done = J

J = 6machines * 12 days * CR (where CR is the constant rate)

Now to do the job in 8 days, let the total number of machines be x. In this case,
J = xmachines * 8 days * CR

Equating since the same job is being done.
xmachines * 8 days * CR = 6machines *12 days *CR

x machines = 6 * 12/8 = 3 * 3 = 9 machines.

So additional machines required is 9-6 = 3.
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 [#permalink] New post 10 Oct 2005, 07:54
Another B

I solved it by 1/72 = (1/8)/x


Brendan.
  [#permalink] 10 Oct 2005, 07:54
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6 machines, all working at a constant rate, together can do

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