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Six machines, each working at the same constant rate [#permalink]

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05 Feb 2012, 23:28

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Six machines, each working at the same constant rate, together can complete a job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?

Welcome to GMAT Club. Below is a solution for the question. Hope it helps

Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days? A. 2 B. 3 C. 4 D. 6 E. 8

Let \(x\) be the time needed for 1 machine to complete the job, so rate of one machine is \(\frac{1}{x}\) (rate is the reciprocal of time) --> rate of 6 machines would be \(\frac{6}{x}\).

As \(job=time*rate\) --> \(1=12*\frac{6}{x}\) --> \(x=72\) days needed for 1 machine to complete the job.

To complete the job in 8 days \(\frac{72}{8}=9\) machines are needed.

Six machines, each working at the same constant rate, together c [#permalink]

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30 Apr 2013, 23:27

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njss750 wrote:

6 machines each working @ the same constant rate together can complete a certain job in 12 days. How many additional machines are reqd (each working at same constant rate) to finish the job in 8 days? 2 3 4 6 8

Pl xplain with wkgs

6 x 12 = (6+x) x 8

9 = 6+x

x=3 _________________

Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

Re: Six machines, each working at the same constant rate [#permalink]

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12 Aug 2013, 11:52

srkaleem wrote:

Six machines, each working at the same constant rate, together can complete a job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?

A. 2 B. 3 C. 4 D. 6 E. 8

Somehow, I found below method to be working for me - 6 Machines in 12 Days do 1 Work i.e. in notation form 6M * 12D = 1W, so 1M * 1D = [1/(6*12)] W How many Machines in 8 Days will complete 1 Work? i.e. xM * 8D = [(x*8)/(6*12)] W = 1W

Solving for x in [(x*8)/(6*12)] = 1, we get x = 9 i.e. 9 Machines in 8 Days will do 1 Work. So 9-6=3 Machines more are required.

Re: Six machines, each working at the same constant rate [#permalink]

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12 Aug 2013, 12:50

srkaleem wrote:

Six machines, each working at the same constant rate, together can complete a job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?

Re: Six machines, each working at the same constant rate [#permalink]

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14 Jan 2016, 09:38

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Re: Six machines, each working at the same constant rate [#permalink]

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19 Feb 2016, 10:18

Not sure if this helps or over-complicates but 6 = 3*2 12 = 3*2*2

so if you move one of the 2s over to the 12 side and one of the three over to the 6 side, you would have 9*8. Basically, if one part of the multiplication goes down, the other part must go up in order to equal the same job. Might be overkill in this simple question but could come in handy on a harder one.

Re: Six machines, each working at the same constant rate [#permalink]

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11 Jul 2016, 16:02

This is an inversely Proportion exercise. Days=K/Machines (as number of machines goes up, number of days goes down). 12=K/6, so K=72. Then using the same formula: 8=K/X, 8=72/X, X=9 machines. You need to add 3 to the initial 6 machines to get the job done in 8 days.

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