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by fambrini » Mon Oct 10, 2016 5:44 pm
Machine X and machine Y, working at the same time, produce w widgets in k hours. Machines A, B and C, together produce w widgets in k - 2 hours. If all the machines have the same productivity, what is the value of K?

A) 2
B) 3
C) 6
D) 8
E) 9

OA: C
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by GMATGuruNY » Mon Oct 10, 2016 6:36 pm
fambrini wrote:Machine X and machine Y, working at the same time, produce w widgets in k hours. Machines A, B and C, together produce w widgets in k - 2 hours. If all the machines have the same productivity, what is the value of K?

A) 2
B) 3
C) 6
D) 8
E) 9
Let the rate for each machine = 1 widget per hour.

Combined rate for X and Y = 2 widgets per hour.
Combined rate for A, B and C = 3 widgets per hour.

At a rate of 2 widgets per hour, the amount of work produced by X and Y in k hours = rt = 2k.
At a rate of 3 widgets per hour, the amount of work produced by A, B and C in k-2 hours = rt = 3(k-2) = 3k-6.

Since the amount of work in each case is THE SAME, we get:
2k = 3k-6
6 = k.

The correct answer is C.
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by DavidG@VeritasPrep » Mon Oct 10, 2016 7:07 pm
fambrini wrote:Machine X and machine Y, working at the same time, produce w widgets in k hours. Machines A, B and C, together produce w widgets in k - 2 hours. If all the machines have the same productivity, what is the value of K?

A) 2
B) 3
C) 6
D) 8
E) 9

OA: C
And there's always good gold-fashioned algebra. If the rate for one machine is 'r,' and all the rates are the same, then we can knock out some simple equations.

X + Y working together: r + r = w/k --> 2r = w/k ---> r = w/2k

A + B + C working together: r + r + r = w/(k-2) ---> 3r = w/(k-2)

Substitute w/2k in place of r in the second equation: 3w/2k = w/k-2 --> 3/2k = 1/(k-2) ---> cross-multiply: 3k - 6 = 2k --> k = 6; Answer is C
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