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If machine A and machine B can finish the task in 4 hours when working

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If machine A and machine B can finish the task in 4 hours when working  [#permalink]

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New post 06 Jul 2017, 07:59
4
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

76% (01:01) correct 24% (01:27) wrong based on 123 sessions

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If machine A and machine B can finish the task in 4 hours when working together at their constant rates, in how many hours can machine B finish the task alone?
1) Machine A can finish the task in 6 hours alone
2) The hours that it takes machine B to finish the task alone is 6 hours longer than the hours that it takes machine A to finish the task alone
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Re: If machine A and machine B can finish the task in 4 hours when working  [#permalink]

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New post 06 Jul 2017, 10:53
Given: Ra +Rb = 1/4

A. Ra= 1/6 --> 1/6 + 1/Tb = 1/4 --> Suff
B. Tb = Ta + 6
Rb = 1/Tb
Ra = 1/(Tb -6)
1/(Tb-6) + 1/Tb = 4
==> (Tb-12)(Tb-2)=0
Tb=2 is invalid for Ta cannot be negative (Ta=Tb -6)
Ans D
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Re: If machine A and machine B can finish the task in 4 hours when working  [#permalink]

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New post 06 Jul 2017, 12:08
By the work equation, we have

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{4}\)

Since the question is asking for \(y\), we only need to eliminate one of the variables to solve the problem.

(1) \(\frac{1}{6} + \frac{1}{y} = \frac{1}{4}\)

Sufficient: This gives us \(x\), so we can now solve for \(y\).

(2) \(\frac{1}{x} + \frac{1}{x+6} = \frac{1}{4}\)

Sufficient: This gives us \(y\) in terms of \(x\) so we can now solve for \(x\) and substitute to solve \(y = x + 6\).

Therefore, D
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Re: If machine A and machine B can finish the task in 4 hours when working  [#permalink]

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New post 26 Oct 2018, 04:56
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Re: If machine A and machine B can finish the task in 4 hours when working   [#permalink] 26 Oct 2018, 04:56
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