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Working alone at its constant rate, machine A can produce 240 pins in

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Working alone at its constant rate, machine A can produce 240 pins in  [#permalink]

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New post 10 Jan 2018, 01:02
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[GMAT math practice question]

Working alone at its constant rate, machine A can produce 240 pins in x hours. Working alone at its constant rate, machine B can produce 240 pins in y hours. How many hours will it take machines A and B to produce 240 pins, when working together at their respective constant rates?

1) \(\frac{xy}{(x+y)}=4\)
2) \(x=20\)

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Re: Working alone at its constant rate, machine A can produce 240 pins in  [#permalink]

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New post 10 Jan 2018, 03:02
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MathRevolution wrote:
[GMAT math practice question]

Working alone at its constant rate, machine A can produce 240 pins in x hours. Working alone at its constant rate, machine B can produce 240 pins in y hours. How many hours will it take machines A and B to produce 240 pins, when working together at their respective constant rates?

1) \(\frac{xy}{(x+y)}=4\)
2) \(x=20\)


Since (1) looks like a very technical expression, we'll look for a technical, equation-intensive solution.
This is a Precise approach.

Machine A produces 240/x pins per hour and machine B produces 240/y pins per hour.
Together they produce 240/x + 240/y pins per hour which simplifies to 240(x+y)/xy pins per hour.
Since rate * time = work then time = work/rate
Therefore, to make 240 pins, they need to work for a total of 240 / (240(x+y)/xy) hours.
This simplifies to xy/(x+y) hours.
Therefore (1) is sufficient to answer the question and (2) is not.

(A) is our answer.
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Re: Working alone at its constant rate, machine A can produce 240 pins in  [#permalink]

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New post 12 Jan 2018, 00:49
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

We have \(\frac{1}{t} = \frac{1}{x} + \frac{1}{y} = \frac{(x+y)}{xy}\) or \(t = \frac{xy}{(x+y)}\), where t is the time that machines A and B take to produce 240 pins, when working together. The question asks for \(\frac{xy}{(x+y)}\).

Thus, only condition 1) is sufficient. Condition 2) is not sufficient as it only gives us information about \(x\).

Therefore, the answer is A.
Answer: A
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Re: Working alone at its constant rate, machine A can produce 240 pins in  [#permalink]

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New post 13 May 2019, 23:33
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Re: Working alone at its constant rate, machine A can produce 240 pins in   [#permalink] 13 May 2019, 23:33
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