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# If x + y is not equal to 0. What is the value of (ax + ay)/(x + y)?

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Intern
Joined: 04 Feb 2011
Posts: 49
Location: US
If x + y is not equal to 0. What is the value of (ax + ay)/(x + y)?  [#permalink]

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11 Mar 2011, 07:30
1
4
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Difficulty:

15% (low)

Question Stats:

72% (00:41) correct 28% (00:52) wrong based on 138 sessions

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If x + y is not equal to 0. What is the value of (ax + ay)/(x + y)?

(1) x = 4 and y = 5
(2) a = 6
Math Expert
Joined: 02 Sep 2009
Posts: 53792
Re: If x + y is not equal to 0. What is the value of (ax + ay)/(x + y)?  [#permalink]

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11 Mar 2011, 07:34
1
Lolaergasheva wrote:
If x+y is not equal to 0. What is the value of (ax+ay)/(x+y)

1) x=4 and y=5
2) a=6

$$\frac{ax+ay}{x+y}=\frac{a(x+y)}{x+y}=a$$, so we can see that the value of the fraction given doesn't depend on the values of $$x$$ and $$y$$ as $$x+y$$ is reduced.

(1) x=4 and y=5. Not sufficient.
(2) a=6. Sufficient.

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Re: If x + y is not equal to 0. What is the value of (ax + ay)/(x + y)?  [#permalink]

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02 Jan 2019, 07:17
Top Contributor
Lolaergasheva wrote:
If x + y is not equal to 0. What is the value of (ax + ay)/(x + y)?

(1) x = 4 and y = 5
(2) a = 6

Target question: What is the value of (ax + ay)/(x + y)?
This is a good candidate for rephrasing the target question.

Take the expression: (ax + ay)/(x + y)
Factor the numerator to get: a(x + y)/(x + y)
Simplify the fraction to get: a
Since (ax + ay)/(x + y) = a, we can rephrase the target question...
REPHRASED target question: What is the value of a?

Aside: the video below has tips on rephrasing the target question

Now that we've REPHRASED the target question, it should be easy to analyze the two statements.

Statement 1: x = 4 and y = 5
Since we have no information about a, we cannot answer the REPHRASED target question with certainty.
So, statement 1 is NOT SUFFICIENT

Statement 2: a = 6
PERFECT - this is exactly what we need.
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

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Re: If x + y is not equal to 0. What is the value of (ax + ay)/(x + y)?   [#permalink] 02 Jan 2019, 07:17
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