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If x+y/n < x+z/n, is n positive?

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If x+y/n < x+z/n, is n positive?  [#permalink]

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New post 04 Aug 2019, 10:56
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If \(\frac{x+y}{n}<\frac{x+z}{n}\), is n positive?

1) \(x^2-z+y<0\)

2) \(xy<xz\)

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If x+y/n < x+z/n, is n positive?  [#permalink]

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New post Updated on: 05 Aug 2019, 22:23
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\(\frac{x+y}{n}-\frac{x+z}{n}<0\)

\(\frac{x+y-x-z}{n}<0\)

\(\frac{y-z}{n}<0\)

Hence if we know the sign of (y-z), we can tell the sign of n

Statement 1-
\(x^2\)< \(z-y\)
As \(x^2≥0\), hence z-y>0
or y-z<0
and n>0

Sufficient

Statement 2
x(y-z)<0
We have no idea about sign of x, hence we can't determine the sign of y-z or n

Insufficient.




GMATPrepNow wrote:
If \(\frac{x+y}{n}<\frac{x+z}{n}\), is n positive?

1) \(x^2-z+y<0\)

2) \(xy<xz\)

Originally posted by nick1816 on 04 Aug 2019, 16:24.
Last edited by nick1816 on 05 Aug 2019, 22:23, edited 1 time in total.
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Re: If x+y/n < x+z/n, is n positive?  [#permalink]

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New post 05 Aug 2019, 07:10
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GMATPrepNow wrote:
If \(\frac{x+y}{n}<\frac{x+z}{n}\), is n positive?

1) \(x^2-z+y<0\)

2) \(xy<xz\)


Target question: Is n positive?
This is a good candidate for rephrasing the target question.

Given: \(\frac{x+y}{n}<\frac{x+z}{n}\)

Rewrite as: \(\frac{x}{n}+\frac{y}{n}<\frac{x}{n}+\frac{z}{n}\)

Subtract \(\frac{x}{n}\) from both sides to get: \(\frac{y}{n}<\frac{z}{n}\)

There are two possible cases to consider: n is POSITIVE and n is NEGATIVE

If n is POSITIVE, we can multiply both sides by n to get: \(y < z\)
If n is NEGATIVE, we can multiply both sides by n to get: \(y > z\)

REPHRASED target question: Is y < z?

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

Statement 1: \(x^2-z+y<0\)
Add z to both sides to get: \(x^2+y<z\)
Subtract y from both sides to get: \(x^2<z-y\)
Since 0 ≤ x², we can write 0 ≤ x² < z - y
This tells us that: 0 < z - y
Add y to both sides to get: y < z
So, the answer to the REPHRASED target question is YES, y IS less than z
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT


Statement 2: \(xy<xz\)
There are several values of x, y, z and n that satisfy statement 2 (and the given information). Here are two:
Case a: x = 1, y = 2, z = 3 and n = 1. In this case, the answer to the REPHRASED target question is YES, y IS less than z
Case b: x = 1, y = 3, z = 2 and n = -1. In this case, the answer to the REPHRASED target question is NO, y is NOT less than z
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

RELATED VIDEO FROM MY COURSE


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Brent
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Re: If x+y/n < x+z/n, is n positive?   [#permalink] 05 Aug 2019, 07:10
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