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CEO  V
Joined: 12 Sep 2015
Posts: 3922
Location: Canada
If x+y/n < x+z/n, is n positive?  [#permalink]

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Top Contributor
3 00:00

Difficulty:   75% (hard)

Question Stats: 49% (02:11) correct 51% (02:07) wrong based on 39 sessions

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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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Director  P
Joined: 19 Oct 2018
Posts: 791
Location: India
If x+y/n < x+z/n, is n positive?  [#permalink]

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2
$$\frac{x+y}{n}-\frac{x+z}{n}<0$$

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

$$\frac{y-z}{n}<0$$

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.
CEO  V
Joined: 12 Sep 2015
Posts: 3922
Location: Canada
Re: If x+y/n < x+z/n, is n positive?  [#permalink]

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Top Contributor
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

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