Is |x|+|y| > |x-y|? (1) |x| > |y| (2) |x-y| < |x|

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Is |x|+|y| > |x-y|?

(1) |x| > |y|
(2) |x-y| < |x|

What is the better approach to resolve this inequality?

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by Jay@ManhattanReview » Thu Apr 06, 2017 12:30 am
ziyuenlau wrote:Is |x|+|y| > |x-y|?

(1) |x| > |y|
(2) |x-y| < |x|

What is the better approach to resolve this inequality?
Hi ziyuenlau,

I think testing extreme values would work for statement 1, while conceptual approach would do for statement 2.

Question: Is |x|+|y| > |x-y|

Statement 1: |x| > |y|

Case 1: Say x = 3 and y =2.

=> |x|+|y| > |x-y| => |3|+|2| ? |3-2| => 5 > 1. Answer is Yes.

Case 1: Say x = -3 and y =2.

=> |x|+|y| > |x-y| => |-3|+|2| ? |-3-2| => 5 = 5. Answer is No. Insufficient.

Statement 2: |x-y| < |x|

=> |x| > |x-y|

Since |y| is a non-negative quantity, and |x| > |x-y|, thus, |x| + |y| > |x-y|. The answer is Yes. Suffcient.

The correct answer: B

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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by GMATGuruNY » Thu Apr 06, 2017 3:06 am
ziyuenlau wrote:Is |x|+|y| > |x-y|?

(1) |x| > |y|
(2) |x-y| < |x|
|x|+|y| = nonnegative + nonnegative = nonnegative.
|x-y| = nonnegative.
Since both sides of the question stem are nonnegative, we can SQUARE THE INEQUALITY:
(|x|+|y|)² > |x-y|²
x² + y² + 2|x||y| > x² + y² - 2xy
2|x||y| > -2xy
|x||y| > -(xy).

The inequality in blue will hold true only if -(xy) is negative.
For -(xy) to be negative, xy must be POSITIVE.
Question stem, rephrased:
Do x and y have the SAME SIGN?

Statement 1: |x| > |y|

If x=2 and y=1, then x and y have the same sign.
If x=-2 and y=1, then x and y do NOT have the same sign.
INSUFFICIENT.

|a-b| = the distance between a and b.
|a| = the distance between a and 0.

Statement 2: |x-y| < |x|
In words:
The distance between x and y is less than the distance between x and 0.
Put another way:
x is closer to y than to 0.

For x to be closer to y than to 0, x and y must be TO THE SAME SIDE OF 0, as in the following:
y--x----0------
x--y----0------
------0----y--x
------0----x--y

If x and y are on opposite sides of 0, then x will NOT be closer to y than to 0:
x-----0--y-----
Here, x is closer to 0 than to y.

Since x and y must be to the same side of 0, they have the SAME SIGN.
SUFFICIENT.

The correct answer is B.
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by GMATGuruNY » Thu Apr 06, 2017 3:23 am
ziyuenlau wrote:What is the better approach to resolve this inequality?
There is no such thing as a "better" approach or the "best" approach.
The best approach for one test-taker might not be the best approach for another.
Whereas some test-takers will fare better with an algebraic approach, others will find it easier and faster to test cases.
You need to determine the best approach for YOU.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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