Is |x|+|y| > |x-y|?
(1) |x| > |y|
(2) |x-y| < |x|
What is the better approach to resolve this inequality?
Is |x|+|y| > |x-y|? (1) |x| > |y| (2) |x-y| < |x|
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Hi ziyuenlau,ziyuenlau wrote:Is |x|+|y| > |x-y|?
(1) |x| > |y|
(2) |x-y| < |x|
What is the better approach to resolve this inequality?
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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|x|+|y| = nonnegative + nonnegative = nonnegative.ziyuenlau wrote:Is |x|+|y| > |x-y|?
(1) |x| > |y|
(2) |x-y| < |x|
|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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There is no such thing as a "better" approach or the "best" approach.ziyuenlau wrote:What is the better approach to resolve this inequality?
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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