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

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

by guerrero » Wed Feb 13, 2013 7:17 am
Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|



IS there a way we can solve it without plugging in the numbers . Please help.

OA E
Source: — Data Sufficiency |

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by GMATGuruNY » Wed Feb 13, 2013 10:18 am
guerrero wrote:Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|
Is |x+y| > |x-y|?
When there is absolute value notation on each side, we can square the inequality.

(x+y)² > (x-y)²
x² + 2xy + y² > x² - 2xy + y²
4xy > 0
xy > 0.

Question rephrased: Do x and y have the same sign?

Statement 1: |x| > |y|
Here, x and y could have the same sign or different signs.
INSUFFICIENT.

Statement 2: |x-y| < |x|
Squaring both sides, we get:
(x-y)² < x²
x² - 2xy + y² < x²
y² < 2xy
xy > y²/2.
Since the square of a value cannot be negative, y²/2 cannot be negative.
Thus, xy>0, implying that x and y have the same sign.
SUFFICIENT.

The correct answer is B.
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by ceilidh.erickson » Thu Feb 21, 2013 1:31 pm
We can certainly solve without plugging in numbers... but that doesn't necessarily mean that we have to do the algebra, either! Just think CONCEPTUALLY:

For what numbers will the absolute value of their sum be greater than the absolute value of the difference? Think positives/negatives:
- if both positive, this will be true: |3 + 2| > |3 - 2|
- if both negative, this will be true |-3 + -2| > |-3 - -2|
- if one positive and one negative, this will not be true: |-3 + 2| < |-3 - 2|
(we didn't really have to test numbers to see that, I'm just using them to illustrate)

In other words, the question is asking: do x and y have the SAME SIGN? (This is the same question we'd end up with by using algebra, as Mitch showed).

(1) |x| > |y|
This tells us nothing about the signs. Insufficient.

(2) |x - y| < |x|
For what kinds of numbers will this be true? Think conceptually about positives/negatives.
- if both positive, this will be true
- if both negative, this will be true
- if one positive and one negative, this will not be true

This tells us that x and y must have the same sign. Sufficient.

The answer is B.
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Is |x+y| > |x-y|?

by GMATGuruNY » Thu Feb 21, 2013 3:00 pm
Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|
We could also treat this as a DISTANCE/NUMBER LINE problem.
|x-y| = the distance between x and y.
|x+y| = the distance between x and -y.
|x| = the distance between x and 0.

Is |x-y| < |x+y|?
In words:
Is the distance between between x and y less than the distance between x and -y?
For x to be CLOSER TO Y than to -y, x and y must be TO THE SAME SIDE OF 0.
To illustrate:
x.....y..............0.............-y
.......y.....x.......0.............-y
As these examples show, when x and y are to the same side of 0, x is closer to y than to -y.

Question rephrased: Are x and y to the same side of 0?

Statement 1: |x| > |y|
x and y could be to the same side of 0 or on opposite sides of 0.
INSUFFICIENT.

Statement 2: |x-y| < |x|
In words:
The distance between x and y is less than the distance between x and 0.
This will not be true if x and y are on opposite sides of 0.
To illustrate:
x.........0...........y
y.........0..........x
In each case here, the distance between x and y is GREATER than the distance between x and 0.
Thus, to satisfy statement 2, x and y must be TO THE SAME SIDE OF 0.
SUFFICIENT.

The correct answer is B.
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