mode inequality

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by Atekihcan » Thu May 23, 2013 1:42 am
a = 2 and b = -1 ---> |a| + |b| = 2 + 1 = 3 > |a + b| = |2 - 1| = 1 ---> YES
a = -2 and b = -1 ---> |a| + |b| = 2 + 1 = 3 = |a + b| = |-2 - 1| = |-3| = 3 ---> NO

Both of the above cases satisfy both the statements but the answer to question is different in either cases.

So, both statements together is also not sufficient.

Answer : E

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by GMATGuruNY » Thu May 23, 2013 2:36 am
vipulgoyal wrote:Is | a | + | b | > | a + b| ?
(1) a^2 > b^2
(2) | a | × b < 0

My take E
Because both sides of the inequality in the question stem have absolute value, we can rephrase the question stem by squaring the inequality:
(|a| + |b|)² > |a+b|²
a² + 2|a||b| + b² > a² + 2ab + b²
2|a||b| > 2ab
|a||b| > ab.
The resulting inequality will be true only if a and b have different signs.

Question rephrased: Do a and b have different signs?

Both statements are satisfied if a=-2 and b=-1.
Both statements are satisfied if a=2 and b=-1.
Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.
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