Pls Help :Tough Manhattan DS

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Pls Help :Tough Manhattan DS

by himu » Fri Oct 17, 2014 9:15 pm

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by GMATGuruNY » Sat Oct 18, 2014 2:33 am
If ab ≠ 0 and a + b ≠ 0, is 1/(a+b) < 1/a + 1/b?

(1) |a| + |b| = a + b

(2) a > b
Statement 1: |a| + |b| = a + b
The equation above is valid only if a and b are BOTH POSITIVE.

Case 1: a=2, b=1
Substituting a=2 and b=1 into 1/(a+b) < 1/a + 1/b, we get:
1/(2+1) < 1/2 + 1/1
1/3 < 1/2 + 1.
YES.

Case 1 illustrates that -- since a>0 and b>0 -- 1/(a+b) < 1/a + 1/b can be rephrased as follows:
1/(greater positive denominator) < 1/(smaller positive denominator) + 1/(smaller positive denominator)
smaller value < bigger value + bigger value.

Since the left-hand side will always be less than the right-hand side, the answer to the question stem is YES.
SUFFICIENT.

Statement 2: a > b
Case 1 also satisfies statement 2.
In Case 1, the answer to the question stem is YES.

Case 2: a=2, b=-1
Substituting a=2 and b=-1 into 1/(a+b) < 1/a + 1/b, we get:
1/(2-1) < 1/2 + 1/-1
1 < -1/2.
NO.

Since the answer to the question stem is YES in Case 1 but NO in Case 2, INSUFFICIENT.

The correct answer is A.
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