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Is \(a > |b|?\)
(1) \(2^{a-b} > 16\)
(2) \(|a - b| < b\)
OA C
Source: e-GMAT
(1) \(2^{a-b} > 16\)
(2) \(|a - b| < b\)
OA C
Source: e-GMAT
Since this question contains an absolute value, we must be sure to think about NEGATIVE as well as positive possibilities.Is \(a > |b|?\)
First, get like bases:(1) \(2^{a-b} > 16\)
\(2^{a-b}>2^4\)
We can infer:
\(a-b>4\)
\(a>b+4\)
Since a is greater than b+4, it must also be greater than b itself. But be careful! That doesn't mean a>|b|.
Think of examples where a and b are both negative:
a = -1
b = -6
This satisfies \(a>b+4\) but not \(a > |b|\)
This tells us that b must be positive, because it's greater than some absolute value. But this could be true whether a is greater than or less than b:(2) \(|a - b| < b\)
a = 3
b = 2
|3 - 2| < 2
\(a > |b|?\) --> yes
a = 2
b = 3
|2 - 3| < 3
\(a > |b|?\) --> no
This is insufficient.
(1) and (2) together
If \(a>b+4\) and \(b>0\), then a must be a positive number greater than b. Thus, the answer to \(a > |b|?\) is yes.
The answer is C.
