Is \(a > |b|?\)

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Is \(a > |b|?\)

by BTGmoderatorDC » Wed May 22, 2019 3:22 am

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Is \(a > |b|?\)

(1) \(2^{a-b} > 16\)
(2) \(|a - b| < b\)


OA C

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by ceilidh.erickson » Sat May 25, 2019 12:05 pm

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Is \(a > |b|?\)
Since this question contains an absolute value, we must be sure to think about NEGATIVE as well as positive possibilities.
(1) \(2^{a-b} > 16\)
First, get like bases:
\(2^{a-b}>2^4\)

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 \(a > |b|\)
(2) \(|a - b| < b\)
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:
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.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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by ceilidh.erickson » Sat May 25, 2019 12:10 pm

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Ceilidh Erickson
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Harvard Graduate School of Education