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Re: Integers a and b are such that a - b > 0. Is |a| > |b| ? [#permalink]
B is correct, the key here is a + b = 12.
since a > b => a >6 > b and |a| = a
now, the question is a > |b|?

we have, a = 12 - b, if b < 0, then a = 12 + |b| => a > |b| => solved
if b > 0, then a > b > 0 => a > |b|
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Re: Integers a and b are such that a - b > 0. Is |a| > |b| ? [#permalink]
alphonsa wrote:
Integers a and b are such that a - b > 0. Is |a| > |b| ?

(1) ab > 0
(2) a + b = 12

Source : 4Gmat


#1
ab can be both -ve or +ve
not sufficient
#2
given a-b>0 so a>b
so a+b=12
means
|a| > |b|
sufficient
IMO B
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Re: Integers a and b are such that a - b > 0. Is |a| > |b| ? [#permalink]
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