Is |a| > |b|? (1) b < -a (2) a < 0

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Is |a| > |b|? (1) b < -a (2) a < 0 [#permalink]

27 Dec 2006, 10:39

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Is |a| > |b|?

(1) b < -a

(2) a < 0

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Re: Absolutes DS [#permalink]

27 Dec 2006, 10:52

edgetech wrote:

Is |a| > |b|?

(1) b < -a

(2) a < 0

I go for E.

(1) b < -a

If a = -2, b <-(-2) which could be 1 or -10. So |a| is not necessarily greater than |b|

If a = 2, b<-2, meaning b <= -3 indicating that |b| is greater than |a|

(2) a<0 which does not tell us anything.

Combining, b<-a where a<0.

Say a = -2, b < 2, which could again be 1 or -10.So |a| is not necessarily greater than |b|.
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Re: Absolutes DS [#permalink]

27 Dec 2006, 10:57

(1) b < -a

Now, let's say a=-5; therefore b<5
but we don't know the range of b so we don't know |a| > |b|
if a =5 then b < -5
so |a| > |b|
(1) is in insufficient.

(2) a < 0 This doesn't say anything about b so insifficient.

Combining (1) and (2) we know that a is a negetive number. But we don't know |a| < |b|

(E) is my answer.
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[#permalink]

29 Dec 2006, 15:05

another E..

(-1,-2), (-1,0) for (a,b)

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[#permalink]

31 Dec 2006, 00:58

Yep. ItÂ´s E. If it had been b<a in (1), then the answerÂ´d have been C.

[#permalink] 31 Dec 2006, 00:58

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Is |a| > |b|? (1) b < -a (2) a < 0

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Is |a| > |b|? (1) b < -a (2) a < 0

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