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If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x [#permalink]
14 Dec 2007, 21:37

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

Please explain how to the first one. And is there a fast way to do the second?

If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x

------------- If x and y are integers, is xy + 1 divisible by 3? (1) When x is divided by 3, the remainder is 1. (2) When y is divided by 9, the remainder is 8.

Re: Absolute value DS [#permalink]
15 Dec 2007, 01:42

aliensoybean wrote:

Please explain how to the first one. And is there a fast way to do the second?

If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x

(D) too

|x| < 1 ?

Stat 1 x^2 < 1
<=> sqrt(x^2) < 1 as sqrt(x) inscreases when x increases
<=> |x| < 1

SUFF.

Stat 2 |x| < 1/x

Implies that x > 0 as 0 =< |x| < 1/x and x != 0

So,
|x| < 1/x
<=> 0 < x < 1/x as x > 0.... then multiplication by x
<=> x^2 < 1 as x > 0
<=> sqrt(x^2) < 1 as sqrt(x) inscreases when x increases
<=> |x| < 1

Re: Absolute value DS [#permalink]
15 Dec 2007, 01:43

but the question si whether XY+1 is divisible by 3 or not . So add 1 to the value of XY. ==>8+1=9 ... so all terms in the equation are divisible by 3. Hence the answer should be C

gmatclubot

Re: Absolute value DS
[#permalink]
15 Dec 2007, 01:43

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