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

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If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x [#permalink]

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New post 14 Dec 2007, 21:37
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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.






**************
OA: D, C
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New post 14 Dec 2007, 21:45
Question #1

A. if X^2 <1 that means that X is a fraction between -1 and 1. It can't be -1, 1 or 0 so |x| with any fraction between -1 and 1 is less than 1.

SUFFICIENT

B. Again, for |x| < 1/x to be true then X needs to be a fraction. In this case it has to be a fraction between 0 and 1.

SUFFICIENT

Answer D


Question #2

1. this means that X is 4, 7, 10, 13, 16, 19, etc

INSUFFICIENT

2. this means that Y is 8, 17, 26, 35, 44, etc

INSUFFICIENT

now taken together...

grab some numbers from each list and find out:

(4*8)+1=33 which is divisible by 3
(7*8)+1=57 which is divisible by 3
(10*17) +1 = 171 which is divisible by 3

good enough for me

Answer C
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Re: Absolute value DS [#permalink]

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New post 14 Dec 2007, 22:58
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


(1) if x^2<1, x is a fraction. so suff.
(2) if |x| < 1/x, x is a +ve fraction. also suff
D.
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Re: Absolute value DS [#permalink]

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New post 14 Dec 2007, 23:23
aliensoybean wrote:
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.


C.

(1) x = 3k + 1
(2) y = 9m + 8

togather:
xy + 1 = (3k + 1) (9m + 8) = 27km+24k+9m+8

so the reminder is 2.
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New post 14 Dec 2007, 23:29
Tiger,

How did you get remainder of 2 from your equation?
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Re: Absolute value DS [#permalink]

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New post 15 Dec 2007, 00:30
GMAT TIGER wrote:
aliensoybean wrote:
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.


C.

(1) x = 3k + 1
(2) y = 9m + 8

togather:
xy + 1 = (3k + 1) (9m + 8) = 27km+24k+9m+8

so the reminder is 2.


in "27km+24k+9m+8" all except 8 are divisible by 3. so only consider 8, 2 is reminder if 8 is divided by 3.
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Re: Absolute value DS [#permalink]

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New post 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
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Re: Absolute value DS [#permalink]

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