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Given X/|X| < X. Which of the following must be true [#permalink ]
06 Aug 2006, 13:35

Given

X/|X| < X. Which of the following must be true about X?

A) X > 1

B) X > -1

C) |X| < 1

D) |X| = 1

E) |X|^2 > 1

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I will go for /x/<1

Senior Manager

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"I will go for /x/<1"
and why would you do that?

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X/|X| < X
divide by X on both sides 1/|x| < 1
multiply with |x| on both sides
1<|x| i.e. |x|>1
Hence will go with E

Director

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I will go with A
take examples X > 1 then any value > x 1.1 or 1 milllion will be both positive and the answer will be 1 so 1 < any value > x

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I will go for E
X/|X| < X implies
X/X < X or X/(-X) > X
so, x > 1 or x <-1. The only option which satisfies this condition is 'E'

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shoonya wrote:

Given X/|X| < X. Which of the following must be true about X? A) X > 1 B) X > -1 C) |X| < 1 D) |X| = 1 E) |X|^2 > 1

rewrite as x < x * |x|

or x < x*n - where n is any +ve number(integer or fraction)

only possible solution is when n > 1

A for me.

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shoonya wrote:

X/|X| < X. Which of the following must be true about X?A) X > 1 B) X > -1 C) |X| < 1 D) |X| = 1 E) |X|^2 > 1

straight A. no matter the value of x, x > x/lxl. since x>1 and x/lxl is always 1. so its pretty clear that x>x/lxl.

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X/|X| < X (

X!=0 )

1. if X > 0 then 1 < X (X/X < X => 1<X)

2. if X < 0 then -1 < X <0 (X/-X < X => -1 < X)

Combine 1 and 2

-1<X<0 and X>1

Thus A.

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A
Agree with proff. explanation

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Clear winner A.
A satisfies the condition perfectly

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Can any of you explain if what I did was actually wrong ?
i.e.
X/|X| < X
divide by X on both sides 1/|x| < 1
multiply with |x| on both sides
1<|x| i.e. |x|>1
Hence will go with E

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A is the only one which works
Take X = 3
3/|3|=1<3
Take x=-3
-3||-3|=-1>X
Take X =1
Then LHS = RHS
X>1 is the only one that works

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shoonya wrote:

Given X/|X| < X. Which of the following must be true about X? A) X > 1 B) X > -1 C) |X| < 1 D) |X| = 1 E) |X|^2 > 1

x< x|x|

when x> 0 |x| = x which implies x< x^2 or x(x-1) > 0 i.e x>1

when x<0 |x| = -x which implies x<-x^2 or x(x+1) < 0 i.e -1<x<0

Hence A

Heman

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gmatornot wrote:

Can any of you explain if what I did was actually wrong ? i.e. X/|X| < Xdivide by X on both sides 1/|x| < 1 multiply with |x| on both sides 1<|x| i.e. |x|>1 Hence will go with E

You can not divide by x, because you don't know the sign of x. If x is -ve you would need to flip the ineqality sign.

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Path wrote:

gmatornot wrote:

Can any of you explain if what I did was actually wrong ? i.e. X/|X| < Xdivide by X on both sides 1/|x| < 1 multiply with |x| on both sides 1<|x| i.e. |x|>1 Hence will go with E

You can not divide by x, because you don't know the sign of x. If x is -ve you would need to flip the ineqality sign.

agree that you cannot divide both sides by x. ok, lets suppose you are when x is +ve, but if x is a fraction x/lxl>x.

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I'll try the old plug numbers routine
2/l2l<2
and so on so my answer is A

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actually freethinking has the best explanation...and this how u shud approach || questions...

freetheking wrote:

X/|X| < X (X!=0 ) 1. if X > 0 then 1 < X (X/X < X => 1<X) 2. if X < 0 then -1 < X <0 (X/-X < X => -1 < X) Combine 1 and 2 -1<X<0 and X>1 Thus A.