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Question

If x is not equal to 0, is |x| less than 1? (1) x/|x| < x (2) |x| > x OA C

austin · Answer

If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x

Stem translation: -1<x<1????

1. |x| is always +ve. x can be positive or negative

x = 3; x/|x| = 3/3 = 1< x (=3); x lies away from -1 and 1
x = -1/2; x/|x| = (-1/2)/(1/2) = -1< x (=-1/2); x lies between -1 and 1
INCONSISTENT

2. x = -3
|-3|>-3; x lies way from -1 and 1
x = -1/2
|-1/2| > -1/2; x lies between -1 and 1

Combining,

x/|x| < x and |x| > x, x has to be negative and x has to lie between 1 and -1.

x = -1/2; x/|x| = (-1/2)/(1/2) = -1< x (=-1/2); x lies between -1 and 1
x = -1/2; |-1/2| > -1/2; x lies between -1 and 1

x = -3/4; x/|x| = (-3/4)/(3/4) = -1< x (=-3/4); x lies between -1 and 1
x = -3/4; |-3/4| > -3/4; x lies between -1 and 1

jimmiejaz · Answer

One more approach.
we have to find Is |x| < 1?

from 1 we have
x/|x| < x
If x>0, |x| = x
x/x < x or x>1
If x<0, |x| = -x
x/-x<x or x>-1, but we taken x<0 so, limit is -1<x<0

But still we get either x>1 or -1<x<0 hence insuff..

From 2 we have
|x|>x which is only possible if x is negative.
this eqn gives us the range x<0. But x can be -1,-2....... -1000 and so on.
Hence insuff.

Combining we get the range as -1<x<0
Hence suff.
C.

vittalgmat · Answer

Here is my approach,
can somebody let know if this is correct.

Stmt 1: x/|x| < x
Since |x| is always positive, we can cross multiply without affecting the sign.
So we get
x < x * |x|

**Not sure of this step***
1 < |x|
ie |x| > 1

case 1: x is +ve
x > 1
case 2: x is -ve
-x > 1
x < -1

So x >1 or x < -1 ----------(1)
Not sufficient.

Stmt 2:
|x| > x
Ths is possible only if x is -ve.

not sufficient.

Combining,
x < -1

So any -ve number < -1 ie -2, ... -infinity will suffice.

So to answer the question |x| < 1, plug in -ve number less than 1 and see the result.
|-2| = 2 NOT < 1.. This is true for all negative numbers.

So sufficient.
ans is C

Ian Stewart · Answer

vittalgmat wrote:
So we get
x < x * |x|

**Not sure of this step***
1 < |x|
ie |x| > 1

Yes, that last step isn't quite right. You arrived correctly at the inequality:

x < x*|x|

but you then divided by x on both sides. You can't do that, because you don't know if x is positive or negative; if x is negative, you'd need to reverse the inequality. We can divide by x as long as we split the inequality into two cases:

1 < |x| if x is positive (so if x is positive, x must be greater than 1)
1 > |x| if x is negative (so if x is negative, x must be between 0 and -1)

So this statement is insufficient: |x| could be less than 1 if x is negative, and could be greater than 1 if x is positive. Since statement 2 guarantees that x is negative, the two statements together are sufficient, and give a 'yes' answer to the question.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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