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by akash singhal » Thu Oct 29, 2015 4:35 am
If 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

Please Explain?
OE B
I doubt that
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by GMATGuruNY » Thu Oct 29, 2015 4:59 am
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by theCEO » Thu Oct 29, 2015 5:03 am
akash singhal wrote:If 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

Please Explain?
OE B
I doubt that
x/|x|<x

If x<0, the expression becomes:
x/|x|<x = x/-x<x = -1<x which is choice B

If x>0, the expression becomes
x/|x|<x = x/x<x = 1<x which is choice A

when x<0, -1<x<0
when x>0, 1<x

only (B) is always true

A. x>1 : we can have x=-1/2 which would make this wrong
B. x>-1: pick any value over -1 (except for 0!)
C. |x|<1: we can have |x|=2 which would make this wrong
D. |x|=1: we can have |x|=2 which would make this wrong
E. |x|^2>1: we can have |x|=-1/4 which would make this wrong

ans = b
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by [email protected] » Thu Oct 29, 2015 11:11 am
Hi akash singhal,

This question is 'quirky' in that the correct answer choice does NOT describe every possible value of X. The correct answer does "contain" all of the possible values of X though (as well as some values that X CANNOT be).

Here's how you can get to the correct answer by TESTing VALUES.

We're told that X/|X| < X

IF...
X = 2
2/|2| = 1 and 1 < 2
So X COULD be 2
Eliminate C and D

Since the inequality includes an Absolute Value, there's a high likelihood that there are NEGATIVE solutions...

IF...
X = -2
-2/|-2| = -1 BUT -1 is NOT < -2
So -2 is NOT a possible solution

Looking at the fraction, you can deduce that ANY negative value that you TEST will make X/|X| = -1, so we should be looking for a negative value that is BIGGER than -1...

IF...
X = -1/2
(-1/2)/|-1/2| = -1 and -1 < -1/2
So -1/2 is also a possible solution
Eliminate A and E

There's only one answer left....

Final Answer: B

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by Matt@VeritasPrep » Fri Oct 30, 2015 12:05 am
I like algebra here: let's consider two cases.

x/|x| < x

Since |x| is positive, we can multiply both sides by |x|:

x < |x|*x

Now we've got our two cases: x is positive or x is negative. If x is positive, we can divide both sides by x, and get

1 < |x|

Since we assumed x is positive, |x| = x, and we have x > 1.

Now let's consider x's being negative. If x is negative, we divide both sides by x and get

1 > |x|

or 1 > |x| > 0

Since x itself is negative, we have -1 < x < 0.

In either case, x > -1, so B must be true.
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