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Inequalities Question

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by bpolley00 » Tue Jun 04, 2013 2:17 pm
IF |X|>3, which of the following must be true?
A) X>3
B) X^2>9
c) |X-1|>2
I only
II only
I and II only
II and III only
I, II, and III

How is this II and III and not just 2?

If you work out C it comes out to being X<-1 X>3. So, if X= -2 then the first condition would not be true, correct?

Can someone please touch on this; although, I think after further review it is just saying that condition 3 can also be true as all the values are also in value one?
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by vivekchandrams » Tue Jun 04, 2013 9:41 pm
Hi bpolley00,

Hope you are fine with I and II.
Now consider III.

|x-1|>2.

Method A): Assign random values to x such that the stem is true.
So, let x be 3.1
Now |x|>3. Now substitute that in 'C'. The statement turns true.
Now lets choose x as -3.1. Substitute it and even that works out.

Method B: It's the one which you followed, by simplifying the given inequality.

(x-1)<-2 and (x-1)>2. That implies what you said i.e., 3<x<-1.

Now comes your issue with -2. I agree with you that x=-2 definitely makes III true but that value of x doesn't satisfy the stem, which says |x|>3. Hence x=-2 is not applicable.

Hope it answers you.

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by GMATGuruNY » Wed Jun 05, 2013 3:33 am
bpolley00 wrote:IF |X|>3, which of the following must be true?
A) X>3
B) X^2>9
c) |X-1|>2
I only
II only
I and II only
II and III only
I, II, and III
Constraint: |x| > 3
This means that the distance between x and 0 is greater than 3.
Any value in the two red ranges below satisfies this constraint:
<----(-3).......(3)---->

I: x>3
The red range on the left illustrates that x does not have to be greater than 3.
Eliminate A, C, and E.

III: |x-1| > 2.
This statement implies that the distance between x and 1 must be greater than 2.
Every value in the red ranges above is more than 2 places away from 1.
Thus, statement III must be true.
Eliminate B.

The correct answer is D.
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