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Inequalities...

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accounts
 Post subject: Inequalities...
 Post Posted: Fri Aug 20, 2010 7:42 am 
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Forum Guests


Posts: 9
If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
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Viswanathan.harsha
 Post subject: Re: Inequalities...
 Post Posted: Wed Sep 08, 2010 12:31 am 
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Course Students


Posts: 46
Is the answer E?
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rohinivt
 Post subject: Re: Inequalities...
 Post Posted: Wed Sep 08, 2010 10:09 pm 
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Students


Posts: 3
I do not know why didn't this question get much importance... Is this problem very simple?

My explanation is as follows. Correct me if I am wrong.

|X| < X^2. This means, +ve value of X is less than the square of X. This means that +ve value of X is not a fraction. {any fraction squared will result in smaller value}. For example: X = 0.5 > 0.25 = X^2

This also says that, X cannot be a fraction between -1 and 0. Then |X| will be greater than X^2. For example: X = -0.5. |-0.5| = 0.5 > 0.25 = (-0.5)^2. This means, X has to be less than -1 for the statement |X| < X^2 to be correct.

So the most likely answer is E.

But I feel that something may not correct. I have a feeling that only III (C) only correct. Any clarification?
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trang.kieu.phung
 Post subject: Re: Inequalities...
 Post Posted: Wed Sep 08, 2010 11:42 pm 
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Students


Posts: 14
accounts wrote:
If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only


I solve it this way:
|x| < x^2 <--> x^2 < x^4 <--> x^4 - x^2 > 0
<--> x^2(x^2 - 1) > 0
Now, bcuz x^2 > 0, we get: x^2 - 1 > 0 ---> I is true
When we solve this equation: x > 1 or x < -1 ---> III is true

So answer is E
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shrikrishna.u
 Post subject: Re: Inequalities...
 Post Posted: Mon Sep 13, 2010 12:35 pm 
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Students


Posts: 1
Why should option III be true?


Let us say x = 2 ( it is greater than -1),
then |x| = 2 and x^2 = 4 and |x| < x^2.

So it is not a must that x < -1. So E is out.

Is my understanding correct here? OR am I doing something wrong?
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parthatayi
 Post subject: Re: Inequalities...
 Post Posted: Mon Sep 13, 2010 10:54 pm 
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Course Students


Posts: 40
shrikrishna.u wrote:
Why should option III be true?


Let us say x = 2 ( it is greater than -1),
then |x| = 2 and x^2 = 4 and |x| < x^2.

So it is not a must that x < -1. So E is out.

Is my understanding correct here? OR am I doing something wrong?


Hi Shrikrishna,

Consider this:
give any modulus problem, consider two cases.
1) x<0
2) x>0

In this problem lets consider x>0

=> x<x^2
=> x<0 and x>1 -----(i)

Next consider the case x<0

=> -x < x^2
=> x>0 and x<-1 -----(ii)

Combining the results in i and ii
we get
x<-1 and x>1
x<-1 which is option 3 and
x>1 => x^2 > 1 which is option 1

Hence the answer is both I and III

thanks and Regards,
Partha.
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tayhalla
 Post subject: Re: Inequalities...
 Post Posted: Wed Sep 15, 2010 6:11 pm 
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Prospective Students


Posts: 1
I'm sorry, but maybe I'm just not getting this thread.

The Answer is A

It can't be option III because in the question it uses the word "must" not "may"

While option III may be true, it is not a universal condition given the original EQ.
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RonPurewal
 Post subject: Re: Inequalities...
 Post Posted: Thu Sep 16, 2010 8:03 am 
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ManhattanGMAT Staff


Posts: 7146
this is a pretty cool problem.

Quote:
If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1


one key fact that you can use here:
x^2 = |x| * |x|
this statement is ALWAYS true.

therefore, you can rephrase and simplify the prompt question in the following way:
|x| < |x| * |x|

now, realize that x can't be 0 (since this inequality would be false if x were 0). therefore, |x| can't be 0, and so it is safe to divide by |x|.
therefore, divide by |x| on both sides, giving
1 < |x|

therefore, the statement in the prompt just means |x| > 1
which can also be written
x > 1 OR x < -1.

once you figure this out, it can be seen at once that (i) must be true, while (ii) and (iii) don't.

--

by the way

if you don't arrive at the above simplification, it shouldn't be very difficult to solve this problem by picking a few numbers and watching what happens.
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