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Inequality

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by sivaelectric » Sat May 28, 2011 1:50 am
Thanks to all.
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by Rastis » Wed Jan 25, 2012 11:04 am
Mitch,

How did you know to use 1/2 and -1/2 as the numbers you plugged in?
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by ronnie1985 » Thu Apr 26, 2012 10:56 am
S1. x^2 > x^3
This is true for x<0 or 0<x<1
For x<0, x^2>x but for 0<x<1, x^2<x. Not Sufficient

S2. x^2>x^4
This is true for -1<x<1
x^2>x for negative x in the domain
x^2<x for positive x in the domain
Not Sufficient

(E)
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by karthikpandian19 » Sun Jun 03, 2012 11:42 pm
This is much simple to understand with plugging numbers
GMATGuruNY wrote:
gmatusa2010 wrote:Is x^2 greater than x ?
(1) x^2 is greater than x^3.
(2) x^2 is greater than x^4.
Statement 1: x^2 is greater than x^3
x = 1/2 works, because (1/2)^2 > (1/2)^3. Is (1/2)^2 > 1/2? No.
x = -1/2 works, because (-1/2)^2 > (-1/2)^3. Is (-1/2)^2 > -1/2? Yes.
Since the answer can be both No and Yes, insufficient.

Statement 2: x^2 is greater than x^4
x = 1/2 works, because (1/2)^2 > (1/2)^4. Is (1/2)^2 > 1/2? No.
x = -1/2 works, because (-1/2)^2 > (-1/2)^4. Is (-1/2)^2 > -1/2? Yes.
Since the answer can be both No and Yes, insufficient.

Since 1/2 and -1/2 satisfy both statements, even when the 2 statements are combined, insufficient.

The correct answer is E.
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by anujan007 » Mon Aug 06, 2012 6:07 am
I plugged in 1/2 and -2 for both the options. Did not evaluate a negative fraction and therein made the mistake. I got B which is wrong.

The explanations above are really good. The plugging in approach is most suitable but one needs to be careful in selecting numbers so as to not miss any choice.

Good one.
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by sathishkumarjva9888 » Sat Aug 11, 2012 8:15 pm
Answer is E. Plugging in numbers makes life easier with inequality problems. :)
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by rajeshsinghgmat » Tue May 07, 2013 12:22 am
E the answer.

x=1/2 and x=-1/2
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by smvjkumar » Mon Jul 22, 2013 4:02 am
please tell me why the below approach is wrong

Dividing both the side of the inequality with same value makes no difference in the relation

So

(i) divide both The side by x

(x^2)/x > (x^3)/x => x > x^2
which gives answer for the question

similarly the same result is possible for (ii) as well
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by Brent@GMATPrepNow » Mon Jul 22, 2013 6:48 am
smvjkumar wrote:please tell me why the below approach is wrong

Dividing both the side of the inequality with same value makes no difference in the relation

So

(i) divide both The side by x

(x^2)/x > (x^3)/x => x > x^2
which gives answer for the question

similarly the same result is possible for (ii) as well
Your approach is based on the concept that "Dividing both the side of the inequality with same value makes no difference in the relation." However, this is not true.

Let's look at an example.

Start with the inequality: 6 < 18 (true)
Divide both sides by 2 to get: 3 < 9 (this new inequality is still true)
Divide both sides by -3 to get: -1 < -3 (this new inequality is NOT true)

If we divide (or multiply) both sides of an inequality by a NEGATIVE value, the inequality sign must be reversed. If we divide (or multiply) both sides of an inequality by a POSITIVE value, the inequality sign stays the same.

So, if we take the inequality, x^2 > x^3, and divide both sides by x (as you suggest), we need to first ask, "Is x positive or negative?" Since we don't know whether x is positive or negative, we have no idea what to do with the inequality sign (reverse it, or keep it the same), so we cannot definitively conclude that dividing both sides by x will yield the inequality x > x^2

I hope that helps.

Cheers,
Brent
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