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Inequality
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sivaelectric
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Thanks to all.
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ronnie1985
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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)
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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karthikpandian19
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This is much simple to understand with plugging numbers
GMATGuruNY wrote:Statement 1: x^2 is greater than x^3gmatusa2010 wrote:Is x^2 greater than x ?
(1) x^2 is greater than x^3.
(2) x^2 is greater than x^4.
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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Karthik
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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.
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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smvjkumar
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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
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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Brent@GMATPrepNow
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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.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
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
Brent Hanneson - Creator of GMATPrepNow.com
