That's what I got as well.

Here's the OA explanation. Anurag, Isn't the error in their method is that they divided by x^2? because that eliminates answer? When can you divide by X^2 and when can't you? I guess in an equation and you know for sure its not zero?



Explanation: x2 is greater than x for all numbers except for those values
of x between 0 and 1. Thus, we need to know whether or not x falls in that
range.

Statement (1) is insu¢ cient. To simplify, divide both sides by x2, resulting
in 1 > x. If thatÂ’s true, x could be between 0 and 1, but it could also be less
than 0.

Statement (2) is also insu¢ cient. Again, simplify by dividing by x2, which
gives you 1 > x2. Thus, x could be any number between -1 and 1. Again, it
could be between 0 and 1, but it could also be between -1 and 0.
Taken together, it’s still insu¢ cient. Both statements allow for the possi-
bility that x is between 0 and 1, but both statements also make it possible that
x is between -1 and 0. Choice (E) is correct.

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.

gmatusa2010 wrote:Is x^2 greater than x ?
(1) x^2 is greater than x^3.
(2) x^2 is greater than x^4.

x^2 > x ?

Statement 1:
x^2 is greater than x^3.

x = -2, x^2 = 4, x^3 = -8 => x^2 > x
x = 1/2, x^2 = 1/4, x^3 = 1/8 => x^2 < x -- Insufficient

Statement 2:
x^2 is greater than x^4.

x = 1/2, x^2 = 1/4, x^4 = 1/16 => x^2 < x
x = -1/2 , x^2 = 1/4, x^4 = 1/16 => x^2 > x --- Insufficient

Combining 1 and 2:
x^2 is greater than x^3 and x^4 :

x = 1/2, x^2 = 1/4, x^3= 1/8, x^4 = 1/16 => x^2 < x
x = -1/2 , x^2 = 1/4,, x^3 = -1/8, x^4 = 1/16 => x^2 > x -- Still Insufficient

Hence, E

Anurag@Gurome wrote: => |x| < 1 except x = 0

Sufficient

The correct answer is B.

|x| < 1, doesn't mean x^2 < x
as when x = 1/2, x^2 < x
but when x =-1/2, x^2 > x

Mitch you are right. I drew the X^2>x wrong. X^2>x => X<0 or X>1. E is right. But can you really divide by x^2? I know that's a legal algebra operation but does it eliminate possibilities?

gmatusa2010 wrote:Mitch you are right. I drew the X^2>x wrong. X^2>x => X<0 or X>1. E is right. But can you really divide by x^2? I know that's a legal algebra operation but does it eliminate possibilities?

Not here, as we for sure know, x!=0 based on the two statements [x^2>x^3 & x^2 >x^4].

I don't quite understand, can you expand on that thought.

I guess this is veering off but can you speak on 1) when can u divide and 2) when can't you?

anshumishra wrote:

gmatusa2010 wrote:Mitch you are right. I drew the X^2>x wrong. X^2>x => X<0 or X>1. E is right. But can you really divide by x^2? I know that's a legal algebra operation but does it eliminate possibilities?

Not here, as we for sure know, x!=0 based on the two statements [x^2>x^3 & x^2 >x^4].

gmatusa2010 wrote:Mitch you are right. I drew the X^2>x wrong. X^2>x => X<0 or X>1. E is right. But can you really divide by x^2? I know that's a legal algebra operation but does it eliminate possibilities?

When a DS question involves a variable and an inequality -- and we don't know that the variable is positive -- I discourage simplifying by multiplication or division. The risk is too great. I find plugging in values far safer.

gmatusa2010 wrote:I don't quite understand, can you expand on that thought.

I guess this is veering off but can you speak on 1) when can u divide and 2) when can't you?

anshumishra wrote:

gmatusa2010 wrote:Mitch you are right. I drew the X^2>x wrong. X^2>x => X<0 or X>1. E is right. But can you really divide by x^2? I know that's a legal algebra operation but does it eliminate possibilities?

Not here, as we for sure know, x!=0 based on the two statements [x^2>x^3 & x^2 >x^4].

Lets take an example (equality, to be simple..In inequalities you have to be extra careful) :
x^4-x^2 = 0
=> x^2(x^2 -1) = 0
=> x^2*(x+1)*(x-1) = 0
All the possible roots = 0,1,-1

If someone just knows that x is non-zero, then you can divide both sides by x^2 , and are not losing this solution (x=0, useless as we know it is invalid).
Also, dividing both sides by x^2 is only allowed when x^2 is non-zero, otherwise in right side you are doing 0/0 in this case which is not defined.

So, in simple terms, when x^2 ! = 0, you should.

GMATGuruNY wrote:When a DS question involves a variable and an inequality -- and we don't know that the variable is positive -- I discourage simplifying by multiplication or division. The risk is too great. I find plugging in values far safer.

I definitely agree with this statement......I just have to be careful and organized with the lists and charts I make personally.

Some simple rules:
1. Positive fraction reduces in value when its power is increased (x > x^2 > x^3...)
2. Positive integer increases in value when its power is increased (x < x^2 < x^3....)
3. A number raised to an EVEN power is always positive
4. Negative no will maintain its sign when raised to an ODD power (x^1, x^3 etc.)
5. +ve no > -ve no

STEM - Is x^2 > x?

From Rule #3 - x^2 is always positive.
From Rule #2 and #5 - x needs to be positive integer _or_ negative no (integer or fraction) for this to be true.

STAT #1 - x^2 is greater than x^3.

From Rule #1 - it could be +ve fraction (because x^3 < x^2)
From rule #2 - x is NOT a positive integer (else x^3 would be > x^2)
From Rule #3 - x^2 is always positive
From Rule #4 - x^3 can be +ve or -ve based on sign of x
From rule #5 - x^2 > x^3 when x is negative

TAKEAWAY #1 - So x is either a positive fraction or a negative fraction or negative no.
Doesn't completely meet the requirement given above

INSUFFICIENT

STAT #2 - x^2 is greater than x^4

Say y=x^2 ==> y > y^2
From Rule #3 - y is positive (x can be +ve or -ve) and y^2 is positive
From rule #1 - y is a fraction ==> x is a fraction

TAKEAWAY #2 - So x is either a positive fraction or negative fraction
Doesn't completely meet the requirement given above

INSUFFICIENT

TOGETHER

Even together, TAKEAWAY #1 and #2 - x is either a positive fraction or a negative fraction.
From requirement above, this would be true in some cases (negative fraction) and not in other cases (positive fraction)

INSUFFICIENT

hence E