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by Anurag@Gurome » Mon Jan 17, 2011 8:21 pm
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 > x^3
Implies, (x^3 - x^2) < 0
=> (x^2)(x - 1) < 0
Thus either x < 1 except x = 0.
For x < 0, x^2 > x
For 0 < x < 1, x^2 < x

Not sufficient

Statement 2: x^2 > x^4
Implies, (x^4 - x^2) < 0
=> (x^2)(x - 1)(x + 1) < 0
Thus -1 < x < 1 except x = 0.
=> |x| < 1 except x = 0
For, -1 < x < 0, x^2 > x
For 0 < x < 1, x^2 < x

Not sufficient

1 & 2 Together: Common region satisfying both the statements is |x| < 1 except x = 0. But that is same as statement 2.

Not sufficient.

The correct answer is E.
Last edited by Anurag@Gurome on Mon Jan 17, 2011 8:46 pm, edited 1 time in total.
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by gmatusa2010 » Mon Jan 17, 2011 8:32 pm
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.
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by GMATGuruNY » Mon Jan 17, 2011 8:34 pm
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.
Last edited by GMATGuruNY on Mon Jan 17, 2011 8:41 pm, edited 2 times in total.
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by anshumishra » Mon Jan 17, 2011 8:35 pm
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
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by anshumishra » Mon Jan 17, 2011 8:37 pm
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
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by gmatusa2010 » Mon Jan 17, 2011 8:40 pm
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?
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by anshumishra » Mon Jan 17, 2011 8:47 pm
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].
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by Anurag@Gurome » Mon Jan 17, 2011 8:50 pm
While dealing with inequalities with exponents always remember there are some traps,
  • 1. The trend is not similar for |x| > 1 and |x| < 1
    2. For negative x, the powers alternately change their signs too.
Here is another approach which tackles the problem logically. While dealing with exponents of variables always remember the following scenarios. Also try to understand them. Then you can solve this kind of problems just by looking at it.

For any +ve number x,
  • 1. If x < 1, then repeated multiplication of the number with itself results in smaller quantity. Hence x > x^2 > x^3 > x^4 > ...
    2. If x >1, then repeated multiplication of the number with itself results in larger quantity. Hence x < x^2 < x^3 < x^4 < ...
However for any -ve number x the scenario is a bit complicated due to alternate sign change.
  • 1. If x < 1, then repeated multiplication of the number with itself results in smaller quantity but even powers are positive. Hence x^2 > x^4 > ... > x > x^3 > x^5 > ...
    2. If x >1, then repeated multiplication of the number with itself results in larger quantity but even powers are positive. Hence ... x^5 < x^3 < x < x^2 < x^4 ...
Now the problem asks for is x^2 > x?
If you understood the scenario, you can easily identify that this is possible if either x is negative or x greater than 1.

Now even combining both the statements we can't answer the question in yes or no.
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by gmatusa2010 » Mon Jan 17, 2011 8:50 pm
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].
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by Anurag@Gurome » Mon Jan 17, 2011 8:51 pm
anshumishra wrote:|x| < 1, doesn't mean x^2 < x
as when x = 1/2, x^2 < x
but when x =-1/2, x^2 > x
Sorry guys.
I realized the mistake while typing the other solution. Edited it.
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by GMATGuruNY » Mon Jan 17, 2011 8:58 pm
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.
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by anshumishra » Mon Jan 17, 2011 9:00 pm
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.
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by chendawg » Fri Feb 25, 2011 2:24 pm
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.
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by thebigkats » Wed Mar 02, 2011 12:58 pm
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
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