Is x^2 greater than x ?
(1) x^2 is greater than x^3.
(2) x^2 is greater than x^4.
Inequality
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Statement 1: x^2 > 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.
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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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.
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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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.
Last edited by GMATGuruNY on Mon Jan 17, 2011 8:41 pm, edited 2 times in total.
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x^2 > x ?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 = -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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|x| < 1, doesn't mean x^2 < xAnurag@Gurome wrote: => |x| < 1 except x = 0
Sufficient
The correct answer is B.
as when x = 1/2, x^2 < x
but when x =-1/2, x^2 > x
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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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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?
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While dealing with inequalities with exponents always remember there are some traps,
For any +ve number 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.
- 1. The trend is not similar for |x| > 1 and |x| < 1
2. For negative x, the powers alternately change their signs too.
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 < ...
- 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 ...
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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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?
I guess this is veering off but can you speak on 1) when can u divide and 2) when can't you?
anshumishra wrote: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?
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Sorry guys.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
I realized the mistake while typing the other solution. Edited it.
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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: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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Lets take an example (equality, to be simple..In inequalities you have to be extra careful) :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: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?
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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I definitely agree with this statement......I just have to be careful and organized with the lists and charts I make personally.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.
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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
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