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Is x > y?

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Source: — Data Sufficiency |

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by Anurag@Gurome » Fri Jan 28, 2011 2:10 pm
diaca wrote:Is x > y?

(1) √x > y
(2) x³ > y
Whenever powers of variables are concerned, check for values greater than 1 and less than 1.

Statement 1: √x > y
If x > 1, then x > √x > y => x > y
If x < 1, then √x > x => x may not be greater than y

Not sufficient

Statement 2: x³ > y
If x > 1, then x³ > x but x may or may not be greater than y
If x < 1, then x > x³ > y

Not sufficient

1 & 2 Together: If x > 1, x³ > x > √x > y => x > y
And if x < 1, √x > x > x³ > y => x > y

Sufficient.

The correct answer is C.
Last edited by Anurag@Gurome on Sat Aug 20, 2011 7:00 pm, edited 1 time in total.
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by aleph777 » Fri Jan 28, 2011 2:15 pm
The thing about a sqrt is that it can only ever reveal the absolute value of a number.

Statement 1 says that sqrt x > y. I think the question expects you to think, ok, let's square both sides, and then we have the sqrt canceled out and: x > y^2, which means x must be greater than y. But when you square a sqrt, you end up with |x|. In other words, you can't tell whether the natural value of x is positive or negative. Thus, x could be 5 and y could be 2, in which case, x > y, but x could also equal -5. The absolute value hides the sign of x, so we can't tell just yet. Insufficient.

Statement 2 says x^3 > y. That could mean -2^3 > -20, or it could mean that 2^3 > 3. Etc. Insufficient, as well.

Combined, however, sqrt x > y tells us that the absolute value of x is greater than y, and x^3 > y tells us that even if x is negative, its negative value cubed is still greater than y. Therefore, -2^3 > y, 2^3 >y 0^3 > y, etc.

Sufficient. C
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by jaguar123 » Sat Aug 20, 2011 5:59 am
Will the GMAT Test us for root of negative fractions?
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by Anurag@Gurome » Sat Aug 20, 2011 6:01 am
jaguar123 wrote:Will the GMAT Test us for root of negative fractions?
GMAT does not require to deal with imaginary numbers. Hence roots of any negative number is not defined for GMAT.
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by AVbyT » Sat Aug 20, 2011 9:20 am
Anurag@Gurome wrote:
diaca wrote:Is x > y?

(1) √x > y
(2) x³ > y
Whenever powers of variables are concerned, check for values greater than 1 and less than 1.

Statement 1: √x > y
If x > 1, say x = 4, then x > √x > y => x > y
If x < 1, say x = 1/4, then √x > x => x may not be greater than y

Not sufficient

Statement 2: x³ > y
If x > 1, say x = 8, then x³ > x > y => x > y
If x < 1, say x = 1/8, then x > x³ => x may not be greater than y

Not sufficient

1 & 2 Together: If x > 1, x³ > x > √x > y => x > y
And if x < 1, √x > x > x³ > y => x > y

Sufficient.

The correct answer is C.

Hello Anurag@Gurome - I think there's a slight error in your explanation for Statement 2:

Statement 2: x³ > y
If x>1, say x = 2 => x³= 8. Now y=6 or 1 so in this condition x may or may not be greater than y
However, if x<1, say x = 1/2 => x³ = 1/8 and in this case x (will always be greater than) > x³ and since x³>y so will x > y.

You got the right answer but reversed the explanation for the conditions in Statement 2.

Yea?

- Amit
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by Anurag@Gurome » Sat Aug 20, 2011 6:58 pm
AVbyT wrote:You got the right answer but reversed the explanation for the conditions in Statement 2.
Thanks Amit.
Editing my reply.
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