Is x y? (1) x^2 y^2 (2) x - y 0

Question

Is x > y?

(1) x^2 > y^2
(2) x - y > 0

Bunuel · Answer

Is x > y?

(1) x^2 > y^2 -->  |x|>|y|  --> so,  x  may or may not be greater than  y . Consider:  x=-2  and  y=1  for a NO answer and  x=2  and  y=1  for an YES answer. Not sufficient.

(2) x - y > 0 -->  x>y . Directly answers the question. Sufficient.

Answer: B.

Hope it's clear.

sarb · Answer

thanks for the help

cyberjadugar · Answer

Hi, x>y? Using (1), x^2 > y^2 or x^2 - y^2>0 or (x-y)(x+y)>0 so, either x-y>0 & x+y>0 or x-y<0 & x+y<0 so, x may or may not be less than y. Insufficient. Using (2), x-y>0 or x>y, so answer to question is Yes. Sufficient Thus, answer is (B) Regards,

sarb · Answer

thanks a lot for helping me

Ralphcuisak · Answer

As the title suggests,i can't figure out when to be wary of the "could be positive or negative,so we can't say" rule in algebra.Please help in understanding this. Acc. to the books,if x^2 is lets say 25,x could be positive or negative both so we cant say what the result could be when we use x in any other calculation.That is fine.Great. :roll:

Next is √x^2= |x| Coz we take square of x and x becomes +ve and then ROOT INSTRUCTS US TO TAKE ONLY THE POSITIVE VALUE OF X. Is x > y? (1) x^2> y^2 So,here if we take the root on both sides of the equation,will we not get x>y???Please correct me if i am wrong or missing something here. :cry:

(2) x - y > 0 Answer is B. Bunuel , insights please!!

EMPOWERgmatRichC · Answer

Hi Ralphcuisak,

You bring up some important examples/math concepts that are worth remembering for Test Day. However, they both involve an EQUALS sign, while this question involves an INEQUALITY. It's a different situation, so you have use different rules.

Since "squaring" a positive has the same effect as "squaring" a negative (the result is POSITIVE in both options), you have to think about the "relative" value of X and Y when you're told....

X^2 > Y^2

X and/or Y COULD be positive OR negative (or even 0).....this really means the same thing as |X| > |Y|

IF....
X = 2
Y = 1
(2)^2 > (1)^1
2 > 1
|2| > |1|

IF....
X = -2
Y = 1
(-2)^2 > (1)^2
-2 is NOT > 1
|-2| > |1|

GMAT assassins aren't born, they're made,
Rich

Ralphcuisak · Answer

It's clear now..Thanks EMPOWERgmatRichC..

EMPOWERgmatRichC · Answer

Hi Ralphcuisak,

I'm happy to help. Remember that math rules don't exist on their own - they have to "work" with real numbers. That means there's always a practical way to "test out" your ideas to make sure that they're correct: TEST VALUES.

For example, what happens when you square a negative fraction?

You don't have to theorize over this (and rubbing your chin and staring at the ceiling is a waste of time) - choose a negative fraction, write it down, then square it.

(-1/2)^2 = (-1/2)(-1/2) = + 1/4

Now you have PROOF of the rule. Keep this idea in mind anytime you feel stuck on a question or "your answer" doesn't match any of the 5 given answers. You should be able to prove any rule that you're using.

GMAT assassins aren't born, they're made,
Rich

stonecold · Answer

here statement 2 is obviously correct
but 1 is not as we cannot square until we know the signs are the same

hiranmay · Answer

Is x > y?

(1) x^2 > y^2 ->  insuff: if x=3 & y=2, then x>y, but if x=-3 & y=-2, then x<y 
(2) x - y > 0 -->  suff: x-y >0 => x>y 
  Answer: B

Basshead · Answer

Is x > y?

(1) x^2 > y^2

Inconclusive. Could or could not be true. INSUFFICIENT.

(2) x - y > 0

x > y. SUFFICIENT.

Is x > y? (1) x^2 > y^2 (2) x - y > 0

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