Is x^2 y^2? (1) x^2 − x y^2 − y (2) x y

Question

Is x^2 > y^2?

(1) x^2 − x > y^2 − y

(2) x > y

DavidTutorexamPAL · Answer

1) We'll start by finding an option for which the answer is "yes": x=2, y=1
2^2-2>1^2-1 ==> 4-2 > 1-1 ==> 2>0 - condition works, and 2^2 > 1^2 ==> 4 > 1- answer to question is "yes"!
Now let's see if we can find an option in which the condition is also met, but where the answer is "no": x=-2. y=2:
(-2)^2 - (-2) > 2 ==> 4+2 >2 ==> 6 >2 - condition is met! as for the question: (-2)^2 = 2^2 - answer is "no"! Insufficient! A and D are eliminated

2) answer "yes": x=2, y=1 ==> 2^2 > 1 - "yes"!
answer "no": x = 2, y = -10 ==> 2^2 < (-10)^2 ==> 4 <100 - "no"! insufficient!

Combined: using the examples we're already used, the conditions are met and the answer is "yes" when x=2,y=1. Can the answer be "no"? well the example we used above (x=-2, y=2) won't work because it violates (2). Logically, so is any option for x to be positive and y to be negative - they must be the same sign... we've already seen if they're positive the answer is yes. can they both be negative? let's see: x=-1, y=-2 ==> (-1)^2-(-1) = 1+1 = 2 > 6 =4+2 = (-2)^2-(-2) - NOPE! won't work...
The answer must be "yes" - sufficient.

fskilnik · Answer

{x^2}\,\,\mathop > \limits^? \,\,{y^2}

\left( 1 \right)\,\,{x^2} - x > {y^2} - y\,\,\,\,\left\{ \matrix{
 \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr 
 \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.

\left( 2 \right)\,\,x > y\,\,\,\,\left\{ \matrix{
 \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr 
 \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.

\left( {1 + 2} \right)\,\,\,\,{x^2} - {y^2}\,\,\,\mathop > \limits^{\left( 1 \right)} \,\,\,x - y\,\,\,\mathop > \limits^{\left( 2 \right)} \,\,\,0\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^2} > {y^2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

rahul16singh28 · Answer

From both Statement we have -

(x-y)(x+y-1) > 0  ---- From Statement I
 (x-y) > 0  ----- From Statement II

As  (x - y) > 0 , we have  (x+y - 1) > 0

Hence,  x^2> y^2 .

Bismarck · Answer

OA: C

Question : Is  x^2 > y^2?

(1)  x^2 − x > y^2 − y

Taking  x = 2  and  y = 1 .These values satisfy  x^2 − x > y^2 − y

Answer to Question: Is  x^2 > y^2?:  Yes as  2^2>1^2

Taking  x = -3  and  y = 3 .These values satisfy  x^2 − x > y^2 − y

Answer to Question: Is  x^2 > y^2?:  No as  (-3)^2=3^2

Statement 1 alone is insufficient.

(2)  x > y

Taking  x = 2  and  y = 1 .These values satisfy  x > y

Answer to Question: Is  x^2 > y^2?:  Yes as  2^2>1^2

Taking  x = -3  and  y = -4 .These values satisfy  x > y

Answer to Question: Is  x^2 > y^2?:  No as  (-3)^2<(-4)^2

Statement 2 alone is insufficient.

Combining (1) and (2), we get 
 \quad x^2 − x> y^2 − y 
  + \qquad x>y 
We get  x^2>y^2 , Combining (1) and (2) is sufficient to answer the question :  Is   x^2 > y^2?

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions