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Intern  S
Joined: 03 Oct 2018
Posts: 14
Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y  [#permalink]

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9
1
2 00:00

Difficulty:   75% (hard)

Question Stats: 43% (01:41) correct 57% (01:51) wrong based on 56 sessions

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Is x^2 > y^2?

(1) x^2 − x > y^2 − y

(2) x > y

Originally posted by piyush26 on 04 Nov 2018, 04:24.
Last edited by Bunuel on 13 Jan 2020, 07:32, edited 3 times in total.
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Re: Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y  [#permalink]

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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.
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Re: Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y  [#permalink]

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1
piyush26 wrote:
Is x^2 > y^2?

(1) x^2 − x > y^2 − y

(2) x > y

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

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

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

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

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

Regards,
Fabio.
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Re: Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y  [#permalink]

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piyush26 wrote:
Is x^2 > y^2?

(1) x^2 − x > y^2 − y

(2) x > y

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, $$x^2> y^2$$.
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Re: Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y  [#permalink]

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1
piyush26 wrote:
Is x^2 > y^2?

(1) x^2 − x > y^2 − y

(2) x > y

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 $$x^2 > y^2?$$
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Re: Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y  [#permalink]

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_________________ Re: Is x^2 > y^2? (1) x^2 − x > y^2 − y (2) x > y   [#permalink] 13 Jan 2020, 07:12
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