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

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Is x^2 > y^2? (1) x > y (2) y > 0 [#permalink]

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New post 11 Mar 2018, 20:57
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A
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C
D
E

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  25% (medium)

Question Stats:

92% (00:36) correct 8% (00:33) wrong based on 53 sessions

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Re: Is x^2 > y^2? (1) x > y (2) y > 0 [#permalink]

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New post 11 Mar 2018, 20:59
Bunuel wrote:
Is x^2 > y^2?

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


9. Inequalities



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Re: Is x^2 > y^2? (1) x > y (2) y > 0 [#permalink]

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New post 11 Mar 2018, 22:07
Bunuel wrote:
Is x^2 > y^2?

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


Statement 1
x > y is no guarantee that x^2 will be > y^2.
Eg., if x=1, y=0.5 then x > y and x^2 > y^2
But if x=1, y=-2, then x > y But x^2 < y^2
So this statement is not sufficient.

Statement 2
is clearly not sufficient on its own since no comparison between x and y is given.

Combining the statements:
y is positive and x>y so x is also positive. So both x/y are positive and given that x > y. Then definitely x^2 will be > y^2.
(Since in inequality x > y both terms are non negative we can raise this to an even power without changing the sign of inequality)

Hence C answer
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Is x^2 > y^2? (1) x > y (2) y > 0 [#permalink]

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New post 12 Mar 2018, 03:50
Bunuel wrote:
Is x^2 > y^2?

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


(1) x > y

No sign for x and y

Let x =-2 & y =-3...........4 >9 ............Answer is NO

Let x = 3 & y = 2..........9 > 4..............Answer is yes

Insufficient

(2) y > 0

No info about x

Insufficient

Combine 1 & 2

x > y >0 ...Alll positive numbers

Hence it is always YES

Sufficient

Answer: C
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Is x^2 > y^2? (1) x > y (2) y > 0 [#permalink]

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New post 12 Mar 2018, 11:54
Bunuel wrote:
Is x^2 > y^2?

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


This implies
x^2-y^2>0
or (x-y)(x+y)>0
For this to be true either both terms are positive or both terms are negative

1) x>y - this implies (x-y) is positive. No information about (x+y), this could be positive, negative or zero (Insufficient)
2) y>0 - No information about x. (Insufficient)

Put them both together,
(x-y) positive from statement 1
(x+y) positive since x>y>0

Therefore the answer is C
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Re: Is x^2 > y^2? (1) x > y (2) y > 0 [#permalink]

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New post 12 Mar 2018, 22:18
Statement 1 says that X>y. but there can be two case for this. If x= 2 and y=-3, then x^2<y^2. However if x=2 and y=1, then x^2>Y^2. Therefore statement 1 is not sufficient.

Statement 2 only tells us that Y>0, hence clearly not sufficient.

Combining the two statements, we can clearly infer that both numbers are positive and x>y. Therefore x^2>y^2. Hence correct answer is C.
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Re: Is x^2 > y^2? (1) x > y (2) y > 0 [#permalink]

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New post 13 Mar 2018, 03:33
Basically, The question stem is asking us whether |x|>|y|?

St.1 tells us that x>y.
Now, if x & y have same sign, this statement is sufficient. Whereas, if their signs are opposite, then Insufficient.

St.2 indicates that y is positive.

St1+2 confirms that both x and y are positive and value of x is greater than y. Sufficient.

Hence, Ans C

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Re: Is x^2 > y^2? (1) x > y (2) y > 0   [#permalink] 13 Mar 2018, 03:33
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