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Is x^2 > y^2?

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Is x^2 > y^2?  [#permalink]

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New post Updated on: 25 Sep 2016, 02:20
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Is x^2 > y^2?

(1) x + y = 2
(2) x > y

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Originally posted by MathRevolution on 24 Sep 2016, 21:48.
Last edited by Bunuel on 25 Sep 2016, 02:20, edited 2 times in total.
Renamed the topic and edited the question.
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Is x^2 > y^2?  [#permalink]

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New post 25 Sep 2016, 02:31
1
MathRevolution wrote:
Is x^2 > y^2?

(1) x + y = 2
(2) x > y



Given x^2 > y^2 and we can write (x+y)(x-y) > 0 and need to find this.

Stat 1: Given x+y = 2..

then we get (2)(x-y) > 0 and here we are not sure whether (x-y) > 0 or (x-y) < 0 ..Insufficient.

Stat 2: we have only (x-y) > 0 information and we are not sure about (x+y) > 0 or (x+y) < 0...Insufficient..

Both : x + y = 2 which is positive and (x-y) > 0

then we can say (2)(x-y) > 0...Sufficient...

Option C.
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Re: Is x^2 > y^2?  [#permalink]

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New post 25 Sep 2016, 03:34
1
1
(1) x + y = 2

we don't have information about x or y.

x=1 , y =1 or x =3 , y =1 -- Not sufficient

(2) x > y

x = -3 and y = -4 or x = 2 and y =1 -- Not sufficient.

But if combine (1) and (2). If x>y and x+y=2. X need to be positive and greater than Y. so , \(X^2 > Y^2\)
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Re: Is x^2 > y^2?  [#permalink]

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New post 27 Sep 2016, 08:09
==> In the original condition, C is highly likely to be an answer since the variable becomes two (x,y). 1) & 2), x^2-y^2>0?, in (x-y)(x+y)>0? you get x-y>0, and thus x+y=2>0. The answer is always yes, therefore suffi. Hence, the answer is C.

answer: C
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Re: Is x^2 > y^2?  [#permalink]

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New post 23 Feb 2017, 03:55
PROMPT ANALYSIS
x and y are real numbers.
SUPERSET
The answer will be either yes or no.

Translation
In order to find the answer we need:
1# exact value of x and y
2# any equation to conclude the condition

STATEMENT ANALYSIS
ST 1: x+y = 2. Take the value of (x, y) = (1,1), (1.5,0.5) and (0.5, 1.5). For these set of values, the condition the different in each case.INSUFFICIENT. HEnce option a and d eliminated

St 2: x>y. Take (x,y) = (2,-1) and (1,-2). Both the cases have different answer.INSUFFICIENT. Option b eliminated.

St 1 & St 2: Combining both, we get x>1. Therefore, for all values that suffice the 1st equation, y<1, hence x^2>y^2.

Option C
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Re: Is x^2 > y^2?  [#permalink]

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New post 28 Feb 2017, 18:41
is x^2 > y^2 ?
is x^2 - y^2 > 0 ?
is (x+y)(x-y) > 0?

stmt-1:
x+y = 2 but we also need to know whether x-y is +ve or negative. in fact x=y is a possiblity in which case (x+y)(x-y) = 0.
insuff.

stmt-2:
x > y
x-y > 0 but we also want to know whether x+y is +ve or -ve. because if y is -ve then also x-y>0 but x+y might not.

stmt-1 + stmt-2:
x+y=2
x-y>0

it means (x+y)(x-y) > 0 because 2*(anything greater than zero) > 0 only.
sufficient.
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Is x^2>y^2?  [#permalink]

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New post 18 Apr 2017, 00:58
Is \(x^2>y^2\)?

1) x+y=2
2) x>y
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Re: Is x^2>y^2?  [#permalink]

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New post 18 Apr 2017, 01:44
1
x^2-y^2>0
(x+y)(x-y)>0

We have to prove the above:

a). x+y=2; satisfies (x+y) > 0; Insufficient

b). x>y; satisfies (x-y)>0; Insufficient

Together both (x+y) & (x-y) > 0; Sufficient

Answer should be C.

It is showing D. How?
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Re: Is x^2>y^2?  [#permalink]

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New post 19 Apr 2017, 10:58
MathRevolution wrote:
Is \(x^2>y^2\)?

1) x+y=2
2) x>y



why is it D. I think it should be C.
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Re: Is x^2>y^2?  [#permalink]

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New post 19 Apr 2017, 11:00
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Re: Is x^2>y^2?  [#permalink]

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New post 20 Apr 2017, 00:37
==> If you modify the original condition and the question, you get\(x^2>y^2\)??, or\(x^2-y^2\)?0?, or (x-y)(x+y)>0?. There are 2 variables (x,y) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), you get x+y>2 and x-y>0, hence yes, it is always sufficient.

Therefore, the answer is C.
Answer: C
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Re: Is x^2>y^2?  [#permalink]

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New post 26 Apr 2017, 21:54
Is x2>y 2?

1) x+y=2
2) x>y

is (x+y)(x-y) >0
statement 1 -> we dont know (x-y) hence insufficient
statement 2- > x>y -> (x-y) > 0 but we dont know (x+y) hence insufficient

combining statemetn 1 and 2 answer is C
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Re: Is x^2 > y^2?  [#permalink]

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