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What is the value of y?

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What is the value of y?

(1) x^2 - y^2 = 5
(2) x and y are each positive integers
[Reveal] Spoiler: OA

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New post 16 Dec 2012, 02:52
JJ2014 wrote:
what is the value of y?

(1) x^2 - y^2 = 5
(2) x and y are each positive integers



Hi,

From 1, we get (x-y)(x+y)=5 i.e (x-y)=1,5 or (x+y)= 5,1...solving we get 2 values of y (one positive and one negative)

From 2 alone we get nothing

combining 1 & 2 we get X and Y as postive and we get one solution
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New post 16 Dec 2012, 08:06
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mridulparashar1 wrote:
JJ2014 wrote:
What is the value of y?

(1) x^2 - y^2 = 5
(2) x and y are each positive integers



Hi,

From 1, we get (x-y)(x+y)=5 i.e (x-y)=1,5 or (x+y)= 5,1...solving we get 2 values of y (one positive and one negative)

From 2 alone we get nothing

combining 1 & 2 we get X and Y as postive and we get one solution


The red part is not correct. From x^2 - y^2 = 5 we cannot say that x-y=1 and x+y=5 (or vise-versa) because for (1) we don't know whether x and y are integers. So, for example it's possible that x-y=10 and x+y=1/2. Even if we knew that x and y are integers, still from (x+y)(x-y)=5 it follows that x+y=5 and x-y=1 (or vise versa) OR x+y=-5 and x-y=-1 (or vise versa).

What is the value of y?

(1) x^2 - y^2 = 5. Infinitely many solutions exist for x and y. Not sufficient.

(2) x and y are each positive integers. Not sufficient.

(1)+(2) Since \(x\) and \(y\) are positive integers then \(x+y=integer>0\) and \(x-y=integer\) AND \(x+y>x-y\). Thus from \(x^2 - y^2 =(x+y)(x-y)= 5\) we'd have that \(x+y=5\) and \(x-y=1\), from which it follows that \(y=2\). Sufficient.

Answer: C.
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(1)\(x^{2}\)-\(y^{2}\)=5
(2) x and y are each positive integers.
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Re: What is the value of y?? [#permalink]

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New post 28 Feb 2013, 22:30
Did you want an explanation?
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New post 28 Feb 2013, 23:28
NonYankee wrote:
Did you want an explanation?


nope just posted the question! If you think its too easy please tag this question as sub 600
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Re: What is the value of y?? [#permalink]

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rajathpanta wrote:
NonYankee wrote:
Did you want an explanation?


nope just posted the question! If you think its too easy please tag this question as sub 600


Ok. Thanks for contributing this problem! As for the difficulty, I have no gauge for the level of questions. :lol:
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Re: What is the value of y?? [#permalink]

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New post 28 Feb 2013, 23:36
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NonYankee wrote:
rajathpanta wrote:
NonYankee wrote:
Did you want an explanation?


nope just posted the question! If you think its too easy please tag this question as sub 600


Ok. Thanks for contributing this problem! As for the difficulty, I have no gauge for the level of questions. :lol:



:) we mere mortals talk in terms of difficulty level :) For you, its a different ball game! you should ask is this a sub-800 question?? LOL :P
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New post 28 Feb 2013, 23:46
rajathpanta wrote:
:) we mere mortals talk in terms of difficulty level :) For you, its a different ball game! you should ask is this a sub-800 question?? LOL :P


Hahaha that made me literally laugh out loud!
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New post 01 Mar 2013, 01:27
NonYankee wrote:
rajathpanta wrote:
NonYankee wrote:
Did you want an explanation?


nope just posted the question! If you think its too easy please tag this question as sub 600


Ok. Thanks for contributing this problem! As for the difficulty, I have no gauge for the level of questions. :lol:


I'd say that the difficulty level of this question is ~650.
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New post 01 Mar 2013, 07:29
Hi Bunuel,

It sounds stupid but can u tell me about this step:

X+y=5 and X-y=1. I always used to think that it will mean x+y=5 or x-y=5.

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New post 04 Mar 2013, 04:50
shreerajp99 wrote:
Hi Bunuel,

It sounds stupid but can u tell me about this step:

X+y=5 and X-y=1. I always used to think that it will mean x+y=5 or x-y=5.


When we consider the two statements together we have that both \(x\) and \(y\) are positive integers, thus \(x+y=integer>0\) and \(x-y=integer\) AND \(x+y>x-y\).

Next, we also have that \(x^2 - y^2 =(x+y)(x-y)= 5\), so we have that the product of two multiple, x+y and x-y is equal to 5, a prime number. Since \(x+y=integer>0\) and \(x-y=integer\), then x+y must be 5 AND x-y must be 1: 5*1=5.

Hope it's clear.
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