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# What is the value of y?

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Intern
Joined: 06 Sep 2012
Posts: 38
Concentration: Social Entrepreneurship
What is the value of y?  [#permalink]

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15 Dec 2012, 10:33
3
6
00:00

Difficulty:

25% (medium)

Question Stats:

68% (01:12) correct 32% (01:05) wrong based on 200 sessions

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What is the value of y?

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

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Joined: 25 Apr 2012
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Re: what is the value of y?  [#permalink]

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16 Dec 2012, 01:52
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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Re: what is the value of y?  [#permalink]

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16 Dec 2012, 07:06
4
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.

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Re: What is the value of y?  [#permalink]

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01 Mar 2013, 06:29
Hi Bunuel,

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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Re: What is the value of y?  [#permalink]

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04 Mar 2013, 03:50
1
shreerajp99 wrote:
Hi Bunuel,

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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Re: What is the value of y?  [#permalink]

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20 Jan 2019, 13:04
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Re: What is the value of y?   [#permalink] 20 Jan 2019, 13:04
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