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If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?

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If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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If 0 < x < y, what is the value of \(\frac{(x + y)^2}{(x- y)^2}\)?


(1) \(x^2 + y^2 = 3xy\)

(2) \(xy = 3\)

Originally posted by Val1986 on 23 Mar 2013, 20:49.
Last edited by Bunuel on 04 Jul 2018, 00:01, edited 3 times in total.
Edited the question.
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 24 Mar 2013, 01:40
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If 0 < x < y, what is the value of (x + y)^2/( x- y)^2?

\(\frac{(x + y)^2}{( x- y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}\)

(1) x^2 + y^2 = 3xy. Substitute: what is the value of \(\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{3xy+2xy}{3xy-2xy}=\frac{5xy}{xy}=5\). Sufficient.

(2) xy = 3. Not sufficient.

Answer: A.
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 23 Mar 2013, 23:29
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1)
x^2+y^2=3xy => x^2+y^2-2xy=xy => (x-y)^2=xy

So you can replace : (x+y)^2/xy

And then just finish the work : (x+y)^2/xy => (x^2+y^2+2xy)/xy => (3xy+2xy)/xy => 5

1 is enough

2) not enough. (x+y)^2/(x-y)^2 => (x^2+y^2+2xy)/(x^2+y^2-2xy) => (x^2+y^2+6)/(x^2+y^2-6) => you can't know the value of x^2 or y^2

Hope it helps.
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 02 Jun 2014, 23:15
In Second statement: xy=3 and 0<x<y. Is it possible to take x=1 & y=3 and make the second statement sufficient.
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 02 Jun 2014, 23:34
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kapil20 wrote:
In Second statement: xy=3 and 0<x<y. Is it possible to take x=1 & y=3 and make the second statement sufficient.

Got it,we also have to take fraction nos.x=0.5 y=6
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 26 Nov 2015, 11:35
Can someone explain how to tackle (2)?

I started testing numbers with x=1 and y=3, result is 16. Then I went to x=\(\frac{1}{2}\) and y=6, but I quickly got bogged down in math and I was unable to prove that it is insufficient.

What is the shortcut I am missing to see that xy=3 is insufficient?

Thanks!
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 26 Nov 2015, 11:54
Dondarrion wrote:
Can someone explain how to tackle (2)?

I started testing numbers with x=1 and y=3, result is 16. Then I went to x=\(\frac{1}{2}\) and y=6, but I quickly got bogged down in math and I was unable to prove that it is insufficient.

What is the shortcut I am missing to see that xy=3 is insufficient?

Thanks!


You can clearly see that you ar egiven that x<y and that both x,y are >0. Statement 2 tells you that 2 positive number s(integers or not) give you a product of 3. Clearly, you can have different combinations that give you xy=3 but will definitely give you different values of x+y and let alone for (x+)^2/(x-y)^2. This makes this statement not sufficient.

With x=1 and y=3, the value should be 4 and not 16 (besides the point). As for x=.5 and y =6, you get (x+y)^2 = 6.5^2 and (x-y)^2 = 5.5^2 ---> 6.5^2/5.5^2 = (13/11)^2 = a value close to 1. Thus you get 2 different values for the expression under consideration, making this statement not sufficient.

Hope this helps.
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 05 Apr 2016, 21:28
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Attached is a visual that should help.
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Screen Shot 2016-04-05 at 10.27.13 PM.png
Screen Shot 2016-04-05 at 10.27.13 PM.png [ 151.45 KiB | Viewed 14757 times ]


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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 25 Nov 2016, 15:07
Hi,

Im not getting why st2 is not correct. This is how I am solving this.

(x+y)^2 x^2+2xy+y^2
------- => ------------- => x^2+2xy+y^2- x^2+2xy-y^2 => 4xy
(x−y)^2 x^2−2xy+y^2

So as per the above method, Only st2 is sufficient. What am I doing wrong?

Thanks for your help!

H.



Bunuel wrote:
If 0 < x < y, what is the value of (x + y)^2/( x- y)^2?

\(\frac{(x + y)^2}{( x- y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}\)

(1) x^2 + y^2 = 3xy. Substitute: what is the value of \(\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{3xy+2xy}{3xy-2xy}=\frac{5xy}{xy}=5\). Sufficient.

(2) xy = 3. Not sufficient.

Answer: A.
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 25 Nov 2016, 22:24
HarveyKlaus wrote:
Hi,

Im not getting why st2 is not correct. This is how I am solving this.

(x+y)^2 x^2+2xy+y^2
------- => ------------- => x^2+2xy+y^2- x^2+2xy-y^2 => 4xy
(x−y)^2 x^2−2xy+y^2

So as per the above method, Only st2 is sufficient. What am I doing wrong?

Thanks for your help!

H.



Bunuel wrote:
If 0 < x < y, what is the value of (x + y)^2/( x- y)^2?

\(\frac{(x + y)^2}{( x- y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}\)

(1) x^2 + y^2 = 3xy. Substitute: what is the value of \(\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{3xy+2xy}{3xy-2xy}=\frac{5xy}{xy}=5\). Sufficient.

(2) xy = 3. Not sufficient.

Answer: A.


First of all please read and follow: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Thank you.

Next, how is the read part above correct? Does 3/2 equals to 3-2=1???
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Re: If 0 < x < y, what is the value of (x + y)^2/(x- y)^2?  [#permalink]

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New post 28 Mar 2017, 13:07
let's rewrite the given value: (x^2+2xy+y^2)/(x^2-2xy+y^2)

1) we see that x^2+y^2=3xy, so if we plug that in, it is 5xy/xy=5 Sufficient
2) alone insufficient

Answer is A
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