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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 \(\frac{(x + y)^2}{(x y)^2}\)? (1) \(x^2 + y^2 = 3xy\) (2) \(xy = 3\)
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Originally posted by Val1986 on 23 Mar 2013, 21:49.
Last edited by Bunuel on 04 Jul 2018, 01:01, 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?
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24 Mar 2013, 02:40




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Re: If 0 < x < y, what is the value of (x + y)^2/(x y)^2?
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24 Mar 2013, 00:29
1) x^2+y^2=3xy => x^2+y^22xy=xy => (xy)^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/(xy)^2 => (x^2+y^2+2xy)/(x^2+y^22xy) => (x^2+y^2+6)/(x^2+y^26) => 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?
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03 Jun 2014, 00:15
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?
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03 Jun 2014, 00:34
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?
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26 Nov 2015, 12:35
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?
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26 Nov 2015, 12:54
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/(xy)^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 (xy)^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?
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05 Apr 2016, 22:28
Attached is a visual that should help.
Attachments
Screen Shot 20160405 at 10.27.13 PM.png [ 151.45 KiB  Viewed 13123 times ]
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Re: If 0 < x < y, what is the value of (x + y)^2/(x y)^2?
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25 Nov 2016, 16:07
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+2xyy^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?
\(\frac{(x + y)^2}{( x y)^2}=\frac{x^2+2xy+y^2}{x^22xy+y^2}\)
(1) x^2 + y^2 = 3xy. Substitute: what is the value of \(\frac{x^2+2xy+y^2}{x^22xy+y^2}=\frac{3xy+2xy}{3xy2xy}=\frac{5xy}{xy}=5\). 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?
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25 Nov 2016, 23:24



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Re: If 0 < x < y, what is the value of (x + y)^2/(x y)^2?
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28 Mar 2017, 14:07
let's rewrite the given value: (x^2+2xy+y^2)/(x^22xy+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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Re: If 0 < x < y, what is the value of (x + y)^2/(x y)^2?
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