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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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Dondarrion
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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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 85186 times ]

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

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

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

= \(\frac{(3xy + 2xy)}{(3xy - 2xy)}\) = \(\frac{5xy}{xy}\) = \(5\)

Hence, Statement 1 alone is sufficient.


(2) xy=3

We will not be able to answer the question stem only with statement xy =3. Hence Statement 2 alone is insufficient.

Option A is the correct answer.

Hope this helps,
Clifin J Francis,
GMAT SME
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Bunuel
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.

Hi Bunuel,
What is the point of giving this information in the question stem: 0 < x < y?
I didn't need it in both statements and thought I did the question wrong even though my steps were correct. The only thing I could get from this info is that both x and y are positive numbers. But I don't see how x<y is a needed information. Am I missing any point here?
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Val1986
If 0 < x < y, what is the value of \(\frac{(x + y)^2}{(x- y)^2}\)?


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

(2) \(xy = 3\)





Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

[email protected]
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