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77. E Explanation: This question rigorously tests your familiarity with common binomials. The only way to do anything with statement (1) is to subtract 2xy from both sides: x^2 - 2xy + y^2 = 1 (x - y)(x - y) = 1 x - y = 1 or x - y = -1 There are an in nite number of possible solutions for x and y, so theres no way to determine a speci c value for x^2 + y^2. Statement (2) is insufficient for similar reasons: x^2 + y^2 = 4 - 2xy x^2 + 2xy + y^2 = 4 (x + y)(x + y) = 4 x + y = 2 or x + y = -2 Again, there are an in nite number of possibilities for x and y. Taken together, you still dont have enough information. If you had exactly two equations (such as x-y = 1 and x+y = 2), you could solve, but you have two diferent pairs of possible equations, which is insufficient to find the values of the variables. Choice (E) is correct.

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

Answer: C. OA must be wrong.

Hope it's clear.

Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

Answer: C. OA must be wrong.

Hope it's clear.

Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.

Can you please post their solution?
_________________

(1) x^2+y^2=2xy+1 --> \(x^2-2xy+y^2=1\) --> \((x-y)^2=1\). If \(x=1\) and \(y=0\), then \(x^2+y^2=1\) but if \(x=2\) and \(y=1\), then \(x^2+y^2=5\). Not sufficient.

(2) x^2+y^2=4-2xy --> \(x^2+2xy+y^2=4\) --> \((x+y)^2=4\). If \(x=2\) and \(y=0\), then \(x^2+y^2=4\) but if \(x=1\) and \(y=1\), then \(x^2+y^2=2\). Not sufficient.

(1)+(2) Sum the equations: \(2(x^2+y^2)=5\) --> \(x^2+y^2=2.5\). Sufficient.

Answer: C. OA must be wrong.

Hope it's clear.

Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.

Thanks Bunuel!!! (I feel like a VIP having you respond to one of my questions!) haha

"C" is what I thought! I got worried and couldn't see what I was doing wrong even after reading the OA description.

Can you please post their solution?

Just put it in the main post

Their solution is not correct.

Four pairs of (x,y) satisfy (x-y)^2=1 and (x+y)^2=4: (-3/2, -1/2), (-1/2, -3/2), (3/2, 1/2) and (1/2, 3/2). For each pair x^2+y^2=10/2.
_________________

This DS problem is from the Jeffrey Sackmann Challenge Algebra set, Quadratics. Not sure about the difficulty level.

What is the value of x^2 + y^2 ? 1) x^2 + y^2 = 2xy + 1 2) x^2 + y^2 = 4 - 2xy

The answer is (E) But I'm not sure why we didn't just use the two statements together > add them > 2xy cancels out and the answer would be: x^2 + y^2 = 5/2 ?

This DS problem is from the Jeffrey Sackmann Challenge Algebra set, Quadratics. Not sure about the difficulty level.

What is the value of x^2 + y^2 ? 1) x^2 + y^2 = 2xy + 1 2) x^2 + y^2 = 4 - 2xy

The answer is (E) But I'm not sure why we didn't just use the two statements together > add them > 2xy cancels out and the answer would be: x^2 + y^2 = 5/2 ?

Thanks in advance

Merging similar topics. please refer to the discussion above.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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