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by yellowho » Mon Feb 28, 2011 1:36 am
What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


OA is E. What?
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by manpsingh87 » Mon Feb 28, 2011 1:52 am
yellowho wrote:What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


OA is E. What?
1) can be written as x^2+y^2-2xy=1;

(x-y)^2=1; from here two different cases are possible;------------a)
x-y=1; x-y= -1;

2) x^2+y^2+2xy= 4;
(x+y)^2 = 2^2; similarly as a) we have;

x+y=2; x+y=-2;

By using 1 and 2 alone we cannot find the value of x^2+y^2 by combining 1 and 2 also we will have different possible combination and hence different values.

Hence the value x^2+y^2 cannot be found uniquely by using given conditions.

i hope it helps..!!!
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by sanju09 » Mon Feb 28, 2011 2:00 am
yellowho wrote:What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


OA is E. What?

(1) If x^2 + y^2 = 2 x y + 1 then (x – y) ^2 = 1 or x – y = ±1, which cannot give us the value of x^2 + y^2. Insufficient

(2) If x^2 + y^2 = 4 – 2 x y then (x + y) ^2 = 4 or x + y = ±2, which cannot give us the value of x^2 + y^2. Insufficient

Taken together, we have different systems of linear equations in two variables, so [spoiler]no exclusive x and/or y could be determined that could answer x^2 + y^2 exclusively.

E[/spoiler]
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by yellowho » Mon Feb 28, 2011 2:01 am
(1) x^2 + y^2 = 2xy + 1 => x^2+y^2-1= 2xy
(2) x^2 + y^2 = 4 - 2xy => x^2+y^2-4=-2xy=> -x^2-y^2+4=2xy

Set them equal:

x^2+y^2-1=-x^2-y^2+4

2(X^2+y^2)=5

x^2+y^2=5/2

Since it deals with quadratic there could be other solutions? is that why its e?
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by yellowho » Mon Feb 28, 2011 2:03 am
Sanju,

Can you explain this a little more:

"Taken together, we have different systems of linear equations in two variables, so [spoiler][b]no exclusive x and/or y could be determined that could answer x^2 + y^2 exclusively."

I'm not familiar with math language.


[quote="sanju09"][quote="yellowho"]What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


OA is E. What?[/quote]


(1) If x^2 + y^2 = 2 x y + 1 then (x – y) ^2 = 1 or x – y = ±1, which cannot give us the value of x^2 + y^2. Insufficient

(2) If x^2 + y^2 = 4 – 2 x y then (x + y) ^2 = 4 or x + y = ±2, which cannot give us the value of x^2 + y^2. Insufficient

Taken together, we have different systems of linear equations in two variables, so [spoiler][b]no exclusive x and/or y could be determined that could answer x^2 + y^2 exclusively.

E[/b][/spoiler][/quote]
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by GMATGuruNY » Mon Feb 28, 2011 2:22 am
yellowho wrote:What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


OA is E. What?
The OA is incorrect.

Combining the 2 statements:
x-y = ±1.
x+y = ±2.

Solving for x-y = 1 and x+y = 2, we get x=3/2, y=1/2.
Solving for x-y = 1 and x+y = -2, we get x=-1/2, y=-3/2.
Solving for x-y = -1 and x+y = 2, we get x=1/2, y=3/2.
Solving for x-y = -1 and x+y = -2, we get x=-3/2, =-1/2.

In each case, x²+y² = 10/4 = 5/2.
Sufficient.

Alternatively, we could solve as yellowho correctly did above.

The correct answer is C.
Last edited by GMATGuruNY on Mon Feb 28, 2011 2:26 am, edited 1 time in total.
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by sanju09 » Mon Feb 28, 2011 2:24 am
yellowho wrote:Sanju,

Can you explain this a little more:

"Taken together, we have different systems of linear equations in two variables, so [spoiler]no exclusive x and/or y could be determined that could answer x^2 + y^2 exclusively."

I'm not familiar with math language.


I have worked these out separately and got the same answer for x^2 + y^2.


[spoiler]C[/spoiler]
Last edited by sanju09 on Mon Feb 28, 2011 2:31 am, edited 1 time in total.
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by yellowho » Mon Feb 28, 2011 2:28 am
GMATGuruNY,

Is my method correct? With quadratic, I'm always afraid of weird things that makes regular algebra illegal.
Or is your method the only way we know for sure?
[quote="GMATGuruNY"][quote="yellowho"]What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


OA is E. What?[/quote]

The OA is incorrect.

Combining the 2 statements:
x-y = ±1.
x+y = ±2.

Solving for x-y = 1 and x+y = 2, we get x=3/2, y=1/2.
Solving for x-y = 1 and x+y = -2, we get x=-1/2, y=-3/2.
Solving for x-y = -1 and x+y = 2, we get x=1/2, y=3/2.
Solving for x-y = -1 and x+y = -2, we get x=-3/2, =-1/2.

In each case, x²+y² = 10/4 = 5/2.
Sufficient.

Alternatively, we could solve as yellowho correctly did above.

The correct answer is [spoiler]C[/spoiler].[/quote]
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by GMATGuruNY » Mon Feb 28, 2011 2:33 am
yellowho wrote:GMATGuruNY,

Is my method correct? With quadratic, I'm always afraid of weird things that makes regular algebra illegal.
Or is your method the only way we know for sure?
GMATGuruNY wrote:
yellowho wrote:What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


OA is E. What?
The OA is incorrect.

Combining the 2 statements:
x-y = ±1.
x+y = ±2.

Solving for x-y = 1 and x+y = 2, we get x=3/2, y=1/2.
Solving for x-y = 1 and x+y = -2, we get x=-1/2, y=-3/2.
Solving for x-y = -1 and x+y = 2, we get x=1/2, y=3/2.
Solving for x-y = -1 and x+y = -2, we get x=-3/2, =-1/2.

In each case, x²+y² = 10/4 = 5/2.
Sufficient.

Alternatively, we could solve as yellowho correctly did above.

The correct answer is C.
Your methodology is fine. We can always add two equations together, whether the equations are linear or quadratic.

If the question had asked for the values of x and y individually, the correct answer would be E. But since the question asks only for the value of x²+y², the correct answer is C.
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by HSPA » Mon Feb 28, 2011 10:09 pm
What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


IMO C
Using both....
Here 2xy+1 = 4-2xy => 2xy = 1.5
So I got x^2+y^2 = 1.5+1 or (4-1.5)

Ans: 2.5
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by sanju09 » Mon Feb 28, 2011 10:59 pm
HSPA wrote:What is the value of x^2 + y^2 ?
(1) x^2 + y^2 = 2xy + 1
(2) x^2 + y^2 = 4 - 2xy


IMO C
Using both....
Here 2xy+1 = 4-2xy => 2xy = 1.5
So I got x^2+y^2 = 1.5+1 or (4-1.5)

Ans: 2.5
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