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by neelgandham » Mon Sep 10, 2012 2:46 am
Is x^2 + y^2 > 100
1. 2xy < 100
xy < 50
Case 1: If x = 0 and y = 0, then xy(=0) < 50 but x^2 + y^2(=0)<100
Case 2: If x = 10 and y = 1, then xy(=10) < 50 but x^2 + y^2(=101)>100
Since we don't have a definite answer, statement I is insufficient to answer the question.
2. (x + y)^2 > 200
x^2 + y^2 + 2xy > 200.
(x^2 + y^2) + 2xy > 200.
Let (x^2 + y^2) be A and 2xy be B.
A + B > 200.

In general we know that (x-y)^2 > 0.
x^2 + y^2 - 2xy > 0.
x^2 + y^2 > 2xy
A>B

Since A>B, and A + B > 200,
A + A > 200.
2A > 200.
A > 100.
x^2 + y^2 > 100.

IMO B is the answer.
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by Anurag@Gurome » Mon Sep 10, 2012 2:52 am
confuse mind wrote:Is x^2 + y^2 > 100

1. 2xy < 100
2. (x + y)^2 > 200

[spoiler]OA - C
IMO - B will post my complete solution later[/spoiler]
(1) 2xy < 100
If x = y = 0, then x² + y² = 0 < 100
If x = 10, y = -10, then x² + y² = 100 + 100 = 200 > 100
No definite answer; NOT sufficient.

(2) (x + y)² > 200
x² + 2xy + y² > 200
Since the square of any number is more than or equal to zero, so (x - y)² more than or equal to zero.
So, x² + y² more than or equal to 2xy implies x² + (x² + y²) + y² > 200
2(x² + y²) > 200
x² + y² > 100; SUFFICIENT.

The correct answer is B.
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by GMATGuruNY » Mon Sep 10, 2012 4:45 am
confuse mind wrote:Is x^2 + y^2 > 100

1. 2xy < 100
2. (x + y)^2 > 200
Statement 1: 2xy < 100.
Thus, xy < 50.
If x=1 and y=1, then x²+y² < 100.
If x=2 and y=10, then x²+y² > 100.
Insufficient.

Statement 2: (x+y)² > 200.
x² + 2xy + y² > 200.
Since the question stem asks about x²+y², we want to eliminate 2xy from statement 2.
One approach: (x+y)² + (x-y)² = (x² + 2xy + y²) + (x² - 2xy + y²) = 2x² + 2y² = 2(x²+y²).
Since the square of a value cannot be negative, (x-y)² ≥ 0.
Thus, we can add together (x+y)² > 200 and (x-y)² ≥ 0:
(x+y)² + (x-y)² > 200+0.
2(x²+y²) > 200.
x² + y² > 100.
Sufficient.

The correct answer is B.
The OA is incorrect.
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by Brent@GMATPrepNow » Mon Sep 10, 2012 6:39 am
confuse mind wrote:Is x^2 + y^2 > 100

1. 2xy < 100
2. (x + y)^2 > 200
Target Question: Is x^2 + y^2 > 100?

Statement 1: 2xy < 100
The only way to show that this statement is not sufficient is by counter-example.
There are several pairs of values that satisfy the condition that 2xy < 100
case a: x=0 y=0, in which case x^2 + y^2 is less than 100
case b: x=0 y=100, in which case x^2 + y^2 is greater than 100
So, statement 1 is NOT SUFFICIENT

Statement 2: (x+y)^2 > 200
Expand to get: x^2 + 2xy + y^2 > 200
Rearrange to get: x^2 + y^2 + 2xy > 200

Aside: (this is the tricky part)
Notice that for any values of x and y, it must be true that (x-y)^2 > 0
Expand to get x^2 - 2xy + y^2 > 0
Rearrange to get x^2 + y^2 > 2xy
Now add x^2 + y^2 to both sides to get: x^2 + y^2 + x^2 + y^2 > x^2 + y^2 + 2xy
Simplify to get: 2(x^2 + y^2) > x^2 + y^2 + 2xy

Now notice that the right-hand side of this inequality matches the left-hand side of the inequality we derived from statement 2.
So, we can write: 2(x^2 + y^2) > x^2 + y^2 + 2xy > 200
Now ignore the middle part to get: 2(x^2 + y^2) > 200
Divide both sides by 2 to get x^2 + y^2 > 100
Since we can answer the target question with certainty, statement 2 is SUFFICIENT and the answer is B

Cheers,
Brent
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by confuse mind » Mon Sep 10, 2012 6:33 pm
Thanks a ton.
My reasoning for B was same, A.M. >= G.M.
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by subhakam » Sun Jan 20, 2013 3:15 pm
GMATGuruNY wrote:
confuse mind wrote:Is x^2 + y^2 > 100

1. 2xy < 100
2. (x + y)^2 > 200
Statement 1: 2xy < 100.
Thus, xy < 50.
If x=1 and y=1, then x²+y² < 100.
If x=2 and y=10, then x²+y² > 100.
Insufficient.

Statement 2: (x+y)² > 200.
x² + 2xy + y² > 200.
Since the question stem asks about x²+y², we want to eliminate 2xy from statement 2.
One approach: (x+y)² + (x-y)² = (x² + 2xy + y²) + (x² - 2xy + y²) = 2x² + 2y² = 2(x²+y²).
Since the square of a value cannot be negative, (x-y)² ≥ 0.
Thus, we can add together (x+y)² > 200 and (x-y)² ≥ 0:
(x+y)² + (x-y)² > 200+0.
2(x²+y²) > 200.
x² + y² > 100.
Sufficient.

The correct answer is B.
The OA is incorrect.
GMAT Guru - how can you add both statements when the inequality signs are not same i.e.
(x-y)² ≥ 0
And (x+y)² > 200

One is greater than equal to and another is greater than - please suggest .... if that matters?

Thanks
Subhakam
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by GMATGuruNY » Sun Jan 20, 2013 6:11 pm
subhakam wrote:GMAT Guru - how can you add both statements when the inequality signs are not same i.e.
(x-y)² ≥ 0
And (x+y)² > 200

One is greater than equal to and another is greater than - please suggest .... if that matters?

Thanks
Subhakam
Given (x-y)² ≥ 0, the LEAST POSSIBLE VALUE of (x-y)² is 0.
If we add (x-y)² = 0 to (x+y)² > 200, we get:
(x+y)² + (x-y)² > 200+0
(x+y)² + (x-y)² > 200.

If (x-y)² > 0, it still must be true that (x+y)² + (x-y)² > 200.
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