Seems easy...but
Is x^2 + y^2 > 6
(1) (x+y)^2 > 6
(2) xy = 2
Scoretop No. 4 - Q 15 - Squares
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My answer is E
1. (x+y)^2 = x^2 + y^2 + 2xy > 6. We can't determine the value of 2xy -> Not sufficient
2. xy = 2 . x=1, y=2 -> x^2+y^2 < 6 buy x=1/6, y=6 absolutely has bigger value -> Not sufficient
1&2. x^2 + y^2 > 2 but it doesn't mean x^2+y^2 > 6 or <6 -> Not sufficient
1. (x+y)^2 = x^2 + y^2 + 2xy > 6. We can't determine the value of 2xy -> Not sufficient
2. xy = 2 . x=1, y=2 -> x^2+y^2 < 6 buy x=1/6, y=6 absolutely has bigger value -> Not sufficient
1&2. x^2 + y^2 > 2 but it doesn't mean x^2+y^2 > 6 or <6 -> Not sufficient
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(1) (x + y)² > 6 implies x² + y² + 2xy > 6akay wrote:Seems easy...but
Is x^2 + y^2 > 6
(1) (x+y)^2 > 6
(2) xy = 2
If x² + y² = 5 and 2xy = 3, then x² + y² < 6
If x² + y² = 10 and 2xy = 3, then x² + y² > 6
No definite answer; NOT sufficient.
(2) xy = 2
If x = y = √2, then x² + y² = 4 < 6
If x = 10, y = 1/5, then x² + y² = 100 + 1/25 > 6
No definite answer; NOT sufficient.
Combining (1) and (2), x² + y² + 2xy > 6
x² + y² + 2(2) > 6
x² + y² + 4 > 6
x² + y² > 2
If x² + y² = 5, then x² + y² < 6
If x² + y² = 10 , then x² + y² > 6
No definite answer; NOT sufficient.
The correct answer is E.
Anurag Mairal, Ph.D., MBA
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