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17 Jul 2008, 06:41
Kudos
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N
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Is xy > \(x^2y^2\)?
(1) \(14x^2 = 3\)
(2) \(y^2 = 1\)
(1) \(14x^2 = 3\)
(2) \(y^2 = 1\)
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17 Jul 2008, 06:58
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pmenon
E for me.
we need to know if
\(xy - x^2 * y^2 > 0\)
= \(xy(1-xy) > 0\)
both 1 & 2 give us values of \(x^2\) & \(y^2\)
x and y can be both +ve, both -ve, one positive other negative, fractions
thus E
17 Jul 2008, 07:01
Kudos
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I think it's C....Alpha, you're right. I forgot to think about negatives. I thought of fractions, but not negatives.
My first thought is that we need to know if x and y are fractions. If only one is a fraction, it's not enough to answer the question with certainty.
(1) Is Insufficient because we know that \(x = \sqrt{\frac{3}{14}}\) but that doesn't tell us if Y is a fraction or not (or 1 as we'll see with statement 2)
(2) Insufficient because we know y = 1, we don't know the value of x.
x could be 10 so then we'd have (1)(10) > (10^2)(1^2) = 10 > 100...not true or x could be 1/2 so (1/2)(1) > (1/4)(1) = 1/2 > 1/4 true. Since we can answer yes or no, it is insufficient.
Together, if we know that \(x = \sqrt{\frac{3}{14}}\) and y = 1, then \(xy > x^2y^2\) always.
EDIT: Correct statement is \(x = +/-\sqrt{\frac{3}{14}}\)
My first thought is that we need to know if x and y are fractions. If only one is a fraction, it's not enough to answer the question with certainty.
(1) Is Insufficient because we know that \(x = \sqrt{\frac{3}{14}}\) but that doesn't tell us if Y is a fraction or not (or 1 as we'll see with statement 2)
(2) Insufficient because we know y = 1, we don't know the value of x.
x could be 10 so then we'd have (1)(10) > (10^2)(1^2) = 10 > 100...not true or x could be 1/2 so (1/2)(1) > (1/4)(1) = 1/2 > 1/4 true. Since we can answer yes or no, it is insufficient.
Together, if we know that \(x = \sqrt{\frac{3}{14}}\) and y = 1, then \(xy > x^2y^2\) always.
EDIT: Correct statement is \(x = +/-\sqrt{\frac{3}{14}}\)
pmenon
