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marcodonzelli
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Is xy > x^2*y^2? (1) 14x^2 = 3 (2) y^2 = 1 19 Jan 2008, 06:15
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Is xy > x^2*y^2?

(1) 14x^2 = 3
(2) y^2 = 1



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Is xy > x^2*y^2? (1) 14x^2 = 3 (2) y^2 = 1 19 Jan 2008, 14:01
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Statement 1 OR

if X is positive X >
if X is negative X <

INSUFFICIENT

Statement 2 y = 1 OR -1

if Y is positive Y =
if Y is negative Y <

INSUFFICIENT

Taken together they are still insufficient. Here are all the possibilities:

Both X and Y are positive
Both X and Y are negative
X is positive, Y is negative
X is negative, Y is positive

Now here's what happens with each possibility

X and Y positive: XY >
X and Y negative: XY >
X positive, Y negative: XY <
X negative, Y positive: XY <

Since we know nothing of the signs in this problem we cannot come to a conclusive answer.

Answer E
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Is xy > x^2*y^2? (1) 14x^2 = 3 (2) y^2 = 1 19 Jan 2008, 14:28
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marcodonzelli
Is xy > x^2*y^2?

(1) 14x^2 = 3
(2) y^2 = 1
Show more

I get C.

both x and y positive
Is sqrt(3/14) * 1 > 3/14 -----NO

x -ve and y +ve
Is -sqrt(3/14) * 1 > 3/14 * 1-----NO

x +ve and y -ve

Is sqrt(3/14) * (-1) > 3/14* (-1)^2-------NO

both are negative
Is (-1)sqrt(3/14) * (-1) > 3/14------NO

Can be answered no for all situations.
C
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walker
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Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
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Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Posts: 2,396
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Is xy > x^2*y^2? (1) 14x^2 = 3 (2) y^2 = 1 19 Jan 2008, 14:37
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ashkrs
marcodonzelli
Is xy > x^2*y^2?

(1) 14x^2 = 3
(2) y^2 = 1

I get C.

both x and y positive
Is sqrt(3/14) * 1 > 3/14 -----NO

x -ve and y +ve
Is -sqrt(3/14) * 1 > 3/14 * 1-----NO

x +ve and y -ve

Is sqrt(3/14) * (-1) > 3/14* (-1)^2-------NO

both are negative
Is (-1)sqrt(3/14) * (-1) > 3/14------NO

Can be answered no for all situations.
C
Show more

sqrt(3/14) * 1 > 3/14 is true.

to not calculate you can remember rule:
for X>1
\(1<x^{\frac15}<x^{\frac14}<x^{\frac13}<x^{\frac12}<x<x^2<x^3<x^4<x^5....\)

for 0<X<1
\(1>x^{\frac15}>x^{\frac14}>x^{\frac13}>x^{\frac12}>x>x^2>x^3>x^4>x^5....\)
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ashkrs
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GMAT 1: 680 Q48 V35
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Is xy > x^2*y^2? (1) 14x^2 = 3 (2) y^2 = 1 19 Jan 2008, 14:48
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walker
ashkrs
marcodonzelli
Is xy > x^2*y^2?

(1) 14x^2 = 3
(2) y^2 = 1

I get C.

both x and y positive
Is sqrt(3/14) * 1 > 3/14 -----NO

x -ve and y +ve
Is -sqrt(3/14) * 1 > 3/14 * 1-----NO

x +ve and y -ve

Is sqrt(3/14) * (-1) > 3/14* (-1)^2-------NO

both are negative
Is (-1)sqrt(3/14) * (-1) > 3/14------NO

Can be answered no for all situations.
C

sqrt(3/14) * 1 > 3/14 is true.

to not calculate you can remember rule:
for X>1
\(1x^{\frac15}>x^{\frac14}>x^{\frac13}>x^{\frac12}>x>x^2>x^3>x^4>x^5....\)
Show more

silly me..i solved in my notes using 14/3 ..and got ans
yes E



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!