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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 [#permalink]  19 Jan 2008, 05:15
Is xy > x^2*y^2?

(1) 14x^2 = 3
(2) y^2 = 1
Director
Joined: 12 Jul 2007
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Re: inequalities [#permalink]  19 Jan 2008, 13:01
Statement 1 $formdata=x+=+sqrt{\frac{3}{14}}$ OR $formdata=-sqrt{\frac{3}{14}}$

if X is positive X > $formdata=x^2$
if X is negative X < $formdata=x^2$

INSUFFICIENT

Statement 2 y = 1 OR -1

if Y is positive Y = $formdata=y^2$
if Y is negative Y < $formdata=y^2$

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 > $formdata=x^2y^2$
X and Y negative: XY > $formdata=x^2y^2$
X positive, Y negative: XY < $formdata=x^2y^2$
X negative, Y positive: XY < $formdata=x^2y^2$

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

Director
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Re: inequalities [#permalink]  19 Jan 2008, 13:28
marcodonzelli wrote:
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
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Re: inequalities [#permalink]  19 Jan 2008, 13:37
Expert's post
ashkrs wrote:
marcodonzelli wrote:
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
$$1<x^{\frac15}<x^{\frac14}<x^{\frac13}<x^{\frac12}<x<x^2<x^3<x^4<x^5....$$

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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Director
Joined: 08 Jun 2007
Posts: 584
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Re: inequalities [#permalink]  19 Jan 2008, 13:48
walker wrote:
ashkrs wrote:
marcodonzelli wrote:
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
$$1<x^{\frac15}<x^{\frac14}<x^{\frac13}<x^{\frac12}<x<x^2<x^3<x^4<x^5....$$

for 0<X<1
$$1>x^{\frac15}>x^{\frac14}>x^{\frac13}>x^{\frac12}>x>x^2>x^3>x^4>x^5....$$

silly me..i solved in my notes using 14/3 ..and got ans
yes E
Re: inequalities   [#permalink] 19 Jan 2008, 13:48
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