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# Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1

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Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

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Updated on: 16 Feb 2012, 21:37
6
00:00

Difficulty:

55% (hard)

Question Stats:

59% (02:00) correct 41% (01:54) wrong based on 169 sessions

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Is xy > x^2*y^2?

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

Originally posted by jade3 on 24 Nov 2009, 20:56.
Last edited by Bunuel on 16 Feb 2012, 21:37, edited 1 time in total.
Edited the question and added the OA
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Re: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1  [#permalink]

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16 Feb 2012, 21:44
3
3
Baten80 wrote:
if as follows:
xy - x^2*y^2 > 0
xy(1- xy) > 0
so, xy>0
or, x>0 or, y>0
and 1-xy > 0
xy<1
Thus no statements are sufficient.
Ans. E

What are wrongs with this approach?

Is xy > x^2*y^2?

Is $$xy>x^2*y^2$$? --> is $$0<xy<1$$? (the same way as a>a^2 means 0<a<1)

(1) 14*x^2 = 3. Clearly insufficient, since no info about y.
(2) y^2 = 1. Clearly insufficient, since no info about x.

(1)+(2) If $$x$$ and $$y$$ have the same sign then $$xy=\sqrt{\frac{3}{14}}$$ and the answer will be YES but if $$x$$ and $$y$$ have the opposite signs then $$xy=-\sqrt{\frac{3}{14}}$$ and the answer will be NO. Not sufficient.

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24 Nov 2009, 21:33
E

both statements dont reference both X and Y

and statement 2 leaves Y with both positive and negative values thus it can be greater or less. (also the square of X is less than X but that doesnt matter if the products can be positive or negative)
Joined: 20 Aug 2009
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Updated on: 27 Nov 2009, 04:10
Let's consider both statements together

$$xy$$ could be positive or negative;

if $$xy<0$$, clearly $$xy \leq x^2y^2$$, no calculations are needed. Inequality in question stem is false

If $$xy>0$$, than $$y=1; x=sqrt{3/14}$$ or $$y=-1; x=-sqrt{3/14}$$. once more no need to calculate this, just note that $$xy<1$$. In this case $$(xy)^2<xy<1$$

[strike]So, in both cases $$xy<x^2y^2$$[/strike]

Originally posted by shalva on 26 Nov 2009, 00:21.
Last edited by shalva on 27 Nov 2009, 04:10, edited 1 time in total.
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26 Nov 2009, 19:43
shalva wrote:
Let's consider both statements together

$$xy$$ could be positive or negative;

if $$xy<0$$, clearly $$xy \leq x^2y^2$$, no calculations are needed. Inequality in question stem is false

If $$xy>0$$, than $$y=1; x=sqrt{3/14}$$ or $$y=-1; x=-sqrt{3/14}$$. once more no need to calculate this, just note that $$xy<1$$. In this case (xy)^2<xy<1

So, in both cases xy<x^2y^2

(C)

I don't get this, in the blue part, you mentioned that (xy)^2<xy, then your conclusion (the red part) was in both cases xy<x^2y^2.

Anyway, IMO, E is the answer because (xy)^2 = 3/14.
If xy<0, clearly that xy<(xy)^2.
If xy>0, because (xy)^2 =3/14 <1 => xy>(xy)^2.

Hence E.
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27 Nov 2009, 04:11
fall2009 wrote:
I don't get this, in the blue part, you mentioned that (xy)^2<xy, then your conclusion (the red part) was in both cases xy<x^2y^2.

Anyway, IMO, E is the answer because (xy)^2 = 3/14.
If xy<0, clearly that xy<(xy)^2.
If xy>0, because (xy)^2 =3/14 <1 => xy>(xy)^2.

Hence E.

You're absolutely right In one case $$xy<x^2y^2$$, in other - $$xy>x^2y^2$$
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Re: Is xy > x2y2? (1) 14x2 = 3 (2) y2 = 1  [#permalink]

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16 Feb 2012, 19:42
if as follows:
xy - x^2*y^2 > 0
xy(1- xy) > 0
so, xy>0
or, x>0 or, y>0
and 1-xy > 0
xy<1
Thus no statements are sufficient.
Ans. E

What are wrongs with this approach?
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

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29 Apr 2015, 22:46
Bunuel wrote:
Baten80 wrote:
if as follows:
xy - x^2*y^2 > 0
xy(1- xy) > 0
so, xy>0
or, x>0 or, y>0
and 1-xy > 0
xy<1
Thus no statements are sufficient.
Ans. E

What are wrongs with this approach?

Is xy > x^2*y^2?

Is $$xy>x^2*y^2$$? --> is $$0<xy<1$$? (the same way as a>a^2 means 0<a<1)

(1) 14*x^2 = 3. Clearly insufficient, since no info about y.
(2) y^2 = 1. Clearly insufficient, since no info about x.

(1)+(2) If $$x$$ and $$y$$ have the same sign then $$xy=\sqrt{\frac{3}{14}}$$ and the answer will be YES but if $$x$$ and $$y$$ have the opposite signs then $$xy=-\sqrt{\frac{3}{14}}$$ and the answer will be NO. Not sufficient.

Hi Bunuel,

Can you explain how you got this step? Thanks!

Is xy>x2∗y2? --> is 0<xy<1? (the same way as a>a^2 means 0<a<1)
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Posts: 53066
Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

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30 Apr 2015, 02:30
amnesia604 wrote:
Bunuel wrote:
Baten80 wrote:
if as follows:
xy - x^2*y^2 > 0
xy(1- xy) > 0
so, xy>0
or, x>0 or, y>0
and 1-xy > 0
xy<1
Thus no statements are sufficient.
Ans. E

What are wrongs with this approach?

Is xy > x^2*y^2?

Is $$xy>x^2*y^2$$? --> is $$0<xy<1$$? (the same way as a>a^2 means 0<a<1)

(1) 14*x^2 = 3. Clearly insufficient, since no info about y.
(2) y^2 = 1. Clearly insufficient, since no info about x.

(1)+(2) If $$x$$ and $$y$$ have the same sign then $$xy=\sqrt{\frac{3}{14}}$$ and the answer will be YES but if $$x$$ and $$y$$ have the opposite signs then $$xy=-\sqrt{\frac{3}{14}}$$ and the answer will be NO. Not sufficient.

Hi Bunuel,

Can you explain how you got this step? Thanks!

Is xy>x2∗y2? --> is 0<xy<1? (the same way as a>a^2 means 0<a<1)

$$xy>x^2*y^2$$;
$$xy>(xy)^2$$;
$$xy(1-xy)>0$$;
$$0<xy<1$$.

Hope it helps.
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

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13 Mar 2016, 07:34
Here nothing is specified about the values of x adn y being >0 or <0
so => not sufficient.
hence E
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

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06 Jun 2016, 23:41
Is xy > x^2*y^2?

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

xy > $$x^2$$*$$y^2$$

xy - $$x^2$$*$$y^2$$ > 0

xy (1-xy) > 0

0<xy < 1 --> This is our revised problem

Statement 1 --> clearly not sufficient, as no info about y.

Statement 2 --> clearly not sufficient, as no info about y.

Combining (1) and (2)

$$x^2$$*$$y^2$$ = 3/14.

One of the solution of xy lies between 0 and 1;

Other solution does not lie between 0 and 1 .

So not sufficient.

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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

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21 Jun 2017, 11:53
Is xy > x^2*y^2?

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

This is definitely a trap question in DS.

Given: xy > x^2y^2

(1) 14 * x^2 = 3

x^2 = 3/14 =====> x = +/-\sqrt{3/14} ==== Clearly Not Sufficient as we do not know value of y.

(2) y^2 = 1

y = +/- 1 =====> Clearly Not Sufficient as we do not know value of x.

Combining both (1) & (2)

We are still not sure of the value of x & y will be positive or negative. Hence, for the question Is xy > x^2*y^2, we get both the answers YES & NO.

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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1  [#permalink]

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10 Feb 2019, 09:19
Is xy > x^2*y^2?

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

KeyWord x and y can be +ive or -ive

from 1 x^2 = 3/14, y can take any value

from 2 y^2 = 1, x can take any value

Now both x and y can be +ive or -ive

different signs of x and y, can result in different answers for the given inequality

E
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Re: Is xy > x^2*y^2? (1) 14*x^2 = 3 (2) y^2 = 1   [#permalink] 10 Feb 2019, 09:19
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