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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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New post Updated on: 16 Feb 2012, 21:37
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A
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E

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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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New post 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.

Answer: E.
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Re: xy > x2y2?  [#permalink]

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New post 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)
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Re: xy > x2y2?  [#permalink]

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New post 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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Re: xy > x2y2?  [#permalink]

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New post 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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Re: xy > x2y2?  [#permalink]

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New post 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 :oops: 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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New post 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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New post 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.

Answer: E.


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

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New post 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.

Answer: E.


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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 06 Jun 2016, 23:41
jade3 wrote:
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.

(E) is the answer.
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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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New post 21 Jun 2017, 11:53
jade3 wrote:
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.

Hence, Answer is 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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New post 10 Feb 2019, 09:19
jade3 wrote:
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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