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Is |xy| > x^2*y^2 ?

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Is |xy| > x^2*y^2 ? [#permalink]  17 Aug 2013, 11:51
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Question Stats:

74% (02:12) correct 26% (01:09) wrong based on 189 sessions
Is |xy| > x^2*y^2 ?

(1) 0 < x^2 < 1/4
(2) 0 < y^2 < 1/9

Source: GMAT Prep Question Pack 1
Difficulty: Medium

--------------------
[Reveal] Spoiler:
Can someone please explain what to do with |xy| > x^2y^2 before we look into the equations?

I got |xy| > (xy)^2 but I didn't know how to interpret the inequality from here. Thanks in advance
[Reveal] Spoiler: OA

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Re: Is |xy| > x^2*y^2 ? [#permalink]  17 Aug 2013, 12:01
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DelSingh wrote:
|xy| > x^2y^2 ?

1) 0 < x^2 < 1/4

2) 0 < y^2 < 1/9

IMO C

$$|xy| > x^2y^2$$
since both sides are positive square both sides
$$(xy)^2 > (xy)^4$$
$$(xy)^2((xy)^2-1)<0$$
since $$(xy)^2>0$$ therefore $$(xy)^2-1<0$$
$$(xy)^2<1$$
or -1<xy<1
......so finally this is question.

finally you need both x and y to come to conclusion

STATEMENT 1==>ONLY X HENCE INSUFFICIENT.
$$0 < x^2 < 1/4$$
$$-1/2<x<1/2$$

STATEMENT 2 ==>ONLY Y HENCE INSUFFICIENT
$$0 < y^2 < 1/9$$
$$-1/3<y<1/3$$
now combining both clearly $$-1<xy<1$$
hence C

HOPE IT HELPS
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Last edited by blueseas on 17 Aug 2013, 14:16, edited 1 time in total.
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Re: Is |xy| > x^2*y^2 ? [#permalink]  17 Aug 2013, 13:45
blueseas wrote:
DelSingh wrote:
|xy| > x^2y^2 ?

1) 0 < x^2 < 1/4

2) 0 < y^2 < 1/9

$$|xy| > x^2y^2$$
since both sides are positive square both sides
$$(xy)^2 > (xy)^4$$
$$(xy)^2((xy)^2-1)<0$$
since $$(xy)^2>0$$ therefore $$(xy)^2-1<0$$
$$(xy)^2<1$$
......so finally this is question.

finally you need both x and y to come to conclusion

STATEMENT 1==>ONLY X HENCE INSUFFICIENT.
STATEMENT 2 ==>ONLY Y HENCE INSUFFICIENT
hence D

HOPE IT HELPS

If both are insufficient OA is either C or E. Could you please elaborate?

for me the OA is C

Case 1: -> -1/2 < x < 1/2 and x <> 0

we dont know if |xy|>x^2y^2 as we dont know about Y

case 2; -> -1/3 < y < 1/3 and y <> 0

we dont know if |xy|>x^2y^2 as we dont know about X

Combining both

for any values of x & Y , |xy| > x^2y^2
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Re: Is |xy| > x^2*y^2 ? [#permalink]  17 Aug 2013, 14:17
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If both are insufficient OA is either C or E. Could you please elaborate?

for me the OA is C

THANKS .
that was mistake.
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Is |xy|>x^2y^2 [#permalink]  24 Jul 2014, 19:33
Q. $$Is |xy|>x^{2}y^{2}$$

1.$$0<x^{2}<1/4$$
2. $$0<y^{2}<1/9$$

[Reveal] Spoiler:
In the answer explanation, the question is boiled down to is x^2 *y^2< 1..
Where as I solved it by saying that since x^2/geq{0} and y^2/geq{0} and thus not equal to zero the expression is true..I don't see a reason to prove less than 1 because if value of x and y are less than 1 then surely x^2y^2 will be less than 1...am I correct ??

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Re: Is |xy|>x^2y^2 [#permalink]  24 Jul 2014, 19:49
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Here, it helps to know that numbers greater than 1, when squared, are larger. Numbers between 0 and 1, when squared, are smaller.

Once you establish that, you're just saying:

Statement 1: x^2 = smaller; y^2 = unknown (smaller or large) ==> Don't know if product is larger or smaller because you don't know magnitude
Statement 2: y^2 = smaller; x^2 = unknkown (smaller or larger) ==> Don't know if product is larger or smaller because you don't know magnitude

Statement 1 and 2: x^2 = smaller; y^2 = smaller ==> product is smaller because both numbers are smaller.

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Re: Is |xy|>x^2y^2 [#permalink]  24 Jul 2014, 20:00
I will go with C .

| xy | is a positive and x^2 y^2 must be positive . x and y are positives or negatives . it does not matter . Only way to satisfy the condition is that both X & Y must be fractions .

basically we are asked that whether both x and y are fraction ?

1) it tells us x is a fraction because the highest possible value of x can be 1/2 . no info about y hence not sufficient.

2) it tells us , y is a fraction but no info about x . not sufficient

(1) + (2) , now both x & y are fractions so

| xy | will always be greater than x ^2 y ^2 .

hence sufficient . so answer is C

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Re: Is |xy|>x^2y^2 [#permalink]  24 Jul 2014, 21:39
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WoundedTiger wrote:
Q. $$Is |xy|>x^{2}y^{2}$$

1.$$0<x^{2}<1/4$$
2. $$0<y^{2}<1/9$$

[Reveal] Spoiler:
In the answer explanation, the question is boiled down to is x^2 *y^2< 1..
Where as I solved it by saying that since x^2/geq{0} and y^2/geq{0} and thus not equal to zero the expression is true..I don't see a reason to prove less than 1 because if value of x and y are less than 1 then surely x^2y^2 will be less than 1...am I correct ??

Two variables are confusing you.

Note that the question is just this:

Is $$|xy|> x^{2}y^{2}$$
Is $$|xy|> |xy|^2$$
Is $$z > z^2$$ where $$z = |xy|$$

When is z greater than z^2? When z lies between -1 and 1 or we can say between 0 and 1 when z is positive.

1.$$0<x^{2}<1/4$$
This tells you that 0 < |x| < 1/2. Doesn't tell you anything about y so you don't know anything about z.

2. $$0<y^{2}<1/9$$
This tells you that 0 < |y| < 1/3. Doesn't tell you anything about x so you don't know anything about z.

Both together, you know that |x|*|y| is less than 1 i.e. z is less than 1. Hence z WILL BE greater than z^2.

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Re: Is |xy| > x^2*y^2 ? [#permalink]  24 Jul 2014, 23:28
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Expert's post
WoundedTiger wrote:
Q. $$Is |xy|>x^{2}y^{2}$$

1.$$0<x^{2}<1/4$$
2. $$0<y^{2}<1/9$$

[Reveal] Spoiler:
In the answer explanation, the question is boiled down to is x^2 *y^2< 1..
Where as I solved it by saying that since x^2/geq{0} and y^2/geq{0} and thus not equal to zero the expression is true..I don't see a reason to prove less than 1 because if value of x and y are less than 1 then surely x^2y^2 will be less than 1...am I correct ??

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Merging topics.
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Re: GMAT prep question pack 1 [#permalink]  18 Aug 2014, 22:26
If (xy)^2 is positive, (xy)^2>(xy)4 or 1> (xy)^2
I and 2 are insufficient because in each statement other value is missing.

By combining
We know that (xy)^2 is positive. So,

(xy)^2< (1/4)*(1/9)
Clearly, 1> (xy)^2
Is my reasoning correct? Need expert help
Re: GMAT prep question pack 1   [#permalink] 18 Aug 2014, 22:26
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