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

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Is |xy| > x^2*y^2 ? [#permalink]

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17 Aug 2013, 12:51
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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]

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17 Aug 2013, 13:01
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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, 15:16, edited 1 time in total.

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Re: Is |xy| > x^2*y^2 ? [#permalink]

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17 Aug 2013, 14:45
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]

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17 Aug 2013, 15:17
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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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24 Jul 2014, 20:33
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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24 Jul 2014, 20:49
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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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24 Jul 2014, 21:00
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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24 Jul 2014, 22:39
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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]

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25 Jul 2014, 00:28
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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 ??

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Merging topics.
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Re: GMAT prep question pack 1 [#permalink]

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18 Aug 2014, 23:26
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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

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Re: Is |xy| > x^2*y^2 ? [#permalink]

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Re: Is |xy| > x^2*y^2 ? [#permalink]

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15 Dec 2015, 02:31
VeritasPrepKarishma wrote:
When is z greater than z^2? When z lies between -1 and 1

VeritasPrepKarishma :
Except for z=0 right?

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Re: Is |xy| > x^2*y^2 ? [#permalink]

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15 Dec 2015, 22:15
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

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

When you modify the original condition and the problem, |xy|>|xy|^2?, 0>|xy|^2-|xy|?, 0>|xy|(|xy|-1)?. That is, 0<|xy|<1? --> xy=/0 and 1<xy<1?.
There are 2 variables(x,y), which should match with the number of equations. So, you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make c the answer. In 1)&2), x=/0 and -1/2<x<1/2, y=/0 and -1/3<y<1/3, xy=/0 and -1/6<xy<1/6, which is always yes and sufficient. Therefore, the answer is C.

-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is |xy| > x^2*y^2 ? [#permalink]

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14 Feb 2017, 11:03
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Re: Is |xy| > x^2*y^2 ?   [#permalink] 14 Feb 2017, 11:03
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