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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

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 _________________

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 ??

_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: Is |xy|>x^2y^2 [#permalink]
24 Jul 2014, 19:49

2

This post received KUDOS

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.

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

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.

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 ??

Re: Is |xy| > x^2*y^2 ? [#permalink]
24 Aug 2015, 04:44

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