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\(|xy| > x^2y^2\) 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

\(|xy| > x^2y^2\) 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.”

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.

| 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 ??

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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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