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

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Manager
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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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Updated on: 17 Aug 2013, 15:16
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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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Originally posted by blueseas on 17 Aug 2013, 13:01.
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, 20:49
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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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Joined: 03 Jul 2013
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 11 Sep 2013 Posts: 151 Concentration: Finance, Finance Re: GMAT prep question pack 1 [#permalink] ### Show Tags 18 Aug 2014, 23:26 1 This post received KUDOS 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 Manager Joined: 15 Dec 2015 Posts: 51 Re: Is |xy| > x^2*y^2 ? [#permalink] ### Show Tags 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? Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5255 GMAT 1: 800 Q59 V59 GPA: 3.82 Is |xy| > x^2*y^2 ? [#permalink] ### Show Tags Updated on: 14 Dec 2017, 21:23 Expert's post 2 This post was BOOKMARKED Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question. $$|xy| > x^2 \cdot y^2$$ $$⇔ |xy| > |xy|^2$$ $$⇔ |xy|^2 -|xy| < 0$$ $$⇔ |xy|(|xy|-1) < 0$$ $$⇔ 0 < |xy| < 1$$ Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. Conditions 1) & 2) $$0 < x^2 < 1/4$$ $$⇔ 0 < |x| < 1/2$$ $$0 < y^2 < 1/9$$ $$⇔ 0 < |y| < 1/3$$ Thus $$0 < |xy| < 1/6$$. Since the range of the question includes the range of the conditions, both conditions together are sufficient. Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) Since we don't know y, the condition 1) is not sufficient. Condition 2) Since we don't know x, the condition 2) is not sufficient, either. Therefore, the answer is C. Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 3 month Online Course"
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Originally posted by MathRevolution on 15 Dec 2015, 22:15.
Last edited by MathRevolution on 14 Dec 2017, 21:23, edited 1 time in total.
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Re: Is |xy| > x^2*y^2 ? [#permalink]

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12 Dec 2017, 04:15
To solve this you dont need to solve at all.

Just look at the question stems. It is a relationship between X & Y. Which means you would need info of both X & Y to determine if the stem is true or not.

Statement 1- Only provides info on X
Statement 2- Only provides info on Y

Whcih elements ABD, You are on to C or E. By doing a bit of reasning you will Establish the answer is C
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Re: Is |xy| > x^2*y^2 ? [#permalink]

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14 Dec 2017, 21:21
DelSingh wrote:
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

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

$$|xy| > x^2 \cdot y^2$$
$$⇔ |xy| > |xy|^2$$
$$⇔ |xy|^2 -|xy| < 0$$
$$⇔ |xy|(|xy|-1) < 0$$
$$⇔ 0 < |xy| < 1$$

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first.

Conditions 1) & 2)

$$0 < x^2 < 1/4$$
$$⇔ 0 < |x| < 1/2$$

$$0 < y^2 < 1/9$$
$$⇔ 0 < |y| < 1/3$$

Thus $$0 < |xy| < 1/6$$.

Since the range of the question includes the range of the conditions, both conditions together are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since we don't know y, the condition 1) is not sufficient.

Condition 2)
Since we don't know x, the condition 2) is not sufficient, either.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Is |xy| > x^2*y^2 ?   [#permalink] 14 Dec 2017, 21:21
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