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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  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
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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 \((xy)^2 > (xy)^4\) \((xy)^2((xy)^21)<0\) since \((xy)^2>0\) therefore \((xy)^21<0\) \((xy)^2<1\) 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 CHOPE 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 \((xy)^2 > (xy)^4\) \((xy)^2((xy)^21)<0\) since \((xy)^2>0\) therefore \((xy)^21<0\) \((xy)^2<1\) ......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 DHOPE 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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maaadhu wrote:
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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Re: Is xy>x^2y^2 [#permalink]
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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.
Therefore, correct answer is (C).



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Re: Is xy>x^2y^2 [#permalink]
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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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Re: Is xy>x^2y^2 [#permalink]
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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. \(0<y^{2}<1/9\) 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. Answer (C)
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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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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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Is xy > x^2*y^2 ? [#permalink]
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Updated on: 14 Dec 2017, 21:23
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. \(xy > x^2 \cdot y^2\) \(⇔ xy > xy^2\) \(⇔ xy^2 xy < 0\) \(⇔ xy(xy1) < 0\) \(⇔ 0 < xy < 1\) 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.
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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  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. \(xy > x^2 \cdot y^2\) \(⇔ xy > xy^2\) \(⇔ xy^2 xy < 0\) \(⇔ xy(xy1) < 0\) \(⇔ 0 < xy < 1\) 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.
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