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If xy ≠ 0, is x^3 + y^3 > 0 ?
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24 Jun 2018, 11:23
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If xy ≠ 0, is x^3 + y^3 > 0 ? (1) x + y > 0 (2) xy > 0 NEW question from GMAT® Official Guide 2019 (DS15561)
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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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25 Jun 2018, 02:40
Bunuel wrote: If xy ≠ 0, is x^3 + y^3 > 0 ? (1) x + y > 0 (2) xy > 0 NEW question from GMAT® Official Guide 2019 (DS15561) Hi... Is \(x^3+y^3>0\)? 1) x+y>0 This means atleast one of x and y is positive and that too higher numeric value.. Say x is positive then x>y And if y is positive then y>x And if both are positive, then it doesn't matter which has greater numeric value. This is the reason why x^3+y^3 will also be >0 Sufficient 2) xy>0 This means both x and y are os same sign.. So if both are positive, \(x^3+y^3>0\).. And if both are negative, \(x^3+y^3<0\) Insufficient A
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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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25 Jun 2018, 01:59
Bunuel wrote: If xy ≠ 0, is x^3 + y^3 > 0 ? (1) x + y > 0 (2) xy > 0 NEW question from GMAT® Official Guide 2019 (DS15561) From the question \(x^3+y^3>0\). I think of the possible cases that will make the equation is true which are 1. X and Y are positive value 2. X is positive value with higher magnitude than Y e.g. 2 + (1) > 0 (1) x + y > 0 Let us know that x+y is positive value which let us know the possible case of x and y are 1. X and Y are positive value 2. X is positive value with higher magnitude than Y which is the same as our previous analysis > Sufficient (2) xy > 0 This statement show us that xy is positive value which let us know the possible case of x and y are 1. X and Y are positive value > e.g. 2*3>0 2. X and Y are negative value > e.g. 2*3>0 This statement is not sufficient since \(x^3+y^3\) can result in both positive or negative values > insufficient So my answer is "A"




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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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24 Jun 2018, 23:12
Bunuel wrote: If xy ≠ 0, is x^3 + y^3 > 0 ? (1) x + y > 0 (2) xy > 0 NEW question from GMAT® Official Guide 2019 (DS15561) 1) Suff, if the sum is >0 , sum of cubes will also be >0 , we can test cases , only way for sum to be positive is by having greater magnitude of positive number ,hence the sum of cube will be more positive than negative cube , hence always be positive 2) Not suff, this tells that both the numbers have same sign, but for negative cases the sum of cube will be more negative , hence not suff A



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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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16 Sep 2018, 04:24
QuestionIf xy ≠ 0, is x^3 + y^3 > 0 ? can be further drill down : In order to x^3 + y^3 > 0, Is X, and Y  Both are positive, or X is positive, and greater than Y, if Y is negative
1) x + y > 0 With the help of this statement we can say that either Both X, and Y are Positive, or X is positive, and greater than Y, if Y is negative Sufficient
2) xy>0 Means both either positive, or Negative.
Hence, Insufficient



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If xy ≠ 0, is x^3 + y^3 > 0 ?
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27 Jun 2018, 04:52
I did it slightly different than the methods above for option A (correct ans)
x^3 + y^3 >0 > x^3 > y^3
if we can arrange A in the same pattern and cube it, we get x^3 > y^3
B ans yes/ no as x,y can be +ve or ve



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If xy ≠ 0, is x^3 + y^3 > 0 ?
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Updated on: 30 Sep 2019, 06:39
\(x^3+y^3= (x+y)(x^2+y^2xy)\) = \((x+y)[x^2+\frac{(y^2)}{4}xy+\frac{3(y^2)}{4}]\)
= \((x+y)[(x\frac{y}{2})^2}+\frac{3(y^2)}{4}]\) as\([(x\frac{y}{2})^2}+\frac{3(y^2)}{4}]\)will always be positive as x,y≠0 Hence \(x^3+y^3\) will take the same sign as (x+y) Option A is correct
Originally posted by nick1816 on 20 Apr 2019, 23:22.
Last edited by nick1816 on 30 Sep 2019, 06:39, edited 2 times in total.



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If xy ≠ 0, is x^3 + y^3 > 0 ?
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Updated on: 25 Jun 2018, 05:50
If xy ≠ 0, is x^3 + y^3 > 0 ? (1) x + y > 0 (2) xy > 0 Statement 1 is insuff. as x and y can be either both positive number or one positive and other negative. Statemtn 2 just tell us that the sign of both varibale are same, hence insuff combining both we get to know that both x and y has same sign hence the answer is C. updating realized my mistake, A is correct.
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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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24 Jun 2018, 11:38
A for me, because it is sufficient to know that x + y > 0 to draw the conclusion that x^3+y^3>0



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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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25 Jun 2018, 01:42
Answer should be A. If x+y>0, then atleast one number is poistive and the magnitude of this number will be greater than the magnitude of the other number. Hence the cube of this number will also be greater than teh cube of the other number. Hence x^3+Y^3>0. Sufficient



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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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03 Jul 2018, 08:00
Bunuel wrote: If xy ≠ 0, is x^3 + y^3 > 0 ? (1) x + y > 0 (2) xy > 0 NEW question from GMAT® Official Guide 2019 (DS15561) here is my reasoning (1) x + y > 0 if x + y > 0 take numbers 2: 1 this implies that X must be bigger and positive, in case Y is negative Answer is Always YES. Sufficient. (2) xy > 0 here both numbers can be negative say X is 4 and y is 2 anserr is NO , if we take X is 4; and Y is 2 answer is YES. Insufficient



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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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11 Jul 2018, 20:04
The last sentence of the official explanation for statement (1) has some typos and should instead be as follows: Therefore, x^3 > (y)^3, or x^3 > y^3, or x^3 + y^3 > 0. As written, this sentence contradicts the last sentence of the previous paragraph, which offers an alternative solution.
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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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15 Sep 2018, 13:12
Hi, For someone like me who is not good at number picking ( i must improve this as it costs me a lot), here is alebric approach We know that \((x+y)^3= x^3+y^3+3xy(x+y)\) or \(x^3+y^3= (x+y)^3 3xy(x+y)\) \(x^3+y^3= (x+y)((x+y)^2 3xy)\) \(x^3+y^3= (x+y)((x^2+y^2 +2xy  3xy)\) \(x^3+y^3= (x+y)((x^2+y^2 xy)\) so if we get to know that either (x+y) & \((x^2+y^2 xy)\)is >0 or both are <0 then we can conclude about \(x^3+y^3\) Now statement A : x+y >0 this means x>0, y>0 so xy>0 hence we can see that (x+y) >0 . Ok what about this exp \((x^2+y^2 xy)\) . For x>0 and y>0 we will have x^2>0, y^2>0 and xy>0 but we can always see that x^2+y^2> xy so we have \((x^2+y^2 xy)\) also >0 Hence we can conclude that \(x^3+y^3\)>0 or x>0, y<0 and x>y , so xy would be <0 , so we can see that (x+y) >0 & \((x^2+y^2 xy)\) >0 hence we can say that \(x^3+y^3\)>0 or x<0, y>0 and y>x , so xy would be <0 , so we can see that (x+y) >0 & \((x^2+y^2 xy)\) >0 hence we can say that \(x^3+y^3\)>0 Hence A is sufficient. Now Statement B: its says xy>0 all i can infer is either x>0&y>0 or x<0&y<0 but for this (x+y) can be either >0 or the exp (x+y) <0 So really cant conclude that the exp \(x^3+y^3\)>0 We have already seen that if x+y>0 then we can conclude that \(x^3+y^3\)>0 But if x+y<0 , then the exp \((x^2+y^2 xy)\) must also be <0 for\(x^3+y^3\)>0 But we can have that \(x^2+y^2\)<xy. in that case we will have \(x^3+y^3\)>0 or\(x^2+y^2\)>xy in that case we will have \(x^3+y^3\)<0 hence B is insufficient. Probus
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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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20 Apr 2019, 22:39
Hi Bunuel, Could you help explain why my approach resulted in wrong answer here: with (xy)^2 >= 0, we have x^2 + y^2 >= 2xy. Apply this to the equation as below: x^3 + y^3 = (x+y)*(x^2  xy + y^2) >= (x+y)*(2xy  xy) = (x+y)*xy. This one is >0 if both (x+y) > 0 and xy>0 or (x+y) <0 and xy < 0 Due to the above, I chose C. What did I do wrong here?



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If xy ≠ 0, is x^3 + y^3 > 0 ?
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Updated on: 01 Oct 2019, 05:43
Statement I is sufficient to answer the equation.
Originally posted by NoMatterWhat on 30 Sep 2019, 06:31.
Last edited by NoMatterWhat on 01 Oct 2019, 05:43, edited 1 time in total.



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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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30 Sep 2019, 15:52
(1) x + y > 0 (2) xy > 0
Statement 1 is sufficient as the relationship of x^3 + y^3 dictates that x and y will remain in proportion and will always be >0 if we are told this. Test numbers and you'll see.
Statement 2 is insufficient  we are simply told x and y are the same sign If x and y are negative then x=1 y=2 1^3 + 2^3 = 9 x+y <0
If positive then x^3+y^3 > 0
Insufficient



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If xy ≠ 0, is x^3 + y^3 > 0 ?
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Updated on: 01 Oct 2019, 05:41
dcummins wrote: (1) x + y > 0 (2) xy > 0
Statement 1 is sufficient as the relationship of x^3 + y^3 dictates that x and y will remain in proportion and will always be >0 if we are told this. Test numbers and you'll see.
Statement 2 is insufficient  we are simply told x and y are the same sign If x and y are negative then x=1 y=2 1^3 + 2^3 = 9 x+y <0
If positive then x^3+y^3 > 0
Insufficient I agree with your solution
Originally posted by NoMatterWhat on 01 Oct 2019, 05:09.
Last edited by NoMatterWhat on 01 Oct 2019, 05:41, edited 1 time in total.



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If xy ≠ 0, is x^3 + y^3 > 0 ?
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01 Oct 2019, 05:23
NoMatterWhat wrote: dcummins wrote: (1) x + y > 0 (2) xy > 0
Statement 1 is sufficient as the relationship of x^3 + y^3 dictates that x and y will remain in proportion and will always be >0 if we are told this. Test numbers and you'll see.
Statement 2 is insufficient  we are simply told x and y are the same sign If x and y are negative then x=1 y=2 1^3 + 2^3 = 9 x+y <0
If positive then x^3+y^3 > 0
Insufficient is x^3 + y^3 > 0 ? I = x+y >0 x=y=1 x^3 + y^3 > 0? ; IT YIELDS YESor x=y=0.75 But when we enter the value in eq; x^3 + y^3 > 0?(0.75)^3 + (0.75)^3 > 0 ; IT YEILDS NO . How Option A is Sufficient?Mate what are you talking about? (3/4)^3 is positive and greater than 0 so (3/4)^3 x2 is greater than 0 Reread your working. Its wrong. Even if you try x=2 and y = 1 x +y still needs to be greater than 0 per the statement, and since cubic roots maintain the sign of their base x^3+y^3 will be proportionately greater than 0



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Re: If xy ≠ 0, is x^3 + y^3 > 0 ?
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01 Oct 2019, 05:40
dcummins wrote: NoMatterWhat wrote: dcummins wrote: (1) x + y > 0 (2) xy > 0
Statement 1 is sufficient as the relationship of x^3 + y^3 dictates that x and y will remain in proportion and will always be >0 if we are told this. Test numbers and you'll see.
Statement 2 is insufficient  we are simply told x and y are the same sign If x and y are negative then x=1 y=2 1^3 + 2^3 = 9 x+y <0
If positive then x^3+y^3 > 0
Insufficient is x^3 + y^3 > 0 ? I = x+y >0 x=y=1 x^3 + y^3 > 0? ; IT YIELDS YESor x=y=0.75 But when we enter the value in eq; x^3 + y^3 > 0?(0.75)^3 + (0.75)^3 > 0 ; IT YEILDS NO . How Option A is Sufficient?Mate what are you talking about? (3/4)^3 is positive and greater than 0 so (3/4)^3 x2 is greater than 0 Reread your working. Its wrong. Even if you try x=2 and y = 1 x +y still needs to be greater than 0 per the statement, and since cubic roots maintain the sign of their base x^3+y^3 will be proportionately greater than 0 OH MY LORD I was trying to ask whether the equation is more than 1 or not, which was not what asked in the question. My blunder....I do apologize. the question got me and I couldn't see my blunder and made 2 posts...I think I should take some rest...HAHAAHA



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If xy ≠ 0, is x^3 + y^3 > 0 ?
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01 Oct 2019, 15:08
NoMatterWhat wrote: OH MY LORD
I was trying to ask whether the equation is more than 1 or not, which was not what asked in the question.
My blunder....I do apologize.
the question got me and I couldn't see my blunder and made 2 posts...I think I should take some rest...HAHAAHA Yea no worries. A positive fraction will only decrease in value as you raise it to a positive exponent power greater than 1 anyway 1/2 is 0.5 away from 1 1/4 is 0.25 away from 1 e.g. (1/2)^4 = 1/16 =0.0625 (15/16 away from 1) (3/4)^2 = 9/16 =0.565 (15/16 away from 1)




If xy ≠ 0, is x^3 + y^3 > 0 ?
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