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If x and y are nonzero integers and x + y = 32, what is [#permalink]
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If x and y are nonzero integers and x + y = 32, what is xy? (1) 4x – 12y = 0 (2) x – y = 16 please explain how A is sufficient I got C. on solving A i get 2 possible values of x and y and thus 2 values of XY
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devinawilliam83 wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16
please explain how A is sufficient I got C. on solving A i get 2 possible values of x and y and thus 2 values of XY If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x  12y = 0\) > \(x+3y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs > so either \(x=x\) and \(y=y\) OR \(x=x\) and \(y=y\) > either \(x+y=x+y=3y+y=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\) OR \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A.
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Re: If x and y are nonzero integers [#permalink]
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21 Sep 2013, 22:32
Bunuel wrote: devinawilliam83 wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16
please explain how A is sufficient I got C. on solving A i get 2 possible values of x and y and thus 2 values of XY If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x  12y = 0\) > \(x+3y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs > so either \(x=x\) and \(y=y\) OR \(x=x\) and \(y=y\) > either \(x+y=x+y=3y+y=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\) OR \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A. Hi Bunel, i have a doubt in ur explanation x+y=32 For this the modulus sign wont have four cases? (x,+y), (+x,y), (+x,+y), (x,y) Please clarify me. Thanks in Advance, Rrsnathan.



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Re: If x and y are nonzero integers [#permalink]
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22 Sep 2013, 03:35
rrsnathan wrote: Bunuel wrote: devinawilliam83 wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16
please explain how A is sufficient I got C. on solving A i get 2 possible values of x and y and thus 2 values of XY If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x  12y = 0\) > \(x+3y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs > so either \(x=x\) and \(y=y\) OR \(x=x\) and \(y=y\) > either \(x+y=x+y=3y+y=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\) OR \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A. Hi Bunel, i have a doubt in ur explanation x+y=32 For this the modulus sign wont have four cases? (x,+y), (+x,y), (+x,+y), (x,y) Please clarify me. Thanks in Advance, Rrsnathan. Generally yes. But from \(x=3y\) we can get that \(x\) and \(y\) have opposite signs, so we are left only with two cases (+, ) or (, +). Hope it's clear.
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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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17 Oct 2013, 07:45
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If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16
(1) 4x  12y = 0 multiply by (1) 4x + 12y = 0 4(x + 3y) = 0 x + 3y = 0 we are also told that x + y = 32, so the only values for x and y that satisfy both equations are x = 24 y = 8 or x = 24 y = 8 in both cases xy is the same (24)(8) = (24)(8) ==> Sufficient.
(2) x  y = 16 multiple values for xy possible, for example x = 24 y = 8 or x = 24 y = 8 ==> Not sufficient.
Answer: A



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Re: If x and y are nonzero integers [#permalink]
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02 May 2014, 02:44
Bunuel wrote: devinawilliam83 wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16
please explain how A is sufficient I got C. on solving A i get 2 possible values of x and y and thus 2 values of XY If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x  12y = 0\) > \(x+3y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs > so either \(x=x\) and \(y=y\) OR \(x=x\) and \(y=y\) > either \(x+y=x+y=3y+y=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\) OR \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A. Where did I go wrong.. x  y = 32 > 3y  y = 32 > 3y  y = 32 > y = 32 3y y= (32  3y)> y= 16 y= 32  3y > y= 8 Guess I am wrong here...But I dont understand why...Absolute value confuses me a lot...been thru GMAT CLub Book...not sufficient I guess
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Re: If x and y are nonzero integers [#permalink]
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JusTLucK04 wrote: Bunuel wrote: devinawilliam83 wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16
please explain how A is sufficient I got C. on solving A i get 2 possible values of x and y and thus 2 values of XY If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x  12y = 0\) > \(x+3y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs > so either \(x=x\) and \(y=y\) OR \(x=x\) and \(y=y\) > either \(x+y=x+y=3y+y=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\) OR \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A. Where did I go wrong.. x  y = 32 > 3y  y = 32 > 3y  y = 32 > y = 32 3y y= (32  3y)> y= 16 y= 32  3y > y= 8 Guess I am wrong here...But I dont understand why...Absolute value confuses me a lot...been thru GMAT CLub Book...not sufficient I guess It's \(x + y = 32\), not x  y = 32. From (1): \(x=3y\) > \(x + y = 32\) > \(3y + y = 32\) > \(3y + y = 32\) > \(4y=32\) > \(y=8\) > \(y=8\) or \(y=8\). If \(y=8\), then \(x=3y=24\) > \(xy=(24)8\). If \(y=8\), then \(x=3y=24\) > \(xy=24(8)\). Both cases give the same value of xy. Hope it's clear.
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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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03 Aug 2014, 16:06
Bunuel wrote: devinawilliam83 wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16
please explain how A is sufficient I got C. on solving A i get 2 possible values of x and y and thus 2 values of XY If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x  12y = 0\) > \(x+3y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs > so either \(x=x\) and \(y=y\) OR \(x=x\) and \(y=y\) > either \(x+y=x+y=3y+y=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\) OR \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient. (2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient. Answer: A. Hi Bunuel, Couple of things to clarify: I understand that you're saying that x and y have opposite signs, but in the equation above, if the absolute value sign is around the equation in the question stem, how can the "minus" be brought outside for either x OR y? meaning, how does it become 4y = 32 vs. 4y = 32. I'm not sure how you can just drop the abs value sign? Thanks



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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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01 Oct 2014, 07:41
Note that one need not determine the values of both x and y to solve this problem; the value of product xy will suffice. (1) SUFFICIENT: Statement (1) can be rephrased as follows: 4x – 12y = 0 4x = 12y x = 3y If x and y are nonzero integers, we can deduce that they must have opposite signs: one positive, and the other negative. Therefore, this last equation could be rephrased as x = 3y We don’t know whether x or y is negative, but we do know that they have the opposite signs. Converting both variables to absolute value cancels the negative sign in the expression x = 3y. We are left with two equations and two unknowns, where the unknowns are x and y: x + y = 32 x – 3y = 0 Subtracting the second equation from the first yields 4y = 32 y = 8 Substituting 8 for y in the original equation, we can easily determine that x = 24. Because we know that one of either x or y is negative and the other positive, xy must be the negative product of x and y, or 8(24) = 192. (2) INSUFFICIENT: Statement (2) also provides two equations with two unknowns: x + y = 32 x  y = 16 Solving these equations allows us to determine the values of x and y: x = 24 and y = 8. However, this gives no information about the sign of x or y. The product xy could either be 192 or 192. The correct answer is A.
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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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25 May 2015, 22:04
Hi All, This question has some interesting 'patternmatching shortcuts' built into it that you can take advantage of IF you take enough notes. We're told that X and Y are NON0 INTEGERS and X + Y = 32. We're asked for the value of (X)(Y). Before dealing with the two Facts, it's worth noting that the prompt provides a significant limitation on the possible values of X and Y. Since the absolute values will turn negative results into positive ones, and the variables CANNOT be 0, there really are NOT that many possible values for X and Y. +1 and +31 +2 and +30 +3 and +29 Etc. We can use this to our advantage when dealing with the two Facts.... Fact 1: 4X  12Y = 0 Here, we can do some algebra to simplify the equation... 4X = 12Y X = 3Y This tells us that one variable MUST be POSITIVE and one MUST be NEGATIVE. When we include the absolute value equation given in the prompt, we know that one absolute value must be 3 times the value of the other.... That leaves us with two options for X and Y.... X = +24 and Y = 8 or X = 24 and Y = +8 Either way, we get the SAME PRODUCT: 192 Fact 1 is SUFFICIENT Fact 2: X – Y = 16 Here, we can use the work we did in Fact 1 to save us some time (and help us create some new possibilities)... IF... X = +24, Y = 8, the answer to the question is 192 IF... X = +24, Y = +8, the answer to the question is +192 Fact 2 is INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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21 May 2016, 03:48
Bunuel wrote: Bunuel wrote: devinawilliam83 wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) \(4x  12y = 0\) > \(x+3y=0\) > \(x=3y\) > \(x\) and \(y\) have opposite signs > so either \(x=x\) and \(y=y\) OR \(x=x\) and \(y=y\) > either \(x+y=x+y=3y+y=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\) OR \(x+y=xy=3yy=4y=32\): \(y=8\), \(x=24\), \(xy=24*8\), same answer. Sufficient.
(2) \(x  y = 16\). Sum this one with th equations given in the stem > \(2x=48\) > \(x=24\), \(y=8\). \(xy=24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient.
Answer: A. Hi Bunel, i have a doubt in ur explanation x+y=32 For this the modulus sign wont have four cases? (x,+y), (+x,y), (+x,+y), (x,y) Please clarify me. Thanks in Advance, Rrsnathan. Generally yes. But from \(x=3y\) we can get that \(x\) and \(y\) have opposite signs, so we are left only with two cases (+, ) or (, +). Hope it's clear. But how did u eliminate the other two cases (+x,+y) and (x,y) Solving x+y=32 and x+3y=0 gives x=48,y=16 Solving xy=32 and x+3y=0 gives x=48, y=16 this also says both x and y are of opposite signs. please help!



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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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If x and y are nonzero integers and x + y = 32, what is [#permalink]
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25 May 2016, 21:49
I am going to only examine statement #1
x+y=32. statement #1: x=3y. Let us substitute this in the prompt. 3y+y = 32 if y> 0 then (3*y)+y=32 (3*y is negative inside absolute value, so to obtain positive we need to multiply by 1) => y = 8. substituting to get x = 24 and so xy=192
what if y<0 then 3y+y=32 => 3*y y=32 (here we retained the 3y outside because y is negative already) => 4y=32 => y=8. x=24 and again xy=192 In both cases, we get y=8.
so only one possible value of x*y . 
another way when x>0 and y>0 x+y=32 so now x=3y 3y+y=32=> y=16. so this is not possible
when x>0 and y<0 xy=32 and x=3y=> y=8 ok then x= 24. x*y = 192
when x<0 y >0 yx=32, substituting x=3y=> y=8 . So then x = 24. x*y = 192
when x<0, y<0 yx=32. and x=3y => y(3y) = 32 =>2y = 32 => y=16. not feasible so the only answer possible is x*y = 192 .. Sufficient



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