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If x and y are nonzero integers and x + y = 32, what is [#permalink]
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08 Aug 2009, 07:32
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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
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If x and y are nonzero integers and x + y = 32, what is [#permalink]
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01 Nov 2009, 05:25
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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If x and y are nonzero integers and x + y = 32, what is [#permalink]
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03 Feb 2010, 05:07



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Re: Absolute Value [#permalink]
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08 Feb 2010, 12:17
Given: X + Y = 32, so 4 equations: X + Y = 32 (I) X + Y = 32 (II) X  Y = 32 (III) X  Y = 32 (IV) From (1) 4X = 12y, so X = 3Y or if X is neg, then Y is pos and viceversa This information means equations (II) and (III) above can be eliminated. Hence, substitute X = 3Y in (I) and (IV) yields Y = 8 or 8. if Y = 8, then X = +24, so xy = 192 if Y = +8, then X = 24, so xy = 192 the same as above. (1) is true From (II) X  Y = 16 and given X + Y = 32, subtract both equations yields 2X = 16, so X = 8 or 8. Y is thus 24 or 24... but the combination can be the following (X,Y): (8, 24) (8, 24) (8, 24) (8, 24) and xy will yield either 192 or +192
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Re: Absolute Question  Please help [#permalink]
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22 Mar 2010, 07:02
A 1) 4x12y=0 > x = 3y > 3y+y = 32 > y = 8 > we have two solutions: (x=24, y=8) and (x=24, y=8). In both cases xy is the same. Sufficient 2) We don't even need to solve it. In the question and second statement we have expressions that include absolute values of both x and y. In other words, there is always at least two possible value for x and y: negative and positive. So, xy can be also negative and positive. Insufficient. It is possible to take a square but I wouldn't recommend especially for absolute value  inequality combination because it is easy to miss something. Did you see this post? It could be helpful. mathabsolutevaluemodulus86462.html
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Re: Absolute Question  Please help [#permalink]
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22 Mar 2010, 08:01
Hey nice explanation walker. I still have one doubt. How did you reach to below conclusion. Should not we consider 4 cases here when 3y is + and when y is +? walker wrote: 3y+y = 32 > y = 8
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Re: Absolute Question  Please help [#permalink]
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22 Mar 2010, 11:55
Here I use two properties: 1) y=y 2) ay = ay if a is a nonnegative number. 3y+y = 32 3y+y = 32 3y+y = 32 4y = 32 y = 8 And only now I open a modulus: y=+/8 Of course, we can open two modules at the beginning, use 4 cases and get the same answer. Trick here is that y and 3y have the same key point: 0. So, if y>0, 3y is negative and vice versa. In other words, y>0 and 3y>0 can't be true simultaneously and we have only 2 cases: 3y  y = 32 and 3y + y = 32 or y = +/8
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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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29 Oct 2010, 04:53
Bunuel wrote: lbsgmat wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16 If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x12y=0\) > \(x=3y\) > \(x\) 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. hey bunnel, i know this is basic stuff, but could you explain the absolute equation y=y, what does this mean, is it "y" inside the modules is negative, and "x+y=x+y=3y+y=4y=32" how did you manage to get a negative coefficient for the "x" when it is in the module.....thanks in advance,,bunnel



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Re: If x and y are nonzero integers and x + y = 32, what is [#permalink]
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29 Oct 2010, 05:03
satishreddy wrote: Bunuel wrote: lbsgmat wrote: If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0
(2) x – y = 16 If x and y are nonzero integers and x + y = 32, what is xy? (1) \(4x12y=0\) > \(x=3y\) > \(x\) 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. hey bunnel, i know this is basic stuff, but could you explain the absolute equation y=y, what does this mean, is it "y" inside the modules is negative, and "x+y=x+y=3y+y=4y=32" how did you manage to get a negative coefficient for the "x" when it is in the module.....thanks in advance,,bunnel Absolute value is always nonnegative: \(x\geq{0}\), so: When \(x\leq{0}\) then \(x=x\) (note that in this case \(x=negative=positive\)); When \(x\geq{0}\) then \(x=x\). \(x=3y\) implies that \(x\) and \(y\) have opposite sign, so when \(x\) is negative then \(y\) is positive. In this case (when \(x\) is negative and \(y\) is positive): \(x=x\) and \(y=y\) and \(x+y=32\) becomes \(x+y=32\). For more check: mathabsolutevaluemodulus86462.htmlHope it helps.
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Re: DSmod [#permalink]
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30 Jul 2011, 03:44
The question is asking us if we have enough information to determine xy if x + y = 32, and x and y are nonzero integers. From statement (1), x = 3y. Solving this with x + y = 32 gives us x=8, y =24 or y=8, x=24. In either case, xy is the same (8*24 or 192). Even though we get two sets of values of x and y, their product is the same in both cases. Sufficient. From statement (2), x  y = 16. Solving x  y = 16 and x + y = 32 gives us the solution x = 24 and y = 8. => The solutions are x=24,24 and y=8,8. If we use x=24 and y=8, we get xy = 24*8. If we use x=24 and y=8, we get xy=24*8. Using just these two cases we can see that we are not getting a unique value for xy. Therefore statement (2) alone is insufficient. The 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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10 Jul 2013, 14:02
If x and y are nonzero integers and x + y = 32, what is xy?
(1) 4x – 12y = 0 4x = 12y x = 3y
x + y = 32 3y + y = 32 4y = 32 y=8
x = 3y x = 3(8) x=24
xy = (24)*(8) SUFFICIENT
(2) x – y = 16
x – y = 16 x + y = 32 2x=48 x=24 x=24, x=24
x=24 x – y = 16 24  y = 16 24  y = 16  y = 8 y = 8 y=8 OR y=8
If x is 24, y could = 8 or 8 meaning xy could be positive or negative. INSUFFICIENT
(A)



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Re: If x and y are nonzero integers [#permalink]
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21 Sep 2013, 23:32
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, 04:35
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, 08:45
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, 03:44
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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02 May 2014, 10:21
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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01 Oct 2014, 08:41
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, 23:04
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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