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Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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Updated on: 05 Nov 2015, 12:45
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QUANT 4PACK SERIES Problem Solving Pack 3 Question 2 What is the positive difference...If \(x^{2}\) + xy  32 = 0, and x and y are integers, then y could equal each of the following except…? A) 31 B) 14 C) 2 D) 4 E) 14 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation
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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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04 Nov 2015, 20:43
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 3 Question 2 What is the positive difference...If \(x^{2}\)  xy  32 = 0, and x and y are integers, then xy could equal each of the following except…? A) 31 B) 14 C) 2 D) 4 E) 14 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation hi rich, there can be three answers for this Q. You may have to change the options... straight method would be to substitute the value of xy in the equation and find if x^2 is an integer, as x itself is an integer.. eqn..\(x^{2}\)  xy  32 = 0 or \(x^{2}\) = xy + 32 A) 31... x^2=1.. ok B) 14... x^2=18, x will not be an integer..not possible C) 2... x^2=34, x will not be an integer..not possible D) 4... x^2=36.. ok E) 14... x^2=46, x will not be an integer..not possible so B,C and E all are contenders for the possible answer...
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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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05 Nov 2015, 02:17
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. If x2  xy  32 = 0, and x and y are integers, then xy could equal each of the following except…? A) 31 B) 14 C) 2 D) 4 E) 14 Hi rich, all the choices except (A) are impossible. So you should change the question as "xy could be...." Here is my methods. Since 32 = x^2 – xy= x *(xy) and x and y are integers. x and xy should be the factors of 32. So (x, xy) has following possibilities (1, 32), (2, 16), (4, 8), (8, 4), (16, 2), (32, 1), (1, 32), (2, 16), (4, 8), (8, 4), (16, 2), (32, 1) After tedious calculations we have (x, y)=(1, 31), (2, 14), (4, 4), (8, 4), (16, 14), (32, 31), (1, 31), (2, 14), (4, 4), (8, 4), (16, 14), (32, 31). So xy can be –31, 28, 16, 32, 224, 992. So only (A) is possible. And To Chetan2u D) 4... x^2=36.. ok.> By your explanation we have x= 6 or –6. But we cannot find integer y satisfying xy=4. So (D) is also impossible.
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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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05 Nov 2015, 03:29
MathRevolution wrote: Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
If x2  xy  32 = 0, and x and y are integers, then xy could equal each of the following except…?
A) 31 B) 14 C) 2 D) 4 E) 14
Hi rich, all the choices except (A) are impossible. So you should change the question as "xy could be...."
Here is my methods. Since 32 = x^2 – xy= x *(xy) and x and y are integers. x and xy should be the factors of 32. So (x, xy) has following possibilities (1, 32), (2, 16), (4, 8), (8, 4), (16, 2), (32, 1), (1, 32), (2, 16), (4, 8), (8, 4), (16, 2), (32, 1)
After tedious calculations we have (x, y)=(1, 31), (2, 14), (4, 4), (8, 4), (16, 14), (32, 31), (1, 31), (2, 14), (4, 4), (8, 4), (16, 14), (32, 31).
So xy can be –31, 28, 16, 32, 224, 992.
So only (A) is possible.
And To Chetan2u
D) 4... x^2=36.. ok.> By your explanation we have x= 6 or –6. But we cannot find integer y satisfying xy=4. So (D) is also impossible. hi buddy, thanks, in the hurry, i forgot to test the integer value of y.... and ofcourse we dont have to get into the tedious calculations but substitue xy and look for integer values for both x and y..
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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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05 Nov 2015, 09:13
MathRevolution wrote: Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
If x2  xy  32 = 0, and x and y are integers, then xy could equal each of the following except…?
A) 31 B) 14 C) 2 D) 4 E) 14
Hi rich, all the choices except (A) are impossible. So you should change the question as "xy could be...."
Here is my methods. Since 32 = x^2 – xy= x *(xy) and x and y are integers. x and xy should be the factors of 32. So (x, xy) has following possibilities (1, 32), (2, 16), (4, 8), (8, 4), (16, 2), (32, 1), (1, 32), (2, 16), (4, 8), (8, 4), (16, 2), (32, 1)
After tedious calculations we have (x, y)=(1, 31), (2, 14), (4, 4), (8, 4), (16, 14), (32, 31), (1, 31), (2, 14), (4, 4), (8, 4), (16, 14), (32, 31).
So xy can be –31, 28, 16, 32, 224, 992.
So only (A) is possible.
And To Chetan2u
D) 4... x^2=36.. ok.> By your explanation we have x= 6 or –6. But we cannot find integer y satisfying xy=4. So (D) is also impossible. I think exactly like you I only got the letter A corrected. xˆ2  xy  32 = 0 xˆ2 = xy + 32 (picking the numbers) xˆ2 = 31 + 32 x = +/ 1 if I choose +1 or 1, in both of this choices i can get an integer value for Y. Therefore, only (A) is correct.



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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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05 Nov 2015, 12:47
Hi All, Thanks for pointing out the inconsistency in the prompt. It had 2 small typos that have since been fixed. Sorry for the inconvenience. GMAT assassins aren't born, they're made, Rich
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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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07 Nov 2015, 15:47
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 3 Question 2 What is the positive difference...If \(x^{2}\) + xy  32 = 0, and x and y are integers, then y could equal each of the following except…? A) 31 B) 14 C) 2 D) 4 E) 14 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation Hi All, This question is based on a common Algebra pattern  Quadratics. Since x and y are INTEGERS, we have to determine the limited number of possible values for x and y and determine which of the 5 answer choices is NOT a possible value for y. To start, we're given the Quadratic: \(x^{2}\) + xy  32 = 0 Using the "32", we can determine the various 'pairs' of values for x and y... 32 and +1 32 and 1 16 and +2 16 and 2 8 and +4 8 and 4 Given the above 'pairs', we can quickly determine the possible values of y (and you don't technically have to do the entire FOIL calculation)... IF the quadratic is (x  32)(x + 1), then the FOILed term is \(x^{2}\) 31x  32 = 0 so y COULD be 31. IF the quadratic is (x  16)(x + 2), then the FOILed term is \(x^{2}\) 14x  32 = 0 so y COULD be 14. IF the quadratic is (x +8)(x  4), then the FOILed term is \(x^{2}\) +4x  32 = 0 so y COULD be +4. IF the quadratic is (x + 16)(x  2), then the FOILed term is \(x^{2}\) +14x  32 = 0 so y COULD be +14. The only answer that is NOT possible under these conditions is 2. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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08 Nov 2015, 03:03
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 3 Question 2 What is the positive difference...
If \(x^{2}\) + xy  32 = 0, and x and y are integers, then y could equal each of the following except…?
\(x^{2} + xy  32 = 0\) Or, \(x^{2} + xy = 32\) Now check the options  A) 31 \(x^{2} 31x  32 =0\) x = 1 and 32 This quadratic equation can be solved.B) 14 \(x^{2} 14x  32 =0\) x = 2 and 16 This quadratic equation can be solved.C) 2 \(x^{2} +2x  32 =0\) This quadratic equation can not be solved, we can stop here and mark this as our answer.D) 4 \(x^{2} +4x  32 =0\) x = 4 and 8 This quadratic equation can be solved.E) 14 \(x^{2} +14x  32 =0\) x = 2 and 16 This quadratic equation can be solved.Hence our answer is (C)
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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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22 Jan 2016, 13:27
QUANT 4PACK SERIES Problem Solving Pack 3 Question 2 What is the positive difference...
If x2  xy  32 = 0, and x and y are integers, then xy could equal each of the following except…?
Hi rich, all the choices except (A) are impossible. So you should change the question as "xy could be...."
X2xy32=0 (x16) (x+2)=0 X=16v X=2
X31xy32=0 X=31
IF the quadratic is (x  32)(x + 1), then the FOILed term is x2 31x  32 = 0 so y COULD be 31.
IF the quadratic is (x  16)(x + 2), then the FOILed term is x2 14x  32 = 0 so y COULD be 14.
IF the quadratic is (x +8)(x  4), then the FOILed term is x2 +4x  32 = 0 so y COULD be +4.
IF the quadratic is (x + 16)(x  2), then the FOILed term is x2 +14x  32 = 0 so y COULD be +14.
The only answer that is NOT possible under these conditions is 2



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Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0....
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16 Apr 2017, 18:34
there is an easy approach to this problem, remember to put kudos for me :D
x^2 +xy  32 = 0 => 4x^2 +4xy +y^2 = 128 + y^2 => (2x + y)^2 = 128 + y^2 Then, try each options, starting at the smallest => C is the answer
Pls dont forget to feed me kudos :D




Re: Problem Solving Pack 3, Question 2 If x^2  xy  32 = 0.... &nbs
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