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Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0....

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EMPOWERgmat Instructor
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Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink]

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04 Nov 2015, 18:17
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QUANT 4-PACK 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 Window
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll Down For Official Explanation
[Reveal] Spoiler: OA

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Last edited by EMPOWERgmatRichC on 05 Nov 2015, 11:45, edited 2 times in total.  EMPOWERgmat Discount Codes Jamboree Discount Codes Math Revolution Discount Codes Math Expert Joined: 02 Aug 2009 Posts: 5662 Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink] Show Tags 04 Nov 2015, 19:43 1 This post received KUDOS Expert's post EMPOWERgmatRichC wrote: QUANT 4-PACK 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 Window Earn 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 4-Pack series Scroll 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... _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html BANGALORE/- Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 4908 GPA: 3.82 Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink] Show Tags 05 Nov 2015, 01:17 1 This post received KUDOS Expert's post 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 *(x-y) and x and y are integers. x and x-y should be the factors of 32. So (x, x-y) 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Find a 10% off coupon code for GMAT Club members. “Receive 5 Math Questions & Solutions Daily” Unlimited Access to over 120 free video lessons - try it yourself See our Youtube demo Math Expert Joined: 02 Aug 2009 Posts: 5662 Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink] Show Tags 05 Nov 2015, 02:29 1 This post received KUDOS Expert's post 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 *(x-y) and x and y are integers. x and x-y should be the factors of 32. So (x, x-y) 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.. _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html BANGALORE/- Manager Joined: 02 Jun 2015 Posts: 91 Location: Brazil Concentration: Entrepreneurship, General Management GPA: 3.3 Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink] Show Tags 05 Nov 2015, 08:13 1 This post received KUDOS 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 *(x-y) and x and y are integers. x and x-y should be the factors of 32. So (x, x-y) 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. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11074 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink] Show Tags 05 Nov 2015, 11:47 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
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Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink]

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07 Nov 2015, 14:47
EMPOWERgmatRichC wrote:
QUANT 4-PACK 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 Window
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll 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.

[Reveal] Spoiler:
C

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Rich
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Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink]

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08 Nov 2015, 02:03
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EMPOWERgmatRichC wrote:
QUANT 4-PACK 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.

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Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink]

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22 Jan 2016, 12:27
QUANT 4-PACK 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...."

X2-xy-32=0
(x-16) (x+2)=0
X=16v X=-2

X-31xy-32=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.... [#permalink]

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26 Mar 2017, 02:03
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Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink]

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16 Apr 2017, 17:34
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....   [#permalink] 16 Apr 2017, 17:34
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