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• ### $450 Tuition Credit & Official CAT Packs FREE January 15, 2019 January 15, 2019 10:00 PM PST 11:00 PM PST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### The winning strategy for a high GRE score January 17, 2019 January 17, 2019 08:00 AM PST 09:00 AM PST Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL. # Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0....  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13325 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink] ### Show Tags Updated on: 05 Nov 2015, 11:45 1 7 00:00 Difficulty: 45% (medium) Question Stats: 66% (02:22) correct 34% (02:10) wrong based on 167 sessions ### HideShow timer Statistics 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 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 _________________ 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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Originally posted by EMPOWERgmatRichC on 04 Nov 2015, 18:17.
Last edited by EMPOWERgmatRichC on 05 Nov 2015, 11:45, edited 2 times in total.
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Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0....  [#permalink]

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04 Nov 2015, 19:43
1
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
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...
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Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0....  [#permalink]

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05 Nov 2015, 01:17
1
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.
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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Expert Joined: 02 Aug 2009 Posts: 7200 Re: Problem Solving Pack 3, Question 2 If x^2 - xy - 32 = 0.... [#permalink] ### Show Tags 05 Nov 2015, 02:29 1 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.. _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor Current Student Joined: 02 Jun 2015 Posts: 81 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 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: 13325 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 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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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.

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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
1
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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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.... &nbs [#permalink] 16 Apr 2017, 17:34
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