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

Question

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

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This question is part of the Quant 4-Pack series

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chetan2u · Accepted Answer

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...

MathRevolution · Answer

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.

chetan2u · Answer

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..

douglasvg · Answer

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.

EMPOWERgmatRichC · Answer

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

EMPOWERgmatRichC · Answer

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:  C

GMAT assassins aren't born, they're made,
Rich

Abhishek009 · Answer

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)

Lorenzo3688 · Answer

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

chesstitans · Answer

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

GMATGuruNY · Answer

x² = 32 - xy
x² must be a perfect square less than 32.

If x²=1, then xy=31, implying that x=±1 and y=±31.
Since A is a possible value for y, eliminate A.
If x²=4, then xy=28, implying that x=±2 and y=±14.
Since B and E are possible values for y, eliminate B and E.
If x²=16, then xy=16, implying that x=±4 and y=±4.
Since D is a possible value for y, eliminate D.

C

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

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

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