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If x and y are non-zero integers

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If x and y are non-zero integers

by gmatdriller » Fri Jan 13, 2012 4:46 pm
If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(i) -4x - 12y = 0
(ii) |x| - |y| = 16

Please explain the approach without to use without stripping the absolute
signs expressions to the various forms.

Source is MANHATTANGMAT CAT6

OA A
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by GMATGuruNY » Fri Jan 13, 2012 6:35 pm
I posted a solution here:

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by Anurag@Gurome » Fri Jan 13, 2012 7:50 pm
gmatdriller wrote:If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(i) -4x - 12y = 0
(ii) |x| - |y| = 16

Please explain the approach without to use without stripping the absolute
signs expressions to the various forms.

Source is MANHATTANGMAT CAT6

OA A
(1) -4x - 12y = 0 ---> x = -3y
This means if x and y have different signs.

Now, two cases are possible,
1. x > 0 and y < 0 ---> x - y = 32 ---> -3y - y = 32 ---> y = -8 and x = 24
1. x < 0 and y > 0 ---> -x + y = 32 ---> 3y + y = 32 ---> y = 8 and x = -24

In both of the above cases, xy = -(8*24)

Sufficient

(2) |x| - |y| = 16
Combining this with |x| + |y| = 32, we have 2|x| = 48 ---> x = ±24 ---> y = ±8

Hence, xy = ±(8*24)

Not sufficient

The correct answer is A.
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by ArunangsuSahu » Sat Jan 14, 2012 8:58 am
When 2 equations have 2 variables with mod it's bound to generate different solution..

There is no determinant

So (B) is always INSUFFICIENT

(A) is SUFFICIENT
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by gmatdriller » Sat Jan 14, 2012 9:13 am
I liked the substituting "x = -3y" into
|x| + |y| = 32 as explained by GmatGuruNY.

Also, since both x and y are non-zero integers, the new
equation shows that both have different signs.
Flipping the sign of x in the original equation was interesting.


Thanks to both GmatGuryNY and Anurag@Gurome
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