sanju09 wrote:If x and y are non-zero integers and |x| + |y| = 32, what is x y?

(1) -4 x - 12 y = 0.
(2) |x| - |y| = 16.


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|x| + |y| = 32
(1) -4 x - 12 y = 0.
x=-3y
|-3y|+|y| = 32
when y>0
3y+y = 32
y = 8, x= -24
when y<0,
-3y-y=32
y=-8, x- 24
Hence xy = 24*-8
SUFF

(2) |x| - |y| = 16.
When x>0, y>0
x-y = 16
x+y = 32
x=24, y= 8

When x>0, y<0
x+y = 16
x-y=32
x=24, y=-8

xy is different in both cases.
INSUFF

pick A

kvcpk wrote:
|x| + |y| = 32
(1) -4 x - 12 y = 0.
x=-3y
|-3y|+|y| = 32
when y>0
3y+y = 32
y = 8, x= -24
when y<0,
-3y-y=32
y=-8, x- 24
Hence xy = 24*-8
SUFF

Isn't there a flaw in the logic? Still answer A would not change..

|-3y|+|y| = 32
when y>0
-3(+y)+ (+y) = 32
i.e -3y + y = 32 --> y = -16, x is 48

and when y < 0
-3(-y)+(-y) = 32
3y - y = 32 --> y = 16 , x is -48

thirst4edu wrote: Isn't there a flaw in the logic? Still answer A would not change..

|-3y|+|y| = 32
when y>0
-3(+y)+ (+y) = 32
i.e -3y + y = 32 --> y = -16, x is 48

and when y < 0
-3(-y)+(-y) = 32
3y - y = 32 --> y = 16 , x is -48

Hi,

I see what you are saying, but I think, my logic was right.
when y>0,
3y>0
-3y<0
Hence |-3y|=-(-3y) = 3y

when y<0, |y| = -y
3y<0
-3y>0
|-3y| = -3y

Hope this helps!!

kvcpk wrote:
Hi,

I see what you are saying, but I think, my logic was right.
when y>0,
3y>0
-3y<0
Hence |-3y|=-(-3y) = 3y

when y<0, |y| = -y
3y<0
-3y>0
|-3y| = -3y

Hope this helps!!

Yeah, I see, that makes sense too.. so there are more than one way to get there!

thirst4edu wrote:

kvcpk wrote:
Hi,

I see what you are saying, but I think, my logic was right.
when y>0,
3y>0
-3y<0
Hence |-3y|=-(-3y) = 3y

when y<0, |y| = -y
3y<0
-3y>0
|-3y| = -3y

Hope this helps!!

Yeah, I see, that makes sense too.. so there are more than one way to get there!

I revisted my logic, and re-read MGMT tutorial on abosulute values, because I realized there shouldnt be two different values to reach to the solution of the same equation.
There is a mistake in my logic. So, correcting it for the records -

y>0
|-3y| + |y| = 32
3(+y) + (+y) = 32 <---- drop the negative sign from -3 because you are taking absolute value, only thing you are not sure about is y. Here y is positive.
4y = 32
y = 8 , x = -24

y<0
|-3y| + |y| = 32
3(-y) + (-y) = 32 <---- drop the negative sign from -3 because you are taking absolute value, only thing you are not sure about is y. Here y is negative.

-3y -y = 32
-4y = 32
y = -8, x = 24


Thanks again kvcpk !

thirst4edu wrote: Thanks again kvcpk !

You are welcome Buddy!!

|-3y| can be rephrased as 3|y|

3|y|=3y if y>0

3|y|=3(-y) if y<0