CAT DS: Pls suggest flaw in reasoning:

[shriwantchoudhary]

DS From CAT-2 (MGMAT)

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

(1) -4x - 12y = 0

(2) |x| - |y| = 16

Consider the question stem: |x| + |y| = 32
Does this equation not mean the following simultaneous equations:
x + y = 32
-x -y = 32
-x + y = 32
x -y = 32

Now using Statement -1 we have x = -3y using the info from the q stem and statement -1 we can deduce that:

y = +/-8 & y = +/-16 for these values of y we can get corresponding four different values for x i.e. x = +/- 24 & +/-48

But then a question comes, does the original question stem get satisfied with all set of values of x & y?

we see that only two pairs of x,y satisfy the question stem hence xy is obtained. so Statement-1 is sufficient.

[700+]

Let me try helping you out with this one

According to the question |x| + |y| = 32. Lets just keep it like this for now. We just need to remember that |x| is the absolute value of x & it is always positive.

According to statement one, -4x - 12y = 0 i.e. x = -3y

Lets make a table

----y--------x--------Is |x| + |y| = 32?--------xy--------
----8------(-24)-------------Yes-------------(-192)------
---(-8)------24--------------Yes-------------(-192)------
----16-----(-48)------------- No---------------NA-------
---(-16)----(48)--------------No---------------NA-------

Since the question is asking for the value of xy, from the above table it can b seen that the value of xy is -192.

Therefore A is sufficient.

We need not consider values where y = + or - 16 because it doesn't satisfy both the conditions |x| + |y| = 32 & x = (-3y).

Hope I was able to clear your doubt

Changes shown in Blue color

[shriwantchoudhary]

"We need not consider values where y = + or - 16 because it doesn't satisfy the condition |x| + |y| = 32."


I suspected the same thing.

Thanks,
SC

[jnelson0612]

"We need not consider values where y = + or - 16 because it doesn't satisfy the condition |x| + |y| = 32."


I suspected the same thing.

Thanks,
SC

Correct! :-)