DS MGMAT question

[nqhussain]

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

(1) -4x – 12y = 0

(2) |x| – |y| = 16


answer is 1 is sufficient.

please advise

[mandy12]

From the given equation
x + y = 32 --- eq 1

or

-x -y = 32 ---- eq 2

Stmt 1 says -4x -12y = 0 or x + 3y = 0 ---- eq 3

Solving eq 1 and 3 we get x = 48 and y=-16

solving eq 2 and 3 we get x = -48 and y = 16

In both the above cases xy is same hence stmt 1 is suff.


Doing the same exercise for stmt 2 gives different possibilities for x and y
hence insuff.

[netigen]

Your approach is incomplete, in this case you can have 4 different possible equations

1. x+y = 32
2. -x+y = 32
3. x-y = 32
4 -x-y = 32

but this is not the best approach to solve this problem.

Lets look at (A)

-4x – 12y = 0
x = -3y -------------------> EQ1

substitute in the original eq

|-3y| + |y| = 32
4|y| = 32
|y| = 8
|x| = 24

from eq one we know that when x is +ve y is -ve and when x is -ve y is +ve hence,

we know that xy = -(24 x 8)

[leovonp]

I actually got D as an answer.

Reasoning is that if you square both the original equation and the equation in stat (2), you end up with a viable result.

I.e.

Original equation (|x|+|y|)^2=32^2
equals x^2+2xy+y^2=32^2

Statement (2) (|x|-|y|)^2=16^2
equals x^2-2xy+y^2=16^2

Now if we subtract stat (2)^2 from the original equation^2 we should end up with 4xy = 32^2-16^2

Could you please advise me if there is any flaw in this logic? Otherwise the answer should be amended to D.

[netigen]

I actually got D as an answer.

Reasoning is that if you square both the original equation and the equation in stat (2), you end up with a viable result.

I.e.

Original equation (|x|+|y|)^2=32^2
equals x^2+2xy+y^2=32^2

Statement (2) (|x|-|y|)^2=16^2
equals x^2-2xy+y^2=16^2

Now if we subtract stat (2)^2 from the original equation^2 we should end up with 4xy = 32^2-16^2

Could you please advise me if there is any flaw in this logic? Otherwise the answer should be amended to D.

You will not get 2xy but 2 |x| |y|

[gogetter08]

eq(1) says x = -3y meaning x and y have opp. sign

without loss of generality, given eq can be re-written as

x-y=16

with 2 eqns you can solve for x,y


eq(2) along wiht given eq will only yield |x| and |y| and will not tell you anything about their sign


so eq(1) is sufficient!

[psriniva]

Hi Gogetter08,

How did you come up with x-y = 16 without loss of generality from the equation |x| + |y| = 32.
Please explain.

Appreciate your feedback.

Thanks

Paddy