Try this one from MGMAT

This topic has expert replies
Source: — Data Sufficiency |

Master | Next Rank: 500 Posts
Posts: 438
Joined: Mon Feb 12, 2007 9:44 am
Thanked: 26 times

by chidcguy » Wed May 21, 2008 11:27 am
Is the answer E?

Junior | Next Rank: 30 Posts
Posts: 12
Joined: Tue May 20, 2008 6:05 pm
Location: california
Thanked: 1 times

by vinod_ece66 » Wed May 21, 2008 11:51 am
i think the answer should be C
the solution goes like this:
|x|+|y|=32
|x|-|y|=16

==>2|x|=48 or |x|=24
Hence x can be 24 or -24;

also x = -3y;

hence if x =24 then y = -8
and if x = -24 then y = 8

either way xy = -(24* 8)

Hence answer is c

Correct me if i am wrong

Master | Next Rank: 500 Posts
Posts: 438
Joined: Mon Feb 12, 2007 9:44 am
Thanked: 26 times

by chidcguy » Wed May 21, 2008 12:02 pm
Oh well I stopped calculating XY when I chose E. God knows when I will star getting things as asked in the Question.

When ever I read |X|, I also keep in my mind that its the distance from 0 on the number line towards either side

(1)
|X| + |Y| =32
X=-3Y

|3Y| + |Y| =32 -> |Y| = 8

Y = 8 X =-24
Y =-8 X= 24

XY =-192 both ways B, C & E out

(2)

|X| + |Y| =32

|X| - |Y| =16

|X|= 24

X=24 means |Y|=8, Y =8 /-8 XY =192/-192

X =-24 means |Y| =8, Y =8/ -8 XY=192/-192

We have both 192 and -192. Can't be D

Hence A.

Probably there is more to it as you asked how to solve two equations involving two equations with absolute variables. Let us know

Legendary Member
Posts: 631
Joined: Mon Feb 18, 2008 11:57 pm
Thanked: 29 times
Followed by:3 members

by netigen » Wed May 21, 2008 1:26 pm
OA is A and I like your approach more than the one suggested by MGMAT