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
DS MGMAT question
This topic has expert replies
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.
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.
-
- Legendary Member
- Posts: 631
- Joined: Mon Feb 18, 2008 11:57 pm
- Thanked: 29 times
- Followed by:3 members
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)
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)
-
- Junior | Next Rank: 30 Posts
- Posts: 14
- Joined: Thu Jun 05, 2008 10:33 am
- Thanked: 2 times
- GMAT Score:710
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.
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.
-
- Legendary Member
- Posts: 631
- Joined: Mon Feb 18, 2008 11:57 pm
- Thanked: 29 times
- Followed by:3 members
You will not get 2xy but 2 |x| |y|leovonp wrote: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.
-
- Junior | Next Rank: 30 Posts
- Posts: 14
- Joined: Mon Jun 02, 2008 12:25 pm
- Thanked: 2 times
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!
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!