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Absolute values

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Source: — Data Sufficiency |
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by mental » Thu Oct 09, 2008 9:51 am
imo b
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by Gmatss » Thu Oct 09, 2008 10:00 am
IMO B
lxl=lyl?
1) x-y=6
x=3 y=-3 yes condition satifies 1 and the answer is yes but
if x=9 y=3 No hence insuffient
2)x+y=0
x=-y
so Yes Suffienct
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by vivek.kapoor83 » Thu Oct 09, 2008 11:59 am
I think it diff way and i just want to know whether my way is correct or not. I think for proving mod x = mod y
we need to prove, if we take out mode... x = y ( if both are > 0) and -x = -y ( if <0)
coz by def of mode , mod x = x if x >0
mod x = -x if x<0
So, do we need to prove both condition in the ques below..or not.

Going by my method.
1.x-y =6
....>>> Not Suff
2. x+y=0, x=-y
utting back...mod(-y) = mod(y ) ....

... y = y ( If Y >00
-y = y( if y<0 )...not true. ....so B Insuff


taking together

x-y=6
x+y=0

x =3,y =-3
Here again comes my ques.
Do we have to put values and consider mode by >0 and <0


Pls f any1 can explain
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by Ian Stewart » Thu Oct 09, 2008 12:14 pm
vivek, I don't really follow your logic, but if:

|x| = |y|

then either x = y, or x = -y.

|x| is just the distance between 0 and x on the number line. If |x| = |y|, then x and y are both the same distance from zero. They're either both on the same side of zero, in which case they're equal, or they're on opposite sides of zero, in which case one is the negative of the other.
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by vivek.kapoor83 » Thu Oct 09, 2008 12:20 pm
but i just wanted to know..do we need to prove both the equations.
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by Ian Stewart » Thu Oct 09, 2008 2:34 pm
vivek.kapoor83 wrote:but i just wanted to know..do we need to prove both the equations.
No, not quite. If |x| = |y|, there are four possibilities (you do not seem to be considering the last two below):

x, y both positive --> x = y
x, y both negative --> x = y
x positive, y negative --> x = -y
x negative, y positive --> x = -y

Which is why I said above that |x| = |y| is true if x = y or x = -y. Note that -x = -y is exactly the same equation as x=y (just multiply both sides by -1), so there is no need to consider it separately.
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