Is /x-y//x-z/? 1) /y//z/ 2) x

Question

Is /x-y/>/x-z/?  

1) /y/>/z/ 
2) x<0

OA is E How do you tackle this problem without plugging in numbers?

anhlukas · Answer

Hi apollo168,

I don't think my solution is the best way to solve the problem but that's all I can think of right now:(

|x-y| > |x-z|
If we square both sides, we have:

(x-y)^2 > (x-z)^2
<=> x^2 - 2xy + y^2 > x^2 - 2xz + z^2
<=> y^2 - z^2 > 2xy - 2xz
<=> (y-z)(y+z) > 2x(y-z)

In case (y-z)>0, we divide both sides by (y-z), we have: y+z > 2x (i)
In case (y-z) <0, we divide both sides by (y+z), we have: y+z < 2x (ii)
-> We need to know the values of (y-z), (y+z) and x.

From (1): |y|> |z| -> y^2 - z^2 = (y-z)(y+z)>0
-> If (y-z)>0 => (y + z )> 0
-> If (y-z) < 0 => (y+z) < 0
-> insufficient. (1) doesn't give much information for us to determine the value of (y-z) and (y+z) or x

(2): x<0 ->2x <0
-> insufficient

(1)+ (2):
(i): We have to prove that with (y-z)>0, y+z > 2x
We combine (1) and (2) we have:
(y + z )> 0 AND 2x <0
-> (i) is always true: because y+z >0>2x

(ii) We have to prove that with (y-z) <0, y+z < 2x
Combine (1) and (2): , y+z <0, 2x < 0 --> doesn't tell anything

-> both (1) and (2) are insufficient

-> E

apollo168 · Answer

Hey thanks for the help. At least you were able to think of something hehe. I just arrived at the answer thru trial and error :-D

anhlukas · Answer

hehe..i don't know if you can understand my solution or not...I've tried my best to make it understandable but I'm not quite sure if I've succeeded

But usually when you see the absolute problem like this, you should square both sides of the (in)equation

I also hope someone can come up with a better solution to this very interesting problem :wink:

kevincan · Answer

Is x further from y than it is from z?

(1) y is further from 0 than is z (Nothing is said about x) NOT SUFF
(2) x is less than 0 (Nothing is said about y or z) NOT SUFF

(1) and (2) see atttached NOT SUFF

apollo168 · Answer

Thanks Kevin,

I now understand the logic of your answer. I researched about absolute value and this website said that for 700 type questions, the best way to attack absolute value problems is to think in terms of distance from something. Thanks