Is |x-y| > |x| - |y| ?

[GMAT Fever]

Is |x-y| > |x| - |y| ?

(1) y<x

(2) xy<0

Can anyone advise on the quickest/most efficient way to solve? Thanks!

[guest]

Substituting values should be a good way. My guess is answer should be B.

[GMAT Fever]

Substituting values should be a good way. My guess is answer should be B.

Yes you are correct the answer is B. And I eventually resorted to guessing numbers for this one, I was wondering if this could quickly be solved algebraically.

[RonPurewal]

Substituting values should be a good way. My guess is answer should be B.

Yes you are correct the answer is B. And I eventually resorted to guessing numbers for this one, I was wondering if this could quickly be solved algebraically.

the best way to solve this problem is to notice that its subject matter is POSITIVES AND NEGATIVES. how do you know this? because it deals only with absolute values and inequality signs - no other numbers or non-absolute values in sight to mess things up.

there is no way to 'quickly solve' this one algebraically, unfortunately. in fact, even at the highest levels of mathematics, the best (and really the only) way to solve problems like these is case-wise, considering the different possibilities for + and - one case at a time.

[Guest]

Is |x-y| > |x| - |y| ?

(1) y<x

(2) xy<0

I approached part of the problem by plugging nos. and the other part geometrically.

A) y < x

I want a case for >

x = 2, y = -3
LHS = 5 > RHS = -1

Case for = or <
x = anything, y = 0
LHS = RHS
Insufficient

B ) xy < 0. x and y are on opposite sides of 0 on the number line

| x - y | - distance of x from y
|x| - distance of x from 0
|y| - distance of y from 0

If you imagine a number line
like this
x--------0-----------y
or
y--------0-----------x

you can conclude that the distance between x and y is greater than the difference between x,0 and y,0.


HTH

[cutlass]

Just to clarify, I wasn't trying to suggest that the geometric way is the quickest. I just wanted to point out another approach that works for absolute problems.

[RonPurewal]

B ) xy < 0. x and y are on opposite sides of 0 on the number line

| x - y | - distance of x from y
|x| - distance of x from 0
|y| - distance of y from 0

If you imagine a number line
like this
x--------0-----------y
or
y--------0-----------x

you can conclude that the distance between x and y is greater than the difference between x,0 and y,0.


HTH

well played.

note to other users reading this post: it is well worth your time to learn the intuitive interpretations of the absolute value, especially interpretations of it as distances on the number line. although these sorts of problems can always be solved by number plugging, it is sometimes much faster and easier to apply the distance interpretations.