Absolute value

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Absolute value

by gmatusa2010 » Sun Dec 12, 2010 10:06 pm
How do you evaluate absolute value when there's no value? (All variables) I keep seeing it pop up in MGMAT materials, but they way OA solves them is making massive charts. I don't think this is efficient, can somebody chime in on how they handle this? I'll throw out a couple of examples. Also, are there signals as to when what kinds of numbers are best to test if it comes down to it?

1) |X-Y|>|X+Y|
2) |X-Y|>|Y-X|
3) |X-Y|> |X|
4) |X-Y|>|X|/|Y|

(These are totally made up, although the have appear in pieces peppered in MGMAT, the actually problem does not require solving them, they were mere red herring.)

I usually do number line method but can that be done with all variables?
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by gmatusa2010 » Mon Dec 13, 2010 12:48 am
Ok, so I just did them out number line way. Can experts please evaluate my method. I know MGMAT is harder than the real thing but is this too time consuming to be real gmat? It took me over 2 mintues to just evaluate 1 expression. Is there an intuitive way to do this? What about algebraic?

1) |X-Y|>|X+Y| : Translation Distance between X and Y greater than Distance between X and -Y?

Since X changes sign here, I will fix X and move Y around. X is positive so I will just assign 2.

a) If Y = 1, then -Y= -1, Thus distance between X and Y is less than distance between X and -Y. Meaning: |X+Y|>|X-Y| when Y<X but greater than 0.
b) If Y= -1, then -Y =1, the reverse is true. Meaning |X+Y|<|X-Y| when Y<X and less than 0
c) If Y=3, then -Y=3, Thus distance between X and Y is less than distance between X and -Y. Meaning: |X+Y|>|X-Y| when Y>X
d) If Y=2, then -Y=-2, thus distance between X and -Y is greater than the distance between X and Y. Meaning |X+Y|>|X-Y| when X=Y
e) If Y=0, the distance is the same.

Conclusion: |X-Y|>|X+Y| only when 0<X<Y

I did the same analysis for the others.