siddharthasingh wrote:
Nice explanation Chris.
But I don't think that picking up numbers is the best strategy. Can you tell how to decide which method , mathematics or picking numbers , to be chosen after seeing the question.
Moreover can you tell me my mistake in the post just before your post.
Hi siddharthasingh,
I think choosing between solving a problem conceptually, algebraically, or by picking numbers is really up to what you are comfortable with. For myself, when it comes to these types of questions with lots of absolute values, equations, and , I know I figure it out quicker either conceptually or by picking numbers. It's basically the best strategy for myself. I basically came to this conclusion after doing lots of practice problems with the help of my
error log.
Now, keep in mind I probably would not have done ALL those calculations in my post. I'm not aiming to solve the question, I just want to see if I have enough information to solve the question. Once I came up with a "yes" AND a "no" for each statement, I would have stopped. I just wrote it all out to explain.
|a|>|b| => either i)a>b ii)a<-bActually, there is a multitude of possibilities. What this statement means is that a is farther away from 0 than b is.
Both positive: a>b
both negative: a<b
a neg, b pos: a<b where 0-a>b-0
a pos, b neg: a>b where a-0>0-b
So I personally think it would be easier to view the statement conceptually, knowing that |a|>|b| means the distance of a is greater than the distance of b.
Now, please keep in mind that the above scenarios are POSSIBILITIES. Only 1 is true. We want to figure out if they're both positive or not.
1) a<0
On combining i and 1, b<a<0. Therefore a-b is positive. a|b| is negative.
On combining ii and 1)
Two cases are emerging.
a-------0-------(-b)
a-------(-b)------0We don't know if b<a. we only know that |a|>|b|. knowing that a is a NEGATIVE gets rid of a few possiblities above.
A<0, and be can be EITHER neg, pos, or both.
i) b=pos while |a|>|b|
(neg)|pos|<(neg)-(pos)....neg<pos... YES
ii)b=neg while |a|>|b|
(neg)|neg|<(neg)-(neg)....neg<neg....YES Keep in mind that in this case a|b| will be a bigger negative than a-b
iii)b=0
(neg)|0|<(neg)-0....0<neg.... NO.
1 is insufficient.
Not sure if this helped clear anything up....