ildude02 wrote:I would still wish if Ian or Stuart can explain the concept of solving such MOD arithmetic and MOD in general since they both do the best job at breaking it down.
Well, to be honest, I almost never do absolute value questions algebraically. Absolute value measures distances on the number line, so if I see an absolute value equation or inequality on the GMAT, I'll ask myself 'what does this say about distances?' Perhaps because I'm used to thinking this way, I find it very fast, and very easy to understand (since it only involves visualization on a number line). The other great advantage is that almost all of the very difficult absolute value questions are much more easily answered if you understand the distance interpretation of absolute value. And, of course, you can fall back on the algebraic approach if you have to, though I don't often need to.
So, I'll try to walk through how I'd do this question in detail- the explanation may seem longer than the algebraic solutions above, but it took about ten seconds to do.
First the theory. Recall that:
|x| measures the distance from x to 0;
|a-b| measures the distance between a and b on the number line;
|a+b| is the distance between a and -b, because |a+b| = |a-(-b)|.
I'll do an easier question first:
If |x-3| = 5
what is x?
We know |a-b| is the distance between a and b. So the above equation just says: "the distance between x and 3 is equal to 5". In other words, "x is five away from 3". Draw the number line, draw 3 on it, and x is either five to the left of 3 (so x = -2), or five to the right of 3 (so x = 8).
Now to the question above. We are asked:
Is |x| < 1?
This says in words: 'is the distance between x and zero less than 1?' Or, is x less than one away from zero? In other words, is it true that -1 < x < 1?
At this point, I'd draw a number line and label the points -1 and 1: they're certain to be important.
1) |x+1| = 2|x-1|
In words, this equation says: "The distance between x and -1 is equal to twice the distance between x and 1". I find that easier to understand if I phrase it this way: "x is twice as far from -1 as it is from 1". Look at the number line: where could x be? There's one value x could have which is somewhere between -1 and 1 (and while we really don't care what that value is- it's a DS question- you can probably see it should be x = 1/3). There's also one value x might have which is larger than 1. Looking ahead to Statement 2 it's useful to notice that this value is x=3. So not sufficient.
2) |x-3| not= 0
So x is not 3. Not sufficient on its own.
1) + 2) We had two values for x from statement 1. With statement 2, we rule out one of these solutions. We can find x, and indeed it is true that |x| < 1. Sufficient.
