OE


As with all absolute value equations or inequalities, here you must solve twice:


\(\begin{array}{rlrl}
|x-2| & >3 & & \\
x-2 & >3 & \text { or } & \\
x-2 & <-3 \\
x & >5 & \text { or } & \\
x<-1
\end{array}\)


Even better, you could express the possible values of x on a number line:




Quantity A is equal to the minimum possible value of |x − 3.5|. Another way to think of |x − 3.5| is the distance on a number line from x to 3.5. Look at 3.5 on the number line above and note the nearest possible distance that is greater than 5 (x may not be exactly 5, but it could be 5.000001, for instance, since there is no requirement that it be an integer).

Therefore, since the distance from 3.5 to greater than 5 is greater than 1.5, Quantity A is equal to greater than 1.5. That is, the minimum possible value of |x − 3.5| is 1.5 plus any very small amount—for instance, 1.5000001 would be a legal value. Quantity B can be conceived as the smallest distance from x to 1.5. Look at 1.5 on the number line—the nearest value is less than −1, which is more than 2.5 units away. Thus, the minimum possible value of |x − 1.5| is greater than 2.5.

If Quantity A’s minimum is just greater than 1.5 and Quantity B’s minimum is just greater than 2.5, Quantity B is larger. The answer is (B). You could also solve this problem by plugging in values, rather than using a number line. First, solve the inequality as above to get x > 5 or x < −1. Now try plugging in greater than 5, less than −1, as well as very small and very large numbers—that is, the extremes of both ranges for x (although you may be able to use a bit of logic beforehand to tell that you only want values very close to 1.5 and 3.5)—to make sure that you generate the smallest possible value for each quantity. Quantity A’s smallest value will be smaller than Quantity B’s smallest value.

The answer is (B).