how the SAT changed in 2006

I'm tossing old papers, or trying to, and in my rummaging came across this Times article from 2005.

I was tickled to see that one of the problem types added to the revised test was the absolute value inequality word problem, a category I had never seen or imagined until I encountered one in Dr. Chung's SAT Math:

For pumpkin carving, Mr. Sephera will not use pumpkins that weigh less than 2 pounds or more than 10 pounds. If x represents the weight of a pumpkin, in pounds, he will not use, which of the following inequalities represents all possible values of x?
a. | x - 2 | > 10
b. | x - 4 | > 6
c. | x - 5 | > 5
d. | x - 6 | > 4
e. | x - 10 | > 4

Pop Quiz; New and (Maybe) Improved
Published: November 7, 2004

Talk about inflexible knowledge. Somehow I had concluded that absolute value inequality calculations were just that: calculations. Arithmetic. I was stunned to discover that you could have an absolute value inequality word problem.

Wonders never cease.

Elizabeth King's explanation of these problems is excellent.

 

help desk - absolute value

Problem 85 in Perfect 800: SAT Math:

In an amusement park, regulations require that a child be between 30" and 50" tall to ride a specific attraction. Which of the following inequalities can be used to determine whether or not a child's height h satisfies the regulation for this ride.

(A) | h -10 | < 50
(B) | h - 20 | < 40
(C) | h - 30 | < 20
(D) | h - 40 | < 10
(E) | h - 45 | < 5

update: The book's answer is C, which is wrong.

Allison explains absolute value inequalities

Catherine: also, I have completely forgotten how to set up and solve a simple inequality involving absolute value

Allison: You need to think about what it means.

The absolute value of any number equal to or greater than 0 is itself. x-->x

The abs value of any number less than 0 is the number*-1, or the number with its sign dropped: x --> -x.

So break down the inequality you see into other inequalities.

|x| - 2 > 3

means
|x| > 5:

allowable positive values of x satisfy
x > 5

allowable negative values of x satisfy
-x > 5

you "solve" this by switching sides (do you know why you're allowed to do that?)
and that becomes
x < -5

Now, you do |x - 3| < 5 in a similar way. To make yourself less confused, write (x-3) = y
and work on
|y| < 5

Then after you've got that into equations without the abs value, sub back in the x-3.