Sunday, July 24, 2011

Trust Us, We're The ETS (And we have a nice bridge for sale in Brooklyn, too!)


This was today's (July 24, 2011) SAT question of the day from our good friends at the College Board and Educational Testing Service:

Read the following SAT test question and then click on a button to select your answer. 

If a, b, and c are numbers such that a / b = 3 and b / c = 7, then (a+b) / (b+c) is equal to which of the following?

A. 7 / 2

B. 7 / 8

C. 3 / 7

D. 1 / 7

E. 21


It's an old problem. I remember it from the '80s (some of the problems that appear in the Problem of the Day go back as far as the '70s). Nothing particularly wrong with it, nothing particularly great about it. But here's what bothers me, though it doesn't surprise me based on experience. Here's the explanation they offer:

Explanation

From a / b = 3 is implied that (a+b) / b = 4 (1)

And from b / c = 7 is implied that b / (b+c) = 7 / 8 (2)

If we multiply (1) and (2) together, we have that (a+b) / (b+c) = 4 times (7 / 8) = 7 / 2 ( b is canceled out).


Hmm. Well, maybe I don't know high school kids all that well after tutoring SAT math for 30-odd years, but I don't think very many of them would get that explanation or think to do the problem that way, assuming they tried to solve it. And that's one of my gripes with many of the ETS solutions to their own problems: they're generally not the quickest or most intuitive way to get at the answer.

My approach was to solve for a and c in terms of b, the variable that's common to both of the equations. It follows that a = 3b. Substituting for a in the expression (a + b) yields 3b + b = 4b. The solving for b in the second statement gives c = b/7 and substituting for c in the expression (b + c) yields b + b/7 or 8b/7. So the entire expression (a + b) / (b + c) in terms of b is 4b/ (8b/7). Indeed, the b's simplify to 1 (it's implied in the original problem that none of the variables are zero), and the rest simplifies to 4 * 7/8 or 7/2.

Maybe that's not "better," that what ETS offers, but I think it's more clear as to what's going on and why the approach works. The ETS explanation feels a bit like magic to me, looking at it from a student's perspective. That kind of thing in textbooks always frustrates me, and I know it loses a host of kids at the starting line. While this sort of answer from them may not be their most egregious sin, it is one example of why I warn students to not assume that when ETS explains something in math (or anything else) that it's always going to be coin of the realm. It's not wrong, but how helpful is it, really, for most kids?