Every day, the ETS/College Board kindly e-mails to subscribers an "SAT Problem of the Day": about one in every four is a math problem. Twisted soul that I am, I frequently start my day by answering the problem of the day, and while I'm generally able to nail better than 98% of them, I particularly look forward to the math problems, and even more so to ones that relatively few people who've attempted them have gotten right (yes, I'll admit to still having a sad little place in my soul that is stroked by such silliness).
This morning's poser was definitely challenging for a majority of respondents. As of this writing, with nearly 80,000 folks attempting the problem, only 40% had answered it correctly. Before we continue, here it is for your smoking enjoyment:
The stopping distance of a car is the number of feet that the car travels after the driver starts applying the brakes. The stopping distance of a certain car is directly proportional to the square of the speed of the car, in miles per hour, at the time the brakes are first applied. If the car’s stopping distance for an initial speed of miles per hour is feet, what is its stopping distance for an initial speed of miles per hour?
(A) feet (B) 51 feet (C) 60 feet (D) 68 feet (E) 85 feet
Okey-doke. I'll give you some time to think about it, preferably before you look further down the screen for the official answer and explanation. And it's that explanation I want to address. No peeking now! After all, your self-esteem is riding on how well you do here (not to mention your entire future).
All done? Great. Now before we see the official explanation/answer, let's take this opportunity to do what the folks in Princeton and Berkeley (East and West Coast headquarters of the ETS) apparently don't want us to do: think!
First of all, were this problem to appear in an actual test (despite my enormous collection of past actual SAT tests, I don't recall having seen this one before), my best guess is that it would be one of the last ten problems or so in a section of mathematics problems, based on the 40% correct response rate (I've seen math problems where fewer than 10% of students attempting it got the right answer; as I have no way of knowing what percent of the folks who answered this one today are actually high school students, this might actually be a bit harder for the intended audience than the numbers indicate). It seems fair to suggest that this isn't something that is ridiculously difficult, but that many high school students would nonetheless get wrong for various reasons.
One reason, of course, is literacy: if students read this carelessly or are unable to comprehend the point, they may overlook or simply fail to understand the significance of the words "directly proportional to the square of the speed of the car." If so, one obvious trap answer, a distractor, if you will, that will draw a disproportionately high number of incorrect replies, is (A). It's relatively easy to miss/ignore the word "square" and figure that since 40 mph is twice the speed 20 mph, the stopping distance will be twice that of the slower car and hence 2 x 17 ft = 34 ft.
But if you're on your toes, even if you're not a math whiz you should be suspicious of this answer. It's just too damned easy to come up with. A late-appearing SAT or ACT math problem isn't going to be quite so simple. And in fact, any simple equation you set up here that ONLY involves the given numbers and an unknown will be wrong. That "square" thing isn't put in for show, and it's going to come into play somehow or other.
Sometimes it's possible on problems at this level of difficulty to do more reasoning and eliminate additional choices after avoiding the "big trap" answer. I've not thought this one through further other than to solve it, but perhaps interested readers can offer ideas as to why the other three wrong answers were chosen by the test-makers as possible replies. For me, the problem itself spoke to such a simple solution that I just cut to the chase and answered it. And then, I eagerly awaited learning what the experts had to say as to why my answer was (of course) correct. ;^) So with no further ado, here is the official explanation from the ETS/College Board:
The stopping distance is directly proportional to the square of the initial speed of the car. If represents the initial speed of the car, in miles per hour, and represents the stopping distance, you have that the stopping distance is a function of and that , where is a constant. Since the car’s stopping distance is feet for an initial speed of miles per hour, you know thatTherefore, , and the car's stopping distance for an initial speed of miles per hour is
Say what? Gosh, but that seems like a hell of a lot of work to me, and quite frankly, 0.0425 never came into play in my calculations. Having done (way too) many SAT/ACT math problems, when I see something that involves ratios (usually geometry problems, but the underlying issue here is the same), I think about dimensions: am I looking at a linear/linear ratio (if it's a challenging problem, probably not)? If not, is it linear/square? Linear/cubic? Some other ratio where the dimensions change? In this case we're looking at a proportion in which we have a linear ratio (speed to speed) that is being set equal to another linear ratio (braking distance to braking distance), but we're told that the second varies as the SQUARE of the first. So we've got (20/40)^2 which simplifies to (1/2)^2 or 1/4. Thus, the other side of this proportion needs a number on the bottom (conveniently, our unknown) that is 4 times the number on top (17) - of course, you might set things up another way. As 4 x 17 = 68, that's the right answer. Doable, I might add, in your head if you have a bit of facility with mental arithmetic. On the other hand, I challenge you to do the number-crunching in the official explanation in your head. Not impossible, of course, but not exactly as simple as squaring (1/2) and then multiplying 17 by 4 (one way is to take (15 + 2) x 4 = 60 + 8 to get a total of 68.
My point is, of course, that the best place to go for explanations on SAT/ACT math problems is not the test-makers. Considering that these are timed tests, they will NEVER do a "process of elimination" approach, which is necessary when you don't know how to proceed (and when you have to deal with certain question structures, both in math and elsewhere, where it's much easier to eliminate wrong answers than arrive at the right one). They'll never talk about the errors they presume you'll make or the whole concept of distractors. Nope, that would be a bit too helpful. And it would undermine the nice illusion that want us all to harbor that these tests are simply logical extensions of the sorts of things we teach/learn in school.
But of course, a word to the wise guy should suffice: these aren't school tests, never have been, never will be. They're part of a game that's been set up to reward those most adept at cutting through the baloney. It's not absolutely necessary to be able to do that in order to do well, but it certainly helps, particularly on a timed test, where the longer you take to do any given problem, the less you can spend on others, some of which may take more than the average allotted time per problem (about 60 seconds on math). The better you are at cutting to the chase, the more time you buy for the ones that you really need a little more time to figure out, and the more likely you are to actually be able to see and think about all the problems. That's an edge every student needs, but few actually get. And if everyone follows the "official" explanations, few ever will.