MORE Bad SAT Advice From ETS

No they didn't!
(Yes, they did.)

Another day, another SAT math Problem of the Day with bad advice for students from the ETS. I just can't figure out why they continue to foist these horrid explanations on kids.

Consider the following problem from May 31, 2013:

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) 34 feet (B) 51 feet (C) 60 feet (D) 68 feet (E) 85 feet

Take some time, think about it, and keep in mind that on average you have less than a minute to do math problems on the SAT and ACT. I don't know where this particular problem would appear in a given math section, but let's assume it would be in the last fifth of a section for argument's sake (and to help novices apply a little test wisdom).

Okay, so, what did you decide? First of all, the "sucker" answer is (A). The problem states (though not until the SECOND sentence) that stopping distance is directly proportional to the SQUARE of the speed of the car in MPH, etc. If you missed that or didn't know what to do with it, you very likely concluded that since 40 is twice as much as 20, then the correct answer should be twice as much as 17 feet, hence the conveniently placed first answer choice! Of course, it's wrong.

If you avoided that trap, you've got a shot at the right answer. If not, well, you didn't and don't.

Another trap answer is (C), which you might come up with if you added the two speeds and then figured that somehow things like distance and rates of speed were pretty much the same (or didn't bother to figure beyond "Hey, two numbers! I'll ADD them! That always works!")

Not sure where the other two wrong answers, (B) and (E) come from, but trust that someone will have an idea posted here before I see it myself and offer a revision.

Regardless, the correct answer, by process of elimination, is (D) 68 feet.

So before I give my approach to the solution, let's see what our helpful friends at ETS suggest students do:

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 that . Therefore, , and the car's stopping distance for an initial speed of  miles per hour is  feet.

SUPER! Got that? And if you did it that way without any help from the ETS, I'm hoping you're a VERY quick calculator and/or were able to get that typed into your electronic device accurately and promptly. So first we compute this utterly unnecessary constant, c, and then we bring decimals in by taking a not particularly pretty fraction, 17/20, and squaring the denominator, to get 17/400, which we THEN turn into a decimal so we can multiply it by 40^2. Uh, huh. That's great. If you're not in much of a hurry.

How about this instead? The ratio of the speeds is 20/40 = 1/2; But the stopping distance varies by the square of that ratio, which is 1/4. That means that the faster car will go 4 times as far before stopping: 4 x 17 = 68. Game over.

To generalize, distance 2 = (distance 1) (rate 1 / rate 2)^2), so, yes, had the speeds been, say, 20 and 60 mph, the distance for car 2 to stop would be 17 x 9 = 153 feet. If you're a habitual tailgater, this concept's for YOU, pal. (Full disclosure: I routinely howitzer people who tailgate on highways at speeds of 70 mph and upwards. You've been warned).

Honestly, someone should recommend that the ETS stop publishing these "helpful" explanations. Or that students learn to read them with great skepticism about their efficacy on actual timed tests. Sure, once in a Purim a problem really requires some tricky or involved calculations; sometimes you have to really think, though the computation part isn't difficult if you've thought accurately and carefully. But this is just b.s. designed to mislead kids into thinking that there is no reason to think. And the sharpest kids KNOW better. They do this kind of thing in their sleep, and are good at mental arithmetic (honestly, I never put pencil to paper for this, and I read it at 5:45 AM or thereabouts), and rarely get sucked into longwinded computation.

Don't get me wrong: as I always say, if you can get a problem with more computation than I did, but in a workmanlike manner, go forth and be well. There IS time for SOME of that sort of thing. Just not on every problem. Not even on every reasonably challenging problem. Too much of that sort of thing and you almost certainly will find yourself short on time at the end of math sections. As I know from painful experience, c. 1967 - 68, and again in 1973 (my first time on the GRE). Somehow, though, between '73 and '91 (second GRE), I managed to go from a 640 on the math to a 780. And it wasn't that I knew all that much more math (though I did). Rather, it was that I had the right math and the right ideas and techniques at my fingertips. Lots of time spent thinking and passing along what I'd learned to high school kids in metropolitan NYC/NJ. Twenty-two years further down the road and I still marvel at the bad advice the ETS gives out free. I wonder what you learn if you PAY for their courses? I have a bad feeling that they don't give away the good stuff, even for $$.