The ETS Says, "Why Make Math Problems Easy?"
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Today's SAT Problem of the Day is a true classic, the one that Princeton Review made its bones with. Please remember before answering that this is from a late 1970s SAT (though a student-generated (free) response version with different numbers appeared within the last decade. It will be obvious in a bit that a non-multiple choice version requires actually knowing how to calculate this accurately). Calculators did not become available on the SAT until 1995. The section on which this appeared contained either 25 or 35 problems, and it was the last one in the section (meaning that it would be among the most difficult for students, based on previously-acquired empirical evidence during the screening process ETS uses for its test items and section construction).
A woman drove to work at an average speed of 
miles per hour and returned along the same route at 
miles per hour. If her total traveling time was 
hour, what was the total number of miles in the round trip?
miles per hour and returned along the same route at miles per hour. If her total traveling time was hour, what was the total number of miles in the round trip?A. 
B. 
C. 
D. 
E. 
If your algebra skills are solid, you're comfy with fractions, and you're good at word problems, you might be able to figure this out in the average 45-60 seconds per math problem you can afford on a timed test from ETS or ACT. If not, well, then, you're probably SOL*
Here's what, according to the ETS, you SHOULD have done:
Let t represent the time it took the woman to drive to work. Since her total traveling time was hour, the time it took her to return home is 1 - t. The distance in miles can be found using the formula distance = rate x time. The distance traveled to work is 40t, and the return distance is 30(1 - t) . Since the distance to work is the same as the return distance, 40t = 30(1 - t) . Solving for t yields. So the distance one way is
miles. The total number of miles in the round trip is
, or
miles.
Got all that? I thought not. First, let me ask ETS, Why in hell would you solve for time in this problem when the question asks for the distance? Why not cut to the chase a bit? So if I were teaching this in a mindless, math-teacherish way, I'd say, "Kids, if distance = rate x time, than time = distance/rate. We know the total time for the round trip is 1 hour and that the two legs take d/30 and d/40 hours, respectively. [Pause to point out that if the distance weren't the same for both legs, we'd be dealing with a different animal, in all likelihood.] Finding a common denominator and simplifying the equation 1 = d/30 + d/40 yields 1 = 7d/120, which yields d = 17 1/17. Doubling gives the round trip distance of 34 2/7.
Not a HUGE improvement, but at least the focus is on the answer we're asked to come up with. The problem can be slightly simplified by substituting d/2 for d, and if complex fractions don't bother you, a step is saved at the end.
But frankly, relatively few people would come up with this quickly enough to get the answer in a reasonable amount of time. So let's get real. What can you do in 60 seconds?
As is often the case, a combination of being test-smart and using a bit of common sense will completely solve this problem without the complicated formulas and calculations in any of the above explanations.
First, note that this is either problem 25 of 25 or 35 of 35. Hence, anything you can figure out in a second or two will be dead-wrong. Most students (and quite a few adults) will jump to the conclusion that the correct choice is the arithmetic mean of 30 and 40 (after all, "half" the trip is spent at 30 mph and "half" at 40 mph, the average of which is. . . 35 mph! And look! The thoughtful problem-poser has give 35 as choice D.
I will bet you pesos to prunes that if you give this problem to random citizens that the vast majority will pick D (assuming you're not asking random citizens of, say, the Institute for Advanced Study). It's just so. . . obvious. But the test-wise student knows better than to fall for this distractor. That frees her/him up to think a bit more. Choices A & E fail the test-wisdom criterion that says, "The correct answer to a challenging problem on a standardized exam is almost never a number that appears in the statement of the problem." And so, just on test-smarts, we've gotten it down to B & C.
Pause to note that I've given groups of students the question, after we've eliminated D, "Which other two answer choices just don't make sense?" and been told with alarming frequency that B & C are illogical. Why? Because (wait for it). . . kids don't like mixed numbers and fractions. They can't imagine how sevenths could arise here. So instead, they are willing to state publicly that 30 and 40 make more sense.
Now, if we stop using "test-wisdom" and try a little common sense, we see from the jump that D is wrong because if you drive a certain distance at 30 mph and then drive the SAME distance at 40 mph, the second leg takes LESS time than the first. So more than half the time is spent driving at the slower rate, bringing the total distance to LESS than the mean of the two rates, or in other words, less than 35 miles. D is eliminated. And if you spend an hour driving either at 30 or 40 mph, you can't go as few as 30 miles or as many as 40, eliminating A & E. So common sense alone brings us to the same spot as test-wisdom. And that spot is, we know the right distance is either 30 1/7 miles or 34 2/7 miles.
If these were VERY close to one another, say, 34 2/7 and 34 3/14, well, it would be coin-flip time (unless you could reason more mathematically as to why one of these is less likely). But lucky for us, the given choices for B & C are quite different from an estimation perspective. B is claiming that driving about half an hour at 30 mph and driving about half an hour at 40 mph takes you only 1/7 miles further than driving an hour at 30 mph. That's crazy. Whereas C implies that you wind up going only 5/7 mile LESS than if you had driven for an hour at 35 mph. . . which is actually plausible. Either way, C is our best choice.
An experience test-taker with some reasonable estimation-skills, test-wisdom, and common sense will get something like this right most of the time. At worst, s/he would have a 50% chance of guessing correctly in a situation where blind guessing offers only a 20% chance of being right, and a (mis-)educated guess of D has exactly a 0% chance of being correct (ditto for A & E, of course).
And so, once again, looking back at the "official" explanation, we see that the ETS rarely, if ever, gives us insight into math questions it offers from the perspective of test-wisdom and/or common sense. It pushes implicitly the viewpoint that to get these problems right, you have to instantly latch onto the/a "correct" formulaic method that even if arrived at in a timely manner is time-consuming and clunky. 30+ years after federal hearings determined that test-prep helps for these things, the ETS is STILL pretending that their math problems are strictly a measure of your grasp of the math you're taught in school, and ONLY that. Those who do well obviously have mastered all the right math (despite the demonstrable fact that some of the more intriguing SAT, GRE, and GMAT math problems involve ideas that are barely touched upon in K-12 mathematics, if they're mentioned at all, such as elementary ideas from combinatorics, abstract algebra, etc.), and that the most advanced ideas generally taught in high school math (e.g., topics from calculus or even precalculus) are rarely, if ever, required for getting a perfect SAT (or even GRE basic test) math score.
I don't want to imply that given the fact that the ETS offers to give you on-line test prep, for a fee, that the sort of math "help" you're likely purchasing isn't worth it (any opportunity to see more authentic standardized test problems is potentially useful to the thoughtful student), but frankly, I'd never advise anyone to pay for ETS on-line course services if reasonable alternatives are available (and they nearly always are).
Happy holidays from all of me here at RME to all of you.
Not a HUGE improvement, but at least the focus is on the answer we're asked to come up with. The problem can be slightly simplified by substituting d/2 for d, and if complex fractions don't bother you, a step is saved at the end.
But frankly, relatively few people would come up with this quickly enough to get the answer in a reasonable amount of time. So let's get real. What can you do in 60 seconds?
As is often the case, a combination of being test-smart and using a bit of common sense will completely solve this problem without the complicated formulas and calculations in any of the above explanations.
First, note that this is either problem 25 of 25 or 35 of 35. Hence, anything you can figure out in a second or two will be dead-wrong. Most students (and quite a few adults) will jump to the conclusion that the correct choice is the arithmetic mean of 30 and 40 (after all, "half" the trip is spent at 30 mph and "half" at 40 mph, the average of which is. . . 35 mph! And look! The thoughtful problem-poser has give 35 as choice D.
I will bet you pesos to prunes that if you give this problem to random citizens that the vast majority will pick D (assuming you're not asking random citizens of, say, the Institute for Advanced Study). It's just so. . . obvious. But the test-wise student knows better than to fall for this distractor. That frees her/him up to think a bit more. Choices A & E fail the test-wisdom criterion that says, "The correct answer to a challenging problem on a standardized exam is almost never a number that appears in the statement of the problem." And so, just on test-smarts, we've gotten it down to B & C.
Pause to note that I've given groups of students the question, after we've eliminated D, "Which other two answer choices just don't make sense?" and been told with alarming frequency that B & C are illogical. Why? Because (wait for it). . . kids don't like mixed numbers and fractions. They can't imagine how sevenths could arise here. So instead, they are willing to state publicly that 30 and 40 make more sense.
Now, if we stop using "test-wisdom" and try a little common sense, we see from the jump that D is wrong because if you drive a certain distance at 30 mph and then drive the SAME distance at 40 mph, the second leg takes LESS time than the first. So more than half the time is spent driving at the slower rate, bringing the total distance to LESS than the mean of the two rates, or in other words, less than 35 miles. D is eliminated. And if you spend an hour driving either at 30 or 40 mph, you can't go as few as 30 miles or as many as 40, eliminating A & E. So common sense alone brings us to the same spot as test-wisdom. And that spot is, we know the right distance is either 30 1/7 miles or 34 2/7 miles.
If these were VERY close to one another, say, 34 2/7 and 34 3/14, well, it would be coin-flip time (unless you could reason more mathematically as to why one of these is less likely). But lucky for us, the given choices for B & C are quite different from an estimation perspective. B is claiming that driving about half an hour at 30 mph and driving about half an hour at 40 mph takes you only 1/7 miles further than driving an hour at 30 mph. That's crazy. Whereas C implies that you wind up going only 5/7 mile LESS than if you had driven for an hour at 35 mph. . . which is actually plausible. Either way, C is our best choice.
An experience test-taker with some reasonable estimation-skills, test-wisdom, and common sense will get something like this right most of the time. At worst, s/he would have a 50% chance of guessing correctly in a situation where blind guessing offers only a 20% chance of being right, and a (mis-)educated guess of D has exactly a 0% chance of being correct (ditto for A & E, of course).
And so, once again, looking back at the "official" explanation, we see that the ETS rarely, if ever, gives us insight into math questions it offers from the perspective of test-wisdom and/or common sense. It pushes implicitly the viewpoint that to get these problems right, you have to instantly latch onto the/a "correct" formulaic method that even if arrived at in a timely manner is time-consuming and clunky. 30+ years after federal hearings determined that test-prep helps for these things, the ETS is STILL pretending that their math problems are strictly a measure of your grasp of the math you're taught in school, and ONLY that. Those who do well obviously have mastered all the right math (despite the demonstrable fact that some of the more intriguing SAT, GRE, and GMAT math problems involve ideas that are barely touched upon in K-12 mathematics, if they're mentioned at all, such as elementary ideas from combinatorics, abstract algebra, etc.), and that the most advanced ideas generally taught in high school math (e.g., topics from calculus or even precalculus) are rarely, if ever, required for getting a perfect SAT (or even GRE basic test) math score.
I don't want to imply that given the fact that the ETS offers to give you on-line test prep, for a fee, that the sort of math "help" you're likely purchasing isn't worth it (any opportunity to see more authentic standardized test problems is potentially useful to the thoughtful student), but frankly, I'd never advise anyone to pay for ETS on-line course services if reasonable alternatives are available (and they nearly always are).
Happy holidays from all of me here at RME to all of you.
*SOL is a technical psychometrics term developed by experts at the Educational Testing Service that stands for "Slow Ordinary Loser."
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