There are two kinds of proportionality (some call these problems “variation” problems, but I’m sticking with proportionality) problems that you might see on the SAT: direct and inverse. I’m going to cover both here since I’m in the business of preparing you for any eventuality, but you should know that the the former is much more prevalent than the latter. Don’t sweat inverse proportions all that much.
There are a few ways to represent direct proportionality mathematically. The Blue Book likes to say that when x and y are directly proportional, y = kx for some constant k. This definition is correct, of course, but I find it to be less useful since it introduces an extra value into the mix and doesn’t lend itself as easily to the kinds of questions you’ll usually be asked on the SAT. I much prefer to say:
When x and y are directly proportional:
Note that k is still in there, but we don’t have to deal with it directly anymore. I like to streamline.
In a direct proportion, as one value gets bigger, the other gets bigger by the same factor. As one gets smaller, the other gets smaller by the same factor*. Observe:
p and q are directly proportional
p = 4, q = 10
p = 8 (⇑), q = 20 (⇑)
p = 40 (⇑), q = 100 (⇑)
p = 2 (⇓), q = 5 (⇓)
p = 1 (⇓), q = 2.5 (⇓)
For easy direct proportion questions, all you’ll need to do is plug values into the proportion above, and solve. And the “hard” direct proportion questions won’t actually be much harder.
Let’s see an example
- If y is directly proportional to x2, and y = 8 when x = 4, what is y when x = 5?
What makes this question tricky is that y is proportional to x2, but we’re given values of x. Don’t get freaked out. Just square your x values before you plug them into the proportion. 42 = 16, and 52 = 25. Watch:
No big deal, right? You should get y2 = 12.5, which is answer choice (C).
Inverse proportions are much less common on the SAT, but as I said above, they are fair game and so you should know what to do if you actually do encounter one. Again, the Blue Book’s definition of inverse proportions (y = k/x for some constant k) involves a constant and is therefore not the most expedient definition for test-day deployment. Instead, think of it this way:
When x and y are inversely proportional:
Again, note that k hasn’t disappeared; k is equal to both sides of the equation. I’ve just given you one less value to keep track of and have a name for.
In an inverse proportion, as one value gets bigger, the other gets smaller, and vice versa*:
p and q are inversely proportional
p = 4, q = 10
p = 8 (⇑), q = 5 (⇓)
p = 10 (⇑), q = 4 (⇓)
p = 2 (⇓), q = 20 (⇑)
p = 1 (⇓), q = 40 (⇑)
So…how about another example?
- If u and w are inversely proportional, and u = 11 when w = 5, what is u when w = 110?
Drop your values into the formula, and you’re good to go:
Solve, and you’ll find that u2 (uns…dos…tres…CATORCE!) is 0.5, so choice (E) is correct. Note that choice (A) is there in case you misread the question and set up a direct proportion instead. Note also that in the first example above, choice (A) would have been the solution to an inverse proportion. I’m not doing that to be a jerk. I’m doing that because the SAT will. Read the question carefully.
* Of course, these statements assume a positive k. If k is negative (in the realm of possibility I suppose, but I’ve never seen it on the SAT) you’d have to revise them to be about absolute values.
Test your might.
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