Alternative Solutions to the Official Practice Tests

Like many of my fellow SAT teachers, I have been spending a fair amount of time going over the four released rSATs – just the math. Then I looked at the official solutions. The solutions all look right to me, but I couldn’t help noticing:

  1. They only present the relentlessly algebraic solutions.
  2. There is something enervating about their style of writing about math. ["Enervating": one of those obscure words kids used to learn for the SAT...]
  3. For somewhere between half and 2/3 of the problems, I found myself thinking: hey wait -- there's another way (and sometimes several).

Depending on how fluent your algebra is, you may find the alternative way easier at least sometimes. I was writing them all up to include in my book revision, but I decided to post them here on my blog as well:

http://wp.me/P4uvY7-9X

Since I would end up posting a lot of these here one thread at a time, I figured why not share them all now. Eventually, as more students prep for this test, the internet will saturate with alternative solutions like it did with the Blue Book. For now, this can help you get going*. The links to the four sets of solutions are at the bottom of the page. And there is some general advice about how to prepare as well.

*Really, they are not step-by-step solutions. More like the kind of hint a math teacher gives you when they want you to know how to get started but still have to finish on your own.

@pckeller
Test 2, Sec. 4 #30: I recommend simply adding the formula for the area of an equilateral triangle of side length s (A = s^2 sqrt(3)/4) to our list of formulas. We could derive it by drawing an altitude, but why do that every time?

Test 3, Sec. 3 #9 You can avoid a lot of algebra by noticing that the second equation is represented by x+y = 3. So without determining a,b, we know that a+b = 3.

Test 3, Sec. 3 #13 I just did 16/2 = 8 because determining the sum of the roots of a polynomial is fairly easy (for a quadratic ax^2 + bx + c, the sum of the roots is -b/a). The only caveat is that you have to check that 4 is not a double root.

Test 4, Sec. 3 #16: You can just shortcut and say that the height of the bottle is 3/6 (or 1/2) of 18 in, since the ratio of the height to the edge lengths (x, 2x, 3x) is constant.

Correction: “Test 3” should be replaced with “Test 2.”

Here’s another one:

Test 4, Sec. 3, #15 You can do it via the quadratic formula and some algebra, or by picking numbers as @pckeller suggested, but a slight variant involves using the sum of the roots equaling -b/a (as above!). The sum of the roots is k/2, so only (A) or (B) could possibly be correct. Then here I might try picking numbers, e.g. k = 0 and p = 2. Here x = ± 2 is the desired solution, and only (B) works. Alternatively, determine which choice is equivalent to x = sqrt(2p). Tbh, I likely would’ve done it this way if I was very low on time.

The reason why I brought that up was because picking numbers immediately might lead to picking k = 0, but k = 0 and any choice for p wouldn’t work since (B) and (D) would be indistinguishable. Picking a nonzero k works but is probably slower.