College Board released a bunch of sample questions this week for the new PSAT and SAT, which will make their debuts in October 2015 and March 2016, respectively. Over the next few days, I’ll be making posts working through each question, a few at a time, and commenting on them when I feel like I have something insightful to say.

In this post, I’ll deal with questions 7 through 12 in the “calculator not permitted” section. For the “calculator permitted” section, see questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, 21 through 26 here, and 27 through 30 here. You can see questions 1 through 6 from the “calculator not permitted” section here, and questions 13 through 18 here

Question 7 (link)

Difficulty: Medium
Is this new? Kinda. The vertex form of a parabola has shown up (rarely) on the current SAT.

And here we see that the new SAT will expect you to know the vertex form of a parabola. Only one of the answer choices is in vertex form, so if you know what that looks like, you don’t even need to confirm that it actually has the right numbers in it—you just need to recognize the structure. Not sure how I feel about that.

Anyway, the vertex form of a parabola is this:

For constants ah, and k, the parabola y=a(x-h)^2+k will have its vertex at (h, k). If you’re nervous about whether y=(2x-4)(x-4) and y=2(x-3)^2+(-2) are equivalent, you can do some algebra (CB walks you through that in the answer explanation) but I’m sticking to my guns—if they’re going to make it so that only one choice is actually in the right form, then you don’t need to bother checking.

Question 8 (link)

Difficulty: Medium
Is this new? Very much so.

Imaginary. Numbers. On the SAT. My whole world is turned upside-down.

No, seriously though, this stuff is easy if you know how to FOIL. All you need to do is remember, at the end, that i^2=-1.

(14)(7) + (14)(12i) + (-2i)(7) +(-2i)(12i)
98 + 168i -14i -24i^2
98 + 154i -24i^2

Now remember that i^2=-1

98 + 154i +24
122 + 154i

See? Easy.

Question 9 (link)

Difficulty: Medium
Is this new? Not really. The current SAT doesn’t usually ask questions that require this kind of equation solving, but it’s not unheard of.

Good ol’ fashioned algebra, here. Finally!


First, simplify the numerators…


Now cross multiply!



Question 10 (link)

Difficulty: Medium
Is this new? Kinda. Systems of equations on the current SAT can almost always be solved easily in one or two steps. This is not that kind of systems question.

More algebra! I’m so happy! Buckle up…

First, get the first equation into a useful form:


Now multiply it by 2:


Now add that to the second equation to eliminate the y terms and find x!


So you know x = 2. From there, it’s easy to find y:


So what’s xy? Well, it’s \left(2\right)\left(\dfrac{1}{4}\right)=\dfrac{1}{2}.

Question 11 (link)

Difficulty: Medium
Is this new? No.

After the last two questions, I kinda feel like they’re joking with this one.

\dfrac{1}{2}x + \dfrac{1}{3}y = 4

Multiply that all by 6 and, well, you’re done.

6\left(\dfrac{1}{2}x + \dfrac{1}{3}y\right) = 6(4)

3x + 2y = 24

Question 12 (link)

Difficulty: Hard
Is this new? Yes. Trig doesn’t appear at all on the current SAT.

And we get to finish this post up with some trig! Wahoo!

unit circle
From Wikipedia

There are a few ways to go here. One is to know your unit circle well enough to know that choices A, B, and D don’t make sense. Honestly, that’s how I thought through this when I first looked at it.

There’s a better way, though, which I’m ashamed not to have recognized at first. It’s what makes this an SAT question, and not just a math test question. The graphs of sine and cosine are in phase, such that \sin x =\cos \left(\dfrac{\pi}{2}-x\right)! Therefore:




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