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 13 through 18 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 7 through 12 here.

Difficulty: Hard
Is this new? Kinda. (This basic concept is tested on the current SAT, but the algebra in this question is a bit more involved.)

The thing you need to know here is that the only way two linear equations will have no solution is when they create parallel lines (and parallel lines have equal slopes). So put both equations into form, and then see what a needs to be to make the slopes the same:   So the slope we’re looking for is 2. Proceeding to the next equation:   So you know . That means a must equal 6.

Difficulty: Hard
Is this new? Yes. The current SAT generally stays away from work/rate problems.

This is another one of the “new” questions that appeared in the first document we got about the new SAT months ago. I hated it then, and I still hate it now.

The basic idea is that you need to see, from the right-hand side of the equation, that the equation is telling you how much the printers will print in 1 hour, which is of the job.

From there, you need to recognize that each fraction on the left represents one of the two printers. From there, hopefully it’s intuitive that the represents the slow printer, and the represents the fast one.

Difficulty: Hard
Is this new? Not really—It’s basically a right triangle question. The only thing new is that the current SAT doesn’t have chord as part of its vocabulary.

This question doesn’t immediately look like a right triangle question, but it’s totally a right triangle question—and one where plugging in will work, at that! Say r = 3, so that AB = 6 and CD = 4. Now draw some lines: Of course, since PD is a radius, it has a length of 3 just like AP and PB.   Only one answer choice has in it, so we can feel pretty good about choice D without even doing the last step of the plugging in process, which is to put 3 in for r in each answer choice. Of course, when you do, D is the only answer to give you .

Difficulty: Hard
Is this new? Yes. There is no trigonometry on the current SAT.

A calculator would be nice here—plug in, graph, and you’re done. Without a calculator, you need to know the relationship between angles with opposite sines. Since the sine function has a period of 2π, subtracting π from x inside the function will result in a negated sine. You might recognize that this really ends up being a graph translation problem. The red graph shifts the blue graph π units to the right (if we had added π and shifter π units to the left we would have had the same result).

Difficulty: Hard
Is this new? Yes. There are no circle equation questions on the current SAT. This could appear on the current SAT Math Subject Tests, though.

To solve this one, you need to know the general equation of a circle, which is this:

A circle with radius r and a center of (ab) has the equation .

You also need to know how to complete the square, because otherwise you’re not going to be able to wrangle what you’re given into that form. Do a little rearranging: Now, what binomial square begins with ? What binomial square begins with ?

• is the beginning of the binomial square
• is the beginning of the binomial square.

So, to complete those squares, we need to add 9 and 16 to both sides of the circle equation!   So we have a circle centered at (3, –4) with a radius of 13. The question asked for the diameter, though, so the answer is 26.

Difficulty: Hard
Is this new? Not really. The algebra is trickier than you’d see on most current SAT questions, but I would not be surprised to see this as a #18 in the current SAT grid-in section. This is just algebra. Let’s do it. First, get a common denominator: Hopefully you recognize that the denominator is a difference of two squares:   Now you can move that denominator over: And set everything equal to 0: And finally, factor: x could equal 7 or 5.