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 27 through 30 in the “calculator permitted” section. See questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, and 21 through 26 here.
Question 27 (link)
Awesome question. You could spend some time wrestling this into y = mx + b form, but you don’t need to if you’re paying attention. Look at the choices! One of them is a parabola. There’s no term in , so you know you’re not going to get a parabola. Cross off D. The other thing you should notice is that A and C have non-zero y-intercepts. Since the given equation is doesn’t have any constant terms, it’s going to go through the origin. You can cross off A and C, and you’re left with B. This is one of the short-cuttiest questions yet!
Question 28 (link)
OK, don’t be intimidated here. All you need to do is take some of the information you’re given, and use it correctly. Put the point (–4, 0) into the given equation, and you’ll be able to solve for c. Simple as that!
Is it weird that they tell you that c is a constant but don’t feel the need to tell you that p is a constant? Yes, I suppose. But do you need p at all to solve for c? Nope. Not at all. Is this question slightly harder because you had to puzzle with that for a minute before getting to work? You bet.
Question 29 (link)
This question is kinda fun, although I bet a lot of students who encounter it will disagree with me. When the question says that those two expressions are equivalent forms, what that means is that you can set them equal to each other.
So let’s do some algebra to get A in terms of x!
Do you see the difference of two squares up top there?
Question 30 (link)
Ah, finally some trig!
You’re going to want to drop an altitude segment down from the top vertex. That’s going to end up being a perpendicular bisector and make two right triangles. (You knew that, because you know those congruent base angles mean this is an isosceles triangle, right? RIGHT???)
Remember your SOH-CAH-TOA. We’re dealing with cosine here, which is the adjacent leg over the hypotenuse.
The cosine of x, then, is just .