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 21 through 26 in the “calculator permitted” section. See questions 1 through 5 here, 6 through 11 here, 12 through 15 here, 16 through 20 here, and 27 through 30 here.

This one’s just about as easy as a medium question can come, but it requires careful reading of both the question and the graph. The thing you need not to miss in the question is that the dishes are 10 square centimeters in area. So when bacteria cover 7 square centimeters of the dish, they also cover 70% of the 10 square centimeter dish.

Let’s just look at each choice one at a time:

• A: Nope. At time t = 0, neither dish is even close to being 100% covered.
• B: Yes, this is the answer. Dish 2 has 2 square centimeters covered, and Dish 1 has 1 square centimeter covered. Because the dishes are 10 square centimeters in area, those percentages are correct.
• C: Nope. Dish 2 covers twice as much as Dish 1 at t = 0. That’s not 50% more, that’s 100% more.
• D: Nope. The opposite is true—Dish 1 grows much more quickly in the first hour than does Dish 2.

Pay SUPER close attention to units here. They’re your lifeblood. First, figure out what 11.2 gigabits is in megabits:

11.2 gigabits × 1024 megabits per gigabit = 11,468.8 megabits

Now figure out how many seconds are in 11 hours:

11 hours × 60 minutes per hour × 60 seconds per minute = 39,600 seconds

Now figure out how many megabits can be transmitted in that period based on the 3 megabits per second rate:

3 megabits per second × 39,600 seconds  = 118,800 megabits

Finally, figure out how many images that’ll let the tracking station receive:

118,800 megabits / 11,468.8 megabits per image ≈ 10.4

That’s 10 full images.

This is one of the most old-school SAT questions we’ve seen–all you need to do is make one quick substitution and solve, but make sure you don’t go one step too far! The question asks for , not !     I hope you don’t mind me plagiarizing myself here—the next three question were floated by College Board way back when they announced the new SAT, and I solved them in a couple posts then. This section and the following two are reprints from here and here.

To get this, first find the area of the hexagon, and then subtract the area of the circle. Once you’ve done that, you can multiply by 1 centimeter to get the volume. Then you can use the given density to find the mass. Sound like fun???

A regular hexagon can be divided into 6 equilateral triangles, like so: The area of an equilateral triangle is . (Derive that for yourself—it’s good practice! Or see here.) For each of those equilateral triangles, then, the area is . There are 6 of those, so the area of the hexagon is square centimeters.

The hole drilled through the middle has a diameter of 2, so it has a radius of 1, and therefore an area of π square centimeters. So the area of the hexagonal face of the nut, minus the hole drilled through it, is square centimeters.

Multiply that area by the height of the nut, 1 centimeter, and you get the volume: cubic centimeters.

Now, you’re told that density = mass/volume, you’re given the density 7.9 grams per cubic centimeter), and we just found the volume. We’re asked for the mass to the nearest gram.   Since we’re asked for the answer to the nearest gram, we write 57.

Yikes, right?

If the bank converts Sara’s purchase to dollars, and adds a 4% charge, then the $9.88 she’s charged doesn’t convert directly to rupees. First we need to get that 4% fee out of there. If x = the dollar cost of her purchase without the fee, then 1.04x = 9.88. x = = 9.5 So her purchase was worth$9.50, which means the exchange rate that day, in rupees per dollar, is: Round that to the nearest whole number and the answer is 63.