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 6 through 11 in the “calculator permitted” section. See questions 1 through 5 here12 through 15 here, 16 through 20 here, 21 through 26 here, and 27 through 30 here.

Question 6 (link)

More of this business where a scenario is described and you need to select the equation that accurately captures it. I actually think this is really important and something worth testing, but the fact that we’ve seen two of these in only 6 questions makes me think I’m going to be pretty tired of them pretty quickly once the new test arrives.

The trick here is that the $5 fee is untaxed, so you need to calculate the room fee, at $99.95 per night, apply tax to that, and then add the $5. You probably work with percents often enough to know that the 1.08 in each answer choice* represents 108%, which is what we need to multiply the room fee by to add 8% tax.

If you stay at the hotel for x nights, then the per-night fee will be 99.95x. Add 8% tax to that and you get 1.08(99.95x). Then add the $5 untaxed fee, and you get answer choice B: 1.08(99.95x) + 5.

* Where are the trap answers? If you put 1.08 in each answer choice, you’re making something a lot of students struggle with, percents, too easy! This ends up being a question that’s more about testing whether students caught the word “untaxed” than whether they really know how to work with percents.

Question 7 (link)

Jeez. The figure here might intimidate you for a second, but look at the question! All it’s asking is how many solutions the system has. Remember this well, as I imagine it’ll appear on this new test fairly often: When the graphs of each function in a system converge in one intersection point, that’s a solution to the system. That happens twice in this figure, so there are two solutions.

Math_Sample_Question__7___SAT_Suite_of_Assessments

Question 8 (link)

Ooh—a grid-in! It’s all about reading a table, though. Nothing too exciting.

There are 7 metalloids in the solids and liquids columns, and there are 92 total elements that are solids or liquids. The fraction you want here is 7/92.

Question 9 (link)

Now this question should feel pretty reminiscent of the current SAT to anyone who’s been prepping with me. It’s a solving for expressions question, through and through! How do you go from what you’re given to what you want? You’re given info about -3t+1, and you want to know about 9t-3? Multiply everything by –3!

Oh, and don’t forget that when you multiply or divide a negative through an inequality, you need to flip the direction!

-\dfrac{9}{5} < -3t+1 < -\dfrac{7}{4}

\left(-\dfrac{9}{5}\right)\left(-3\right)>\left(-3t+1\right)\left(-3\right) >\left(-\dfrac{7}{4}\right)\left(-3\right)

\dfrac{27}{5}> 9t-1 > \dfrac{21}{4}

27/5 = 5.4 and 21/4 = 5.25, so any number between those works just fine. Note that while this question is probably tougher than anything else we’ve seen yet, it’s still classified by CB as easy.

Question 10 (link)

More table reading! Here, we’re asked which age group had the highest percentage of voters. Easy enough—just divide the number of voters by the number of survey respondents in each row! You’ll see that the 55- to 74-year-olds were the best voters: 43,075/59,998 ≈72%.

Note that you might be able to eyeball the first two age groups and see that they’re not in competition, but you’re going to need to calculate the latter two. The people aged 75 and over had a 70% voting rate—too close to 72% to call by eyeball.

Question 11 (link)

More with the voters, and again, whether you get this question right will really hinge on how carefully you read it, rather than how well you can solve an equation. 287/500 voters in the 18- to 34-year-old group voted for Candidate A. Assuming that the random sample of 500 of those folks is perfectly representative of the whole group, that means we can set up and solve a proportion:

\dfrac{287}{500}=\dfrac{\text{Candidate A voters}}{30,329,000}

17,408,846=\text{Candidate A voters}

Of course, the sample won’t really be perfectly representative, which is why we just say about 17 million people in that age group voted for Candidate A.

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