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 1 through 6 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 7 through 12 from the “calculator not permitted” section here, and questions 13 through 18 here.

Math_Sample_Question__1___SAT_Suite_of_AssessmentsQuestion 1 (link)

Difficulty: Easy
Is this new? No.

This is an interesting question, because it’s really just testing whether you know that when a line is translated, its slope doesn’t change at all! The slope of the line in the figure is -\dfrac{3}{2}, which is easy to see from the two darkened points—the line moves down 3, and right 2. Therefore, the translated line will also have a slope of -\dfrac{3}{2}.

Question 2 (link)

Difficulty: Easy
Is this new? Kinda. The content is familiar, but the way the question is asked is new.

Heart of Algebra questions (the easy ones, anyway) will usually look like this, apparently. The thing to pay attention to in this equation is where the variable, x, is. If x represents the number of years since 2004, we can conclude that the average must have been 27.227 in 2004, and has increased slightly every year since then, at a roughly constant rate. The 0.8636 number, which is multiplied by x, must be the number that’s added to the average every year—the estimated annual increase in the average number of students per classroom.

Question 3 (link)

Difficulty: Easy
Is this new? Not really, although they’re calling it easy here and this would not be an “easy” question on the current test.

In this “Passport to Advanced Math” question (I promise that I’ll get over these crazy category names soon, but I’m not quite there yet) we’re being tested on whether we recognize the opportunity to complete a binomial square. That a^2+14a bit is most of what we get when we square the quantity a+7. What’s missing?

(a+7)^2=a^2+14a+49

Do you see the opportunity to substitute? We already know thata^2+14a=51!

(a+7)^2=51+49

(a+7)^2=100

Because the question tells us that a is positive, we can ignore the negative square root and just say:

a+7=10

(If this weren’t the “calculator not permitted” section, we also could have solved this by graphing very quickly.)

Question 4 (link)

Difficulty: Medium
Is this new? No.

Just do a little cross-multiplying here:

\dfrac{2}{a-1}=\dfrac{4}{y}

2y=4a-4

y=2a-2

Question 5 (link)

Difficulty: Medium
Is this new? No.

They’re calling this one a medium difficulty question, but I just don’t buy it, unless they think the simple presence of a cubic term is going to send kids running for the hills.*

If y=x^3+2x+5 and z=x^2+7x+1, then:

2y+z=2\left(x^3+2x+5\right)+x^2+7x+1  =2x^3+4x+10+x^2+7x+1  =2x^3+x^2+11x+11

*With the elimination of the guessing penalty on the new SAT, there’s less reason than ever before to go running for the hills.

Question 6 (link)

Difficulty: Medium
Is this new? No.

I have to say that so far I’m liking the “calculator not permitted” questions a lot better. I don’t really know why, since most of the “calculator permitted” questions weren’t really calculator questions either, but it seems like we’ll get to do our fun algebraic manipulation work mostly in the “calculator not permitted” section.

A few ways to go here, but what I usually do with a question like this is just take it one step at a time until I’ve isolated the thing I want to isolate. In this case, that means I’m going to first raise both sides to the –1 power, and then square both sides.

a^{-\frac{1}{2}}=x

\left(a^{-\frac{1}{2}}\right)^{-1}=x^{-1}

a^{\frac{1}{2}}=x^{-1}

\left(a^{\frac{1}{2}}\right)^{2}=\left(x^{-1}\right)^2

a=x^{-2}

That’s not an answer choice, so let’s apply what we know about negative exponents:

a=\dfrac{1}{x^{2}}

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