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Perhaps you’ve heard by now (or perhaps you knew first, because you took the thing yourself) that there was a printing issue with some test forms during the June 6 SAT administration. Some test booklets told students that they would have 25 minutes section 8 or 9, even though those are always 20-minute sections. Some proctors gave students 25 minutes; others gave them 20 minutes. This was a widespread issue: all students in the United States are affected.

College Board has issued a statement that they will be able to produce valid and reliable scores for all June 6 test takers, even without sections 8 and 9, so they’re just going to do that. If you took the June 6 SAT, your reading and math scores will be based only on your performance on the two 25-minute sections. Your writing score will still be based on your essay and both multiple choice writing sections.

I’m sure some people, whose energy flagged towards the end of the test, will be thrilled at this revelation. Others who might have felt they did their best on those sections will be upset. Nobody will ever know how they did on those sections. College Board will probably take some heat in the press.

I can’t help but pile on a little bit. In their statement, College Board says the following:

Q: How is it possible to not score a whole section and still have valid scores?

To accommodate the wide range of incidents that can impact a testing experience, the SAT is designed to collect enough information to provide valid and reliable scores even with an additional unscored section. From fire drills and power outages to mistiming and disruptive behavior, school-based test administrations can be fragile, so our assessments are not.

We have deliberately constructed both the Reading and the Math Tests to include three equal sections with roughly the same level of difficulty. If one of the three sections is jeopardized, the correlation among sections is sufficient to be able to deliver reliable scores.

That’s some pretty next-level spin, there. How they can boast that their test is so reliable because they “deliberately constructed both the Reading and the Math Tests to include three equal sections,” even as the new SAT will unceremoniously do away with that structure, is beyond me. Are we to believe that power outages will never occur again after March 2016? Are school-based test administrations suddenly going to become less “fragile” next year? Or should we conclude that the new SAT is inherently less robust than the current test?

Anyway, if you’ve been affected by this, I’d love to hear from you in the comments. Did you think you nailed sections 8 and 9? Are you relieved because you think you bombed them?

This kind of thing is going to be featured much more heavily on the new SAT than it is on the current one, so I figured I’d put a quick quiz together.

Struggling with any of these? Discuss them in the comments!

For each of the items below, choose the mathematical statement that is equivalent to the given sentence.

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The College Board has released, at long last, its first full practice test since announcing sweeping changes to the PSAT and SAT last year. You may now finally, if you’re one of a small group of merry misfits that actually enjoy this kind of thing, sit down for 3 hours and take a prototype test from start to finish. Then you can correct it. You cannot, however, score it yet. Unsurprisingly, College Board still hasn’t fully figured out how it’s going to scale these things.

You can download the test here, and the answer key and explanations here.

Believe it or not, the answer document is longer than the test document.

Anyway, I just spent 3 hours of my beautiful Sunday afternoon with this thing, so I figured I’d jot down a few of my initial reactions in bullet form.

Reading

  • At an hour long, the reading section is a real slog. And it’s the first thing students will do.
  • As advertised, there’s less emphasis on tough vocabulary.
  • The passages, for the most part, feel familiar. Many of the questions would be just as at home on the current SAT.
  • The stuff that’s new is…fine, I guess. The questions that ask students to select the best justification from the passage for their previous answer are annoyingly formatted. Because the sections from the passage aren’t recreated in the choices—it’s the familiar old SAT-format, e.g. Lines 7–9 (“Thanks . . . life”)—there’s a LOT of necessary flipping back and forth.
  • The graphs that appear seem tossed in after the fact, without much thought put into integrating them into the passage.
  • I guess the “Great Global Conversation” piece in this test is the excerpt from Carnegie’s “Wealth”? Two thoughts: 1) Blech. 2) Students with some contextual knowledge of Carnegie’s life will be at an advantage answering those questions.
  • Strong readers will probably welcome the changes—they’ll need to worry much less that, despite their comprehension skills, they might encounter vocabulary words they don’t know on test day. Weak readers will be at a disadvantage, but, well, it’s a reading test.

Writing and Language

  • The writing is, basically, ACT writing. I know I’m not making an original observation here, but there’s just no other way to describe it.
  • In this one test, the difference between their, they’re, and there, AND the difference between its and it’s are both tested.
  • There’s punctuation, too. Students will love that. </sarcasm>
  • Something I unsarcastically do love is an emphasis on clear, concise prose over both colloquialisms and awkward usage of fancy words. In one question, a student must choose between “prosaic directives,” “simple directions,”  “bare-bones how-tos,” and “facile protocols.” I am fully on board with discouraging people from using phrases like “prosaic directives” and “facile protocols” whenever possible.

Math

  • Right off the bat, it’s obvious this test is different—more information is supplied on the first page than there used to be. Formulas for the volume of a sphere, the volume of a right cone, and the volume of a right pyramid are now provided. Also, the note that used to say that all numbers on the test are real numbers now says: “All variables and expressions used represent real numbers unless otherwise indicated.” This sample PSAT contains no spheres, cones, pyramids, or imaginary numbers, but they’re fair game now.
  • It’s probably too early to say this, but the answer choices seem less intentionally devious. On one question, I rescued myself from forgetting a negative sign because the answer I arrived at wasn’t a choice. Phew! The old SAT might not have been so kind.
  • The “Heart of Algebra” questions that come early in the section are kinda fun, if you know what you’re doing.
  • With the exception of one basic trigonometry question, a question about graphing inequalities, and the last two question about shooting an arrow in the air (more on that later), pretty much everything in this test felt like it could have been fair game on the last test. This surprised me a bit, but it probably shouldn’t have. I didn’t start working in test prep until a year after the last big change in 2005, but I’m told that the same thing happened—early promises of a revolution were greatly exaggerated. Then again, it’s early yet.
  • My friend Akil worries that the calculator allowed/no calculator allowed sections will result in unfairness, or at the very least confusion; I agree with him to an extent. However, I liked that I was forced to solve some questions by hand that I normally would have turned to my calculator for. It’s good to work out unused muscles.
  • There are still only 8 grid-in questions, but now they’re split among the two math sections—4 in the no calculator section, and 4 in the calculator allowed section.
  • It’s possible that there isn’t a stronger emphasis on the advanced topics showcased in previous question releases because this is a PSAT and they’re actually going to start saving the tough stuff for the SAT. We’ll have to wait and see on that one.
  • The last two grid-ins come as a pair—two questions about the same mathematical scenario. In earlier released questions, we were shown questions about a traveler dealing with exchange rates and bank fees—pretty tough stuff. Here, we’re dealing with an arrow being shot from the ground, which is tough stuff if you haven’t taken physics yet, but pretty standard if you have. What was interesting to me about the questions, though, was the amount of information provided that you don’t need at all. You are given 3 equations, and you need only one of them—the same one—to answer both questions. The old SAT wouldn’t have done that. We’ll have to keep an eye on how often the new one will.

Holy cow, I rambled on for a good long while. If you’ve got the stomach to sit down with the test, I’m very curious to hear what you think. Please, make liberal use of the comments section on this post. 🙂

Even though very little information has been released to date—not even a full practice test yet!—I’m hard at work trying to get a head start on the next iteration of the Math Guide for the new SAT (coming about a year from now—March 2016). When I write questions, I have a very hard time keeping them to myself, so if you’re curious what I’m thinking some of the tougher topics the new SAT might throw at you will look like, well, they might look something like this.

(Once you’re done here, you can try my last set of questions in the style of the new SAT.)

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7% got 5 right
13% got 4 right
18% got 3 right
26% got 2 right
24% got 1 right
12% got 0 right

 
 

You can redeem them if you wish, either for a PWN the SAT sticker or for DonorsChoose.org gift cards.

As you may or may not know, I donate 10% of the royalties I receive from book sales to educational charities. I spread it around to a few different charities, but most of that money goes to DonorsChoose.org, because I think it’s a really cool organization that addresses a really pressing need.

One of my favorite things to do every month is choose classrooms to donate to, but I’ve always thought that it would be cool if I could involve you guys, the users of this site and the readers of my books, in the process. When you redeem your quiz points for gift cards, I make the donation, but you get to decide where it goes.

Anyway, you can use the form below to redeem quiz points you’ve earned for stuff. Going forward, I may add more rewards if this ends up being a feature people like.

Oh—last thing! If you redeem points for a sticker, I need your shipping address to mail it. If you redeem points for a gift card, it’ll go to the email address I have for you. Therefore, before you redeem points, click here to make sure the right information is in your account profile.


 

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You don’t need to find this post again to redeem points—you can always do so by clicking stats in the upper right hand corner of the page when you’re logged in, and then going to the “Points and Rewards” tab.

The five questions in this short quiz would all, I’m fairly sure, fit into the “Heart of Algebra” category that will be so heavily emphasized on the new SAT. Obviously, without even a full practice test released yet, it’s tough to know for sure if I’ve got the style right, but based on what I’ve seen, you’d better be able to cruise through these if you’re planning to take the SAT after January 2016.

(UPDATE: Try another set of questions in the style of the new SAT here.)

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How’s everyone else doing on this quiz?

36% got 5 right
29% got 4 right
14% got 3 right
12% got 2 right
7% got 1 right
2% got 0 right

Here’s a new quiz on patterns. For one week, it’ll be available to all site members, but then it’ll only be available to Math Guide owners. If you want the extra patterns practice, make sure you take this by 2/14 (Valentine’s Day)! How’s everyone else doing on this quiz?…

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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.

Question 13 (link)

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 y=mx+b form, and then see what a needs to be to make the slopes the same:

\dfrac{1}{2}x-\dfrac{1}{4}y=5

-\dfrac{1}{4}y=-\dfrac{1}{2}x+5

y=2x-20

So the slope we’re looking for is 2. Proceeding to the next equation:

ax-3y=20

-3y=-ax+20

y=\dfrac{a}{3}x-\dfrac{20}{3}

So you know \dfrac{a}{3}=2. That means a must equal 6.

Question 14 (link)

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 \dfrac{1}{5} 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 \dfrac{1}{x} represents the slow printer, and the \dfrac{2}{x} represents the fast one.

Question 15 (link)

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:

test_specifications_for_the_redesigned_sat_na3_pdf 8

Of course, since PD is a radius, it has a length of 3 just like AP and PB.

2^2+QP^2=3^2

4+QP^2 = 9

QP=\sqrt{5}

Only one answer choice has \sqrt{5} 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 \sqrt{5}.

Question 16 (link)

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).

Question 17 (link)

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 (x-a)^2+(y-b)^2=r^2.

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.

x^2+y^2-6x+8y=144

Do a little rearranging:

(x^2-6x)+(y^2+8y)=144

Now, what binomial square begins with x^2-6x? What binomial square begins with y^2+8y?

  • x^2-6x is the beginning of the (x-3)^2=x^2-6x+9 binomial square
  • y^2+8y is the beginning of the (y+4)^2=y^2+8y+16 binomial square.

So, to complete those squares, we need to add 9 and 16 to both sides of the circle equation!

(x^2-6x+9)+(y^2+8y+16)=144+9+16

(x-3)^2+(y+4)^2=169

(x-3)^2+(y+4)^2=13^2

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.

Question 18 (link)

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.

\dfrac{24}{x+1}-\dfrac{12}{x-1}=1

This is just algebra. Let’s do it. First, get a common denominator:

\dfrac{24(x-1)}{(x+1)(x-1)}-\dfrac{12(x+1)}{(x+1)(x-1)}=1

Hopefully you recognize that the denominator is a difference of two squares:

\dfrac{24x-24}{x^2-1}-\dfrac{12x+12}{x^2-1}=1

\dfrac{24x-24-(12x+12)}{x^2-1}=1

\dfrac{12x-36}{x^2-1}=1

Now you can move that denominator over:

12x-36=x^2-1

And set everything equal to 0:

0=x^2-12x+35

And finally, factor:

0=(x-7)(x-5)

x could equal 7 or 5.

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 7 through 12 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 questions 13 through 18 here

Question 7 (link)

Difficulty: Medium
Is this new? Kinda. The vertex form of a parabola has shown up (rarely) on the current SAT.

And here we see that the new SAT will expect you to know the vertex form of a parabola. Only one of the answer choices is in vertex form, so if you know what that looks like, you don’t even need to confirm that it actually has the right numbers in it—you just need to recognize the structure. Not sure how I feel about that.

Anyway, the vertex form of a parabola is this:

For constants ah, and k, the parabola y=a(x-h)^2+k will have its vertex at (h, k). If you’re nervous about whether y=(2x-4)(x-4) and y=2(x-3)^2+(-2) are equivalent, you can do some algebra (CB walks you through that in the answer explanation) but I’m sticking to my guns—if they’re going to make it so that only one choice is actually in the right form, then you don’t need to bother checking.

Question 8 (link)

Difficulty: Medium
Is this new? Very much so.

Imaginary. Numbers. On the SAT. My whole world is turned upside-down.

No, seriously though, this stuff is easy if you know how to FOIL. All you need to do is remember, at the end, that i^2=-1.

(14-2i)(7+12i)
(14)(7) + (14)(12i) + (-2i)(7) +(-2i)(12i)
98 + 168i -14i -24i^2
98 + 154i -24i^2

Now remember that i^2=-1

98 + 154i +24
122 + 154i

See? Easy.

Question 9 (link)

Difficulty: Medium
Is this new? Not really. The current SAT doesn’t usually ask questions that require this kind of equation solving, but it’s not unheard of.

Good ol’ fashioned algebra, here. Finally!

\dfrac{5(k+2)-7}{6}=\dfrac{13-(4-k)}{9}

First, simplify the numerators…

\dfrac{5k+3}{6}=\dfrac{9+k}{9}

Now cross multiply!

(9)(5k+3)=(6)(9+k)
45k+27=54+6k
39k=27

k=\dfrac{27}{39}=\dfrac{9}{13}

Question 10 (link)

Difficulty: Medium
Is this new? Kinda. Systems of equations on the current SAT can almost always be solved easily in one or two steps. This is not that kind of systems question.

More algebra! I’m so happy! Buckle up…

First, get the first equation into a useful form:

4x-y=3y+7
4x-4y=7

Now multiply it by 2:

8x-8y=14

Now add that to the second equation to eliminate the y terms and find x!

8x-8y=14
+(x+8y=4)
9x=18

So you know x = 2. From there, it’s easy to find y:

2+8y=4
8y=2
y=\dfrac{1}{4}

So what’s xy? Well, it’s \left(2\right)\left(\dfrac{1}{4}\right)=\dfrac{1}{2}.

Question 11 (link)

Difficulty: Medium
Is this new? No.

After the last two questions, I kinda feel like they’re joking with this one.

\dfrac{1}{2}x + \dfrac{1}{3}y = 4

Multiply that all by 6 and, well, you’re done.

6\left(\dfrac{1}{2}x + \dfrac{1}{3}y\right) = 6(4)

3x + 2y = 24

Question 12 (link)

Difficulty: Hard
Is this new? Yes. Trig doesn’t appear at all on the current SAT.

And we get to finish this post up with some trig! Wahoo!

There are a few ways to go here. One is to know your unit circle well enough to know that choices A, B, and D don’t make sense. Honestly, that’s how I thought through this when I first looked at it.

There’s a better way, though, which I’m ashamed not to have recognized at first. It’s what makes this an SAT question, and not just a math test question. The graphs of sine and cosine are in phase, such that \sin x =\cos \left(\dfrac{\pi}{2}-x\right)! Therefore:

\sin\dfrac{\pi}{5}=\cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{5}\right)

\sin\dfrac{\pi}{5}=\cos\left(\dfrac{5\pi}{10}-\dfrac{2\pi}{10}\right)

\sin\dfrac{\pi}{5}=\cos\dfrac{3\pi}{10}

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}}

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 ymxb 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 x^2 term in y-x=k(x+y), 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!

f(x)=2x^3+3x^2+cx+8

f(-4)=2(-4)^3+3(-4)^2+c(-4)+8

0=-128+48-4c+8

72=-4c

-18=c

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.

\dfrac{4x^2}{2x-1}=\dfrac{1}{2x-1}+A

So let’s do some algebra to get A in terms of x!

\dfrac{4x^2}{2x-1}-\dfrac{1}{2x-1}=A

\dfrac{4x^2-1}{2x-1}=A

Do you see the difference of two squares up top there?

\dfrac{(2x+1)(2x-1)}{2x-1}=A

2x+1=A

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 \dfrac{16}{24}=\dfrac{2}{3}.

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.

Question 21 (link)

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.

Question 22 (link)

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.

Question 23 (link)

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 x^2, not x!

x^2+y^2=153

y = -4x

x^2 + (-4x)^2 = 153

17x^2 = 153

x^2 = 9

Question 24 (link)

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 \left(\dfrac{1}{2}\right)\left(b\right)\left(\dfrac{b\sqrt{3}}{2}\right). (Derive that for yourself—it’s good practice! Or see here.) For each of those equilateral triangles, then, the area is \left(\dfrac{1}{2}\right)\left(2\right)\left(\dfrac{2\sqrt{3}}{2}\right)=\sqrt{3}. There are 6 of those, so the area of the hexagon is 6\sqrt{3} 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 6\sqrt{3}-\pi square centimeters.

Multiply that area by the height of the nut, 1 centimeter, and you get the volume:6\sqrt{3}-\pi 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.

7.9=\dfrac{m}{6\sqrt{3}-\pi}

7.9\left(6\sqrt{3}-\pi\right)=m

57.2706… = m

Since we’re asked for the answer to the nearest gram, we write 57.

Question 25 (link)

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 = \dfrac{9.88}{1.04} = 9.5

So her purchase was worth $9.50, which means the exchange rate that day, in rupees per dollar, is:

\dfrac{602\text{ rupees}}{9.5\text{ dollars}}\approx 63.36\text{ rupees per dollar}

Round that to the nearest whole number and the answer is 63.

Question 26 (link)

If Sara buys the prepaid card, she doesn’t have to pay the 4% fee her credit card charges her, but she could lose money if she doesn’t spend the whole card. The exchange rate doesn’t actually matter anymore—we can actually forget about dollars entirely at this point.

The question is really how much Sara has to spend on her credit card, with the 4% fee, before she’d save money by buying the prepaid card, even if she doesn’t spend it all. Say y = the amount of rupees Sara spends:

7,500 = 1.04y
7,211.54 = y

If Sara spends 7,211.54 rupees on her credit card, it costs her the same as buying a 7,500 rupee prepaid card. Therefore, if Sara spends 7212 rupees or more on her prepaid card, she gets a better deal than if she spends the same amount on her credit card.

 

 

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

Question 16 (link)

Math_Sample_Question__16___SAT_Suite_of_AssessmentsWell, this is embarrassing for CB, but now there have been 2 questions in a row where an answer choice (and the correct answer choice, no less) is printed incorrectly. (See Question 15 in my previous post.) In this case, choices A and C, screengrabbed at right, should have || symbols, not blank squares. Someone didn’t check his or her font library carefully enough before choosing a math font!

Typo aside, this is a good question. The thing you need to know about similar triangles is that their corresponding angles are congruent. Since you have vertical angles at point C, which must be congruent to each other, you know how the corresponding angles are laid out. I’ve color coded them below.

 

Do you see what’s going on at points B and D, and points A and E? Those are alternate interior angles of lines AB and DE, so if they’re congruent, then those lines must be parallel!

Question 17 (link)

Another Heart of Algebra question, another scenario described that we need to identify an equation for. For this one, it’ll be helpful to use units. You know you want to end up with gallons, so make sure all the miles and hours units in your equation cancel out!

Hopefully it’s obvious that you need to start with “17 –” something. That right there eliminates choices C and D, amazingly.

Each remaining choice has a 50t in it. The 50, we know, is a speed, in miles per hour. The t is time, in hours. So the units of 50t is miles per hour × hours = miles. Remember that we want our final units to be gallons.

The units of 21 is miles per gallon. So between choices A and B, we’re choosing between \dfrac{\text{miles per gallon}}{\text{miles}} and \dfrac{\text{miles}}{\text{miles per gallon}}. Which of those simplifies to gallons? B does.

Question 18 (link)

They’re not even asking you to solve systems of equations anymore—they’re just asking you to be able to write them. This is just the first step on medium and hard questions on the current test, but on this medium question it’s the only step. There are x cars and y trucks, so you xy should equal the total number of vehicles, 187. You also know how much each kind of vehicle has to pay in tolls. Cars pay $6.50 and trucks pay $10. So 6.5x + 10y should tell you the total amount collected in tolls, $1338.

Your equations, then, are xy = 187, and 6.5x + 10y = 1338. That’s choice C.

Question 19 (link)

OK. There a few ways we might go here. One is to just plug the (d, p) coordinates we know (9, 18.7) and (14, 20.9) into each answer choice and see which one gives us a true equation. Since calculators are allowed on this question, that might be the fastest way. You might also enter each equation into your graphing calculator and use the table function to see which equation gives you the points you want, like so:

See how Y2 has the points you want? That means the equation entered into Y2 is the right equation!

Of course, you can also use the points you’re given to find the equations algebraically. First you find the slope:

\dfrac{20.9-18.7}{14-9}=0.44

…then you find the intercept using one of the points:

18.7 = 0.44(9) + b
14.74 = b

That’ll tell you that the equation you want is p = 0.44d + 14.74. If you ask me, though, using the calculator is the way to go here.

Question 20 (link)

This question is essentially asking for the slope of the line of best fit, so our job is to find 2 points on the graph, and then use the slope formula: slope = \dfrac{y_2-y_1}{x_2-x_1}.

Math_Sample_Question__20___SAT_Suite_of_Assessments 2Remember that it’s fine to approximate here! The answer choices are really far apart, so even if our approximations aren’t very good, we’ll know the right answer. Using my numbers, here’s the slope calculation:

\dfrac{3000-2000}{2003-1996}=\dfrac{1000}{7}\approx 143

143 isn’t an answer choice, but it’s awfully close to 150, and not very close at all to any other answer choice. 150, therefore, is the right answer.

 

 

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

Question 12 (link)

Now we’re finally getting into some fun stuff! We know the company will bring in 12n dollars for n items sold, and we know it costs the company 7n + 350 to make n items. To figure out how many items need to be sold to make a profit, solve the following inequality:

12n > 7n + 350
5n > 350
n > 70

Again, the focus here is not on the algebra itself—it’s on whether you can figure out how to set the algebra up. That, I suppose, is what the College Board means by “Heart of Algebra.”

Question 13 (link)

I’m flummoxed here because they’re calling this a medium difficulty question, and it seems far easier than most of the other easy questions we’ve seen together so far, as long as you remember that the number of males doesn’t necessarily equal the number of females. (Again, this is a reading question as much as it is a math question.) If the average age of the males is 15, and the average age of the females is 19. If the number of males equalled the number of females, then the average age would be exactly 17, but we don’t know that the preserve has the same number of males and females.

Question 14 (link)

In spite of myself, I kinda like this question. It’s all about being cautious and modest with data. The only statement that data about 2000 16-year-olds in the US can support is a statement specifically about 16-year-olds in the US. If you want to draw conclusions about people in the world, then you have to sample the whole world. So cross off B and D.

Then, you have to remember something that will be drilled into your head in any social science class (and hopefully has already been drilled into your head a bit): correlation does not imply causation! Just because exercise and sleep are positively correlated (this question uses the word associated, which is a more general term, but association does not imply causation lacks that alliterative flair) we cannot conclude that one causes the other.

Question 15 (link)

Before we start discussing this, note that the CB, at the time of my writing this post, has F’d up a bit. Answer choice D, which should be the right answer, is misprinted. It looks like that over there on the right, but it should look like this: P=50(2)^{\frac{n}{12}}. That’s a big boo-boo to have a typo in the right answer—hopefully they’ll fix it soon.

Anyhoo, you can plug in here if you like. You know the population is supposed to double every 12 years, and you know the starting population is 50. So if you say n = 12, the right answer should resolve to 100. Let’s check:

A: 12 + 50(12) \neq 100
B: 50 + 12(12) \neq 100
C: 50(2)^{12(12)}\neq 100
D: 50(2)^{\frac{12}{12}}= 100

Only D works!

If you want to understand why D works, you need to first understand exponential growth in general. The basic setup of exponential growth (compound interest in a savings account is one classic application of this) is that you have a starting value (a), a growth rate over a certain period (r), and a number of periods (p). Where you’ll be after p periods is ar^p. Why?

Well, look at just the first 3 periods, one period at a time. You start with a. After 1 period, a grows by a factor of r, so you have ar. After a second period, ar grows by a factor of r again, so you have ar^2. After a third period, ar^2 grows by a factor of r again, and you get ar^3. Et cetera. I guess if this is going to start appearing on the new SAT, I should make a whole post about it. 🙂

Anyway, in this case, the growth period is 12 years, but we’re given n in years. That’s why we need our exponent to be \dfrac{n}{12}. That way, in 24 years, when there have only been 2 growth periods, the exponent \dfrac{24}{12} will equal 2, as it should.

 

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.

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.