Posts filed under: main blog

This is becoming tradition at this point (see results from October and November). As long as you guys keep responding and finding this useful, I’ll keep it up. If you took the December 2015 SAT, and want to help inform the argument about how lenient or punishing the scoring table might be with some data, answer this simple poll.

Relative to other tests you’ve taken (real and/or practice), how hard did you find the December 2015 SAT?

This poll is over, but you can view the results here.

I’ve received a few questions in my email asking about the Beta I’m running (full info here). Specifically, people want to know what they get if they sign up.

I figured that the best way to answer that question is just to post one of the chapters up here in public. If you join the Beta, you get access to chapters of the new book as I finish them just like you’ll see below. You also get the opportunity to, if you spot a typo, submit it to me in exchange for a $5 gift card.

Registered Math Guide owners can already access the Beta for free. For everyone else, it’s $16.99. When the actual book is released, paying Beta customers will be able to buy it at a big discount.

Update 2/7/16: Thanks to everyone who joined! The Beta is still running for existing members, but is now closed to new signups as I finish out the book. Original post below…


 

As many of you know, I’ve been working for some time now on putting together a Math Guide for the new SAT. It’s been slower going than I’d hoped, but I am making progress despite the fact that this is in many ways a ground-up rewrite.

I’m at a place now where I’m keen to start sharing some things and getting feedback from you guys. So, today I’m officially launching the Beta program, through which you’ll be able to read the book as I continue to write it.

Here are the important details.

  • Once you’re in the Beta, you can read the chapters I’ve uploaded in your browser. (You cannot print or download them.)
  • The Beta will officially end on March 5, the first day the new SAT is given. When the Beta ends, Beta memberships will convert to Math Guide owner memberships.
  • When the physical book is finally available for sale, everyone who paid for Beta access will receive a code to purchase the book at a discount.
  • If you spot a typo and report it to me (a misspelling, a missing negative sign, a grammar mistake) I will reward you with a $5 Amazon gift card. There is no limit to how many of these you can get. If you spot 10 errors, you’ll end up with $50 in gift cards.
  • Access to the Beta costs $16.99 USD. Sign up here by selecting the bottom option, or use the button below. (If you’re already a registered owner of the current Math Guide, though, you get it for free. You do not need to sign up, just go to the Math Guide Owners Area and look for the link.)

At the time I’m writing this (11/21/15), there are 13 chapters available, plus solutions to all the drills in those chapters. By the end of this week, I expect there will be 14 or 15 chapters available. After that, well, I am just going to keep working until it’s done. Come along for the ride!

This survey resulted in some useful (or at least interesting) data when I did it for the October SAT, so I figured I’d do it again. Relative to other tests you’ve taken (real and/or practice), how hard did you find the November 2015 SAT?

This survey is over, but you can still view the results.

…but if you took the test, you probably already knew that. In all likelihood, if you’re reading this blog, you were probably up clicking refresh on the College Board website at 5am so that you could see how you did the very second scores went live.

I really, truly hope that when you did see your scores, you were thrilled.

If you weren’t, of course, then you’re probably acutely aware that you only have 3 more chances to take the SAT in its current form before it’s changed forever.

Here’s the good news: the current test is very intimately known by those of us who work in test prep. It’s been around for a little over 10 years, after all. If you want to improve and you have the will to devote yourself to some good prep, there’s a very good chance that you’ll be able to better your score.

There’s no time to waste, though. So dust yourself off, convert your disappointment to motivation, and start transforming your weaknesses into strengths.

Good luck to all of you taking the brand new PSAT tomorrow! I look forward to hearing your reactions to it.

Relative to other tests you’ve taken (real and/or practice), how hard did you find the SAT today?

This survey is over, but you can still view the results.

2015-09-20 20.22.06

(An actual fortune cookie I got a few days ago.)

I know a LOT of you are taking the October SAT. By the time I’m posting this, hopefully many of you will already be in bed, sleeping soundly in confident anticipation of your morning victories.

I wish you well.

Those of you who have been using the site regularly over the past few days might have noticed a few site hiccups—either images not loading, or the whole site not loading, logins not working, etc. Let’s be clear: all that happened because I’m a tutor, not a web developer, but I still insist on doing pretty much all of this by myself. Hubris.

Anyway, I think I’ve got it all fixed now. Further, I think that this site should be faster and more stable than it’s ever been. So…yay!

Please let me know if you’re running into any problems—just leave a comment in this post.

keep-calm-and-randomize-on-1The new SAT will occasionally ask you questions about experimental design—whether the results of an study conducted by some students, for example, can be generalized to an entire population, or whether some experimental intervention has a causal impact. These questions will not be rocket science, and will not require any math at all, even though they’re in the math section. They will require some critical thinking and careful reading, though.

Randomization

They will also require you to be on the lookout for randomization. Generally speaking, the more randomized an experiment is, the stronger the conclusions that can be drawn.

If you’re doing research, and you want to be able to generalize your findings over an entire population, then you have to randomly select the subjects for your experiment from that entire population. Say you want to know, on average, how likely a new driver in the United States is to have an accident within one year of getting his or her driver license. If you want to be able to generalize your findings to the entire population of new drivers in the United States, then you need to do a random selection from that whole population. That’s not going to be easy to do, of course, but good research is hard! If you want to generalize about all new drivers in the US, you’re going to have to find someone who has the data you need. Maybe start calling the Department of Motor Vehicles in every state, or car insurance companies.

If you tried to take the easy way out and just went to your local shopping mall and asked the first hundred 17-year-olds you saw whether they’d been in an accident in the first year of having their licenses, then you could really only generalize your findings about the population of new drivers who visit that mall. Maybe going to that mall requires driving on a highway, which not every new driver in the country does. Maybe that mall is in a city, where people are more likely to get into a fender bender than new drivers who live in rural areas. Maybe people who shop in malls are more likely to get into car accidents than people who do most of their shopping online. You see?

Things get a bit more complicated when you’re dealing with experiments in which an intervention of some kind occurs. If you want to be able to generalize your results to an entire population, you need to select randomly from that population. If, further, you want to be able to argue that some intervention is causing some difference between experimental groups, you need to make sure that the intervention is assigned to subjects randomly. Again, in practice, this is not easy to do. That’s why good research is valuable: it’s hard to do!

Say a cognitive scientist is trying to determine whether a certain intervention can be used to cause infants to exhibit object permanence earlier than they usually do. Say, further, that this researcher managed to obtain a truly random sample of infants (this is very difficult, since parents would need to sign off on such a thing, and maybe there’s some difference between children of parents who would sign off and children of parents who wouldn’t…). If the data shows that babies who received the intervention did, on average, exhibit object permanence earlier than those who did not receive the intervention, the cognitive scientist could only claim that the intervention caused the accelerated object permanence if the babies given the intervention were selected at random.

Holy cow—that’s a lot of text about this stuff! My longwindedness notwithstanding, though, experimental design questions really aren’t hard—I promise. Just remember: if the question you get asks you to find a weakness in the design of an experiment, chances are very good that the answer will be something that’s obviously not random.

Other elements of experimental design

Remember that we use statistics to make generalizations about large populations based on observations we make of small groups from those populations. The small groups we use (randomly, if we’re doing it right) are called samples. The number of members of the sample is called the sample size. Generally speaking, the bigger the sample size, the more likely it is that it’s representative of the whole population.

Experimental design and interpretation questions on the SAT will usually provide a sample size. In the released tests, there’s no question where a too-small sample size is an experimental design problem, but I suppose it’s possible that something like that could appear in the future. Rule of thumb: if your sample size is bigger than 100, it’s probably fine. Only consider sample size a problem if it’s comically small.

This kind of question will also sometimes make mention of a study’s margin of error or confidence interval. If you’ve taken a statistics course, then you’ve probably had to calculate these things. You’ll never have to do that on the SAT. You’ll just have to know, in the most basic sense, what they are and why they’re important.

When we generalize what we’ve observed in a sample to draw a conclusion about an entire population, we have to remember that there’s a small possibility that we’re wrong. For example, after measuring the heights of 300 randomly selected 33-year-old American males, we might say that we are 95% sure that the true average height height of 33-year-old American males is within 3 inches of the 70 inch average we found. 3 inches is our margin of error, 67 to 73 is our confidence interval. As our sample size increases and we become more confident that the sample is representative of the population, our margin of error and confidence interval will shrink: we’ll be able to, with the same level of certainty, that the average we’re finding is closer and closer to the true population average.

One more thing: the 95% in this example is called the confidence level—you’ll probably only ever see confidence levels of 95% (maybe 99% once in a while). We include a confidence level to acknowledge that, even if we’ve designed our study carefully—randomized properly, made sure our sample size was sufficient, etc.—it’s mathematically possible that our sample just won’t represent the overall population well.

You need to be registered and logged in to take this quiz. Log in or Register

The new SAT places a heavy emphasis on the “Heart of Algebra,” which is a bizarre and tortured euphemism for, mostly, working with linear equations. One of the kinds of questions you know you’re going to see, probably more than once, on your SAT is solving systems of linear equations. For example:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

There are a bunch of good ways to solve such a problem, and I think you should know all of them. So, I’m going to talk about all of them. That’s what you come here for, after all.

Substitution

To solve by substitution, first you’ll want to get one of the variables in one of the equations alone. It doesn’t matter which variable, and it doesn’t matter which equation, so pick the one that looks easiest to isolate. I’m gonna get the x from the first equation by itself. First, subtract the y-term from each side…

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{3}x=12-\dfrac{1}{6}y

Then multiply by 3 to get x alone…

3\left(\dfrac{1}{3}x\right)=3\left(12-\dfrac{1}{6}y\right)

x=36-\dfrac{3}{6}y

x=36-\dfrac{1}{2}y

Once I’ve got x alone, I’m going to substitute what I now know x equals into the other equation. This is important! You won’t get anywhere if you substitute back into the equation you just manipulated—you’ll eventually just end up at 0 = 0. Right, so:

x=36-\dfrac{1}{2}y

2x+5y=21

2\left(36-\dfrac{1}{2}y\right)+5y=21

Then I’m going to solve that equation for y:

72-y+5y=21

72+4y=21

4y=-51

y=-\dfrac{51}{4}

Once I’ve got my y I can actually look at the answer choices and see that I’m able to eliminate every choice but A. If I were in a hurry, I’d bubble that and move on. Since we’re practicing, though, I’m going to finish this process by putting the y I just found into one of the previous equations to get x. I like to use the equation I already solved for x to save me a step or two:

x=36-\dfrac{1}{2}y

x=36-\left(\dfrac{1}{2}\right)\left(-\dfrac{51}{4}\right)

x=36+\dfrac{51}{8}

x=\dfrac{288}{8}+\dfrac{51}{8}

x=\dfrac{339}{8}

So, there you have it. Choice A is for sure the answer: \left(\dfrac{339}{8},\dfrac{-51}{4}\right) is a solution for that system of equations.

Elimination

Substitution is the classic, the mainstay. Most people know how to do it and are comfortable with it. Elimination, to my constant amazement, seems less universally known. I say to my amazement because elimination is awesome.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

To solve the same problem with elimination (I’ve pasted it in above so you don’t have to scroll to see it), what you’ll want to do is instead of trying to isolate any variables, find a way to multiply one of the equations by something so that either the x-coefficients in each equation are equal, or the y-coefficients are equal. I see an easy way to do that:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

6\left(\dfrac{1}{3}x+\dfrac{1}{6}y\right)=6(12)

\dfrac{6}{3}x+\dfrac{6}{6}y=72

2x+y=72

That’s what I’m talking about! I’ve got a 2x there, and a 2x in the other equation: 2x+5y=21. Now all I need to do is subtract one equation from the other, eliminating the 2x making it very easy to solve for y:

     \begin{align*} 2x+y &=72\\ -(2x+5y&=21)\\ -4y&=51\\ y&=-\dfrac{51}{4} \end{align*}

Easy, right? Now to find x, we just substitute -\dfrac{51}{4} in for y in either equation:

2x+5y=21

2x+5\left(-\dfrac{51}{4}\right)=21

2x-\dfrac{255}{4}=21

2x-\dfrac{255}{4}=\dfrac{84}{4}

2x=\dfrac{339}{4}

x=\dfrac{339}{8}

Cool, right? Unsurprisingly, because math works, we landed on the same answer with both methods. But we’re not done yet, folks. There are still more ways to solve a system of linear equations!

Graphing

Always remember, as you’re taking the SAT, that when two graphs intersect, they do so at a point (x,y) that is a solution set for the two equations that make those graphs. If you’re in the section where calculators are allowed (and with the numbers in this question we’re playing with, you would be), then you have still more weapons for solving systems of equations at your disposal.  Let’s look at that question again.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

To solve a system of equations by graphing, first get each equation into y= form:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{6}y=12-\dfrac{1}{3}x

y=72-2x

2x+5y=21

5y=21-2x

y=\dfrac{21}{5}-\dfrac{2}{5}x

Pop those into your calculator and graph! If your window is currently set to standard zoom, you’ll probably only see one line. You should expect this to happen from time to time—the writers of the SAT would love to see you doing the algebra, not taking graphing shortcuts. Try zooming out a bunch by setting your window to go from –100 to 100 in both axes, instead of the standard –10 to 10.

Now just use your calculator’s intersect function! On a TI-83 or TI-84, you’re going to hit [2nd][TRACE] to open up the CALC menu, and then select “intersect.” Now just hit ENTER on the first line (your calculator asks for the “first curve,” but that’s only because this same function will also find intersections of non-linear functions), then hit ENTER on the second line. The calculator asks for a guess next—you don’t need to do anything here but hit ENTER. You should see something like this:

Of course, those are the decimal values of the answers we already know are correct:

\left(\dfrac{339}{8},-\dfrac{51}{4}\right)=(42.375,-12.75)

Backsolving

Finally, a multiple choice question like this can be solved by backsolving. All you have to do is try answer choices by making sure they work in both equations. Only the correct answer choice will result in true outcomes when substituted into the given equations.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

Here’s what happens when you backsolve with a wrong answer choice (I’ll use choice B):

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{3}(18)+\dfrac{1}{6}(36)=12

6 + 6=12

12=12

That worked for the first equation. But let’s see what happens in the second equation:

2x+5y=21

2(18)+5(36)=21

36 + 180 =21

216=21

Nope—that’s not true at all!

The right answer, though, will work beautifully:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\left(\dfrac{1}{3}\right)\left(\dfrac{339}{8}\right)+\left(\dfrac{1}{6}\right)\left(-\dfrac{51}{4}\right)=12

\dfrac{113}{8}-\dfrac{17}{8}=12

\dfrac{96}{8}=12

12=12

Yep!

2x+5y=21

2\left(\dfrac{339}{8}\right)+5\left(-\dfrac{51}{4}\right)=21

\dfrac{339}{4} - \dfrac{255}{4} =21

\dfrac{84}{4}=21

21=21

Yep again! That’s a good answer.

Of course, I’m showing a lot of intermediate steps you won’t need to do yourself. All you’ll want to do is type the substituted left-side of the equation into your calculator and see what you get. For example, you might enter

2(339/8)+5(-51/4)

When you hit ENTER and get 21, which is what the question said you should get, then you know that ordered pair works for that equation.

Now you try

Think you’ve got all this? I bet you do. But just so that we can both sleep a little better tonight, why don’t you try a few more questions. Because this is practice and all, you might consider trying to solve all of these questions all 4 ways.

You need to be registered and logged in to take this quiz. Log in or Register

As I said in my last post, I’m busy working on the new edition of the Math Guide for the new SAT. Here’s another quick pair of graph questions for you to try out.

You don’t need to be a site member to take this little quiz, but I’d encourage it. Site membership is free, and if you’re logged in all the quizzes you take will be stored. If you take enough of them, that adds up to useful data for you to see where your strengths and weaknesses are.

The next two questions refer to the following information.

hair salon website visitors

The owner of a hair salon is able to use his website’s analytics feature to track roughly how far visitors are from his hair salon, and how long those visitors stay on the website. Above is a scatterplot the salon owner made of the first 40 visitors to his website on a particular Saturday, and the line of best fit for that data. (You can click the graph to see a bigger version in a new window.)

You need to be registered and logged in to take this quiz. Log in or Register

I’m in full-on writing mode for the next edition of the Math Guide, which will of course be aimed at the new SAT that debuts March 2016 (and the new PSAT, which debuts in October of this year). The new test is pretty different, so it’ll be a major overhaul—some chapters in the current guide will go away, other chapters will be brand new. Obviously, that means lots of new questions to write.

To that end, I’ll be posting practice questions on the blog with some regularity over the next few months. This will hopefully benefit those of you who are already prepping for the new exam. Your feedback will also help me fine-tune the questions for clarity and difficulty. It’s a win-win!

Anyway—here’s a question pair for you to play with. You don’t need to be a site member to take this little quiz, but I’d encourage it. Site membership is free, and if you’re logged in all the quizzes you take will be stored. If you take enough of them, that adds up to useful data for you to see where your strengths and weaknesses are.

The next two questions refer to the following information.
average global temperature 2
The graph above shows the average global temperatures measured in degrees Celsius during the month of May over two 15 year periods, and their respective trend lines. (You can click the graph to see a bigger version in a new window.)

You need to be registered and logged in to take this quiz. Log in or Register

 

Data source: GISTEMP Team, 2015: GISS Surface Temperature Analysis (GISTEMP). NASA Goddard Institute for Space Studies. Dataset accessed 2015-06-15 at http://data.giss.nasa.gov/gistemp/.

By Alex E. Proimos (http://www.flickr.com/photos/proimos/4199675334/) [CC BY 2.0], via Wikimedia Commons

 

Am I missing something here? To me, this falls somewhere between not-a-huge-deal and kinda-a-big-deal. Of course, the nature of the paper SAT has always enabled people willing to risk being caught cheating to keep thinking about a question after time is called and revisit it during the next section, but now what you have is a section where calculators are prohibited followed immediately by a section where calculators are allowed.

There are, by design, questions in the no calculator section that would be very easy with a calculator. For example, look at question #18 on page 36 of Practice Test 4:

That’s a bit tricky without a calculator, no? But it’s a 2-second problem if you graph it. The function is at 0 when x = 5. No factoring or tricky algebra necessary. This is one of the hardest questions in the no calculator section.

If they know they’ll have their calculators and be able to graph the equation just a few minutes later, some kids are going to memorize the equation (or just flip back to it surreptitiously) and graph it when they get their calculators.

What boggles my mind about this issue is that it goes away if the calculator section comes first. I have to assume it wasn’t just a random choice to put the sections in the order they’re in, but I can’t imagine why this particular decision was made.

 

Hello Mike, I want to know about the new SAT of 2016. I am in high school and will be giving the SAT in 2016. Is there any thing(except without calculator section) or topic of math section that is new and can I start practicing for the new SAT now? And is new “writing” section like improving paragraph questions. I want to go for a good university or an ivy, do you recommend giving the essay section?

Thanks.

Yes, there’s a lot that’s new with the new SAT math—some of it is just different flavors (more algebra, less geometry) and some of it is totally new content (basic trigonometry, complex numbers).

College Board recently released 4 practice tests. You can download them here. You can also take them online at Khan Academy, but since you’ll take the real thing on paper, you might as well practice as you’ll play.

Impossible to say for sure about the Essay section yet, but I’m betting that the most selective schools will expect you to do it. If you’ve got Ivy aspirations, you should plan for that.