Posts filed under: Q and A

How come you don’t have a Kindle version of your book?

Formatting. There’s so much that’s not simple text in the book (e.g., fractions, graphs, geometrical figures). None of that stuff plays nicely with a Kindle. However, I do have a digital version of the book available through Google Play. That version reads exactly like a physical copy of the book.

Would you please work #10 in section 3 of PSAT Test 1?

Sure. For me, the trick to solving questions where the function isn’t defined the simplest way (e.g., f(x)=...) is thinking a little abstractly about what a function is. Basically, when you see f(x-1)=2x+3, you know that some combination of mathematical operations must happen to x-1 to transform it into 2x+3.

What could those operations be? Well, there’ll have to be some multiplication—that’s the only way we’re getting the x to turn into 2x. So let’s just start with that as an experiment; let’s say this function is just f(x)=2x and see what we’d get by plugging in x-1 as the argument:

    \begin{align*}f(x)&=2x\\f(x-1)&=2(x-1)\\f(x-1)&=2x-2\end{align*}

Of course, that’s not exactly what we want, but it is close. All we need to do now is add 5 to turn -2 into +3. That tells us the function should really be f(x)=2x+5. Let’s just double-check:

    \begin{align*}f(x)&=2x+5\\f(x-1)&=2(x-1)+5\\f(x-1)&=2x-2+5\\f(x-1)&=2x+3\end{align*}

Yep, that works! Once we know our function is f(x)=2x+5, all we need to do is drop -3 in there:

    \begin{align*}f(x)&=2x+5\\f(-3)&=2(-3)+5\\f(-3)&=-6+5\\f(-3)&=-1\end{align*}

 

Hello Mike,

I used your book(not the hard copy) for the old test and raised my score from a shaky 650 to a solid 750, I was wondering if the google play edition that you had on sale for the old test is also available for the new test. I am looking into buying that for my brother who is outside the United States.

Sidenote: I am reading your blog posts and thinking about trying the new test just for fun.

That’s a nice improvement you’ve got there…great work! Yes, the version currently for sale on Google Play is fully updated for the new SAT.

If you are thinking of trying the new test just for fun, then you are part of a pretty exclusive club! Most people outside the club think we’re loony. 🙂

I don’t know how to navigate your site yet, so please forgive me if you have already answered this question. Regarding question number 8 in your book, will you further explain why y=180-x is the same angle degree as the unmarked angle? I know y=2x because of geometry, but I do not understand how y can also equal x?

You’re talking about this question from the triangles chapter, right?

You’ve actually hit on a totally legit way to solve the question: If y=2x due to the exterior angle theorem, and y=180-x, then you can substitute and solve for x:

    \begin{align*}2x&=180-x\\3x&=180\\x&=60\end{align*}

Once you know that x=60, you can determine that the triangle is equilateral, so you need a perimeter that’s a multiple of 3.

To answer your question more directly, look at just the top part of the figure:

See how you’ve got a straight line there? That means the angle that’s NOT marked y^\circ (i.e., the angle inside the triangle) has a measure of (180-y)^{\circ}. Since the question tells you that y=180-x, you can substitute:

    \begin{align*}180-(180-x)=180-180+x=x\end{align*}

 

Hello Mike! Could you please tell me, what is the answer of Question 16, Section 4, Test 3? Thank you veery much for your help and time.
Katerina

The answer is -2, and here’s why. They’re asking you to find a value of x for which f(x)+g(x)=0. That can only happen when f(x)=-g(x). So you’re looking for a place on the graph where f curve and the g curve are the exact same distance from the x-axis.

The vertices of the graphs of f(x) and g(x) are (-2,-2) and (-2,2), respectively. In other words, when x=-2, the graphs of f(x) and g(x) are the same distances from the x-axis. In other words, f(-2)=-2, g(-2)=2, and therefore f(-2)+g(-2)=0

 

Are you still doing the daily PWN practice question. It looks like question #110 was the last one I received back on April 2nd.

For now, #110 is the last one, although I might start adding more over the summer. To be clear, the way the Daily PWN works is that everyone who signs up starts at the beginning, so someone who signs up today will get a question a day for 110 days—more than 3 and a half months. As it turns out, the data suggests that most people unsubscribe or just stop answering questions before then.

I do love writing the questions, though. If and when I do add more, assuming you’re still subscribed, you’ll be one of the first to know.

 

Can you explain #15 in Section 3 of test 4? I understand how to do the problem but was wondering if there is a shortcut or trick to seeing the correct answer?

Great question! First, I don’t know if this is really a shortcut but for me this question becomes much easier to tackle when I multiply by 2 to eliminate the fraction. So the quadratic I’m working with after multiplying and rearranging is:

    \begin{align*}x^2-\dfrac{k}{2}x&=2p\\2x^2-kx&=4p\\2x^2-kx-4p&=0\end{align*}

So, for quadratic formula purposes, a=2, b=-k, and c=-4p.

The real shortcut, for this question and ones like it, is to focus on the differences in the answer choices rather than doing the whole quadratic formula process and trying to match a choice.

In this question, for example, you might notice that you only have two choices for the first term: \dfrac{k}{2} or \dfrac{k}{4}. If you start by focusing only on that term, you can eliminate half the choices right away!

The quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. That means the first term is \dfrac{-b}{2a}. In this question, then, the first term is \dfrac{-(-k)}{2(2)}=\dfrac{k}{4}. So we know the answer must be A or B.

From there, we know the bit under the radical must either be \sqrt{k^2+2p} or \sqrt{k^2+32p}.

Since the radical bit equals \sqrt{b^2-4ac}, we can plainly see that \sqrt{k^2+32p} is the way to go. Choice B is the answer.

The full algebra for anyone who came here looking for it:

    \begin{align*}2x^2-kx-4p&=0\\\dfrac{-(-k)\pm\sqrt{k^2-4(2)(-4p)}}{2(2)}&=x\\\dfrac{k\pm\sqrt{k^2+32p}}{4}&=x\\\dfrac{k}{4}\pm\dfrac{\sqrt{k^2+32p}}{4}&=x\end{align*}

Can you please explain Test 3 Section 4 #23?

Yep! The key to getting this one is recognizing that when they tell you that \sin \left(a^{\circ}\right)=\cos \left(b^{\circ}\right), they’re telling you that a+b=90. That’s just a little trig fact you should keep in mind when you’re taking the SAT. It can be derived pretty easily—just picture any right triangle with acute angles measuring a^{\circ} and b^{\circ}:

Of course, since it’s a right triangle, a+b=90. And if you do your SOH-CAH-TOA, you’ll see that \sin \left(a^{\circ}\right)=\cos \left(b^{\circ}\right) and \cos \left(a^{\circ}\right)=\sin \left(b^{\circ}\right).

Anyway, back to the question. Once we recognize that a+b=90, the rest is just substitution.

    \begin{align*}a+b&=90\\(4k-22)+(6k-13)&=90\\10k-35&=90\\10k&=125\\k&=12.5\end{align*}

would you please explain #12 in test 5, section 3.

 

Yeah, this one is trickier than it looks. I think most people get turned around on it because they try to get out of fractional exponent notation too early, and then they get stuck. It’s a LOT easier, in my opinion, to do a little manipulation in exponent form before you jump to radicals. In this case, that means making a 9=3^2 substitution, which will allow you to reduce the fraction.

    \begin{align*}&9^{3/4}\\=&\left(3^2\right)^{3/4}\end{align*}

Remember than when you raise a power to a power, you multiply those powers. So:

    \begin{align*}&9^{3/4}\\=&\left(3^2\right)^{3/4}\\=&3^{6/4}\\=&3^{3/2}\end{align*}

Now you can convert way more comfortably to radical notation. Remember, with fractional exponents, the numerator (top number in the fraction) is the power and the denominator (bottom number in the fraction) is the root.

    \begin{align*}&3^{3/2}\\=&\sqrt{3^3}\\=&\sqrt{3\times 3\times 3}\\=&3\sqrt{3}\end{align*}

Would you please explain the solution to Problem #31 in the calculator-allowed section of Practice Test #6?

This one is all about translating carefully between words and math. You can write two equations. Let’s use f for the original number of friends. If all f friends go on the trip, the cost per person is $20 less than if f-2 friends go on the trip. Cost per Person with f-2 Friends minus Cost per Person with f Friends is $20.

    \begin{align*}\dfrac{800}{f-2}-\dfrac{800}{f}&=20\end{align}

To solve, multiply by f(f-2):

    \begin{align*}f(f-2)\left(\dfrac{800}{f-2}\right)-f(f-2)\left(\dfrac{800}{f}\right)&=20f(f-2)\\800f-800(f-2)&=20f^2-40f\\800f-800f+1600&=20f^2-40f\\0&=20f^2-40f-1600\\0&=f^2-2f-80\\0&=(f-10)(f+8)\end{align}

That tells you that the number of friends originally in the group is either 10 or –8. Since the answer can’t be negative, it must be 10.

BONUS: The other way to go on this one is the way I actually did it the first time: just dividing 800 by a bunch of numbers in a row until I saw a difference of 20:

800/4 = 200
800/5 = 160
800/6 = 133.333
800/7 = 114.286
800/8 = 100
800/9 = 88.889
800/10 = 80

There you have it: 100 and 80 are 20 apart, and that’s the difference between splitting 800 over 10 people vs. over 8 people. That was fast!

Problem #25 in the calculator-allowed section of Practice Test #4, please.

This is a fun question! First, note that the polynomials provided have a little something in common:

    \begin{align*}f(x)&=2x^3+6x^2+4x\\g(x)&=x^2+3x+2\end{align*}

Do you see it? How about now:

    \begin{align*}f(x)&=2x(x^2+3x+2)\\g(x)&=x^2+3x+2\\f(x)&=2x[g(x)]\end{align*}

Takeaway lesson from that: when a question at first looks like it might take a very long time to do (in this case, when it looks at first like you’re going to have to do long division possibly four times) slow down for a minute and think about whether there’s a shortcut built into the problem.

You’re looking for a way to combine f(x) and g(x) so that they’re divisible by 2x+3. Once you see that you can pull a 2x (as in, the same 2x that’s part of 2x+3) out of f(x) and get g(x), you know to use that as your base of operations. Let’s just see what happens if we multiply g(x) by 2x+3:

    \begin{align*}&(2x+3)g(x)\\=&2x[g(x)]+3g(x)\end{align*}

Ready for the last step? All we need to do is substitute because we know f(x)=2x\times g(x):

    \begin{align*}&(2x+3)g(x)\\=&2x[g(x)]+3g(x)\\=&f(x)+3g(x)\end{align*}

That matches choice B, and we know that it’s divisible by 2x+3 because we made it by multiplying by 2x+3!

The problem with which I’m having some difficulty, taken from the chapter on the Passport to Advanced Math section in The Official SAT Study Guide book (page 229), is as follows:

(y^5) – (2y^4) – (cxy) + (6x)

In the polynomial above, “c” is a constant. If the polynomial is divisible by “y – 2,” what is the value of “c”?

Specifically, how does one factor “y – 2” from “- (cxy) + (6x)”?

Thank you for all your help, Mike.

First, an apology: while I aim to answer Q&A questions within 24 hours, I’m late on this one. Sorry!

This is a tricky one, and pretty unlike anything I’ve seen on a real test. For that reason, I wouldn’t worry too much about it. As you note, factoring y-2 out of the first part is easy enough.

    \begin{align*}y^5-2y^4-cxy+6x\\y^4(y-2)-cxy+6x\end{align*}

From there, you have to get creative. The way I think through it is that I know y-2 must be a factor of -cxy+6x. That means I know (y-2)(\text{SOMETHING})=-cxy+6x.

Figuring out the SOMETHING requires recognizing a few things:

  • It must contain x (both terms on the right contain x).
  • It must contain -3 (the second term gets multiplied by -2 and ends up positive 6x).

So what if we say that SOMETHING equals -3x ?

    \begin{align*}(y-2)(-3x)&=-cxy+6x\\-3xy+6x&=-cxy+6x\\-3xy&=-cxy\\3&=c\end{align*}

Does that help?

Can you explain question #24 in Practice Test #8 Section 4?

The key to this one is to eliminate every choice that makes an unjustified (i.e., not directly supported by the question) assumption.

You can eliminate choice A because we don’t know anything about the preferences of the viewers who didn’t vote. Generally speaking, don’t draw conclusions about what could have happened, draw conclusions about what DID happen.

You can eliminate choice B because the question mentions nothing about the ages of voters. Don’t let your own assumptions about who texts and who uses social media cloud your judgment!

You can eliminate choice C because it’s mathematically false (more on this below) but you should also lean towards eliminating it just because it’s using something that didn’t happen as a basis for its conclusion. As before, resist the urge to make conclusions about what could have happened when you only know what DID happen.

Choice D is totally supported by the data: 70% of social media voters preferred Contestant 2, and only 40% of text message voters preferred Contestant 2. Therefore, social media voters were more likely to prefer Contestant 2 than text message voters.

Now, as for why C isn’t mathematically true…let’s plug in! Say there were 100 voters. We know 30% of the votes came in via social media, so that’s 30 votes. The other 70 votes must have come in via text message.

Of the 30 social media votes, Contestant 2 got 70% of them, or 21 votes. Of the 70 text message votes, Contestant 2 got 40% of them, or 28 votes.

So of the 100 votes, Contestant 2 only got 21 + 28 = 49 votes! That’s not enough to win the contest.

I just bought your book and with it in tow, I am learning a lot of helpful, useful ways to do SAT math. Thank you.
Here are my questions, all calculator related:
1. Since I am new to using a calculator, would you recommend the TI83 plus for the SAT?
2. If not, which calculator do you recommend.
3. Is there a book, website, or other media source you would recommend to learn calculator skills with ease and accuracy?
4.Which calculator functions do you recommend mastering before taking the SAT?

I like the TI-83 Plus. That’s the calculator I use personally; all the calculator screenshots you’ll find in my book are from my TI-83 Plus. So if you’re using my book to help you learn to use the calculator, that’s a good thing. The TI-84 series calculators are also great and have basically the same menus, etc.

I don’t have a particular calculator source other than my own book. The calculator will help you lots, but my advice is to search out specific lessons as you find you need them rather than spending too much time learning stuff you won’t need from a thorough book, video series, etc. Basically, when you come across something in my book that I recommend solving with a calculator, then you should make sure you learn how to do that thing. If the examples in my book aren’t enough, search out some videos for that function.

Quick list off the top of my head of what you should master:

  • Graphing functions
  • Zooming and changing window settings to see the important parts of functions
  • Resetting your view to standard zoom
  • Finding intersections of two functions
  • Finding minimums and maximums of curved functions
  • Finding zeros (x-intercepts) of functions
  • Viewing and manipulating the table of function values
  • Converting from decimals to fractions
  • Solving basic trig equations in both radians and degrees

 

 

Could you please explain Test #8, calculator-allowed section, number 28?

Thanks!

You have to evaluate both range and standard deviation here. The former you can calculate. The latter you just evaluate visually; You will never have to actually calculate a standard deviation on the SAT.

Let’s talk about range first. To calculate range you just subtract the smallest number in the set from the biggest number. In this case, that means that r_1=88-56=32 and r_2=112-80=32. The numbers in the second set are obviously higher, but the range in both sets is the same: the highest number is 32 greater than the lowest number. Therefore, you really only need to look at choices A and D—the only choices that say r_1=r_2.

Like range, standard deviation is a measure of variability—the bigger the standard deviation, the more spread out the values in a set are from the mean. Visually, you can look at both dot plots in this question and see that the distributions are quite different. In the first plot, there are a couple outliers, but generally the pulse rates are pretty tightly concentrated around the mean of 72.  In the second plot, the dots are much more spread out—if you picked a dot randomly in the second plot, you’d be just as likely to land on an extreme data point like 80 or 112 as you would on the mean of 96. Therefore, you can conclude that the standard deviations of the two sets are not the same (it will be bigger in the second plot). That makes choice D the only option.