Posts filed under: Q and A

Hi. Would it be possible for you to explain #13 on SAT practice test 5 on calculator inactive (the tea bag problem)? I think the main reason I found it confusing was the wording, and the SAT explanation for the answer was also a little bit wordy.

I’ll try to be as un-wordy as possible! 🙂

I think plugging in numbers helps here. Say the restaurant wants to make 20 cups of tea one night, so n=20. The equation t=n+2 tells us that they’d use t=22 tea bags to make 20 cups of tea.

Now, what if they wanted to make one more cup of tea (in our example, 21 cups instead of 20 cups)? Well, the equation tells us they’d use 23 tea bags to do that:


So they needed 22 bags to make 20 cups, and 23 bags to make 21 cups. That’s one extra bag for one extra cup.

The other way to think about this is that they’re just giving you a linear equation and asking for the rate of change (i.e., the slope). Think of the equation t=n+2 as y=x+2. That’s a line with a slope of 1 (and a y-intercept of 2, but we don’t need that for this question). A slope of 1 means for every change in x, you get an equal change in y. In other words, for every change in cups of tea, you get an equal change in tea bags.

Do either of those explanations help?

Note that questions like this don’t appear on the SAT, but they might appear on a SAT Subject Test.

For the first digit, you have four choices: 2, 3, 4, or 5.

Once you choose that first digit, you have three choices for the second digit.

Once you’ve chosen the first and second digits, you only have two choices for the third digit.

By the final digit, you actually only have one choice, because only one digit is left–you’ve already used up the other three.

What you want to do here is multiply these choices to figure out the total number of possibilities.


It takes a minute, but you can also get this just by listing the numbers. Start by listing the lowest one you can (2345) and go in order until you get to the highest one (5432). The ones that begin with 2 look like this:

2345, 2354, 2435, 2453, 2534, 2543

Then you list the ones that begin with 3:

3245, 3254, 3425, 3452, 3524, 3542

And so on:

4235, 4253, 4325, 4352, 4523, 4532
5234, 5243, 5324, 5342, 5423, 5432

Sure enough, 24 possibilities.

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Hello! Could you please explain how to answer Question #16 from Section #3 from official practice SAT test #5? Thank you in advance!

She only has to take the water safety course ($10) once, so you can really think about her budget as the $280 the question provides minus the cost of the safety course: $280 – 10 = $270.

The question is asking how many whole hours Maria can rent the boat, at $60 per hour, and stay within her budget. In other words, how many times does 60 go into 270?

60\times 4=240 <– Works.

60\times 5=300 <– Too big!

Therefore, Maria can afford to rent the boat for 4 hours—the answer is 4.

When would you use a semicolon vs a colon when dealing with explanations?
For example, in the following sentence, which would be better?

Walker was not only a trailblazer in the medical field; she was also a relentless visionary.
Walker was not only a trailblazer in the medical field: she was also a relentless visionary.

I’ve actually been asked about this one before. I haven’t seen the question in the wild myself, so I’m assuming it’s on Khan Academy?

I doubt something exactly like your example would appear on a real, test-day SAT. If it did, though, I’d go with the semicolon because it’s got the conditions a semicolon requires (independent clauses on both ends). Because the second clause completes the “not only___” thought, though, I can see why people would argue that we’re dealing with an explanation that allows a colon.

For an extremely clueless student who is unsure where to start, how would you recommend preparing for the SAT subject test for Maths 2? Thank you for your time!!

First, get this book. It contains four old tests, which is awesome. There used to only be two!

Take the first test and see where you land. After you’ve taken it, break the questions into the following categories:

  1. Got it right and should always get questions like it right
  2. Got it right but felt rusty doing it
  3. Got it wrong but I know the concept
  4. Got it wrong and have never seen anything like it before

Read the solutions in the book for EVERY question. Feel free to submit questions to me in the Q&A for any you’d like additional help on. Seek out help from your math teacher in school, too. Spend most of your time shoring up categories 2 and 3.

Once you feel like you’ve addressed your biggest problem areas with content review, take the next test and repeat.

A few important things to keep in mind about the Math 2:

  • You can get an 800 on the Subject Test even if you miss a lot of questions! If you only have one or two questions in category 4, you might choose to just ignore them. Be strategic!
  • Unlike the SAT, on Subject Tests there is still a penalty of -0.25 raw score points for each wrong answer. Don’t worry too much about this, and don’t let it discourage you from picking an answer when you’re not 100% sure. For basically every student I’ve ever tutored, this penalty has been a nonissue. For more information, read this old post of mine from back when the regular SAT had the guessing penalty.

Can you please explain Test 7 section 4 question #22 ?

Sure. To find the median value of the 7 states you’re given, first place the numbers in order:

19.5%, 21.9%, 25.9%, 27.9%, 30.1%, 36.4%, 35.5%

Because there are 7 items in the list (an odd number), the median is just the middle number once they’re listed in order. In this case, that’s 27.9%.

The median for all 50 states, the question tells us, is 26.95%.

The question asks us for the difference between the median for the 7 states we’re given (we just calculated that to be 27.9%) and the median for all 50 states (the question gives this to us, 26.95%).

27.9% – 26.95% = 0.95%

Hi Mike…SAT 7, Section 4, Q6: I now see the shortcut here (that both sides of the equation are perfect squares,) but if I did expand and FOIL the left side, wouldn’t I still get the correct “a” values even though it takes longer? I can’t get it to work !! Can you please show the alternate path math steps? Or is recognizing the perfect squares the ONLY way to solve this one ? Thanks!

FOIL should work, sure. Plug in –3 for x and go to town:


That tells you that a could equal –1 or 3, and –1 is an answer choice.

You can also get this one by backsolving, though. That would actually be my first choice. You’re given x = –3. Try the answer choices for a until one works! Conveniently, I usually start with C when I backsolve, so I finish this one very quickly:

Choice C:



36=36 <– checks out

If Janice reads x pages every 30 minutes, then she reads 2x pages per hour because there are 2 30-minute periods every hour. Because Janice reads for 4 hours, she reads 4(2x) = 8x pages.

Likewise, if Kim reads y pages every 15 minutes, then she reads 4y pages per hour. Because Kim reads for 5 hours, she reads 5(4y) = 20y pages.

So the answer probably looks something like 8x + 20y.

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Dear Mike,
Do you happen to know of TI-84 programs! If so, which ones do you recommend to have programmed into our TI-84s?

I think everyone should have a good Quadratic Formula program, but otherwise I think programs aren’t really worth the trouble for the SAT. Programs are most useful when 1) you have to do the same process over and over again, and 2) it’s time consuming or complicated.

The SAT asks a lot of different kinds of questions; it also tests single topics in many different ways. There just aren’t a lot of complex processes you need to repeatedly perform that the calculator won’t already do without a program. (I think, for example, that you should be good at graphing functions and finding intersections, but your calculator does that without a program.)

If f and g are functions, where f(x) = x^3 -10x^2 +27x – 18 and g(x) = x^3 – x^2 – 6x, which of the following gives a relationship between f and g?

A) g(x) = 3 f(x)
B) g(x) = f(x) – 3
C) g(x) = f(x) + 3
D) g(x) = f(x – 3)
E) g(x) = f(x + 3)

Can you solve w/o graphing?


Note that the first term in both functions is x^3, so the relationship isn’t multiplication. Eliminate choice A.

Note also that it’s obvious that g(x) isn’t just 3 bigger or 3 smaller than f(x), so you can also eliminate choices B and C without any real work.

From there, it gets a little tricky. Focus on the constant terms, though. Note that f(x) has a –18 in it that goes away in g(x). Note also that when we plug x-3 or x+3 in for x, we’ll have a bunch of binomials to expand, but the only way that –18 goes away is if those binomial expansions spit out a positive 18. Without doing all the math, you should be confident that choice E is the one that will do that. To test it quickly, though, set x=0 (since this does need to be true when x=0).

    \begin{align*}g(0)&=f(0+3)\\g(0)&=f(3)\\0^3 - 0^2 - 6(0)&=3^3 -10(3)^2 +27(3) - 18\\0&=27-90+81-18\\0&=0\end{align*}

Yep! That worked nicely.

Here are a couple questions from the old official SAT Subject Test Math I practice exam:

The function f is defined by f(x) is x^4 – 4x^2 + x + 1 for -5 ≤ x ≤ 5. In which of the following intervals does the minimum value of f occur?
A) -5 ≤ x ≤ -3
B) -3 ≤ x ≤ -1
C) -1 ≤ x ≤ 1
D) 1 ≤ x ≤ 3
E) 3 ≤ x ≤ 5

Can you solve w/o graphing?

Yes, but just before beginning I think it’s important to stress HOW USEFUL a graphing calculator is for the math Subject Tests. Anyone who’s prepping for those that doesn’t own one and can’t borrow one from school should find another way to get their hands on one they know how to use for test day.

To get this without graphing, you’re going to want to plug in values Just the integers should work. Note that x^4 will always be 0 or positive and -4x^2 will always be 0 or negative, but x will be both positive and negative, you should start by plugging in negative numbers.







From there, you probably see that the minimum is around –1. So plug in a couple more values to see whether you want to choose B or C.



Yeah…gonna want to go with B.

Again, the graph is SO helpful here:

With that, you immediately see that the minimum is between –1 and –2.

This is a counting question, which used to appear on the old SAT (pre-2016) but don’t appear on the current SAT. I just wanted to point that out before getting into it because I didn’t want to scare anyone. I assume you’re prepping for a Subject Test (or maybe even GRE or something like that).

My advice is to draw the points, draw the segment, and count AS YOU DRAW (not after you draw). Start with a point, draw every segment that can be drawn from that point to another point. When you’ve drawn every segment you can from that one point, go to the next one. Like so:



You know you’re done when you’ve got a nice enclosed star design like above. You drew 5 + 4 + 3 + 2 + 1 = 15 segments.

From where I sit, I really think that if you care about always getting questions like this right on whatever standardized test you’re prepping for, you should practice doing it by drawing and counting like I did above.

However, you can also solve this with combinations. Each segment connects two points. How many combinations of two items can you choose from a list of six items? If you know nCr notation, 6C2 = 15. If you don’t, well, you’ve always got factorials…


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Could you please explain SAT Past paper 3, section 4, question 12? I don’t really understand why the double root is considered a ‘distinct’ zero.

Thank you for your time!

Sure. Would it make you feel better if “distinct” were replaced with “unique”? Those terms are interchangeable here.

When the question says the function has “five distinct zeros,” that really just means there are five values that 1) don’t equal each other, and 2) make the function equal zero.

In other words, f(a) = 0, f(b) = 0, f(c) = 0, f(d) = 0, and f(e) = 0 for unique values abcd, and e.

In OTHER other words, the function f touches the x-axis five times. Sidestepping all of the above, all you really need to do is count the number of times each graph touches the x-axis.

Choices A and B touch the x-axis four times, and choice C touches the x-axis six times. Only choice D touches it five times.

Hi Mike, Can you please explain Question 11, Test 6, section 3 ? I know the parabola opens downward, but I’m confused after that. Thanks.

You’ve already got the first part—when you add a negative to the front of a parabola you’ll flip it vertically. Since this parabola begins facing up, the negative flips it so it faces down.

From there, I think it’s helpful to think about this in terms of how functions shift. For some function f(x) that’s graphed on the xy-plane:

  • f(x)+1 ⇒ (graph moves UP one)
  • f(x)–1 ⇒ (graph moves DOWN one)
  • f(x+1) ⇒ (graph moves LEFT one)
  • f(x–1) ⇒ (graph moves RIGHT one)

That applies here. You’re basically taking the ax^2 function and applying a few shifts (and a flip). Check it out–if you start by assuming a function p(x)=ax^2, you can build the function in this question by shifting that original function:


-p(x-b)+c flips the function p(x) upside-down, then shifts it b units to the right and c units up.

If you hate everything I’ve just said, I have good news! You can also just memorize the vertex form of a parabola (which we basically just derived).

If you have a parabola in the form f(x)=a(x-h)^2+k, then you know it has its vertex at (h, k) and that the sign of a tells you whether the parabola opens up or down. This question basically gives you the vertex form, only it uses b and c instead of h and k. Recognize that and you know right away that the parabola’s vertex is at (b, c).

Hi Mike, I’m asking about SAT 8, Section 3, Number 7. Is it always best to immediately plug in answer choices on questions like this? For the algebraic practice, I rewrote the equation as a quadratic and solved for (x=5) and (x= -1). Then sub’d each value back into the given equation to find that only (x=5) worked. OK…but what a time-killer at #7 out of 20. Any other solution path to consider? Thanks!

For questions like this (extraneous solution quadratics) the fastest solution IS to backsolve from the answer choices. If you look at the choices right away, you see that the only possible answers are 0, –1, and 5. Check those three out and you’re done in no time. Solve algebraically and you have to check the answers you get anyway to make sure they’re not extraneous!

To drive it home, here’s all the work you need to do if you backsolve:




Obviously only the last one of those worked out correctly, so the solution set contains only 5.

If you solve algebraically, as you point out, you eventually end up at –1 and 5 as possible answers (work below for anyone still reading), but because you had to square everything to get there, you need to check for extraneous solutions by doing the step you could have started with if you just backsolved from the beginning!