The article will deep dive into the 10 most important formulae, concepts and strategies that you need to know to ace the SAT Math section.

1. Highlight the Question Stem

The most common SAT Math Test mistakes students make are “silly” mistakes It doesn’t matter if you are a math whiz, or totally hate math, one thing’s guaranteed: these seemingly small mistakes add up in a big, big, way and are an SAT score killer.

A lot of students brush off silly mistakes because they know how to solve a problem correctly. Sure, you “could have” gotten the problem right” if you didn’t make that silly mistake,” but then again, a panda bear could have invented the first car if he went back in time, turned himself into a human, and beat Henry Ford to the chase. Could haves are stupid.

The primary source of these kinds of errors is students confusing questions with prompts. On the SAT, the question and the prompt are two entirely different things.

PROMPT: The entire set of words, numbers, and phrases that make up a math problem.
QUESTION: The precise question you have to answer. It usually comes at the end of the prompt.

In the example below, the prompt is in orange font and the question is in green font.

If the radius of a certain circle is three times the radius of another circle, and if the radius of the larger circle is equal to the area of a square with a side length as long in inches as the square root of 169, then what is the area of the smaller circle?

The question is almost much simpler and shorter than the prompt. Most students are so focused on the prompt that they forget to answer the question. So, always underline the question and ensure you answered what the question is asking you.

2. Verify Each Step

Most students check their work only once they finish all the questions on the SAT. However, no student in the history of the test has ever found a mistake after they have completely finished taking the test.

Okay, that is a bit too dramatic, but here’s the point: students rarely find errors after they have completed the SAT Math Test because, at that point, they are typically only “confirming” answers rather than really scrutinizing them for mistakes.

Additionally, you generally have no more than five minutes left at the end, and this causes students to rush even more while they are checking their work.

Therefore, you need to check your work as you work through each problem. As you are solving a problem, verify each step. For example, verify that 3 x 12 actually does equal 36 while you are solving a problem. Verify that the formula for the area of a triangle really is ½ bh.

3. Use a Calculator for Arithmetic

Your calculator should be used as a tool, not as a crutch. Always use your calculator for ARITHMETIC (the stuff you’d so using your fingers to count), but never use your calculator for MATH.

Calculators are godsends when it comes to doing the arithmetic. They make you faster and more accurate. Never do arithmetic in your head. Sure, 2+3 is fine, but if the calculation takes you longer than half a second, you need your calculator.

Don’t waste your time or your mental energy doing arithmetic in your head. Yes, yes, we’re all proud that you can add and multiply so quickly in your own brain - now quit it.

While we encourage you to do arithmetic with your calculator, do not EVER, EVER do MATH on your calculator. Here are some things we consider math:

• Graphing two equations on the line graphing tool to see where their intersection point.
• Trying solving a multi-variable equation.
• Figuring out what equations to use in the first place.

Your calculator will never be able to figure out how to solve a problem faster than you will, especially when you’re well trained on the SAT. If you rely on your calculator, you’ll never learn the right tricks and strategies necessary to truly understand this section.

4. Know Your Formulas

You need to know all the facts, formulae, and figures tested on the SAT Math Test.

Imagine the SAT Math Test as a house and math facts as your building materials. It doesn’t matter if you’re the best architect in the universe; without the right materials, you’re not going to be able to build much of a home.

Hopefully, you already know most of the material, and we’ve provided you with a formula sheet in the next chapter. However, the formula sheet does not cover all the material since the SAT requires you know more than 100 math facts and formulas.

A better way to test your knowledge is to scan through mistakes on your past mock tests and see if there are any math facts that trouble you. You can then document them in the notes pages in this book.

This process can be exhausting at times, but it is well worth it.

5. Plug in Numbers (PIN)

Many SAT Math Test problems don’t deal in concrete numbers - they deal in variables. That makes things unnecessarily difficult, and who likes when things are difficult?

Math is about numbers. When you do math without numbers, it’s hard. When you do math with numbers, it’s easy. With that in mind, our goal is to use real numbers whenever we possibly can and to make them up if they don’t exist.

Always follow this Golden Rule: If you don’t need to know the value of anything in any given math problem, make the value up. It doesn’t matter whether it’s the value of x, the side length of a square, the number of students in a class, whatever. If you don’t need to know it specifically, then make it up and plug it in.

Question: When the positive integer x is divided by 7, the remainder is 4. Which of the following expressions will yield a remainder of 3 when it is divided by 7?
A. X + 2 Let x = 4. 4+2=6. 6 divided by 7 has a remainder of 6. Doesn’t work
B. X + 3 Let x = 4. 4+3=7. 7 divided by 7 has a remainder of 0. Doesn’t work
C. X + 4 Let x = 4. 4+4=8. 8 divided by 7 has a remainder of 1. Doesn’t work
D. X + 6 Let x = 4. 4+6=10. 10 divided by 7 has a remainder of 3. Works

6. Ultimate Percent Change Formula

For some reason, the SAT is absolutely obsessed with percentage problems. If you have any trouble with percentages, the following formula will help eradicate that problem once and for all.

(Original Number)(1.00 +- Percent Change) = Final Number

Question: John goes to a store and sees a jacket for 30% off. He decides to buy it. When he gets to the counter, the clerk takes another 20% off the already discounted price. The clerk rings the jacket up, and there is a 9% tax rate. John ends up paying $390. What was the original price (rounded to the nearest dollar) of the jacket before any discounts or taxes?
A. USD 639
B. USD 688
C. USD 701
D. USD 709

Solution: The first thing you need to realize is that there isn’t one operation that happens to this jacket’s price - there are three. We, therefore, need to use the formula three times - you cannot just add % changes together and do them in one step.

Let’s start backwards. Since we know the final price of the jacket is $390, our first task is to find the price pre tax.

Tax transaction: (Purchase Price) (1.00 + .09) = $390
So before taxes, the Purchase Price = $357.80

Now let’s use that number for our next step:
‍

Second discount: (Original Price) (1.00 - 0.20) = $357.80
Original Price = $447.25
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Now we have one last transaction:
First discount: (Truly original price) (1.00 - 0.30) = $447.25
Truly original price = $ 638.93

The correct answer is A.

7. Ratios are Different to Fractions

Ratio-based problems make up a huge share of all SAT Math test problems, and they can be tricky. Here’s the first way that the SAT will try to force you to slip up.

They’ll ask a question like this: If there are half as many chickens are there are pigs on a farm, then what fraction of the animals on the farm are chickens?

One of the answer choices you’re offered will inevitably be this: ½. That might seem to make sense. After all, the ratio of chickens to pigs is 1:2, so aren’t half of the animals just chickens?

NO, NO, NO! - The actual answer is 1/3.

THINK ABOUT IT: If there is one chicken, that means there must be two pigs. So, in total, 1 chicken + 2 pigs = 3 animals. Chickens are 1/3 of the animals on the farm.

REMEMBER: Ratios are NOT quite the same thing as fractions. They serve a very different role. Just knowing this distinction will help you.

From now on, every time you think about, see, or write ratios without imagining an “implied x.”

If you hear that the ratio of girls to boys in a school is 3 to 5, you should see this in your head as 3x : 5x. At this point, you have no idea how many boys or girls there actually are.

There could be three girls or 300,000,000,000.

Once you understand that ratios “MOVE IN GROUPS,” this entire concept will become much easier for you.

In our above example, 3x + 5x = 8x, we know that boys and girls will come in “groups” of 8. You can have 8 students, or 32, or 48, or 64 – but never have just 15 students.

8. Number of Solutions

Often you will be asked to find the number of solutions to simultaneous linear equations. Number of solutions is just a fancy way of asking the number of times both equations will intersect each other.

There are three possible outcomes:

1. No Solutions - When the lines are distinct but parallel. They have the same slope but different y-intercepts.
2. One Solution - When the lines are distinct and not parallel. They have different slopes and y-intercepts.
3. Infinite Solutions - When the two lines are infact the same line. They have the same slopes and y-intercepts.

9. Disguised Quadratics

The SAT will often attempt to disguise quadratic equations by putting them in forms that do not quite look like the traditional form of \(ax^2 + bx + c = 0\) .

Here is a very common “disguised” form for a quadratic: \(3w^2 = 6w\)

This is certainly a quadratic equation. However, it is very tempting to try to solve this equation without thinking of it as a quadratic. This classic mistake looks like this:

In solving this equation without factoring it like a quadratic, we have missed one of the solutions! Let us now solve it by factoring it as a quadratic equation:

\(3w^2 = 6w\)
\(3w^2-6w=0\) 
\(w(3w - 6) = 0\)
\(w = 0 \ or\  3w - 6 = 0\)
\(w = 0\ or\  w = 2\)

In recognizing that \(3w^2 = 6w\) is a disguised quadratic, we have found both solutions instead of accidentally missing one (in this case, the solution w = 0).

10. Growth and Decay

Increasingly, a lot of SAT Math problems deal with the exponential growth or decline of populations of humans/bacteria/etc. They are modeled on the following equations:

To solve these questions, the trick is to plug in numbers from the equation or the graphs in the answer choices to see what gives you the correct answer.

Question: The amount of mice during a plague doubled every 10 days. If their original number is A, which of the following equation gives their final number, N, that are after t days.
A. \(N = A\times(2)^t\) Substituting t = 10, gives us \(N = A\times(2)^{10}\) . We need N = 2A
B. \(N = A \times (2)^{10t}\) Substituting t = 10, gives us \(N = A \times (2)^{100}\) . Eliminate again
C. \(N = A \times (2)^{t/10}\) Substituting t = 10, gives us N = A x 2. This is what we need. Correct
D. \(N = A \times (2)^{2t}\) Substituting t = 10, gives us \(N = A \times (2)^{20}\)

11. Absolute Value with Inequalities

The Absolute Value of a number is its distance from zero (on the number line). Don’t freak out when they see something like this:

Solve for X in | X + 2| < 5

The best way to understand how to solve this problem is to create a number line. For a simple inequality such as | x | < 5, this would say “x must be less than 5 units from zero on the number line.”

Therefore, here -5 < x < 5. Easy.

Now, let’s try to do it |x + 2| First, we need to create a number line for “x + 2.” We can see that this expression as a whole must be between - 5: and 5:

In other words, x + 2 must be less than 5 units away from zero on the number line. We cannot stop there - we must graph for x alone. How does the “ + 2 ” change our graph? It forces us to shift the entire graph by 2, because the absolute value expression will be equal to zero when x = - 2. Thus, the graph for x alone will look like this:

Notice that x is at the center, and it’s still 5 points away from the endpoints (-7 & 3). Based on the graph, the following is the solution for the inequality: -7 < x < 3

12. Compound Functions

A compound function (a function inside a function) is identical to any other function question. All you have to do is to apply the function operation twice.

For example, if \(f(x) = x^3 + g(x) \ and \ g(x) = 4x - 3, what\  is\ f(g(3))?\)

The expression f(g(3)), pronounced “f of g of 3”, looks ugly, but the key to solving compound function problems is to WORK FROM THE INSIDE OUT. In this case, start with g(3).

g(3) = 4(3) - 3 = 12 - 3 = 9

Use the result from the inner function g as the new input variable for the outer function f:

f(g(3)) = f(9) = (9)3 + √9 = 729 + 3 = 732

Question: What is f(g(x−2)) when f(x) = x2 − 6 and g(x) = 3x + 4?
A. 3x − 2
B. \(3x^2 + 12x-6\)
C. \(9x^2 + 24x + 10\)
D. \(9x^2 -12x-2\)

Because g(x) is nested the deepest, we must start with g(x-2),
g(x) = 3x + 4
g(x−2) = 3(x−2) + 4            Replace the x in the g(x) function with x-2
g(x−2) = 3x − 2

Now, putting this in f(x)
\(f(x) = x^2 - 6\)
\(f(g(x-2)) = f(3x - 2) = (3x-2)^2 - 6\)
\(= 9x^2 - 12x - 2\)

The correct answer is D.

13.Two Special Triangles

Certain right triangles appear over and over on the SAT, so instead of using the Pythagorean Theorem to solve for the lengths of the sides of these common right triangles, you should be able to calculate them from memory.

1. THE 45-45-90 TRIANGLE. By 45-45-90, we mean the three angles of a right triangle are 45°, 45°,and 90°. The lengths of the legs of every 45-45-90 triangle have a specific ratio.

2. THE 30-60-90 TRIANGLE. The most popular of all right triangles is the 30-60-90 triangle.

14.Radians and Degrees

It's important that you know how to convert between RADIANS and DEGREES. Use this formula:

For example, if we need to convert 50 degrees to radians, we need to multiply the degree measure by \(\frac{\pi}{180}\) , i.e. \(50^{\circ} = 50 \times\frac{\pi}{180} rad. = \frac{5\pi}{180}rad.\)

Some important conversions are:

15.Equation of a Circle

The standard equation for a circle is in the form \(\textbf{(x-h)}^\textbf{2} \textbf+ \textbf{(y-k)}^\textbf{2} = \textbf{r}^\textbf{2}\) , where the circle with radius r has its center at (h, k).

Question: Find the centre and radius of the circle \(x^2 + y^2 + 8x + 7 = 0\).
A. Centre is (-4,0) and radius is 9
B. Centre is (0,-4) and radius is 9
C. Centre is (-4,0) and radius is 3
D. Centre is (0,-4) and radius is 3

Solution: There is no Y term, so we got to complete the squares only for the X terms \(x^2 + 8x +(\ \ \ )+ y^2 = -7\)

We need to fill in the (     ) to complete the square.

The middle term is 8 and therefore the last variable should be 16.

Adding 16 on both sides: \(x^2 + 8x + 16 + y^2 = -7 + 16\)
\((x + 4)^2 + y^2 = 9\)
We get (h,k) -> (-4,0) and r = 3.

The answer is C.

16.Unit Circles

We know that in a right triangle, an angle cannot be larger than 90° (as the sum of all angles is 180°.

So, whenever you’ve been given an angle greater than 90° in a trigonometry problem, draw it in the coordinate plane.

‍Consider the following angles:

We can conclude that        

• Angle 140° is the same as angle 40°, except that angle 140° is in the second quadrant        
• Angles 220° and 320° are the same as angle 400, except that angles 220° and 320° are drawn in the third and fourth quadrant respectively.

This means, in terms of the trigonometric functions, the magnitudes of 400 and all other drawn angles are the same, but the sign of the trigonometric functions are different, depending on the quadrant. The sings of the trigonometric functions, as per the quadrant are given below:

Now, when θ = 30°, sinθ is 0.5.
Now, what value of θ, will sinθ = -0.5?

In the diagram on the previous page, we can see sinθ is negative in the third and fourth quadrant, so there are two possible values of θ i.e.
θ = 180° + 30°       &    θ = 360° - 30°
θ = 210°                 &     θ = 330°

17.Conditional Probability

Conditional probability is the chance of an event happening given that another event or condition has already happened. It’s still simple probability - desired outcomes over total outcomes - but figuring out the correct number of desired vs. total outcomes can be a little tricky.

‍Example: There are 100 people working on a performance: 52 dancers, 12 stage technicians, and 36 musicians. Among the dancers, 14 are ballet dancers, 20 are jazz dancers, and 18 are modern dancers. What is the probability of selecting a ballet dancer from those working on the performance, given that the person selected is a dancer?

It might seem like this is asking you the probability of selecting a ballet dancer (of which there are 14) from everyone working on the performance (of which there are 100).

But actually, it’s asking you the probability of selecting a ballet dancer from the dancers, because we are accepting as a given (as a condition) that the person we are randomly selecting is a dancer. We can tell this from the phrase “given that the person selected is a dancer.”

Thus, we must calculate the probability of selecting a ballerina given that the person we select will be from among the 52 dancers. So, the answer is \(\frac{14}{52}\)

18.Standard Deviation

STANDARD DEVIATION (SD) is used to describe the spread, or variation, of the data in a set.It indicates how far from the average (mean) the data points typically fall‍

• A small SD indicates that a set is clustered closely around the average (mean) value
• A large SD indicates that the set is spread out widely

Example: Consider the sets {5, 5, 5, 5}, {2, 4, 6, 8}, and {0, 0, 10, 10}. These sets all have the same mean value of 5.

However, let’s look at their respective SDs:
1. The first set has an SD of zero (no spread at all),
2. The second set has a moderate SD,
3. The third set has a large SD.

‍You do not need to know how to calculate an exact SD. If you just pay attention to what the average spread is doing, you’ll be able to answer all SAT standard deviation problems. Just remember that the more spread out the numbers, the larger the SD.

If you see a problem on the test that focuses on changes in the SD, ask yourself whether the changes move the data closer to the mean, farther from the mean, or neither. If you see a problem requiring comparisons, ask yourself which set is more spread out from its mean.

Question: Consider the following sets: L = {3, 4, 5, 5, 6, 7}; M = {2, 2, 2, 8, 8, 8}; N = {15, 15, 15, 15, 15, 15}. Rank those three sets from the least standard deviation to the greatest standard deviation.
A. L, M, N
B. M, L, N
C. M, N, L
D. N, L, M

Solution: When all the members of a set are identical, the standard deviation is zero, which is the smallest possible standard deviation. So, automatically, N must have the lowest. Right away, we can eliminate A, B, and C. Only D. The correct answer is D.

19.Causation vs Correlation

The difference between correlation and causation is where the SAT stumps a lot of students. For example, just because students who do attempt more mock tests get better scores on the SAT doesn’t mean that completing more mock tests cause an improvement in SAT scores.

Rather, completing mock tests is associated with an improvement in SAT scores.

Perhaps students who attempt more mock tests have more discipline, or they have more demanding parents who make them study harder. But, due to the way the experiment was designed, we can’t tell what the underlying factor is.

‍Question: Researchers conducted an experiment to determine whether eating junk packaged food increases body weight. They randomly selected 500 people who eat processed packaged food at least once a week, and 500 people who do not eat processed packaged food at all.

After tracking the people’s weight for a year, the researchers found that the people who eat processed packaged food at least once a week had experienced weight gain significantly higher than people who do not eat processed packaged food at all. Based on the design and results of the study, which of the following is an appropriate conclusion?
A. Eating processed packaged food at least once a week is likely to increase body weight.
B. Eating processed packaged food three times a week increases body weight more than eating processed packaged food just once a week.
C. Any person who starts eating processed packaged food at least once a week will increase his or her body weight.
D. There is a positive association between eating processed packaged food and increased weight gain.

Solution: This question deals with a classic case of correlation vs. causation. Just because people who eat processed packaged food had a higher body weight doesn’t mean that eating processed packaged food causes an increase in body weight. It’s just associated with an increase in body weight.

A. This is wrong because it implies causation.
B. This is wrong because it not only implies causation but also implies that the frequency of eating processed packaged food matters, something that wasn’t tracked in the experiment.
C. This is wrong because it suggests a completely certain outcome. Even if eating processed packaged food DID increase body weight, not every single person who starts eating processed packaged food will increase his body weight.
D. This is the answer because there is a positive association between eating processed  packaged food and an increase in body weight.

20.Word Problems: Steal Answers

The difficulty many students face when dealing with long word problems is that you get lost in the details. If a problem throws six different variables, two equations, and a bunch of other details at you, it’s easy to get confused.

For most word problems, the correct answer is sitting right in front of you. Why would you ever go through the horrible work of actually solving a math problem or working harder than you need to when you could just cherry-pick the right answer instead?

‍Question: Andrea manages a company that currently has 116 customers, which is 8 more than twice the number of customers the company had 1 year ago. How many customers did the company have 1 year ago?
A. 50
B. 54
C. 61
D. 66

Solution: The best way to approach this problem is to take all the answer choices, double them, and add 8. See which one works. For example, Option A would be 50 x 2 = 100. Add 8 to it that would be 108. Therefore, A is not the answer. Option B would be 54 x 2 = 108. + 8 is 116. There we have it. B is the answer (there can’t be two right answers to a math problem). That was easy!‍‍