# What Math Concepts Are on the SAT?

The SAT has been around for nearly a century. In that time, it has undergone numerous revisions. The abandonment of the paper and pencil format is the most recent and possibly most radical of the SAT changes. The move to a digital platform has allowed for many beneficial changes, such as computer adaptivity in scoring, unlimited calculator use in the Math section, an overhaul of the Reading and Writing section, and a shortened test length. Important to note is that there is only one change to the SAT math concepts that are tested.

In this article, we will introduce you to the Math section of the digital SAT. We’ll begin by looking at the structure and format of the section. Then, we’ll discuss the major and minor math topics you can expect to see. In addition, we’ll focus on some of the SAT math concepts that you are almost certain to encounter.

## Here are the topics we’ll cover:

Before we concentrate on SAT math, let’s get an overview of the digital exam as a whole.

## Overview of the Digital SAT

The digital SAT, often referred to as dSAT, is broken down into 2 sections:

• Reading and Writing: 2 modules, each with 27 questions. You’ll have 32 minutes to complete each module. Reading passages are shorter than on the old SAT, and each passage has only one question. In addition to passage questions, you will encounter vocabulary and grammar/punctuation questions.
• Math: 2 modules, each with 22 questions. You’ll have 35 minutes to complete each module. About 75% of the questions are multiple-choice questions with 4 answer choices. The remaining questions are student-produced response (SPR) questions, in which you type in your calculated answer. You’ll have access to an online Desmos graphing calculator, or you may use your own.

Next, let’s discuss the adaptive nature of the digital SAT.

Because of its digital platform, the dSAT is able to personalize the exam to each test-taker. Here’s how “section adaptivity” works, using the Math section as an example.

In Math module 1, you’ll answer questions of varying difficulty levels: 6 easy, 8 medium, 6 hard, and 2 unscored. After you finish module 1, the computer scores your responses. If you have scored 60% or better, the questions in module 2 will be more challenging. Accordingly, correct answers will earn you more points. However, if your score is less than 60%, you will encounter easier questions in module 2. As a result, you will not be able to earn as high a score.

If you perform well on Math module 1, your possible Math section score range is 450 to 800. If you score less than 60% on Math module 1, your possible score range for the Math section is 200 to 600. It is well worth your while to ensure that you earn a great score on the first module!

Section adaptivity works in a similar manner for the dSAT Reading and Writing Section.

KEY FACT:

Section adaptivity allows each student to answer questions geared to his or her ability.

Next, let’s discuss how the dSAT Math section differs from previous SAT Math sections.

### Changes to the Math Section

You may have classmates who took the SAT in 2023 or earlier. If so, don’t depend on them to give you current information about the SAT. The dSAT, effective Spring 2024, has quite a few differences. Let’s look at the key changes to the Math section:

• Figures are now drawn to scale.
• Student-produced response questions can now yield a negative answer.
• There is no longer a “No Calculator” section. You have access to an online graphing calculator for both modules. If you wish, you may use your own calculator during the exam.
• Imaginary and complex numbers are no longer tested.

Now that we’re familiar with what’s new about the digital SAT, let’s discuss the math topics you’ll face.

## The Topics Tested on the SAT Math Section

The College Board (maker of the SAT) has provided a list of 4 overarching math categories with relevant topics. We have bulked up their list to provide more detail about what’s tested in each category. Note that there are hundreds of math topics tested, so we can’t mention every one of them. So, use this list as a general overview rather than a comprehensive list.

• Algebra (35% of Math questions, or 13-15 questions)
• Linear equations in 1 or 2 variables
• Examples: y = 3x + 6, 3x + 4y = 81
• Systems of 2 linear equations in 2 variables
• Example: If 3x + 5y = 7 and 2x – y = 3, solve for x.
• Linear inequalities in 1 or 2 variables
• Examples: x < 5, y > 2x + 5
• Word problems solved with linear equations
• Advanced Math (35% of Math questions, or 13-15 questions)
• Functions and function notation
• Quadratic and higher order polynomials
• Factoring and graphing
• Parabolas
• Vertex, line of symmetry, intercepts, min/max values
• Exponents
• Absolute value
• Square root equations
• Rewrite algebraic expressions as equivalent expressions
• Exponential growth versus linear growth
• Systems of equations in 2 variables
• Problem-Solving and Data Analysis (15% of Math questions, or 5-7 questions)
• Ratios, proportional relationships, unit conversion
• Rates
• Percentages
• Statistics: data distributions
• Mean, median, mode
• Charts and graphs
• Scatterplots
• Probability and conditional probability
• Inference from samples and evaluating claims
• Margin of error
• Observational studies and experiments
• Geometry and Trigonometry (15% of Math questions, or 5-7 questions)
• Areas and volumes
• Lines, angles, and triangles
• Pythagorean theorem
• Right triangles and trigonometry
• Circles
• Area and circumference
• Algebraic expression of circles
• Example: x^2 + y^2 = 9

Now that we have a good idea of what we will encounter on the digital SAT, let’s focus on some high-value concepts that are consistently tested. Mastery of these will give you a head start on getting a great score on the exam!

## Concept 1: “Rental” Problem (Linear Word Problem)

Recall that the equation of a line can be expressed as y = mx + b, where m is the slope of the line and b is the y-intercept of the line. This equation can be used for a variety of word problems, including what are sometimes referred to as “rental” problems. Some examples might be renting a car, paying a toll, or even hiring a tutor!

The key to these rental problems is that you pay a flat fee plus a variable fee. Thus, your total payment is based on the flat fee plus a payment based on the number of hours, days, or miles that the item was used.

For example, you rent a pressure washer for $27 plus$6 for each additional hour after the first 3 hours. If you use the pressure washer for 5 hours, you will pay the flat fee of $27 plus the variable fee of$6 for the additional 2 hours. Thus, your total rental cost will be $27 +$12 = $39. Let’s take a look at an example problem. ### Example 1: Word Problem (Linear Model) June rents a horse trailer for a flat rate of$50 plus $19 for each day after the second day. If she rents the trailer for 7 days, how much will she pay for the rental? #### Solution: The cost of the rental includes the flat fee and the variable fee. We can express this as: Total cost = flat fee + variable fee If we let d = the number of days she rents the trailer, we see that (d – 2) is the number of days she’ll have to pay the variable rate of$19. So, for example, if she rents the trailer for a total of d = 5 days, then she pays the variable rate for (d – 2) = (5 – 2) = 3 days. Thus, the equation for renting the horse trailer for any number of days is:

Total cost = 50 + (d – 2)(19)

So, if June rents the trailer for d = 7 days, she will pay:

Total cost = 50 + (7 – 2)(19)

Total cost = 50 + (5)(19)

Total cost = 50 + 95 = $145 June will pay$145 for the 7-day rental.

This “rental” technique can be used for any type of problem where you pay a flat fee plus a variable fee. For a car rental, you might pay $20 per day plus$0.30 for each additional mile after the first 100 miles. For a song on an internet music service, you might earn $0.80 plus$0.003 for each download after the first 100.

In each of these cases, there is a “flat fee” part and a “variable fee” part. The words might change, but the math stays the same.

KEY FACT:

A classic word problem is the “rental” problem, in which you pay both a flat fee and a variable fee.

Next, let’s discuss solving a system of linear equations.

## Concept 2: Solving a System of Linear Equations

The phrase “system of linear equations” sounds daunting, but with practice, you’ll be able to solve them quickly and confidently. On the SAT, you can expect to see several questions asking you to solve a system of linear equations. So, this is a concept that you must include in your wheelhouse.

On the SAT, you might simply be given a set of 2 linear equations. Some questions might give you a word problem that requires you to create the 2 equations. In either case, you need to be familiar with both setting up and solving a linear system. Let’s take a look at an example in which we both set up and solve the system.

### Example 2: System of Linear Equations

Nitish bought avocados and bananas at the market. Avocados cost $1.45 each, and bananas cost$0.90 each. He bought 23 pieces of fruit and spent a total of $25.10. How many avocados did he buy? #### Solution: First, let’s create two equations. Then we will solve for the values of the variables. Let’s let a = the number of avocados Nitish bought and b = the number of bananas that he bought. Since he bought 23 pieces of fruit, our first equation is: a + b = 23 Our second equation represents the amount of money he spent. Since he bought a avocados for$1.45 each, he spent 1.45a on avocados. Similarly, he bought b bananas for $0.90 each, so he spent 0.90b on bananas. The total amount was$25.10, so our second equation is:

1.45a + 0.90b = 25.10

To eliminate the decimals, let’s multiply the entire equation by 100:

145a + 90b = 2510

You might recall learning the “substitution method” for solving systems of linear equations. We’ll use that method here. First, let’s look at our system:

a + b = 23   (equation 1)

145a + 90b = 2510   (equation 2)

To use the substitution method, we isolate one variable in one equation and then substitute into the other equation. Let’s isolate a in equation 1:

a = 23 – b

We can now substitute 23 – b for a in equation 2 and solve for b:

145(23 – b) + 90b = 2510

3335 – 145b + 90b = 2510

-55b = -825

b = 15

So, Nitish bought 15 bananas. Since he bought a total of 23 pieces of fruit, we know that the number of avocados is 23 – 15 = 8.

TTP PRO TIP:

You can solve many systems of linear equations with the substitution method.

You might recall a second technique used to solve a system of linear equations: the combination method. You can review this method and learn all the math you need for SAT success in our SAT self-study course.

Next, let’s discuss how to evaluate functions.

## Concept 3: Evaluating Functions

Too often, when students see an equation like f(x) = 5x – 12, they scratch their heads and move on to the next question. They don’t understand that “function notation” f(x) is actually quite straightforward and just needs a little love. Let’s give some attention to this oft-maligned math concept!

First, understand that “f(x)” is notation. While there’s (obviously) more to it than that, for SAT purposes, just know that f(x) (pronounced “f of x”) is just notation. The letter f just stands for “function.” Also note that f(x) is NOT multiplication. In fact, you can often replace f(x) with y, as shown here:

f(x) = 5x – 12 can generally be re-expressed as y = 5x – 12

Let’s get back to function notation and learn how to evaluate a function. When you see a number instead of the “x” inside the parentheses, it just means to substitute that number for x in the equation. For example, if you are given f(x) = 5x – 12 and you are asked to evaluate f(4),it means to let x = 4 and plug in 4 wherever you see x in the equation. Thus, you have:

f(x) = 5x – 12

f(4) = 5(4) – 12 = 8

Thus, you would say that f(4) = 8, and you can think of it as meaning “The value of the function when x equals 4 is 8.”

Let’s solve a few practice questions about functions.

### Example 3: Evaluating Functions

1. Evaluate f(3) for f(x) = 7x^2 – 30
2. Evaluate f(-1) for f(x) = 4x + 12
3. If f(x) = 8x – w and f(4) = 12, what is the value of w?

#### Solution:

A. We substitute 3 for x, so we have:

f(3) = 7(3)^2 – 30 = 63 – 30 = 33

B. We substitute -1 for x, so we have:

f(-1) = 4(-1) + 12 = -4 + 12 = 8

C. This function requires two steps to solve. First, substitute 4 for x in the equation:

f(4) = 8(4) – w

f(4) = 32 – w

Next, since we are given that f(4) = 12 we can substitute 12 for f(4) and solve for w:

f(4) = 32 – w

12 = 32 – w

-20 = -w

20 = w

KEY FACT:

Remember that f(x) is function notation. If you are asked, for example, to evaluate f(5), just substitute the number 5 for x in the equation.

## Concept 4: Factoring a Quadratic Trinomial

You may recall that the FOIL technique allows you to multiply two binomials. For example, if you have f(x) = (x + 3)(x + 2), you can FOIL the two binomials (expressions inside the parentheses):

f(x) = (x + 3)(x + 2) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6

The FOIL technique is easy to learn and easy to remember. But going in reverse — performing an operation called factoring — is not so straightforward. Factoring involves looking at the trinomial x^2 + 5x + 6 and finding the two binomials that multiply together to produce it.

So, how do we go backward? How do we determine the two binomials (x + ? ) and (x + ??) that multiply together to give us f(x) = x^2 + 5x + 6? Let’s take a look.

### The Steps for Factoring a Trinomial

To learn how to factor a trinomial, let’s use the example given earlier:  (x + 2)(x + 3) =  x^2 + 5x + 6.

Step 1: Make sure that the x^2 term has a coefficient of 1. What this means is that you have to make sure you don’t have 5x^2 or (1/8)x^2. You just want x^2. In our example, we have just plain x^2.

Step 2: Look at the constant at the end. In our case, for x^2 + 5x + 6, the constant is 6.  Next, look at the middle term, which is 5x in our example. The coefficient (number in front) is 5. Our goal is to list all the whole numbers that multiply to give us 6 (the constant) and sum to give us 5 (the coefficient of x).

Let’s first look at all pairs of values that could multiply to give us 6:

1 x 6      2 x 3     (-1) x (-6)     (-2) x (-3)

Next, we look at the pairs of numbers to determine which pair also gives us a sum of 5. Of these pairs of numbers, only one pair also gives us a sum of 5: 2 and 3.

Step 3: Based on the two numbers found in Step 2, write the two binomials that work. We see that the two binomials that will multiply together to give us x^2 + 5x + 6 are (x + 2) and (x + 3).

Let’s try another factoring question.

### Example 4: Factoring a Trinomial

Factor f(x) = x^2 + 4x – 12.

#### Solution:

Step 1: We see that x^2 has a coefficient of 1.

Step 2: The constant term is -12, and the coefficient of the middle term is 4. Thus, we have to find two numbers that multiply to give us -12 and add together to give us 4. Let’s first list all pairs of numbers that multiply to yield -12:

1 x (-12)      2 x (-6)     3 x (-4)     (-1) x 12       -2) x 6      (-3) x 4

Of these pairs, only -2 and 6 yield a sum of 4.

Step 4: The two binomials are (x – 2) and (x + 6). Let’s double-check to see if this is the correct factoring by FOILing the two binomials.

(x – 2)(x + 6) = x^2 + 6x – 2x – 12 = x^2 + 4x – 12

The FOILed answer matches the original trinomial, x^2 + 4x – 12. Thus, we know that our factoring was correct.

TTP PRO TIP:

Use the 3-step procedure to efficiently factor a basic trinomial.

Next, let’s discuss an important statistical concept: the margin of error.

## Concept 5: Margin of Error

You don’t need to have taken AP Statistics to know how to answer a question about the margin of error. Read on to learn all you need to know to answer a margin of error question.

### Take a Sample and Get the Results

Consider a large high school with 1,000 students who have a driver’s license. You want to determine the percentage of students who drive to school at least once a week. You don’t have time or energy to ask all 1,000 students about their driving habits. Instead, you choose a random sample of 100 students.You find that 62 of the 100 students say they drive to school at least once a week. The corresponding proportion is 62/100, or 62%.

Now, does this result mean that exactly 62% of all the school’s 1,000 students drive to school at least once a week?

Maybe, maybe not. Because you didn’t ask all 1,000 students, you cannot trust that 62% is the exact percentage. In fact, if you took a different sample and asked the same question, you might obtain a result of 51% or even 83%. There is always a bit of “wiggle room” when you take a sample and try to generalize to the entire group. This is where the margin of error comes into play.

### How to Use the Margin of Error

We might say, for example, that the true percentage of students who drive to school at least once a week is most likely 62% +/- 4%. This “plus or minus” value of 4% is called the margin of error. We add that margin of error to the percentage we obtained, and we subtract it from that value. Thus, using the margin of error, we can say with some amount of confidence that the true percentage of students who drive to school at least once a week is between 62% – 4% =  58% and 62% + 4% = 66%. So, for our entire school population of licensed students, we could say that between 580 and 660 students drive to school at least once a week.

You might wonder where the 4% value came from in the above example. For SAT purposes, you don’t need to know how it is calculated. The only fact you need to know beyond the calculations already provided is this: if you want to decrease your margin of error, you can increase the number of people you ask.

Let’s look at a question about the margin of error that you might encounter on the SAT.

### Example 5: Margin of Error

Before an election in a small town, a random sample of residents was selected and asked if they intended to vote for Candidate X. The proportion of those asked who said that they intended to vote for Candidate X was 43%, with a margin of error of 5%. The total number of voters expected to vote in the election is 3,510. Between what two values would one plausibly expect the number of votes for Candidate X to be?

#### Solution:

The sample results for Candidate X were 43% +/- 5%, so we can reasonably expect that between 38% and 48% of the 3,510 voters will vote for Candidate X.

38% of the voters is (0.38)(3510) = 1,333.8, which rounds to 1,334.

48% of the voters is (0.48)(3510) = 1,684.8, which rounds to 1,685.

Thus, we can plausibly expect that between 1,334 and 1,685 voters will vote for Candidate X.

KEY FACT:

The margin of error is a “plus or minus” value used to calculate an interval of values that a statistical measure might be expected to fall within.

Next, let’s discuss an important geometry topic: the algebraic expression of circles.

## Concept 6: Algebraic Expression of Circles

We all know what a circle looks like. However, it’s another thing to express the equation of a circle algebraically. The SAT will test you on the algebra, so let’s see how it works.

### A Circle Centered on the Origin

First, let’s look at a basic circle in the xy coordinate plane with its center at the origin (0, 0) and a radius of 3. We use the standard version of the equation of a circle: x^2 + y^2 =  r^2. So for our circle with radius of 3, we have:

x^2 + y^2 = 3^2

x^2 + y^2 = 9

That’s it! The equation of a circle centered at the origin with a radius of 3 is x^2 + y^2 = 9.

Let’s solve a quick review question.

### Example 6A: Equation of a Circle Centered on the Origin

What is the radius of a circle whose equation is x^2 + y^2 = 16?

#### Solution:

Using the standard version of the equation of a circle, x^2 + y^2 = r^2, we see that since r^2 = 16, the radius of the circle is 4.

Next, let’s review the general form of a circle when we don’t know that it is centered on the origin.

### The General Form of a Circle

A circle with center (a, b) and radius r will have the equation (x – a)^2 + (y – b)^2 = r^2. For example, a circle with radius 6 whose center is at the point (3, 5) will have the equation:

(x – 3)^2 + (y – 5)^2 = 36

Let’s try a straightforward example.

### Example 6B: The Equation of a Circle Not Centered on the Origin

What is the equation of the circle with radius 8 and center at the point (3, -4)?

Using the general equation (x – a)^2 + (y – b)^2 = r^2, we substitute 3 for a, -1 for b, and 8 for r:

(x – 3)^2 + (y – (-1))^2 = 8^2

(x – 3)^2 + (y + 1)^2 = 64

KEY FACT:

The equation of a circle in the coordinate plane with center (a, b) and radius r is (x – a)^2 + (y – b)^2 = r^2.

For easy and medium circle questions on the SAT, the two formulas introduced here will suffice. However, you might encounter more difficult questions in which you will need to use the technique called “completing the square” to determine the equation of the circle. This technique is beyond the scope of this article, but you can find it covered in detail in our self-study SAT course.

## Key Takeaways

In this article, we covered the key changes from previous versions of the SAT to the digital SAT:

• The digital SAT is computer-adaptive at the section level.
• Students can now use a calculator on all math modules.
• Reading passages now have only 1 question per passage.

We also reviewed the 4 broad content areas covered in the dSAT Math section:

• Algebra (35% of the Math section)
• Problem-Solving and Data Analysis (15%)
• Geometry/Trigonometry (15%)

Finally, we discussed 6 high-value concepts that you are likely to encounter in the SAT Math section:

• Linear word problems
• Systems of linear equations
• Evaluating functions