How to Solve SAT® Math Problems
Tricks and Strategies
The SAT® Math Section tests a wide range of topics, covering multiple years' worth of your math education, and that breadth can make studying for the exam a daunting task. Thankfully, you can simplify this task by learning the categorizations that the College Board® uses to group SAT math question types. This grouping method aids advanced math students in focusing their studies on unfamiliar domains and enables them to learn SAT Math Section strategies more effectively by tackling subjects one at a time rather than all at once.
Below, we’ve broken down the SAT Math Section’s four official domains, Heart of Algebra, Problem Solving and Data Analysis, Passport to Advanced Math, and Additional Topics in Math. Continue below, for example, questions, strategies, and tricks for each.
Heart of Algebra
Heart of Algebra focuses on linear functions, equations, and inequalities. These questions use equations, inequalities, or systems of equations without any exponents or word problems that reference rates, like growth in size, price per unit, or speed.
Note that the Digital SAT renames this domain simply "Algebra," but the content remains unchanged.
This domain accounts for about a third of all math questions on any given exam on both the penandpaper SAT and the Digital SAT.
Tricks and strategies to ace SAT Algebra questions
The key to correctly approaching linear functions is understanding what they represent. Begin by drawing a graph or marking up one that is already there. Graphs of linear equations help visualize their slopes and intercepts, and the intersection of two such graphs provides the solution to a system of linear equations. If you’re unsure how you would plot a graph for a given Algebra question, rewrite the equation in slopeintercept form. When no equation is provided, create one by setting the output of the situation equal to the sum of the starting value and the effective input (e.g., total price = base price + [price per unit * units]).
Heart of Algebra examples
In the xyplane, the graph of which of the following equations is a line that is perpendicular to the line shown in the graph above?
A. y = ─ 3 / 2 x + 6  
B. y = ─ 2 / 3 x + 1 / 3  
C. y = 2 / 3 x + 3  
D. y = 3 / 2 x + 6 
Hint: Perpendicular lines have slopes that are negative reciprocals of each other.
Perpendicular lines have slopes that are negative reciprocals of each other. To find the equation of a line that is perpendicular to the line given in the graph, first find the slope of the given line.
To find the slope, identify the coordinates of the two points marked on the graph, and divide the "rise" (vertical distance) by the "run" (horizontal distance).
Note: It is also possible to plug the given points into the slope formula y_{2}  y_{1} / x_{2}  x_{1} 
The slope of the given line is  3 / 2 .
The negative reciprocal of  3 / 2 is 2 / 3 , so any line with a slope of 2 / 3 is perpendicular to the given line.
Each answer choice is the equation of a line in slopeintercept form y = mx + b, where m is the slope and b is the yintercept.
Any line that is perpendicular to the given line has a slope of 2 / 3 , so its equation must have a slope equal to 2 / 3 .
Eliminate Choices A, B, and D because they do not have slope m = 2 / 3 .
Only Choice C includes a slope equal to 2 / 3 , so the equation must be y = 2 / 3 x + 3 .
Note: All lines with slope 2 / 3 are perpendicular to the given line. Therefore, the yintercept can be any value, and it is not necessary to consider the yintercepts of the choices to answer the question.
Elimination Strategy
The line in the given graph decreases vertically as values on the horizontal axis increase, so it has a negative slope. Therefore, any line perpendicular to it must have a positive slope.
It is possible to eliminate Choices A and B because they do not have positive slopes.
(Choice A) y = ─ 3 / 2 x + 6 is the equation of a line that has the same slope as the given line. Lines with the same slope are parallel, but the question asks for the equation of a line that is perpendicular to the given line.
(Choice B) y = ─ 2 / 3 x + 1 / 3 may result from the misconception that perpendicular lines have slopes that are only reciprocals, instead of negative reciprocals.
(Choice D) y = 3 / 2 x + 6 may result from the misconception that perpendicular lines have slopes that are only opposite in sign, instead of negative reciprocals.
Things to remember:
 Perpendicular lines have slopes that are negative reciprocals of each other (ex. 2 / 1 and  1 / 2 ).
 The slope of a line is equal to rise / run , where the "rise" is the vertical distance between any two points and the "run" is the horizontal distance between the same two points.
 The slopeintercept form of a line is y = mx + b, where m is the slope and b is the yintercept of the line.
y ≥ x + 4
4x + 5y ≤ 20
Which of the following has a shaded region that represents the solution set in the xyplane to the system of inequalities above?
A.  
B.  
C.  
D. 
Hint: The graph of a linear inequality of the form y > mx + b or y ≥ mx + b has shading above the boundary line.
The graph of a linear inequality of the form y < mx + b or y ≤ mx + b has shading below the boundary line.
Notice that the graphs in all four answer choices have the same boundary lines and differ only in the shading (solution sets). One of the boundary lines has a positive slope and the other has a negative slope.
Therefore, it is possible to identify each inequality from only the slope of its boundary line and then analyze the inequality signs to answer the question by process of elimination.
For each graph, one line must represent y = x + 4 and the other must represent 4x + 5y = 20.
Compare y = x + 4 to slopeintercept form to see that the slope (1) is positive. The line with a positive slope must be the boundary line for y ≥ x + 4, and the line with a negative slope must be the boundary line for 4x + 5y ≤ 20.
The inequality y ≥ x + 4 means that all values of y are greater than or equal to (≥) the boundary line, so the shading is above this line.
Eliminate Choices A and D because they have shading below the line with a positive slope.
The remaining choices have shading above the boundary line for y ≥ x + 4, so they represent solutions to this inequality.
To determine which one represents the solution set to the entire system, isolate y in the other inequality 4x + 5y ≤ 20 and determine whether the shading lies above or below that line.
4x + 5y ≤ 20  
5y ≤ −4x + 20  Subtract 4x from both sides 
y ≤ − 4 / 5 x + 4  Divide both sides by 5 
The inequality y ≤ − 4 / 5 x + 4 means that all values of y are less than or equal to (≤) the boundary line, so the shading is below this line.
Eliminate Choice C because the shading is above the line with a negative slope.
By process of elimination, the shaded area in Choice B must represent the solution set to the given system of inequalities.
(Choice A) This graph represents solutions for 4x + 5y ≤ 20 but not for y ≥ x + 4, because the shading is below the line that has a positive slope.
(Choice C) This graph may result from the misconception that dividing or multiplying both sides of an inequality by a number flips the inequality sign, but that is true only if the number is negative. The shading is above the line that has a negative slope.
(Choice D) This graph does not represent solutions for either given inequality. The shading is below the line that has a positive slope and above the line that has a negative slope.
Things to remember:
 The graph of a linear inequality of the form y > mx + b or y ≥ mx + b has shading above the boundary line.
 The graph of a linear inequality of the form y < mx + b or y ≤ mx + b has shading below the boundary line.
Alternate Method :
Alternately, graph each inequality and then determine which shaded region from the choices matches the solution set. First graph the boundary line of each inequality, and then determine where the shading lies.
The equations for the two boundary lines are y = x + 4 and 4x + 5y = 20. Rewrite 4x + 5y = 20 in slopeintercept form y = mx + b, where m is the slope and b is the yintercept of the line.
4x + 5y = 20  
5y = −4x + 20  Subtract 4x from both sides 
y = − 4 / 5 x + 4  Divide both sides by 5 to rewrite in slopeintercept form 
Note: The sign of the inequality y ≤ − 4 / 5 x + 4 remains ≤ (less than or equal to) because both sides of the boundary line equation were divided by a positive number (4), not a negative number.
Now graph the inequalities y ≥ x + 4 and y ≤ − 4 / 5 x + 4.
Combine these graphs to see that points in the intersection of both regions are solutions to both inequalities.
The intersection of the shaded regions represents the solutions to the system of inequalities, which matches the graph in Choice B.
Things to remember:
 The slopeintercept form of a linear equation is y = mx + b, where m is the slope and b is the yintercept.

To graph a linear inequality:
 Write the boundary line equation in slopeintercept form, and then graph the line. Note: Multiplying or dividing both sides of the equation by a negative number flips the sign of the inequality.
 Shade above the line if the inequality is of the form y > mx + b or y ≥ mx + b.
 Shade below the line if the inequality is of the form y < mx + b or y ≤ mx + b.
A research study conducted on a group of adult males determined that the formula H = 0.98A + 44.09 can be used to approximate the height H, in millimeters, of an adult male in this group, based on his arm span A, in millimeters. What is the meaning of 0.98 in this context?
A. The approximate increase in a man's height, in millimeters, for each onemillimeter increase in arm span.  
B. The approximate increase in a man's height, in millimeters, for each increase of 44.09 millimeters in arm span.  
C. The approximate increase in a man's arm span, in millimeters, for each onemillimeter increase in height.  
D. The approximate increase in a man's arm span, in millimeters, for each increase of 44.09 millimeters in height. 
Hint: The linear model is in slopeintercept form y = mx + b, where m is the rate of change in quantity y per quantity x.
The given equation is a linear equation in slopeintercept form y = mx + b, where m is the rate of change (slope) and b is the initial value (yintercept).
The rate of change of a quantity y per quantity x is equal to the change in y divided by the change in x.
Compare the given equation H = 0.98A + 44.09 to y = mx + b to see that H corresponds to y and A corresponds to x.
Identify that 0.98 is the rate of change m of quantity H per quantity A.
The rate of change m = 0.98 is the change in H per change in A. The number 0.98 is equivalent to 0.98 / 1 , so a change in H of 0.98 millimeters corresponds to a change in A of 1 millimeter.
Therefore, the number 0.98 represents the estimated increase in a man's height, in millimeters, for each onemillimeter increase in arm span.
(Choice B) "The approximate increase in a man's height, in millimeters, for each increase of 44.09 millimeters in arm span" is represented by 0.98 / 44.09 , but the question asks for the meaning of 0.98.
(Choice C) "The approximate increase in a man's arm span, in millimeters, for each onemillimeter increase in height" may result from the misconception that m (0.98) is a rate of change in quantity A (arm span) per quantity H (height) instead of the rate of change in quantity H per quantity A.
(Choice D) "The approximate increase in a man's arm span, in millimeters, for each increase of 44.09 millimeters in height" is represented by 44.09 / 0.98 , but the question asks for the meaning of 0.98.
Things to remember:
 The slopeintercept form of a linear equation is y = mx + b, where m is the rate of change (slope) and b is the initial value (yintercept).
 The rate of change of a quantity y per quantity x is equal to the change in y divided by the change in x.
Problem Solving and Data Analysis
While all SAT Math problems require problemsolving, this category refers to solving problems that rely on ratios, percentages, and the collection and use of one or two variables. Recognize these questions by their reference to statistical terms like mean, median, and mode; their use of scatter plots and lines of best fit; their description of experiments and data collection or sampling methods; and their inclusion of percent or ratio relationships. Also, though only present on some exams, boxandwhiskers plot questions on the SAT belong to this section.
Note that the Digital SAT continues the penandpaper SAT's use of this category of questions in name and relevant content.
This domain accounts for just under a third of all math questions on the penandpaper SAT on any given exam. This category accounts for much less on the Digital SAT—only about a sixth of all math questions.
Tricks and strategies to ace SAT Problem Solving and Data Analysis questions
As you prepare for this section, keep track of definitions for technical terms and, when they refer to values or sets of values, how to calculate them. SAT math questions frequently include wrong answers derived from common errors or misconceptions (e.g., if a question asks about a dataset's mean, you can be sure the answer choices will include the median). Once you've mastered the relevant vocabulary, you'll approach this domain much more confidently and answer the question with less distraction.
Another common trap in this section is that the data points in these questions appear randomly rather than in numerical order. When deriving statistical values like the median, numerical order is critical, so reordering any values is always worth taking a few seconds.
Problem Solving and Data Analysis examples
Mr. Staine is one of the marine biologists at Coastal University, which has 1,816 students. He selected 50 of his marine biology students at random and asked each whether they enjoy swimming. Of the 50 students surveyed, 39 responded that they enjoy swimming. Based on the design of the study, which of the following is the largest group of students to whom the results of Mr. Staine's survey can be generalized?
A. All of Mr. Staine's marine biology students at Coastal University  
B. All of the marine biology students at Coastal University  
C. All of the students at Coastal University  
D. All of the marine biology students in the country 
Hint: The results of a study can be generalized only to a population from which the participants were randomly selected.
The results of a study can be generalized only to a population from which participants were randomly selected.
It is given that Mr. Staine randomly selected 50 of his marine biology students, so the population is all of Mr. Staine's marine biology students.
Therefore, all of Mr. Staine's marine biology students is the largest population to which the results can be generalized.
Notice that only one choice describes this population, while the others all describe populations that are larger.
The largest population to which the results can be generalized is all of Mr. Staine's marine biology students at Coastal University.
Things to remember:
The results of a study can be generalized only to a population from which participants were randomly selected.
Hint: A box plot is a graph of a data distribution and is based on five numbers: the minimum, 1st quartile (Q_{1}), median, 3rd quartile (Q_{3}), and maximum.
A box plot is a graph of a data distribution and is based on five numbers: the minimum, 1st quartile (Q_{1}), median, 3rd quartile (Q_{3}), and maximum.
The two outermost vertical lines are the minimum and maximum values of the data. The median splits the data into equal halves, and each of the four quartiles (quarters) represent approximately 25% of the data.
Identify the median of each box plot and use the number line to find each value. Then subtract to find the number of miles the median of Truck A exceeds the median of Truck B.
The labels on the number line increase by increments of 20, so the unlabeled marks represent increments of 10.
The median number of miles driven by Truck A is 60 and the median number of miles driven by Truck B is 50, so Truck A exceeds Truck B by 60 − 50 = 10 miles.
Things to remember:
 A box plot is a graph of a data distribution and is based on five numbers: the minimum, 1st quartile (Q_{1}), median, 3rd quartile (Q_{3}), and maximum.
 A box plot shows how spread apart the data are but does not show each individual data value.
 It is not possible to determine the number of data points or the mean of a data set from a box plot.
Two units of volume used to measure dry goods in the United States are bushels and pecks. A grocer purchased 160 bushels of apples. If 1 peck is equivalent to 0.25 bushels, which of the following best approximates the volume of the apples in pecks?
A. 0.025  
B. 40  
C. 160  
D. 640 
Hint: To convert from one unit to another, multiply by a conversion factor that cancels the old unit and replaces it with the new unit.
To convert from one unit to another, multiply by a conversion factor that cancels the old unit and replaces it with the new unit.
To convert the payment from bushels to pecks, multiply the given volume in bushels by a fraction with pecks in the numerator and the equivalent number of bushels in the denominator.
It is given that the volume of apples is 160 bushels, and the conversion is 1 peck equivalent to 0.25 bushels. Multiply 160 bushels by the conversion factor 1 peck / 0.25 bushels to find the volume in pecks.
The answer choice that best approximates the volume of the apples in pecks is 640.
Note: It is possible to rewrite the conversion factor 1 peck / 0.25 bushels as an equivalent fraction of whole numbers 4 pecks / 1 bushel and then multiply to calculate the volume in pecks.
(Choice A) 0.025 may result from dividing (instead of multiplying) the conversion factor ( 1 peck / 0.25 bushels) by 160.
(Choice B) 40 may result from incorrectly writing the conversion factor as 0.25 bushels / 1 peck instead of 1 peck / 0.25 bushels.
(Choice C) 160 is the volume of the apples in bushels instead of pecks.
Things to remember:
To convert from one unit to another, multiply by a conversion factor that cancels the old unit and replaces it with the new unit. For example, to convert 5 feet to inches: (
5 feet
/
1
) (
12 inches
/
1 foot
) = 60 inches.
Passport to Advanced Math
As its name might suggest, this domain covers a wide range of topics that go beyond the “Heart of Algebra” into Algebra II and, depending on how your classes are structured, potentially precalculus. Specifically, this category includes any nonlinear functions, expressions, or equations (e.g., quadratics, exponential and square root functions, and systems of equations with at least two variables). Recognize these questions by their references to quadratics, in any form, expressions with exponents or roots, and function notation, especially composite functions (e.g., g(f(x))).
The Digital SAT shortens the name of this content type to “Advanced Math,” but the content remains unchanged.
This domain accounts for just under a third of all math questions on the penandpaper SAT on any given exam. This category accounts for more than a third of all math questions on the Digital SAT.
Tricks and strategies to ace SAT Advanced Math questions
This category heavily focuses on higherorder functions and expressions, so familiarize yourself with how to manipulate the various expressions you encounter. For example, because different formats reveal different information about quadratics (standard form shows the yintercept, factored form shows the xintercepts and vertex form shows the vertex), learning to solve or rewrite such functions by factoring, completing the square, and the quadratic formula will significantly increase your comfort and consistency with such questions. Similarly, understanding how to rewrite exponents and roots in terms of one another will simplify and even trivialize many equivalent expression questions.
Passport to Advanced Math examples
h(x) = (x + 1)(x + 9)
The function h is defined above and the graph of h in the xyplane is a parabola. Which of the following intervals contains the xcoordinate of the vertex of the graph of h ?
A. −10 < x < −9  
B. −9 < x < 1  
C. 1 < x < 9  
D. 9 < x < 10 
Hint: The vertex of a parabola is the point at which the minimum or maximum of a quadratic function occurs.
The graph of a quadratic function is a parabola. The vertex of a parabola is the point at which the minimum or maximum of the function occurs.
A parabola has a vertical axis of symmetry that passes through its vertex (h, k ). The axis of symmetry is halfway between any two points on the parabola that have the same yvalue.
The quadratic function h (x ) = (x + 1) (x + 9) is in factored form f (x ) = a (x − p ) (x − q ), where p and q are the xintercepts. The xintercepts have the same yvalue (0), so the axis of symmetry is halfway between the xintercepts.
Every point on a vertical line has the same xcoordinate, so the xcoordinate of the vertex is also halfway between the xintercepts. Compare the equation for h (x ) to factored form to determine the xintercepts.
The xintercepts are −1 and −9, so the xcoordinate of the vertex is halfway between x = −1 and x = −9.
Add these xvalues and divide by 2 to find the xvalue that is halfway between −1 and −9.
1 + 9 / 2  Value halfway between −1 and −9 
 10 / 2  Add in numerator 
5  Divide 
The xvalue that is halfway between −1 and −9 is −5, so the vertex of the graph of h (x ) has an xcoordinate of −5 (graph). The only answer choice that contains x = −5 is the interval −9 < x < 1.
Note: The only interval from the choices that includes values between the xintercepts is −9 < x < 1, so it is not necessary to calculate the value of the xcoordinate of the vertex for this particular question.
(Choice A) −10 < x < −9 may result from the combination of the error described in Choice D and a sign error.
(Choice C) 1 < x < 9 may result from mistaking the xintercepts to be 1 and 9 instead of −1 and −9. This may also result from incorrectly using b / 2a instead of x =  b / 2a as the vertex formula (see alternate method).
(Choice D) 9 < x < 10 may result from the misconception that the xintercepts are displayed as constants in the standard form of a quadratic function f (x ) = ax^{2} + bx + c instead of the factored form f (x ) = a (x − p ) (x − q ).
Things to remember:
 The vertex of a parabola is the point at which the minimum or maximum of a quadratic function occurs.
 A parabola has a vertical axis of symmetry that passes through its vertex. The axis of symmetry is halfway between any two points on the parabola that have the same yvalue.
 The factored form of a quadratic function is f (x ) = a (x − p ) (x − q ), where p and q are the xintercepts.
Alternate Method :
It is also possible to rewrite the given function in standard form f (x ) = ax^{2} + bx + c and then use the vertex formula to find the xcoordinate of the vertex.
Distribute each term in x + 1 to each term in x + 9 to rewrite the given equation h (x ) = (x + 1) (x + 9) in standard form.
h (x ) = (x + 1) (x + 9)  
h (x ) = x^{2} + 9x + x + 9  Distribute 
h (x ) = x^{2} + 10x + 9  Combine like terms 
Compare h (x ) = x^{2} + 10x + 9 to standard form f (x ) = ax^{2} + bx + c to identify the values of a and b. The term x^{2} has coefficient 1, so rewrite it as 1x^{2}.
Plug a = 1 and b = 10 into the vertex formula and simplify.
x =  b / 2a  xcoordinate of the vertex 
x =  10 / 2 ( 1 )  Plug in a = 1 and b = 10 
x = 5  Simplify 
The xcoordinate of the vertex of the graph of h is −5 (graph). The only answer choice that contains x = −5 is the interval −9 < x < −1.
Things to remember:
The vertex formula defines the xcoordinate of the vertex of a quadratic equation in standard form y = ax^{2} + bx + c:
Daisy modeled the growth of monarch butterflies during the caterpillar stage by estimating the length of a monarch on each day of the caterpillar stage. She estimated that a monarch measures 1.77 millimeters on the first day of the caterpillar stage, with a 28% daily increase in length until it reaches the pupa stage. Which of the following functions models L(t), the length of a monarch t days after the first day of the caterpillar stage?
A. L(t) = 1.77^{t}  
B. L(t) = 1.77^{1.28t}  
C. L(t) = 1.77(0.72)^{t}  
D. L(t) = 1.77(1.28)^{t} 
Hint: Model the length of a monarch butterfly over time with an exponential equation of the form y = a(b)^{x}, where a is the initial value and b is the growth (b > 1) or decay (b < 1) factor.
The length of a monarch butterfly in the caterpillar stage is estimated to increase by a percentage of its length each day, which is an exponential increase.
Model the length of a monarch over time with an exponential function of the form y = a (b )^{x}, where a is the initial value and b is the growth factor. The length of a monarch L (t ) corresponds to y and time t corresponds to x.
The initial value is the length of a monarch on the first day of the caterpillar stage, so a is 1.77. The growth factor b is the factor by which the length of the monarch changes each day.
It is given that the length of the monarch increases by 28% each day in the caterpillar stage, so the growth factor is the percent from the previous stage (100%) plus the percent of the increase (28%) in decimal form.
100% + 28%  Percent of first day length (100%) plus increase (28%) 
128%  Simplify 
1.28  Convert the percent to a decimal: 128% = 128 / 100 
The growth factor b is 1.28.
Substitute a = 1.77 and b = 1.28 into the model above to write an equation for L (t ).
L (t ) = a(b)^{t}  Exponential equation for L (t ) 
L (t ) = 1.77(1.28)^{t}  Substitute a = 1.77 and b = 1.28 
Therefore, the function that models the length of a monarch butterfly t days after the first day of the caterpillar stage is L (t ) = 1.77(1.28)^{t}.
(Choices A and B) L (t ) = 1.77^{t} and L (t ) = 1.77^{1.28}^{t} may result from using incorrect general forms of an exponential equation.
(Choice C) L (t ) = 1.77(0.72)^{t} is the function that models a 28% daily decrease in the length of a monarch, but the question states that there is a 28% daily increase in the length.
Things to remember: In an exponential equation of the form y = a (b )^{x}, a is the initial value and b is the growth (b > 1) or decay (b < 1) factor.
Alternate Method :
It is given that the length of a monarch butterfly is estimated to increase by 28% each day after the first day of the caterpillar stage.
To increase a value by p%, multiply it by ( 1 + p / 100 ) (proof). Therefore, the estimated length of a monarch 1 day after the first day of the caterpillar stage is (1.77) multiplied by ( 1 + 28 / 100 ).
1.77 ( 1 + 28 / 100 )  First day length increased by 28% 
1.77(1.28)  Simplify: 1 + 28 / 100 = 1 + 0.28 
The estimated length of a monarch 1 day after the first day is 1.77(1.28). Each day after the first, the length of the monarch increases by a factor of 1.28.
Therefore, the length 2 days after the first day is 1.77(1.28)(1.28), the length 3 days after the first day is 1.77(1.28)(1.28)(1.28), and so on.
Therefore, the function that models the length of a monarch butterfly t days after the first day of the caterpillar stage is L(t) = 1.77(1.28)^{t}.
Things to remember:
To increase a value by p%, multiply it by ( 1 +
p
/
100
).
For the function h in the table above, values of h(x) are shown in terms of constants a, b, c, and d. If a < b < c < d, which of the following could NOT be the graph of y = h(x) in the xyplane?
A.  
B.  
C.  
D. 
Hint: The table gives values of h(x) for increasing values of x. Identify the corresponding points and use the given relationship between a, b, c, and d to determine which could NOT be the graph of h(x).
The table gives values of h(x) for increasing values of x. On the graph of function h, these correspond to the points (1, a), (2, b), (3, c), and (4, d).
It is given that a < b < c < d, which means that a is smaller than b, b is smaller than c, and c is smaller than d. Therefore, the value of h(x) increases as x increases within the given interval.
Any graph that increases between x = 1 and x = 4 could be the graph of y = h(x).
The question asks which graph could NOT be y = h(x), so eliminate Choices A, B, and D because they each contain a graph that increases in this interval.
By process of elimination, the graph in Choice C could NOT be the graph of y = h(x) because it decreases between x = 1 and x = 4.
Things to remember:
 If the value of a function f(x) increases as x increases, the function is increasing.
 If the value of a function f(x) decreases as x increases, the function is decreasing.
Additional Topics in Math
“Additional Topics in Math” suggests something of a catchall, but the category focuses heavily on geometry, with the remaining questions in the section covering complex numbers. Recognize these questions by their references to, and diagrams or equations of, shapes and angles, as well as by using the imaginary constant i.
The Digital SAT renames this category to “Geometry and Trigonometry, and the content follows suit, retaining the geometric questions but not those related to complex numbers.
On the penandpaper SAT, this domain accounts for a tenth of all math questions on any given exam. This domain accounts for a little more on the Digital SAT—about a sixth of all math questions.
Tricks and strategies to ace SAT Additional Topics in Math questions
As Geometry makes up the core of this domain, familiarity with geometric rules and theorems is vital. In particular, be sure you can recognize how angle values relate to each other within sets of parallel lines or triangles, noting which angles must be congruent (equivalent) or supplementary (summing to 180 degrees). Circle theorems comprise a small proportion of this section, so they do not appear on every exam. Still, many students need to familiarize themselves with how to arrive at the size and angles of a given circle, so once you're confident in the other categories, take time to memorize these applications.
Separately, for those taking the penandpaper SAT, the difficulty of complex number questions depends almost entirely on your comfort with the constant i. If you study the exponential powers of i and practice FOILing (multiplying two binomials) expressions that use them, you're likely to master these questions.
Additional topics in Math examples
In the figure above, line m is parallel to line n. If y = 75 and z = 50, what is the value of x ?
A. 50  
B. 55  
C. 60  
D. 65 
Hint: When two parallel lines are intersected by a third line (called a transversal), they create pairs of corresponding angles that are equal in measure.
First label the figure with the given information y = 75 and z = 50, as well as m  n.
When parallel lines are intersected by another line (called a transversal), the pairs of corresponding angles that they form are equal in measure.
It is given that lines m and n are parallel, so the corresponding angles indicated below must both have a measure of 50°.
Notice that the x° angle forms a triangle with the 75° and 50° angles. The measures of the interior angles of a triangle sum to 180°, so set the sum of these three measures equal to 180 and solve for x.
x + 75 + 50 = 180  Sum of measures of interior angles of a triangle is 180° 
x + 125 = 180  Simplify 
x = 55  Subtract 125 from both sides 
(Choice A) 50 is a result of mistaking the z° and x° angles for vertical angles and therefore equal in measure.
(Choice C) 60 is a result of the mistaken assumption that the triangle is equilateral and therefore all interior angles are equal in measure.
(Choice D) 65 is a result of a subtraction error: 180 − 125 ≠ 65.
Things to remember:
 If parallel lines are cut by a transversal, the pairs of corresponding angles that they form are equal in measure.
 The measures of the interior angles of a triangle sum to 180°.
Alternative Method:
An alternate way to find the value of x is to use congruent alternate interior angles rather than congruent corresponding angles.
First label the figure with the given information y = 75 and z = 50, as well as m  n.
When parallel lines are intersected by another line (called a transversal), the pairs of alternate interior angles that they form are equal in measure.
It is given that lines m and n are parallel, so the alternate interior angles indicated below must both have a measure of 75°.
Notice that the 50°, 75°, and x° angles combine to form a straight line (line q). Angles that combine to form a straight line have measures that sum to 180°, so set the sum of these measures equal to 180 and solve for x.
50 + 75 + x = 180  Angles that combine to form a straight line have measures that sum to 180° 
125 + x = 180  Simplify 
x = 55  Subtract 125 from both sides 
Things to remember:
 If parallel lines are cut by a transversal, the pairs of alternate interior angles that they form are equal in measure.
 Angles that combine to form a straight line have measures that sum to 180°.
In the figure above, AB = BC. What is the value of x ?
A. 90  
B. 95  
C. 100  
D. 105 
Hint: First label the figure with the given information that AB = BC, which means that triangle ABC is isosceles.
The base angles of an isosceles triangle are equal in measure, so ∠BAC and ∠BCA have equal measures.
First label the figure with the given information that AB = BC, which means that triangle ABC is isosceles.
The base angles of an isosceles triangle are equal in measure, so let the base angles (∠BAC and ∠BCA) of triangle ABC both have a measure of a°.
The measures of the interior angles of a triangle sum to 180°. Set the sum of the measures of the angles of triangle ABC equal to 180 to find the value of a.
a + 100 + a = 180  The measures of the interior angles of triangle ABC sum to 180° 
2a + 100 = 180  Simplify 
2a = 80  Subtract 100 from both sides 
a = 40  Divide both sides by 2 
To calculate x, use the rightmost triangle that contains both the x° and 40° angles. The third angle in this triangle and the 130° angle are supplementary because they form a straight line, so its measure is 180° − 130° = 50°.
Now set the sum of the measures of the interior angles of triangle FGC equal to 180 and solve for x.
50 + x + 40 = 180  The measures of the interior angles of triangle FGC sum to 180° 
x + 90 = 180  Simplify 
x = 90  Subtract 90 from both sides 
(Note: After the measure of angle BCA is found, the value of x is more efficiently calculated using the exterior angle of a triangle theorem.)
(Choice B) 95 is a result of the mistaken assumption that the two smaller triangles are congruent.
(Choice C) 100 is a result of the mistaken assumption that segments AB and FG are parallel.
(Choice D) 105 may result from the misconception that x° must be greater than the measure of angle ABC.
Things to remember:
 The measures of the interior angles of a triangle sum to 180°.
 The base angles of an isosceles triangle are equal in measure.
 Angles that combine to form a straight line are supplementary (measures sum to 180°).
Which of the following complex numbers is equivalent to (1 − i) − (4 − 3i) ? (Note: i = √1 )
A. −3 − 4i  
B. −3 + 2i  
C. 5 − 4i  
D. 5 + 2i 
Hint: To add or subtract complex numbers, add or subtract their real and imaginary parts separately.
The expression a + bi represents a complex number, where a is the real part and bi is the imaginary part.
To add or subtract complex numbers, add or subtract the real and imaginary parts separately.
Distribute the negative sign to (4 − 3i) to subtract, and then combine the real and imaginary terms.
(1 − i) − (4 − 3i)  
1 − i − (4) − (−3i)  Distribute the negative sign to (4 − 3i) 
1 − i − 4 + 3i  Simplify: −(−3i) = 3i 
(1 − 4) + (−i + 3i)  Group the real and imaginary terms 
−3 + 2i  Add real and imaginary terms separately 
Note: For complex number questions on the SAT, the note "i = √1 will always be included to specify that the question uses complex numbers, regardless of whether i needs to be substituted with √1 or not.
(Choice A) −3 − 4i is a result of not distributing the negative sign to −3i.
(Choice C) 5 − 4i is the sum of the two complex numbers.
(Choice D) 5 + 2i is a result of adding the real parts 1 and 4 instead of subtracting them.
Things to remember:
Add or subtract complex numbers by collecting the like terms of the real and imaginary parts separately.
How To Fill the GridIns on SAT Math
Most SAT Math test questions come in the form of multiplechoice questions. However, some questions do not provide answer choices and require you to come up with an answer on your own and fill it in on the answer sheet. These are called Gridin questions. This type of question provides a grid where you write the answer and fill in bubbles for the corresponding digits or symbols (like decimals or fractions). Here are some points to remember:
 Make sure that when you write an answer, you also bubble it in. Do not bubble more than one circle per column. Providing a written answer or a bubble will not be considered for scoring.
 Remember that decimal answers cannot start from zero. They must be written and bubbled in starting from the decimal itself, such as .77, .56, .08, etc.
 If your decimal answer is larger than four digits, you may round up or fill in an abbreviated version. For example, .777992 may be bubbled in as .778 or .777.
 If a question has more than one correct answer, be sure to only bubble in one answer per question.
 You can provide your answer as a decimal or fraction for some questions, but be sure to convert any mixed numbers to improper fractions. However, you do not need to reduce fractions.
 Gridin questions will not provide you with a negative sign, so negative answers are not possible for these questions.
Key Takeaways
Studying for the SAT Math Section can feel overwhelming due to the variety of question types. However, you can tackle these questions with less intimidation by dividing them into the exam's four domains. As you practice SAT math questions, group them into these categories and, when possible, study one domain at a time until you’re comfortable and rarely miss a related question. By arranging your study plan this way, you can gradually master the subject one step at a time as the topics within a domain progressively build upon each other. This approach prevents you from overwhelming yourself by tackling the subject simultaneously.
If you are unfamiliar with the domains’ content, uncomfortable with classifying the questions on your own, or would simply prefer a more organized way to study the Section, UWorld’s SAT Math QBank does this grouping for you and allows you to easily build a test of questions and comprehensive explanations for each domain, one subsection at a time.
Frequently Asked Questions (FAQs)
What are the most common types of SAT math problems?
As previously mentioned, the Heart of Algebra domain constitutes most of the SAT Math Section. Consequently, linear functions, equations, and inequalities are the most prevalent SAT math question types on the exam.
What are some common mistakes to avoid while solving SAT math problems?
Most SAT Math mistakes, including those deemed “silly mistakes,” often occur due to stress or unfamiliarity with the content. The most effective approach is to practice with similar questions to prevent these common errors. Take the time to identify potential traps in the questions and write out your stepbystep process for each problem. As you become more familiar with a particular question type, you will naturally avoid miscalculating or misreading parts of the question.
What should I do if I get stuck on a particularly difficult SAT math problem?
Remember that each question carries equal weight in your score, so focus on answering the questions you feel more confident about first. Due to the limited time allocated for each SAT math question (approximately a minute and a half, on average), don’t be afraid to skip questions you don’t know where to begin or feel you’d only be able to answer through a lengthy process. Once you finish the questions you’re more comfortable with, you can return to the challenging questions with less pressure.
If you encounter a challenging question while practicing, be sure to study its explanation and note it for later review (if you’re practicing with UWorld) or discuss it with peers, a tutor, or your math teacher (if you’re studying on your own). When possible, follow up on that study by answering similar questions; in doing so, you’ll apply the lessons you learned from the first explanation and cement those methods into your memory.