Digital SAT Math

Provider · College Board Format · Digital · section-adaptive Chapters · 11 Explorers · 6 interactive

Digital SAT Format & Adaptive Testing Strategy

📌 Overview: Exam Strategy · Digital SAT Section-Adaptive Engine · Module 1 Routing & Hard Module 2 · Question Timing & SPR Grid-In Rules

The Digital SAT Math section is section-adaptive and delivered on the Bluebook application: your performance on Module 1 dynamically selects whether you receive the easier or harder version of Module 2. The built-in Desmos graphing calculator is available throughout.

Level 1 · Foundations — Exam Blueprint & Scoring

Understand the foundational architecture of the Digital SAT: 44 questions across two 35-minute modules, testing four core math domains.

Delivery

digital · adaptive

Calculator

Desmos, all questions

Questions

multiple choice + SPR

Score Scale

200–800

📊 Digital SAT Math — Exam Blueprint

Algebra

~35% SAT

Advanced Math

~35% SAT

Problem-Solving & Data Analysis

~15% SAT

Geometry & Trigonometry

~15% SAT

Source: College Board — SAT Suite of Assessments Math Test Specification

▲ The 4 Core Digital SAT Commandments

1. Always answer what is asked: The SAT frequently solves for $2x$ , $x+y$ , or $b$ , rather than $x$ . Highlight the target quantity.
2. Use regression for constant fitting: Type points into a Desmos table and use y1 ~ mx1 + b or y1 ~ a(x1-h)^2 + k to solve unknown parameters instantaneously.
3. Backsolve from the choices: On single-variable equations with complicated structures, substitute choice B or C to eliminate options rapidly.
4. Mind the reference sheet: Geometry formulas (spheres, cones, pyramids, special right triangles) are always available via the in-app reference icon — never guess a standard volume formula.

Level 1 (Foundations) — Worked Example

Problem: On the 44-question Digital SAT Math section, Algebra (~35%) and Advanced Math (~35%) comprise approximately 70% of the scored questions. How many total questions come from these two domains combined?

Solution: Multiply total questions by combined percentage: $44 \times 0.70 = 30.8 \approx 31\text{ questions}.$ Approximately 31 of 44 questions test algebra and advanced math.

Level 1 · Foundations

The Digital SAT Math section allocates 70 minutes across two modules of equal length. If each module contains 22 questions, what is the exact average time available per question in seconds?

Step 1: Calculate total seconds per module: $35\text{ minutes} \times 60\text{ s/min} = 2100\text{ seconds}.$

Step 2: Divide by the 22 questions in the module: $\frac{2100\text{ seconds}}{22\text{ questions}} = \frac{1050}{11} \approx 95.45\text{ seconds (about 1 min 35 s per question)}.$

Level 2 · Medium — Section-Adaptive Pacing & Question Formats

Module 1 contains a calibrated mix of easy, medium, and hard questions. Pacing and precision dictate whether you unlock the upper score bracket.

⏱️ Digital SAT Math Pacing & Module Routing

The Digital SAT Math section consists of two 35-minute modules of 22 questions each (~95 seconds per question):

Module 1 (Routing): A broad mix of easy, medium, and hard questions. High accuracy here routes you to the harder Module 2, unlocking scores up to 800.
Module 2 (Scoring): Your difficulty tier is locked. Scoring well on the harder module ensures a top percentile score. If routed to the easier module, maximum scores are capped (~600).
Pacing Rule: Never spend more than 2 minutes on a single question in Module 1. Flag it, make an educated guess, and move forward to secure all straightforward points.

⚠️ Digital SAT Traps & Pitfalls — Grid-Ins & Desmos Time Traps

Over-calculating with Desmos: While Desmos is accessible throughout, entering complex equations for basic mental algebra burns valuable time. Use algebraic shortcuts when obvious.
Grid-In Formatting Errors: Student-Produced Responses accept up to 5 characters (including negative sign - and decimal . or fraction slash /). Never enter mixed numbers like 3 1/2 (read as 31/2); enter 7/2 or 3.5. Never round repeating decimals prematurely: 2/3 must be entered as 2/3, .6666, or .6667 (never .66 or .67).
Leaving Questions Blank: There is zero penalty for incorrect guesses. Always fill every single bubble or grid-in before the timer expires.
Calculator Angle Mode: For trigonometry questions, check whether the angle is given in degrees or radians. Desmos defaults to radians; switch to degree mode in the wrench settings when working with degree values.

Level 2 (Medium) — Worked Example

Problem: A student completes the first 16 questions of Module 1 in an average of 75 seconds per question. How many seconds per question remain on average for the final 6 questions?

Solution: Total time is $35 \times 60 = 2100$ seconds. Time spent on first 16: $16 \times 75 = 1200$ seconds. Remaining time: $2100 - 1200 = 900$ seconds. Average time for remaining 6 questions: $\frac{900}{6} = 150\text{ seconds (2 min 30 s per question)}.$

Level 2 · Medium

A student solves a linear equation and obtains the fraction $x = -\frac{7}{4}$ . What are the valid ways to enter this answer into the Digital SAT Student-Produced Response (SPR) grid?

The Digital SAT SPR allows up to 5 characters including negative signs and decimal points:

As a fraction: -7/4 (4 characters).
As a terminating decimal: -1.75 (5 characters).
Invalid: Mixed number -1 3/4 is read as -13/4 and will be graded incorrect.

Level 3 · Advanced — Hard Module 2 Optimization & 750+ Strategy

Securing a 750–800 score requires a near-perfect Module 1 to guarantee placement in the hard Module 2, followed by targeted strategies for multi-step questions.

▲ The 4 Core Digital SAT Commandments for 750+

1. Always answer what is asked: The SAT frequently solves for $2x$ , $x+y$ , or $b$ , rather than $x$ . Highlight the target quantity.
2. Use regression for constant fitting: Type points into a Desmos table and use y1 ~ mx1 + b or y1 ~ a(x1-h)^2 + k to solve unknown parameters instantaneously.
3. Backsolve from the choices: On single-variable equations with complicated structures, substitute choice B or C to eliminate options rapidly.
4. Mind the reference sheet: Geometry formulas (spheres, cones, pyramids, special right triangles) are always available via the in-app reference icon — never guess a standard volume formula.

Level 3 (Advanced) — Worked Example

Problem: Module 1 contains 22 questions, including 2 non-scored pretest/experimental questions. Psychometric analysis indicates that answering at least 15 of 20 operational questions correctly guarantees routing to the hard Module 2. If a student is 100% confident on 14 questions, what is the optimal strategy for the remaining 8 questions?

Solution: Since there is no penalty for guessing, the student should quickly eliminate impossible choices on remaining questions, guess on any remaining items, and verify that all 14 high-confidence questions have zero arithmetic or copy errors to secure the hard Module 2 pathway.

Level 3 · Advanced

In the hard Module 2, questions 18–22 are typically high-difficulty multi-step problems with parameters $k$ . If a student has 7 minutes remaining for the last 3 questions, which approach optimizes score yield: attempting all 3 partially, or using Desmos regression and choice elimination to guarantee complete accuracy on 2 of them?

Optimal Strategy: On the Digital SAT's Item Response Theory (IRT) scoring model, fully correct answers yield substantially higher raw theta weights than rushed guesses. Allocating 3.5 minutes each to 2 questions using Desmos graphing/regression guarantees 2 correct hard items (+40–60 scaled points), while placing a rapid educated guess on the 3rd question preserves time with zero guessing penalty.

📝 Chapter Quiz

Questions Level

Linear Equations in One Variable & Systems of Equations

📌 Unit 1 · Algebra · Linear equations in one variable · Systems of two linear equations in two variables · Systems with zero, one, or infinitely many solutions

Linear equations and systems form the core of Unit 1: solving single-variable equations, modeling relationships with slope and intercept, and evaluating system intersections.

Level 1 · Foundations — Linear Equations in One Variable

Master isolating variables, distributing terms, combining like terms, and solving foundational linear equations.

Slope-intercept

y=mx+b

Point-slope

y-y_1=m(x-x_1)

Slope

\dfrac{y_2-y_1}{x_2-x_1}

System

substitute / eliminate

Level 1 (Foundations) — Worked Example

Solve for $x$ : $3x + 5 = 20.$

Subtract 5 from both sides: $3x = 15$ . Divide by 3: $x = 5.$

Level 1 · Foundations

If $5(2x - 3) = 3(x + 4) + 6$ , what is the value of $7x - 15$ ?

Expand both sides:

10x - 15 = 3x + 12 + 6 \implies 10x - 15 = 3x + 18.

7x = 33 \implies x = \frac{33}{7}.

Substitute into the target expression $7x - 15$ :

7\left(\frac{33}{7}\right) - 15 = 33 - 15 = 18.

Level 2 · Medium — Systems of Two Linear Equations

Solve systems of two linear equations using substitution, elimination, and graph intersection points in Desmos.

The system

y = 2x - 1

and

y = -x + 5

has exactly one unique solution at

(2, 3)

where the two lines intersect.

Level 2 (Medium) — Worked Example

Solve the system: $\begin{cases} x + y = 10 \\ x - y = 2 \end{cases}$

Add the equations together to eliminate $y$ : $2x = 12 \implies x = 6.$

Substitute into the first equation: $6 + y = 10 \implies y = 4$ . Solution is $(6, 4)$ .

Level 2 · Medium

A theater sold a total of 250 adult and student tickets for an evening performance, collecting $2,400. If adult tickets cost $12 each and student tickets cost $8 each, how many student tickets were sold?

Let $a$ be adult tickets and $s$ be student tickets:

\begin{cases} a + s = 250 \\ 12a + 8s = 2400 \end{cases}

Multiply the first equation by 12: $12a + 12s = 3000$ .

Subtract the second equation: $(12a + 12s) - (12a + 8s) = 3000 - 2400 \implies 4s = 600 \implies s = 150.$

The theater sold 150 student tickets.

Level 3 · Advanced — Unknown Constants & Zero / Infinitely Many Solutions

Analyze systems with parameters $k$ and determine exact conditions for parallel lines (no solution) and coincident lines (infinite solutions).

⚠️ SAT Traps & Pitfalls — Systems with Zero or Infinitely Many Solutions

No Solution (Parallel Lines): Occurs when lines have the same slope but different $y$ -intercepts ( $m_1 = m_2, b_1 \neq b_2$ ). In standard form $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ .
Infinitely Many Solutions (Same Line): Occurs when the equations are scalar multiples: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ .
Distributing Negative Signs: In expressions like $-(3x - 4)$ , students often write $-3x - 4$ instead of $-3x + 4$ .
Target Variable Trap: The SAT frequently asks for $2x - 1$ , $x + y$ , or $y/x$ rather than $x$ . Never stop solving until you answer the specific expression requested!

Level 3 (Advanced) — Worked Example

Problem: In the system below, for what value of constant $k$ does the system have infinitely many solutions? $\begin{cases} 4x + 6y = 12 \\ 6x + ky = 18 \end{cases}$

Solution: For infinitely many solutions, the ratio of all coefficients and constants must be equal: $\frac{4}{6} = \frac{6}{k} = \frac{12}{18}.$ Cross-multiply $\frac{2}{3} = \frac{6}{k} \implies 2k = 18 \implies k = 9$ .

Level 3 · Advanced

In the system of equations below, $k$ is a constant. For what value of $k$ does the system have no solution? $\begin{cases} kx - 6y = 15 \\ 4x - 3y = 9 \end{cases}$

For a system of two linear equations in standard form $Ax + By = C$ to have no solution, the lines must be parallel (equal ratios of $x$ - and $y$ -coefficients, but unequal ratio of constants):

\frac{k}{4} = \frac{-6}{-3} \neq \frac{15}{9}

\frac{k}{4} = 2 \implies k = 8.

Check constant ratio: $\frac{15}{9} = \frac{5}{3} \neq 2$ , confirming the lines never intersect. Thus, $k = 8$ .

📝 Chapter Quiz

Questions Level

Linear Functions & Linear Inequalities

📌 Unit 1 · Algebra · Linear functions & rate of change · Linear inequalities in one or two variables · Systems of linear inequalities & feasible regions

A linear function models a constant rate of change; a linear inequality describes an entire half-plane region of solutions.

Level 1 · Foundations — Linear Functions, Slope & Intercepts

Understand slope as rate of change, evaluate linear functions, and identify intercepts on the coordinate plane.

Function

f(x)=mx+b

x-intercept

set

y=0

Parallel

equal slopes

Perpendicular

m_1m_2=-1

⊙ Explorer · Slope & interceptsinteractive

Slope m = 1

Intercept b = 1

slope—

y-intercept—

x-intercept—

The y-intercept is $(0,b)$ ; the x-intercept solves $mx+b=0$ .

Level 1 (Foundations) — Worked Example

If $f(x) = 3x + 1$ , find $f(2) - f(0)$ .

Evaluate $f(2) = 3(2) + 1 = 7$ , and $f(0) = 3(0) + 1 = 1$ .

f(2) - f(0) = 7 - 1 = 6.

Level 1 · Foundations

Find the $x$ -intercept and $y$ -intercept of the linear equation $3x - 6y = 18$ .

$x$ -intercept: Set $y = 0$ : $3x - 6(0) = 18 \implies 3x = 18 \implies x = 6. \text{ Intercept is } (6, 0).$
$y$ -intercept: Set $x = 0$ : $3(0) - 6y = 18 \implies -6y = 18 \implies y = -3. \text{ Intercept is } (0, -3).$

Level 2 · Medium — Contextual Interpretation & Rate of Change Traps

Interpret slope as unit rate and $y$ -intercept as initial condition in real-world scenarios, avoiding scaling pitfalls.

A real-world linear model

C(t) = 25t + 50

where the vertical intercept

(0, 50)

represents a fixed initial fee ($50) and the slope

m = 25

represents the hourly rate ($25/hour).

⚠️ SAT Traps & Pitfalls — Rate of Change & Inequality Half-Planes

Rate of Change Multiplier Trap: When a question states 'the depth increases by $3$ cm for every $10$ minutes', the slope is $\frac{3}{10} = 0.3$ , NOT $3$ . Pay close attention to the denominator units!
Initial Value Confusion: The $y$ -intercept ( $x = 0$ ) represents the starting value before any change occurs. If a word problem measures time $t$ in years since 2010, the year 2015 corresponds to $t = 5$ , NOT $t = 2015$ .
Dashed vs. Solid Boundary Lines: Strict inequalities ( $<$ or $>$ ) require dashed lines (boundary points excluded); inclusive inequalities ( $\le$ or $\ge$ ) use solid lines (boundary included).
Test-Point Method for Inequalities: Always test $(0, 0)$ in the inequality. If $(0,0)$ satisfies the statement, shade the region containing $(0,0)$ ; if false, shade the opposite half-plane.

Level 2 (Medium) — Worked Example

Problem: A plumber charges a flat consultation fee plus an hourly service rate. The total cost is modeled by $C(h) = 45h + 80$ , where $h$ is the hours worked. What do 45 and 80 represent?

Solution: The constant $80$ is the $y$ -intercept $(h=0)$ , representing the initial flat consultation fee ($80). The slope $45$ represents the hourly service rate ($45 per hour).

Level 2 · Medium

An ecologist models the population $P$ of trout in a lake as $P(t) = 1400 - 35t$ , where $t$ is the number of years after 2015. What is the best interpretation of the number $35$ in this model?

The model is in linear form $P(t) = mt + b$ , where $b = 1400$ is the initial population in year 2015, and $m = -35$ is the rate of change. The negative sign indicates a decrease. Therefore, $35$ represents the estimated decrease in the trout population for each one-year increase in time.

Level 3 · Advanced — Inequality Systems & Perpendicular Constraints

Solve systems of linear inequalities, identify feasible region vertices, and compute lines with perpendicular slope constraints.

Level 3 (Advanced) — Worked Example

Problem: Line $L_1$ is given by $2x - 3y = 7$ . Line $L_2$ is perpendicular to $L_1$ and passes through $(4, 1)$ . What is the equation of $L_2$ in slope-intercept form?

Solution: The slope of $L_1$ is $m_1 = \frac{2}{3}$ . The perpendicular slope is the negative reciprocal: $m_2 = -\frac{3}{2}.$ Using point-slope form with $(4, 1)$ : $y - 1 = -\frac{3}{2}(x - 4) \implies y - 1 = -\frac{3}{2}x + 6 \implies y = -\frac{3}{2}x + 7.$

Level 3 · Advanced

Which ordered pair $(x, y)$ is a solution to the system of inequalities $y > 2x - 3$ and $y \le -x + 4$ ?

Test the candidate point $(1, 2)$ :

First inequality: $2 > 2(1) - 3 \implies 2 > -1$ (True).
Second inequality: $2 \le -(1) + 4 \implies 2 \le 3$ (True).

Since $(1, 2)$ satisfies both inequalities simultaneously, it lies within the feasible solution region.

📝 Chapter Quiz

Questions Level

Equivalent Expressions & Quadratic Equations

📌 Unit 2 · Advanced Math · Factoring & equivalent algebraic expressions · Quadratic equations & vertex form · Discriminant & nonlinear system intersections

Unit 2 tests advanced algebra through quadratics: factoring, vertex form, parabolas, and determining real solutions with the discriminant.

Level 1 · Foundations — Factoring, Roots & Parabola Features

Master quadratic factoring, finding zeros/roots, identifying the vertical intercept, and recognizing difference of squares.

Quadratic formula

\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}

Discriminant

b^{2}-4ac

Vertex

x=-\dfrac{b}{2a}

Difference of squares

a^{2}-b^{2}

Parabola

f(x) = x^2 - 4x + 3 = (x - 1)(x - 3) = (x - 2)^2 - 1

showing the vertex

(2, -1)

, axis of symmetry

x = 2

, roots

(1, 0)

and

(3, 0)

, and vertical intercept

(0, 3)

Level 1 (Foundations) — Worked Example

Solve for $x$ : $x^2 - 5x + 6 = 0.$

Factor the quadratic: $(x - 2)(x - 3) = 0$ .

x = 2 \quad\text{or}\quad x = 3.

Level 1 · Foundations

What are all real solutions to the quadratic equation $2x^2 + 7x - 15 = 0$ ?

Factor by grouping (find two numbers whose product is $2(-15) = -30$ and sum is 7: 10 and -3):

2x^2 + 10x - 3x - 15 = 0 \implies 2x(x + 5) - 3(x + 5) = 0 \implies (2x - 3)(x + 5) = 0.

x = \frac{3}{2} \quad\text{or}\quad x = -5.

Level 2 · Medium — Vertex Form, Extreme Values & Vieta's Formulas

Convert standard form to vertex form $y = a(x - h)^2 + k$ , identify maximum/minimum values, and apply Vieta's shortcuts.

⚠️ SAT Traps & Pitfalls — Quadratics & Vertex Form

Minimum/Maximum Value Trap: When asked for the minimum or maximum value of a quadratic, the answer is the $y$ -coordinate of the vertex ( $k$ ), NOT the $x$ -value ( $h$ ) where it occurs.
Vieta's Formula Shortcuts: For $ax^2 + bx + c = 0$ , the sum of roots is $-b/a$ and the product of roots is $c/a$ . Memorizing these saves minutes over factoring or using the quadratic formula!
Discriminant Zero / Tangency: 'Exactly one real solution' or 'intersects at exactly one point' means $b^2 - 4ac = 0$ . 'No real solutions' means $b^2 - 4ac < 0$ .
Vertex Form Signs: In $y = a(x - h)^2 + k$ , the vertex is $(h, k)$ . For $y = 3(x + 5)^2 - 4$ , the vertex is $(-5, -4)$ , NOT $(+5, -4)$ .

Level 2 (Medium) — Worked Example

Find the vertex and state the minimum value of $y = (x - 3)^2 + 2$ .

In vertex form $y = a(x - h)^2 + k$ , the vertex is $(h, k) = (3, 2)$ . Since $a = 1 > 0$ , the parabola opens upward and the minimum value of the function is $y = 2$ .

Level 2 · Medium

What is the sum of the solutions to $3x^2 - 18x + 7 = 0$ , and what is the minimum value of $f(x) = 3x^2 - 18x + 7$ ?

Sum of solutions: By Vieta's formulas, $\text{Sum} = -\frac{b}{a} = -\frac{-18}{3} = 6$ .
Minimum value: The vertex $x$ -coordinate is $h = -\frac{b}{2a} = -\frac{-18}{2(3)} = 3$ . The minimum value is $f(3) = 3(3)^2 - 18(3) + 7 = 27 - 54 + 7 = -20$ .

Level 3 · Advanced — Linear–Quadratic Systems & Tangent Intersections

Solve nonlinear systems of equations, set up discriminant conditions for tangency (one intersection), and evaluate absolute value quadratics.

|f(x)|=k

Two cases:

f(x)=\pm k

|x|<k

-k<x<k

|x|>k

x<{-k}

x>k

No solution

|f(x)|=\text{negative}

Worked example 3.D — Line–parabola system

Find all intersections of $y=x^{2}$ and $y=x+2$ .

Set equal: $x^{2}=x+2\Rightarrow x^{2}-x-2=0\Rightarrow(x-2)(x+1)=0$ .

x=2,\;y=4\qquad\text{and}\qquad x=-1,\;y=1.

Level 3 (Advanced) — Worked Example

Find all points of intersection between $y = x^2$ and $y = x + 2$ .

Set equal: $x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0$ .

x = 2 \implies y = 4 \quad\text{and}\quad x = -1 \implies y = 1.

The intersection points are $(2, 4)$ and $(-1, 1)$ .

Level 3 · Advanced

How many real solutions does the system of equations $y = x^2 - 4x + 4$ and $y = 2x - 3$ have?

Substitute the linear expression into the quadratic:

x^2 - 4x + 4 = 2x - 3 \implies x^2 - 6x + 7 = 0.

Compute the discriminant $\Delta = b^2 - 4ac$ :

\Delta = (-6)^2 - 4(1)(7) = 36 - 28 = 8.

Since $\Delta > 0$ , the quadratic has two distinct real roots. Thus, the system has two real solutions.

📝 Chapter Quiz

Questions Level

Exponentials, Radicals, Polynomials & Rational Expressions

📌 Unit 2 · Advanced Math · Laws of exponents & radical equations · Exponential growth & decay models · Polynomial factors, remainders & rational equations

Master exponential curves, exponent and radical properties, polynomial operations, and solving rational equations.

Level 1 · Foundations — Exponent Rules & Radical Operations

Apply exponent product, quotient, and power rules, simplify radicals, and evaluate fractional exponents.

Product

a^{m}a^{n}=a^{m+n}

Power

(a^{m})^{n}=a^{mn}

Exponential

y=a\cdot b^{x}

Simplify

cancel common factors

Exponential models

y = a \cdot b^x

with

a = 1

. Base

b = 2 > 1

represents rapid exponential growth, while

0 < b = 0.5 < 1

represents asymptotic decay toward

y = 0

Level 1 (Foundations) — Worked Example

Part 1: Solve $2^x = 16$ . Since $16 = 2^4$ , $x = 4$ .

Part 2: Simplify $\dfrac{6x^2}{2x} = 3x$ (for $x \neq 0$ ).

Level 1 · Foundations

Which expression is equivalent to $\frac{x^{5/3}}{x^{2/3}} \cdot \sqrt{x}$ for all positive values of $x$ ?

Step 1: Subtract powers for the quotient: $\frac{x^{5/3}}{x^{2/3}} = x^{5/3 - 2/3} = x^{3/3} = x^1.$

Step 2: Multiply by $\sqrt{x} = x^{1/2}$ : $x^1 \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2} = \sqrt{x^3}.$

Level 2 · Medium — Exponential Growth & Decay Models

Formulate and interpret exponential functions $y = a(b)^x$ , distinguishing growth factors from percentage rates and scaling time exponents.

⊙ Explorer · Exponential modelinteractive

Base b = 1.5

behaviour—

value at x = 1—

Each $+1$ in $x$ multiplies $y$ by $b$ ; the start value $(0,2)$ is amber.

⚠️ SAT Traps & Pitfalls — Exponentials, Polynomials & Radicals

Growth Rate vs. Growth Factor: 'Increases by $12\%$ each year' means the base is $b = 1 + 0.12 = 1.12$ , NOT $0.12$ . Similarly, 'decreases by $12\%$ ' means $b = 1 - 0.12 = 0.88$ .
Time Interval Exponent Scaling: If a population doubles every $4$ hours, the exponent is $t/4$ , yielding $P(t) = P_0(2)^{t/4}$ . Writing $2^{4t}$ means it doubles $4$ times per hour (16-fold increase per hour)!
Extraneous Roots from Squaring: Squaring both sides of a radical equation can introduce false solutions (e.g. $(-3)^2 = 3^2$ ). Always check prospective roots in the original radical equation.
Fractional Exponent Rule: $x^{a/b} = \sqrt[b]{x^a} = (\sqrt[b]{x})^a$ . The denominator is always the root index, and the numerator is the power.

Level 2 (Medium) — Worked Example

Problem: A computer depreciates in value by 18% each year. If its initial purchase price was $1,500, write a function modeling its value $V(t)$ after $t$ years.

Solution: The initial value is $a = 1500$ . An 18% decrease means the annual decay factor is $b = 1 - 0.18 = 0.82$ . The model is: $V(t) = 1500(0.82)^t.$

Level 2 · Medium

A sample of a radioactive substance decays such that its mass is halved every 15 days. If the initial mass is 480 grams, which function models the mass $M(t)$ , in grams, remaining after $t$ days?

The initial mass is $a = 480$ . Halving corresponds to a decay factor of $b = \frac{1}{2}$ . Since halving occurs every 15 days, the exponent must count how many 15-day intervals have elapsed, which is $\frac{t}{15}$ :

M(t) = 480\left(\frac{1}{2}\right)^{\frac{t}{15}}.

Level 3 · Advanced — Radical & Rational Equations with Extraneous Roots

Solve equations involving square roots and rational algebraic fractions, checking domain constraints and identifying extraneous solutions.

▲ Rational Equations

Multiply through by the LCD to clear fractions. Identify domain restrictions (values making any denominator zero) before solving. Discard any solution that equals a restricted value.

Level 3 (Advanced) — Worked Example

Solve for $x$ : $\frac{3}{x - 1} + \frac{1}{x + 1} = \frac{4}{x^2 - 1}.$

Note restrictions: $x \neq \pm 1$ . Multiply both sides by LCD $(x - 1)(x + 1)$ :

3(x + 1) + (x - 1) = 4 \implies 4x + 2 = 4 \implies 4x = 2 \implies x = \frac{1}{2}.

Check restriction: $x = 1/2 \neq \pm 1$ , so the solution is valid.

Level 3 · Advanced

What is the set of all real solutions to the radical equation $\sqrt{2x - 5} = x - 4$ ?

Square both sides:

2x - 5 = (x - 4)^2 = x^2 - 8x + 16 \implies x^2 - 10x + 21 = 0 \implies (x - 3)(x - 7) = 0.

Check prospective roots in original equation:

For $x = 3$ : $\sqrt{2(3) - 5} = \sqrt{1} = 1$ , but right side is $3 - 4 = -1$ . Since $1 \neq -1$ , $x = 3$ is extraneous.
For $x = 7$ : $\sqrt{2(7) - 5} = \sqrt{9} = 3$ , and right side is $7 - 4 = 3$ . Valid!

The only real solution is $x = 7$ .

📝 Chapter Quiz

Questions Level

Nonlinear Functions & Function Notation

📌 Unit 2 · Advanced Math · Function notation & evaluation · Graph transformations: f(x - h) + k · Piecewise & absolute value functions

Evaluating, interpreting, and transforming functions runs through the Advanced Math domain: notation, shifts, composite functions, and piecewise branches.

Level 1 · Foundations — Function Notation, Evaluation, Domain & Range

Master substituting numerical and algebraic inputs into function rules and determining valid domain constraints.

Evaluate

f(a)

Composite

f(g(x))

Shift up

f(x)+k

Shift right

f(x-h)

Level 1 (Foundations) — Worked Example

If $f(x) = x^2 - 1$ , find $f(4)$ .

Substitute $x = 4$ : $f(4) = 4^2 - 1 = 16 - 1 = 15.$

Level 1 · Foundations

For the function $f(x) = 2x^2 - 3x + 5$ , what is the value of $f(-3)$ ?

Substitute $x = -3$ into the expression, taking care with negative signs:

f(-3) = 2(-3)^2 - 3(-3) + 5 = 2(9) + 9 + 5 = 18 + 9 + 5 = 32.

Level 2 · Medium — Graph Transformations & Composite Functions

Understand horizontal shifts $f(x - h)$ , vertical shifts $f(x) + k$ , reflections, and nested evaluations $f(g(x))$ .

The graph of

g(x) = (x - 3)^2 + 2

is obtained by shifting the parent parabola

f(x) = x^2

right by 3 units and up by 2 units, placing its vertex at

(3, 2)

⚠️ SAT Traps & Pitfalls — Function Transformations & Notation

Horizontal Shift Direction: $f(x - c)$ shifts the graph $c$ units to the right, whereas $f(x + c)$ shifts it $c$ units to the left.
Input vs. Output: For $f(x) = 12$ , you are finding the input $x$ that yields 12 (solving $f(x) = 12$ ), NOT calculating $f(12)$ !
Composition Evaluation Order: In $f(g(x))$ , evaluate the innermost expression $g(x)$ first, then pass the resulting value into $f$ .
Undefined Points / Domain Restrictions: Radical functions require non-negative radicands ( $x \ge 0$ ), and rational functions are undefined where the denominator is zero.

Level 2 (Medium) — Worked Example

For $f(x) = 3x - 5$ and $g(x) = x^2 + 2$ , evaluate $g(f(2))$ .

Evaluate inner function: $f(2) = 3(2) - 5 = 1$ .

Substitute into outer function: $g(1) = 1^2 + 2 = 3$ .

Level 2 · Medium

Let $f(x) = x - 4$ and $g(x) = 2x^2 + 1$ . If $h(x) = g(f(x))$ , what is the value of $h(1)$ ?

Step 1: Evaluate inner function at 1: $f(1) = 1 - 4 = -3.$

Step 2: Pass result into $g(x)$ : $g(-3) = 2(-3)^2 + 1 = 2(9) + 1 = 18 + 1 = 19.$

Level 3 · Advanced — Piecewise Functions & Boundary Continuity

Evaluate piecewise-defined functions by identifying the active interval, and determine continuity at domain transitions.

⊙ Explorer · Evaluating a piecewise functioninteractive

evaluate at x: -1

active piecex²

f(x)1.000

This is exactly Worked example 8.B's f(x): x² for x < 2 (indigo), 3x − 1 for x ≥ 2 (amber). Drag x to −1 or 4 to reproduce the worked example's f(−1) = 1 and f(4) = 11, then drag across x = 2 to see the two pieces don't actually meet there.

Level 3 (Advanced) — Worked Example

Let $f(x) = \begin{cases} x^2 & x < 2 \\ 3x - 1 & x \ge 2 \end{cases}$ . Find $f(-1)$ and $f(4)$ .

For $x = -1 < 2$ , use $x^2$ : $(-1)^2 = 1$ .

For $x = 4 \ge 2$ , use $3x - 1$ : $3(4) - 1 = 11$ .

Level 3 · Advanced

For the piecewise function $g(x) = \begin{cases} 2x + 5 & x \le -1 \\ x^2 - 3 & x > -1 \end{cases}$ , what is the value of $g(-1) + g(3)$ ?

Evaluate $g(-1)$ : Since $-1 \le -1$ , use the first piece: $g(-1) = 2(-1) + 5 = 3.$

Evaluate $g(3)$ : Since $3 > -1$ , use the second piece: $g(3) = 3^2 - 3 = 6.$

Sum: $g(-1) + g(3) = 3 + 6 = 9.$

📝 Chapter Quiz

Questions Level

Ratios, Rates, Proportions, Percentages & Scatterplots

📌 Unit 3 · Problem-Solving & Data Analysis · Ratios, rates & unit conversions · Percentages & percent change · Scatterplots & line of best fit

Unit 3 tests contextual reasoning: ratios, rates, multi-step percentages, dimensional analysis, and reading linear trends from data scatterplots.

Level 1 · Foundations — Ratios, Rates, Proportions & Percent of a Number

Convert units, solve proportional relationships, compute unit rates, and calculate percentages.

Percent of

\tfrac{p}{100}\times N

Unit rate

per one unit

Mean

\dfrac{\text{sum}}{\text{count}}

Probability

\dfrac{\text{favourable}}{\text{total}}

Level 1 (Foundations) — Worked Example

Part 1: Find $20\%$ of 150: $0.20 \times 150 = 30.$

Part 2: A $40 item is marked down by $30\%$ . Sale price: $40 \times (1 - 0.30) = 40 \times 0.70 = \$28.$

Level 1 · Foundations

On an architectural blueprint, a scale of 0.5 inches represents an actual distance of 6 feet. What actual distance, in feet, is represented by a line 3.5 inches long on the blueprint?

Set up a proportion where $x$ is the actual distance in feet:

\frac{0.5\text{ in}}{6\text{ ft}} = \frac{3.5\text{ in}}{x\text{ ft}} \implies 0.5x = (3.5)(6) = 21 \implies x = \frac{21}{0.5} = 42\text{ feet}.

Level 2 · Medium — Multi-Step Percentages, Line of Best Fit & Residuals

Model trend lines, calculate residuals ( $\text{Actual } y - \text{Predicted } \hat{y}$ ), and evaluate successive percentage changes.

Scatterplot illustrating positive linear association between study hours and exam performance. Points above the trend line have positive residuals (

y - \hat{y} > 0

); points below have negative residuals.

⊙ Explorer · Line of best fitinteractive

Trend slope = 0.5

sum of squared residuals—

The line pivots through the data's centre; the best-fit slope here is about $0.95$ .

⚠️ SAT Traps & Pitfalls — Rates, Percentages & Data Analysis

Percent Change Base Denominator: $\text{Percent Change} = \frac{\text{New} - \text{Old}}{\text{Old}} \times 100\%$ . The denominator is always the original starting value. If an item drops from $80 to$ 60, the percent decrease is $\frac{20}{80} = 25\%$ , NOT $\frac{20}{60} = 33.3\%$ .
Square and Cubic Unit Conversions: Linear conversions must be raised to the appropriate power. $1\text{ yard} = 3\text{ feet} \implies 1\text{ sq yard} = 3^2 = 9\text{ sq feet}$ .
Residual Sign: $\text{Residual} = \text{Actual } y - \text{Predicted } \hat{y}$ . Points above the line of best fit have positive residuals; points below have negative residuals.
Causation vs. Association: Only randomized experiments can establish cause-and-effect. Observational surveys can only demonstrate correlation/association.

Level 2 (Medium) — Worked Example

Problem: A trend line for study hours $x$ and score $y$ is $\hat{y} = 7x + 45$ . A student studied 5 hours and scored 84. What is the residual?

Solution: Predicted score: $\hat{y} = 7(5) + 45 = 80$ . Actual score: $y = 84$ . $\text{Residual} = y - \hat{y} = 84 - 80 = +4.$

Level 2 · Medium

A retailer reduces the price of a winter jacket by 25%. Two weeks later, the retailer reduces the sale price by an additional 20%. The final price is $96. What was the original price of the jacket before any discounts?

Let $P$ be the original price. After the first 25% discount, the price is $P(1 - 0.25) = 0.75P$ . After the additional 20% discount on the sale price, the multiplier is $(1 - 0.20) = 0.80$ :

(0.80)(0.75)P = 96 \implies 0.60P = 96 \implies P = \frac{96}{0.60} = 160.

The original price was $160.

Level 3 · Advanced — Non-Linear Trends, Dimensional Scaling & Study Design

Analyze non-linear trends in two-variable data, scale square/cubic dimensional units, and distinguish causal inferences from observational associations.

Level 3 (Advanced) — Worked Example

Problem: A container has a volume of $4 \text{ cubic yards}$ . What is the volume of the container in cubic feet?

Solution: Since $1\text{ yd} = 3\text{ ft}$ , the volume conversion factor must be cubed: $1\text{ yd}^3 = (3\text{ ft})^3 = 27\text{ ft}^3.$ $4\text{ yd}^3 \times 27\frac{\text{ft}^3}{\text{yd}^3} = 108\text{ cubic feet}.$

Level 3 · Advanced

A researcher randomly assigns 50 students to a tutoring programme and 50 to a control group, then compares test scores. Which conclusion is best supported if the tutored group scores significantly higher?

(A) Tutoring causes higher scores for the population sampled.
(B) There is an association, but causation cannot be determined.
(C) Results apply to all high school students nationwide.
(D) The sample size is too small to draw any valid conclusion.

Correct Answer: (A).

Because participants were randomly assigned to the treatment and control groups (a randomized controlled experiment), a causal conclusion can be established. Because the sample was drawn from a specific student population, the result generalizes to that sampled population (not nationwide, eliminating C).

📝 Chapter Quiz

Questions Level

One-Variable Statistics, Distributions & Margin of Error

📌 Unit 3 · Problem-Solving & Data Analysis · Mean, median, mode, range & IQR · Standard deviation & distribution shape · Margin of error, sample size & population inferences

Analyze one-variable distributions: center (mean, median), spread (standard deviation, IQR, range), skewness, and survey margins of error.

Level 1 · Foundations — Center, Spread & Box Plots

Calculate mean, median, mode, and range, interpret five-number summary box plots, and evaluate outlier effects.

Mean

\bar x=\dfrac{\sum x_i}{n}

— sensitive to outliers

Median

Middle value (sorted) — resistant

IQR

Q_3-Q_1

— resistant to outliers

Conf. interval

estimate

\pm

margin of error

Box plot displaying Min = 10, Q1 = 25, Median = 40, Q3 = 60, Max = 85 with an outlier at 95. The interquartile range is

\text{IQR} = Q3 - Q1 = 35

Level 1 (Foundations) — Worked Example

Eight salaries (thousands): 30, 32, 35, 36, 37, 38, 40, 120.

Mean $= 368/8 = 46$ . Median $= (36+37)/2 = 36.5$ .

The outlier ($120k) inflates the mean. The median better represents the typical salary.

Level 1 · Foundations

A list of test scores is: 68, 72, 75, 78, 82, 85, 90, 94. What is the interquartile range (IQR) of this dataset?

Step 1: Find the median of the 8 scores (average of 4th and 5th): $(78 + 82)/2 = 80$ .

Step 2: Find $Q_1$ (median of lower half {68, 72, 75, 78}): $(72 + 75)/2 = 73.5$ .

Step 3: Find $Q_3$ (median of upper half {82, 85, 90, 94}): $(85 + 90)/2 = 87.5$ .

Step 4: Compute $\text{IQR} = Q_3 - Q_1 = 87.5 - 73.5 = 14$ .

Level 2 · Medium — Standard Deviation & Distribution Skewness

Compare standard deviation (dispersion from the mean) across distributions and recognize how skewness pulls the mean away from the median.

⊙ Explorer · Mean vs. median under a moving outlierinteractive

eighth salary: 120

mean46.00

median36.50

The seven grey points (30, 32, 35, 36, 37, 38, 40) are fixed; the amber point is the eighth, exactly Worked example 9.A's outlier at its default 120. Drag it down toward the rest of the data and watch the mean (indigo dashed line) fall back in line with the median (amber dashed line) — the mean is sensitive to outliers, the median resists them.

⚠️ SAT Traps & Pitfalls — Statistics, Sampling & Margin of Error

Margin of Error Depends on Sample Size, NOT Population: Increasing the sample size ( $n$ ) reduces the margin of error (proportional to $1/\sqrt{n}$ ). The total population size has negligible impact on margin of error.
Generalization Boundaries: A random sample can only be generalized to the population from which the sample was drawn. A survey of visitors at an athletic store cannot generalize to all city residents.
Outlier Pull: Extreme high outliers pull the mean upward without significantly shifting the median ( $\text{Mean} > \text{Median}$ for right-skewed data). For skewed data, the median and IQR are preferred over the mean and standard deviation.
Standard Deviation Definition: Standard deviation measures the spread or dispersion of values around the mean. Two datasets can have the same mean and range but very different standard deviations if values are clustered tightly vs. spread toward extremes.

Level 2 (Medium) — Worked Example

Set A: {10, 10, 10, 10}. Set B: {4, 8, 12, 16}. Both have mean 10.

Set A has $\text{SD} = 0$ (all values equal the mean). Set B has a substantially larger standard deviation because its values are dispersed farther from 10.

Level 2 · Medium

Two data sets have the same mean of 50 and the same sample size of 100. Dataset A has a standard deviation of 3.2, and Dataset B has a standard deviation of 8.9. Which statement must be true?

(A) Dataset A values are more spread out from the mean than Dataset B values.
(B) Dataset B has greater variability about the mean than Dataset A.
(C) Both datasets have identical minimum and maximum values.
(D) Dataset B has more outliers than Dataset A.

Correct Answer: (B).

Standard deviation measures the typical distance of data values from their mean. A larger standard deviation directly corresponds to greater variability or dispersion about the mean.

Level 3 · Advanced — Margin of Error & Generalization Boundaries

Calculate confidence intervals from survey estimates, understand how sample size $n$ scales margin of error ( $1/\sqrt{n}$ ), and restrict conclusions to sampled populations.

Level 3 (Advanced) — Worked Example

Problem: A poll of 600 randomly selected registered voters finds that 54% support a proposition, with an associated margin of error of $\pm 3\%$ . Can we conclude that a majority of all voters nationwide support the proposition?

Solution: The confidence interval is $54\% \pm 3\% = [51\%, 57\%]$ . While every value in the interval exceeds 50% (indicating a probable majority), the conclusion generalizes only to the specific population of registered voters sampled, not to all voters nationwide unless the sample was drawn from that exact national population.

Level 3 · Advanced

A school administrator randomly surveys 120 high school seniors at Oakwood High School. In the sample, 65% report planning to attend a four-year college, with a margin of error of $\pm 4\%$ . Which of the following conclusions is most statistically sound?

(A) Exactly 65% of all Oakwood High seniors will attend college.
(B) It is plausible that between 61% and 69% of all Oakwood High seniors plan to attend a four-year college.
(C) Between 61% and 69% of all high school seniors in the state plan to attend college.
(D) To reduce the margin of error by half, the administrator should survey 240 seniors.

Correct Answer: (B).

The margin of error creates a plausible confidence interval $65\% \pm 4\% = [61\%, 69\%]$ . Because the random sample was drawn exclusively from Oakwood High seniors, the conclusion generalizes only to that school's senior class (eliminating C). To halve the margin of error, sample size must quadruple ( $4 \times 120 = 480$ ), making D incorrect.

📝 Chapter Quiz

Questions Level

Probability, Conditional Probability & Two-Way Tables

📌 Unit 3 · Problem-Solving & Data Analysis · Simple probability & complement rule · Two-way frequency tables · Conditional probability P(A | B) & independence

Master calculating probabilities from two-way contingency tables: simple probability, restricted sample spaces (given that), and testing for independence.

Level 1 · Foundations — Basic Probability & Two-Way Frequency Tables

Read marginal and joint frequencies from contingency tables, find sample spaces, and apply the complement rule $P(\text{not } A) = 1 - P(A)$ .

Basic probability

\dfrac{\text{favourable}}{\text{total}}

Conditional

P(A\mid B)

\dfrac{n(A\cap B)}{n(B)}

— restrict to B's row/col

Independence test

P(A\mid B)=P(A)

Complement

1-P(A)

Two-Way Frequency Table: Preferred Learning Format
	Online	In-person
Grade 11	48	72
Grade 12	32	48

Two-way contingency table showing survey preferences. The highlighted cell (Grade 11 Online, 48) represents the intersection count used for conditional probability

P(\text{Online} \mid \text{Grade 11}) = 48/120 = 40\%

Level 1 (Foundations) — Worked Example

Using the survey table of 200 students (80 Grade 11, 120 Grade 12; 80 prefer online, 120 prefer in-person):

What is the probability that a randomly chosen student prefers online instruction?

P(\text{Online}) = \frac{\text{Total preferring online}}{\text{Grand total}} = \frac{80}{200} = 0.40\; (40\%).

Level 1 · Foundations

A bag contains 14 red marbles, 18 blue marbles, and 8 green marbles. If one marble is drawn at random, what is the probability that the marble is NOT green?

Step 1: Find the grand total: $14 + 18 + 8 = 40\text{ marbles}.$

Step 2: Use the complement rule: $P(\text{not green}) = 1 - P(\text{green}) = 1 - \frac{8}{40} = 1 - 0.20 = 0.80 = \frac{4}{5}.$

Level 2 · Medium — Conditional Probability & Denominator Restrictions

Compute conditional probabilities $P(A \mid B)$ by restricting the denominator to the row or column specified by 'given that' or 'of those who'.

⚠️ SAT Traps & Pitfalls — Conditional Probability & Tables

Denominator Restriction in 'Given That': The phrase 'given that the student is in Grade 11' restricts the entire sample space to the Grade 11 total row only ($120$). Dividing by the grand total ($200$) is the most common distractor on the SAT!
'Of Which' / 'Of Those' Phrasing: Questions phrasing like 'What fraction of the students who passed were tutored?' restricts the denominator to the passed column total.
Independent vs. Mutually Exclusive: If two events with positive probabilities are mutually exclusive (cannot happen together, $P(A \cap B) = 0$), they can never be independent, because $P(A \mid B) = 0 \neq P(A)$.
'At Least One' Complement Rule: To find $P(\text{at least 1})$, always compute $1 - P(\text{none})$.

Level 2 (Medium) — Worked Example

From the table of 200 students (Grade 11: 48 online, 72 in-person; Grade 12: 32 online, 48 in-person):

Find $P(\text{Online} \mid \text{Grade 11})$ :

The condition restricts the denominator to Grade 11 students only (120 total):

P(\text{Online} \mid \text{Grade 11}) = \frac{48}{120} = 0.40\; (40\%).

Level 2 · Medium

A study of 100 students recorded test results based on exam prep:

• Studied: 54 passed, 6 failed (60 total)
• Did not study: 18 passed, 22 failed (40 total)

What is the probability that a randomly chosen student passed the exam, given that the student studied?

The condition 'given that the student studied' restricts the denominator to the 60 students who studied:

P(\text{Passed} \mid \text{Studied}) = \frac{54}{60} = 0.90 = 90\%.

Level 3 · Advanced — Independence Testing & Inverted Conditional Subsets

Verify event independence using $P(A \mid B) = P(A)$ , and solve complex conditional probabilities involving subset intersections.

Level 3 (Advanced) — Worked Example

Problem: In a clinical trial of 250 patients, 160 received Medication X and 90 received a Placebo. Of those receiving Medication X, 120 reported relief. Of those receiving the Placebo, 30 reported relief. If a patient who reported relief is chosen at random, what is the probability they received Medication X?

Solution: The condition restricts to patients who experienced relief: $\text{Total with relief} = 120 + 30 = 150.$ $P(\text{Medication X} \mid \text{Relief}) = \frac{120}{150} = \frac{4}{5} = 0.80 = 80\%.$

Level 3 · Advanced

Using the student prep table (Studied: 54 passed, 6 failed; Did not study: 18 passed, 22 failed; Total passed: 72 out of 100), are the events 'passed' and 'studied' independent?

Step 1: Calculate marginal probability: $P(\text{Passed}) = \frac{72}{100} = 0.72.$

Step 2: Compare with conditional probability: $P(\text{Passed} \mid \text{Studied}) = \frac{54}{60} = 0.90.$

Since $P(\text{Passed} \mid \text{Studied}) = 0.90 \neq 0.72 = P(\text{Passed})$ , the events are not independent (studying is positively associated with passing).

📝 Chapter Quiz

Questions Level

Area, Volume, Triangles, Circles & Trigonometry

📌 Unit 4 · Geometry & Trigonometry · Area, perimeter & 3D volume formulas · Special right triangles & trigonometry (SOH CAH TOA) · Circle equations, arc length & sector area

Unit 4 covers high-yield spatial geometry: area, perimeter, volume, parallel lines and angles, right triangle trigonometry, and circle equations.

Level 1 · Foundations — Area, Volume Formulas & Pythagorean Theorem

Apply standard 2D and 3D formulas from the College Board reference sheet: prisms, cylinders, cones, spheres, and right triangles.

Right triangle

ABC

with side lengths 3, 4, 5. For angle

\theta

at vertex

A

\sin\theta = \frac{3}{5}

\cos\theta = \frac{4}{5}

, and

\tan\theta = \frac{3}{4}

Triangle area

\tfrac12 bh

Pythagoras

a^{2}+b^{2}=c^{2}

Circle

\pi r^{2},\ 2\pi r

Trig

SOH-CAH-TOA

Worked example 6.B

Find the area of a circle with radius $3$ .

\pi(3)^{2}=9\pi.

Level 1 (Foundations) — Worked Example

Part 1: A right triangle has legs 6 and 8. Hypotenuse: $c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10.$

Part 2: A cylinder has radius 3 and height 5. Volume: $V = \pi r^2 h = \pi(3)^2(5) = 45\pi.$

Level 1 · Foundations

A cone-shaped storage tank has a base radius of 4 feet and a height of 9 feet. What is the volume of the tank in cubic feet, rounded to the nearest tenth?

Using the cone volume formula from the reference sheet:

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (4)^2 (9) = \frac{1}{3}\pi (16)(9) = 48\pi \approx 150.8\text{ cubic feet}.

Level 2 · Medium — Parallel Lines, Special Right Triangles & Circles

Use transversal angle theorems, 30-60-90 / 45-45-90 triangle ratios, basic SOH CAH TOA, and arc length / sector area formulas.

Worked example 6.C — Cylinder volume

A cylinder has radius 3 and height 5.

V=\pi(3)^{2}(5)=45\pi.

Worked example 6.D — Parallel lines

Two parallel lines are cut by a transversal. One angle is 65°. Find its co-interior angle.

180°-65°=115°.

Level 2 (Medium) — Worked Example

A circle has radius $r = 6$ and a central angle $\theta = 120^\circ$ .

\text{Arc length} = \frac{120}{360}(2\pi \cdot 6) = \frac{1}{3}(12\pi) = 4\pi.

\text{Sector area} = \frac{120}{360}(\pi \cdot 6^2) = \frac{1}{3}(36\pi) = 12\pi.

Level 2 · Medium

In a right triangle $ABC$ where angle $C = 90^\circ$ , $\cos(A) = \frac{5}{13}$ . What is the value of $\tan(A)$ ?

In right triangle $ABC$ , $\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{13}$ . Using Pythagoras: $\text{opposite} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12.$ $\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{5} = 2.4.$

Level 3 · Advanced — Circle Equations, Cofunction Identities & Radians

Complete the square to find circle centers and radii, apply cofunction identities $\sin(x) = \cos(90^\circ - x)$ , and scale 3D volumes.

⚠️ SAT Traps & Pitfalls — Geometry & Trigonometry

Cofunction Identity Shortcut: $\sin(x^\circ) = \cos(90^\circ - x^\circ)$ . When an SAT question states $\sin(3x + 10^\circ) = \cos(2x - 5^\circ)$ , set $(3x + 10) + (2x - 5) = 90^\circ \implies 5x + 5 = 90 \implies x = 17^\circ$ .
Circle Equation Center Signs: For $(x - h)^2 + (y - k)^2 = r^2$ , the center is $(h, k)$ . For $(x + 4)^2 + (y - 7)^2 = 25$ , the center is $(-4, 7)$ and the radius is $r = \sqrt{25} = 5$ (not $25$ ).
Radians vs. Degrees in Arc Length: $\text{Arc length} = r\theta$ and $\text{Sector area} = \frac{1}{2}r^2\theta$ require $\theta$ to be in radians! If given degrees, multiply by $\frac{\theta^\circ}{360^\circ}$ .
Similar Figures Scale Factors: If two similar shapes have linear ratio $k$ , their areas scale by $k^2$ and their volumes scale by $k^3$ .

Level 3 (Advanced) — Worked Example

Problem: If $\sin(3x + 10^\circ) = \cos(2x - 5^\circ)$ , what is the value of $x$ ?

Solution: The acute angles of a right triangle are complementary, so $\sin(A) = \cos(B) \iff A + B = 90^\circ$ :

(3x + 10) + (2x - 5) = 90 \implies 5x + 5 = 90 \implies 5x = 85 \implies x = 17^\circ.

Level 3 · Advanced

The equation of a circle in the $xy$ -plane is $x^2 - 8x + y^2 + 6y = 24$ . What are the coordinates of the center and the radius of the circle?

Complete the square for both $x$ and $y$ :

(x^2 - 8x + 16) + (y^2 + 6y + 9) = 24 + 16 + 9

(x - 4)^2 + (y + 3)^2 = 49.

In standard form $(x - h)^2 + (y - k)^2 = r^2$ , the center is $(h, k) = (4, -3)$ and the radius is $r = \sqrt{49} = 7$ .

📝 Chapter Quiz

Questions Level

Desmos Graphing Calculator Tactics

📌 Digital SAT Strategy · Root finding & system intersection points · Parameter sliders for unknown constant k · Table regressions (y1 ~ mx1 + b & y1 ~ ax1^2 + bx1 + c)

The built-in Desmos graphing calculator is the Digital SAT's greatest asset: quickly solve systems, evaluate extreme values, and fit parameters with zero algebraic mistakes.

Level 1 · Foundations — Graphical Root Finding & Intercepts

Graph single equations, click gray dots to immediately identify roots, vertices, and intercepts, and format student-produced response entries.

Graph

type the equation

Intersect

click the crossing point

Table

plot data points

Slider

solve for a parameter

Level 1 (Foundations) — Worked Example

Problem: Find the roots and vertex of $y = x^2 - 4x - 5$ in Desmos.

Solution: Type y = x^2 - 4x - 5. Click the gray dot on the vertex to read $(2, -9)$ . Click the gray dots on the $x$ -axis to read the roots $(-1, 0)$ and $(5, 0)$ .

Level 1 · Foundations

What is the positive integer solution to the equation $2^{x-1} = x + 1$ ?

Solution via Desmos:

Type y = 2^(x - 1) into Line 1.
Type y = x + 1 into Line 2.
Click the intersection point in the first quadrant: Desmos displays $(3, 4)$ . The positive $x$ -solution is $x = 3$ .
Verification: $2^{3-1} = 2^2 = 4$ , and $3 + 1 = 4$ . Exact match!

Level 2 · Medium — Solving Systems & Tangency with Sliders

Graph pairs of equations to find simultaneous intersections, and use parameter sliders to identify values of $k$ creating tangency.

Graphing

y = x^2 - 4x + 3

and

y = 2x - 6

in Desmos. The parabola and line touch at exactly one tangent intersection point

(3, 0)

, providing a rapid graphical solution without factoring.

⚠️ Digital SAT Traps & Pitfalls — Desmos Traps & Limitations

Zoom Boundary Blindness: If an intersection or intercept occurs outside the default $[-10, 10]$ window (e.g. $x = 120$ ), Desmos will appear to show no intersection! Use the wrench tool to expand axis limits to match the problem's scale.
Variable Renaming: Desmos only graphs in terms of $x$ and $y$ . If a problem uses variables $m$ and $n$ , or $p$ and $q$ , immediately substitute $x$ and $y$ .
Exact Roots vs. Decimals: If question choices contain radicals like $\frac{3 \pm \sqrt{17}}{4}$ , calculate the decimal values of the choices in Desmos ( $1.78$ and $-0.28$ ) and match them to the graph's gray intercept dots.
Slider Ranges: When using sliders for 'for what value of $k$ does the line intersect at one point', expand the slider boundaries beyond $[-10, 10]$ if needed.

Level 2 (Medium) — Worked Example

Problem: For what value of constant $k$ is the line $y = 2x + k$ tangent to the parabola $y = x^2$ ?

Solution: Graph y = x^2 and y = 2x + k with a slider for $k$ . Move the slider until the line touches the parabola at exactly one point: $k = -1$ at tangent point $(1, 1)$ . Alternatively, $x^2 - 2x - k = 0 \implies \Delta = 4 + 4k = 0 \implies k = -1$ .

Level 2 · Medium

How many points of intersection exist between the line $y = 2x - 6$ and the parabola $y = x^2 - 4x + 3$ ?

Graph both equations in Desmos. Inspect their crossing: the line touches the parabola at exactly one tangent point $(3, 0)$ . Algebraically, $x^2 - 4x + 3 = 2x - 6 \implies x^2 - 6x + 9 = (x - 3)^2 = 0$ , confirming exactly 1 solution.

Level 3 · Advanced — Table Regressions & Rapid Curve Fitting

Enter coordinate tables and use the tilde (~) regression operator to solve multi-variable parameters without writing systems of equations.

Level 3 (Advanced) — Worked Example

Problem: An exponential model $y = a(b)^x$ passes through $(1, 12)$ and $(3, 108)$ . Find $a$ and $b$ .

Solution via Desmos: Create a table with $x_1 = \{1, 3\}$ and $y_1 = \{12, 108\}$ . Type y1 ~ a*b^x1. Desmos instantly outputs $a = 4$ and $b = 3$ . Function is $y = 4(3)^x$ .

Level 3 · Advanced

A quadratic function $f(x) = ax^2 + bx + c$ passes through the points $(-2, 11)$ , $(0, 3)$ , and $(2, 7)$ . What is the value of $a + b + c$ ?

Solution via Desmos Table Regression:

Click the + button and add a Table. Enter $x_1 = \{-2, 0, 2\}$ and $y_1 = \{11, 3, 7\}$ .
In the next expression line, type the regression template: y1 ~ a*x1^2 + b*x1 + c.
Desmos instantly calculates the parameters: $a = 1$ , $b = -1$ , and $c = 3$ .
Calculate the target value: $a + b + c = 1 + (-1) + 3 = 3$ .

📝 Chapter Quiz

Questions Level

Practice set

Fifty SAT-style problems across all four domains — algebra, advanced math, problem-solving & data, and geometry & trigonometry — tagged by difficulty. Click any problem for a full worked solution.

Solve $4x-3=13$ .

Basic

4x=16,\qquad x=4.

If $y=2x-3$ , find $y$ when $x=4$ .

Basic

2(4)-3=5.

A line has slope 3 and passes through $(0,2)$ . Find $y$ when $x=2$ .

Intermediate

$y=3x+2$ :

3(2)+2=8.

If $2x+y=7$ and $x+y=4$ , find $x$ .

Intermediate

Subtract the equations:

x=3.

Find the slope of the line through $(1,2)$ and $(5,10)$ .

Intermediate

\frac{10-2}{5-1}=2.

Solve $2x+3=4x-7$ .

Intermediate

$10=2x$ :

x=5.

If $5x-2y=10$ and $x=4$ , find $y$ .

Intermediate

$20-2y=10$ :

y=5.

A line is parallel to $y=2x+1$ . What is its slope?

Intermediate

Parallel lines share a slope:

2.

Solve $0.5x+1.5=4$ .

Advanced

$0.5x=2.5$ :

x=5.

Find the x-intercept of $y=2x-6$ .

Intermediate

Set $y=0$ : $2x=6$ :

x=3.

If $f(x)=3x+1$ , find $f(4)$ .

Intermediate

3(4)+1=13.

Solve the system $3x+y=7$ and $x+y=3$ .

Advanced

Subtract: $2x=4 \Rightarrow x=2$ , then $y=1$ :

x=2,\ y=1.

The cost is $C=5n+20$ . Find $C$ when $n=6$ .

Intermediate

5(6)+20=50.

Solve $x^{2}=49$ .

Basic

x=\pm 7.

Factor $x^{2}+7x+12$ .

Intermediate

(x+3)(x+4).

Solve $x^{2}+x-12=0$ .

Intermediate

$(x+4)(x-3)=0$ :

x=3,\ -4.

Find the vertex of $y=(x+1)^{2}-4$ .

Intermediate

(-1,\,-4).

How many real solutions does $x^{2}+2x+5=0$ have?

Advanced

Discriminant $=4-20=-16<0$ :

\text{none.}

Expand $(x+2)^{2}$ .

Intermediate

x^{2}+4x+4.

Solve $2x^{2}=18$ .

Intermediate

$x^{2}=9$ :

x=\pm 3.

Solve $x^{2}+6x+9=0$ .

Advanced

$(x+3)^{2}=0$ :

x=-3.

Find the positive solution of $x^{2}-9=0$ .

Intermediate

x=3.

Find the sum of the roots of $x^{2}-7x+10=0$ .

Advanced

Roots $2$ and $5$ (or $-\tfrac{b}{a}$ ):

7.

Find the x-intercepts of $y=x^{2}-4$ .

Intermediate

x=\pm 2.

Solve $x^{2}-8x+15=0$ .

Intermediate

$(x-3)(x-5)=0$ :

x=3,\ 5.

Find $x^{2}$ when $x=-5$ .

Intermediate

(-5)^{2}=25.

Solve $3^{x}=27$ .

Intermediate

3^{3}=27\Rightarrow x=3.

Simplify $x^{2}\cdot x^{3}$ .

Intermediate

x^{5}.

Simplify $(x^{3})^{2}$ .

Intermediate

x^{6}.

If $2^{x}=32$ , find $x$ .

Advanced

2^{5}=32\Rightarrow x=5.

Simplify $\dfrac{8x^{3}}{2x}$ .

Intermediate

4x^{2}.

A population of $100$ doubles three times. Find the result.

Advanced

$100\times2^{3}$ :

800.

Find $25\%$ of $80$ .

Basic

0.25\times80=20.

Share $40$ in the ratio $3:5$ . Find the larger part.

Intermediate

Eight parts, each $5$ ; larger is 5 parts:

25.

Find the mean of $5,\,10,\,15,\,20$ .

Intermediate

\frac{50}{4}=12.5.

Find the median of $4,\,8,\,15,\,16,\,23$ .

Intermediate

Middle of the ordered list:

15.

What percent of $120$ is $30$ ?

Intermediate

\frac{30}{120}=25\%.

A car travels at $60$ mph for $2.5$ hours. Find the distance.

Intermediate

60\times2.5=150\ \text{mi}.

Increase $80$ by $25\%$ .

Intermediate

80\times1.25=100.

If $4$ pens cost \$6, find the cost of $10$ pens.

Advanced

Each pen costs \$1.50:

10\times1.5=\$15.

A fair die is rolled. Find $P(\text{even})$ .

Intermediate

\frac{3}{6}=\tfrac12.

Find the range of $3,\,7,\,2,\,9,\,5$ .

Intermediate

9-2=7.

A jacket costs \$50 after a $20\%$ discount. Find the original price.

Advanced

\$50 is $80\%$ of the original:

\frac{50}{0.8}=\$62.50.

A car goes $150$ miles on $3$ gallons. Find the miles per gallon.

Intermediate

\frac{150}{3}=50\ \text{mpg}.

Find the area of a rectangle $8\times5$ .

Basic

8\times5=40.

A triangle has angles $30^{\circ}$ and $90^{\circ}$ . Find the third.

Intermediate

180^{\circ}-120^{\circ}=60^{\circ}.

A right triangle has legs $5$ and $12$ . Find the hypotenuse.

Intermediate

\sqrt{25+144}=13.

Find the circumference of a circle with $r=7$ . Take $\pi=\tfrac{22}{7}$ .

Intermediate

2\pi r=2\times\tfrac{22}{7}\times7=44.

Evaluate $\sin 30^{\circ}$ .

Advanced

\tfrac12.

Find the area of a triangle with base $10$ and height $6$ .

Intermediate

\tfrac12\times10\times6=30.

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Digital SAT Math

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Level 1 · Foundations — Exam Blueprint & Scoring

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Level 2 · Medium — Section-Adaptive Pacing & Question Formats

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Linear Equations in One Variable & Systems of Equations

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Level 1 · Foundations — Function Notation, Evaluation, Domain & Range

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Ratios, Rates, Proportions, Percentages & Scatterplots

Level 1 · Foundations — Ratios, Rates, Proportions & Percent of a Number

Level 2 · Medium — Multi-Step Percentages, Line of Best Fit & Residuals

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One-Variable Statistics, Distributions & Margin of Error

Level 1 · Foundations — Center, Spread & Box Plots

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Probability, Conditional Probability & Two-Way Tables

Level 1 · Foundations — Basic Probability & Two-Way Frequency Tables

Level 2 · Medium — Conditional Probability & Denominator Restrictions

Level 3 · Advanced — Independence Testing & Inverted Conditional Subsets

Area, Volume, Triangles, Circles & Trigonometry

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