March 2026 Digital SAT Practice Test Form C Math Study Notes

Practice Test Administration and Scoring

• General Recommendation: Students are strongly encouraged to set aside time to take the practice test as a primary method of preparation for the Digital SAT.

• Form Identification: The document corresponds to the March Int 2026 Digital SAT Form (C).

Digital SAT Math: Module 1

• Linear Equation Solving (Problem 1):     * Given Equation: $6x = 60$.     * Required: Find the value of the expression $7x + 3$.     * Procedure: Divide both sides by $6$ to find $x = 10$. Substitute into the expression: $7(10) + 3 = 70 + 3 = 73$.     * Correct Option: (A) $73$.

• Systems of Linear Equations (Problem 2):     * Given System:         * Equation 1: $3x + 8y = 93$         * Equation 2: $y = 9$     * Procedure: Substitute $y = 9$ into Equation 1. $3x + 8(9) = 93 \Rightarrow 3x + 72 = 93$. Subtract $72$ from both sides: $3x = 21$. Divide by $3$: $x = 7$.

• Data Interpretation from Tables (Problem 3):     * Context: A sample of $21$ youth softball pitchers' average speeds in miles per hour (mph).     * Distribution:         * $[30, 35)$ mph: $11$ pitchers.         * $[35, 40)$ mph: $5$ pitchers.         * $[40, 45)$ mph: $4$ pitchers.         * $[45, 50)$ mph: $1$ pitcher.     * Requirement: Determine the number of pitchers with speeds of at least $30$ mph but less than $35$ mph.     * Correct Option: (A) $11$.

• Function Evaluation (Problem 4):     * Definition: $f(x) = 5(3x)$.     * Requirement: Evaluate $f(x)$ when $x = 4$.     * Procedure: $f(4) = 5(3 \times 4) = 5(12) = 60$. Note: Looking at transcript options, there may be a typo in the transcript text for $f(x)=5(3^x)$ vs $f(x)=5(3x)$. Calculating for $f(x)=5(3^x)$: $f(4) = 5(3^4) = 5(81) = 405$.     * Correct Option (based on transcript options): (D) $405$.

• Geometry and Parallel Lines (Problem 5):     * Properties: Line $p$ is parallel to line $s$. Line $s$ has a slope of $8$. Line $p$ passes through the point $(0, 7)$.     * Logic: Parallel lines have equal slopes. Therefore, line $p$ has slope $m = 8$. The point $(0, 7)$ is the $y$-intercept ($b = 7$).     * Equation: $y = 8x + 7$.     * Correct Option: (C).

• Saline Solution Mixture Equations (Problem 6):     * Variables: $x$ liters of $4\%$ solution and $y$ liters of $7\%$ solution mixed to get a $6\%$ solution.     * Assumption: Volumes are additive ($x + y$).     * Conversion: Percentage to decimal: $4\% = 0.04$, $7\% = 0.07$, $6\% = 0.06$.     * Equation: $0.04x + 0.07y = 0.06(x + y)$.     * Correct Option: (C).

• Right Triangle Geometry (Problem 7):     * Given: Rectangle diagonal = $\sqrt{145}$. Side 1 = $8$.     * Procedure: Use Pythagorean theorem to find Side 2 ($b$). $8^2 + b^2 = (\sqrt{145})^2 \Rightarrow 64 + b^2 = 145 \Rightarrow b^2 = 81 \Rightarrow b = 9$.     * Perimeter Calculation: $P = 2(\text{length} + ext{width}) = 2(8 + 9) = 2(17) = 34$.     * Correct Option: (B) $34$.

• Combining Like Terms (Problem 8):     * Expression: $7x^3 + 16x^3 - 10x^3 = bx^3$.     * Procedure: Combine coefficients: $7 + 16 - 10 = 13$.     * Result: $b = 13$.

• Solving Quadratics (Problem 9):     * Equation: $x^2 - 8x = 0$.     * Procedure: Factor the equation: $x(x - 8) = 0$. Solutions are $x = 0$ or $x = 8$.     * Correct Option: (B) $8$.

• Geological Mineral Volume Calculation (Problem 10):     * Equation: $3.00x + 4.00y = 98.0$.     * Given: $y = 12.8\,cm^3$.     * Procedure: $3.00x + 4.00(12.8) = 98.0 \Rightarrow 3.00x + 51.2 = 98.0 \Rightarrow 3.00x = 46.8 \Rightarrow x = 15.6\,cm^3$.

• Statistics and Standard Deviation (Problem 11):     * Concept: Standard deviation measures the spread of data points from the mean.     * Data Sets (for constant $p$):         * (A) ${p - 4, p, p, p, p + 4}$         * (B) ${p - 1, p - 1, p, p + 1, p + 1}$         * (C) ${p, p, p, p, p}$ (Zero deviation)         * (D) ${p - 5, p - 4, p, p + 4, p + 5}$     * Conclusion: Set (D) has data points furthest from the mean ($p$), resulting in the largest standard deviation.     * Correct Option: (D).

• Trigonometry (Problem 12):     * Given: Right triangle where the side adjacent to angle $A$ is $22$ and the hypotenuse is $41$.     * Definition: $\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{22}{41}$.     * Correct Option: (C).

• Quadratic Vertex Form Modeling (Problem 13):     * Context: Object launched from $132$ feet. Reaches a maximum height of $276$ feet at $t = 3$ seconds.     * Form: Vertex form is $f(t) = a(t - h)^2 + k$, where $(h, k)$ is the maximum (vertex).     * Values: $(h, k) = (3, 276)$.     * Equation: $f(t) = -16(t - 3)^2 + 276$.     * Correct Option: (D).

• Function Transformations (Problem 14):     * Function: $f(x) = x^3 - 2x^2 - 9x + 4$.     * Transformation: $h(x) = f(x) + 7$.     * Intercept Calculation: The $y$-intercept of $f(x)$ is at $f(0) = 4$. Translating it up $7$ units results in a $y$-intercept of $4 + 7 = 11$.     * Correct Option: (C).

• Angle Conversion (Problem 15):     * Measure of angle $S = \frac{9\pi}{11}$ radians.     * Measure of angle $T = 3 \times S = \frac{27\pi}{11}$.     * Conversion factor: $\frac{180}{\pi}$.     * Expression: $\frac{27\pi}{11} \times \frac{180}{\pi}$.     * Correct Option: (B) (matches the components in transcripts).

• Ratio and Proportions (Problem 16):     * Given: $\frac{x}{y} = \frac{5}{6}$. Also $x = 7t$.     * Substitution: $\frac{7t}{y} = \frac{5}{6}$.     * Solve for $y$: $5y = 42t \Rightarrow y = \frac{42t}{5}$.     * Correct Option: (D).

• Complex Percentage Relationships (Problem 17):     * Equation 1: $m = 0.20(m + q + r)$.     * Equation 2: $q = 0.30(q + r)$.     * Constant: $r = 1,526$.     * Step 1 (Solve for $q$): $q = 0.30q + 0.30(1,526) \Rightarrow 0.70q = 457.8 \Rightarrow q = 654$.     * Step 2 (Solve for $m$): $m = 0.20(m + 654 + 1,526) \Rightarrow m = 0.20(m + 2,180) \Rightarrow m = 0.20m + 436 \Rightarrow 0.80m = 436 \Rightarrow m = 545$.

• Exponential Growth Interpretation (Problem 18):     * Model: $P(t) = 67 \left(\frac{5}{4}\right)^t$.     * Base Interpretation: The base $\frac{5}{4}$ means that for every $4$ units in current size, the size becomes $5$ units in the next time period (multiplication by $1.25$).     * Correct Option: (A).

• Triangle Congruence (Problem 19):     * Setup: Lines $m \parallel n$. Lines $AE$ and $CD$ intersect at $B$. Angles $\angle BAC$ and $\angle BED$ are alternate interior angles. Vertical angles at $B$ are equal.     * Congruence Criteria: To prove $riangle ABC \cong \triangle EBD$ via AAS or ASA, a side length is necessary.     * Property: If $AB = 15$ and $EB = 15$, then a side and angles are established.     * Correct Option: (B).

• Solving Linear Equalities (Problem 20):     * Equation: $11n - 8 = 5n + 16$.     * Procedure: $11n - 5n = 16 + 8 \Rightarrow 6n = 24 \Rightarrow n = 4$.     * Correct Option: (A).

• Real-world Linear Modeling (Problem 21):     * Equation: $y = 30x + 200$ ($y$ is cost, $x$ is shirts).     * Target: Total cost $y = 650$.     * Procedure: $650 = 30x + 200 \Rightarrow 450 = 30x \Rightarrow x = 15$.     * Correct Option: (B).

• Proportional Movement (Problem 22):     * Rate: $0.80\,meters$ per $3.0\,seconds$.     * Total Time: $18\,seconds$.     * Calculation: $( rac{0.80\,m}{3.0\,s}) \times 18\,s = 0.80 \times 6 = 4.8\,meters$.

Digital SAT Math: Module 2

• Scatterplot and Line of Best Fit (Problem 1):     * Visual Observation: The line has a positive $y$-intercept ($+9.5$) and a negative slope (moving downward as $x$ increases).     * Correct Option: (B) $y = 9.5 - 0.4x$.

• Function Coordinate Points (Problem 2):     * Context: Graph passes through $(9, 8)$.     * Meaning: By definition of a function graph, $f(x) = y$, so $f(9) = 8$.     * Correct Option: (D).

• Exponential Doubling (Problem 3):     * Property: $g(x)$ doubles for every $+2$ in $x$.     * Given: $g(4) = 9$.     * Calculation: At $x = 4 + 2 = 6$, value doubles to $9 \times 2 = 18$.

• Linear Model for Sunspots (Problem 4):     * Variable $x$: Months since Dec 2013. Dec 2014 is $x = 12$.     * Graphical analysis: Locate $x = 12$ on the graph and find corresponding $y$-value.     * Estimate: Approximately $71$.     * Correct Option: (B).

• Slope-Intercept Intercepts (Problem 5):     * Given: Slope $m = -\frac{4}{7}$, point $(0, 12)$.     * Equation: $y = -\frac{4}{7}x + 12$.     * Find $x$-intercept: Set $y = 0$. $0 = -\frac{4}{7}x + 12 \Rightarrow \frac{4}{7}x = 12 \Rightarrow 4x = 84 \Rightarrow x = 21$.     * Point: $(21, 0)$.     * Correct Option: (D).

• Linear Inequalities (Problem 6):     * Visual Analysis: Dotted line (indicates strictly $<$ or $>$). Region shaded above the line means $y >$.     * Key components: Intercept at $(0, 1)$, slope is $\frac{1}{4}$.     * Correct Equation: $y > \frac{1}{4}x + 1$.     * Correct Option: (C).

• System of Parallel Lines (Problem 7):     * System: $y = 5x + 18$ and $y = 5x - 18$.     * Observation: Both lines have the same slope ($5$) but different $y$-intercepts ($18$ and $-18$). They are parallel and will never intersect.     * Solutions: Zero.     * Correct Option: (A).

• Trigonometric Identities (Problem 8):     * Given: Right triangle $ABC$. Angles $A$ and $C$ are complementary ($A + C = 90^\circ$).     * Identity: $\cos(A) = \sin(90^\circ - A) = \sin(C)$.     * Result: If $\cos(A) = 0.65$, then $\sin(C) = 0.65$.     * Correct Option: (B).

• Linear Function Solving (Problem 9):     * Function: $f(x) = 3x - 7$.     * Condition: $f(b) = \frac{4}{5}$.     * Procedure: $3b - 7 = 0.8 \Rightarrow 3b = 7.8 \Rightarrow b = 2.6$ (or $\frac{13}{5}$).

• Absolute Value Sum (Problem 10):     * Equation: $|x - 7| + 5 = 9 \Rightarrow |x - 7| = 4$.     * Cases: $x - 7 = 4 \Rightarrow x = 11$; $x - 7 = -4 \Rightarrow x = 3$.     * Sum: $11 + 3 = 14$.     * Correct Option: (A).

• Quadratic Forms for Minimums (Problem 11):     * Function: $f(x) = x^2 - 4x - 780$.     * Requirement: Form showing the minimum value ($y$-coordinate of vertex) as a constant.     * Vertex Form: $f(x) = a(x - h)^2 + k$. Completing the square: $f(x) = (x^2 - 4x + 4) - 4 - 780 = (x - 2)^2 - 784$.     * Correct Option: (B).

• Quadratic Constants Evaluation (Problem 12):     * Equation: $y = 6x^2 + bx + c$.     * Analysis: Root at $x = 2$ and $y$-intercept at $(0, -12)$ from graph visual context. If $c$ is the $y$-intercept, $c = -12$.     * Logic: Using roots and vertex, calculated product $bc = -36$.

• Linearity through Points (Problem 13):     * Points: $(9, 1)$ and $(0, 8)$.     * Slope: $m = \frac{8 - 1}{0 - 9} = -\frac{7}{9}$.     * $Y$-intercept: $b = 8$.     * Equation: $y = -\frac{7}{9}x + 8$.     * Solve for $c$ at $(c, 0)$: $0 = -\frac{7}{9}c + 8 \Rightarrow \frac{7}{9}c = 8 \Rightarrow 7c = 72 \Rightarrow c = \frac{72}{7} \approx 10.29$.

• Sealant Cost Modeling (Problem 14):     * Area: $d\,sq. ft$.     * Job: Two coats ($2d$).     * Coverage: $350\,sq. ft/gal$.     * Gallons needed: $\frac{2d}{350} = \frac{d}{175}$.     * Cost per gallon: $\$19$.     * Total Cost: $C = 19 \times \left(\frac{d}{175}\right)$.     * Correct Option: (C).

• Net Percentage Increase (Problem 15):     * Initial Value ($V$).     * 2012 Value: $V(1 + 1.66) = 2.66V$.     * 2013 Value: $2.66V(1 - 0.14) = 2.66V(0.86) = 2.2876V$.     * Net Change: $2.2876 - 1 = 1.2876 = 128.76\%$.     * Correct Option: (A).

• Quadratic Real Solutions (Problem 16):     * Equation: $4x^2 - px + w = -83 \Rightarrow 4x^2 - px + (w + 83) = 0$.     * Condition: Exactly one real solution occurs when discriminant $D = 0$. $D = b^2 - 4ac = (-p)^2 - 4(4)(w + 83) = 0$.     * Analysis: $p^2 = 16(w + 83)$. Since $p$ is an integer, $(w + 83)$ must be a perfect square. Testing options reveals which $w$ does not fit.     * Correct Option: (C) $36$.

• Unit Conversion Acceleration (Problem 17):     * Value: $12.60\,m/s^2$.     * Conversion facts: $1\,mi = 1,609\,m$; $1\,min = 60\,s$.     * Calculation: $12.60 \times (\frac{1\,mi}{1,609\,m}) \times (\frac{60\,s}{1\,min})^2 = \frac{12.60 \times 3,600}{1,609} \approx \frac{45,360}{1,609} \approx 28.19$.     * Answer (Rounded): $28.2$.

• Saline Mixture Redux (Problem 18):     * Equation for $2\%$ and $7\%$ yielding $6\%$:     * $0.02x + 0.07y = 0.06(x + y)$.     * Correct Option: (C).

• Polynomial Coefficients (Problem 19):     * Expression: $(3x + 5)(7x - 8) = 21x^2 - 24x + 35x - 40 = 21x^2 + 11x - 40$.     * Identify constants: $a = 21, b = 11, c = -40$.     * Calculation: $a + b = 21 + 11 = 32$.

• Quadratic Area Application (Problem 20):     * Dimensions: Length $= x$, Width $= x - 3$.     * Area: $x(x - 3) = 28 \Rightarrow x^2 - 3x - 28 = 0$.     * Factoring: $(x - 7)(x + 4) = 0$.     * Solution: $x = 7$ (negative length impossible).     * Correct Option: (B).

• Inequality Boundaries (Problem 21):     * Condition: $x \le 3y - 27$.     * Given: $y = 8$.     * Calculation: $x \le 3(8) - 27 \Rightarrow x \le 24 - 27 \Rightarrow x \le -3$.     * Greatest value: $-3$.

• Sampling and Margin of Error (Problem 22):     * Population: $8,000$ scales. Sample: $370$ scales.     * Estimate: $9\%$ inaccurate. Margin of error: $2.9\%$.     * Confidence Interval: $9\% \pm 2.9\% = [6.1\%, 11.9\%]$.     * Population Estimate: $0.061 \times 8,000 = 488$ and $0.119 \times 8,000 = 952$.     * Conclusion: It is plausible that between $488$ and $952$ scales are inaccurate.     * Correct Option: (A).