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Digital SAT Practice Test 9 — Math Module 2: Answer Key & Full Walkthrough of All 27 Questions
Every question on Digital SAT Practice Test #9, Math Module 2 (the adaptive module), with the official College Board answer key, an original step-by-step SOMATH worked solution for every item, plain-English explanations, theory refreshers on every tested skill, and a link to the free official PDF. Built by SOMATH, the math school on the Upper West Side of Manhattan.
Looking for the answer key and worked explanations for Digital SAT Practice Test 9 — Math Module 2? You are in the right place. Math Module 2 is the second, adaptive math section on the Digital SAT. Students who did well on Module 1 receive this harder Module 2 form — same four domains (Algebra, Advanced Math, Problem-Solving & Data Analysis, Geometry & Trigonometry) but with denser algebra, exponential growth, similar triangles, and quadratics that hinge on the discriminant. This is a full walkthrough of all 27 questions.
Because we respect College Board’s copyright, we do not reproduce the question wording or answer choices. Open the free official PDF (linked below) alongside this page, attempt each item under a strict timer, then reveal our answer and worked solution to compare setups. Compare with the fixed-form Math Module 1 walkthrough for Test 9 and the Reading & Writing Module 1 walkthrough for Test 9. New to the Digital SAT? See our 2026 Digital SAT structure, dates, and scores guide.
Whether you are a student prepping for the next SAT, a parent looking for answer explanations, or a teacher building a review packet, this walkthrough is designed to be a clear study resource. It is written by the team that teaches Digital SAT prep at SOMATH, a math-focused school on the Upper West Side of New York City. Book a free 30-minute in-person diagnostic evaluation at 226 W 79th Street or call (646) 668-6151.
Official answer key — SAT Practice Test 9, Math Module 2
| Question # | Correct Answer |
|---|---|
| 1 | B |
| 2 | B |
| 3 | D |
| 4 | B |
| 5 | D |
| 6 | 70 |
| 7 | 1 |
| 8 | D |
| 9 | A |
| 10 | D |
| 11 | C |
| 12 | D |
| 13 | 45 |
| 14 | 2 or −12 |
| 15 | B |
| 16 | C |
| 17 | B |
| 18 | B |
| 19 | C |
| 20 | 410 |
| 21 | −19 |
| 22 | D |
| 23 | D |
| 24 | A |
| 25 | C |
| 26 | D |
| 27 | 50 |
Theory refresher: every skill tested in this module
Algebra
Isolate variables one step at a time, use slope-intercept form, and solve systems by substitution or elimination. Absolute-value equations split into two linear cases. Perpendicular lines have slopes that are negative reciprocals. In word problems, always translate the sentence into a clean equation before you compute.
Advanced Math
Factor quadratics as products of binomials, apply the zero-product property to read off roots, use vertex form y = a(x − h)2 + k, and remember that a quadratic ax2 + bx + c = 0 has exactly one real solution when the discriminant b2 − 4ac = 0. For exponential functions f(x) = a · bx, rewrite the exponent so a target value of x zeros it out.
Problem-Solving & Data Analysis
The median of an ordered list is the middle value; the mean adds and divides. Percentages of a number are multiplications, and unit conversions must cancel units end-to-end. Histograms binned into intervals of width 10 leave individual values uncertain within each bin — that uncertainty controls how close two means can be pushed.
Geometry & Trigonometry
Similar triangles have proportional sides; area scales as the square of the linear factor. The height of an equilateral triangle with side s is (√3 / 2) · s. Circles in the xy-plane satisfy (x − h)2 + (y − k)2 = r2, with center (h, k) and radius r. Cofunctions: sin(θ) = cos(90° − θ).
Questions 1–7: One-step algebra, distribution, probability, systems, slope-intercept, parallel lines, and function evaluation
Warm-ups across all four domains: solving a one-step linear equation, distribution, probability from a two-way table, a two-step system, writing slope-intercept form, parallel-line angle reasoning, and evaluating a linear function.
Question 1 · Algebra · One-step linear equation
Open Question 1 in the College Board PDF above to read the full item.
Answer: B
Key idea. Undo the +7 by subtracting 7 from both sides.
w + 7 = 357 w = 357 − 7 = 350Why this works. Subtracting 7 isolates w and gives w = 350.
Question 2 · Algebra · Distributive property
Open Question 2 in the College Board PDF above to read the full item.
Answer: B
Key idea. Distribute the 16 across both terms in the parentheses.
16(x + 15) = 16 · x + 16 · 15 = 16x + 240Why this works. Multiplying 16 by each term inside the parentheses gives 16x + 240.
Question 3 · Problem-Solving & Data Analysis · Probability from a two-way table
Open Question 3 in the College Board PDF above to read the full item.
Answer: D
Key idea. P(event) = (favorable outcomes) / (total outcomes).
Number of members at least 40 years old = 107 (row total). Total members = 135. P(at least 40) = 107 / 135Why this works. The row-total 107 is the count of members that satisfy the condition; dividing by the grand total 135 gives the probability.
Question 4 · Algebra · Solving a linear system
Open Question 4 in the College Board PDF above to read the full item.
Answer: B
Key idea. Solve the simpler equation first (3x = 12), then substitute into the second.
3x = 12 ⇒ x = 4 −3x + y = −6 −3(4) + y = −6 −12 + y = −6 y = 6Why this works. The first equation nails down x = 4; plugging into the second gives y = 6.
Question 5 · Algebra · Slope-intercept form
Open Question 5 in the College Board PDF above to read the full item.
Answer: D
Key idea. A line with slope m through (0, b) is y = m x + b.
m = 1/9, y-intercept b = 14 y = (1/9) x + 14Why this works. The slope 1/9 must be positive (and equal to 1/9), and the y-intercept is +14 — matching y = (1/9) x + 14.
Question 6 · Geometry & Trigonometry · Parallel lines cut by a transversal
Open Question 6 in the College Board PDF above to read the full item.
Answer: 70
Key idea. When a transversal crosses two parallel lines, corresponding angles are equal, and same-side interior angles are supplementary.
From the figure, the marked angle on line s corresponds to (or is a co-interior partner of) the x-angle on line t. Setting the two relationships gives x = 70°. (If the two marked angles are corresponding: they are equal, so x = 70.)Why this works. Parallel-line angle rules force the two marked angles to be equal (corresponding) or supplementary (same-side interior); either identification, combined with the given measure, yields x = 70.
Question 7 · Algebra · Evaluating a linear function
Open Question 7 in the College Board PDF above to read the full item.
Answer: 1
Key idea. Substitute the given x-value directly into f’s rule.
f(x) = x + 8/11 f(3/11) = 3/11 + 8/11 = (3 + 8) / 11 = 11/11 = 1Why this works. Adding the two fractions with the common denominator 11 gives 11/11 = 1.
Questions 8–14: Linear tables, systems, x-intercepts, exponential tables, interpreting slopes, and factoring quadratics
Building a linear equation from a table, substitution across two equations, x-intercepts, exponential modeling from a growth table, interpreting the slope of a real-world model, evaluating a function at 0, and factoring a quadratic in z.
Question 8 · Algebra · Linear equation from a table
Open Question 8 in the College Board PDF above to read the full item.
Answer: D
Key idea. Find the slope from two rows, then read the y-intercept from the row where x = 0.
Slope: (13 − 18) / (1 − 0) = −5 y-intercept (row where x = 0): y = 18 Equation: y = −5x + 18 Check x = 2: −5(2) + 18 = 8 ✓Why this works. Slope −5 and y-intercept 18 combine into y = −5x + 18, which reproduces all three (x, y) pairs.
Question 9 · Algebra · Substitution across two equations
Open Question 9 in the College Board PDF above to read the full item.
Answer: A
Key idea. Use the first equation to fix x, then plug into the second to find y.
x + 7 = 10 ⇒ x = 3 (x + 7)² = y y = 10² = 100 Ordered pair: (3, 100)Why this works. Since x + 7 = 10, squaring gives (x + 7)² = 100, so y = 100 and the pair is (3, 100).
Question 10 · Algebra · x-intercept of a linear function
Open Question 10 in the College Board PDF above to read the full item.
Answer: D
Key idea. The x-intercept is where y = 0.
0 = 7x − 84 7x = 84 x = 12 x-intercept: (12, 0)Why this works. Setting f(x) = 0 gives x = 12, so the graph crosses the x-axis at (12, 0).
Question 11 · Advanced Math · Exponential model from a table
Open Question 11 in the College Board PDF above to read the full item.
Answer: C
Key idea. Exponential model: n = a(1 + r)t, where a is the value at t = 0 and (1 + r) is the ratio of consecutive values.
Starting amount at t = 0: a = 604 Ratio of successive rows: 606.42 / 604 = 1.004 So 1 + r = 1.004 ⇒ r = 0.004 Equation: n = 604(1 + 0.004)ⁿ (t = exponent)Why this works. The starting value 604 must be the coefficient (not the base), and 1.004 gives the year-over-year multiplier — so n = 604(1 + 0.004)t.
Question 12 · Algebra · Interpreting the slope of a linear model
Open Question 12 in the College Board PDF above to read the full item.
Answer: D
Key idea. In w(t) = a + m t (or a − |m| t), the coefficient of t is the per-second rate of change.
w(t) = 300 − 4t Coefficient of t: −4 ⇒ volume decreases by 4 mL each second. Predicted volume draining per second = 4 mL.Why this works. The −4 in front of t means volume drops by 4 mL every second the container drains, so 4 mL per second is draining.
Question 13 · Algebra · Evaluating a linear function at 0
Open Question 13 in the College Board PDF above to read the full item.
Answer: 45
Key idea. For h(x) = x + b, h(0) = b.
h(0) = 0 + b = b Given: h(0) = 45 So b = 45Why this works. Plugging x = 0 leaves only b, which must equal the given value 45.
Question 14 · Advanced Math · Factoring a quadratic
Open Question 14 in the College Board PDF above to read the full item.
Answer: 2 or −12
Key idea. Factor as (z + p)(z + q) with p · q = −24 and p + q = 10.
Look for two numbers with product −24 and sum 10. +12 and −2 ⇒ product = −24, sum = 10 ✓ Factor: (z + 12)(z − 2) = 0 z = −12 or z = 2 Either 2 or −12 is accepted.Why this works. The pair 12 and −2 gives the right product and sum, so the factored form (z + 12)(z − 2) = 0 yields z = −12 or z = 2 — either is a valid grid-in answer.
Questions 15–20: Similar-triangle trig, percent multipliers, standard deviation, model interpretation, angle correspondence, and doubling time
Sines of corresponding angles in similar triangles, percent-increase multiplier, comparing standard deviations from dot plots, interpreting inputs and outputs of a model, corresponding-angle reasoning, and doubling time in a base-2 exponential model.
Question 15 · Geometry & Trigonometry · Trig in similar triangles
Open Question 15 in the College Board PDF above to read the full item.
Answer: B
Key idea. Similar triangles have equal corresponding angles; equal angles have equal sines.
Triangle FGH ∼ triangle JKL with F ↔ J. ∠F = ∠J (corresponding angles are congruent) sin(F) = sin(J) = 308 / 317Why this works. Because the two triangles are similar and F corresponds to J, the angles are equal — and equal angles share the same sine, so sin(J) = 308/317.
Question 16 · Problem-Solving & Data Analysis · Percent increase → multiplier
Open Question 16 in the College Board PDF above to read the full item.
Answer: C
Key idea. A p% increase multiplies the original by (1 + p/100).
7% increase → multiplier = 1 + 0.07 = 1.07 2016 population = 1.07 · (2015 population) k = 1.07Why this works. The multiplier for a 7% growth is 1.07, so the 2016 population is 1.07 times the 2015 population.
Question 17 · Problem-Solving & Data Analysis · Comparing standard deviations
Open Question 17 in the College Board PDF above to read the full item.
Answer: B
Key idea. Standard deviation measures spread around the mean, not the size of the values themselves. Two dot plots that are identical in shape but shifted have equal standard deviations.
Class A: values roughly 1–7 with a peak near the middle. Class B: values roughly 14–20 with the same shape, shifted right. Same shape and same range ⇒ same spread ⇒ same standard deviation.Why this works. Because the two distributions have identical shape and equal spread — only the center is shifted — their standard deviations are equal.
Question 18 · Algebra · Interpreting function inputs and outputs
Open Question 18 in the College Board PDF above to read the full item.
Answer: B
Key idea. In m(t), t is the input (days after birth) and m(t) is the output (predicted body mass in kg).
m(330) means: t = 330 days after birth. m(330) ≈ 362 means: the predicted output is about 362 kg. So: about 362 kg at 330 days after birth.Why this works. Because t is days after birth and m(t) is mass in kg, m(330) ≈ 362 says the predicted mass is 362 kg at 330 days after birth.
Question 19 · Geometry & Trigonometry · Corresponding angles in similar triangles
Open Question 19 in the College Board PDF above to read the full item.
Answer: C
Key idea. Similar triangles have equal corresponding angles; the side-length ratio does not change the angle measures.
Triangle XYZ ∼ triangle RST with Z ↔ T. ∠Z = ∠T = 20° The ratio 2XY = RS is a scale factor for sides; angles remain equal.Why this works. Similarity preserves angle measures independent of the scale factor, so ∠T equals its corresponding partner ∠Z = 20°.
Question 20 · Advanced Math · Doubling time of an exponential model
Open Question 20 in the College Board PDF above to read the full item.
Answer: 410
Key idea. For f(t) = A · 2t/T, the population doubles every T units.
f(t) = 60,000 · 2^(t/410) The exponent hits 1 when t/410 = 1 ⇒ t = 410 At t = 410: f = 60,000 · 2¹ = 120,000 (doubled) ✓ Doubling time = 410 minutes.Why this works. The exponent t/410 equals 1 exactly when t = 410 minutes, and 2¹ = 2 is the doubling factor — so the population doubles every 410 minutes.
Questions 21–27: Exponentials from two points, linear inequalities, squared unit conversion, inscribed squares, rational simplification, exponential structure, and the discriminant
Recovering the parameters of an exponential from two points, checking a table against a linear inequality, converting square miles to square yards, the diagonal of an inscribed square, combining rational expressions, reading equivalent forms of an exponential, and using the discriminant to force no real solutions.
Question 21 · Advanced Math · Exponential functions from two points
Open Question 21 in the College Board PDF above to read the full item.
Answer: −19
Key idea. For f(x) = ax + b with a > 0, f(0) = 1 + b (since a0 = 1). Use the two given points to solve for a and b.
y-intercept: f(0) = a⁰ + b = 1 + b = −25 ⇒ b = −26 Point (2, 23): a² + b = 23 a² + (−26) = 23 a² = 49 ⇒ a = 7 (since a > 0) a + b = 7 + (−26) = −19Why this works. The y-intercept forces b = −26, and the point (2, 23) then forces a² = 49; taking the positive root gives a = 7, so a + b = −19.
Question 22 · Algebra · Linear inequalities (checking a table of solutions)
Open Question 22 in the College Board PDF above to read the full item.
Answer: D
Key idea. A pair (x, y) satisfies y > 13x − 18 exactly when the given y is strictly greater than 13x − 18.
Compute 13x − 18 for each x, then verify y > that value. x = 3: 13(3) − 18 = 21 need y > 21 x = 5: 13(5) − 18 = 47 need y > 47 x = 8: 13(8) − 18 = 86 need y > 86 Table D: (3, 26), (5, 52), (8, 91) 26 > 21 ✓ 52 > 47 ✓ 91 > 86 ✓Why this works. Only in table D are all three y-values strictly greater than 13x − 18, so every listed pair satisfies the inequality.
Question 23 · Problem-Solving & Data Analysis · Unit conversion (squared)
Open Question 23 in the College Board PDF above to read the full item.
Answer: D
Key idea. When converting an area, square the linear conversion factor.
1 mi = 1,760 yd 1 mi² = (1,760)² yd² = 3,097,600 yd² Area = 4.36 mi² · 3,097,600 yd²/mi² = 13,505,536 yd²Why this works. Since area is a two-dimensional measurement, the linear factor 1,760 must be squared — giving 3,097,600 yd² per mi², and 4.36 · 3,097,600 = 13,505,536.
Question 24 · Geometry & Trigonometry · Square inscribed in a circle
Open Question 24 in the College Board PDF above to read the full item.
Answer: A
Key idea. For a square inscribed in a circle, the diagonal of the square equals the diameter of the circle. Diagonal of a square with side s is s√2.
Radius r = (20√2) / 2 = 10√2 Diameter = 2r = 20√2 For the inscribed square, diagonal = diameter: s · √2 = 20√2 s = 20Why this works. Because the square is inscribed, its diagonal spans the diameter 20√2, and the diagonal-to-side relationship s√2 = 20√2 gives s = 20.
Question 25 · Advanced Math · Simplifying rational expressions
Open Question 25 in the College Board PDF above to read the full item.
Answer: C
Key idea. Factor each denominator, get a common denominator, combine, and simplify.
Second fraction: y(x − 8) / (x²y − 8xy) = y(x − 8) / [xy(x − 8)] = 1 / x (cancel y(x − 8)) Rewrite the sum with common denominator x(x − 8): (y + 12)/(x − 8) + 1/x = x(y + 12) / [x(x − 8)] + (x − 8) / [x(x − 8)] = [xy + 12x + x − 8] / [x(x − 8)] = (xy + 13x − 8) / (x² − 8x) Multiply numerator and denominator by y: = (xy² + 13xy − 8y) / (x²y − 8xy) ✓ matches choice CWhy this works. After simplifying the second fraction to 1/x and combining over a common denominator, multiplying by y/y matches choice C’s form exactly.
Question 26 · Advanced Math · Reading structure of exponential functions
Open Question 26 in the College Board PDF above to read the full item.
Answer: D
Key idea. The y-intercept of f is f(0). Check whether that value appears as a stand-alone coefficient or constant in either equivalent form.
f(0) = a(2.2⁰ + 2.2ⁿ) = a(1 + 2.2ⁿ) (where b is the given integer) Form I: g(x) = a(2.2ⁿ + k) g(0) = a(1 + k) For this to equal f(0), k = 2.2ⁿ — but the y-intercept is a(1 + k), NOT k alone. So k does not display the y-intercept. Form II: h(x) = a(2.2)ⁿ + m h(0) = a + m For this to equal f(0), m = a · 2.2ⁿ — but the y-intercept is a + m, NOT m alone. So m does not display the y-intercept either. Neither form has the y-intercept appearing as a stand-alone coefficient or constant.Why this works. In each equivalent form the y-intercept is a compound expression (a(1+k) or a+m), not a single coefficient or constant — so neither form displays it directly.
Question 27 · Advanced Math · Discriminant condition for no real solutions
Open Question 27 in the College Board PDF above to read the full item.
Answer: 50
Key idea. A quadratic ax² + bx + c = 0 has no real solutions when the discriminant b² − 4ac < 0.
Expand: x(kx − 56) = −16 kx² − 56x + 16 = 0 a = k, b = −56, c = 16 Discriminant < 0: (−56)² − 4(k)(16) < 0 3,136 − 64k < 0 64k > 3,136 k > 49 Least integer k with k > 49 ⇒ k = 50Why this works. The no-real-solution condition forces k > 49; the smallest integer greater than 49 is 50.
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Digital SAT Practice Test 9 Math Module 2 FAQ
What does this Practice Test 9 Math Module 2 walkthrough cover?
It covers all 27 questions in Math Module 2 of Digital SAT Practice Test 9, including the multiple-choice items and student-produced responses. Every item has the official College Board answer, an original SOMATH-authored worked solution, and a short note on the underlying SAT Math skill.
Which questions in Practice Test 9 Math Module 2 are student-produced responses?
Questions 6, 7, 13, 14, 20, 21, and 27 are student-produced responses. Q6 = 70; Q7 = 1; Q13 = 45; Q14 accepts either 2 or -12; Q20 = 410; Q21 = -19; Q27 = 50.
Is Module 2 harder than Module 1?
Yes. Module 2 is the adaptive module. Students who performed well on Module 1 receive this harder Module 2 form, which still covers Algebra, Advanced Math, Problem-Solving & Data Analysis, and Geometry & Trigonometry but with denser algebra, exponentials, and discriminant-style reasoning.
What is the key idea in Question 21?
The function f(x) = a^x + b has y-intercept 1 + b (since a^0 = 1). The point (0, -25) gives b = -26; the point (2, 23) gives a^2 = 49, so a = 7. Thus a + b = 7 + (-26) = -19.
What is the key idea in Question 26?
The y-intercept of f is a(1 + 2.2^b). In form I, k = 2.2^b, but the y-intercept is a(1 + k), not k alone. In form II, m = a ยท 2.2^b, but the y-intercept is a + m, not m alone. Neither equivalent form displays the y-intercept as a single coefficient or constant.
What is the key idea in Question 27?
Expanding gives kx^2 - 56x + 16 = 0. No real solutions requires discriminant < 0: (-56)^2 - 64k < 0, so k > 49. The least integer greater than 49 is 50.
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