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Digital SAT Practice Test 9 — Math Module 1: Answer Key & Full Walkthrough of All 27 Questions

Every question on Digital SAT Practice Test #9, Math Module 1, 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.

· By the SOMATH team · 226 W 79th St, UWS · (646) 668-6151

Looking for the answer key and worked explanations for Digital SAT Practice Test 9 — Math Module 1? You are in the right place. Module 1 is the fixed-form math section every student sees first — 27 questions across Algebra, Advanced Math, Problem-Solving & Data Analysis, and Geometry & Trigonometry, with a mix of multiple-choice items and grid-in student-produced responses. This SOMATH walkthrough gives you the official College Board answer and a full worked solution for every one of those 27 items.

Because we respect College Board’s copyright, we do not reproduce the question text or answer choices. Open the free official PDF (linked below) alongside this page — then, question by question, attempt the item under a timer, reveal our worked solution, and check where your setup diverged. Compare with our companion Math Module 2 walkthrough for Test 9 and the Reading & Writing Module 1 walkthrough for Test 9 to see how the adaptive Digital SAT scales up. New to the Digital SAT structure? Start with 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.

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Official College Board PDFOpen Digital SAT Practice Test 9 to view every question and passage.
Open PDF →
How to use this walkthrough: Open the official PDF alongside this page. Attempt each question first under a strict timer, then reveal our answer and worked solution and compare your setup. Note any skill tag whose approach felt unfamiliar — that is your study list.

Official answer key — SAT Practice Test 9, Math Module 1

Question #Correct Answer
1B
2C
3B
4A
5A
69
7224
8A
9C
10B
11A
12B
1340
1414
15C
16D
17B
18D
19D
2052
21−3
22B
23D
24A
25D
26B
271,260

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: Perimeter, equivalent equations, rates, quadratics, factoring, mean, and profit

Warm-up items across all four Math domains: perimeter of a triangle, equivalent linear equations, distance-rate-time, evaluating and solving a quadratic, factoring by GCF, mean of a small data set, and a linear profit function.

Question 1 · Geometry & Trigonometry · Perimeter of a triangle

Open Question 1 in the College Board PDF above to read the full item.

Answer: B

Key idea. The three side lengths must add to the perimeter.

Two sides given: 4 cm and 6 cm. Let the third side = s. 4 + 6 + s = 18 s = 18 − 10 = 8 cm

Why this works. Subtracting the two known sides from the perimeter leaves the third side, 8 cm.

Question 2 · Algebra · Equivalent forms of a linear equation

Open Question 2 in the College Board PDF above to read the full item.

Answer: C

Key idea. Isolate the 16x term by subtracting 30 from both sides.

16x + 30 = 190 Subtract 30 from both sides: 16x = 160

Why this works. Both sides lose the +30, giving an equation with the same solution as the original.

Question 3 · Problem-Solving & Data Analysis · Rates (distance = rate × time)

Open Question 3 in the College Board PDF above to read the full item.

Answer: B

Key idea. Time = distance / rate.

Distance to cover ≥ 24 km Rate = 4 km/hr Minimum time = 24 / 4 = 6 hours

Why this works. At 4 km per hour, covering 24 km takes exactly 6 hours; anything less falls short of the goal.

Question 4 · Advanced Math · Evaluating a function / solving a quadratic

Open Question 4 in the College Board PDF above to read the full item.

Answer: A

Key idea. Set the rule equal to 25 and solve for x.

g(x) = x² + 9 = 25 x² = 16 x = ±4 Only +4 appears in the choices.

Why this works. Isolating x² and taking the square root gives x = 4 (the choice that appears).

Question 5 · Advanced Math · Factoring / GCF

Open Question 5 in the College Board PDF above to read the full item.

Answer: A

Key idea. Pull the greatest common factor out of both terms.

9x² + 5x Common factor of both terms: x = x(9x + 5)

Why this works. Distributing x(9x + 5) returns 9x² + 5x, confirming the factorization.

Question 6 · Problem-Solving & Data Analysis · Mean of a data set

Open Question 6 in the College Board PDF above to read the full item.

Answer: 9

Key idea. Mean = (sum of values) / (number of values).

Values: 6, 10, 13, 2, 15, 22, 10, 4, 4, 4 Sum = 6 + 10 + 13 + 2 + 15 + 22 + 10 + 4 + 4 + 4 = 90 Count = 10 Mean = 90 / 10 = 9

Why this works. Adding the ten heights gives 90 cm total; dividing by 10 plants gives an average of 9 cm.

Question 7 · Algebra · Solving a linear equation in context

Open Question 7 in the College Board PDF above to read the full item.

Answer: 224

Key idea. Set the profit function equal to 900 and solve for x.

p(x) = 5x − 220 = 900 5x = 1,120 x = 224

Why this works. 224 posters generate 5(224) = 1,120 dollars in revenue-share; subtracting the 220 in costs leaves the 900-dollar profit.

Questions 8–14: Rate translations, payment plans, literal equations, exponentials, arcs, systems, and frequency tables

Setting up equations from words, isolating a variable, modeling compound growth, working with arcs formed by two diameters, substitution in a small system, and reading a frequency table.

Question 8 · Algebra · Translating a rate word problem

Open Question 8 in the College Board PDF above to read the full item.

Answer: A

Key idea. Distance = rate × time for each activity, then add.

Walking: 3 mph · w hours = 3w miles Running: 5 mph · r hours = 5r miles Combined miles: 3w + 5r = 14

Why this works. Each product is (mph)(hours) = miles, so the two distances add directly to the 14-mile total.

Question 9 · Algebra · Translating a linear word problem

Open Question 9 in the College Board PDF above to read the full item.

Answer: C

Key idea. Total paid = fixed down payment + (payment amount)(number of payments).

Down payment: $37 (one-time, no p) Monthly: 16 dollars per month for p months = 16p Total: 16p + 37 = 165

Why this works. The 37 is a constant (paid once), and 16 multiplies the variable p (paid each month), so the total is 16p + 37 = 165.

Question 10 · Algebra · Isolating a variable in a literal equation

Open Question 10 in the College Board PDF above to read the full item.

Answer: B

Key idea. Add 57 to both sides to isolate y.

y − 57 = px y = px + 57

Why this works. Adding 57 undoes the −57 on the left, leaving y = px + 57.

Question 11 · Advanced Math · Exponential growth (compound interest)

Open Question 11 in the College Board PDF above to read the full item.

Answer: A

Key idea. Compound-growth model: A(t) = P · (1 + r)t, where P is the starting balance and r is the annual rate as a decimal.

Starting balance P = 36,100.00, so the coefficient must be 36,100.00 (eliminates B and C). The base must exceed 1 for the balance to grow (eliminates D, which has base 0.05 < 1). Check A: 36,100 · (1.05)¹³ ≈ 36,100 · 1.8856 ≈ 68,071.93 ✓

Why this works. Only choice A uses the initial deposit as the coefficient and a growth factor greater than 1, and the 13-year value matches the given balance.

Question 12 · Geometry & Trigonometry · Arc lengths from diameters

Open Question 12 in the College Board PDF above to read the full item.

Answer: B

Key idea. The two diameters divide the circle into four arcs; opposite arcs across a diameter are congruent, and adjacent arcs on the same semicircle sum to half the circumference.

Circumference = 144π, so each semicircle = 72π. Let arc PQ = a. Given arc PS = 2 · arc PQ = 2a. Vertical arcs (across the center) are equal: arc QR = arc PS = 2a arc SR = arc PQ = a Adjacent arcs on the same semicircle add to 72π: arc PQ + arc QR = 72π a + 2a = 72π ⇒ 3a = 72π ⇒ a = 24π arc QR = 2a = 48π

Why this works. Because the two diameters create vertical (equal) arc pairs, the four arcs simplify to 2a, a, 2a, a in order — summing to 6a = 144π, so a = 24π and arc QR = 48π.

Question 13 · Algebra · Solving a system by substitution

Open Question 13 in the College Board PDF above to read the full item.

Answer: 40

Key idea. Substitute y = −2x into the second equation and solve for x.

y = −2x 3x + y = 40 3x + (−2x) = 40 x = 40

Why this works. Replacing y with −2x reduces the second equation to x = 40 — a single-step substitution.

Question 14 · Problem-Solving & Data Analysis · Reading a frequency table

Open Question 14 in the College Board PDF above to read the full item.

Answer: 14

Key idea. The maximum data value is the largest value that appears at least once (nonzero frequency).

Scan the table for the largest value with frequency > 0. Value 13 has frequency 0 (does not appear). Value 14 has frequency 6 (appears 6 times). No value larger than 14 is listed with a frequency. Maximum = 14

Why this works. A frequency of 0 means the value never occurs, so 13 is not in the data set; 14, which appears six times, is the largest value present.

Questions 15–20: Pythagorean theorem, word-problem systems, function shifts, rates, parallel lines, and percent

Pythagorean legs, translating and solving a small system, vertical translations of a function, proportional rates, parallel-line (no-solution) conditions, and percent equations.

Question 15 · Geometry & Trigonometry · Pythagorean theorem

Open Question 15 in the College Board PDF above to read the full item.

Answer: C

Key idea. In a right triangle, leg² + leg² = hypotenuse².

leg² + (43.2)² = (196.8)² leg² = 196.8² − 43.2² = 38,730.24 − 1,866.24 = 36,864 leg = √36,864 = 192 mm

Why this works. Subtracting the squared known leg from the squared hypotenuse leaves 36,864, whose square root is 192 mm.

Question 16 · Algebra · Linear word problem / system

Open Question 16 in the College Board PDF above to read the full item.

Answer: D

Key idea. Translate into two equations, then substitute.

x + y = 106 (total length) x = 4y + 6 ("6 more than 4 times y") Substitute into the first: (4y + 6) + y = 106 5y + 6 = 106 5y = 100 ⇒ y = 20 x = 4(20) + 6 = 86

Why this works. Substituting the second equation into the first gives 5y = 100, so y = 20 and x = 86.

Question 17 · Advanced Math · Vertical translations of a function

Open Question 17 in the College Board PDF above to read the full item.

Answer: B

Key idea. The graph of y = f(x) − 3 shifts every y-value of f down by 3.

Roots of f come from (x + 6)(x + 5)(x − 4): f(−6) = 0, f(−5) = 0, f(4) = 0 For y = f(x) − 3, each y drops by 3: x = −6 → y = 0 − 3 = −3 x = −5 → y = 0 − 3 = −3 x = 4 → y = 0 − 3 = −3

Why this works. Since −6, −5, and 4 are roots of f, the shifted function y = f(x) − 3 has value −3 at each of them.

Question 18 · Problem-Solving & Data Analysis · Proportional rates

Open Question 18 in the College Board PDF above to read the full item.

Answer: D

Key idea. Set up a rate (ounces per minute), then multiply by the new time.

Rate = 88x ounces / (5y minutes) = 88x / (5y) oz per min Ounces in 9y min = rate · time = [88x / (5y)] · (9y) = 88x · 9 / 5 = 792x / 5

Why this works. The y-terms cancel because both the given time and the target time are proportional to y, leaving the constant scaling 9/5 on 88x, or 792x/5.

Question 19 · Algebra · Systems with no solution (parallel lines)

Open Question 19 in the College Board PDF above to read the full item.

Answer: D

Key idea. A linear system has no solution when the two lines have the same slope but different intercepts — i.e., the coefficients of x and y are proportional, but the constants are not.

Rewrite both equations in the form Ax + By = C: Eq 1: 4x − 9y = 9y + 5 ⇒ 4x − 18y = 5 Eq 2: hy = 2 + 4x ⇒ −4x + hy = 2 or 4x − hy = −2 For parallel (no solution), match the y-coefficient: −h = −18 ⇒ h = 18 Constants: 5 vs. −2 (different) → lines are parallel but not the same ✓

Why this works. With h = 18 the two equations become 4x − 18y = 5 and 4x − 18y = −2 — same left side, different right side — so they never intersect.

Question 20 · Problem-Solving & Data Analysis · Percent equations

Open Question 20 in the College Board PDF above to read the full item.

Answer: 52

Key idea. "p% of 25" translates to (p/100) · 25 = 13. Solve for p.

(p / 100) · 25 = 13 p / 100 = 13 / 25 p = (13 / 25) · 100 = 52

Why this works. Dividing 13 by 25 gives 0.52; multiplying by 100 converts to a percent, so p = 52.

Questions 21–27: Quadratics, root shifts, linear growth, vertex shifts, two-coat rates, poll scaling, and similar solids

Quadratic factoring, horizontal function shifts and their effect on roots, linear vs. exponential recognition, vertex-of-a-shifted-parabola reasoning, doubling a proportional rate, scaling a poll to a population, and the linear-area-volume ratios for similar solids.

Question 21 · Advanced Math · Quadratic equations (factoring)

Open Question 21 in the College Board PDF above to read the full item.

Answer: −3

Key idea. Expand, move every term to one side, factor, and take the smaller root.

(x − 2)² = 3x + 34 x² − 4x + 4 = 3x + 34 x² − 7x − 30 = 0 Factor: (x − 10)(x + 3) = 0 x = 10 or x = −3 Smallest = −3

Why this works. Expanding and rearranging produces x² − 7x − 30 = 0, which factors cleanly as (x − 10)(x + 3) = 0; the smaller solution is −3.

Question 22 · Advanced Math · Horizontal translations and roots

Open Question 22 in the College Board PDF above to read the full item.

Answer: B

Key idea. If g(x) = f(x − 1), each root of f shifts right by 1 to become a root of g.

Roots of f from (x + 6)(x + 5)(x + 1): x = −6, x = −5, x = −1 For g(x) = f(x − 1), each root shifts by +1: a = −6 + 1 = −5 b = −5 + 1 = −4 c = −1 + 1 = 0 Sum: a + b + c = −5 + (−4) + 0 = −9

Why this works. The substitution x → x − 1 shifts the graph right by 1, so every root moves up by 1; the three new roots sum to −9.

Question 23 · Algebra · Recognizing linear vs. exponential

Open Question 23 in the College Board PDF above to read the full item.

Answer: D

Key idea. "201% of x" means 2.01·x — a first-degree expression, so f is linear with a positive slope.

f(x) = (201/100) · x = 2.01 x Form: f(x) = m x with m = 2.01 > 0 Linear (highest power of x is 1) and slope positive → increasing linear.

Why this works. Multiplying x by a constant is linear (not exponential), and the constant 2.01 is positive, so the function is an increasing linear function.

Question 24 · Advanced Math · Vertex of a quadratic under a horizontal shift

Open Question 24 in the College Board PDF above to read the full item.

Answer: A

Key idea. The vertex of f(x) = ax² + bx + c sits at x = −b/(2a). If g(x) = f(x + 5), g’s vertex is 5 units to the left of f’s.

For f: a = 4, b = 64 Vertex of f at x = −b/(2a) = −64/8 = −8 g(x) = f(x + 5) shifts the graph LEFT by 5: New vertex x-coordinate = −8 − 5 = −13

Why this works. Replacing x with (x + 5) inside f slides the whole parabola 5 units to the left, so its minimum, previously at x = −8, is now at x = −13.

Question 25 · Problem-Solving & Data Analysis · Proportional reasoning (two coats)

Open Question 25 in the College Board PDF above to read the full item.

Answer: D

Key idea. Each coat of stain needs w/170 gallons; two coats need twice that.

One coat: w square feet ÷ 170 sq ft/gal = w/170 gallons Two coats: 2 · (w/170) = 2w/170 = w/85 So S = w/85

Why this works. Doubling the area to be stained (for a second coat) doubles the gallons needed; simplifying 2/170 gives 1/85.

Question 26 · Problem-Solving & Data Analysis · Proportion / sample-to-population

Open Question 26 in the College Board PDF above to read the full item.

Answer: B

Key idea. Use the poll’s vote-share gap as a proportion, then apply it to the full electorate.

Poll margin: 483 − 320 = 163 votes out of 803 polled. Scale to 6,424 voters: (163 / 803) · 6,424 = 163 · (6,424 / 803) = 163 · 8 = 1,304

Why this works. Since 6,424 is exactly 8 × 803, scaling the polled margin of 163 by 8 gives an expected win margin of 1,304 votes.

Question 27 · Geometry & Trigonometry · Similar solids — area & volume scale factors

Open Question 27 in the College Board PDF above to read the full item.

Answer: 1,260

Key idea. For similar solids, area scales as the square of the linear ratio and volume scales as the cube.

Surface-area ratio Y : X = 1,450 / 58 = 25 Linear ratio = √25 = 5 (Y’s edges are 5× X’s) Volume ratio Y : X = 5³ = 125 Vol_Y = 1,250 cm³ ⇒ Vol_X = 1,250 / 125 = 10 cm³ Sum: 10 + 1,250 = 1,260 cm³

Why this works. The area ratio of 25 gives a linear ratio of 5 and therefore a volume ratio of 125, so the smaller prism’s volume is 10 cm³ and the total is 1,260 cm³.

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Digital SAT Practice Test 9 Math Module 1 FAQ

What does this Practice Test 9 Math Module 1 walkthrough cover?

It covers all 27 questions in Math Module 1 of Digital SAT Practice Test 9. Every item has the official College Board answer plus an original SOMATH-written worked solution, along with a short explanation of the underlying SAT Math skill.

Which questions in Practice Test 9 Math Module 1 are student-produced responses?

Questions 6, 7, 13, 14, 20, 21, and 27 are student-produced responses (grid-ins). Q6 = 9; Q7 = 224; Q13 = 40; Q14 = 14; Q20 = 52; Q21 = -3; Q27 = 1260.

Is Module 1 the same difficulty for every student?

Yes. Module 1 is the fixed-form section that every test taker sees. Performance on Module 1 determines whether the student sees the easier or harder version of Module 2, which is the adaptive section.

What is the key idea in Question 11?

The compound-growth model is A(t) = P(1+r)^t, where P is the starting balance. Only the choice with P = 36,100 and a growth factor greater than 1 (1.05) matches; checking 36,100(1.05)^13 gives about 68,072, matching the given balance.

What is the key idea in Question 22?

If g(x) = f(x-1), each root of f shifts right by 1 to become a root of g. The roots of f are -6, -5, -1, so the roots of g are -5, -4, 0, and their sum is -9.

What is the key idea in Question 27?

For similar solids, area scales as the square of the linear ratio and volume scales as the cube. A surface-area ratio of 25 gives a linear ratio of 5 and volume ratio of 125, so the smaller prism has volume 10 and the total is 1,260 cubic centimeters.

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