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Digital SAT Practice Test 4 — Module 2: Full Walkthrough of All 27 Math Questions with Answers & Explanations

Every question on Digital SAT Practice Test #4, Math Module 2, transcribed verbatim, with the official College Board answer key, a step-by-step worked solution, a plain-English explanation for every choice, theory refreshers on every tested skill, and a free PDF download. 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 answers, explanations, and full walkthrough of Digital SAT Practice Test 4 — Math Module 2? You are in the right place. This is a complete, question-by-question walkthrough of SAT Test #4, Math Module 2 — every one of the 27 questions transcribed verbatim, with the official College Board answer for each item and a plain-English worked solution showing exactly which SAT-Math tool each question wants and how to apply it. Theory refreshers on every skill category are included, plus a free PDF download of the entire test for offline studying.

Module 2 is the second Math module on the Digital SAT, and unlike Module 1, it is adaptive: the specific set of 27 questions you see depends on how you performed on Module 1. Students who do well on Module 1 are routed into a harder Module 2 (with access to the full 200–800 scoring range); students who do not are routed into an easier Module 2 that typically tops out in the mid-600s. The version reproduced here — from the official College Board Practice Test 4 booklet — is the higher, harder form, which makes it an especially good stress test for students aiming at 700+.

Whether you are a student prepping for the next SAT test date, a parent looking for answer explanations, or a teacher building a review packet, this walkthrough is designed to be the most complete, honest, and student-friendly Digital SAT Math walkthrough on the internet. Written by the same team that teaches Digital SAT prep at SOMATH, a math-focused school on the Upper West Side of New York City run by Northwestern-trained cofounder Marcelo Ambrozio and Harvard-trained cofounder Vivianne Wright.

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Download SAT Practice Test 4 (Full PDF) The complete official College Board digital SAT Practice Test 4 booklet, including Math Module 2. Ready to print.
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How to use this set. Read each question carefully, cover the choices, work it out on paper before you click “Show answer & explanation.” The point is not to read solutions — it is to build the reflex of recognizing which SAT-Math tool the question wants (substitution, completing the square, discriminant, right-triangle trig, etc.) before you touch the calculator. If you get stuck for more than 90 seconds, peek at just the correct answer, close the reveal, and try to reconstruct the worked solution on your own.

Sections

  1. Answer key at a glance
  2. Theory refresher · What Math Module 2 tests
  3. Section 1 · Algebra & Linear Systems (Q1–7)
  4. Section 2 · Advanced Math & Functions (Q8–9, Q12–14, Q17–19, Q21)
  5. Section 3 · Problem-Solving & Data Analysis (Q10–11, Q13, Q15, Q23–24, Q27)
  6. Section 4 · Geometry & Trigonometry (Q16, Q20, Q25–26)
  7. How SOMATH prepares NYC students for the Digital SAT
  8. Digital SAT FAQ

Answer key at a glance

All 27 official College Board answers for Practice Test 4, Math Module 2:

QAnswerQAnswerQAnswer
1B) 199410A) Between 0.45 and 0.5319A) Min height 3 in.
2B) 2.911A) 342015/17
3C) 7x³12B) 9/42151
4A) (15, 3)13.3 or 3/1022A) Zero
5A) x>0, y>014223C) 36
615 or −515A) 7,50024C) Median greater, range equal
75016C) 4125D) x²+(y+1)²=49
8B) Table B17B) 1/726B) 8
9D) 27018D) −3/227600

Theory refresher · What Math Module 2 tests

The Digital SAT Math section has two 35-minute modules of 22 questions each (20 scored + 2 pretest). Module 2 is adaptive — the exact set of questions depends on Module 1 performance. The Bluebook Desmos calculator is available on every single question. Roughly 75% of items are multiple choice (four options each) and 25% are student-produced responses (fill-in-the-blank numeric or fractional answers). The four content domains are:

1 · Algebra

Linear systems by substitution. When one equation already isolates a variable (like 5y = x), substitute directly into the other equation rather than doing full elimination.

Absolute value equations. |x − a| = b splits into two linear equations: x − a = b and x − a = −b. Always solve both branches.

Systems of inequalities graphically. The solution region for x > 0, y > 0 is Quadrant I; x > 0, y < 0 is Quadrant IV; x < 0, y > 0 is Quadrant II; x < 0, y < 0 is Quadrant III.

Linear equations with no solution. An equation in one variable has no solution when, after simplifying, it reduces to a false numeric statement (like 0 = 28) — this happens when the coefficient of the variable becomes exactly 0.

Systems of three linear equations. A third line intersects a two-line system's solution point only if that point actually lies on the third line. If it doesn't, the three-equation system has zero solutions.

2 · Advanced Math

Quadratic equations by completing the square or factoring. Both work; factoring is faster when the numbers cooperate. Always check which root the question actually wants (positive, negative, etc.).

Vertex of a parabola. For f(x) = ax² + bx + c, the vertex's x-coordinate is −b/(2a). In factored form f(x) = (x−r₁)(x−r₂), the vertex x-coordinate is the midpoint of the two roots: (r₁+r₂)/2.

Vertex form and minimum/maximum. y = a(x−h)² + k has vertex (h, k). If a > 0 the parabola opens upward and k is the minimum value; if a < 0 it opens downward and k is the maximum.

Discriminant. For ax² + bx + c = 0, the discriminant is b² − 4ac. No real solution ⇒ discriminant < 0. Exactly one real solution ⇒ discriminant = 0. Two real solutions ⇒ discriminant > 0.

Exponential decay and percent change. f(x) = a(r)^(x/k) decreases by 100(1−r)% every k units of x. Watch the exponent's denominator carefully — it changes the time unit of the percent change.

Slope of parallel lines. Parallel lines have identical slopes. If f(x) = mx + b, any line parallel to y = f(x) also has slope m.

3 · Problem-Solving & Data Analysis

Margin of error. A sample estimate with margin of error E gives a plausible range of [estimate − E, estimate + E]. Never conclude the true value is "exactly" the point estimate.

Two-way tables & probability. P(event) = (favorable outcomes)/(total outcomes), read straight from row/column totals.

Median and range under a shift. If every value in a data set is increased by a constant c, the median increases by c too, but the range (max − min) stays exactly the same, since both the max and min shift by the same amount.

Percent "greater than" language. "A is p% greater than B" means A = (1 + p/100)·B. Solve for p by isolating it, not by computing a simple percent difference.

Unit conversion. Multiply by the conversion factor written as a fraction so units cancel: 5,104 yd × (1 mi / 1,760 yd).

4 · Geometry & Trigonometry

Vertical angles & same-side interior angles. Vertical angles (opposite rays at an intersection) are always equal. When a transversal crosses two parallel lines, same-side interior angles are supplementary (sum to 180°).

Right-triangle trig (co-function identity). In a right triangle, the two acute angles are complementary, so cos(one acute angle) = sin(the other acute angle). If you know cos(K) as a ratio of sides, you can find cos(L) using the Pythagorean theorem on the same right triangle.

Equation of a circle under translation. (x−h)² + (y−k)² = r² has center (h, k) and radius r. Shifting the circle by (Δx, Δy) moves the center to (h+Δx, k+Δy) without changing r.

Surface area of composite solids. When two solids are glued along a shared face, the combined surface area equals the sum of their individual surface areas minus twice the area of the glued face (since that face disappears from both solids).

Every question below is transcribed verbatim from College Board’s Practice Test 4 booklet. Try each one on paper, then click Show answer & explanation to see the correct answer, the full worked solution, and a plain-English explanation for every choice.

Section 1 · Algebra & Linear Systems

Linear systems, absolute value equations, and inequality-region graphs — the opening stretch of the module.

Question 1 · Two-Variable Data (Reading a Line Graph)

See figure in the original College Board PDF above.

The line graph shows the estimated number of chipmunks in a state park on April 1 of each year from 1989 to 1999. Based on the line graph, in which year was the estimated number of chipmunks in the state park the greatest?

  • A) 1989
  • B) 1994
  • C) 1995
  • D) 1998

Answer: B) 1994

The estimated number of chipmunks is on the vertical axis. The greatest estimated number of chipmunks is the point with the greatest height on the graph. Reading the graph, the peak (about 150) occurs at 1994 — higher than every other labeled year.

✗ A) 1989
1989 is one of the lowest points on the graph, roughly 50 — not the greatest.

✓ B) 1994
1994 is the tallest point on the line graph, at roughly 150 estimated chipmunks — higher than any other year shown.

✗ C) 1995
1995 dips back down after the 1994 peak — not the greatest year.

✗ D) 1998
1998 is a smaller secondary peak, lower than the 1994 high point.

Question 2 · Ratios/Rates/Proportions (Unit Conversion)

A fish swam a distance of 5,104 yards. How far did the fish swim, in miles? (1 mile = 1,760 yards)

  • A) 0.3
  • B) 2.9
  • C) 3,344
  • D) 6,864

Answer: B) 2.9

Convert yards to miles using the conversion factor (1 mile / 1,760 yards): 5,104 yards × (1 mile / 1,760 yards) = 5,104 / 1,760 miles ≈ 2.9 miles.

✗ A) 0.3
0.3 ≈ 1,760/5,104 — this inverts the conversion, dividing yards by miles instead of the reverse.

✓ B) 2.9
5,104 ÷ 1,760 ≈ 2.9 miles.

✗ C) 3,344
3,344 = 5,104 − 1,760. Subtracting instead of dividing.

✗ D) 6,864
6,864 = 5,104 + 1,760. Adding instead of dividing.

Question 3 · Equivalent Expressions (Like Terms)

Which expression is equivalent to 12x³ − 5x³?

  • A) 7x⁶
  • B) 17x³
  • C) 7x³
  • D) 17x⁶

Answer: C) 7x³

12x³ and 5x³ are like terms (same variable, same exponent). Subtract the coefficients: 12x³ − 5x³ = (12 − 5)x³ = 7x³.

✗ A) 7x⁶
7 is the right coefficient, but the exponent stays 3 when subtracting like terms — it does not become 6.

✗ B) 17x³
17x³ = 12x³ + 5x³ — this adds the terms instead of subtracting them.

✓ C) 7x³
Subtracting the coefficients of the like terms gives (12 − 5)x³ = 7x³.

✗ D) 17x⁶
Adds the coefficients (17) and also incorrectly changes the exponent to 6.

Question 4 · Linear Systems

x + y = 18 and 5y = x. What is the solution (x, y) to the given system of equations?

  • A) (15, 3)
  • B) (16, 2)
  • C) (17, 1)
  • D) (18, 0)

Answer: A) (15, 3)

The second equation gives x = 5y directly. Substitute x = 5y into the first equation: 5y + y = 18 6y = 18 y = 3. Substitute back: x = 5(3) = 15. Solution: (x, y) = (15, 3).

✓ A) (15, 3)
Substituting x = 5y into x + y = 18 gives 6y = 18, so y = 3 and x = 5(3) = 15.

✗ B) (16, 2)
Check the second equation: 5(2) = 10 ≠ 16. Fails.

✗ C) (17, 1)
Check the second equation: 5(1) = 5 ≠ 17. Fails.

✗ D) (18, 0)
Check the second equation: 5(0) = 0 ≠ 18. Fails.

Question 5 · Linear Inequalities (System, Graphical Region)

The point (8, 2) in the xy-plane is a solution to which of the following systems of inequalities?

  • A) x > 0 and y > 0
  • B) x > 0 and y < 0
  • C) x < 0 and y > 0
  • D) x < 0 and y < 0

Answer: A) x > 0 and y > 0

The point (8, 2) has a positive x-coordinate and a positive y-coordinate, so it lies in Quadrant I. Check choice A: x > 0 → 8 > 0 ✓; y > 0 → 2 > 0 ✓. Both true. This system (x > 0, y > 0) represents exactly Quadrant I.

✓ A) x > 0 and y > 0
Substituting (8, 2): 8 > 0 is true and 2 > 0 is true. This system represents Quadrant I, which contains (8, 2).

✗ B) x > 0 and y < 0
This represents Quadrant IV. y = 2 is not less than 0, so (8, 2) fails this system.

✗ C) x < 0 and y > 0
This represents Quadrant II. x = 8 is not less than 0, so (8, 2) fails this system.

✗ D) x < 0 and y < 0
This represents Quadrant III. Neither coordinate of (8, 2) is negative, so this fails.

Question 6 · Nonlinear Equations in 1 Var (Absolute Value)

|x − 5| = 10. What is one possible solution to the given equation?

Student-produced response — enter as a fraction or decimal.

Answer: 15 or −5

By the definition of absolute value, |x − 5| = 10 means: x − 5 = 10 or x − 5 = −10. First branch: x − 5 = 10 → x = 15. Second branch: x − 5 = −10 → x = −5. Either 15 or −5 is an acceptable answer.

Why this works.
Absolute value equations split into two linear cases: x − 5 = 10 gives x = 15, and x − 5 = −10 gives x = −5. Both satisfy the original equation.

Question 7 · Linear Functions (Evaluating in Context)

f(x) = 7x + 1. The function gives the total number of people on a company retreat with x managers. What is the total number of people on a company retreat with 7 managers?

Student-produced response — enter as a fraction or decimal.

Answer: 50

Substitute x = 7 into f(x) = 7x + 1: f(7) = 7(7) + 1 = 49 + 1 = 50.

Why this works.
Substituting x = 7 gives f(7) = 7(7) + 1 = 50, so there are 50 total people on the retreat with 7 managers.

Section 2 · Advanced Math & Functions

Function tables, exponential evaluation, completing the square, quadratics, discriminants, and vertex form.

Question 8 · Nonlinear Functions (Table of Values)

h(x) = x² − 3. Which table gives three values of x and their corresponding values of h(x) for the given function h?

See figure in the original College Board PDF above.

  • A) Table A: x = 1, 2, 3 → h(x) = 4, 5, 6
  • B) Table B: x = 1, 2, 3 → h(x) = −2, 1, 6
  • C) Table C: x = 1, 2, 3 → h(x) = −1, 1, 3
  • D) Table D: x = 1, 2, 3 → h(x) = −2, 1, 3

Answer: B) Table B: x = 1, 2, 3 → h(x) = −2, 1, 6

Substitute each x-value into h(x) = x² − 3. h(1) = 1² − 3 = 1 − 3 = −2. h(2) = 2² − 3 = 4 − 3 = 1. h(3) = 3² − 3 = 9 − 3 = 6. So the correct table is x = 1, 2, 3 → h(x) = −2, 1, 6.

✗ A) Table A: h(x) = 4, 5, 6
This is the table for h(x) = x + 3, not h(x) = x² − 3.

✓ B) Table B: h(x) = −2, 1, 6
h(1) = −2, h(2) = 1, h(3) = 6 — exactly matches x² − 3 for each x.

✗ C) Table C: h(x) = −1, 1, 3
This is the table for h(x) = 2x − 3, not h(x) = x² − 3.

✗ D) Table D: h(x) = −2, 1, 3
The first two values match (−2, 1) but the third value should be 6, not 3 — a calculation error at x = 3.

Question 9 · Nonlinear Functions (Exponential, Evaluating at 0)

The function f is defined by f(x) = 270(0.1)ˣ. What is the value of f(0)?

  • A) 0
  • B) 1
  • C) 27
  • D) 270

Answer: D) 270

Substitute x = 0 into f(x) = 270(0.1)ˣ. f(0) = 270(0.1)⁰ = 270(1) [any nonzero number to the 0 power is 1] = 270.

✗ A) 0
0 would be the value of the input x, not the output f(x).

✗ B) 1
1 is the value of (0.1)⁰ alone — forgets to multiply by the coefficient 270.

✗ C) 27
27 = 270(0.1)¹ — this is f(1), not f(0).

✓ D) 270
(0.1)⁰ = 1, so f(0) = 270 × 1 = 270.

Question 12 · Nonlinear Equations in 1 Var (Completing the Square)

−4x² − 7x = −36. What is the positive solution to the given equation?

  • A) 7/4
  • B) 9/4
  • C) 4
  • D) 7

Answer: B) 9/4

Move everything to one side: −4x² − 7x = −36 0 = 4x² + 7x − 36 (multiply both sides by −1) Factor: 4x² + 16x − 9x − 36 = 0 4x(x + 4) − 9(x + 4) = 0 (4x − 9)(x + 4) = 0. So 4x − 9 = 0 → x = 9/4, or x + 4 = 0 → x = −4. The two solutions are 9/4 and −4. The positive one is 9/4.

✗ A) 7/4
Substituting x = 7/4 into the original equation does not satisfy it — not an actual root.

✓ B) 9/4
Factoring 4x² + 7x − 36 gives (4x − 9)(x + 4) = 0, so x = 9/4 or x = −4. The positive solution is 9/4.

✗ C) 4
Substituting x = 4 into the original equation gives −4(16) − 7(4) = −64 − 28 = −92 ≠ −36. Not a root.

✗ D) 7
Substituting x = 7 gives −4(49) − 7(7) = −196 − 49 = −245 ≠ −36. Not a root.

Question 14 · Linear Functions (Slope of Parallel Line)

f(x) = 2x + 3. For the given function f, the graph of y = f(x) in the xy-plane is parallel to line j. What is the slope of line j?

Student-produced response — enter as a fraction or decimal.

Answer: 2

In y = f(x) = 2x + 3, the equation is in slope-intercept form y = mx + b with m = 2. So the graph of y = f(x) has slope 2. Parallel lines have equal slopes, so line j also has slope 2.

Why this works.
The slope of y = 2x + 3 is 2. Since line j is parallel to this line, and parallel lines share the same slope, line j also has slope 2.

Question 17 · Linear Equations in 1 Var (No Solution)

−3x + 21px = 84. In the given equation, p is a constant. The equation has no solution. What is the value of p?

  • A) 0
  • B) 1/7
  • C) 4/3
  • D) 4

Answer: B) 1/7

Factor the left-hand side: −3x + 21px = −3x(1 − 7p). So the equation is −3x(1 − 7p) = 84. Divide both sides by −3: x(1 − 7p) = −28. Divide both sides by (1 − 7p): x = −28 / (1 − 7p). This equation has NO solution exactly when the denominator is 0 (division by zero → no valid x): 1 − 7p = 0 7p = 1 p = 1/7.

✗ A) 0
If p = 0, the equation becomes −3x = 84, which has the solution x = −28 — a valid solution, not "no solution."

✓ B) 1/7
When p = 1/7, the coefficient (1 − 7p) becomes 0, turning the equation into 0 = −28 (false for any x), so there is no solution.

✗ C) 4/3
Plugging in p = 4/3 leaves a nonzero coefficient on x, so the equation still has a valid solution.

✗ D) 4
Plugging in p = 4 leaves a nonzero coefficient on x, so the equation still has a valid solution.

Question 18 · Nonlinear Functions (Vertex from Factored Form)

The function f is defined by f(x) = (x − 10)(x + 13). For what value of x does f(x) reach its minimum?

  • A) −130
  • B) −13
  • C) −23/2
  • D) −3/2

Answer: D) −3/2

Expand: f(x) = (x − 10)(x + 13) = x² + 3x − 130. This is an upward-opening parabola (coefficient of x² is positive), so it has a minimum at its vertex. Vertex x-coordinate = −b/(2a) = −3/(2·1) = −3/2. Alternate approach: the x-intercepts are x = 10 and x = −13. The vertex x-coordinate is the midpoint of the roots: (10 + (−13)) / 2 = −3/2.

✗ A) −130
−130 is the y-intercept (constant term) of f(x) = x² + 3x − 130, not the x-coordinate of the vertex.

✗ B) −13
−13 is one of the x-intercepts (roots) of f, not the vertex x-coordinate.

✗ C) −23/2
This does not match −b/(2a) = −3/2; likely a sign or arithmetic slip.

✓ D) −3/2
The vertex x-coordinate is −b/(2a) = −3/2, confirmed by taking the midpoint of the roots 10 and −13: (10−13)/2 = −3/2.

Question 19 · Nonlinear Functions (Vertex Form, Interpretation)

The function f(x) = (1/9)(x − 7)² + 3 gives a metal ball's height above the ground f(x), in inches, x seconds after it started moving on a track, where 0 ≤ x ≤ 10. Which of the following is the best interpretation of the vertex of the graph of y = f(x) in the xy-plane?

  • A) The metal ball's minimum height was 3 inches above the ground.
  • B) The metal ball's minimum height was 7 inches above the ground.
  • C) The metal ball's height was 3 inches above the ground when it started moving.
  • D) The metal ball's height was 7 inches above the ground when it started moving.

Answer: A) The metal ball's minimum height was 3 inches above the ground.

f(x) = (1/9)(x − 7)² + 3 is in vertex form y = a(x − h)² + k with a = 1/9, h = 7, k = 3. Since a = 1/9 > 0, the parabola opens upward, so the vertex is the minimum point. Vertex: (h, k) = (7, 3). In context, x = time (seconds) and f(x) = height (inches). The vertex means: at x = 7 seconds, the height reaches its minimum value of 3 inches.

✓ A) The metal ball's minimum height was 3 inches above the ground.
The vertex is (7, 3); since the parabola opens upward, the y-coordinate 3 is the minimum value of f(x) — the ball's minimum height.

✗ B) The metal ball's minimum height was 7 inches above the ground.
7 is the x-coordinate of the vertex (the time in seconds), not the height. The minimum height is 3, not 7.

✗ C) The metal ball's height was 3 inches above the ground when it started moving.
"When it started moving" means x = 0, but the vertex occurs at x = 7, not x = 0.

✗ D) The metal ball's height was 7 inches above the ground when it started moving.
This confuses the vertex's x-coordinate (7 seconds) with a height at x = 0, and neither matches "7 inches at start."

Question 21 · Nonlinear Equations in 1 Var (Discriminant)

−x² + bx − 676 = 0. In the given equation, b is a positive integer. The equation has no real solution. What is the greatest possible value of b?

Student-produced response — enter as a fraction or decimal.

Answer: 51

For ax² + bx + c = 0, the discriminant is b² − 4ac. No real solution ⟺ discriminant < 0. Here a = −1, c = −676. Discriminant = b² − 4(−1)(−676) = b² − 2,704. No real solution requires: b² − 2,704 < 0 b² < 2,704 |b| < √2,704 = 52. So b < 52. Since b is a positive integer, the greatest possible value is 51.

Why this works.
The discriminant b² − 4(−1)(−676) = b² − 2,704 must be negative for no real solution, giving b < 52. The greatest positive integer less than 52 is 51.

Section 3 · Problem-Solving & Data Analysis

Margin of error, budget inequalities, two-way tables, medians and ranges, and percent-change algebra.

Question 10 · Inference from Sample Statistics

To estimate the proportion of a population that has a certain characteristic, a random sample was selected from the population. Based on the sample, it is estimated that the proportion of the population that has the characteristic is 0.49, with an associated margin of error of 0.04. Based on this estimate and margin of error, which of the following is the most appropriate conclusion about the proportion of the population that has the characteristic?

  • A) It is plausible that the proportion is between 0.45 and 0.53.
  • B) It is plausible that the proportion is less than 0.45.
  • C) The proportion is exactly 0.49.
  • D) It is plausible that the proportion is greater than 0.53.

Answer: A) It is plausible that the proportion is between 0.45 and 0.53.

The plausible range for the true proportion = estimate ± margin of error. Estimate = 0.49, margin of error = 0.04. Lower bound: 0.49 − 0.04 = 0.45. Upper bound: 0.49 + 0.04 = 0.53. Plausible range: [0.45, 0.53].

✓ A) It is plausible that the proportion is between 0.45 and 0.53.
Subtracting and adding the margin of error to the estimate gives exactly the interval 0.45 to 0.53.

✗ B) It is plausible that the proportion is less than 0.45.
0.45 is the lower bound of the plausible range — values below it are outside the plausible interval.

✗ C) The proportion is exactly 0.49.
0.49 is only the point estimate. With a nonzero margin of error, the appropriate conclusion is a plausible range, not a single exact value.

✗ D) It is plausible that the proportion is greater than 0.53.
0.53 is the upper bound of the plausible range — values above it are outside the plausible interval.

Question 11 · Linear Inequalities (Weight Constraint)

A moving truck can tow a trailer if the combined weight of the trailer and the boxes it contains is no more than 4,600 pounds. What is the maximum number of boxes this truck can tow in a trailer with a weight of 500 pounds if each box weighs 120 pounds?

  • A) 34
  • B) 35
  • C) 38
  • D) 39

Answer: A) 34

Let b = number of boxes. Combined weight: 500 + 120b ≤ 4,600. Subtract 500: 120b ≤ 4,100. Divide by 120: b ≤ 4,100/120 = 205/6 ≈ 34.17. Since b must be a whole number, the maximum is 34.

✓ A) 34
b ≤ 34.17, and since the number of boxes must be a whole number, the greatest possible value is 34.

✗ B) 35
500 + 120(35) = 4,700 pounds, which exceeds the 4,600-pound limit.

✗ C) 38
500 + 120(38) = 5,060 pounds — far over the limit.

✗ D) 39
500 + 120(39) = 5,180 pounds — far over the limit.

Question 13 · Probability (Two-Way Table)

The table summarizes the distribution of color and shape for 100 tiles of equal area.

RedBlueYellowTotal
Square10202555
Pentagon20101545
Total303040100

If one of these tiles is selected at random, what is the probability of selecting a red tile? (Express your answer as a decimal or fraction, not as a percent.)

Student-produced response — enter as a fraction or decimal.

Answer: .3 or 3/10

Total tiles = 100 (the total number of possible outcomes). Total red tiles = 30 (from the "Total" row under "Red"). P(red) = (favorable outcomes) / (total outcomes) = 30/100 = 3/10 = 0.3.

Why this works.
There are 30 red tiles out of 100 total tiles, so the probability of selecting a red tile is 30/100 = 3/10 = 0.3.

Question 15 · Linear Equations in 1 Var (Word Problem)

A proposal for a new library was included on an election ballot. A radio show stated that 3 times as many people voted in favor of the proposal as people who voted against it. A social media post reported that 15,000 more people voted in favor of the proposal than voted against it. Based on these data, how many people voted against the proposal?

  • A) 7,500
  • B) 15,000
  • C) 22,500
  • D) 45,000

Answer: A) 7,500

Let x = number of people who voted against the proposal. Then 3x = number of people who voted in favor. "15,000 more voted in favor than against" means: 3x − x = 15,000 2x = 15,000 x = 7,500.

✓ A) 7,500
Solving 2x = 15,000 gives x = 7,500 — the number who voted against the proposal.

✗ B) 15,000
15,000 is the difference between the "for" and "against" vote counts, not the number who voted against.

✗ C) 22,500
22,500 = 3(7,500) is the number who voted in favor of the proposal, not against.

✗ D) 45,000
45,000 = 3 × 15,000 — this multiplies incorrectly rather than solving 2x = 15,000.

Question 23 · Percentages (Exponential Decay Rate)

f(x) = 5,470(0.64)^(x/12). The function f gives the value, in dollars, of a certain piece of equipment after x months of use. If the value of the equipment decreases each year by p% of its value the preceding year, what is the value of p?

  • A) 4
  • B) 5
  • C) 36
  • D) 64

Answer: C) 36

For f(x) = a(r)^(x/k), the value decreases by 100(1 − r)% every k units of x, when r < 1. Here a = 5,470, r = 0.64, k = 12 (months). Percent decrease every 12 months = 100(1 − 0.64)% = 100(0.36)% = 36%. Since 12 months = 1 year, the value decreases each year by 36% of its value the preceding year. So p = 36.

✗ A) 4
4 does not match 100(1 − 0.64) = 36. Likely confuses the growth factor's complement incorrectly.

✗ B) 5
5 does not correspond to any direct computation from r = 0.64 and k = 12.

✓ C) 36
Every 12 months (one year), the value multiplies by 0.64, a decrease of 100(1 − 0.64)% = 36%.

✗ D) 64
64 is just 100r — this treats r itself as the percent decrease, forgetting to subtract from 100 first.

Question 24 · One-Variable Data (Median & Range Under a Shift)

See figure (dot plot) in the original College Board PDF above.

The dot plot represents the 15 values in data set A. Data set B is created by adding 56 to each of the values in data set A. Which of the following correctly compares the medians and the ranges of data sets A and B?

  • A) The median of data set B is equal to the median of data set A, and the range of data set B is equal to the range of data set A.
  • B) The median of data set B is equal to the median of data set A, and the range of data set B is greater than the range of data set A.
  • C) The median of data set B is greater than the median of data set A, and the range of data set B is equal to the range of data set A.
  • D) The median of data set B is greater than the median of data set A, and the range of data set B is greater than the range of data set A.

Answer: C) The median of data set B is greater than the median of data set A, and the range of data set B is equal to the range of data set A.

Data set A (from the dot plot, 15 values in ascending order): the 8th value (the middle one) is the median, which is 23. Maximum value in A = 26, minimum = 22 → range of A = 26 − 22 = 4. Data set B = every value in A plus 56. Median of B = median of A + 56 = 23 + 56 = 79 → greater than median of A (23). Max of B = 26 + 56 = 82. Min of B = 22 + 56 = 78. Range of B = 82 − 78 = 4 → same as range of A. So: median of B > median of A, and range of B = range of A.

✗ A) Median equal, range equal.
Adding 56 to every value shifts the median up by 56 — the median of B is NOT equal to A's median.

✗ B) Median equal, range greater.
Both parts are wrong: the median does shift up, and adding a constant to every value never changes the range.

✓ C) Median greater, range equal.
Adding 56 to all 15 values shifts the median from 23 to 79 (greater), while the range stays 4 since max and min both shift by the same 56.

✗ D) Median greater, range greater.
The median comparison is correct, but the range does NOT increase — adding a constant to every value preserves the spread (range) exactly.

Question 27 · Percentages (Percent Greater Than)

210 is p% greater than 30. What is the value of p?

Student-produced response — enter as a fraction or decimal.

Answer: 600

"A is p% greater than B" means A = (1 + p/100)·B. 210 = (1 + p/100)(30). Divide both sides by 30: 7 = 1 + p/100. Subtract 1: 6 = p/100. Multiply by 100: p = 600.

Why this works.
Setting up 210 = (1 + p/100)(30) and solving gives p/100 = 6, so p = 600 — meaning 210 is 600% greater than 30 (i.e., 210 is 7 times 30).

Section 4 · Geometry & Trigonometry

Parallel lines and transversals, right-triangle trigonometry, systems and graphs, circle translations, and composite surface area.

Question 16 · Lines/Angles/Triangles (Parallel Lines & Transversal)

See figure in the original College Board PDF above.

In the figure, lines m and n are parallel. If x = 6k + 13 and y = 8k − 29, what is the value of z?

  • A) 3
  • B) 21
  • C) 41
  • D) 139

Answer: C) 41

Lines t (transversal) and m intersect, creating vertical angles x° and y°. Vertical angles are equal: x = y. 6k + 13 = 8k − 29 13 = 2k − 29 42 = 2k k = 21. Lines m and n are parallel, cut by transversal t. The angles y° and z° are same-side interior angles → supplementary: y + z = 180. y = 8k − 29 = 8(21) − 29 = 168 − 29 = 139. 139 + z = 180 z = 41.

✗ A) 3
3 does not match any correctly computed angle or k value.

✗ B) 21
21 is the value of k, not z — a common mix-up between the solved variable and the requested angle.

✓ C) 41
Using vertical angles to find k = 21, then y = 139°, and since y and z are supplementary same-side interior angles, z = 180 − 139 = 41.

✗ D) 139
139 is the value of angle y (or x), not z. z is the supplement of y, not y itself.

Question 20 · Right Triangles & Trig (Co-function via Pythagorean Theorem)

In triangle JKL, cos(K) = 24/51 and angle J is a right angle. What is the value of cos(L)?

Student-produced response — enter as a fraction or decimal.

Answer: 15/17 or .8824 or .8823

Angle J is the right angle, so the hypotenuse is side KL (opposite J). cos(K) = adjacent/hypotenuse = 24/51 = 8/17 (simplified). So JK (adjacent to K) = 8n and KL (hypotenuse) = 17n for some constant n. Pythagorean theorem: JK² + JL² = KL². (8n)² + JL² = (17n)² 64n² + JL² = 289n² JL² = 225n² JL = 15n. cos(L) = adjacent to L / hypotenuse = JL / KL = 15n / 17n = 15/17.

Why this works.
Since J is the right angle, KL is the hypotenuse. cos(K) = 24/51 = 8/17 gives leg ratios 8:17, and the Pythagorean theorem gives the third side as 15. So cos(L) = 15/17 ≈ 0.8824.

Question 22 · Linear Systems (Graphical, Adding a Third Line)

See figure (graph) in the original College Board PDF above.

If a new graph of three linear equations is created using the system of equations shown and the equation x + 4y = −16, how many solutions (x, y) will the resulting system of three equations have?

  • A) Zero
  • B) Exactly one
  • C) Exactly two
  • D) Infinitely many

Answer: A) Zero

A solution to a system of equations must satisfy every equation in the system, so it must lie on every line's graph. The two lines shown intersect at exactly one point: (8, 2). For the 3-equation system to have a solution, (8, 2) must also lie on the new line x + 4y = −16. Check: substitute x = 8, y = 2 into x + 4y = −16: 8 + 4(2) = 8 + 8 = 16. 16 ≠ −16 → false. Since (8, 2) does NOT satisfy the third equation, and it was the only candidate solution, the three-equation system has zero solutions.

✓ A) Zero
The only point common to the original two lines is (8, 2), and it fails to satisfy x + 4y = −16 (16 ≠ −16), so no point satisfies all three equations.

✗ B) Exactly one
This would require (8, 2) to also lie on the third line, but it does not.

✗ C) Exactly two
A system of linear equations (straight lines) can never have exactly two solutions — lines either meet at zero, one, or infinitely many shared points.

✗ D) Infinitely many
Infinitely many would require all three lines to be identical, which is not the case here.

Question 25 · Circles (Translation)

The equation x² + (y − 1)² = 49 represents circle A. Circle B is obtained by shifting circle A down 2 units in the xy-plane. Which of the following equations represents circle B?

  • A) (x − 2)² + (y − 1)² = 49
  • B) x² + (y − 3)² = 49
  • C) (x + 2)² + (y − 1)² = 49
  • D) x² + (y + 1)² = 49

Answer: D) x² + (y + 1)² = 49

Circle A: x² + (y − 1)² = 49 → center (0, 1), radius 7 (since 49 = 7²). Shifting a circle down 2 units moves the center from (h, k) to (h, k − 2), radius unchanged. New center: (0, 1 − 2) = (0, −1). New equation: (x − 0)² + (y − (−1))² = 49 = x² + (y + 1)² = 49.

✗ A) (x − 2)² + (y − 1)² = 49
This shifts circle A to the right by 2 units, not down.

✗ B) x² + (y − 3)² = 49
This shifts circle A up by 2 units (center becomes (0, 3)), the opposite of "down."

✗ C) (x + 2)² + (y − 1)² = 49
This shifts circle A to the left by 2 units, not down.

✓ D) x² + (y + 1)² = 49
Shifting the center from (0, 1) down 2 units gives (0, −1), so the new equation is x² + (y + 1)² = 49.

Question 26 · Area & Volume (Composite Surface Area)

Two identical rectangular prisms each have a height of 90 centimeters (cm). The base of each prism is a square, and the surface area of each prism is K cm². If the prisms are glued together along a square base, the resulting prism has a surface area of (92/47)K cm². What is the side length, in cm, of each square base?

  • A) 4
  • B) 8
  • C) 9
  • D) 16

Answer: B) 8

Let x = side length of each square base. Surface area of one prism: K = 2x² (two square bases) + 4(90x) (four lateral faces) = 2x² + 360x. Gluing two identical prisms along one square base removes that shared face from BOTH prisms' total surface: Combined surface area = 2K − 2x². It's given that combined surface area = (92/47)K, so: 2K − 2x² = (92/47)K. Multiply everything by 47 to clear the fraction: 94K − 94x² = 92K 2K = 94x² K = 47x². Now substitute K = 2x² + 360x: 2x² + 360x = 47x² 0 = 45x² − 360x 0 = 45x(x − 8). So x = 0 (rejected — a prism can't have side length 0) or x = 8. Side length of each square base = 8 cm.

✗ A) 4
4 does not satisfy 45x² − 360x = 0 (45(16) − 360(4) = 720 − 1,440 ≠ 0).

✓ B) 8
Setting K = 2x² + 360x equal to 47x² (from the combined-surface-area condition) and solving 45x² − 360x = 0 gives x = 8 (rejecting x = 0).

✗ C) 9
9 does not satisfy the equation 45x² − 360x = 0.

✗ D) 16
16 does not satisfy the equation 45x² − 360x = 0.

How SOMATH prepares NYC students for the Digital SAT

SOMATH (School of Math) is a math-focused school on the Upper West Side of Manhattan, cofounded by Marcelo Ambrozio (Northwestern-trained lead math teacher) and Vivianne Wright (Harvard-trained, also runs MBA House, one of the largest international GMAT/EA prep firms). Vivianne’s track record with standardized tests — hundreds of admits to Harvard Business School, Stanford GSB, Wharton, MIT Sloan, and INSEAD — is the same rigor we bring to SAT prep for high schoolers.

Our Digital SAT track:

Manhattan families: If your student is targeting a 1500+ Digital SAT score for Ivy League, MIT, Stanford, or top-15 admissions, book a free diagnostic evaluation or call (646) 668-6151. We are two blocks from the 79th Street 1 train and three blocks from the B/C at the American Museum of Natural History.

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SOMATH course · Grades 9–12

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Digital SAT FAQ

What is on Module 2 of the Digital SAT Math section?

Module 2 is a 35-minute, 20 scored + 2 pretest question module. Unlike Module 1, Module 2 is stage-adaptive: the difficulty of the questions you see depends on your Module 1 performance. It draws from the same four content domains — Algebra, Advanced Math, Problem-Solving & Data Analysis, and Geometry & Trigonometry — but the harder Module 2 form leans more heavily on multi-step algebra, quadratics, and trigonometric ratios. The Bluebook Desmos calculator is available on every question.

Is Digital SAT Math Module 2 harder than Module 1?

It depends on the student. Module 2 is adaptive: students who score well on Module 1 are routed into a harder version of Module 2 that unlocks access to the full 200–800 scoring range, while students who score lower are routed into an easier version that typically caps in the mid-600s. This walkthrough reproduces the higher (harder) form from Practice Test 4.

How is the Digital SAT Math section scored?

The Math section is scored from 200 to 800. Module 1 is the same for every student; Module 2 is stage-adaptive — students who perform well on Module 1 are routed to a harder Module 2 with access to the full 800, and students who did not are routed to an easier Module 2 that typically caps in the mid-600s.

Can I use a calculator on Digital SAT Math?

Yes, on every single question. The Bluebook app includes a built-in Desmos graphing calculator, and you may also bring an approved handheld calculator. Learning to use Desmos efficiently (solving systems, finding zeros, checking answers by graphing) can save 30–60 seconds per question.

What is a good Digital SAT Math score?

As a general reference: 600 ≈ 75th percentile, 700 ≈ 90th percentile, 750+ ≈ top 5%. For MIT, Stanford, Harvard, and the Ivy League, competitive applicants score 780+ on Math. For SUNY Binghamton or UNC, 720+. For most CUNY campuses, 600–680.

Where is SOMATH located?

226 West 79th Street, 1st Floor, New York, NY 10024. Upper West Side of Manhattan, between Amsterdam and Broadway, right by the 1 train at 79th and the B/C at 81st.

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