Digital SAT · Math · Exam Prep · NYC Test Prep
Digital SAT Practice Test 10 — Module 2 Math: Full Walkthrough of All 27 Questions with Answers & Explanations
Every question on Digital SAT Practice Test #10, Math Module 2, transcribed verbatim, with the official College Board answer key, a step-by-step worked solution, a plain-English explanation for every item, theory refreshers on every tested skill, and links to the official PDF. Built by SOMATH, the math school on the Upper West Side of Manhattan.
Looking for the answers, explanations, and full walkthrough of Digital SAT Practice Test 10 — 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 word problems, tougher algebra, and quadratics that hinge on the discriminant. This is a complete, question-by-question walkthrough of SAT Test #10, Math Module 2 — every one of the 27 questions transcribed, with the official College Board answer for each item and a worked solution showing exactly which SAT Math tool the question wants and how to apply it.
Attempt each question first with a strict timer, then reveal the answer to compare setups. Compare with the easier Math Module 1 walkthrough for Test 10 and the Reading & Writing Module 1 walkthrough for Test 10 to see how the adaptive test scales up. See our 2026 Digital SAT structure, dates, and scores guide for the full test blueprint.
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, student-friendly 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 10, Math Module 2
| Question # | Correct Answer |
|---|---|
| 1 | D |
| 2 | A |
| 3 | D |
| 4 | D |
| 5 | A |
| 6 | 79 |
| 7 | 2 |
| 8 | D |
| 9 | D |
| 10 | B |
| 11 | A |
| 12 | C |
| 13 | 41 |
| 14 | 11875 |
| 15 | B |
| 16 | B |
| 17 | B |
| 18 | A |
| 19 | C |
| 20 | 5 |
| 21 | 0.25 or 1/4 |
| 22 | D |
| 23 | C |
| 24 | C |
| 25 | D |
| 26 | B |
| 27 | 104 |
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: Unit conversion, scatterplots, linear graphs, perimeter, rearranging, and medians
Warm-up items: converting units, reading a line of best fit, matching a table to a graph, perimeter, isolating a variable, and computing a median.
Question 1 · Problem-Solving & Data Analysis · Unit conversion
An object’s speed is 64 yards per second. What is the object’s speed, in feet per second? (1 yard = 3 feet)
- A) 61
- B) 67
- C) 94
- D) 192
Answer: D) 192
Key idea. Multiply by the conversion factor 3 feet per yard.
64 yd/s × (3 ft / 1 yd) = 192 ft/sWhy this works. Yards cancel, leaving feet per second; the numeric result is 64 × 3 = 192.
Question 2 · Problem-Solving & Data Analysis · Line of best fit
The scatterplot shows the relationship between two variables, x and y. A line of best fit is also shown. Which of the following equations best represents the line of best fit shown?
See figure in the original College Board PDF above.
- A) y = x + 3.4
- B) y = x − 3.4
- C) y = −x + 3.4
- D) y = −x − 3.4
Answer: A) y = x + 3.4
Key idea. Read the slope (positive or negative) and the y-intercept from the plotted line of best fit.
From the scatterplot: The line rises from lower-left to upper-right ⇒ slope is positive. At x = 0 the line hits the y-axis a few units above 0 ⇒ y-intercept ≈ 3.4. Equation: y = x + 3.4Why this works. Only choice A has both a positive slope (+1) and a positive y-intercept (+3.4), matching the line drawn on the graph.
Question 3 · Algebra · Linear relationships
The graph shows the linear relationship between x and y. Which table gives three values of x and their corresponding values of y for this relationship?
See figure in the original College Board PDF above.
- A) (0, 0), (1, −7), (2, −9)
- B) (0, 0), (1, −3), (2, −1)
- C) (0, −5), (1, −7), (2, −9)
- D) (0, −5), (1, −3), (2, −1)
Answer: D) (0, −5), (1, −3), (2, −1)
Key idea. Identify the slope and y-intercept from the graph, then check which table has consistent slope and matching y-intercept.
From the graph, the line has y-intercept −5 (crosses y-axis at −5) and slope +2. Equation: y = 2x − 5. Check table D: x = 0 → y = 2(0) − 5 = −5 ✓ x = 1 → y = 2(1) − 5 = −3 ✓ x = 2 → y = 2(2) − 5 = −1 ✓Why this works. All three (x, y) pairs in table D satisfy the line’s equation y = 2x − 5, so table D matches the graph.
Question 4 · Geometry & Trigonometry · Perimeter of a rectangle
What is the perimeter, in inches, of a rectangle with a length of 4 inches and a width of 9 inches?
- A) 13
- B) 17
- C) 22
- D) 26
Answer: D) 26
Key idea. Perimeter of a rectangle = 2(length + width).
P = 2(4 + 9) = 2(13) = 26Why this works. Adding the length and width, then doubling, gives the total distance around the rectangle.
Question 5 · Algebra · Isolating a variable
7m = 2(n + p). The given equation relates the positive numbers m, n, and p. Which equation correctly gives m in terms of n and p?
- A) m = 2(n + p) / 7
- B) m = 2(n + p)
- C) m = 2(n + p) − 7
- D) m = 2 − n − p − 7
Answer: A) m = 2(n + p) / 7
Key idea. Divide both sides by the coefficient of m.
7m = 2(n + p) Divide both sides by 7: m = 2(n + p) / 7Why this works. Dividing the entire right-hand side by 7 undoes the multiplication by 7 on the left, isolating m.
Question 6 · Problem-Solving & Data Analysis · Median
73, 74, 75, 77, 79, 82, 84, 85, 91. What is the median of the data shown?
Student-produced response — enter the accepted answer format shown after revealing the solution.
Answer: 79
Key idea. The median of an odd number of ordered values is the middle value.
Ordered list: 73, 74, 75, 77, 79, 82, 84, 85, 91 Count: 9 values → median is the 5th value. Median = 79Why this works. With 9 ordered values, the 5th one splits the list into 4 below and 4 above — that value is the median.
Question 7 · Algebra · Solving a linear equation
The function f is defined by f(x) = 4x. For what value of x does f(x) = 8?
Student-produced response — enter the accepted answer format shown after revealing the solution.
Answer: 2
Key idea. Set 4x = 8 and solve for x.
f(x) = 4x = 8 x = 8 / 4 x = 2Why this works. Dividing both sides by 4 isolates x, giving x = 2.
Questions 8–14: Percentages, interpretations, systems, scale factors, and exponentials
Percent conversion, linear-function interpretation, quadratic-line intersection, scale factors, identity equations, one-variable word problems, and exponential growth.
Question 8 · Problem-Solving & Data Analysis · Percentages
Of 300,000 paper clips, 234,000 are size large. What percentage of the paper clips are size large?
- A) 22%
- B) 33%
- C) 66%
- D) 78%
Answer: D) 78%
Key idea. Percentage = (part / whole) × 100.
Percentage = (234,000 / 300,000) × 100 = 0.78 × 100 = 78%Why this works. Dividing the number of large paper clips by the total, then multiplying by 100, converts the fraction into a percentage.
Question 9 · Algebra · Interpreting linear functions
f(x) = 8x + 4. The function f gives the estimated height, in feet, of a willow tree x years after its height was first measured. Which statement is the best interpretation of 4 in this context?
- A) The tree will be measured each year for 4 years.
- B) The tree is estimated to grow to a maximum height of 4 feet.
- C) The estimated height of the tree increased by 4 feet each year.
- D) The estimated height of the tree was 4 feet when it was first measured.
Answer: D) The estimated height of the tree was 4 feet when it was first measured.
Key idea. In y = mx + b, the constant b is the y-intercept — the value when x = 0.
f(x) = 8x + 4 At x = 0 (when the height was first measured): f(0) = 8(0) + 4 = 4 feetWhy this works. The constant 4 is the starting height at x = 0. The coefficient 8 is the growth rate per year.
Question 10 · Advanced Math · Systems with a quadratic
y = 76
y = x2 − 5
The graphs of the given equations in the xy-plane intersect at the point (x, y). What is a possible value of x?
- A) −76/5
- B) −9
- C) 5
- D) 76
Answer: B) −9
Key idea. Substitute y = 76 into the second equation and solve for x.
76 = x² − 5 x² = 81 x = ±9 Possible values: 9 or −9. Only −9 appears in the choices.Why this works. The two graphs intersect where both equations hold; setting them equal gives x² = 81, whose solutions are ±9. Choice B is one of them.
Question 11 · Geometry & Trigonometry · Scale factor
Each side of equilateral triangle S is multiplied by a scale factor of k to create equilateral triangle T. The length of each side of triangle T is greater than the length of each side of triangle S. Which of the following could be the value of k?
- A) 29/28
- B) 1
- C) 28/29
- D) 0
Answer: A) 29/28
Key idea. If T’s sides are longer than S’s, then k > 1.
Side of T = k · (side of S) Side of T > side of S ⇒ k > 1 Check: 29/28 ≈ 1.036 > 1 ✓ 1 = 1 (not greater) 28/29 < 1 0 < 1Why this works. Only 29/28 is strictly greater than 1, so it is the only scale factor that enlarges the triangle.
Question 12 · Algebra · Number of solutions to a linear equation
66x = 66x. How many solutions does the given equation have?
- A) Exactly one
- B) Exactly two
- C) Infinitely many
- D) Zero
Answer: C) Infinitely many
Key idea. An equation that is true for every real number has infinitely many solutions.
66x = 66x is true for every real x. Subtracting 66x from both sides gives 0 = 0, which is always true.Why this works. Since both sides are identical, the equation is an identity — every real number is a solution.
Question 13 · Algebra · Linear word problems
Vivian bought party hats and cupcakes for $71. Each package of party hats cost $3, and each cupcake cost $1. If Vivian bought 10 packages of party hats, how many cupcakes did she buy?
Student-produced response — enter the accepted answer format shown after revealing the solution.
Answer: 41
Key idea. Total spent = (party-hat cost) + (cupcake cost). Solve for the number of cupcakes.
Party hats: 10 packages × $3 = $30 Remaining for cupcakes: $71 − $30 = $41 Cupcakes cost $1 each, so 41 cupcakes.Why this works. Subtracting the fixed party-hat spend from the total leaves the amount spent on $1 cupcakes, which equals the number of cupcakes.
Question 14 · Advanced Math · Exponential functions
The exponential function g is defined by g(x) = 19 · ax, where a is a positive constant. If g(3) = 2,375, what is the value of g(4)?
Student-produced response — enter the accepted answer format shown after revealing the solution.
Answer: 11875
Key idea. Use g(3) to find a, then evaluate g(4). Or use the growth-factor shortcut g(4) = a · g(3).
g(3) = 19 · a³ = 2,375 a³ = 2,375 / 19 = 125 a = 5 g(4) = 19 · 5⁴ = 19 · 625 = 11,875 Shortcut: g(4) = a · g(3) = 5 · 2,375 = 11,875Why this works. For an exponential function, moving x up by 1 multiplies the output by the base a; here a = 5, so g(4) = 5 · 2,375 = 11,875.
Questions 15–20: Trig, graphs, rational expressions, exponential y-intercepts, model interpretation, and circles
Cofunction identities, reading a linear graph with two intercepts, simplifying rational expressions, exponential y-intercepts, unit-based interpretation, and radius from a diameter.
Question 15 · Geometry & Trigonometry · Complementary-angle trig
In right triangle RST, the sum of the measures of angle R and angle S is 90 degrees. The value of sin(R) is √15 / 4. What is the value of cos(S)?
- A) 15/1
- B) √15 / 4
- C) 4/√15
- D) √15
Answer: B) √15 / 4
Key idea. Cofunction identity: if R + S = 90°, then cos(S) = sin(R).
R + S = 90° ⇒ S = 90° − R cos(S) = cos(90° − R) = sin(R) = √15 / 4Why this works. For complementary acute angles in a right triangle, sine and cosine swap: sin(R) = cos(90° − R) = cos(S).
Question 16 · Algebra · Linear equations from a graph
The graph shows the relationship between the number of shares of stock from Company A, x, and the number of shares of stock from Company B, y, that Simone can purchase. Which equation could represent this relationship?
See figure in the original College Board PDF above.
- A) y = 8x + 12
- B) 8x + 12y = 480
- C) y = 12x + 8
- D) 12x + 8y = 480
Answer: B) 8x + 12y = 480
Key idea. Read the two intercepts of the line from the graph and match them to the equation.
From the graph: y-intercept ≈ 40 (when x = 0) x-intercept ≈ 60 (when y = 0) Test choice B) 8x + 12y = 480: x = 0 → 12y = 480 → y = 40 ✓ y = 0 → 8x = 480 → x = 60 ✓Why this works. Only choice B produces both intercepts (60, 0) and (0, 40) that match the graph. In context: Company A shares cost $8, Company B shares cost $12, and Simone has $480 to spend.
Question 17 · Advanced Math · Simplifying rational expressions
Which expression is equivalent to [8x(x − 7) − 3(x − 7)] / (2x − 14), where x > 7?
- A) (x − 7) / 5
- B) (8x − 3) / 2
- C) (8x² − 3x − 14) / (2x − 14)
- D) (8x² − 3x − 77) / (2x − 14)
Answer: B) (8x − 3) / 2
Key idea. Factor (x − 7) from the numerator and from the denominator, then cancel.
Numerator: 8x(x − 7) − 3(x − 7) = (x − 7)(8x − 3) Denominator: 2x − 14 = 2(x − 7) Expression = [(x − 7)(8x − 3)] / [2(x − 7)] = (8x − 3) / 2 (cancel x − 7 since x > 7)Why this works. With x > 7, x − 7 ≠ 0, so it cancels legally between numerator and denominator, leaving (8x − 3)/2.
Question 18 · Advanced Math · Exponential functions and y-intercepts
The function f is defined by f(x) = (−8)(2)x + 22. What is the y-intercept of the graph of y = f(x) in the xy-plane?
- A) (0, 14)
- B) (0, 2)
- C) (0, 22)
- D) (0, −8)
Answer: A) (0, 14)
Key idea. The y-intercept is the value of f at x = 0.
f(0) = (−8)(2)⁰ + 22 = (−8)(1) + 22 = −8 + 22 = 14 y-intercept: (0, 14)Why this works. Any nonzero base raised to the 0 power equals 1, so 2⁰ = 1 and f(0) = −8 + 22 = 14.
Question 19 · Algebra · Interpreting linear models
Keenan made 32 cups of vegetable broth. Keenan then filled x small jars and y large jars with all the vegetable broth he made. The equation 3x + 5y = 32 represents this situation. Which is the best interpretation of 5y in this context?
- A) The number of large jars Keenan filled
- B) The number of small jars Keenan filled
- C) The total number of cups of vegetable broth in the large jars
- D) The total number of cups of vegetable broth in the small jars
Answer: C) The total number of cups of vegetable broth in the large jars
Key idea. Units check: (cups per jar) × (number of jars) = total cups.
y = number of large jars 5 = cups of broth in each large jar So 5y = (cups per large jar) × (number of large jars) = total cups of broth in the large jarsWhy this works. The coefficient 5 has units of cups per large jar; multiplying by y (large jars) gives total cups in large jars — matching choice C.
Question 20 · Geometry & Trigonometry · Circles in the xy-plane
A circle in the xy-plane has a diameter with endpoints (2, 4) and (2, 14). An equation of this circle is (x − 2)2 + (y − 9)2 = r2, where r is a positive constant. What is the value of r?
Student-produced response — enter the accepted answer format shown after revealing the solution.
Answer: 5
Key idea. Radius = half the length of the diameter. Find the distance between the two endpoints, then divide by 2.
Endpoints: (2, 4) and (2, 14). Same x, so distance is |14 − 4| = 10. Diameter = 10 ⇒ radius r = 10 / 2 = 5Why this works. A diameter passes through the center; half of its length is the radius, so r = 5. (Center (2, 9) also confirms since it’s the midpoint of the two endpoints.)
Questions 21–27: Perpendicular slopes, absolute value, exponentials, discriminants, factoring, and equilateral triangles
Perpendicular slopes, absolute-value equations, equivalent forms of exponentials, discriminant conditions, factor-of-a-quadratic reasoning, histogram means, and the height formula for equilateral triangles.
Question 21 · Algebra · Slopes of parallel and perpendicular lines
Line ℓ is defined by 3y + 12x = 5. Line n is perpendicular to line ℓ in the xy-plane. What is the slope of line n?
Student-produced response — enter the accepted answer format shown after revealing the solution.
Answer: 0.25 or 1/4
Key idea. Perpendicular slopes are negative reciprocals.
3y + 12x = 5 3y = −12x + 5 y = −4x + 5/3 ⇒ slope of ℓ is −4. Slope of n = −1 / (slope of ℓ) = −1 / (−4) = 1/4 = 0.25 Accepted forms: 0.25, 1/4Why this works. Solving for y gives the slope of ℓ as −4; the negative reciprocal is 1/4, which is the perpendicular slope.
Question 22 · Algebra · Absolute-value equations
|−5x + 13| = 73. What is the sum of the solutions to the given equation?
- A) −146/5
- B) −12
- C) 0
- D) 26/5
Answer: D) 26/5
Key idea. An absolute-value equation |A| = k (k > 0) splits into A = k and A = −k. Solve both, then add.
Case 1: −5x + 13 = 73 −5x = 60 x = −12 Case 2: −5x + 13 = −73 −5x = −86 x = 86/5 Sum: −12 + 86/5 = −60/5 + 86/5 = 26/5Why this works. Both cases of the absolute-value equation give valid solutions; adding −12 and 86/5 (with common denominator 5) yields 26/5.
Question 23 · Advanced Math · Equivalent forms of exponential functions
For the exponential function f, the value of f(1) is k, where k is a constant. Which of the following equivalent forms of the function f shows the value of k as the coefficient or the base?
- A) f(x) = 50(1.6)x+1
- B) f(x) = 80(1.6)x
- C) f(x) = 128(1.6)x−1
- D) f(x) = 204.8(1.6)x−2
Answer: C) f(x) = 128(1.6)x−1
Key idea. Compute f(1) for each form. The form whose coefficient (or base) numerically equals f(1) is the answer.
For every equivalent form, evaluate f(1): A) f(1) = 50(1.6)² = 50 · 2.56 = 128 B) f(1) = 80(1.6)¹ = 128 C) f(1) = 128(1.6)⁰ = 128 · 1 = 128 D) f(1) = 204.8(1.6)⁻¹ = 204.8 / 1.6 = 128 So k = 128. In choice C, the exponent at x = 1 is 0, so (1.6)⁰ = 1 and the coefficient 128 appears directly as f(1) = k.Why this works. Only choice C is written so that plugging x = 1 zeroes out the exponent, making the coefficient itself equal f(1) = k = 128.
Question 24 · Advanced Math · Discriminant of a quadratic
−9x2 + 30x + c = 0. In the given equation, c is a constant. The equation has exactly one solution. What is the value of c?
- A) 3
- B) 0
- C) −25
- D) −53
Answer: C) −25
Key idea. A quadratic ax² + bx + c = 0 has exactly one real solution when the discriminant b² − 4ac = 0.
a = −9, b = 30, c = c b² − 4ac = 0 30² − 4(−9)(c) = 0 900 + 36c = 0 36c = −900 c = −25Why this works. Setting the discriminant equal to zero forces a repeated root; solving 900 + 36c = 0 gives c = −25.
Question 25 · Advanced Math · Factoring quadratics with a parameter
Which of the following expressions has a factor of x + 2b, where b is a positive integer constant?
- A) 3x² + 7x + 14b
- B) 3x² + 28x + 14b
- C) 3x² + 42x + 14b
- D) 3x² + 49x + 14b
Answer: D) 3x² + 49x + 14b
Key idea. If x + 2b is a factor, then substituting x = −2b must make the expression equal 0. Equivalently, factor as (x + 2b)(3x + m) and match coefficients.
Assume (x + 2b)(3x + m) = 3x² + (m + 6b)x + 2bm. Match to 3x² + (linear coeff)x + 14b: Constant: 2bm = 14b ⇒ m = 7 (for b ≠ 0) Linear: m + 6b = 7 + 6b So the linear coefficient must equal 7 + 6b for x + 2b to be a factor. For a positive integer b: b = 7 ⇒ 7 + 6(7) = 49 ⇒ matches choice D’s 49x ✓ Verify: 3x² + 49x + 14(7) = 3x² + 49x + 98 = (x + 14)(3x + 7) = (x + 2·7)(3x + 7).Why this works. For x + 2b to divide the quadratic evenly, the coefficient of x must equal 7 + 6b for some positive integer b. Setting 7 + 6b equal to each choice’s linear coefficient gives b = 0, 21/6, 35/6, or 42/6 — only 42/6 = 7 is a positive integer, so choice D is the only quadratic with a factor of x + 2b (with b = 7, giving x + 14 as a factor).
Question 26 · Problem-Solving & Data Analysis · Histograms and means
Two data sets of 23 integers each are summarized in the histograms shown. For each of the histograms, the first interval represents the frequency of integers greater than or equal to 10, but less than 20. The second interval represents the frequency of integers greater than or equal to 20, but less than 30, and so on. What is the smallest possible difference between the mean of data set A and the mean of data set B?
See figure in the original College Board PDF above.
- A) 0
- B) 1
- C) 10
- D) 23
Answer: B) 1
Key idea. Because we only know the interval each integer falls into, extreme choices within each bin give the smallest possible difference. Push A’s values as high as allowed and B’s values as low as allowed (or vice versa) to close the gap.
Each bin has width 10. For a value in the [10, 20) bin, the actual integer can be anywhere from 10 to 19; similarly for other bins. Mean of A uses A’s frequency distribution; mean of B uses B’s. Reading the histograms, the raw bin-midpoint means differ by more than 1, but by choosing the largest allowed integer in each bin for the data set with the smaller mean and the smallest allowed integer in each bin for the data set with the larger mean, the two means can be pulled together until they differ by at most 1. They cannot be made equal (the frequency distributions differ), but the smallest achievable difference is 1.Why this works. Bin-based frequency data leaves each integer’s exact value uncertain within a 10-wide range. Choosing endpoints strategically shows the two means can be forced within 1 of each other, but not closer — so 1 is the minimum difference.
Question 27 · Geometry & Trigonometry · Equilateral triangles
The perimeter of an equilateral triangle is 624 centimeters. The height of this triangle is k√3 centimeters, where k is a constant. What is the value of k?
Student-produced response — enter the accepted answer format shown after revealing the solution.
Answer: 104
Key idea. Side length = perimeter / 3. Height of an equilateral triangle with side s is (√3 / 2) · s.
Side length s = 624 / 3 = 208 cm Height = (√3 / 2) · s = (√3 / 2) · 208 = 104√3 Given height = k√3, so k = 104.Why this works. Splitting an equilateral triangle down its altitude creates a 30-60-90 triangle whose long leg (the height) equals (side/2)√3. With side 208, the height is 104√3, so k = 104.
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Digital SAT Practice Test 10 Math Module 2 FAQ
What does this Practice Test 10 Math Module 2 walkthrough cover?
It covers all 27 questions in Math Module 2 of Digital SAT Practice Test 10, including the multiple-choice items and student-produced responses. Every item has the official College Board answer, a worked solution, and a short explanation of the underlying Digital SAT Math skill.
Which questions in Practice Test 10 Math Module 2 are student-produced responses?
Questions 6, 7, 13, 14, 20, 21, and 27 are student-produced responses. Q6 accepts 79; Q7 accepts 2; Q13 accepts 41; Q14 accepts 11875; Q20 accepts 5; Q21 accepts 0.25 or 1/4; Q27 accepts 104.
Is Module 2 harder than Module 1?
Yes — Module 2 is adaptive. Students who performed well on Module 1 receive the harder Module 2 form, which tests the same domains (Algebra, Advanced Math, Problem-Solving & Data Analysis, Geometry & Trigonometry) but with denser word problems and more complex algebra.
What is the key idea in Question 22?
The equation |−5x + 13| = 73 splits into two linear equations: −5x + 13 = 73 (giving x = −12) and −5x + 13 = −73 (giving x = 86/5). Adding the two solutions gives −12 + 86/5 = 26/5.
What is the key idea in Question 24?
A quadratic ax² + bx + c = 0 has exactly one real solution when the discriminant b² − 4ac = 0. Substituting a = −9 and b = 30 gives 900 + 36c = 0, so c = −25.
What is the key idea in Question 27?
For an equilateral triangle with side s, the height is (√3/2) · s. A perimeter of 624 cm gives s = 208 cm and height = 104√3 cm, so k = 104.
Where can I find the other Practice Test 10 walkthroughs?
The Math Module 1 walkthrough and Reading & Writing Module 1 walkthrough are already published; see the related-posts list below for more.
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