Digital SAT · Math · Exam Prep · NYC Test Prep
Digital SAT Practice Test 4 — Module 1: Full Walkthrough of All 27 Math Questions with Answers & Explanations
Every question on Digital SAT Practice Test #4, Math Module 1, 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.
Looking for the answers, explanations, and full walkthrough of Digital SAT Practice Test 4 — Math Module 1? You are in the right place. This is a complete, question-by-question walkthrough of SAT Test #4, Math Module 1 — 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 practice test for offline studying.
Module 1 is the first Math module on the Digital SAT, and unlike Module 2, it is not adaptive: every student sees the same questions. Your performance on Module 1 is what routes you into the harder or the easier version of Module 2. That means Module 1 is arguably the single most important 35 minutes of the Math section — and drilling it question by question is the fastest way to raise your composite score.
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.
Sections
- Answer key at a glance
- Theory refresher · What Math Module 1 tests
- Section 1 · Algebra & Linear Systems (Q1–14)
- Section 2 · Advanced Math & Functions (Q15–21, Q23, Q24, Q26, Q27)
- Section 3 · Geometry, Trig & Circles (Q9, Q20, Q22, Q25)
- How SOMATH prepares NYC students for the Digital SAT
- Digital SAT FAQ
Answer key at a glance
All 27 official College Board answers for Practice Test 4, Math Module 1, in one table for quick review:
| Q | Answer | Q | Answer | Q | Answer |
|---|---|---|---|---|---|
| 1 | B) 39 | 10 | D) y = 9.4 − 0.9x | 19 | C) w = (x/y)² − 19 |
| 2 | A) 25% | 11 | A) 0 | 20 | 100 |
| 3 | B) 30 | 12 | C) y = −x − 8 | 21 | 361/8 (or 45.12, 45.13) |
| 4 | D) 8x + 3 = 83 | 13 | 1/5 (or .2) | 22 | B) 12 |
| 5 | A) With each monthly deposit, +$25 | 14 | 80 | 23 | D) 45/k |
| 6 | 9 | 15 | D) y = −x/3 + 13 | 24 | C) 6 |
| 7 | 10 | 16 | B) ≈243 dollars in 1962 | 25 | C) 58 + 58√2 |
| 8 | A) f(x) = 3x + 29 | 17 | B) increase by 24.5 units | 26 | D) −12 |
| 9 | B) 72° | 18 | A) f(x) = (x + 44)² | 27 | 5 |
Theory refresher · What Math Module 1 tests
The Digital SAT Math section has two 35-minute modules of 22 questions each (20 scored + 2 pretest). Module 1 is not adaptive — every student sees the same items. 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 equations in one variable. Translate word phrases like “3 more than 8 times a number” into 8x + 3, and solve equations such as x²/25 = 36 by isolating the variable then taking a square root (remember both the positive and negative root exist algebraically, but only one may appear among the choices).
Systems of linear equations. Solve by elimination (line up matching coefficients and subtract) or substitution. For “what is the value of y” type items, you often do not need to fully solve for both variables — sometimes a clever combination isolates just the variable you need in one step.
Interpreting linear models & graphs. In y = mx + b, the slope m is the rate of change and b is the starting value. On a graph, read the y-intercept directly and compute slope as rise over run between two clearly marked points.
Ratios & proportions. A ratio like “length to width is 35 to 10” means length = 3.5 · width. If width changes, length must change by the same factor to preserve the ratio.
2 · Advanced Math
Function notation. In f(t) = 100 + 25t, the constant added to the variable term (25) is the rate of change per unit of t; the constant term (100) is the starting/initial value.
Exponential functions. f(x) = a·b^x: a is the value at x = 0 and b is the growth factor per unit of x. When asked to interpret f(5) ≈ 243, plug the input/output meaning back into the real-world context (year, dollars, etc.) rather than doing new algebra.
Rational exponents & radicals. ⁿ√y^m = y^(m/n). When multiplying expressions with different roots, first convert every radical to a fractional exponent with the SAME base, then add exponents: y^p · y^q = y^(p+q).
Quadratics & the discriminant. A system of one line and one parabola intersects at exactly one point when the resulting quadratic has discriminant b² − 4ac = 0. Vertex form y = a(x−h)² + k converts to standard form by expanding — useful for evaluating a + b + c (which is just y at x = 1).
Factoring with unknown coefficients. If 4x² + bx − 45 = (hx+k)(x+j), expanding the right side and matching constant terms gives kj = −45, which constrains what must be an integer.
3 · Problem-Solving & Data Analysis
Percentages. “What percentage of 300 is 75?” translates to (x/100)·300 = 75. Solve for x directly — do not confuse “percent of” with “percent more than.”
Reading bar graphs & scatterplots. For bar graphs, read the height of the relevant bar against the y-axis scale. For scatterplots, look at the overall trend (increasing/decreasing) to pick the sign of the slope, and check where points sit near x = 0 to estimate the y-intercept.
Rates. Cost = fixed fee + (rate)·(quantity), or simple division for unit rates like dollars per pound.
4 · Geometry & Trigonometry
Similar & right triangles. Corresponding angles of similar triangles are equal. Once two angles of a triangle are known, the third follows from the triangle angle sum of 180°.
Pythagorean theorem & isosceles right triangles. a² + b² = c². In an isosceles right triangle, the two legs are equal, so 2x² = c², giving x = c/√2.
Area of a right triangle. The two legs of a right triangle can serve as base and height: A = ½bh. Watch for the hypotenuse (the longest side) — it is never one of the two legs used for area.
Circles & central angles. The degree measure of an arc equals the degree measure of its central angle (the angle at the circle’s center formed by the two radii to the arc’s endpoints).
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.
Section 1 · Algebra & Linear Systems
Reading data displays, linear equations, systems, ratios, and function interpretation.
Question 1 · Reading a Bar Graph
See figure in the original College Board PDF above.
A group of students voted on five after-school activities. The bar graph shows the number of students who voted for each of the five activities. How many students chose activity 3?
- A) 25
- B) 39
- C) 48
- D) 50
Answer: B) 39
The height of each bar represents the number of students who voted for that activity. The bar for activity 3 has a height between 35 and 40. Of the given choices, only 39 falls between 35 and 40.✗ A) 25
25 is close to the height of activity 1's bar, not activity 3.
✓ B) 39
The bar for activity 3 sits between the 35 and 40 gridlines — 39 students chose activity 3.
✗ C) 48
This is the number of students who chose activity 5, not activity 3.
✗ D) 50
50 is above every bar on the graph — no activity reached this value.
Question 2 · Percentages
What percentage of 300 is 75?
- A) 25%
- B) 50%
- C) 75%
- D) 225%
Answer: A) 25%
Let x represent the percentage of 300 that is 75. (x/100)(300) = 75 3x = 75 x = 25. So 25% of 300 is 75.✓ A) 25%
(25/100)(300) = 75. Confirmed.
✗ B) 50%
50% of 300 is 150, not 75.
✗ C) 75%
75% of 300 is 225, not 75.
✗ D) 225%
225% of 300 is 675, not 75.
Question 3 · Nonlinear Equations in 1 Var
x²/25 = 36. What is a solution to the given equation?
- A) 6
- B) 30
- C) 450
- D) 900
Answer: B) 30
Multiply both sides by 25: x² = 900. Take the square root of both sides: x = 30 or x = −30. Of these, only 30 is offered as a choice.✗ A) 6
6 is a solution to x² = 36, not to the given equation.
✓ B) 30
30² = 900, and 900/25 = 36. Confirmed.
✗ C) 450
450²/25 is far larger than 36 — does not satisfy the equation.
✗ D) 900
900 is x², not x — a common mix-up between the squared value and its root.
Question 4 · Linear Equations in 1 Var
3 more than 8 times a number x is equal to 83. Which equation represents this situation?
- A) (3)(8)x = 83
- B) 8x = 83 + 3
- C) 3x + 8 = 83
- D) 8x + 3 = 83
Answer: D) 8x + 3 = 83
"8 times a number x" → 8x. "3 more than" that quantity → 8x + 3. This is equal to 83: 8x + 3 = 83.✗ A) (3)(8)x = 83
This represents 3 times the quantity 8 times x — a multiplication, not "3 more than" (an addition).
✗ B) 8x = 83 + 3
This says 8 times a number is 3 more than 83 — reverses which side gets the +3.
✗ C) 3x + 8 = 83
This represents "8 more than 3 times a number x" — swaps the roles of 3 and 8.
✓ D) 8x + 3 = 83
8 times x, plus 3 more, equals 83 — exactly matches the wording.
Question 5 · Linear Functions
Hana deposited a fixed amount into her bank account each month. The function f(t) = 100 + 25t gives the amount, in dollars, in Hana’s bank account after t monthly deposits. What is the best interpretation of 25 in this context?
- A) With each monthly deposit, the amount in Hana's bank account increased by $25.
- B) Before Hana made any monthly deposits, the amount in her bank account was $25.
- C) After 1 monthly deposit, the amount in Hana's bank account was $25.
- D) Hana made a total of 25 monthly deposits.
Answer: A) With each monthly deposit, the amount in Hana's bank account increased by $25.
In f(t) = 100 + 25t, t is the number of monthly deposits and 25 is the coefficient of t. Each increase of t by 1 (one more deposit) increases f(t) by 25. So each monthly deposit adds $25 to the account.✓ A) With each monthly deposit, the amount in Hana's bank account increased by $25.
This is exactly what the coefficient of t represents — the rate of change per deposit.
✗ B) Before Hana made any monthly deposits, the amount in her bank account was $25.
Before any deposits (t = 0), the balance was f(0) = 100, not 25.
✗ C) After 1 monthly deposit, the amount in Hana's bank account was $25.
After 1 deposit, f(1) = 100 + 25 = 125, not 25.
✗ D) Hana made a total of 25 monthly deposits.
25 is the rate per deposit, not a count of deposits.
Question 6 · Ratios/Rates/Proportions
A customer spent $27 to purchase oranges at $3 per pound. How many pounds of oranges did the customer purchase?
Student-produced response — enter as a fraction or decimal.
Answer: 9
Pounds purchased = total spent ÷ price per pound = $27 ÷ $3 per pound = 9 pounds.Why this works.
Dividing the total cost by the unit rate ($3/pound) gives the quantity purchased: 27 ÷ 3 = 9 pounds.
Question 7 · Linear Equations in 1 Var
Nasir bought 9 storage bins that were each the same price. He used a coupon for $63 off the entire purchase. The cost for the entire purchase after using the coupon was $27. What was the original price, in dollars, for 1 storage bin?
Student-produced response — enter as a fraction or decimal.
Answer: 10
Cost before coupon = cost after coupon + coupon amount = 27 + 63 = 90. Nasir bought 9 bins, so price per bin = 90 / 9 = 10.Why this works.
Adding back the $63 discount recovers the original total price of $90. Dividing evenly among the 9 bins gives $10 per bin.
Question 8 · Linear Functions
See figure in the original College Board PDF above.
For the linear function f, the table shows three values of x and their corresponding values of f(x): (x=0, f(x)=29), (x=1, f(x)=32), (x=2, f(x)=35). Which equation defines f(x)?
- A) f(x) = 3x + 29
- B) f(x) = 29x + 32
- C) f(x) = 35x + 29
- D) f(x) = 32x + 35
Answer: A) f(x) = 3x + 29
f(x) = mx + b. At x = 0, f(x) = 29: 29 = m(0) + b → b = 29. At x = 1, f(x) = 32: 32 = m(1) + 29 → m = 3. So f(x) = 3x + 29. Check x = 2: 3(2) + 29 = 35 ✓.✓ A) f(x) = 3x + 29
The y-intercept is 29 (value at x = 0), and each increase of x by 1 increases f(x) by 3, so the slope is 3.
✗ B) f(x) = 29x + 32
Swaps the slope and y-intercept — 29 is the intercept, not the slope.
✗ C) f(x) = 35x + 29
Uses 35 (the value at x=2) as the slope, which is far too large.
✗ D) f(x) = 32x + 35
Neither coefficient matches the true slope (3) or intercept (29).
Question 9 · Right Triangles & Trig
See figure in the original College Board PDF above.
Right triangles PQR and STU are similar, where P corresponds to S. If the measure of angle Q is 18°, what is the measure of angle S? (Note: Figures not drawn to scale.)
- A) 18°
- B) 72°
- C) 82°
- D) 162°
Answer: B) 72°
P corresponds to S, and R and U are both right angles (90°), so R corresponds to U. That leaves Q corresponding to T: angle Q = angle T = 18°. We want angle S, which is the third angle of triangle STU. Angles of a triangle sum to 180°: S + T + U = 180 S + 18 + 90 = 180 S = 72°.✗ A) 18°
18° is the measure of angle T (which corresponds to Q), not angle S.
✓ B) 72°
Since angle U = 90° and angle T = 18°, angle S = 180 − 90 − 18 = 72°.
✗ C) 82°
82° does not follow from any correct triangle-angle-sum computation here.
✗ D) 162°
162° = 180° − 18°, which ignores the right angle entirely.
Question 10 · Two-Variable Data
See figure in the original College Board PDF above.
The scatterplot shows the relationship between two variables, x and y. Which of the following equations is the most appropriate linear model for the data shown?
- A) y = 0.9 + 9.4x
- B) y = 0.9 − 9.4x
- C) y = 9.4 + 0.9x
- D) y = 9.4 − 0.9x
Answer: D) y = 9.4 − 0.9x
As x increases, the data points show y decreasing → the model needs a negative slope. Near x = 0, the points are above y = 9 → the y-intercept must be greater than 9. Writing candidates as y = a + bx: only D has a negative slope (−0.9) and a y-intercept (9.4) greater than 9.✗ A) y = 0.9 + 9.4x
Positive slope and a y-intercept under 1 — contradicts both the downward trend and the high starting values.
✗ B) y = 0.9 − 9.4x
Negative slope is right, but the y-intercept (0.9) is far too low — the data starts above 9.
✗ C) y = 9.4 + 0.9x
Correct y-intercept, but a positive slope — the data trends downward, not upward.
✓ D) y = 9.4 − 0.9x
Negative slope (−0.9) matches the downward trend, and y-intercept 9.4 matches the points near x = 0.
Question 11 · Linear Equations in 2 Vars
2.5b + 5r = 80. The given equation describes the relationship between the number of birds, b, and the number of reptiles, r, that can be cared for at a pet care business on a given day. If the business cares for 16 reptiles on a given day, how many birds can it care for on this day?
- A) 0
- B) 5
- C) 40
- D) 80
Answer: A) 0
Substitute r = 16: 2.5b + 5(16) = 80 2.5b + 80 = 80 2.5b = 0 b = 0.✓ A) 0
With 16 reptiles, the reptile term alone (5·16 = 80) already uses the entire capacity, leaving 0 for birds.
✗ B) 5
Does not satisfy 2.5b + 80 = 80 — plugging in 5 gives 92.5, not 80.
✗ C) 40
2.5(40) + 80 = 180 ≠ 80. Too large.
✗ D) 80
This is the total right-hand-side constant, not the value of b.
Question 12 · Linear Functions
See figure in the original College Board PDF above.
What is an equation of the graph shown?
- A) y = −2x − 8
- B) y = x − 8
- C) y = −x − 8
- D) y = 2x − 8
Answer: C) y = −x − 8
y = mx + b, where (0, b) is the y-intercept. The line passes through (0, −8), so b = −8. The line also passes through (−8, 0). Slope: m = (0 − (−8)) / (−8 − 0) = 8 / (−8) = −1. So y = −1·x + (−8) = −x − 8.✗ A) y = −2x − 8
Slope of −2 is too steep — the correct slope through (0,−8) and (−8,0) is −1.
✗ B) y = x − 8
Slope of +1 has the wrong sign — through (−8, 0) and (0, −8) the line falls from left to right, so the slope must be negative (−1), not positive.
✓ C) y = −x − 8
Slope −1 and y-intercept −8 match both points (0, −8) and (−8, 0).
✗ D) y = 2x − 8
Slope of +2 has the wrong sign and magnitude.
Question 13 · Equivalent Expressions
If x/8 = 5, what is the value of 8/x?
Student-produced response — enter as a fraction or decimal.
Answer: 1/5 (or .2)
x/8 = 5 means x = 40. 8/x is the reciprocal of x/8. Since x/8 = 5, the reciprocal 8/x = 1/5. Check: 8/40 = 1/5 = 0.2.Why this works.
8/x is exactly the reciprocal of x/8. Since x/8 = 5, 8/x = 1/5 = 0.2.
Question 14 · Linear Systems
24x + y = 48 and 6x + y = 72. The solution to the given system of equations is (x, y). What is the value of y?
Student-produced response — enter as a fraction or decimal.
Answer: 80
Subtract the second equation from the first: (24x + y) − (6x + y) = 48 − 72 18x = −24 6x = −8 (dividing by 3). Substitute 6x = −8 into the second equation: −8 + y = 72 y = 80.Why this works.
Subtracting the equations eliminates y at first, giving 18x = −24. But it's easier to solve for 6x directly: 6x = −8. Substituting into 6x + y = 72 gives −8 + y = 72, so y = 80.
Section 2 · Advanced Math & Functions
Slope-intercept form, exponential interpretation, ratios in geometry, rational exponents, quadratics, and function transformations.
Question 15 · Linear Functions
Line t in the xy-plane has a slope of −1/3 and passes through the point (9, 10). Which equation defines line t?
- A) y = 13x − 1/3
- B) y = 9x + 10
- C) y = −x/3 + 10
- D) y = −x/3 + 13
Answer: D) y = −x/3 + 13
y = mx + b, with m = −1/3. y = −x/3 + b. Substitute the point (9, 10): 10 = −(9)/3 + b 10 = −3 + b b = 13. So y = −x/3 + 13.✗ A) y = 13x − 1/3
Swaps the slope and intercept — slope should be −1/3, not 13.
✗ B) y = 9x + 10
Uses the x-coordinate of the point (9) as the slope and the y-coordinate (10) as the intercept — both wrong.
✗ C) y = −x/3 + 10
Correct slope, but uses 10 (the y-coordinate of the given point) as the y-intercept instead of solving for b.
✓ D) y = −x/3 + 13
Slope −1/3 and solving 10 = −3 + b gives b = 13.
Question 16 · Nonlinear Functions
The function f(x) = 206(1.034)ˣ models the value, in dollars, of a certain bank account by the end of each year from 1957 through 1972, where x is the number of years after 1957. Which of the following is the best interpretation of “f(5) is approximately equal to 243” in this context?
- A) The value of the bank account is estimated to be approximately 5 dollars greater in 1962 than in 1957.
- B) The value of the bank account is estimated to be approximately 243 dollars in 1962.
- C) The value, in dollars, of the bank account is estimated to be approximately 5 times greater in 1962 than in 1957.
- D) The value of the bank account is estimated to increase by approximately 243 dollars every 5 years between 1957 and 1972.
Answer: B) The value of the bank account is estimated to be approximately 243 dollars in 1962.
x = number of years after 1957, so x = 5 corresponds to the year 1962. f(5) ≈ 243 means: when x = 5 (year 1962), the account value is approximately $243. Translate directly: the account is worth about $243 in 1962.✗ A) 5 dollars greater in 1962 than in 1957.
Confuses the input (5 years) with a dollar difference — 5 is the year offset, not a dollar amount.
✓ B) approximately 243 dollars in 1962.
Correctly reads f(5) = 243 as: 5 years after 1957 (i.e., 1962), the value is about $243.
✗ C) approximately 5 times greater in 1962 than in 1957.
Confuses the input variable 5 with a growth multiplier — that's not what f(5) = 243 states.
✗ D) increase by approximately 243 dollars every 5 years.
Misreads the output 243 as a repeating rate of change rather than a single value at x = 5.
Question 17 · Ratios/Rates/Proportions
For a certain rectangular region, the ratio of its length to its width is 35 to 10. If the width of the rectangular region increases by 7 units, how must the length change to maintain this ratio?
- A) It must decrease by 24.5 units.
- B) It must increase by 24.5 units.
- C) It must decrease by 7 units.
- D) It must increase by 7 units.
Answer: B) It must increase by 24.5 units.
length/width = 35/10 = 3.5, so length = 3.5 · width. If width increases by 7, length must increase by x to maintain the ratio: (length + x) = 3.5(width + 7) length + x = 3.5·width + 24.5 Since length = 3.5·width, this simplifies to: x = 24.5.✗ A) decrease by 24.5 units.
If width increases, length must also increase (same direction) to keep the ratio constant — not decrease.
✓ B) increase by 24.5 units.
Since length = 3.5·width, an increase of 7 in width requires an increase of 3.5·7 = 24.5 in length.
✗ C) decrease by 7 units.
Wrong direction — increasing width requires increasing length, not decreasing.
✗ D) increase by 7 units.
Since the ratio is 3.5 to 1 (not 1 to 1), the length must increase by more than 7 units — specifically 3.5×7 = 24.5.
Question 18 · Nonlinear Functions
Square P has a side length of x inches. Square Q has a perimeter that is 176 inches greater than the perimeter of square P. The function f gives the area of square Q, in square inches. Which of the following defines f?
- A) f(x) = (x + 44)²
- B) f(x) = (x + 176)²
- C) f(x) = (176x + 44)²
- D) f(x) = (176x + 176)²
Answer: A) f(x) = (x + 44)²
Perimeter of square P = 4x. Perimeter of square Q = 4x + 176. Side length of square Q = (4x + 176) / 4 = x + 44. Area of square Q = (side length)² = (x + 44)². So f(x) = (x + 44)².✓ A) f(x) = (x + 44)²
Dividing the perimeter difference (176) by 4 gives a side-length increase of 44, so Q's side length is x + 44.
✗ B) f(x) = (x + 176)²
Forgets to divide 176 by 4 — uses the full perimeter difference as a side-length increase.
✗ C) f(x) = (176x + 44)²
Incorrectly multiplies 176 by x instead of adding a constant side-length increase.
✗ D) f(x) = (176x + 176)²
Both terms are mishandled — neither matches the correct side length x + 44.
Question 19 · Equivalent Expressions
14x/(7y) = 2√(w + 19). The given equation relates the distinct positive real numbers w, x, and y. Which equation correctly expresses w in terms of x and y?
- A) w = √(x/y) − 19
- B) w = √(28x/14y) − 19
- C) w = (x/y)² − 19
- D) w = (28x/14y)² − 19
Answer: C) w = (x/y)² − 19
Simplify the left side first: 14x/(7y) = 2x/y. So the equation becomes 2x/y = 2√(w + 19). Divide both sides by 2: x/y = √(w + 19). Square both sides (valid since all quantities are positive): (x/y)² = w + 19. Subtract 19: w = (x/y)² − 19.✗ A) w = √(x/y) − 19
Takes a square root instead of squaring — the correct operation to undo the square root on the right side is squaring, not rooting again.
✗ B) w = √(28x/14y) − 19
Also takes an extra square root, and uses unsimplified coefficients 28/14 instead of the already-simplified 14/7.
✓ C) w = (x/y)² − 19
Simplifying 14x/7y to 2x/y, then squaring both sides of x/y = √(w+19), correctly isolates w.
✗ D) w = (28x/14y)² − 19
The original equation had coefficients 14 and 7 (14x/7y), not 28 and 14 — this choice introduces coefficients that were never in the given equation.
Question 20 · Circles
Point O is the center of a circle. The measure of arc RS on this circle is 100°. What is the measure, in degrees, of its associated angle ROS?
Student-produced response — enter as a fraction or decimal.
Answer: 100
O is the center, so OR and OS are radii, and angle ROS is a central angle. The degree measure of an arc always equals the degree measure of its associated central angle. Arc RS = 100°, so angle ROS = 100°.Why this works.
A central angle's measure is defined to equal the measure of the arc it subtends. Since arc RS = 100°, angle ROS = 100°.
Question 21 · Equivalent Expressions
The expression 6·⁵√(3⁵x⁴⁵) · ⁸√(2⁸x) is equivalent to axᵇ, where a and b are positive constants and x > 1. What is the value of a + b?
Student-produced response — enter as a fraction or decimal.
Answer: 361/8 (or 45.12, 45.13)
Use the rational exponent rule: ⁿ√(yᵘ) = y^(u/n). ⁵√(3⁵x⁴⁵) = 3^(5/5)·x^(45/5) = 3¹·x⁹ = 3x⁹. ⁸√(2⁸x) = 2^(8/8)·x^(1/8) = 2¹·x^(1/8) = 2x^(1/8). So the full expression is: 6 · 3x⁹ · 2x^(1/8) = 6·3·2 · x⁹ · x^(1/8) = 36 · x^(9 + 1/8) = 36 · x^(73/8). So a = 36 and b = 73/8. a + b = 36 + 73/8 = 288/8 + 73/8 = 361/8.Why this works.
Converting both radicals to fractional exponents and combining constants gives 36x^(73/8), so a + b = 36 + 73/8 = 361/8 = 45.125.
Question 22 · Area & Volume
A right triangle has sides of length 2√2, 6√2, and √80 units. What is the area of the triangle, in square units?
- A) 8√2 + √80
- B) 12
- C) 24√80
- D) 24
Answer: B) 12
A = (1/2)·base·height, where the two legs of a right triangle serve as base and height. √80 is the greatest of the three lengths, so it is the hypotenuse — not a leg. The two legs are 2√2 and 6√2. A = (1/2)(2√2)(6√2) = (1/2)(12·2) [since √2 · √2 = 2] = (1/2)(24) = 12.✗ A) 8√2 + √80
This resembles a perimeter-style sum, not the area formula.
✓ B) 12
Using the two legs (2√2 and 6√2) as base and height: (1/2)(2√2)(6√2) = (1/2)(24) = 12.
✗ C) 24√80
Multiplies incorrectly and leaves in an un-simplified radical form that doesn't represent the area.
✗ D) 24
This is 2×Area (forgetting the 1/2 factor in the triangle area formula) — (2√2)(6√2) = 24, but the area is half that: 12.
Question 23 · Equivalent Expressions
The expression 4x² + bx − 45, where b is a constant, can be rewritten as (hx + k)(x + j), where h, k, and j are integer constants. Which of the following must be an integer?
- A) b/h
- B) b/k
- C) 45/h
- D) 45/k
Answer: D) 45/k
Expand (hx + k)(x + j) = hx² + jhx + kx + kj = hx² + (jh + k)x + kj. Matching to 4x² + bx − 45: h = 4 (coefficient of x²) kj = −45 (constant term) jh + k = b. From kj = −45, divide both sides by k: j = −45/k. Since j must be an integer (given), −45/k must be an integer. Therefore 45/k must also be an integer.✗ A) b/h
There's no requirement that b be divisible evenly by h = 4; b = jh + k need not be a multiple of 4.
✗ B) b/k
Similarly, b is a sum (jh + k), not necessarily divisible by k.
✗ C) 45/h
h = 4 is fixed by matching the x² coefficient, but there's no requirement that 45 be divisible by h.
✓ D) 45/k
Since kj = −45 and j must be an integer, 45/k (equivalently −45/k) must be an integer.
Question 24 · Nonlinear Functions
y = 2x² − 21x + 64 and y = 3x + a. In the given system of equations, a is a constant. The graphs of the equations in the given system intersect at exactly one point, (x, y), in the xy-plane. What is the value of x?
- A) −8
- B) −6
- C) 6
- D) 8
Answer: C) 6
Subtract the second equation from the first: 0 = 2x² − 21x + 64 − 3x − a 0 = 2x² − 24x + (64 − a). For exactly one solution, the discriminant must equal 0: b² − 4ac = 0, where here "a,b,c" of the quadratic are 2, −24, and (64 − a): (−24)² − 4(2)(64 − a) = 0 576 − 8(64 − a) = 0 576 − 512 + 8a = 0 64 + 8a = 0 a = −8. Substitute a = −8 back into 0 = 2x² − 24x + 64 − a: 0 = 2x² − 24x + 64 + 8 0 = 2x² − 24x + 72 0 = 2(x² − 12x + 36) 0 = x² − 12x + 36 = (x − 6)². So x = 6.✗ A) −8
−8 is the value of the constant a, not x — a common mix-up.
✗ B) −6
Sign error somewhere in solving the perfect-square factorization (x − 6)² = 0.
✓ C) 6
Setting the discriminant to 0 forces a = −8, and the resulting quadratic factors as (x − 6)², giving the double root x = 6.
✗ D) 8
8 is close to |a|, but not the correct root of the resulting quadratic.
Section 3 · Geometry, Circles & Advanced Functions
Isosceles right triangles, parabola vertex form, and exponential function transformations — the final stretch of the module.
Question 25 · Right Triangles & Trig
An isosceles right triangle has a hypotenuse of length 58 inches. What is the perimeter, in inches, of this triangle?
- A) 29√2
- B) 58√2
- C) 58 + 58√2
- D) 58 + 116√2
Answer: C) 58 + 58√2
In an isosceles right triangle, the two legs are equal. Let each leg = x. Pythagorean theorem: x² + x² = 58² 2x² = 3364 x² = 1682 x = √1682 = √(841·2) = 29√2. Perimeter = hypotenuse + 2 legs = 58 + 2(29√2) = 58 + 58√2.✗ A) 29√2
This is just the length of one leg, not the full perimeter.
✗ B) 58√2
This is the sum of the two legs alone (2 × 29√2 = 58√2) — it omits the hypotenuse.
✓ C) 58 + 58√2
Perimeter = hypotenuse (58) + both legs (58√2 total) = 58 + 58√2.
✗ D) 58 + 116√2
Doubles the leg contribution incorrectly — 116√2 would be 4 legs' worth, not 2.
Question 26 · Nonlinear Functions
In the xy-plane, a parabola has vertex (9, −14) and intersects the x-axis at two points. If the equation of the parabola is written in the form y = ax² + bx + c, where a, b, and c are constants, which of the following could be the value of a + b + c?
- A) −23
- B) −19
- C) −14
- D) −12
Answer: D) −12
Vertex form: y = a(x − 9)² − 14. Expand: y = a(x² − 18x + 81) − 14 = ax² − 18ax + 81a − 14. So b = −18a and c = 81a − 14. a + b + c = a − 18a + 81a − 14 = 64a − 14. Note a + b + c is just the value of y when x = 1 (since 1² = 1, 1¹ = 1). Since the vertex (9, −14) is below the x-axis and the parabola crosses the x-axis twice, it must open upward, so a > 0. Test choice D: 64a − 14 = −12 → 64a = 2 → a = 2/64 = 1/32, which is positive. ✓ Test choices A, B: solving gives negative a (opens downward, would not cross x-axis twice from a vertex below it). Test choice C: gives a = 0, not a valid parabola.✗ A) −23
Solving 64a − 14 = −23 gives a negative a, meaning the parabola opens downward and never reaches the x-axis from a vertex at y = −14 — contradicts "intersects the x-axis at two points."
✗ B) −19
Also yields a negative a — same contradiction as A.
✗ C) −14
Yields a = 0, which is not a parabola at all (a cannot be zero in y = ax² + bx + c).
✓ D) −12
Solving 64a − 14 = −12 gives a = 1/32 > 0 — the parabola opens upward, consistent with a vertex below the x-axis that crosses it twice.
Question 27 · Nonlinear Functions
Function f is defined by f(x) = −aˣ + b, where a and b are constants. In the xy-plane, the graph of y = f(x) − 15 has a y-intercept at (0, −99/7). The product of a and b is 65/7. What is the value of a?
Student-produced response — enter as a fraction or decimal.
Answer: 5
f(x) = −aˣ + b, so y = f(x) − 15 = −aˣ + b − 15. At the y-intercept, x = 0, y = −99/7: −99/7 = −a⁰ + b − 15 −99/7 = −1 + b − 15 −99/7 = b − 16. Add 16 to both sides: b = 16 − 99/7 = 112/7 − 99/7 = 13/7. Given: a·b = 65/7. a·(13/7) = 65/7 a = (65/7) ÷ (13/7) = 65/13 = 5.Why this works.
Plugging x = 0 into y = f(x) − 15 uses a⁰ = 1, isolating b = 13/7. Then dividing the given product ab = 65/7 by b = 13/7 gives a = 5.
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:
- Small-group in-person classes (6–8 students max) at 226 W 79th Street on the Upper West Side. Same room, same teacher, same students, week after week.
- Structured curriculum aligned to the current Digital SAT framework across all four Math domains and all four Reading & Writing domains.
- Bluebook + Desmos drilling. We do not just teach math — we teach how to use the built-in Desmos calculator to solve problems 2–3× faster than by hand.
- Full-length timed Modules 1 and 2 for both Math and R&W with graded free-response feedback. Every student sees their scores tracked week by week.
- Free 30-minute in-person diagnostic evaluation before enrollment ($99 one-time enrollment fee applies for new students who continue into our program). The evaluation identifies exactly which domain needs the most work and produces a written diagnostic within 48 hours.
Ready to raise your SAT Math score?
Start with a free 30-minute in-person diagnostic at our Upper West Side classroom. Includes a written diagnostic report within 48 hours — yours to keep whether you enroll or not. A one-time $99 enrollment fee applies if you join a course.
Book Free Evaluation → or call (646) 668-6151
Want your child in an SAT Math class at SOMATH?
Small-group Digital SAT Math prep on the Upper West Side. Full Bluebook + Desmos drilling, timed Modules 1 & 2, weekly score tracking, and a free 30-minute in-person diagnostic before enrollment.
Digital SAT FAQ
What is on Module 1 of the Digital SAT Math section?
Module 1 is a 35-minute, 20 scored + 2 pretest question module that every student sees in the same fixed form. It draws from four content domains: Algebra, Advanced Math, Problem-Solving & Data Analysis, and Geometry & Trigonometry. The Bluebook Desmos calculator is available on every question.
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.
How long is the Digital SAT Math section?
70 minutes total, split into two 35-minute modules of 22 questions each (20 scored + 2 pretest). Together with the Reading & Writing section (64 minutes), the full Digital SAT is 2 hours 14 minutes.
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.
What test dates does SOMATH prepare for?
All U.S. Digital SAT dates — August, October, November, December, March, May, and June each year. Our small-group SAT classes run on rolling 12-week cohorts, so a student can start any month.
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.
Related posts
- Continue with the second math module: Digital SAT Practice Test 4 — Math Module 2 Walkthrough (All 27 Questions)
- Companion R&W walkthroughs: Digital SAT Practice Test 4 — Reading & Writing Module 1 Walkthrough and Module 2 Walkthrough
- Digital SAT Practice Test 10 — Math Module 1 Walkthrough (All 27 Questions)
- Digital SAT Practice Test 10 — Module 1 Reading & Writing Walkthrough
- SAT Prep on the Upper West Side of NYC (2026)
- Digital SAT Tutoring in NYC: Structure, Dates & Scores (2026)
- SAT Math vs. School Math: Why They Feel So Different