Digital SAT · Math · Exam Prep · NYC Test Prep

Digital SAT Practice Test 10 — Module 1: Full Walkthrough of All 27 Math Questions with Answers & Explanations

Every question on Digital SAT Practice Test #10, 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.

· 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 10 — Math Module 1? You are in the right place. This is a complete, question-by-question walkthrough of SAT Test #10, 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 walkthrough 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.

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Download SAT Practice Test 10 — Math Module 1 (PDF) Full 27-question walkthrough with figures described, choices, answer key, worked solutions, and per-choice explanations. 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, exponent rules, Pythagorean theorem, sum of roots, 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. Theory refresher · What Math Module 1 tests
  2. Section 1 · Algebra & Linear Systems (Q1–13)
  3. Section 2 · Advanced Math & Functions (Q14–20)
  4. Section 3 · Geometry & Trigonometry (Q21–24)
  5. Section 4 · Problem-Solving, Data Analysis & Advanced Algebra (Q25–27)
  6. Answer key at a glance
  7. How SOMATH prepares NYC students for the Digital SAT
  8. Digital SAT FAQ

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 (roughly 35% of the section)

Linear equations & inequalities in one and two variables. Solve, model, and interpret. Watch for “maximum” or “at most” language which means a ≤ inequality, and “fewer than” or “less than” which means a strict < inequality.

Systems of linear equations. Solve by substitution, elimination, or Desmos (graph both and read the intersection). For “What is x − y?” type items, solve both variables first, then compute.

Interpreting linear models. In y = mx + b, the slope m is the per-unit rate of change and the y-intercept b is the starting value / fixed cost / initial amount.

2 · Advanced Math (roughly 35% of the section)

Quadratic functions. Factored form y = a(x−r₁)(x−r₂) gives the roots directly. Vertex form y = a(x−h)² + k gives the vertex (h, k). Sum of roots of ax² + bx + c = 0 is −b/a; product is c/a.

Exponential functions. The general form is y = a · b^t: a is the starting value and b is the growth factor per unit of time. Doubling ⇒ b = 2. Growth by 10% ⇒ b = 1.10. Half-life ⇒ b = 1/2.

Function notation & transformations. g(x) = f(x) + k shifts the graph up by k. g(x) = f(x−h) shifts right by h. Corresponding sides of congruent triangles are equal; corresponding angles of similar triangles are equal.

3 · Problem-Solving & Data Analysis (roughly 15% of the section)

Percents. An increase of p% means multiply by 1 + p/100. Increasing x by 1,800% means the new value is 19x, not 18x.

Two-way tables & probability. Read the row and column totals carefully. P(event) = (favorable outcomes) / (total outcomes).

Rates & unit conversions. Cancel units to check your work — if the units come out right, the arithmetic usually does too.

4 · Geometry & Trigonometry (roughly 15% of the section)

Pythagorean theorem. In a right triangle with legs a, b and hypotenuse c: a² + b² = c². Common Pythagorean triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25.

Scale factor and area. If lengths scale by a factor of k, area scales by k² and volume scales by k³. Halving all sides of a rectangle produces a rectangle with one quarter the area.

Congruence vs. similarity. Two triangles are similar if all three pairs of angles are equal — but two similar triangles can be different sizes. To upgrade similarity to congruence, you need at least one pair of corresponding side lengths.

Equation of a circle. (x−h)² + (y−k)² = r² gives center (h, k) and radius r. A circle is tangent to the y-axis when |h| = r (touches at exactly one point) and tangent to the x-axis when |k| = r.

Every question below is transcribed verbatim from College Board’s Practice Test 10 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 equations & inequalities, linear systems, and function evaluation — the fastest-paced part of the module.

Question 1 · Reading a Line Graph

See figure in the original College Board PDF above.

The line graph shows the percent of cars for sale at a used car lot on a given day by model year. For what model year is the percent of cars for sale the smallest?

  • A) 2012
  • B) 2013
  • C) 2014
  • D) 2015

Answer: C) 2014

Read the y-value of each labeled dot on the line graph and pick the model year with the lowest point. 2010: ≈12% 2011: ≈12% 2012: ≈12% 2013: ≈10% 2014: ≈4% ← minimum 2015: ≈10%

✗ A) 2012
2012 sits at about 12%, one of the highest points on the graph, not the lowest.

✗ B) 2013
2013 sits at about 10% — lower than 2010–2012 but still well above 2014.

✓ C) 2014
2014 is the single lowest dot on the graph, at about 4%. This is the model year with the smallest percent of cars for sale.

✗ D) 2015
2015 sits at about 10%, back above the 2014 low.

Question 2 · System of Linear Equations (Graphical)

See figure in the original College Board PDF above.

The graph of a system of linear equations is shown. What is the solution (x, y) to the system?

  • A) (4, −5)
  • B) (0, 3)
  • C) (0, −2)
  • D) (−2, 3)

Answer: A) (4, −5)

The solution to a system shown on a graph is the point where the two lines intersect. Read the intersection point from the graph: (x, y) = (4, −5).

✓ A) (4, −5)
The two lines cross at the point four units right of the origin and five units down. That is (4, −5).

✗ B) (0, 3)
(0, 3) is the y-intercept of one line, not the intersection of both lines.

✗ C) (0, −2)
(0, −2) is not a point where either line crosses an axis; not the intersection.

✗ D) (−2, 3)
(−2, 3) is not the intersection point of the two lines.

Question 3 · Writing a Linear Inequality

The total cost, in dollars, to rent a surfboard consists of a $25 service fee and a $10 per hour rental fee. A person rents a surfboard for t hours and intends to spend a maximum of $75 to rent the surfboard. Which inequality represents this situation?

  • A) 10t ≤ 75
  • B) 10 + 25t ≤ 75
  • C) 25t ≤ 75
  • D) 25 + 10t ≤ 75

Answer: D) 25 + 10t ≤ 75

Total cost = fixed service fee + (hourly fee × hours). Total cost = 25 + 10t. Maximum $75 → 25 + 10t ≤ 75.

✗ A) 10t ≤ 75
Ignores the $25 service fee entirely.

✗ B) 10 + 25t ≤ 75
Reverses which number is the fixed fee and which is the hourly rate. $10 is per hour, not the fixed fee.

✗ C) 25t ≤ 75
Treats $25 as the hourly rate and ignores the $10 hourly rate — reversed.

✓ D) 25 + 10t ≤ 75
The $25 service fee is fixed, $10 is per hour for t hours (10t), and the total must be at most $75, so 25 + 10t ≤ 75.

Question 4 · Translating a Graph (Function Transformation)

See figure in the original College Board PDF above.

The graph shown will be translated up 4 units. Which of the following will be the resulting graph?

  • A) Parabola with vertex at (3, 2), opening upward, passing through the x-axis above y = 2.
  • B) Parabola with vertex at (3, −6), opening upward.
  • C) Parabola with vertex at (−3, −2), opening upward.
  • D) Parabola with vertex at (7, −2), opening upward.

Answer: A) Parabola with vertex at (3, 2), opening upward, passing through the x-axis above y = 2.

Translating a graph up by 4 units means every point (x, y) on the original graph moves to (x, y + 4). The original vertex is at about (3, −2). New vertex: (3, −2 + 4) = (3, 2). The shape does not change — only the height.

✓ A) Parabola with vertex at (3, 2), opening upward, passing through the x-axis above y = 2.
The vertex moves from (3, −2) up to (3, 2). Every point on the parabola shifts up by 4 units. This matches choice A.

✗ B) Parabola with vertex at (3, −6), opening upward.
This would be a shift down by 4, not up.

✗ C) Parabola with vertex at (−3, −2), opening upward.
This shifts the parabola left, not up.

✗ D) Parabola with vertex at (7, −2), opening upward.
This shifts the parabola right, not up.

Question 5 · Evaluating a Linear Function

s = 40 + 3t. The equation gives the speed s, in miles per hour, of a certain car t seconds after it began to accelerate. What is the speed, in miles per hour, of the car 5 seconds after it began to accelerate?

  • A) 40
  • B) 43
  • C) 45
  • D) 55

Answer: D) 55

Substitute t = 5 into s = 40 + 3t. s = 40 + 3(5) = 40 + 15 = 55.

✗ A) 40
This is the starting speed at t = 0. The question asks for speed at t = 5 seconds.

✗ B) 43
This is 40 + 3 — it uses t = 1, not t = 5.

✗ C) 45
This is 40 + 5 — it forgets to multiply 5 by the coefficient 3.

✓ D) 55
Substituting t = 5 gives s = 40 + 3(5) = 40 + 15 = 55 mph.

Question 6 · Evaluating a Quadratic Function

The function f is defined by f(x) = x² + x + 71. What is the value of f(2)?

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

Answer: 77

Substitute x = 2 into f(x) = x² + x + 71. f(2) = 2² + 2 + 71 = 4 + 2 + 71 = 77.

Why this works.
Substituting x = 2 gives f(2) = 4 + 2 + 71 = 77.

Question 7 · Linear Inequality (Budget Constraint)

An event planner is planning a party. It costs the event planner a one-time fee of $35 to rent the venue and $10.25 per attendee. The event planner has a budget of $300. What is the greatest number of attendees possible without exceeding the budget?

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

Answer: 25

Let n = number of attendees. Total cost: 35 + 10.25n ≤ 300 10.25n ≤ 265 n ≤ 265 / 10.25 ≈ 25.85. n must be a whole number of attendees, so n = 25.

Why this works.
The budget inequality 35 + 10.25n ≤ 300 gives n ≤ 25.85. Since the number of attendees must be a whole number, the greatest possible value is 25.

Question 8 · Two-Way Table Probability

See figure in the original College Board PDF above.

The table gives the distribution of votes for a new school mascot and grade level for 80 students. If one of these students is selected at random, what is the probability of selecting a student whose vote for new mascot was for a lion?

  • A) 1/9
  • B) 1/5
  • C) 1/4
  • D) 2/3

Answer: C) 1/4

P(Lion) = (Total votes for Lion) / (Total students) = 20 / 80 = 1/4.

✗ A) 1/9
1/9 is not the ratio 20/80. This does not match any row total.

✗ B) 1/5
1/5 = 16/80. There are 20 lion votes, not 16.

✓ C) 1/4
20 students voted Lion out of 80 total → 20/80 = 1/4.

✗ D) 2/3
2/3 ≈ 53/80 — far more than any single mascot received.

Question 9 · Congruent Triangles (Angle Sum)

Triangles ABC and DEF are congruent, where A corresponds to D, and B and E are right angles. The measure of angle A is 18°. What is the measure of angle F?

  • A) 18°
  • B) 72°
  • C) 90°
  • D) 162°

Answer: B) 72°

Corresponding angles of congruent triangles are equal, so angle F corresponds to angle C. Angles in a triangle sum to 180°. In triangle ABC: A + B + C = 180°. 18° + 90° + C = 180° → C = 72°. So angle F = 72°.

✗ A) 18°
18° is the measure of angle A, which corresponds to angle D — not angle F.

✓ B) 72°
Angle F corresponds to angle C. Since A + B + C = 180° and A = 18°, B = 90°, we get C = 72° = F.

✗ C) 90°
90° is the measure of angle B (and its corresponding angle E), not angle F.

✗ D) 162°
162° = 180° − 18°. This ignores the right angle at B.

Question 10 · Scaling a Linear Equation

If 4x + 2 = 12, what is the value of 16x + 8?

  • A) 40
  • B) 48
  • C) 56
  • D) 60

Answer: B) 48

Multiply both sides of 4x + 2 = 12 by 4: 4(4x + 2) = 4(12) 16x + 8 = 48. (No need to solve for x.)

✗ A) 40
40 comes from solving 4x = 10 → x = 2.5, then 16(2.5) = 40 — but this forgets the +8 term.

✓ B) 48
16x + 8 is exactly 4 times 4x + 2, so it equals 4(12) = 48.

✗ C) 56
56 would require 4x + 2 = 14, not 12.

✗ D) 60
60 = 5 × 12, but 16x + 8 is 4 × (4x + 2), not 5 ×.

Question 11 · Exponent Rules (Product of Monomials)

Which expression is equivalent to (m⁴q⁴z⁻¹)(mq⁵z³), where m, q, and z are positive?

  • A) m⁴q²⁰z⁻³
  • B) m⁵q⁹z²
  • C) m⁶q⁸z⁻¹
  • D) m²⁰q¹²z⁻²

Answer: B) m⁵q⁹z²

When multiplying like bases, add the exponents. m: 4 + 1 = 5 q: 4 + 5 = 9 z: −1 + 3 = 2 Result: m⁵q⁹z².

✗ A) m⁴q²⁰z⁻³
This multiplies the q exponents (4 × 5 = 20) instead of adding them, and mishandles the m and z exponents too.

✓ B) m⁵q⁹z²
Add exponents of like bases: m^(4+1) = m⁵, q^(4+5) = q⁹, z^(−1+3) = z². This gives m⁵q⁹z².

✗ C) m⁶q⁸z⁻¹
6 = 4 + 1 + 1 (extra m) and 8 = 4 + 4 (wrong for q). Doesn't match adding once correctly.

✗ D) m²⁰q¹²z⁻²
Multiplies the m exponents (4 × 1 × 5 = 20) — exponents add when multiplying like bases, they don't multiply.

Question 12 · Modeling with Functions (Linear vs Exponential)

An airplane descends from an altitude of 9,500 feet to 5,000 feet at a constant rate of 400 feet per minute. What type of function best models the relationship between the descending airplane's altitude and time?

  • A) Decreasing exponential
  • B) Decreasing linear
  • C) Increasing exponential
  • D) Increasing linear

Answer: B) Decreasing linear

"Constant rate" of change = linear function (same change per unit time). "Descending" means altitude is going down as time increases → decreasing. Combined: decreasing linear.

✗ A) Decreasing exponential
Exponential decay changes by a constant percent, not a constant amount, per unit time. The airplane loses 400 ft each minute (a constant amount), not a constant percent.

✓ B) Decreasing linear
A constant rate of change (400 ft/min every minute) is the signature of a linear function. Since altitude is decreasing, the linear function is decreasing.

✗ C) Increasing exponential
Altitude is decreasing, not increasing.

✗ D) Increasing linear
Altitude is decreasing, not increasing.

Question 13 · Solving a Linear System

3x + 6 = 4y and 3x + 4 = 2y. 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: 1

Both equations start with 3x. Subtract the second from the first: (3x + 6) − (3x + 4) = 4y − 2y 2 = 2y y = 1.

Why this works.
Subtracting the second equation from the first eliminates x and gives 2 = 2y, so y = 1.

Section 2 · Advanced Math & Functions

Quadratics, exponentials, function transformations, and interpreting function notation in context.

Question 14 · Function Transformation (Vertical Shift)

The function f is defined by f(x) = (x − 6)(x − 2)(x + 6). In the xy-plane, the graph of y = g(x) is the result of translating the graph of y = f(x) up 4 units. What is the value of g(0)?

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

Answer: 76

A shift up by 4 units means g(x) = f(x) + 4. First find f(0): f(0) = (0 − 6)(0 − 2)(0 + 6) = (−6)(−2)(6) = 72. Then g(0) = f(0) + 4 = 72 + 4 = 76.

Why this works.
Translating up 4 units means g(x) = f(x) + 4. f(0) = (−6)(−2)(6) = 72, so g(0) = 72 + 4 = 76.

Question 15 · Interpreting Function Notation in Context

The function f(w) = 6w² gives the area of a rectangle, in square feet (ft²), if its width is w ft and its length is 6 times its width. Which of the following is the best interpretation of f(14) = 1,176?

  • A) If the width of the rectangle is 14 ft, then the area of the rectangle is 1,176 ft².
  • B) If the width of the rectangle is 14 ft, then the length of the rectangle is 1,176 ft.
  • C) If the width of the rectangle is 1,176 ft, then the length of the rectangle is 14 ft.
  • D) If the width of the rectangle is 1,176 ft, then the area of the rectangle is 14 ft².

Answer: A) If the width of the rectangle is 14 ft, then the area of the rectangle is 1,176 ft².

In f(w) = 6w², the input w is the width (in ft) and the output f(w) is the area (in ft²). So f(14) = 1,176 means: when the width is 14 ft, the area is 1,176 ft².

✓ A) If the width of the rectangle is 14 ft, then the area of the rectangle is 1,176 ft².
The input 14 represents width in ft, and the output 1,176 represents area in ft². So f(14) = 1,176 means width 14 ft gives area 1,176 ft².

✗ B) If the width of the rectangle is 14 ft, then the length of the rectangle is 1,176 ft.
1,176 is the area, not the length. The length would be 6·14 = 84 ft.

✗ C) If the width of the rectangle is 1,176 ft, then the length of the rectangle is 14 ft.
Reverses input and output: 14 is the input (width), not the output.

✗ D) If the width of the rectangle is 1,176 ft, then the area of the rectangle is 14 ft².
Reverses input and output and swaps units.

Question 16 · Exponential Growth Model (Doubling)

The number of bacteria in a liquid medium doubles every day. There are 44,000 bacteria in the liquid medium at the start of an observation. Which of the following represents the number of bacteria, y, in the liquid medium t days after the start of the observation?

  • A) y = (1/2)(44,000)ᵗ
  • B) y = 2(44,000)ᵗ
  • C) y = 44,000(1/2)ᵗ
  • D) y = 44,000(2)ᵗ

Answer: D) y = 44,000(2)ᵗ

Exponential growth: y = (starting value) · (growth factor)^t. Starting value = 44,000. Doubles each day → growth factor = 2. y = 44,000 · (2)ᵗ.

✗ A) y = (1/2)(44,000)ᵗ
The starting value should be the base 44,000, not (1/2). Also, doubling means multiply by 2, not by (1/2).

✗ B) y = 2(44,000)ᵗ
Uses 44,000 as the base of the exponent, which grows far too fast.

✗ C) y = 44,000(1/2)ᵗ
(1/2)ᵗ means the bacteria are halving each day, but the problem says doubling.

✓ D) y = 44,000(2)ᵗ
Starting with 44,000 and doubling each day gives y = 44,000·(2)ᵗ.

Question 17 · Exponential Function from a Table

See figure in the original College Board PDF above.

The table shows the exponential relationship between the number of years, x, since Hana started training in pole vault, and the estimated height h(x), in meters, of her best pole vault for that year. Which of the following functions best represents this relationship, where x ≤ 4?

  • A) h(x) = 1.12(0.23)ˣ
  • B) h(x) = 1.12(1.23)ˣ
  • C) h(x) = 1.23(0.12)ˣ
  • D) h(x) = 1.23(1.12)ˣ

Answer: D) h(x) = 1.23(1.12)ˣ

Exponential model: h(x) = a · bˣ. At x = 0: h(0) = a · b⁰ = a. Table says h(0) = 1.23, so a = 1.23. Find growth factor b using x = 2: h(2) = 1.23 · b² = 1.54. b² = 1.54 / 1.23 ≈ 1.2520. b ≈ √1.2520 ≈ 1.12. So h(x) = 1.23(1.12)ˣ. Check x = 4: 1.23·(1.12)⁴ ≈ 1.23·1.5735 ≈ 1.94 ✓.

✗ A) h(x) = 1.12(0.23)ˣ
a = 1.12 is wrong: h(0) in the table is 1.23, not 1.12. Base 0.23 would also decrease h, not increase it.

✗ B) h(x) = 1.12(1.23)ˣ
Correct base (1.23) but wrong a: 1.12 doesn't match h(0) = 1.23.

✗ C) h(x) = 1.23(0.12)ˣ
Base 0.12 causes h to shrink rapidly, contradicting the increasing values in the table.

✓ D) h(x) = 1.23(1.12)ˣ
a = h(0) = 1.23, and multiplying by 1.12 each year matches: 1.23→1.54→1.94, consistent with the table.

Question 18 · Intercepts of a Linear Function

The function h is defined by h(x) = 4x + 28. The graph of y = h(x) in the xy-plane has an x-intercept at (a, 0) and a y-intercept at (0, b), where a and b are constants. What is the value of a + b?

  • A) 21
  • B) 28
  • C) 32
  • D) 35

Answer: A) 21

y-intercept: set x = 0 → h(0) = 4(0) + 28 = 28. So b = 28. x-intercept: set h(x) = 0 → 4x + 28 = 0 → x = −7. So a = −7. a + b = −7 + 28 = 21.

✓ A) 21
a = −7 (from 4x + 28 = 0) and b = 28 (from x = 0). Sum: −7 + 28 = 21.

✗ B) 28
28 is just b, the y-intercept — forgets to add a.

✗ C) 32
32 would require a = 4, but 4x + 28 = 0 gives x = −7, not 4.

✗ D) 35
35 = 28 + 7. Uses |a| instead of a. The x-intercept a is negative, not positive.

Question 19 · Linear Inequality — Table Check

See figure in the original College Board PDF above.

y < 5x + 6. For which of the following tables are all the values of x and their corresponding values of y solutions to the given inequality?

  • A) Table A: (3, 17), (5, 27), (7, 37).
  • B) Table B: (3, 17), (5, 35), (7, 37).
  • C) Table C: (3, 25), (5, 35), (7, 45).
  • D) Table D: (3, 21), (5, 31), (7, 41).

Answer: A) Table A: (3, 17), (5, 27), (7, 37).

Compute 5x + 6 for each row and require y < 5x + 6. A: (3,17): 5(3)+6 = 21; 17 < 21 ✓ ; (5,27): 31; 27 < 31 ✓ ; (7,37): 41; 37 < 41 ✓ — all satisfy. B: (3,17): 17 < 21 ✓ ; (5,35): 5(5)+6=31; 35 < 31? No. C: (3,25): 25 < 21? No. D: (3,21): 21 < 21? No — the inequality is strict.

✓ A) Table A: (3, 17), (5, 27), (7, 37).
All three rows satisfy y < 5x + 6: 17<21, 27<31, 37<41.

✗ B) Table B: (3, 17), (5, 35), (7, 37).
At x = 5, 5(5)+6 = 31, and 35 is NOT less than 31. Fails.

✗ C) Table C: (3, 25), (5, 35), (7, 45).
At x = 3, 5(3)+6 = 21, and 25 is NOT less than 21. Fails at the very first row.

✗ D) Table D: (3, 21), (5, 31), (7, 41).
At x = 3, y = 21 and 5(3)+6 = 21. The inequality is strict (<), so 21 < 21 is false.

Question 20 · Solving a Linear System

y = 4x + 1 and 4y = 15x − 8. The solution to the given system of equations is (x, y). What is the value of x − y?

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

Answer: 35

Substitute y = 4x + 1 into 4y = 15x − 8: 4(4x + 1) = 15x − 8 16x + 4 = 15x − 8 x = −12. Then y = 4(−12) + 1 = −47. x − y = −12 − (−47) = −12 + 47 = 35.

Why this works.
Substituting y = 4x + 1 gives 16x + 4 = 15x − 8, so x = −12 and y = −47. Then x − y = −12 − (−47) = 35.

Section 3 · Geometry & Trigonometry

Pythagorean theorem, scale factors and area, equations of circles, and triangle congruence.

Question 21 · Pythagorean Theorem (Simplifying a Radical)

A right triangle has legs with lengths of 24 centimeters and 21 centimeters. If the length of this triangle's hypotenuse, in centimeters, can be written in the form 3√d, where d is an integer, what is the value of d?

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

Answer: 113

Pythagorean theorem: c² = a² + b². c² = 24² + 21² = 576 + 441 = 1,017. c = √1,017. Factor out the largest perfect square: 1,017 = 9 · 113 (113 is prime). c = √(9·113) = 3√113. So d = 113.

Why this works.
c = √(24² + 21²) = √1,017 = √(9·113) = 3√113, so d = 113.

Question 22 · Scale Factor and Area

The floor of a ballroom has an area of 600 square meters. An architect creates a scale model of the floor of the ballroom, where the length of each side of the model is 1/10 times the length of the corresponding side of the actual floor of the ballroom. What is the area, in square meters, of the scale model?

  • A) 6
  • B) 10
  • C) 60
  • D) 150

Answer: A) 6

When you scale all lengths by a factor k, the area scales by k². Here k = 1/10, so k² = 1/100. Model area = 600 · (1/100) = 6 square meters.

✓ A) 6
Sides scale by 1/10, so area scales by (1/10)² = 1/100. 600 · 1/100 = 6.

✗ B) 10
10 = 600 · (1/60). This treats the linear scale factor as an area scale factor incorrectly.

✗ C) 60
60 = 600 · (1/10). This uses the linear scale factor 1/10 on the area directly — but area scales by the SQUARE of the linear factor.

✗ D) 150
150 = 600 · (1/4). This uses (1/2)² instead of (1/10)².

Question 23 · Equation of a Circle (Tangent to Axis)

Which of the following equations represents a circle in the xy-plane that intersects the y-axis at exactly one point?

  • A) (x − 8)² + (y − 8)² = 16
  • B) (x − 8)² + (y − 4)² = 16
  • C) (x − 4)² + (y − 9)² = 16
  • D) x² + (y − 9)² = 16

Answer: C) (x − 4)² + (y − 9)² = 16

In (x−h)² + (y−k)² = r², center = (h, k) and radius = r. A circle intersects the y-axis at exactly one point when it is TANGENT to the y-axis: |h| = r. All four choices have r² = 16, so r = 4. We need |h| = 4. A) center (8, 8): |h| = 8 ≠ 4 → does not touch the y-axis at all. B) center (8, 4): |h| = 8 ≠ 4 → does not touch the y-axis at all. C) center (4, 9): |h| = 4 = r → tangent to the y-axis at exactly one point ✓. D) center (0, 9): |h| = 0 < 4 → crosses the y-axis at two points.

✗ A) (x − 8)² + (y − 8)² = 16
Center (8, 8), radius 4. Distance from center to the y-axis is 8, which is greater than 4, so the circle does not touch the y-axis at all.

✗ B) (x − 8)² + (y − 4)² = 16
Center (8, 4), radius 4. Distance from center to the y-axis is 8, greater than 4, so it does not touch the y-axis.

✓ C) (x − 4)² + (y − 9)² = 16
Center (4, 9), radius 4. The distance from the center to the y-axis equals the radius, so the circle is tangent to the y-axis — it touches at exactly one point.

✗ D) x² + (y − 9)² = 16
Center (0, 9), radius 4. The center is ON the y-axis, so the circle crosses the y-axis at two points, not one.

Question 24 · Triangle Congruence (SSS/SAS/ASA/AAS)

In triangles ABC and DEF, angles B and E each have measure 27° and angles C and F each have measure 41°. Which additional piece of information is sufficient to determine whether triangle ABC is congruent to triangle DEF?

  • A) The measure of angle A
  • B) The length of side AB
  • C) The lengths of sides BC and EF
  • D) No additional information is necessary.

Answer: C) The lengths of sides BC and EF

We already know two pairs of corresponding angles are equal (∠B=∠E=27°, ∠C=∠F=41°). Since the angles of a triangle sum to 180°, the third pair (∠A and ∠D) are automatically equal too — the triangles are SIMILAR. But similarity alone is not congruence: two triangles can be similar without being congruent (one could be a scaled-up version of the other). To pin down congruence we need at least one pair of corresponding SIDES that are equal. That's exactly what choice C gives: BC and EF are corresponding sides (both opposite the 27° angle in their triangle). If BC = EF, the triangles are congruent by AAS (two angles + a corresponding side).

✗ A) The measure of angle A
The measure of angle A adds no new info — we can already compute A = 180° − 27° − 41° = 112° from the two given angles. Still no side information, so still can't confirm congruence.

✗ B) The length of side AB
AB is a side of triangle ABC, but this choice does not tell us anything about DEF's corresponding side. Only one triangle's side is described.

✓ C) The lengths of sides BC and EF
BC (opposite ∠A in triangle ABC) and EF (opposite ∠D in triangle DEF) are CORRESPONDING sides. Combined with the two equal angle pairs already known, this gives AAS congruence — sufficient to determine congruence.

✗ D) No additional information is necessary.
Two pairs of equal angles establish similarity, NOT congruence. The triangles could be different sizes. More information IS necessary.

Section 4 · Problem-Solving, Data Analysis & Advanced Algebra

Percent change, piecewise cost models, and the sum of roots of a quadratic — the last three items on the module.

Question 25 · Percent Increase (Solving Backwards)

The result of increasing the quantity x by 1,800% is 684. What is the value of x?

  • A) 12,996
  • B) 12,312
  • C) 38
  • D) 36

Answer: D) 36

Increasing by 1,800% means adding 18 times the original to itself: new = x + 18x = 19x. 19x = 684. x = 684 / 19 = 36.

✗ A) 12,996
12,996 = 684 × 19. This applies the increase forward instead of undoing it.

✗ B) 12,312
12,312 = 684 × 18. Also applies the increase in the wrong direction.

✗ C) 38
38 = 684 / 18. Forgets to add the original x back — an increase of 1,800% multiplies by 19 total, not 18.

✓ D) 36
An 1,800% increase means the new amount is 19 times the original (1x + 18x). 19x = 684, so x = 36.

Question 26 · Piecewise Linear Cost Function

A window repair specialist charges $220 for the first two hours of repair plus an hourly fee for each additional hour. The total cost for 5 hours of repair is $400. Which function f gives the total cost, in dollars, for x hours of repair, where x ≥ 2?

  • A) f(x) = 60x + 100
  • B) f(x) = 60x + 220
  • C) f(x) = 80x
  • D) f(x) = 80x + 220

Answer: A) f(x) = 60x + 100

The $220 covers the first 2 hours. Beyond that, each extra hour costs some fixed rate r. For 5 hours (3 hours beyond the base 2): total = 220 + 3r = 400 → 3r = 180 → r = $60/hour. So for x hours (x ≥ 2), extra hours = (x − 2): f(x) = 220 + 60(x − 2) = 220 + 60x − 120 = 60x + 100. Check: f(5) = 60(5) + 100 = 400 ✓; f(2) = 220 ✓.

✓ A) f(x) = 60x + 100
Additional hourly rate r satisfies 220 + 3r = 400, so r = 60. Then f(x) = 220 + 60(x − 2) = 60x + 100. Check: f(2) = 220 ✓, f(5) = 400 ✓.

✗ B) f(x) = 60x + 220
f(2) = 60(2) + 220 = 340 ≠ 220. Also f(5) = 520 ≠ 400. Fails both anchor points.

✗ C) f(x) = 80x
f(2) = 160 ≠ 220. Doesn't respect the flat $220 for the first two hours.

✗ D) f(x) = 80x + 220
f(2) = 380 and f(5) = 620 — neither anchor point works.

Question 27 · Sum of Roots (Quadratic)

x(x + 1) − 56 = 4x(x − 7). What is the sum of the solutions to the given equation?

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

Answer: 29/3

Expand both sides: Left: x² + x − 56. Right: 4x² − 28x. Bring everything to one side: x² + x − 56 − 4x² + 28x = 0 −3x² + 29x − 56 = 0. Multiply by −1: 3x² − 29x + 56 = 0. For a quadratic ax² + bx + c = 0, sum of roots = −b/a. Sum = −(−29) / 3 = 29/3.

Why this works.
After expanding and collecting terms, the equation becomes 3x² − 29x + 56 = 0. For a quadratic ax² + bx + c = 0, the sum of the solutions equals −b/a, so −(−29)/3 = 29/3 (≈ 9.667).

Answer key at a glance

All 27 answers in one table for quick review:

QAnswerQAnswerQAnswer
1C) 201410B) 4819A) Table A: (3, 17), (5, 27), (7, 37).
2A) (4, −5)11B) m⁵q⁹z²2035
3D) 25 + 10t ≤ 7512B) Decreasing linear21113
4A) Parabola with vertex at (3, 2), open13122A) 6
5D) 55147623C) (x − 4)² + (y − 9)² = 16
67715A) If the width of the rectangle is 14 24C) The lengths of sides BC and EF
72516D) y = 44,000(2)ᵗ25D) 36
8C) 1/417D) h(x) = 1.23(1.12)ˣ26A) f(x) = 60x + 100
9B) 72°18A) 212729/3

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.

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.

Book Free Evaluation →   or call (646) 668-6151

SOMATH course · Grades 9–12

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.

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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.

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