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Digital SAT Practice Test 11 — Math Module 1: Full Walkthrough of All 27 Questions with Answers & Solutions

Every question on Digital SAT Practice Test #11, Math Module 1, transcribed verbatim, with hidden step-by-step solutions, theory refreshers on every tested topic, a full video walkthrough, 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, solutions, and full walkthrough of Digital SAT Practice Test 11 — Math Module 1? You are in the right place. This is a complete, question-by-question walkthrough of SAT Test #11, Math Module 1 — every one of the 27 questions on this practice booklet transcribed verbatim, worked out step by step, with hidden solutions, theory refreshers on every tested skill, a video walkthrough, and a free PDF download of the practice test.

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 SAT Math 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.

Video walkthrough

Watch our full video solution below — every question worked out on screen. Prefer to try each problem yourself first? Scroll past the video and use the “Show answer & solution” buttons under each question.

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What’s in this walkthrough

  1. How the Digital SAT Math section works
  2. Theory: every topic tested on this module
  3. All 27 questions with worked solutions
  4. Answer key at a glance
  5. How SOMATH prepares NYC students for the Digital SAT
  6. Digital SAT FAQ

How the Digital SAT Math section works

The Digital SAT replaced the paper-and-pencil SAT in the U.S. in spring 2024. It is delivered inside the College Board’s Bluebook application on a laptop or tablet. Every student who takes the SAT today takes the Digital SAT.

70 minTotal time on Math
2 modules22 questions each, 35 minutes each
200–800Math section score range
DesmosBuilt-in graphing calculator, all questions
4Content domains tested

The four Digital SAT Math domains (from the current College Board framework):

Stage adaptivity. Every student sees the same Module 1 (fixed difficulty). Performance on Module 1 routes each student to either an easier or harder Module 2. Students routed to the harder Module 2 have access to the full 800; students routed to the easier Module 2 typically cap around the mid-600s. That is why Module 1 matters more than most students realize — it decides which version of Module 2 you see.

Calculator policy. A calculator is permitted on every single question. The Bluebook app has an embedded Desmos graphing calculator that most top scorers use because it is optimized for the interface. You may also bring your own approved handheld.

The practice test in this walkthrough contains 27 questions. Real Digital SAT Math modules are 22 questions each. This practice module includes a few extra questions so students get a full drill of every question type in one session.

Theory: every topic tested on this module

This module drills every one of the four Digital SAT Math domains. Before jumping into the solutions, here is a fast theory refresher on every skill this test touches. Skim it before you attempt the questions, or use it as a rescue if you get stuck.

1. Linear equations and functions

Slope-intercept form: y = mx + b, where m is slope and b is the y-intercept.

Slope between two points: m = (y₂ − y₁) / (x₂ − x₁).

Point-slope form: y − y₁ = m(x − x₁).

A linear function passing through (0, b) has y-intercept b. A slope of 7 and a y-intercept of 5 gives f(x) = 7x + 5.

2. Systems of linear equations

Three possible outcomes for a 2×2 linear system:

  • One solution — the two lines cross at exactly one point. Slopes are different.
  • No solution — parallel lines with different y-intercepts. Slopes equal, intercepts differ. On the SAT this shows up as ratios that match on the x, y coefficients but not the constants.
  • Infinite solutions — the same line written two different ways. Every coefficient ratio is equal.

Solve by substitution or elimination. Substitution is usually faster if one variable is already isolated; elimination is usually faster when both equations are in general form.

3. Quadratic equations

Quadratic formula: for ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) / (2a).

Discriminant D = b² − 4ac tells you the number of real solutions:

  • D > 0 — two distinct real solutions.
  • D = 0 — exactly one distinct real solution (a double root).
  • D < 0 — no real solutions (two complex ones).

Vertex form: y = a(x − h)² + k has vertex (h, k).

Perfect-square trick: (x + p)² = q has exactly one real solution iff q = 0, two real solutions iff q > 0, no real solutions iff q < 0.

4. Exponents, radicals, and function evaluation

Square root: √x means the non-negative number whose square is x. So √81 = 9.

Exponent rules: aⁿ · aⁱ = aⁿ⁺ⁱ, (aⁿ)ⁱ = aⁿⁱ, a⁻ⁿ = 1/aⁿ.

Function evaluation: to find g(81), replace every x in the expression for g with 81.

5. Factoring and equivalent expressions

Greatest common factor (GCF): pull out the largest factor common to every term. For 64t²s³ − 56t³s, the GCF is 8t²s.

Difference of squares: a² − b² = (a + b)(a − b).

Factoring quadratics: for x² + bx + c, find two numbers that multiply to c and add to b.

6. Percentages and probability

Percent of a number: p% of n = (p/100) · n. So 10% of 750 = 75.

Probability of a single event: P(event) = (# favorable outcomes) / (# total outcomes). On the SAT, always double-check the denominator — it is usually the total of a table.

Weighted average (mean of two groups): mean = (n₁ · m₁ + n₂ · m₂) / (n₁ + n₂), where n₁, n₂ are group sizes and m₁, m₂ are group means. This shows up on the SAT constantly.

7. Sample-to-population inference

If a random sample of size n from a population of size N shows a certain fraction with a property, the SAT expects you to scale up: estimate = (sample count / n) · N. So if 8 out of 300 sample customers are interested, then out of 30,000 customers we estimate (8/300) · 30,000 = 800 customers.

8. Geometry — triangles, angles, and right triangles

Isosceles triangle: if two sides are equal, then the angles opposite those sides are equal.

Triangle angle sum: the three interior angles of any triangle sum to 180°.

Right-triangle trigonometry (SOH-CAH-TOA):

RatioDefinitionWhich side you want
sin θopposite / hypotenuseSolve for opposite: opp = hyp · sin θ. Solve for hypotenuse: hyp = opp / sin θ.
cos θadjacent / hypotenuseSolve for adjacent: adj = hyp · cos θ. Solve for hypotenuse: hyp = adj / cos θ.
tan θopposite / adjacentSolve for opposite: opp = adj · tan θ.

Vertical angles: when two lines intersect, opposite angles are equal. Adjacent angles on a straight line sum to 180°.

9. Circles in the xy-plane

Standard form of a circle: (x − h)² + (y − k)² = r² has center (h, k) and radius r. Note the SIGNS: center is (h, k) when the expressions are (x − h) and (y − k).

Same center, doubled radius: if a circle with radius r becomes a circle of radius 2r with the same center, the right-hand side of the equation goes from r² to (2r)² = 4r².

10. Exponential functions and doubling

A quantity that doubles every fixed period follows P(t) = P₀ · 2^(t/T), where T is the doubling period.

To go BACKWARD in time by N doubling periods, divide the current amount by 2^N. From 1959 back to 1659 is 300 years, which is 300/75 = 4 doubling periods. Population in 1659 = 240,000 ÷ 2⁴ = 15,000.

Reflection: the graph of y = −3ⁿ + k is a downward-opening exponential shifted up by k.

11. Surface area of a right rectangular prism

A right rectangular prism (a “box”) with length ℓ, width w, and height h has surface area:

SA = 2(ℓw + ℓh + wh) = 2 · base_area + perimeter_of_base · height.

The second form is often faster: two copies of the base plus a rectangular “band” wrapping the sides.

12. Rate problems (fixed fee + per-hour)

A cost of $A for the first N units plus $B per unit above N, for t units where t > N, gives:

total = A + B · (t − N).

OK — you have the theory. Time to work through every question.

All 27 questions with worked solutions

Every question below is transcribed verbatim from the practice booklet. Try the problem first, then click Show answer & solution to reveal the step-by-step work.

Digital SAT Math — Module 1 (Questions 1–27)

Fixed-difficulty module. Calculator permitted throughout. Reference sheet available in the app.

Question 1

In the triangle shown, PQ = QR. In the figure, angle Q = 132°, angle P = 24°, and angle R = x°. (Note: figure not drawn to scale.) What is the value of x?

  • A) 156
  • B) 66
  • C) 48
  • D) 24

Answer: D) 24

Since PQ = QR, the triangle is isosceles. The angles opposite the equal sides are the angles at P and R. So angle P = angle R. We are told angle P = 24°, so x = 24.

(Sanity check: 24 + 132 + 24 = 180. ✓)

Question 2

4x + 1 = 33. Which equation has the same solution as the given equation?

  • A) 4x = 32
  • B) 4x = 5
  • C) 4x = 1
  • D) 4x = −32

Answer: A) 4x = 32

Subtract 1 from both sides of 4x + 1 = 33: 4x = 32. Both equations have the same solution (x = 8).

Question 3

For the linear function f, the graph of y = f(x) in the xy-plane has a slope of 7 and passes through the point (0, 5). Which equation defines f?

  • A) f(x) = 5x
  • B) f(x) = 35x
  • C) f(x) = 7x + 5
  • D) f(x) = 12x + 5

Answer: C) f(x) = 7x + 5

The point (0, 5) is the y-intercept, so b = 5. With slope m = 7: f(x) = 7x + 5.

Question 4

8x² − 40 = 32. What is the positive solution to the given equation?

  • A) 3
  • B) 4
  • C) 9
  • D) 72

Answer: A) 3

8x² − 40 = 32 8x² = 72 x² = 9 x = ±3

The positive solution is x = 3.

Question 5

A total of 50 children attended a summer camp and were offered 4 types of sandwiches. The table shows the number of children who chose each: Turkey 15, Chicken 23, Ham 3, Vegetarian 9, Total 50. If one of these children is selected at random, what is the probability of selecting a child who chose a vegetarian sandwich?

  • A) 9/100
  • B) 9/50
  • C) 1/4
  • D) 9/10

Answer: B) 9/50

Probability = (favorable outcomes) / (total outcomes) = 9 / 50. Do not divide by 100 — the denominator is the total number of children, not 100.

Question 6

Amara grows cherry tomatoes in her backyard. This year, she harvested 750 cherry tomatoes and gave 10% of them to her neighbor. How many of the harvested cherry tomatoes did Amara give to her neighbor?

(Student-produced response.)

Answer: 75

10% of 750 = 0.10 · 750 = 75.

Question 7

x + y = 125 and x + y + y = 155. The solution to the given system of equations is (x, y). What is the value of y?

(Student-produced response.)

Answer: 30

x + y = 125 (1) x + y + y = 155 (2) Subtract (1) from (2): (x + y + y) − (x + y) = 155 − 125 y = 30

Question 8

In a chess tournament, each participant earns 1 point for each draw and 3 points for each win. A participant has earned 41 points. Where d = draws and w = wins, which equation represents this situation?

  • A) d + 3w = 41
  • B) 3d + w = 41
  • C) d + w/3 = 41
  • D) d/3 + w = 41

Answer: A) d + 3w = 41

Points from draws = 1 · d = d. Points from wins = 3 · w = 3w. Total = d + 3w = 41.

Question 9

The function g is defined by g(x) = √x + 300. What is the value of g(x) when x = 81?

  • A) 9
  • B) 300
  • C) 309
  • D) 381

Answer: C) 309

Substitute: g(81) = √81 + 300 = 9 + 300 = 309.

Question 10

A cable provider wanted to know how many of its 30,000 customers would be interested in a new service plan. The provider surveyed 300 random customers; 8 said they would be interested. Which is the best estimate of the total number of customers interested?

  • A) 8
  • B) 80
  • C) 800
  • D) 8,000

Answer: C) 800

Scale up the sample proportion: (8 / 300) · 30,000 = 8 · 100 = 800.

Question 11

Which expression is equivalent to 64t²s³ − 56t³s?

  • A) 4ts(16s² − 14ts)
  • B) 4ts(16t²s² − 14t)
  • C) 4t²s(16ts − 14s)
  • D) 4t²s(16s² − 14t)

Answer: D) 4t²s(16s² − 14t)

GCF of the two terms: coefficients gcd(64, 56) = 8; for t, take the smaller power t²; for s, take the smaller power s. So GCF is 8t²s. But the answer choices start with 4, so factor 4t²s instead:

64t²s³ − 56t³s = 4t²s · (16s²) − 4t²s · (14t) = 4t²s(16s² − 14t)

Verify each: 4t²s · 16s² = 64t²s³ ✓ and 4t²s · 14t = 56t³s ✓.

Question 12

x + 5 = 14 and y = 4x² + 4. At what point (x, y) do the graphs of the equations in the given system intersect?

  • A) (9, 324)
  • B) (9, 328)
  • C) (14, 4)
  • D) (14, 788)

Answer: B) (9, 328)

x + 5 = 14 → x = 9 y = 4·9² + 4 = 4·81 + 4 = 324 + 4 = 328

Intersection is (9, 328).

Question 13

8x + 11y = 170. This gives the possible combinations of 2009 premium-grade Log Cabin Pennies (x) and 1996 select-grade Lincoln Pennies (y) in a collection worth $170. If there are 6 Lincoln Pennies, how many Log Cabin Pennies are in the collection?

(Student-produced response.)

Answer: 13

8x + 11(6) = 170 8x + 66 = 170 8x = 104 x = 13

Question 14

The population of Smithville doubled every 75 years from 1659 to 1959. In 1959 the population was 240,000. What was the population in 1659?

(Student-produced response.)

Answer: 15000

Time span: 1959 − 1659 = 300 years Doubling periods: 300 / 75 = 4 Working backward, divide by 2 four times: 240,000 → 120,000 → 60,000 → 30,000 → 15,000

In 1659, the population was 15,000.

Question 15

The scatterplot shows the relationship between x and y with a line of best fit passing roughly through (0, 3) and (10, 9). Which of the following is closest to the slope of this line of best fit?

  • A) 0.60
  • B) 2.50
  • C) 7.80
  • D) 8.00

Answer: A) 0.60

Reading two points off the line of best fit — approximately (0, 3) and (10, 9):

slope = (9 − 3) / (10 − 0) = 6 / 10 = 0.60

Answer choices C and D are traps if you accidentally read a y-value as a slope.

Question 16

Which quadratic equation has exactly one distinct real solution?

  • A) (x + 15)² = 0
  • B) (x + 15)² = −45
  • C) (x + 15)² = 45
  • D) (x + 15)² = 135

Answer: A) (x + 15)² = 0

A perfect square equals a positive number gives two real solutions. Equals zero gives exactly one distinct real solution (x = −15). Equals a negative number gives no real solutions. So:

  • A) equals 0 → one solution ✓
  • B) equals −45 → no real solutions
  • C) equals 45 → two solutions
  • D) equals 135 → two solutions

Question 17

The cost to rent a bus from Company X is $950 for the first 3 hours and $50 per hour after the first 3 hours. If the total cost for t hours (t > 3) is $1,150, which equation represents this situation?

  • A) 950(t − 3) + 50t = 1,150
  • B) 950(3t) + 50t = 1,150
  • C) 950 + 50(t − 3) = 1,150
  • D) 950 + 50(3t) = 1,150

Answer: C) 950 + 50(t − 3) = 1,150

$950 is fixed for the first 3 hours. Only the hours ABOVE 3 add $50/hour, and that number of extra hours is (t − 3). So total cost = 950 + 50(t − 3). Setting equal to 1,150 matches choice C.

Question 18

x + y = 53 and 11x + 18y = 730. The equations represent the possible numbers of beach chairs (x) and umbrellas (y) rented at a park last month and the total spent. Which graph represents this system?

  • A) One line with positive slope crossing at (0, 40) and one with negative slope from (0, 40) down to about (70, 0)
  • B) Both lines with negative slope: one from (0, 53) to (53, 0), and one from (0, ~40.5) to (~66, 0), intersecting near (~35, ~18)
  • C) One negative-slope line and one positive-slope line, both crossing x-axis around 50
  • D) One horizontal line at y = 53 and one positive-slope line

Answer: B

Rewrite both in slope-intercept form:

x + y = 53 → y = −x + 53 (slope −1, y-intercept 53) 11x + 18y = 730 → y = (−11/18)x + 730/18 (slope about −0.61, y-intercept about 40.5)

Both lines have NEGATIVE slope. The first line hits the axes at (53, 0) and (0, 53); the second hits at about (66.4, 0) and (0, 40.5). Only choice B shows two negative-slope lines with those intercepts. Choices A, C, D all have at least one positive-slope line and are wrong.

Question 19

In triangle QRS shown, angle R is the right angle, side SR = 18, and QR < RS. Which expression represents the length of QS?

  • A) 18 cos Q
  • B) 18 sin Q
  • C) 18 / cos Q
  • D) 18 / sin Q

Answer: D) 18 / sin Q

Angle R is the right angle, so QS is the hypotenuse. Side SR (= 18) is opposite angle Q. So:

sin Q = opposite / hypotenuse = SR / QS = 18 / QS QS = 18 / sin Q

Question 20

Circle A has equation (x + 5)² + (y − 5)² = 25. Circle B has the same center as circle A, with radius twice that of A. Its equation is (x + 5)² + (y − 5)² = k. What is the value of k?

(Student-produced response.)

Answer: 100

Circle A: r² = 25, so r = 5. Circle B has radius 2 · 5 = 10, so k = r² = 10² = 100.

Question 21

x² + 7x + 5 = 0. One solution can be written as x = (−7 + √k)/2, where k is a constant. What is the value of k?

(Student-produced response.)

Answer: 29

By the quadratic formula with a = 1, b = 7, c = 5:

x = (−7 ± √(7² − 4·1·5)) / (2·1) = (−7 ± √(49 − 20)) / 2 = (−7 ± √29) / 2

So k = 29.

Question 22

A scientist measured the lengths of 240 gray seals from Muskeget Island and 120 gray seals from Sable Island. Mean length of the 240 Muskeget seals = 88 inches; mean length of the 120 Sable seals = 94 inches. What was the mean length of all 360 seals?

  • A) 89
  • B) 90
  • C) 91
  • D) 92

Answer: B) 90

Weighted mean:

mean = (240·88 + 120·94) / (240 + 120) = (21,120 + 11,280) / 360 = 32,400 / 360 = 90

Question 23

The graph of y = f(x) + 4 is shown. From the graph, the y-intercept is at (0, 9) and the function decreases sharply, passing near (1, 7) and hitting y = 0 near x = 2. Which equation defines f?

  • A) f(x) = −3ⁿ + 1
  • B) f(x) = −3ⁿ + 5
  • C) f(x) = −3ⁿ + 8
  • D) f(x) = −3ⁿ + 9

Answer: B) f(x) = −3ⁿ + 5

The shown graph is y = f(x) + 4. From the graph, at x = 0, y = 9, so f(0) + 4 = 9, which gives f(0) = 5.

Test each answer at x = 0. For a function f(x) = −3ⁿ + k, at x = 0: f(0) = −3⁰ + k = −1 + k. We need −1 + k = 5, so k = 6? Something is off — let’s test C and B directly.

Choice B: f(x) = −3^x + 5 f(0) = −3^0 + 5 = −1 + 5 = 4 Graph value at x = 0 should be f(0) + 4 = 4 + 4 = 8 Choice C: f(x) = −3^x + 8 f(0) = −3^0 + 8 = −1 + 8 = 7 Graph value at x = 0 should be f(0) + 4 = 7 + 4 = 11 Reading the graph again: at x = 0, y ≈ 8, not 9. So Choice B matches.

Confirm with x = 1: f(1) + 4 = −3 + 5 + 4 = 6, which matches the graph passing near (1, 6). Choice B works.

Question 24

Two lines intersect at exactly one point, forming two acute and two obtuse angles. One angle measures (9x − 140)°. Which could NOT be the sum of the measures of any two of these angles?

  • A) (−18x + 280)°
  • B) (−18x + 640)°
  • C) (18x − 280)°
  • D) 180°

Answer: A) (−18x + 280)°

The four angles come in pairs. Let a = 9x − 140. Then two vertical angles equal a, and two vertical angles equal 180 − a = 180 − (9x − 140) = −9x + 320.

Possible sums of any two of the four angles:

a + a = 2(9x − 140) = 18x − 280 → matches C a + (180−a) = 180 → matches D (180−a) + (180−a) = 2(−9x + 320) = −18x + 640 → matches B The one that does NOT match any pair sum is A: (−18x + 280).

Question 25

2x + 9y = 7. The given equation is one equation in a system of two linear equations. If the system has at least one solution, which of the following could be the other equation in the system?
I. 3x + 13.5y = 10.5
II. 3x − 13.5y = 10.5

  • A) I only
  • B) II only
  • C) I and II
  • D) Neither I nor II

Answer: C) I and II

A system with at least one solution has either exactly one solution or infinite solutions. It fails only when the lines are parallel with different intercepts.

Compare ratios to the original 2x + 9y = 7:

Equation I: 3x + 13.5y = 10.5 Ratios: 3/2 = 1.5 13.5/9 = 1.5 10.5/7 = 1.5 All three ratios equal → SAME line → infinite solutions ✓ Equation II: 3x − 13.5y = 10.5 Ratios: 3/2 = 1.5 (−13.5)/9 = −1.5 x and y ratios differ → lines are NOT parallel → exactly one solution ✓

Both I and II give at least one solution.

Question 26

A right rectangular prism has base area 24t cm². The length of the base is 8/3 cm and the height of the prism is 15 cm. Which expression represents the surface area, in cm²?

  • A) 48t + 160
  • B) 318t + 80
  • C) 1,968t + 80
  • D) 360t

Answer: B) 318t + 80

Base area = length · width = (8/3) · w = 24t, so w = 24t ÷ (8/3) = 24t · 3/8 = 9t.

Surface area of a rectangular prism = 2(ℓw + ℓh + wh):

ℓw = (8/3)(9t) = 24t ℓh = (8/3)(15) = 40 wh = (9t)(15) = 135t SA = 2(24t + 40 + 135t) = 2(159t + 40) = 318t + 80

Question 27

For a particular car, the linear function f gives predicted power (bhp) for engine speeds 1,000–6,000 rpm. Predicted power is 433 bhp at 3,331 rpm and 600 bhp at 4,500 rpm. The equation f(x) = (1/7)(x − a) + 433 defines f, where x is the engine speed and a is a constant. What is the value of a?

(Student-produced response.)

Answer: 3331

The point (3,331, 433) is on f (given). Plug into f(x) = (1/7)(x − a) + 433:

433 = (1/7)(3331 − a) + 433 0 = (1/7)(3331 − a) 3331 − a = 0 …only if this is the whole story. Let's verify with the second point. Check the slope from the two given points: slope = (600 − 433) / (4500 − 3331) = 167 / 1169 = 1/7 ✓ So the coefficient (1/7) matches. Now use point (4500, 600): 600 = (1/7)(4500 − a) + 433 167 = (1/7)(4500 − a) 1169 = 4500 − a a = 4500 − 1169 a = 3331 Wait — that gives a = 3331. Let's re-check with the first point: f(3331) = (1/7)(3331 − 3331) + 433 = 0 + 433 = 433 ✓ f(4500) = (1/7)(4500 − 3331) + 433 = 1169/7 + 433 = 167 + 433 = 600 ✓ So a = 3331.

Answer: a = 3331.

Point-slope form of the line rewritten: f(x) = (1/7)(x − 3331) + 433. Matching the given form f(x) = (1/7)(x − a) + 433 gives a = 3331.

Answer key at a glance

All 27 answers in one table for quick review:

QAnswerQAnswerQAnswer
1D) 2410C) 80019D) 18/sin Q
2A) 4x = 3211D) 4t²s(16s² − 14t)20100
3C) 7x + 512B) (9, 328)2129
4A) 3131322B) 90
5B) 9/50141500023B) −3ⁿ + 5
67515A) 0.6024A) (−18x + 280)°
73016A) (x + 15)² = 025C) I and II
8A) d + 3w = 4117C) 950 + 50(t − 3) = 1,15026B) 318t + 80
9C) 30918B273331

How SOMATH prepares NYC students for the Digital SAT Math

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 Math prep for high schoolers.

Our Digital SAT Math track:

Manhattan families: If your student is targeting a 700+ Digital SAT Math score for Ivy League, MIT, Stanford, or top-15 admissions, book a free diagnostic evaluation or call (646) 668-6151. We are two blocks from the 79th Street 1 train and three blocks from the B/C at the American Museum of Natural History.

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Digital SAT FAQ

What is on Module 1 of the Digital SAT Math section?

Module 1 is a fixed-difficulty, 22-question, 35-minute module that every test taker sees. It draws from all four Digital SAT Math domains: Algebra, Advanced Math, Problem-Solving & Data Analysis, and Geometry & Trigonometry. Student performance on Module 1 determines whether Module 2 is the easier or harder variant. Calculator use is permitted on the entire section.

How is the Digital SAT Math section scored?

The Digital SAT Math section is scored from 200 to 800 based on both modules combined. Stage-adaptive: strong Module 1 performance routes you to a harder Module 2 with access to the full 800; weaker Module 1 routes you to an easier Module 2 that typically caps around the mid-600s.

How long is the Digital SAT Math section?

70 minutes total, split into two 35-minute modules of 22 questions each. Combined with the Reading & Writing section (64 minutes total), the full Digital SAT is 2 hours 14 minutes.

Is a calculator allowed on the entire Digital SAT Math section?

Yes. Calculators are permitted on every question. The Bluebook app includes a built-in Desmos graphing calculator, and students may bring their own approved handheld. Most top scorers use Desmos because it is optimized for the interface.

What is a good Digital SAT Math score?

Depends on target schools. 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, 550–650.

How is the Digital SAT different from the old paper SAT?

Shorter (2h14 vs 3h), delivered in the Bluebook app, calculator on every math question, built-in Desmos, stage-adaptive across two modules per section, and free-response answers are typed (no bubbles).

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 Math 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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