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Digital SAT Practice Test 11 — Math Module 2: Full Walkthrough of All 27 Questions with Answers & Solutions

Every question on Digital SAT Practice Test #11, Math Module 2, 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 2? You are in the right place. This is a complete, question-by-question walkthrough of SAT Test #11, Math Module 2 — 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.

Module 2 is the second math module on the Digital SAT, and it is adaptive: the College Board serves a harder version of Module 2 to students who performed above average on Module 1, and an easier version to students who did not. That means the Module 2 you actually see on test day depends on you — but the underlying skills tested are drawn from the same four content domains either way. This walkthrough covers a full 27-question Module 2 practice set so you drill every question type regardless of which version you land on. If you have not worked through Module 1 yet, start there first — it is the module that decides which Module 2 you get.

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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Download SAT Practice Test 11 — Math Module 2 (PDF) Full 10-page Digital SAT Practice Test #11, Math Module 2 booklet with reference sheet.
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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. Module 2 is the module that makes the Digital SAT “adaptive.” Every student sees the same fixed-difficulty Module 1. Based on how many Module 1 questions a student answers correctly, the testing platform then serves one of two versions of Module 2: a harder variant for students who performed above average, or an easier variant for students who performed below average. 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. This is exactly why Module 1 matters so much — it is the gatekeeper that decides which Module 2 shows up on your screen. This walkthrough works through a full 27-question Module 2 practice set so every question type is covered regardless of which version a student ultimately receives on test day.

Calculator policy. A calculator is permitted on every single question, in both modules. 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), and when a < 0 the vertex is the maximum point of the parabola.

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 √169 = 13.

Exponent rules: aⁿ · aⁱ = aⁿ⁺ⁱ, (aⁿ)ⁱ = aⁿⁱ, a⁻ⁿ = 1/aⁿ, and fractional exponents like a^(2/3) = ³√(a²) convert cube roots and other radicals to exponent form.

Function evaluation: to find f(63), replace every x in the expression for f with 63, or set the expression equal to 63 and solve backward for x.

5. Factoring and equivalent expressions

Combining like terms: only terms with identical variable parts can be combined. 6x + 5x = 11x, but a lone 4y term cannot combine with x-terms.

Difference of squares/cubes: a² − b² = (a + b)(a − b); a³ + b³ = (a + b)(a² − ab + b²).

Factoring quadratics and testing table values: for x² + bx + c, find two numbers that multiply to c and add to b. When a quadratic is given in factored or vertex-adjacent form with an unknown constant, plug in known (x, f(x)) pairs from a table to test each answer choice.

6. Percentages, probability, and percent change

Percent of a number: p% of n = (p/100) · n.

Percent increase/decrease: “b is k% greater than a” means b = a(1 + k/100). “a is k% less than b” means a = b(1 − k/100). Chain these carefully when a problem defines one quantity relative to another, which is itself relative to a third.

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.

7. Data displays: dot plots, histograms, and adding a data point

Dot plots: each dot represents one data value; the height of the stack of dots above a number is the frequency (count) of that value.

Histograms: each bar represents a range (bin) of values; the bar height is the count of data points falling in that bin.

Adding one extreme value to a data set: the mean always shifts toward the new value (it is sensitive to outliers). The median shifts by at most one ranking position and is often unaffected or barely affected — do not assume the median moves just because an extreme value was added.

8. Geometry — triangles, angles, similar triangles, and right triangles

Isosceles and 45-45-90 triangles: if two sides are equal, then the angles opposite those sides are equal. A right triangle with two equal legs is always a 45–45–90 triangle.

Similar triangles from intersecting segments: when two segments intersect and create vertical angles plus a pair of equal angles, the two triangles formed are similar. Corresponding sides are then proportional.

Altitude to the hypotenuse: in a right triangle, the altitude drawn from the right angle to the hypotenuse creates two smaller right triangles, both similar to the original and to each other.

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. Complementary angles: tan(90° − θ) = 1/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. To find the radius from an equation like (x + 2)² + (y + 5)² = 169, take the square root of the right-hand side: r = √169 = 13.

10. Exponential functions, growth/decay, and graph shape

A general exponential model is y = A · a^x, where A is the initial (x = 0) value and a is the growth/decay factor. Setting x = 0 always isolates the initial value, since a^0 = 1.

Repeated halving/doubling: a quantity that is cut in half each round follows p = P₀(1/2)^r; a quantity that doubles each round follows p = P₀(2)^r.

Reading exponential graph shape: a curve that starts nearly flat and then rises steeply is an increasing exponential (not linear, which would be a straight line). A curve that starts steep and flattens as it falls is decreasing exponential.

11. Inequalities, graphing regions, and word-problem setups

Shaded regions on a graph: a horizontal boundary line at y = k with shading above it represents y ≥ k; shading below represents y ≤ k.

Budget/spending inequalities: total cost = (price₁ · quantity₁) + (price₂ · quantity₂); if a fixed budget cannot be exceeded, the relationship uses ≤.

12. Unit conversion for area

When converting an area between units using a linear conversion factor, square the conversion factor. If 1 nmi = 1.852 km, then 1 nmi² = (1.852)² km² ≈ 3.4299 km².

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 2 (Questions 1–27)

Adaptive module — harder if you scored well on Module 1. Calculator permitted throughout. Reference sheet available in the app.

Question 1

Which expression is equivalent to 6x + 5x + 4y?

  • A) 15x
  • B) 15y
  • C) 11x + 4y
  • D) 30x + 4y

Answer: C) 11x + 4y

Combine like terms: 6x + 5x = 11x. The term 4y is not a like term with the x-terms, so it stays as is. Result: 11x + 4y.

Question 2

To study the characteristics of sea stars in a group of tide pools, researchers measured the diameter, to the nearest inch, of each sea star in these tide pools. The results are shown in a dot plot with diameter (inches) on the horizontal axis: the column above 16 inches has 6 dots (the tallest column), the column above 17 has 5 dots, the column above 18 has 4 dots, the column above 19 has 2 dots, and the column above 20 has 3 dots. Based on the dot plot, how many sea stars had a diameter, to the nearest inch, of 16 inches?

  • A) 16
  • B) 6
  • C) 4
  • D) 1

Answer: B) 6

The tallest column in the dot plot sits above the value 16, and it contains 6 dots. Each dot represents one sea star, so 6 sea stars had a diameter of 16 inches.

Question 3

A rectangle has a length of 56 inches and a width of 28 inches. What is the area, in square inches, of the rectangle?

  • A) 28
  • B) 84
  • C) 168
  • D) 1,568

Answer: D) 1,568

Area of a rectangle = length · width: 56 · 28 = 1,568 square inches.

Question 4

10x = 110 and 6x − 63 = y. The solution to the given system of equations is (x, y). What is the value of y?

  • A) 63
  • B) 11
  • C) 10
  • D) 3

Answer: D) 3

10x = 110 → x = 11 6x − 63 = y 6(11) − 63 = 66 − 63 = 3

So y = 3.

Question 5

The function f is defined by f(x) = 9(2x + 3). For what value of x does f(x) = 63?

  • A) 2
  • B) 5
  • C) 7
  • D) 30

Answer: A) 2

9(2x + 3) = 63 2x + 3 = 7 2x = 4 x = 2

Question 6

10x = 86. What value of x is the solution to the given equation?

(Student-produced response.)

Answer: 8.6

Divide both sides by 10: x = 86 / 10 = 8.6.

Question 7

y = 3,600(a)^x. The given equation, where a is a positive constant, gives the predicted number of bacteria, y, in a growth medium x hours after the number of bacteria was initially measured. According to the equation, what was the predicted number of bacteria initially measured in the growth medium?

(Student-produced response.)

Answer: 3600

“Initially” means x = 0. Substitute: y = 3,600 · a⁰ = 3,600 · 1 = 3,600. The constant 3,600 is the initial (y-intercept) value, regardless of the value of a.

Question 8

Leo goes to a packing store to buy containers and tape. Leo has $15. Each container costs $1.87 and each roll of tape costs $2.40. Which inequality represents the relationship between the number of containers, c, and the number of rolls of tape, t, that Leo can buy?

  • A) 1.87c + 2.40t ≤ 15
  • B) 1.87c + 2.40t ≥ 15
  • C) 2.40c + 1.87t ≤ 15
  • D) 2.40c + 1.87t ≥ 15

Answer: A) 1.87c + 2.40t ≤ 15

Total spent = (price per container · number of containers) + (price per roll of tape · number of rolls) = 1.87c + 2.40t. Leo has only $15 available, so his total spending must be at most $15: 1.87c + 2.40t ≤ 15.

Question 9

The graph of the function f is shown, where y = f(x). The curve is nearly flat on the left side of the graph and rises more and more steeply as x increases, sweeping upward toward the top right of the graph. Which of the following best describes the function f?

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

Answer: B) Increasing exponential

The graph is curved, not a straight line, so both linear options are eliminated. The curve rises as x increases, so it is increasing, not decreasing. The distinctive shape — nearly flat, then increasingly steep — is the signature of an increasing exponential function.

Question 10

A graph in the xy-plane shows a horizontal boundary line at y = 36, with the shaded region covering everything at or above that line. The shaded region represents all the solutions to which inequality?

  • A) x ≤ 36
  • B) x ≥ 36
  • C) y ≤ 36
  • D) y ≥ 36

Answer: D) y ≥ 36

The boundary is a horizontal line at y = 36, and the shading covers all points at or above that line — that is, all points with y ≥ 36. Since the boundary is horizontal, the inequality is in terms of y, not x, ruling out A and B.

Question 11

There are 240 players in a tennis competition that includes 4 rounds of matches. Each player in the competition will play a match against another player in round 1. At the end of each round, the player who loses the match is eliminated, and the player who won the match advances to the next round to play a match against another winning player. Which equation gives the number of players, p, eliminated at the end of round r, where r ≤ 4?

  • A) p = 15(1/2)^r
  • B) p = 15(2)^r
  • C) p = 240(1/2)^r
  • D) p = 240(2)^r

Answer: C) p = 240(1/2)^r

Each round eliminates half of the players who entered that round.

Round 1: eliminated = half of 240 = 120 = 240·(1/2)¹ Round 2: eliminated = half of remaining 120 = 60 = 240·(1/2)² Round 3: eliminated = half of remaining 60 = 30 = 240·(1/2)³ Round 4: eliminated = half of remaining 30 = 15 = 240·(1/2)⁴

The pattern is p = 240(1/2)^r.

Question 12

Line k is defined by y = 6x + 4. Line j is parallel to line k in the xy-plane and passes through the point (0, 5). Which equation defines line j?

  • A) y = 6x + 5
  • B) y = −5x + 5
  • C) y = −6x + 5
  • D) y = 5x + 5

Answer: A) y = 6x + 5

Parallel lines have the same slope. Line k has slope 6, so line j also has slope 6, ruling out B, C, and D. Since line j passes through (0, 5), its y-intercept is 5: y = 6x + 5.

Question 13

In the right triangle shown, the right angle is at the bottom-right corner. The two legs of the triangle are labeled 15 and 15, and the angle labeled x° is at the top of the triangle. (Note: figure not drawn to scale.) What is the value of x?

(Student-produced response.)

Answer: 45

A right triangle with two equal legs (15 and 15) is a 45–45–90 isosceles right triangle. Both non-right angles measure 45°, so x = 45.

Question 14

What is the radius of the circle in the xy-plane defined by (x + 2)² + (y + 5)² = 169?

(Student-produced response.)

Answer: 13

Standard form of a circle is (x − h)² + (y − k)² = r². Here r² = 169, so r = √169 = 13.

Question 15

The graph shows the estimated boiling point y, in degrees Celsius, of a normal paraffin with a molecular weight of x grams per mole, for 1 ≤ x ≤ 280. The point (149.02, 186.05) is plotted on the curve, roughly in the middle of the graph. Which statement is the best interpretation of the point (149.02, 186.05)?

  • A) A normal paraffin with a molecular weight of 186.05 grams per mole has an estimated boiling point of 149.02 degrees Celsius.
  • B) A normal paraffin with a molecular weight of 149.02 grams per mole has an estimated boiling point of 186.05 degrees Celsius.
  • C) The minimum estimated boiling point for normal paraffins corresponds to a paraffin with a molecular weight of 149.02 grams per mole and an estimated boiling point of 186.05 degrees Celsius.
  • D) The maximum estimated boiling point for normal paraffins corresponds to a paraffin with a molecular weight of 149.02 grams per mole and an estimated boiling point of 186.05 degrees Celsius.

Answer: B

On this graph, x represents molecular weight and y represents boiling point. So the point (149.02, 186.05) means: molecular weight 149.02 grams per mole corresponds to boiling point 186.05 degrees Celsius — matching choice B. Since 149.02 sits roughly in the middle of the domain 1 ≤ x ≤ 280, it cannot represent a minimum or maximum, ruling out C and D. Choice A simply swaps the roles of x and y.

Question 16

For the polynomial function f, the graph of y = f(x) in the xy-plane passes through the points (−5, 0), (1, 0), and (4, 0). Which of the following must be a factor of f(x)?

  • A) x + 1
  • B) x + 4
  • C) x − 1
  • D) x − 5

Answer: C) x − 1

A zero of a polynomial at x = c corresponds to a factor (x − c). The graph passes through (1, 0), so x = 1 is a zero, meaning (x − 1) must be a factor — choice C. Choice A would require a zero at −1, choice B a zero at −4, and choice D a zero at 5; none of those x-intercepts are given.

Question 17

For the linear function g, the table shows four values of x and their corresponding values of g(x): at x = 1, g(x) = 32; at x = 2, g(x) = 28; at x = 3, g(x) = 24; at x = 4, g(x) = 20. The function can be written as g(x) = mx + b. What is the value of b?

  • A) 4
  • B) 16
  • C) 32
  • D) 36

Answer: D) 36

m = (28 − 32) / (2 − 1) = −4/1 = −4 Using (1, 32): 32 = −4(1) + b 32 = −4 + b b = 36

Check against the table: g(0) = 36, and each successive step down decreases by 4: 36, 32, 28, 24, 20 — matches the table exactly.

Question 18

For the quadratic function f, the table shows three values of x and their corresponding values of f(x): at x = 24, f(x) = −8; at x = 30, f(x) = −8; at x = 32, f(x) = 8. Which equation defines f?

  • A) f(x) = (x − 24)(x − 30) + 4
  • B) f(x) = (x − 24)(x − 30) − 8
  • C) f(x) = (x − 8)(x − 32) + 32
  • D) f(x) = (x − 8)(x − 32) − 32

Answer: B) f(x) = (x − 24)(x − 30) − 8

Test choice B directly against the table:

f(24) = (24−24)(24−30) − 8 = (0)(−6) − 8 = −8 ✓ f(30) = (30−24)(30−30) − 8 = (6)(0) − 8 = −8 ✓ f(32) = (32−24)(32−30) − 8 = (8)(2) − 8 = 16 − 8 = 8 ✓

All three points from the table match, so choice B is correct.

Question 19

A certain open star cluster contains M-type stars and K-type stars. The estimated total mass of the M-type and K-type stars in this open star cluster is 127,882 quettagrams. The graph shows all combinations of (x, y) — number of M-type stars, x, and number of K-type stars, y — that could form this cluster, assuming all M-type stars share the same estimated mass and all K-type stars share the same estimated mass. The graph is a straight line running from about (0, 140) on the y-axis to about (158, 0) on the x-axis. Based on the graph, which of the following is closest to the estimated mass, in quettagrams, of each M-type star in this cluster?

  • A) 811
  • B) 938
  • C) 51,904
  • D) 75,978

Answer: A) 811

The x-intercept of the graph, about x = 158, represents the scenario where all of the total mass is made up entirely of M-type stars (y = 0 K-type stars). So the mass of each M-type star is approximately the total mass divided by that intercept:

mass per M-type star ≈ 127,882 / 158 ≈ 809.4 ≈ 811 (closest choice)

Choice A is closest.

Question 20

³√(p²) = t^(9/7). In the given equation, p > 1 and t > 1. If t = p^(3n − 1), where n is a constant, what is the value of n?

(Student-produced response.)

Answer: 41/81

Rewrite the left side as an exponent: ³√(p²) = p^(2/3). Substitute t = p^(3n−1) into the right side:

t^(9/7) = (p^(3n − 1))^(9/7) = p^(9(3n − 1)/7) Set exponents of p equal (same base p, both sides equal): 2/3 = 9(3n − 1)/7 Cross-multiply: 2·7 = 3·9(3n − 1) 14 = 27(3n − 1) 3n − 1 = 14/27 3n = 14/27 + 1 = 14/27 + 27/27 = 41/27 n = 41/81

So n = 41/81.

Question 21

The number a is 55% less than the number b. The number b is 320% greater than 160. What is the value of a?

(Student-produced response.)

Answer: 302.4

“b is 320% greater than 160” means b equals 160 plus 320% of 160:

b = 160 + 3.20 × 160 = 160 × (1 + 3.20) = 160 × 4.20 = 672

“a is 55% less than b” means a equals b minus 55% of b:

a = b − 0.55b = 0.45b = 0.45 × 672 = 302.4

Question 22

The function f is defined by f(x) = −6x² + 60x − 126. Which of the following equivalent forms of the equation displays the maximum value of the function as a constant or coefficient?

  • A) f(x) = −6x² + 42x + 18x − 126
  • B) f(x) = −6x(x − 7) + 18(x − 7)
  • C) f(x) = −6(x − 5)² + 24
  • D) f(x) = −6(x − 7)(x − 3)

Answer: C) f(x) = −6(x − 5)² + 24

Since the leading coefficient is negative, the parabola opens downward and has a maximum value at its vertex. Vertex form a(x − h)² + k displays the maximum directly as the constant k. Here, −6(x − 5)² + 24 shows the maximum value is 24, achieved at x = 5. Choice A is just a regrouping of the middle term, choice B is a partially factored intermediate step, and choice D is the fully factored (zero-product) form, which displays the roots (x = 7 and x = 3), not the maximum.

Question 23

In the figure shown, segments WZ and XY intersect at point Q. The angle at W and the angle at Y are both equal to a° (so angle W = angle Y), and the angles at Q formed by the intersection are vertical angles (angle WQX = angle ZQY). It is given that YQ = 21, WQ = 70, WX = 60, and XQ = 120. What is the length of segment YZ?

  • A) 18
  • B) 36
  • C) 120
  • D) 200

Answer: A) 18

Because angle W = angle Y, and angle WQX = angle ZQY (vertical angles), triangle WQX is similar to triangle YQZ (angle-angle similarity). Corresponding sides are proportional:

WQ/YQ = XQ/ZQ = WX/YZ WQ/YQ = 70/21 = 10/3 So: WX/YZ = 10/3 60/YZ = 10/3 YZ = 60 · 3/10 = 18

So YZ = 18.

Question 24

52(x³ + 64)(x⁴ − 81) = 0. How many distinct real solutions does the given equation have?

  • A) Exactly two
  • B) Exactly three
  • C) Exactly five
  • D) Exactly seven

Answer: B) Exactly three

Since 52 ≠ 0, the product is zero only when one of the two remaining factors is zero.

Factor 1: x³ + 64 = 0 x³ = −64 x = −4 Factor 2: x⁴ − 81 = 0 x⁴ = 81 = 3⁴ x = ±3

Distinct real solutions: −4, −3, 3 — exactly three.

Question 25

A histogram summarizes data set A: the point totals of 50 players in a game. The histogram bars are roughly: 4 players scoring 30–40 points, 7 players scoring 40–50 points, 27 players scoring 50–60 points (the tallest bar, and the bin containing the median), and 12 players scoring 60–70 points. A new player who scores 18 points is added to data set A to create data set B, which has 51 values. Which of the following must be true?
I. The median of data set B is less than the median of data set A.
II. The mean of data set B is less than the mean of data set A.

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

Answer: B) II only

The mean is sensitive to extreme values. Adding a very low score (18, far below the 50–60 range where most of the data sits) always pulls the mean down, so statement II is guaranteed true.

The median, however, only shifts by at most one ranking position when a single value is added. With 50 values, the original median of A was the average of the 25th and 26th ordered values. With 51 values, the median of B is exactly the 26th ordered value. Because 18 is an extreme low value, it slides into the very first rank and pushes every original value up by one position — so the new 26th-ranked value in B is what used to be the 25th-ranked value in A. This new median is not guaranteed to be strictly less than the old median; it can be equal to it (if the 25th and 26th values in A were the same, for example). So statement I is not guaranteed. Only II is guaranteed — answer B.

Question 26

In triangle XYZ, the measure of angle X is 90°. Point W lies on segment YZ, and segment WX is perpendicular to segment YZ. The length of segment WY is 572, and the length of segment WX is 429. What is the value of tan Z?

  • A) 3/5
  • B) 3/4
  • C) 4/5
  • D) 4/3

Answer: D) 4/3

Dropping the altitude XW from the right angle X onto the hypotenuse YZ creates triangle WXY, a smaller right triangle with the right angle at W. In this smaller triangle, angle Y is shared with the original triangle XYZ.

In right triangle WXY (right angle at W): tan Y = opposite / adjacent = WX / WY = 429 / 572 = 3/4

In triangle XYZ, angle X = 90°, so angles Y and Z are complementary: Y + Z = 90°. Complementary angles have reciprocal tangents: tan Z = 1 / tan Y = 1 / (3/4) = 4/3.

Question 27

An area of 46.00 square nautical miles is equivalent to k square kilometers, where 1 nautical mile = 1.852 kilometers. To the nearest tenth, what is the value of k?

(Student-produced response.)

Answer: 157.8

To convert an area using a linear conversion factor, square the conversion factor:

1 nmi = 1.852 km 1 nmi² = (1.852)² km² = 3.429904 km² k = 46.00 × 3.429904 = 157.7756 ≈ 157.8

So k ≈ 157.8.

Answer key at a glance

All 27 answers in one table for quick review:

QAnswerQAnswerQAnswer
1C) 11x + 4y10D) y ≥ 3619A) 811
2B) 611C) p = 240(1/2)^r2041/81
3D) 1,56812A) y = 6x + 521302.4
4D) 3134522C) −6(x−5)²+24
5A) 2141323A) 18
68.615B24B) Exactly three
7360016C) x − 125B) II only
8A) 1.87c+2.40t≤1517D) 3626D) 4/3
9B) Increasing exponential18B) (x−24)(x−30)−827157.8

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.

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.

See the SAT Math course → Book free evaluation

Digital SAT FAQ

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

Module 2 is a 22-question, 35-minute module that comes right after Module 1. Unlike Module 1, Module 2 is adaptive: the platform serves either a harder or an easier version depending on how the student performed on Module 1. It still draws from all four Digital SAT Math domains: Algebra, Advanced Math, Problem-Solving & Data Analysis, and Geometry & Trigonometry. 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, in both Module 1 and Module 2. 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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