Digital SAT Practice Test 3 Math Walkthrough (2026)

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This page gives you the digital sat practice test 3 math answers explained — not the letters alone. Digital SAT Practice Test 3 has three math modules in Bluebook: Module 1, which everyone sits, and then Module 2A or Module 2B depending on how Module 1 went. The full answer key for all three modules — 22 entries each — is in the recap tables near the end. Before that, 18 actual questions from this form are printed in full and worked line by line, grouped by the four College Board math domains.

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One thing runs through every worked question below: the built-in Desmos calculator. Practice Test 3 leans on absolute values, exponential models and unknown constants — exactly the screens where typing the equation as printed, or handing a constant a slider, changes the economics of the question. Every question on this page carries either a step-by-step Desmos route or a one-line reason the by-hand route wins.

Practice Test 3 Math at a glance

Thing	Detail
Test form	Digital SAT Practice Test 3 (Bluebook), Section 2: Math — Module 1 plus both Module 2 branches.
Answer keys on this page	All three modules, 22 entries each, in the recap tables below — multiple choice letters and grid-in values.
Worked in full here	18 questions: 5 Algebra · 7 Advanced Math · 3 Problem-Solving and Data Analysis · 3 Geometry and Trigonometry. Each one is printed with its full stem and choices, solved by hand and — where it helps — in Desmos.
Module split of the worked set	8 from Module 1, 5 from Module 2A (the easier branch), 5 from Module 2B (the harder branch).
Time limit	35 minutes per module for 22 questions — about 95 seconds a question.
Grid-ins (student-produced responses)	Derived from the key: 6 in Module 1, 6 in Module 2A, 6 in Module 2B. Entry rules are in the FAQ.
Desmos routes	10 of the 18 worked questions get a full keystroke-by-keystroke Desmos route; the other 8 state exactly why the by-hand route is faster.
Hardest questions	Module 1 question 20, Module 2A question 17, Module 2B questions 9, 10 and 20 — the 5 questions labelled Hard in their own headings below.
Figures	Every question worked on this page is pure text — no stimulus needed a figure. This form’s graph-based questions (box plots, scatterplots, a bar graph, two parallel-line diagrams, two histograms) could not be redrawn faithfully from the source, so they were left out rather than approximated; their keys still appear in the recap tables.
What you can do by the end	Name the domain of a math screen on sight, decide hand-versus-Desmos before you touch anything, and grid the quantity the stem actually asked for.

How to use this page

Work one domain section at a time. The first question of each section is fully modelled, the second asks you to make one call before the reveal, and the rest are behind a try-it-first toggle — attempt those cold before you open anything. If you have already sat Practice Test 3, jump to the questions you missed and read the red boxes first: the number you gridded, or the letter you picked, is probably named there.

How the adaptive Math section works on this form

Everyone gets the same Module 1. Bluebook scores it on the spot and then hands you one of two second modules: Module 2A if Module 1 went badly, Module 2B if it went well. The harder branch, 2B, is the one with the higher scoring ceiling — you cannot reach the top scores from 2A.

On this form the two branches are not disjoint: some questions appear on both. The clearest case on this page is the absolute-value equation worked in the Advanced Math section, which sits at question 17 on Module 2A and question 15 on Module 2B with the same key — the same screen, word for word, on both branches. So a familiar screen does not tell you which branch you got. The branch shows in the overall texture instead: set Module 2A’s opener — the one-line percent question at the top of the data-analysis section below — beside the harder branch’s closing stretch, worked at the end of the Advanced Math and Geometry sections, and the gap between the branches stops being abstract.

For other free and paid options, see our guide to the SAT math practice test.

Note → Worth knowing before you score yourself: students on r/Sat consistently report the first four Bluebook practice tests running easier than the current paper. Treat Practice Test 3 as a domain diagnostic — it tells you where you lose points; be more careful trusting the scaled score it prints.

The Desmos moves this page teaches

The Bluebook calculator button opens a built-in Desmos graphing calculator — the same core graphing tool as the free desmos.com version, minus a few menu items you will not miss. You never lose it: it stays available on every math question in both modules. These are the moves the worked questions below actually use:

Type both equations, read the crossings. Any system, typed exactly as printed, becomes two graphs — a grey dot where they cross, or an empty screen when they never do. Worked at question 14 (Module 1) and question 9 (Module 1).
Click the vertex for min-max. Desmos marks the turn of every parabola; the dot carries both the location and the value, and the stem decides which half you need. Worked at question 15 (Module 1).
Define the function, transform it in one line. f(x)=... then g(x)=f(x)+4 — the whole translation in five characters, and g(0) prints the answer. Worked at question 9 (Module 2B).
Sliders for "for what value of…". When a constant is unknown, give it a slider and watch the condition happen — a parabola lifting off the axis, a curve landing on a data point. Worked at question 20 (Module 2B) and question 20 (Module 1).
Roots and big powers as one typed line. 474552^(1/3) is a cube root; 166^2 is a five-digit square. Neither belongs on paper. Worked at question 10 (Module 2B) and question 17 (Module 1).

And the honest limit, because r/Sat is full of students trying to make one tool do everything: regression — the y1~mx1+b move — is for fitting an equation to data points when you have several points and no equation. None of the 18 questions worked on this page needs it. If you catch yourself setting up a regression for a two-point line or a solvable system, you are doing a slower version of a direct move. Learn regression for scatterplot and table-fit questions; do not treat it as the answer to every screen.

Algebra — 5 worked questions

Algebra on the Digital SAT means linear things: equations, systems of lines, slopes and intercepts, and word problems that hide a linear equation. The five below run from a rental-fee model you can assemble by reading labels to a system whose fractional x-coordinate exists purely to shake your nerve. Desmos is at its strongest in this domain, because linear equations type straight in with no rearranging — and because the calculator does not care whether a coordinate is ugly.

Question 8, Module 2A — Algebra: Building Linear Models (Easy)

I do — watch every step

Sean rents a tent at a cost of $11 per day plus a onetime insurance fee of $10. Which equation represents the total cost c, in dollars, for d days? Margin note: "per day" marks the number that rides with d — the slope. "Onetime" marks the number that stands alone — the flat fee. So c = 11d + 10. Two of the four choices swap the 11 and the 10, and the other multiplies the fee by the days — but a onetime fee is paid once, not daily.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

Which equation represents the total cost c for d days?

A) c = 11(d + 10)
B) c = 10(d + 11)
C) c = 11d + 10
D) c = 10d + 11

By hand
Tag each number	$11 — "per day" → multiplies d. $10 — "onetime" → added once.
Assemble	c = 11d + 10. Choice C.
Check with d = 1	11 + 10 = 21 ✓ — one day of tent, one fee.

🖩 Desmos check

Faster by hand here. There is nothing to graph — the whole question is matching two labelled numbers to the slope slot and the intercept slot. If you want a check anyway, plug d = 1 into the choice you like and ask whether the total is one day of tent plus one fee.

Correct: C) c = 11d + 10
The per-day rate multiplies the day count: 11d. The onetime fee is added once: + 10. So c = 11d + 10. Sanity check with one day: 11(1) + 10 = 21, which is one day of tent plus one insurance fee — exactly right.

A) c = 11(d + 10) — the fee dragged inside the parentheses. Distribute it: 11d + 110. This charges $110 of insurance, because the onetime $10 got multiplied by the $11 rate. At d = 1 it predicts $121 for one day of tent.

B) c = 10(d + 11) — swapped numbers, wrong shape. This makes the fee daily and the daily rate a fee — both errors at once. At d = 1: 10(12) = 120, six times the real one-day cost.

D) c = 10d + 11 — the clean swap. The right shape — rate times days plus fee — with the two numbers exchanged. "Per day" belongs to the 11, "onetime" to the 10; the sentence tells you which is which before any algebra starts.

The move — The Intercept Is the Starting Value: In any linear model, the flat, paid-once, before-anything-happens number is the intercept, and the per-day, per-mile, per-popsicle number is the slope riding with the variable. Word problems mark them for you — "onetime" and "per day" here. Match the labels to the slots before touching the choices, and the swap traps stop working.

Question 12, Module 1 — Algebra: Interpreting Linear Models (Easy)

We do — you make one call, then read on

w(t) = 300 − 4t

The function w models the volume of liquid, in milliliters, in a container t seconds after it begins draining from a hole at the bottom. According to the model, what is the predicted volume, in milliliters, draining from the container each second?

What is the predicted volume, in milliliters, draining from the container each second?

A) 300
B) 296
C) 75
D) 4

Note → Your call first. Before any choice: which number in 300 − 4t changes the output when t ticks up by one second — and by how much?

w(t) = 300 − 4t

The function w models the volume of liquid in a container t seconds after it begins draining. What is the predicted volume draining from the container each second? Margin note: "each second" is per-unit-of-t language — it asks for the slope, and the slope is 4 (the volume drops by 4 mL every second). The 300 is the starting volume, sitting in the intercept slot; it answers "how much was in the container", a question nobody asked.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Name what "each second" asks for	A per-second change — the slope of the model.
Read the slope	w(t) = 300 − 4t: each unit of t removes 4.
Answer the asked quantity	4 milliliters drains each second. Choice D.

🖩 Desmos check

Faster by hand here. Reading one coefficient does not need a graph. If the slope-versus-intercept split ever feels shaky, ten seconds in Desmos rebuilds it: graph y = 300 − 4x and watch the line start at 300 and step down 4 for every step right — the two numbers doing their two different jobs.

Correct: D) 4
Each second, t goes up by 1 and w(t) goes down by 4. That per-second change is the slope’s size, and "volume draining each second" is exactly that: 4 milliliters. The minus sign is already spent on the word "draining".

A) 300 — the starting amount. 300 is w(0) — the volume before draining starts. It is the intercept, and it answers "how much liquid was there", not "how fast does it leave".

B) 296 — the volume after one second. w(1) = 296 is what remains at t = 1. The question asks for the amount that left each second — the drop from 300 to 296, which is 4, not the leftover itself.

C) 75 — dividing the two printed numbers. 300 ÷ 4 = 75 is the number of seconds until the container is empty. Interesting, true in the model — and the answer to a different question.

The move — The Slope Is the Per-Step Change: "Each second", "per pound", "for every additional hour" — per-unit language always asks for the slope, the coefficient on the variable. The intercept answers "how much at the start". Interpretation items bank on you grabbing the bigger, rounder number; decide which slot the question names before you look at the numbers at all.

Question 9, Module 1 — Algebra: Systems of Two Linear Equations (Easy)

Your turn — try it before you open the reasoning

y = 2x + 10
y = 2x − 1

At how many points do the graphs of the given equations intersect in the xy-plane?

At how many points do the graphs intersect?

A) Zero
B) Exactly one
C) Exactly two
D) Infinitely many

Try it first — then open the reasoning

y = 2x + 10
y = 2x − 1

At how many points do the graphs intersect? Margin note: same slope, 2 and 2 — the lines run parallel. Different intercepts, 10 and −1 — so they are two separate parallel lines, which never meet. The count is zero. Do not solve for x: subtracting the equations gives 0 = 11, which is the algebra’s way of saying "no crossing" — not an invitation to keep going.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

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By hand
Compare slopes	2 and 2 — equal, so the lines are parallel.
Compare intercepts	10 and −1 — different, so they are distinct lines.
Count	Distinct parallel lines share zero points. Choice A.

🖩 In Desmos

Type y=2x+10 on line 1 and y=2x-1 on line 2, exactly as printed.
Look for a grey crossing dot. There is none — the two lines run side by side across the whole screen.
Zoom out once or twice to convince yourself they never converge. Parallel lines stay the same distance apart forever.
Count the crossings you can see: zero. Choice A.

Which is faster: A tie at about ten seconds either way. By hand it is one slope comparison; in Desmos it is two typed lines and an empty screen where the dot should be. Use whichever check makes you certain — on a question this fast, certainty is the product.

Correct: A) Zero
Both lines have slope 2, so they are parallel. Their intercepts differ — 10 versus −1 — so they are two distinct parallel lines, and distinct parallel lines never intersect. Zero points.

B) Exactly one — the default for two lines. Two random lines usually cross once, and that reflex fills this bubble. But crossing once requires different slopes, and these slopes match. Check slopes before you assume the usual picture.

C) Exactly two — counting the equations. Two equations do not make two crossings — two straight lines can meet at most once (or not at all, or everywhere). "Exactly two" is impossible for a pair of lines no matter what the numbers are.

D) Infinitely many — parallel confused with identical. Infinitely many happens only when the two equations describe the same line — same slope and same intercept. Here the slopes match but 10 ≠ −1, so the lines are neighbors, not twins.

The move — Graph Both Sides, Count the Crossings: "At how many points" and "how many solutions" questions never need the solutions themselves — they need a count. Compare slopes first: different slopes cross once; equal slopes with different intercepts cross zero times; the same line twice crosses everywhere. Or type both into Desmos and literally count the dots. Never solve for coordinates nobody asked for.

Question 14, Module 1 — Algebra: Systems of Two Linear Equations (Medium)

Your turn — try it before you open the reasoning

24x + y = 48
6x + y = 72

The solution to the given system of equations is (x, y). What is the value of y?

What is the value of y?

Student-produced response — no choices. You type the answer. Up to 5 characters for a positive answer, 6 for a negative one; a fraction that fits may be entered as a fraction, or as its decimal equivalent.

Try it first — then open the reasoning

24x + y = 48
6x + y = 72

The solution is (x, y). What is the value of y? Margin note: both equations carry a bare + y, so subtracting kills y in one move: 18x = −24, x = −4⁄3. Ugly — and fine, because the stem wants y, not x. Back-substitute: y = 48 − 24(−4⁄3) = 48 + 32 = 80. The trap is a sign slip in that last multiplication — minus times minus, the 32 gets added.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Subtract the equations	Both have + y: 18x = −24 ⇒ x = −4⁄3.
Back-substitute for y	y = 48 − 24(−4⁄3) = 48 + 32 = 80.
Check in the other equation	6(−4⁄3) + 80 = 72 ✓. Grid 80.

🖩 In Desmos

Type 24x+y=48 on line 1, exactly as printed — no rearranging.
Type 6x+y=72 on line 2.
Click the grey dot where the lines cross. Desmos shows (-1.333, 80).
The stem wants y. Grid 80.

Which is faster: Desmos, clearly — about 15 seconds, and it never has the sign accident. The fractional x that makes the by-hand route slippery costs the calculator nothing. This is the clearest Desmos win on the page.

Correct answer: 80 (accepted entries: 80)
Subtract the second equation from the first: (24x − 6x) + (y − y) = 48 − 72, so 18x = −24 and x = −4⁄3. Then y = 48 − 24x = 48 − 24(−4⁄3) = 48 + 32 = 80. Check in the second equation: 6(−4⁄3) + 80 = −8 + 80 = 72 ✓.

The number students grid instead — losing the sign on the back-substitution. With x = −4⁄3, the term −24x is positive 32. Students who drop the minus compute 48 − 32 = 16 and grid it. The fractional x is the test’s bet that you will bail into arithmetic errors — or waste time doubting a perfectly legal x. Fractions are allowed to be ugly; only y had to be gridded.

The move — Click the Grey Dot: Any system of two equations, typed exactly as printed, becomes two graphs whose crossing Desmos marks with a grey dot. Click it, read the pair, and then — before you grid — re-read the stem for which coordinate it wants. An ugly x-coordinate is not a mistake; it is usually the sign that the question wanted the other coordinate all along.

Question 21, Module 2B — Algebra: Linear Functions from Two Values (Medium)

Your turn — try it before you open the reasoning

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?

Which function f gives the total cost for x hours of repair?

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

Try it first — then open the reasoning

A window repair specialist charges $220 for the first two hours plus an hourly fee for each additional hour. The total for 5 hours is $400. Which function f gives the total cost for x hours, where x > 2? Margin note: from 2 hours to 5 hours is 3 extra hours, costing 400 − 220 = 180, so the hourly fee is $60. But f(x) is not 60x + 220 — the 220 already bought the first two hours. Build from the point: f(x) = 220 + 60(x − 2) = 60x + 100. Check: f(5) = 400 ✓.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Turn the sentences into points	(2, 220) and (5, 400).
Slope	(400 − 220)⁄(5 − 2) = 60 dollars per additional hour.
Point-slope, then simplify	f(x) = 220 + 60(x − 2) = 60x + 100. Choice A.
Verify both facts	f(2) = 220 ✓, f(5) = 400 ✓.

🖩 Desmos check

Faster by hand here. Two plug-ins beat any graph here: the right function must give f(2) = 220 and f(5) = 400, and only choice A passes both. Testing four choices at two inputs is eight mental multiplications — faster than typing a single line, and it uses the stem’s own numbers as the answer key.

Correct: A) f(x) = 60x + 100
The two facts are two points: 2 hours cost $220 and 5 hours cost $400. Slope: (400 − 220)⁄(5 − 2) = 180⁄3 = 60 dollars per additional hour. Point-slope from (2, 220): f(x) = 220 + 60(x − 2) = 60x + 100. Both checks pass: f(2) = 220 ✓, f(5) = 400 ✓.

B) f(x) = 60x + 220 — gluing the fee to the whole clock. Right hourly fee, wrong start: this charges $60 for all x hours and the full $220 on top, double-billing the first two hours. f(5) = 520, not the $400 the stem states. One plug-in kills it.

C) f(x) = 80x — averaging the 5-hour total. 400 ÷ 5 = 80 treats the whole bill as one flat hourly rate. But the pricing is not flat — the first two hours are premium. f(2) = 160, and the stem says two hours cost $220.

D) f(x) = 80x + 220 — the averaged rate plus the fee. Both errors at once: the fake $80 average and the double-billed $220. f(5) = 620 — $220 past the printed total. Every distractor here fails the printed f(5) = 400 check.

The move — Step Along the Slope: When a price schedule gives you one anchor point and a per-step rate, build the line from the point: start value plus rate times steps from the anchor, not from zero. The constant in the simplified form — the 100 here — is not a fee anyone pays; it is where the line would cross x = 0 if the pricing extended there. Anchor at the printed point, then step.

Say it back before you move on

From memory: Without scrolling up: in a linear model, which number answers "each second" questions and which answers "at the start"? What do equal slopes with different intercepts mean for a system’s solution count? And when a price schedule anchors at (2, 220), where does the line start counting from?
Check yourself: If you missed the system grid-in — 14 on Module 1 — the gap is nerve, not algebra: the fractional x is legal, and the stem wanted y anyway. If the two interpretation questions cost you, re-read the slope-versus-intercept rule boxes; both misses come from grabbing the wrong slot, never from arithmetic. Close the gap with the free SAT Math Mixed Review lesson notes.

If the constant-hunt questions above cost you time, that is fixable with volume, not talent — the same few setups repeat across every form. A TTA plan opens the full library of real 2023–2026 Digital SAT administrations at $39.99 a month with a 1-day trial, or $250 a year with a 3-day trial. If you would rather test the water first, the free practice tests need no card.

Advanced Math — 7 worked questions

Advanced Math is the nonlinear domain: quadratics, exponentials, absolute values, and functions being shifted around the plane. This is where Practice Test 3 stacks its expensive questions — a doubling-time constant hiding in an exponent, a translated cubic, a discriminant hunt with an integer boundary — and where Desmos’s function definitions and sliders pay the biggest rent. The seven below include the shared-pool absolute-value screen that appears on both Module 2 branches.

Question 16, Module 2A — Advanced Math: Exponential Functions in Context (Medium)

I do — watch every step

A company opens an account with an initial balance of $36,100.00. The balance is an exponential function A(t), t years after opening. After 13 years it is $68,071.93. Which equation could define A? Margin note: the number in front of an exponential is its value at t = 0 — so the coefficient must be 36,100.00, which eliminates the two choices that lead with 31,971.93 (that number is the 13-year growth, 68,071.93 − 36,100). And the account grows, so the base must sit above 1 — which eliminates the (0.05) bases. One choice survives.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

Which equation could define A?

A) A(t) = 36,100.00(1.05)^t
B) A(t) = 31,971.93(1.05)^t
C) A(t) = 31,971.93(0.05)^t
D) A(t) = 36,100.00(0.05)^t

By hand
Read t = 0	A(0) must be 36,100.00 → the coefficient is 36,100.00. Kills B and C.
Read the direction	The account grows → base > 1 → (1.05), not (0.05). Kills D.
Confirm with the 13-year value	36,100.00 × 1.05¹³ ≈ 68,071.93 ✓. Choice A.

🖩 In Desmos

Type 36100(1.05)^13 on line 1. Desmos returns 68071.93... — the printed 13-year balance, to the cent.
That single line verifies choice A. If you want the picture too, graph y=36100(1.05)^x and click x = 13.
Try the same check on choice D: 36100(0.05)^13 is a decimal dust cloud — nowhere near $68,000.

Which is faster: By hand for the elimination — the t = 0 read takes five seconds and kills three choices — then one typed Desmos line as the confirmation. Doing the arithmetic 1.05¹³ by hand is the only slow route on this screen; never do it.

Correct: A) A(t) = 36,100.00(1.05)^t
Two reads settle it. At t = 0, an exponential equals its front coefficient — the stem says that is $36,100.00. A growing account needs a base above 1, so (1.05), not (0.05). Only choice A has both. The 13-year value confirms it: 36,100.00(1.05)¹³ ≈ 68,071.93 ✓.

B) A(t) = 31,971.93(1.05)^t — the growth amount posing as the start. 31,971.93 is 68,071.93 − 36,100.00 — the interest earned over the 13 years. The front coefficient must be the balance at t = 0, and at t = 0 this function says the account opened with the interest instead of the deposit.

C) A(t) = 31,971.93(0.05)^t — both errors at once. Wrong start and a base of 0.05, which does not mean "5% growth" — it means the balance keeps 5% of itself each year, a 95% annual collapse. By year 2 this account holds about $80.

D) A(t) = 36,100.00(0.05)^t — the rate typed as the base. The right start, but 0.05 is the rate, not the base. Growth at 5% per year needs base 1 + 0.05 = 1.05. As written, this account loses 95% of its value every year while the stem says it earns interest.

The move — The Starting Amount Rides in Front: In any exponential a(b)^t, the coefficient a is the value at t = 0 and the base b carries the change: 1 plus the rate for growth, 1 minus the rate for decay. Read the stem for the starting value and the direction first — those two facts usually eliminate three choices before any computation. A bare rate like 0.05 sitting in the base slot is always a trap.

Question 15, Module 1 — Advanced Math: Quadratic Minimum and the Vertex (Medium)

We do — you make one call, then read on

y = x² − 14x + 22

The given equation relates the variables x and y. For what value of x does the value of y reach its minimum?

For what value of x does y reach its minimum?

Note → Your call first. Before any formula: the parabola opens upward, so its minimum is at the vertex. Say the vertex formula out loud — with the sign of b in this equation.

y = x² − 14x + 22

For what value of x does y reach its minimum? Margin note: minimum means vertex, and the vertex sits at x = −b⁄2a. Here b is −14 — sign included — so x = −(−14)⁄2 = 7. The two grids to refuse: −7, from dropping b’s sign, and 22, which is just the constant along for the ride. The stem wants the x, not the minimum y-value.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Minimum means vertex	Upward parabola → the low point is the vertex.
Apply x = −b/2a	a = 1, b = −14 → x = 14⁄2 = 7.
Answer the asked quantity	The stem wants the x-location. Grid 7.

🖩 In Desmos

Type y=x^2-14x+22 exactly as printed.
Desmos marks the turn of the parabola with a grey dot. Click it: (7, -27).
The stem asks where the minimum happens — the x-coordinate. Grid 7.

Which is faster: A tie, ten seconds either way — but the two routes fail differently. The formula route risks the −7 sign slip; the click route risks gridding −27, the dot’s other coordinate. Whichever you use, the last step is the same: re-read which coordinate the stem asked for.

Correct answer: 7 (accepted entries: 7)
The parabola opens upward (the x² coefficient is positive), so the minimum is the vertex. x = −b⁄2a = −(−14)⁄2(1) = 7. Equivalently, complete the square: y = (x − 7)² − 27, minimum at x = 7.

The number students grid instead — gridding −7 or 22. −7 comes from reading b as 14 instead of −14 — the formula’s leading minus then goes unspent. 22 is the constant term, which controls where the parabola sits vertically, not where it turns. And if you found the actual minimum value, −27, that is the y at the vertex — the stem asked for the x.

The move — The Vertex Answers Min and Max: Minimum, maximum, highest, lowest — every one of those words sends you to the vertex. Click it in Desmos or compute x = −b⁄2a with b’s sign attached. Then match the coordinate to the question: "for what x" wants the first number, "what is the minimum value" wants the second. The formula and the click both hand you the pair; the stem decides which half you grid.

Question 20, Module 1 — Advanced Math: Exponential Growth and Doubling Time (Hard)

Your turn — try it before you open the reasoning

A scientist initially measures 12,000 bacteria in a growth medium. 4 hours later, the scientist measures 24,000 bacteria. Assuming exponential growth, the formula P = C(2)^rt gives the number of bacteria in the growth medium, where r and C are constants and P is the number of bacteria t hours after the initial measurement. What is the value of r?

What is the value of r?

A) 1⁄12,000
B) 1⁄4
C) 4
D) 12,000

Try it first — then open the reasoning

A scientist initially measures 12,000 bacteria. 4 hours later: 24,000 — the population doubled. The formula is P = C(2)^rt. What is r? Margin note: at t = 0, P = C, so C = 12,000. At t = 4: 24,000 = 12,000(2)^4r, so 2^4r = 2¹. Same base — set the exponents equal: 4r = 1, r = 1⁄4. The bait answer is 4: it works only if the exponent were t⁄r, and this formula says rt. Read the exponent’s shape before you answer.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Pin C at t = 0	P(0) = C = 12,000.
Use the 4-hour measurement	24,000 = 12,000(2)^4r ⇒ 2^4r = 2¹.
Set the exponents equal	4r = 1 ⇒ r = 1⁄4. Choice B.

🖩 In Desmos

Type 12000(2)^(rx) and add the slider Desmos offers for r.
Type the target point (4, 24000) on the next line.
Set the slider limits to 0 and 1 with step 0.05, then drag until the curve passes through the point. It lands at r=0.25.
0.25 = 1⁄4. Choice B.

Which is faster: By hand, decisively — the same-base equation solves in two lines. The slider route works but takes longer to set up than the algebra takes to finish; keep it for checking, or for the day the exponent’s shape confuses you and you want to see which r makes the curve hit the data.

Correct: B) 1⁄4
C is the starting count, 12,000. Four hours in: 24,000 = 12,000(2)^4r, so 2^4r = 2 = 2¹. With the bases matched, the exponents must match: 4r = 1, so r = 1⁄4. Sanity check: at t = 4, the exponent rt = 1, giving one doubling — exactly what the two measurements show.

A) 1⁄12,000 — the reciprocal of the head count. Built from the 12,000 because it is the biggest number on screen. But C absorbs the 12,000; r lives in the exponent and only answers "how fast", never "how many".

C) 4 — the doubling time itself. The population does double every 4 hours — but r is not the doubling time in this formula. With r = 4 the exponent at t = 4 would be 16, predicting 2¹⁶ times the start: about 786 million bacteria instead of 24,000. r = 4 answers the exponent shape 2^t⁄r, a formula this stem did not write.

D) 12,000 — the starting count again. 12,000 is C, and the stem already spent it there. A constant cannot hold both jobs; once C = 12,000 is fixed, r is determined by the doubling, not the head count.

The move — Set the Exponents Equal: When both sides of an equation can be written on the same base — 2^4r = 2¹ here — the exponents must be equal, and the equation collapses to one line. Before that, always read the exponent’s shape: rt and t⁄r swap the meaning of r, and the test prints the doubling time among the choices to catch whoever answered the other formula.

Question 17, Module 2A — Advanced Math: Absolute Value Equations (Hard)

Your turn — try it before you open the reasoning

2|4 − x| + 3|4 − x| = 25

What is the positive solution to the given equation?

What is the positive solution to the given equation?

Try it first — then open the reasoning

2|4 − x| + 3|4 − x| = 25

What is the positive solution? Margin note: the same chunk |4 − x| appears twice — two copies plus three copies is five copies, so 5|4 − x| = 25 and |4 − x| = 5. Split the absolute value both ways: 4 − x = 5 gives x = −1; 4 − x = −5 gives x = 9. The stem wants the positive one: 9. The premature grid is 5 — the value of the chunk, not of x.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Combine the copies	2 + 3 = 5 copies: 5\|4 − x\| = 25 ⇒ \|4 − x\| = 5.
Split into both cases	4 − x = 5 ⇒ x = −1; 4 − x = −5 ⇒ x = 9.
Answer the asked solution	Positive → grid 9.

🖩 In Desmos

Type y=2|4-x|+3|4-x| on line 1 — the vertical bars type straight in.
Type y=25 on line 2.
Click the two grey crossing dots: (-1, 25) and (9, 25).
The stem wants the positive solution. Grid 9.

Which is faster: A tie on speed, but Desmos shows both solutions at once — the V-shaped graph makes the "two answers" structure visible, which is exactly what the by-hand route is most likely to forget. If absolute values reliably cost you a case, click instead.

Correct answer: 9 (accepted entries: 9)
Treat |4 − x| as one object: 2 copies + 3 copies = 5 copies, so 5|4 − x| = 25 and |4 − x| = 5. An absolute value of 5 means the inside is 5 or −5: 4 − x = 5 ⇒ x = −1, and 4 − x = −5 ⇒ x = 9. The positive solution is 9. Check: |4 − 9| = 5, and 2(5) + 3(5) = 25 ✓.

The number students grid instead — gridding 5, or stopping at −1. 5 is the chunk’s value — the middle of the solve, not the end. And −1 is a genuine solution that arrives first if you work the + case first; students grid it and move on. The stem’s word "positive" is doing real work: it tells you there are two solutions and names which one pays.

The move — Solve for the Chunk: When the same wrapped expression — an absolute value, a square, a whole parenthesis — appears more than once, it is one unknown wearing a costume. Combine its copies, solve for the chunk first, and only then open up what is inside. And an absolute value equal to a positive number always splits into two cases; the stem’s "positive solution" or "negative solution" tells you it knows there are two, and which one it wants.

Shared-pool sidebar. This exact absolute-value screen appears as question 17 on Module 2A and again as question 15 on Module 2B, keyed 9 both times — verified word for word on the rendered pages of both modules. The two Module 2 branches draw on an overlapping pool, so meeting a familiar screen tells you nothing about which branch you landed in.

Question 18, Module 2A — Advanced Math: Equivalent Expressions and Common Factors (Medium)

Your turn — try it before you open the reasoning

The expression 90y⁵ − 54y⁴ is equivalent to ry⁴(15y − 9), where r is a constant. What is the value of r?

What is the value of r?

Try it first — then open the reasoning

The expression 90y⁵ − 54y⁴ is equivalent to ry⁴(15y − 9), where r is a constant. What is r? Margin note: match one term instead of expanding everything. First terms: r · 15y · y⁴ must equal 90y⁵, so 15r = 90 and r = 6. Confirm on the second terms: r · 9 = 54 → r = 6 ✓ — both terms vote for the same r, so nothing is broken. Do not grid 15 or 9, the coefficients the stem already printed.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Match the first terms	r · 15y⁵ = 90y⁵ ⇒ r = 90⁄15 = 6.
Confirm on the second terms	r · 9y⁴ = 54y⁴ ⇒ r = 54⁄9 = 6 ✓.
Grid	Both terms agree: 6.

🖩 Desmos check

Faster by hand here. Two divisions — 90 ÷ 15 and 54 ÷ 9 — decide it, and the second one is a free error check the question builds in for you. Setting up a graph to compare polynomial expressions works, but it turns a ten-second coefficient match into a two-minute detour.

Correct answer: 6 (accepted entries: 6)
Distribute in your head one term at a time. The y⁵ terms: r × 15 = 90, so r = 6. The y⁴ terms confirm: r × (−9) = −54, so r = 6 again. Both matches agree; grid 6. (Equivalently: factor 6y⁴ out of 90y⁵ − 54y⁴ and get 6y⁴(15y − 9) directly.)

The number students grid instead — gridding a printed coefficient. Under time pressure the eye grabs 15 or 9 — numbers the stem already spent inside the parentheses. r is the number the stem hid: the common factor that connects 15 to 90 and 9 to 54. If your r equals something printed in the parentheses, you answered the wrong slot.

The move — Divide the Coefficients First: When an expression is declared equivalent to a factored form with an unknown constant, the variable parts are passengers — the constant lives entirely in the numbers. Divide matching coefficients to find it, then divide the other pair as a check: a factoring question always offers two independent routes to its constant, and they must agree.

Question 9, Module 2B — Advanced Math: Function Translations (Hard)

Your turn — try it before you open the reasoning

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)?

What is the value of g(0)?

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Try it first — then open the reasoning

f(x) = (x − 6)(x − 2)(x + 6). The graph of g is f translated up 4 units. What is g(0)? Margin note: "up 4" adds 4 to every output: g(x) = f(x) + 4. So g(0) = f(0) + 4, and f(0) drops straight out of the factors: (−6)(−2)(6) = 72. Answer: 72 + 4 = 76. Do not expand the cubic — evaluating at 0 only ever needs the constant terms. And do not grid 72, the pre-shift value.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Translate the translation	Up 4 units ⇒ g(x) = f(x) + 4.
Evaluate f(0) from the factors	(−6)(−2)(+6) = 72 — two negatives cancel.
Add the shift	g(0) = 72 + 4 = 76. Grid 76.

🖩 In Desmos

Type f(x)=(x-6)(x-2)(x+6) on line 1. Desmos accepts the function definition as typed.
Type g(x)=f(x)+4 on line 2 — the whole translation in five characters.
Type g(0) on line 3. Desmos prints 76.
Grid 76.

Which is faster: A tie — and the Desmos route is worth knowing even if you solve by hand, because g(x)=f(x)+4 is the general weapon for every function-transformation screen: shifts, stretches and compositions all type in as one line, and evaluating at any point is one more.

Correct answer: 76 (accepted entries: 76)
Translating a graph up 4 units adds 4 to every output: g(x) = f(x) + 4. Evaluate f at 0 directly from the factors: (0 − 6)(0 − 2)(0 + 6) = (−6)(−2)(6) = 72. So g(0) = 72 + 4 = 76. No expansion, no graphing — three signed numbers multiplied and one addition.

The number students grid instead — gridding 72. 72 is f(0) — the answer to the question with the translation deleted. The shift is worth exactly 4 points of output, and the test put it in the second sentence hoping the first sentence’s cubic would use up your attention. The other slip is a sign miscount: (−6)(−2)(6) has two negatives, so the product is positive 72, not −72.

The move — Slide the Graph, Slide the Output: "Translated up k units" means every output grows by k: the new function is the old one plus k. Write that equation before touching any numbers — g(x) = f(x) + 4 here — then evaluate. And when a polynomial is handed to you in factored form, evaluating at a point never requires expanding: substitute into the factors and multiply the results.

Question 20, Module 2B — Advanced Math: Quadratics and the Discriminant (Hard)

Your turn — try it before you open the reasoning

x(kx − 56) = −16

In the given equation, k is an integer constant. If the equation has no real solution, what is the least possible value of k?

What is the least possible value of k?

Try it first — then open the reasoning

x(kx − 56) = −16

k is an integer constant and the equation has no real solution. What is the least possible value of k? Margin note: collect first: kx² − 56x + 16 = 0. No real solution means a negative discriminant: 56² − 4k(16) < 0, so 3136 < 64k and k > 49. The least integer above 49 is 50. The classic wrong grid is 49 — at k = 49 the discriminant is exactly 0, and the equation has one real solution, which the stem forbids.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Collect to standard form	x(kx − 56) = −16 ⇒ kx² − 56x + 16 = 0.
Demand a negative discriminant	56² − 4k(16) < 0 ⇒ 3136 < 64k.
Solve and take the least integer	k > 49; strict, so the least integer is 50. Grid 50.

🖩 In Desmos

Type y=x(kx-56)+16 — the equation’s left side minus its right — and add the slider for k.
Set the slider from 40 to 60 with step 1. "No real solution" looks like the parabola floating clear of the x-axis.
At k=49 the parabola just kisses the axis — one solution, still illegal. At k=50 it lifts off completely.
The least integer that clears the axis: grid 50.

Which is faster: By hand, slightly — the discriminant line is mechanical once you have collected the quadratic. But the slider is the best boundary-case detector on the test: you can watch the difference between touching the axis at 49 and missing it at 50, which is exactly the distinction the wrong answer fails.

Correct answer: 50 (accepted entries: 50)
Expand and collect: kx² − 56x + 16 = 0. A quadratic has no real solution exactly when its discriminant is negative: (−56)² − 4(k)(16) < 0, that is 3136 − 64k < 0, so k > 3136⁄64 = 49. The inequality is strict, and k must be an integer — the least integer strictly greater than 49 is 50.

The number students grid instead — gridding 49. 49 is the boundary, not a member. At k = 49 the discriminant equals zero and the equation has exactly one real solution — the stem says no real solution, so 49 is disqualified by one point of discriminant. Every "least integer such that" question lives or dies on whether the boundary case counts; here the strict inequality says it does not.

The move — No Real Solutions Means a Negative Discriminant: For a quadratic, "no real solution" translates to b² − 4ac < 0 — collect to standard form first, then write the inequality. Watch two details at the finish: the inequality is strict (the boundary value gives one solution, not none), and if the stem says integer, the answer is the first integer past the boundary, never the boundary itself.

If the discriminant hunt above is the kind of screen where you freeze, it is faster to be talked through it than to read about it. Mr. Osama runs live small-group Digital SAT sessions — groups of 3 to 5, from $15 a session, with the first diagnostic session free — and this question type is a regular on the whiteboard.

Say it back before you move on

From memory: Say the vertex rule from memory: which questions does a parabola’s vertex answer, and what is the by-hand formula for its x-coordinate? Then: what does "no real solution" demand of a quadratic’s discriminant — and does the boundary value count? And in P = C(2)^rt, what job does each constant hold?
Check yourself: If the discriminant grid-in beat you at the boundary — gridding 49 for 50 — the gap is the strict inequality, not the formula. If the doubling-time question got you, you answered the exponent shape t⁄r instead of the printed rt. Both fixes are one careful read of the stem’s own notation. Review both in the free SAT Math Mixed Review lesson notes.

Problem-Solving and Data Analysis — 3 worked questions

This domain is percentages, statistics and reading data displays. The arithmetic is light; the misses come from answering about the wrong quantity — the leftover instead of the part, half an interval instead of both sides. The three below are chosen for exactly that failure mode: two percent translations that bracket the section’s difficulty floor, and the margin-of-error question that punishes the word "exactly".

Question 1, Module 2A — Problem-Solving and Data Analysis: Percentages (Easy)

I do — watch every step

What is 10% of 470? Margin note: "of" means multiply: 0.10 × 470 = 47 — or just slide the decimal point one place left, because 10% is the one percent every student can do without a pencil. The distractors are decimal-slip artifacts: 37 looks like an answer, 423 and 460 are 470 minus things. None of them is a tenth of 470.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

What is 10% of 470?

A) 37
B) 47
C) 423
D) 460

By hand
Translate	10% of 470 = 0.10 × 470.
Slide the decimal	470 → 47.
Reverse-check	47 × 10 = 470 ✓. Choice B.

🖩 Desmos check

Faster by hand here. Typing this into any calculator costs more than doing it: 10% of anything is one decimal slide. The value of this screen is the 20 seconds it hands you — Module 2A opens with it, and the time you bank here pays for the absolute-value equation waiting at question 17.

Correct: B) 47
10% of a number is the number divided by 10 — move the decimal one place left. 470 becomes 47. As a check: 47 × 10 = 470 ✓.

A) 37 — a near-miss that tests estimation. 37 × 10 = 370, not 470. This choice exists for students who "know it starts with the digits" and grab the wrong ones. One reverse multiplication exposes it.

C) 423 — 470 minus 47. This is what remains after removing 10% — the answer to "what is 470 decreased by 10%?". The stem asked for the 10% piece itself, not the leftover.

D) 460 — 470 minus 10. Subtracting the literal number 10 instead of taking 10 percent. Percent is never a subtraction; it is a multiplication by a fraction of 100.

The move — Percent Means Part Over Whole: Every percent sentence is the same equation wearing three disguises: part = percent⁄100 × whole. Whichever piece is missing, translate the sentence literally — "of" is multiplication, "is" is equals — and solve for the empty slot. When the percent is 10, skip even that: slide the decimal one place left and check by multiplying back.

Question 8, Module 1 — Problem-Solving and Data Analysis: Margin of Error (Medium)

We do — you make one call, then read on

To estimate the proportion of a population that has a certain characteristic, a random sample was selected from the population. Based on the sample, it is estimated that the proportion of the population that has the characteristic is 0.49, with an associated margin of error of 0.04. Based on this estimate and margin of error, which of the following is the most appropriate conclusion about the proportion of the population that has the characteristic?

Which is the most appropriate conclusion about the proportion?

A) It is plausible that the proportion is between 0.45 and 0.53.
B) It is plausible that the proportion is less than 0.45.
C) The proportion is exactly 0.49.
D) It is plausible that the proportion is greater than 0.53.

Note → Your call first. Before the choices: add the margin to the estimate, then subtract it. What two numbers do you get — and what do they bracket?

Based on a random sample, the estimated proportion is 0.49, with a margin of error of 0.04. Which is the most appropriate conclusion? Margin note: a margin of error builds an interval both ways around the estimate: 0.49 − 0.04 = 0.45 up to 0.49 + 0.04 = 0.53. The honest conclusion is "plausibly between 0.45 and 0.53" — never "exactly 0.49", which pretends the sample nailed the whole population, and never a one-sided claim, which uses only half the interval.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Build the interval	0.49 − 0.04 = 0.45; 0.49 + 0.04 = 0.53.
Say what it means	The population proportion plausibly lies between 0.45 and 0.53.
Match the sentence	Choice A — the only two-sided, both-endpoints conclusion.

🖩 Desmos check

Faster by hand here. One addition and one subtraction — the work is done before a calculator opens. The question is testing a sentence, not a computation: whether you know a margin of error makes an interval, and that "exactly" and one-sided claims both misuse it.

Correct: A) It is plausible that the proportion is between 0.45 and 0.53.
The margin of error turns a point estimate into an interval: 0.49 ± 0.04 spans 0.45 to 0.53. The appropriate conclusion is that the population proportion plausibly lies inside that interval. Choice A states exactly that — both endpoints, built from both directions of the margin.

B) Plausibly less than 0.45 — living below the interval. 0.45 is the interval’s floor. Values below it are precisely the ones the margin of error rules out as implausible — this choice claims the excluded zone instead of the included one.

C) Exactly 0.49 — the word the margin forbids. "Exactly" is the one word a sample estimate can never earn. If the sample could pin the population value exactly, there would be no margin of error to report. The 0.04 exists because 0.49 is an estimate, not a census.

D) Plausibly greater than 0.53 — living above the interval. The mirror of B: 0.53 is the ceiling, and this choice claims everything past it. Both one-sided choices take a real endpoint and point the conclusion the wrong way from it.

The move — Percent First, Then the Margin: A margin of error always builds a two-sided interval: estimate minus margin up to estimate plus margin. The right conclusion says the true value plausibly lies inside; the wrong ones say "exactly" (too strong), or claim one side only (half the interval), or drift outside it entirely. Compute both endpoints first, then read the choices against them.

Question 3, Module 2B — Problem-Solving and Data Analysis: Percentages (Easy)

Your turn — try it before you open the reasoning

What percentage of 300 is 75?

A) 25%
B) 50%
C) 75%
D) 225%

Try it first — then open the reasoning

What percentage of 300 is 75? Margin note: the sentence translates straight across: p⁄100 × 300 = 75, so p = 75⁄300 × 100 = 25. Keep the roles straight — 300 is the whole (it follows "of"), 75 is the part (it follows "is"). The bait is 75%: it recycles the printed number as the answer, and it reverses the roles — 75% of 300 would be 225.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Assign the roles	"of 300" → whole; "is 75" → part.
Divide	75⁄300 = 0.25.
Convert and check	0.25 = 25%; and 25% of 300 = 75 ✓. Choice A.

🖩 Desmos check

Faster by hand here. 75 over 300 reduces by eye — both divide by 75 — to one quarter. The whole screen is one fraction and a sanity check; the only real hazard is answering above 100%, which the part-inside-whole picture rules out before any arithmetic.

Correct: A) 25%
Part over whole: 75⁄300 = 1⁄4 = 0.25 = 25%. The "of" marks 300 as the whole; the "is" marks 75 as the part. Check: 25% of 300 = 75 ✓.

B) 50% — the halfway reflex. Half of 300 is 150, not 75. This bubble collects students who estimate "75 is a decent chunk of 300" without doing the one division that measures the chunk.

C) 75% — the printed number recycled. The stem’s own 75 dressed as a percent. 75% of 300 is 225 — three times the printed part. When an answer choice repeats a number from the stem, it must survive the reverse check, and this one fails it instantly.

D) 225% — the roles reversed. 300⁄75 × 100... almost: this is what you get flirting with the upside-down fraction (400%) and then "fixing" it toward the printed 225 = 75% of 300. Any answer above 100% says the part outweighs the whole — impossible when 75 sits inside 300.

The move — Percent Means Part Over Whole: Every percent sentence is the same equation wearing three disguises: part = percent⁄100 × whole. Whichever piece is missing, translate the sentence literally — "of" is multiplication, "is" is equals — and solve for the empty slot. The word "of" always marks the whole; if your percent comes out above 100 while the part sits inside the whole, you divided upside down.

Say it back before you move on

From memory: From memory: in a percent sentence, which word marks the whole and which marks the part? And what does a margin of error build around an estimate — one bound or two?
Check yourself: If the margin-of-error question cost you, the miss was a sentence, not a number: "exactly" overclaims and the one-sided choices use half the interval. If either percent question slowed you past fifteen seconds, drill the translation until "of" means multiply without thought. Topic review is in the free SAT Math Mixed Review lesson notes.

Geometry and Trigonometry — 3 worked questions

The smallest domain by question count, and the one where the reference sheet — the Reference button, next to the calculator — quietly hands you the formulas. The three below share one discipline: figure out what the geometry hands you for free before any formula fires. A congruence statement hands you a letter map; a length ratio hands you its square for areas; a cube’s volume hands you the edge, one cube root away.

Question 4, Module 1 — Geometry and Trigonometry: Congruent Triangles and Angle Measures (Easy)

I do — watch every step

Triangles ABC and DEF are congruent, where A corresponds to D, and B and E are right angles. Angle A is 18°. What is the measure of angle F? Margin note: the letter order carries the map — ABC ↔ DEF pairs A↔D, B↔E, and so F pairs with C. Angle C = 180 − 90 − 18 = 72°, and congruence copies it to F. The lazy read grabs 18° — but that is D’s angle, not F’s.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

What is the measure of angle F?

A) 18°
B) 72°
C) 90°
D) 162°

By hand
Map the letters	ABC ↔ DEF: A↔D, B↔E, C↔F.
Find the unprinted angle	Angle C = 180 − 90 − 18 = 72°.
Carry it across	Congruence copies measures: angle F = 72°. Choice B.

🖩 Desmos check

Faster by hand here. No figure, no coordinates — nothing to graph. The entire question is one subtraction guarded by a letter map. Write the correspondence out (A↔D, B↔E, C↔F) before any arithmetic; the three wrong choices all exist for students who skipped that line.

Correct: B) 72°
Congruent triangles named in matching order pair their letters by position: F corresponds to C. In triangle ABC, the three angles total 180°: angle C = 180 − 90 − 18 = 72°. Congruence preserves angle measures, so angle F = 72°.

A) 18° — the printed angle, wrong letter. 18° belongs to A — and therefore to D, its correspondent. F sits third in DEF, pairing with C, the angle nobody printed. The question is testing the letter map, and this choice is the map ignored.

C) 90° — the right angle, wrong letter again. The right angles are B and E — the middle letters. F is not a middle letter. Every letter in a congruence statement has exactly one partner, and F’s partner is C.

D) 162° — 180 minus the printed angle. 180 − 18 = 162 subtracts the one known angle from a straight line instead of from a triangle. The triangle’s total is 180 for all three angles, and one of them here is a right angle that must be paid first.

The move — Triangle Angles Total 180: Two triangle angles given, third asked: subtract both from 180. When congruent triangles enter, do the letter map first — the naming order ABC ↔ DEF pairs first with first, second with second, third with third, and corresponding angles are equal. The test prints the angle of the wrong letter among the choices every time; the map is what stops you from picking it.

Question 17, Module 1 — Geometry and Trigonometry: Similar Figures and Area Scale Factors (Medium)

We do — you make one call, then read on

Square A has side lengths that are 166 times the side lengths of square B. The area of square A is how many times the area of square B?

The area of square A is how many times the area of square B?

Note → Your call first. Before computing anything: if one square’s sides were exactly twice another’s, how many times bigger would its area be? Now apply the same logic to 166.

Square A has side lengths that are 166 times the side lengths of square B. The area of square A is how many times the area of square B? Margin note: lengths scale by 166, so areas — which multiply two lengths — scale by 166² = 27,556. The trap grid is 166 itself, the length factor left unsquared. If the numbers feel too big to trust, shrink the problem: sides 2 times longer make an area 4 times bigger, not 2.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Set up the ratio	Side of A = 166 × side of B.
Square for area	Area of A = (166)² × area of B.
Compute 166²	166² = (166)(166) = 27,556. Grid 27556.

🖩 In Desmos

The setup is by hand — the calculator’s job is only the big square.
Type 166^2. Desmos returns 27556.
Grid 27556 — five characters, exactly the grid-in limit for a positive answer.

Which is faster: Split the labor: the insight (square the length factor) is instant by hand, and the arithmetic (166²) is instant by calculator. Doing 166 × 166 with a pencil invites a carrying error on a question you have already solved conceptually.

Correct answer: 27556 (accepted entries: 27556)
Let square B have side s, area s². Square A has side 166s, so its area is (166s)² = 166²s² = 27,556 · s². The area ratio is 166² = 27,556 — the length factor squared, whatever the actual side lengths are.

The number students grid instead — gridding 166. The length ratio, unsquared. Lengths are one-dimensional and areas are two-dimensional, so every length factor gets used twice on the way to an area. The stem is engineered so the honest computation is a single squaring — miss the squaring and the whole question was for nothing.

The move — Area Scales by the Square: When lengths scale by k, areas scale by k² and volumes by k³ — one factor of k for each dimension. Any "how many times" question about area hands you a length ratio and bets you will forget to square it. If the sizes feel abstract, test the rule on k = 2: double the side, four times the area, eight times the volume.

Question 10, Module 2B — Geometry and Trigonometry: Volume and Surface Area of a Cube (Hard)

Your turn — try it before you open the reasoning

A cube has a volume of 474,552 cubic units. What is the surface area, in square units, of the cube?

What is the surface area, in square units, of the cube?

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Try it first — then open the reasoning

A cube has a volume of 474,552 cubic units. What is the surface area, in square units? Margin note: volume to surface area runs through the edge. Un-cube first: edge = ∛474,552 = 78. Then surface area = 6 × 78² = 6 × 6,084 = 36,504. Two grids to refuse: 6,084, one face instead of six, and 78, the edge itself — each is a stop one step short of the asked quantity.

yellow, dotted = the numbers and words the question turns on · green, solid underline = what actually decides the answer · red, dashed = what pulls you toward a wrong number

By hand
Write the target first	Surface area = 6 × edge². The edge is the missing piece.
Un-cube the volume	edge = ∛474,552 = 78 (78³ = 474,552 ✓).
Finish the target	6 × 78² = 6 × 6,084 = 36,504. Grid 36504.

🖩 In Desmos

Type 474552^(1/3). Desmos returns 78 — the edge. (Raising to the 1⁄3 power is the cube root.)
Type 6*78^2. Desmos returns 36504.
Grid 36504.

Which is faster: Desmos, decisively — nobody should extract a cube root of a six-digit number by hand under time pressure. The route is two typed lines, and the ^(1/3) move works on every root the test can ask for: square, cube or fifth.

Correct answer: 36504 (accepted entries: 36504)
A cube’s volume is edge³, so the edge is the cube root of 474,552, which is 78 (check: 78³ = 474,552 ✓). A cube has 6 square faces, each 78² = 6,084 square units. Surface area = 6 × 6,084 = 36,504.

The number students grid instead — gridding 6084 or 78. 6,084 is one face — the ×6 for the six faces never happened. 78 is the edge — the un-cubing happened and nothing after it. Multi-step grid-ins reward writing the target ("surface area = 6 × edge²") before computing, so every intermediate number knows it is not the answer.

The move — Take the Cube Root First: Volume questions that ask for anything else — surface area, edge, diagonal — run through the edge, and the edge comes from un-cubing: raise the volume to the 1⁄3 power in the calculator. Then rebuild toward the asked quantity, and keep the cube’s two formulas separate: volume = edge³ uses one face’s worth of squaring three ways; surface area = 6 × edge² pays all six faces.

Say it back before you move on

From memory: Without looking: in congruent triangles named ABC and DEF, which angle pairs with F? When lengths scale by k, what do areas scale by? And what is the first move on every cube question that gives you the volume?
Check yourself: If you missed the two grid-ins — 17 on Module 1 and 10 on Module 2B — the gap is stopping one step short: the length factor never got squared, or one face never became six. Write the target formula before computing and the intermediate numbers stop masquerading as answers. Review it in the free SAT Math Mixed Review lesson notes.

Timing note: the back half of Module 2

The pacing complaint that fills r/Sat threads is not the whole section — it is the last stretch of Module 2, where the harder branch stacks its multi-step items. The budget is about 95 seconds a question, but you should not spend it evenly: bank time on the early one-steppers (Module 2A opens with a percent question this page works in under ten seconds) and arrive at the back half with three minutes spare. And decide hand-versus-Desmos before you start writing — switching tools mid-question is how a 95-second budget becomes four minutes. The Desmos routes above are exactly the screens where the calculator buys that time back.

Rules and shortcuts from Practice Test 3 Math

Every rule below was earned by a question on this page, and the question is printed with it. Nothing here is general advice.

1. Click the Grey Dot (taught by question 14 of Module 1)

When it fires: Any system of two equations, in any printed form.
What you do: Type both equations into Desmos exactly as printed and click the grey crossing dot — then re-read the stem for which coordinate it wants.
What it saves: The elimination arithmetic and the back-substitution sign slip, on question 14 of Module 1 and on every system after it.

2. The Vertex Answers Min and Max (taught by question 15 of Module 1)

When it fires: A quadratic and the word minimum, maximum, highest or lowest.
What you do: Graph it and click the vertex dot, or use x = −b⁄2a with b’s sign attached; then check whether the stem wants the location or the value.
What it saves: Formula recall under pressure and the −7-for-7 sign slip, on question 15 of Module 1.

3. Set the Exponents Equal (taught by question 20 of Module 1)

When it fires: Both sides of an equation can be written on the same base.
What you do: Match the bases, set the exponents equal, and read the exponent’s printed shape — rt and t⁄r swap the constant’s meaning.
What it saves: The whole logarithm detour, and the doubling-time trap, on question 20 of Module 1.

4. Solve for the Chunk (taught by question 17 of Module 2A)

When it fires: The same wrapped expression appears more than once in one equation.
What you do: Treat it as a single chunk, combine the copies, solve for the chunk, and only then split open what is inside — both ways, if it is an absolute value.
What it saves: A case-by-case absolute-value slog, on question 17 of Module 2A — and the same move re-fires when the identical screen returns on the other branch.

5. No Real Solutions Means a Negative Discriminant (taught by question 20 of Module 2B)

When it fires: A quadratic is declared to have no real solution and a constant is unknown.
What you do: Collect to standard form, write b² − 4ac < 0, solve the inequality — and keep it strict, because the boundary value gives one solution, not none.
What it saves: The doomed attempt to solve an unsolvable equation, and the off-by-one at the boundary, on question 20 of Module 2B.

6. Slide the Graph, Slide the Output (taught by question 9 of Module 2B)

When it fires: A graph is translated, and a value of the new function is asked.
What you do: Write the transformation as an equation first — up k means plus k on every output — then evaluate from the factored form without expanding.
What it saves: A full cubic expansion nobody asked for, on question 9 of Module 2B.

7. Percent First, Then the Margin (taught by question 8 of Module 1)

When it fires: An estimate is reported with a margin of error.
What you do: Build both endpoints — estimate minus margin, estimate plus margin — and accept only the conclusion that stays two-sided and says "plausible".
What it saves: The "exactly" trap and both one-sided misreads, on question 8 of Module 1.

8. Take the Cube Root First (taught by question 10 of Module 2B)

When it fires: A cube’s volume is given and anything else about the cube is asked.
What you do: Un-cube first — raise the volume to the 1⁄3 power for the edge — then rebuild toward the asked quantity with the target formula already written down.
What it saves: A by-hand cube root and the one-face-for-six slip, on question 10 of Module 2B.

Cheat sheet — screenshot this

17 rows. Every rule name below is printed word for word inside the question it cites, so you can jump from a row straight to the worked example that taught it.

Rule	When it fires	What you do
The Intercept Is the Starting Value (M2a Q8)	A word problem with a per-something rate and a onetime fee.	The rate rides with the variable; the fee stands alone. Match the labels to the slots before reading choices.
The Slope Is the Per-Step Change (M1 Q12)	"Each second" / "per unit" questions about a linear model.	Per-unit language asks for the slope — the coefficient on the variable. The intercept answers "at the start".
Graph Both Sides, Count the Crossings (M1 Q9)	"At how many points" / "how many solutions".	Compare slopes: different cross once, equal-and-distinct cross zero times, identical cross everywhere. Or count the dots in Desmos.
Click the Grey Dot (M1 Q14)	A system of two equations, any form.	Type both into Desmos, click the grey dot, grid the coordinate the stem names — ugly x usually means they wanted y.
Step Along the Slope (M2b Q21)	A price schedule with an anchor point and a per-step rate.	Build from the printed point: start value plus rate times steps from the anchor, never from zero.
The Starting Amount Rides in Front (M2a Q16)	An exponential model with a printed starting value.	The coefficient is the t = 0 value; the base is 1 ± the rate. Those two reads eliminate before you compute.
The Vertex Answers Min and Max (M1 Q15)	Minimum, maximum, or the turn of a parabola.	Vertex: click it in Desmos or use x = −b⁄2a — then grid the coordinate the stem asked for.
Set the Exponents Equal (M1 Q20)	Same base available on both sides of an equation.	Set the exponents equal; read whether the exponent is rt or t⁄r before naming the constant.
Solve for the Chunk (M2a Q17)	One wrapped expression appearing twice.	Combine the copies, solve for the chunk, then split — an absolute value always opens two ways.
Divide the Coefficients First (M2a Q18)	An expression declared equivalent to a factored form with an unknown constant.	Divide matching coefficients; then divide the other pair as the built-in check.
Slide the Graph, Slide the Output (M2b Q9)	A translated graph and a value of the new function.	Up k means + k on every output: write g(x) = f(x) + k first, then evaluate from the factors.
No Real Solutions Means a Negative Discriminant (M2b Q20)	"No real solution" with an unknown constant in a quadratic.	Collect, then demand b² − 4ac < 0 — strictly. The boundary integer is the wrong answer by construction.
Percent Means Part Over Whole (M2a Q1 · M2b Q3)	Any percent sentence.	Part = percent⁄100 × whole; "of" marks the whole, "is" marks the part. Solve for the empty slot.
Percent First, Then the Margin (M1 Q8)	An estimate with a margin of error.	Estimate ± margin builds a two-sided interval; the right conclusion says "plausibly inside", never "exactly".
Triangle Angles Total 180 (M1 Q4)	Congruent triangles and an unprinted angle.	Map the letters by position first, then subtract the knowns from 180 and carry the measure across.
Area Scales by the Square (M1 Q17)	A length ratio and an area question.	Areas scale by the square of the length factor; volumes by its cube. Test the rule on k = 2 if in doubt.
Take the Cube Root First (M2b Q10)	A cube’s volume given, something else asked.	Edge = volume^(1⁄3) in the calculator, then rebuild: surface area = 6 × edge².

Digital SAT Practice Test 3 Math answer key — all three modules

Here is the whole key: every entry for Module 1, Module 2A and Module 2B, straight from the form. Grid-in rows show every accepted entry. The 18 questions worked above are marked in the right-hand column.

Module 1 answer key (22 questions)

Question	Answer	On this page
Question 1	C	—
Question 2	D	—
Question 3	.2 or 1/5	—
Question 4	B	Worked in full above (Geometry and Trigonometry)
Question 5	B	—
Question 6	C	—
Question 7	B	—
Question 8	A	Worked in full above (Problem-Solving and Data Analysis)
Question 9	A	Worked in full above (Algebra)
Question 10	C	—
Question 11	24	—
Question 12	D	Worked in full above (Algebra)
Question 13	C	—
Question 14	80	Worked in full above (Algebra)
Question 15	7	Worked in full above (Advanced Math)
Question 16	A	—
Question 17	27556	Worked in full above (Geometry and Trigonometry)
Question 18	C	—
Question 19	C	—
Question 20	B	Worked in full above (Advanced Math)
Question 21	-3	—
Question 22	C	—

Module 2A answer key — the easier branch (22 questions)

Question	Answer	On this page
Question 1	B	Worked in full above (Problem-Solving and Data Analysis)
Question 2	B	—
Question 3	40	—
Question 4	9	—
Question 5	2	—
Question 6	A	—
Question 7	D	—
Question 8	C	Worked in full above (Algebra)
Question 9	D	—
Question 10	D	—
Question 11	70	—
Question 12	D	—
Question 13	D	—
Question 14	A	—
Question 15	B	—
Question 16	A	Worked in full above (Advanced Math)
Question 17	9	Worked in full above (Advanced Math)
Question 18	6	Worked in full above (Advanced Math)
Question 19	D	—
Question 20	D	—
Question 21	B	—
Question 22	A	—

Module 2B answer key — the harder branch (22 questions)

Question	Answer	On this page
Question 1	A	—
Question 2	D	—
Question 3	A	Worked in full above (Problem-Solving and Data Analysis)
Question 4	9	—
Question 5	D	—
Question 6	B	—
Question 7	A	—
Question 8	D	—
Question 9	76	Worked in full above (Advanced Math)
Question 10	36504	Worked in full above (Geometry and Trigonometry)
Question 11	C	—
Question 12	C	—
Question 13	B	—
Question 14	D	—
Question 15	9	—
Question 16	182	—
Question 17	C	—
Question 18	B	—
Question 19	B	—
Question 20	50	Worked in full above (Advanced Math)
Question 21	A	Worked in full above (Algebra)
Question 22	B	—

What students said about this

Better way to do system of equations problems with Desmos? — yes: type both equations as printed and click the dot. That is the whole method, worked here on a real system from this form whose by-hand route hides a sign trap.
Top 75 math types from BB 4 – 11 — the recurring Bluebook question types. Nearly every entry on that list has an ancestor on this form: the no-real-solution constant, the translated function, the absolute-value chunk.
Desmos regressions — a long thread on the move students most want to over-use. The Desmos section above says exactly when regression is the right tool and when it is a slower detour.

Questions students ask about this test

What are the answers to practice test 3 math?

All three module keys — Module 1, Module 2A and Module 2B, 22 entries each — are printed in the recap tables above, grid-in values included. The 18 questions worked in full on this page are marked in those tables.

Which math domain has the most questions on test 3?

Across forms, Algebra and Advanced Math carry the most questions, with Problem-Solving and Data Analysis and Geometry/Trigonometry smaller — that balance is College Board’s published section design, and it is why this page gives its two biggest sections to those two domains: 5 Algebra and 7 Advanced Math questions worked, against 3 and 3 for the smaller pair.

What is the hardest math question on test 3?

On this page’s spread, question 20 of Module 2B — the no-real-solution discriminant hunt with the integer boundary — is the one that costs the most points, with question 10 of Module 2B and question 17 of Module 2A just behind it. The solve: collect the quadratic, demand a negative discriminant, and refuse the boundary integer. College Board publishes no per-question difficulty, so treat any hardest ranking, including this one, as a reading of the items rather than a measurement.

How do I use Desmos on test 3’s systems questions?

Type each equation exactly as printed — no rearranging — and read the graphs: a grey crossing dot is the solution pair, and an empty screen is the answer "zero solutions". Both cases are worked above, at question 14 of Module 1 and question 9 of Module 1, with every keystroke printed.

How many grid-ins are on test 3?

Derived from this form’s key: 6 on Module 1, 6 on Module 2A, 6 on Module 2B. You type the answer instead of choosing one: up to 5 characters for a positive answer, 6 for a negative one. A fraction that fits may be gridded as a fraction; otherwise enter the decimal — the form’s own key lists .2 or 1/5 as equally accepted entries for one Module 1 grid-in.

What percentage of Digital SAT math is algebra?

College Board’s published section design puts Algebra at about 35% of the questions, with Advanced Math at roughly the same weight and the two smaller domains splitting the rest. On a 22-question module that is seven or eight Algebra screens — which is why the Algebra section above drills the slope-intercept reads until they are automatic.

How much time do I get for each math module?

35 minutes for 22 questions is about 95 seconds each. Do not spend it evenly: the early questions of each module are one-steppers, and the time they release is what the back half of Module 2 runs on.

Is test 3 math a good predictor of my real score?

Not officially — Bluebook forms are built to the same specification and scores are equated, so neither form is designed harder. But students on r/Sat consistently report the first four Bluebook practice tests running easier than the current paper, so read your result as a domain map — where you lose points — rather than a promise about the scaled number.

Keep going

Sit Practice Test 3’s Math section in Bluebook if you have not, then come back, mark your misses in the recap tables, and work the domain section above where they cluster — the Desmos routes are the part that transfers to every later form.

See plans — from $39.99/mo, 1-day trial Work with Mr. Osama — from $15/session

Questions quoted on this page come from the official College Board Digital SAT Practice Test 3, Section 2 (Math), Modules 1, 2A and 2B. Stems, choices and data values are reproduced for study and belong to the College Board; every quoted item was checked against the rendered pages of the official PDF, and questions whose figures or garbled text could not be reproduced faithfully were left out rather than approximated. Every explanation, rule name, Desmos route and annotation on this page is The Test Advantage’s own work.

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Practice Test 3 Math at a glance

How to use this page

How the adaptive Math section works on this form

The Desmos moves this page teaches

Algebra — 5 worked questions

Question 8, Module 2A — Algebra: Building Linear Models (Easy)

Question 12, Module 1 — Algebra: Interpreting Linear Models (Easy)

Question 9, Module 1 — Algebra: Systems of Two Linear Equations (Easy)

Question 14, Module 1 — Algebra: Systems of Two Linear Equations (Medium)

Question 21, Module 2B — Algebra: Linear Functions from Two Values (Medium)

Advanced Math — 7 worked questions

Question 16, Module 2A — Advanced Math: Exponential Functions in Context (Medium)

Question 15, Module 1 — Advanced Math: Quadratic Minimum and the Vertex (Medium)

Question 20, Module 1 — Advanced Math: Exponential Growth and Doubling Time (Hard)

Question 17, Module 2A — Advanced Math: Absolute Value Equations (Hard)

Question 18, Module 2A — Advanced Math: Equivalent Expressions and Common Factors (Medium)

Question 9, Module 2B — Advanced Math: Function Translations (Hard)

See which question types are costing you points

Question 20, Module 2B — Advanced Math: Quadratics and the Discriminant (Hard)

Problem-Solving and Data Analysis — 3 worked questions

Question 1, Module 2A — Problem-Solving and Data Analysis: Percentages (Easy)

Question 8, Module 1 — Problem-Solving and Data Analysis: Margin of Error (Medium)

Question 3, Module 2B — Problem-Solving and Data Analysis: Percentages (Easy)

Geometry and Trigonometry — 3 worked questions

Question 4, Module 1 — Geometry and Trigonometry: Congruent Triangles and Angle Measures (Easy)

Question 17, Module 1 — Geometry and Trigonometry: Similar Figures and Area Scale Factors (Medium)

Question 10, Module 2B — Geometry and Trigonometry: Volume and Surface Area of a Cube (Hard)

Rules and shortcuts from Practice Test 3 Math

Cheat sheet — screenshot this

Digital SAT Practice Test 3 Math answer key — all three modules

Module 1 answer key (22 questions)

Module 2A answer key — the easier branch (22 questions)

Module 2B answer key — the harder branch (22 questions)

Questions students ask about this test

What are the answers to practice test 3 math?

Which math domain has the most questions on test 3?

What is the hardest math question on test 3?

How do I use Desmos on test 3’s systems questions?

How many grid-ins are on test 3?

What percentage of Digital SAT math is algebra?

How much time do I get for each math module?

Is test 3 math a good predictor of my real score?

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