A full ACT Math review built around what differs from the SAT: a tighter clock against rising difficulty, roughly double the geometry, and the four topics the SAT never tests — logarithms, sequences, matrices and complex numbers. Covers percent and absolute value, coordinate and plane geometry, both special right triangles, sectors and similar-figure scaling, the sine and cosine laws, trigonometric graphs, and when to backsolve rather than rearrange.
Subject: ACT Prep · 63 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
ACT Prep · Session 1
What differs from the SAT, and the topics it tests that the SAT never touches
Objectives
If your instincts came from the SAT, three things will surprise you: there is far more geometry, the clock is much tighter, and several topics appear that the SAT does not test at all.
ACT — Description of the Mathematics Test and reporting categories — the reporting categories this deck follows
Section
Section 1
Warm-up
Two minutes before any content.
Discussion prompt
You have done SAT-style math prep. Name two habits from that preparation that you expect to transfer, and one you suspect will not.
Hint: Which of your habits depends on having spare minutes?
Answer:
Transferring well: reading for what is actually asked, and substituting an answer back to check it.
Transferring badly: the pace. There is appreciably less time per question here, so the long, careful method that worked before will not fit.
Also new: a calculator is allowed on the whole section, and the geometry share is roughly double.
ACT — the official test site, including the current test description and free practice — the official test description
Concept
First, the questions get harder roughly in order. The early ones are genuinely quick, and the last stretch is where the section is decided.
Second, a calculator is permitted throughout, which changes what counts as a hard question: arithmetic is never the obstacle, so the difficulty is always in the setup.
Check the current question count and timing against your own admission materials before test day, since the section was revised in 2025.
ACT — Description of the Mathematics Test and reporting categories — and confirm against your registration
Picture it
Not strictly, but closely enough to plan around.
Figure (svg): A rising curve of difficulty against question number, with the first third shaded green and the last third shaded red
Which gives the single most useful pacing rule here: never spend more than about forty seconds on anything in the first third.
Picture it
Approximate shares, but the ordering is the point.
Figure (svg): Grouped bars comparing the ACT and SAT content shares, with geometry and trigonometry far larger on the ACT
So this deck spends its middle third on geometry, which is not where an SAT-shaped revision plan would put it.
Sorting
Worth knowing, because these are the topics your existing preparation will have skipped entirely.
Sort into buckets
On the ACT only, or on both tests?
The ACT-only column is small and completely learnable in one session, which makes it the best return on time in the whole deck.
Pattern
Four rules. The first one is the one that creates the time for the other three.
That last rule is not a strategy so much as arithmetic, and every year students still leave blanks.
ACT — the official test site, including the current test description and free practice — scoring, and the absence of a guessing penalty
Check
Think about it before answering.
Check your understanding
With about a minute per question on average, roughly how long should a question in the first third of the section take?
Answer: A
Why: Every question is worth the same, but the later ones take longer. Spending the average on an easy question means arriving at the hard ones with no reserve, which is how strong students run out of time.
Section
Section 2
Concept
Almost every percent question is one of: what is this percent of that, this is what percent of that, or this is that percent of what. Identifying which one turns it into a single equation.
\[ \text{part} = \text{percent} \times \text{whole} \]
The word of means multiply, and the word is means equals. Translating literally beats trying to remember a rule.
Worked example
Two questions from the same numbers, which are commonly confused.
What percent of 80 is 28?
Why: Translate literally: the unknown percent times 80 equals 28.
\[ p \times 80 = 28 \Rightarrow p = 0.35 \]
A value rises from 80 to 28 higher. What is the percent increase?
Why: Percent change always divides by the original, never by the new value.
\[ \frac{28}{80} = 0.35 \Rightarrow 35\% \]
Now the trap version: it rises from 80 to 108, then falls back to 80
Why: The fall is 28 out of 108, not out of 80.
\[ \frac{28}{108} \approx 0.259 \]
Verify: that the two percentages differ
Why: Up 35 percent then down 25.9 percent returns to the start. If a question offers 35 percent for the decrease, that is the distractor, and it is there every time.
Picture it
The same 28 units, measured against two different bases.
Figure (svg): Two bars showing twenty-eight as thirty-five percent of eighty and as roughly twenty-six percent of one hundred and eight
Concept
An absolute value is a distance from zero, so an absolute-value equation asks which values sit a fixed distance from a point. There are two of them, symmetric about that point.
\[ |2x - 5| = 9 \]
Worked example
Split it into the two cases the absolute value is hiding.
Write both branches
Why: The expression inside is either nine or negative nine. There is no third possibility.
\[ 2x - 5 = 9 \qquad \text{or} \qquad 2x - 5 = -9 \]
Solve each
Why: Two ordinary linear equations.
\[ x = 7 \qquad \text{or} \qquad x = -2 \]
Verify: by substituting both back
Why: Two times seven minus five is nine, and its absolute value is nine. Two times negative two minus five is negative nine, and its absolute value is also nine. Both work.
Picture it
The two solutions sit the same distance either side of 2.5.
Figure (svg): A number line with solutions marked at negative two and seven, each four and a half units from the midpoint at two point five
If the right-hand side had been negative, there would be no solutions at all, since a distance is never negative. That version appears too.
Prediction
No working.
Predict first
How many real solutions does the equation with absolute value of 3x plus 1 equal to negative 4 have?
Correct: None.
Why: An absolute value is a distance and can never be negative, so no value of x makes it equal negative four. Recognising this takes two seconds; splitting into cases and solving takes a minute and produces two wrong answers.
Section
Section 3
Concept
The logarithm answers the question: to what power must the base be raised to get this number. Reading it that way removes almost all the difficulty.
\[ \log_b(x) = y \;\Longleftrightarrow\; b^y = x \]
\[ \log_2(32) = 5 \]
Paul's Online Math Notes — logarithms, sequences and series Algebra, Logarithm Functions
Worked example
Solve for x when three raised to the power x plus one equals eighty-one.
Write both sides as powers of the same base
Why: Eighty-one is three to the fourth. Matching bases avoids logarithms entirely, and on this test that is usually possible.
\[ 3^{x+1} = 3^4 \]
Equate the exponents
Why: If the bases match and are not one, the exponents must match.
\[ x + 1 = 4 \Rightarrow x = 3 \]
Verify: by substituting back
Why: Three to the power four is eighty-one. If the bases had not matched, taking a logarithm of both sides would have been the fallback.
Picture it
Two routes to the same answer, and one is much faster when it is available.
Figure (svg): Two bars comparing the speed of matching bases against taking logarithms of both sides
The three log laws are still worth knowing: a sum of logs is the log of a product, a difference is a quotient, and a coefficient is a power.
Concept
An arithmetic sequence adds a fixed amount each step. A geometric sequence multiplies by a fixed amount. The wording tells you which, exactly as with linear against exponential.
\[ a_n = a_1 + (n-1)d \]
\[ a_n = a_1 r^{n-1} \]
Note the minus one in both. It is where nearly every sequence error comes from.
OpenStax, Precalculus 2e — sequences, logarithms and trigonometric functions Ch. 11 — sequences and series
Worked example
An arithmetic sequence starts at 4 and increases by 6 each step.
Find the twentieth term
Why: Nineteen steps are taken to get from the first term to the twentieth, not twenty.
\[ a_{20} = 4 + 19 \times 6 = 118 \]
Sum the first twenty terms
Why: The sum is the number of terms times the average of the first and last.
\[ S_{20} = \frac{20}{2}(4 + 118) = 10 \times 122 = 1220 \]
Verify: with a small case
Why: The first three terms are 4, 10 and 16, summing to 30. The formula gives three halves times twenty, which is 30. The formula is right, so the large case can be trusted.
Picture it
Both start at 4. One adds 6 each step; the other doubles.
Figure (svg): Bars comparing an arithmetic sequence rising by six each step against a geometric sequence doubling each step
A geometric sequence from the same start with ratio two reaches 384 by the eighth term rather than 46. The wording is the only thing distinguishing the two questions.
Fill the middle
An arithmetic sequence with first term 4 and common difference 6.
Fill in the blanks
a_19 = 4 + 118 \times 6 = ___
Why: Getting from term one to term twenty takes nineteen steps, not twenty. Using twenty gives 124, which is the distractor that appears in the answer choices every time this question is asked.
Concept
The ACT asks very little about matrices. Two facts cover almost all of it: the determinant of a two by two, and when two matrices can be multiplied at all.
\[ \det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc \]
Multiplication requires the inner dimensions to match, and the result has the outer dimensions.
OpenStax, Precalculus 2e — sequences, logarithms and trigonometric functions Ch. 9 — matrices
Check
Solve it on paper before you click.
Check your understanding
What is the determinant of the two by two matrix with first row 3 and 4, and second row 2 and 5?
Answer: A
Why: The determinant is the product of the main diagonal minus the product of the other diagonal: three times five is fifteen, minus four times two which is eight, giving seven.
Concept
The imaginary unit squares to negative one. Every complex-number question on this test reduces to using that fact and then collecting like terms.
\[ i^2 = -1 \]
\[ (2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13 \]
A pair like that is called a conjugate pair, and multiplying conjugates always clears the imaginary part completely.
Error analysis
A classmate simplifies a product of complex numbers.
Annotate
On: \( (2+3i)(2-3i) = 4 - 9i^2 = 4 - 9 = -5 \)
Two sign flips in one step is why this is worth slowing down for. Write the i squared step out rather than doing it in your head.
Section
Section 4
Concept
Distance, midpoint and slope all come from the same right triangle drawn between two points. Learn the picture and you can rebuild all three.
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]
\[ M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \]
Worked example
Between the points two, three and seven, fifteen.
Find the horizontal and vertical gaps
Why: Five across and twelve up. Recognise the pair.
\[ \Delta x = 5, \qquad \Delta y = 12 \]
Distance is the hypotenuse
Why: Five, twelve, thirteen is a Pythagorean triple, so no calculator is needed.
\[ d = \sqrt{25 + 144} = 13 \]
Midpoint is the average of each coordinate
Why: Not the difference. Averaging is what a midpoint means.
\[ M = (4.5,\; 9) \]
Slope is rise over run
Why: Twelve over five, and a perpendicular line would have slope negative five over twelve.
Verify: that the midpoint lies between the two points
Why: Its coordinates sit between 2 and 7, and between 3 and 15. If a midpoint falls outside the two points, the formula was applied as a difference rather than an average.
Picture it
The dashed legs give the gaps; the solid line is the distance.
Figure (svg): Two points on the coordinate plane joined by a line, with dashed horizontal and vertical legs of five and twelve and a hypotenuse of thirteen
Memorise the triples: three-four-five, five-twelve-thirteen, eight-fifteen-seventeen, seven-twenty-four-twenty-five. They turn up constantly and each one saves twenty seconds.
Matching
Two lines in the plane.
Match the pairs
Why: The perpendicular condition is often stated as the negative reciprocal, which is the same thing: a slope of twelve fifths pairs with negative five twelfths, and their product is negative one.
Section
Section 5
Concept
The forty-five forty-five ninety and the thirty sixty ninety appear constantly, often hidden inside a square or an equilateral triangle. Knowing their side ratios turns a derivation into a glance.
Picture it
Two ratios. Everything else in this section leans on them.
Figure (svg): The forty-five forty-five ninety triangle with sides one, one and root two, beside the thirty sixty ninety triangle with sides one, root three and two
That last sentence is how to recover them if memory fails under pressure.
Worked example
A circle of radius six has a central angle of sixty degrees. Find the arc length and the sector area.
Find what fraction of the circle the angle is
Why: Sixty out of three hundred and sixty is one sixth.
\[ \frac{60}{360} = \frac{1}{6} \]
Take that fraction of the circumference
Why: The full circumference is two pi times six, which is twelve pi.
\[ \text{arc} = \frac{1}{6} \times 12\pi = 2\pi \]
Take the same fraction of the area
Why: The full area is pi times thirty-six.
\[ \text{sector} = \frac{1}{6} \times 36\pi = 6\pi \]
Verify: that both are a sixth of the whole
Why: Twelve pi over six is two pi, and thirty-six pi over six is six pi. One rule covers both, so there are not two formulas to remember.
Picture it
One fraction, applied to two different whole-circle quantities.
Figure (svg): A circle of radius six with a sixty degree sector shaded, labelled with an arc length of two pi and a sector area of six pi
Trap
A figure is enlarged so its sides are five thirds as long. Its area was six.
Multiply the area by five thirds
Why: The scale factor was applied once, as though area were a length.
\[ 6 \times \frac{5}{3} = 10 \]
Report an area of 10
Why: The correct answer is nearly seventeen, so this is not close.
Area scales by the square of the length factor.
\[ 6 \times \left(\frac{5}{3}\right)^2 = 6 \times \frac{25}{9} = \frac{50}{3} \]
Use the dimension as the exponent
Why: One for length, two for area, three for volume. That single rule replaces three separate facts.
Sanity-check the size
Why: Making something two thirds longer should make it well over half again as big in area, and fifty thirds is nearly triple six.
Picture it
One scale factor, three different exponents.
Figure (svg): A small rectangle and its enlargement, annotated to show sides scaling by five thirds, areas by twenty-five ninths and volumes by one hundred and twenty-five twenty-sevenths
This is asked as often about volumes of similar solids as about areas of similar figures, and the cube catches even more people than the square.
Estimation
A regular octagon.
Predict first
Roughly how large is each interior angle?
Correct: About 135 degrees.
Why: The interior angles of an n-sided polygon sum to n minus two, times 180. For eight sides that is six times 180, which is 1080, and dividing by eight gives exactly 135. Estimating first also rules out the two small options instantly, since the angle of a many-sided polygon must be obtuse.
Section
Section 6
Concept
In a right triangle the sine is opposite over hypotenuse, the cosine is adjacent over hypotenuse, and the tangent is opposite over adjacent. Most trigonometry questions on this test need nothing more.
Which side counts as opposite or adjacent depends on which angle you are working from, and mixing that up is the commonest error here.
Concept
When the triangle has no right angle, one of two laws applies, and which one depends on what you were given.
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
\[ c^2 = a^2 + b^2 - 2ab\cos C \]
The second reduces to the Pythagorean theorem when the angle is ninety degrees, because its cosine is zero. That is a good way to remember it is not an unrelated formula.
OpenStax, Precalculus 2e — sequences, logarithms and trigonometric functions Ch. 10
Worked example
A triangle has sides of 5 and 7 with an angle of sixty degrees between them. Find the third side.
Choose the law by what you were given
Why: Two sides and the angle between them is exactly the law of cosines. The law of sines cannot start here, because it needs an angle opposite a known side.
\[ c^2 = 5^2 + 7^2 - 2(5)(7)\cos 60^{\circ} \]
Substitute the cosine
Why: The cosine of sixty degrees is one half, which is worth knowing without a calculator.
\[ c^2 = 25 + 49 - 70 \times 0.5 = 39 \]
\[ c = \sqrt{39} \approx 6.24 \]
Verify: against the triangle inequality
Why: The third side must lie between two and twelve, and it should sit between five and seven because the angle is moderate. Six point two four passes both checks.
Picture it
The choice is decided entirely by what the question hands you.
Figure (svg): A decision flow showing that two sides and the included angle calls for the law of cosines while an angle opposite a known side calls for the law of sines
Concept
For a sine or cosine curve written with a coefficient in front and a coefficient inside, the outside one is the amplitude and the inside one compresses the period.
\[ y = a\sin(bx): \quad \text{amplitude } |a|, \quad \text{period } \frac{2\pi}{b} \]
OpenStax, Precalculus 2e — sequences, logarithms and trigonometric functions Ch. 6
Picture it
The outside coefficient stretches vertically; the inside one squeezes horizontally.
Figure (svg): A sine curve with amplitude three and period pi, with the amplitude and one full period marked
The inside coefficient behaving backwards is the same idea as the horizontal shift in function transformations. One principle, several appearances.
Prediction
Without drawing anything.
Predict first
What is the period of the curve with amplitude coefficient 5 and inside coefficient 4?
Correct: Pi over 2.
Why: The period is two pi divided by the inside coefficient, so two pi over four is pi over two. The amplitude coefficient of five has no effect on the period at all, which is what the fourth option is testing.
Section
Section 7
Concept
Because every question is multiple choice, the answers are data. Two techniques exploit that, and knowing which to reach for is most of the benefit.
backsolving — Try an answer choice in the question and see whether it works. Best when the choices are plain numbers.
plugging in numbers — Pick an easy value for the variable, compute the answer, then see which choice matches. Best when the choices contain variables.
ACT — the official test site, including the current test description and free practice — the official practice materials, for questions to drill these on
Picture it
Four situations, four responses.
Figure (svg): A decision flow listing backsolving for numeric choices, plugging in numbers for algebraic choices, and marking up or drawing the figure for geometry
Backsolve from the middle choice: if it is too large you have eliminated it and everything above it in one test.
Worked example
A question asks for the value of x satisfying an equation you would rather not rearrange. The choices are 2, 4, 6 and 8.
Start with the second-largest or second-smallest, not the first
Why: If the choices are ordered, testing a middle one tells you the direction as well as the answer.
Substitute and compare
Why: You are not solving; you are checking. That is much faster and it cannot go wrong algebraically.
Use the direction to eliminate
Why: If the value came out too large, every larger choice is gone too, so one test can eliminate three options.
| choice tried | result | what it eliminates |
|---|---|---|
| 6 | too large | 6 and 8 |
| 4 | correct | done in two tests |
Verify: that the surviving choice actually works
Why: Backsolving only proves a choice fits; always confirm the one you keep rather than inferring it by elimination alone.
Picture it
Ordered numeric choices make elimination compound.
Figure (svg): Two bars comparing the time to rearrange and solve algebraically against the time to backsolve two answer choices
On a clean equation, rearranging is faster. The technique is a tool for the hard end of the section, not a replacement for algebra.
Discrimination
Read only the answer choices, not the question.
Sort into buckets
Backsolve, or plug in a number?
Pattern
Seven rules. The first and the last are the two that move a score on their own.
| situation | do this |
|---|---|
| a question in the first third | forty seconds maximum, then move |
| cannot start within fifteen seconds | skip it and come back |
| any question at all, at the end | answer it; there is no penalty for guessing |
| percent change | divide by the original, never the new value |
| a sequence | count steps, not terms, and remember the minus one |
| similar figures | the exponent is the dimension: one, two or three |
| a triangle with no right angle | between means cosines, opposite means sines |
| ugly algebra, numeric choices | backsolve from the middle |
ACT — Description of the Mathematics Test and reporting categories
Check
Solve it on paper before you click.
Check your understanding
A triangle has sides of 9 and 12 with an angle of 40 degrees between them, and you need the third side. Which approach works?
Answer: A
Why: Two sides with the angle between them is exactly the configuration the law of cosines handles. Substituting gives the third side directly.
Check
Solve it on paper before you click.
Check your understanding
Two similar solids have corresponding edges in the ratio 2 to 3. What is the ratio of their volumes?
Answer: A
Why: Volume is three-dimensional, so the ratio is the cube of the edge ratio: two cubed to three cubed, which is eight to twenty-seven.
Exit ticket
One honest answer, and it sets the next session.
Predict first
Which of these would you least want to meet cold?
Correct: Whichever you named opens the next session.
Why: If the answer is the second option, that is the best possible one to pick: those four topics are small, self-contained, and completely absent from SAT preparation, which makes them the highest return on an hour of work anywhere in this deck.
Connect it up
Twenty minutes, handwritten, one side of a card.
Draw it
Write down, from memory: the distance and midpoint formulas, both special triangle ratios, the sector rule, both sequence formulas, the two triangle laws, and the amplitude and period rules. Circle anything you had to look up.
The circled ones are the session plan. Bring the card.
Recap
Seven sections, and the two that most change a score are the clock and the four topics your SAT preparation never covered.
| question | answer here |
|---|---|
| what percent of 80 is 28 | 35 percent |
| absolute value of 2x minus 5 equals 9 | x = 7 or x = -2 |
| 3 to the power x plus 1 equals 81 | x = 3 |
| 20th term, start 4, step 6 | 118, and the sum is 1220 |
| distance from (2,3) to (7,15) | 13 |
| arc and sector, r = 6, angle 60 | 2 pi and 6 pi |
| third side, 5 and 7 with 60 between | root 39, about 6.24 |
ACT — the official test site, including the current test description and free practice — official free practice, for drilling any of the above
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