ACT Math · ClipSAT

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ACT Math

Provider · ACT, Inc. Format · 45 questions · 50 min Chapters · 8 Explorers · 3 interactive
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Test format & strategy

🗻 ACT Math — Official Reporting Categories (enhanced ACT: 45 questions, 50 min)

Reporting Category% of TestKey Topics
Preparing for Higher Mathematics80%Broken into 5 sub-areas below
   · Number & Quantity10–12%Integers, ratios, percents, complex numbers, vectors, matrices
   · Algebra17–20%Linear/quadratic equations, systems, inequalities, absolute value
   · Functions17–20%Definition, notation, graphs, transformations, composition, inverse
   · Geometry17–20%Area, volume, angle relationships, trig, coordinate geometry
   · Statistics & Probability12–15%Mean/median/mode, data interpretation, probability, expected value
Integrating Essential Skills20%Multi-step problems spanning rates, percentages, proportions, measurement, area/volume

The enhanced ACT Math test (from September 2025) is 45 multiple-choice questions in 50 minutes — about 67 seconds each — and a calculator is allowed throughout. 41 questions are scored; 4 are unscored field-test items you can't identify, so answer every one. Questions run roughly easy to hard, drawing on everything from pre-algebra to trigonometry.

What to expect

Four answer choices per question (the enhanced ACT dropped the fifth). There is no formula sheet, so the key formulas must be memorized. The content spans number & quantity, algebra, functions, geometry, and statistics & probability.

Questions
45 · multiple choice
Time
50 minutes
Choices
4 per question
Calculator
allowed throughout
▲ Three tactics that score

Don't linger — flag a hard item and move on so easy points aren't lost. Backsolve from the answer choices, and plug in convenient numbers for variables when a question is abstract.

⏱ 3-Tier Pacing Strategy (45 Questions / 50 Minutes)

  • Questions 1–15 (Easy): Spend ~40 seconds per question (10 minutes total). Bank time early while building confidence.
  • Questions 16–30 (Medium): Spend ~65 seconds per question (16 minutes total). Watch for multi-step algebra and word problems.
  • Questions 31–45 (Hard): Spend ~85 seconds per question (21 minutes total). These feature advanced geometry, trigonometry, and novel combinations.
  • Final 3 Minutes: Check flagged items and ensure EVERY single question has a bubbled answer!

▲ Zero Guessing Penalty & 4-Choice Advantage
  • No Negative Marking: There is absolutely no penalty for wrong answers on the ACT. An unanswered question is guaranteed 0 points; an educated guess gives a 25% to 50% chance of a point.
  • 4-Choice Format: Enhanced ACT has 4 options (A, B, C, D) instead of 5, boosting blind guessing accuracy from 20% to 25%. Eliminating just one obvious wrong answer raises your odds to 33.3%, and eliminating two gives a 50% coin-flip.
  • Letter of the Day: If running out of time on the last 3–4 questions, pick one consistent letter (e.g., C) for all remaining items rather than alternating randomly.
Bar chart: ACT Math Reporting Category Weighting (%) with categories: Algebra (19), Geometry (19), Functions (19), Statistics (13), Number & Quantity (10), Essential Skills (20)05101520Algebra19Geometry19Functions19Statistics13Number & Quantity10Essential Skills20Approximate Percentage of Test (%)ACT Math Reporting Category Weighting (%)
ACT Math reporting category distribution. Preparing for Higher Mathematics comprises 80% (Algebra 19%, Geometry 19%, Functions 19%, Stats/Prob 13%, Number & Quantity 10%), while Integrating Essential Skills makes up the remaining 20%.
On the enhanced ACT Math test (45 questions in 50 minutes), a student allocates 10 minutes to the first 15 questions and 16 minutes to questions 16 through 30. How many seconds per question remain for the final 15 questions, assuming 3 minutes are reserved at the end to check answers?
Time spent so far=10+16=26 minutes\text{Time spent so far} = 10 + 16 = 26\text{ minutes}. Leaving 3 minutes for review leaves 50−26−3=21 minutes50 - 26 - 3 = 21\text{ minutes} for the final 15 questions. Time per question=21×60 seconds15 questions=126015=84 seconds per question\text{Time per question} = \frac{21 \times 60\text{ seconds}}{15\text{ questions}} = \frac{1260}{15} = 84\text{ seconds per question}.
📝 Chapter Quiz
1

Numbers, Percents & Ratios

🗻 Number & Quantity 7–10%  ·  Integers & primes  ·  Fractions, decimals, percents  ·  Ratios & proportions  ·  Real number properties

Pre-algebra fundamentals: fractions and decimals, percentages, ratios and proportions, factors and multiples.

▲ Reference — Proportional reasoning

Percent of a number: p100×N \tfrac{p}{100}\times N . A ratio a:b a:b splits a total into a+b a+b equal parts. Set up a proportion ab=cd \tfrac{a}{b}=\tfrac{c}{d} and cross-multiply.

Number line from -4 to 6 with marked points-4-3-2-10123456-305/25
Real number line showing integers and rational fractions. On the ACT, distance between two coordinates a a and b b is given by the absolute difference ∣a−b∣ |a - b| .
Percent of
p100×N \tfrac{p}{100}\times N
Percent change
new−oldold \tfrac{\text{new}-\text{old}}{\text{old}}
Ratio parts
total ÷(a+b) \div (a+b)
Proportion
cross-multiply
⚠️ Classic ACT Traps: Percents, Ratios & Number Properties
  • Percent Change Base Value: Always divide the difference by the original starting value, never the new value: Percent Change=New−OldOld×100% \text{Percent Change} = \frac{\text{New} - \text{Old}}{\text{Old}} \times 100\% .
  • Successive Percentage Changes Do Not Add: A 20% discount followed by a 10% discount is NOT a 30% discount. The total multiplier is 0.80×0.90=0.72 0.80 \times 0.90 = 0.72 , meaning an effective 28% total discount.
  • Ratio Parts vs. Whole: If boys to girls is 3:5 3:5 , boys represent 33+5=38 \frac{3}{3+5} = \frac{3}{8} of the total students, NOT 35 \frac{3}{5} . Always add ratio terms to find total parts.
  • Number Property Facts: 2 2 is the only even prime number; 1 1 is neither prime nor composite; 0 0 is an even integer (neither positive nor negative).
Worked example 1.A

What is 15% 15\% of 200 200 ? 0.15×200=30. 0.15\times200=30.

Worked example 1.B

Split 30 30 in the ratio 2:3 2:3 . Find the smaller part.

Five parts, each 6 6 ; smaller is 2 parts:

12. 12.

A jacket costs $120 after a 25% discount. What was the original price?
Let original = xx. 0.75x=120⇒x=1600.75x = 120 \Rightarrow x = 160. Original price: $160.
A merchant buys a gadget for $50, marks it up by 40%, and later sells it at a 20% discount off the marked price. What was the final selling price and the merchant's net profit?
Marked price: 50×1.40=$7050 \times 1.40 = \$70. Sale price after 20% discount: 70×0.80=$5670 \times 0.80 = \$56. Net profit: 56−50=$656 - 50 = \$6.
📝 Chapter Quiz
2

Elementary & Intermediate Algebra

🗻 Algebra 12–15%  ·  Linear equations & inequalities  ·  Systems of equations  ·  Quadratics  ·  Absolute value equations

Linear equations, quadratics, exponents, radicals and absolute value — the algebraic core of the test.

▲ Reference — Algebra essentials

Quadratic formula x=−b±b2−4ac2a x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a} ; difference of squares a2−b2=(a−b)(a+b) a^{2}-b^{2}=(a-b)(a+b) . Exponents: aman=am+n a^{m}a^{n}=a^{m+n} . Absolute value ∣x∣=c |x|=c gives x=±c x=\pm c .

Quadratic Parabola: Vertex, Roots & Axis of Symmetry — Coordinate plane with graph of y = x² - 4x + 3; marked points Vertex (2, -1) (2, -1), Root (1, 0) (1, 0), Root (3, 0) (3, 0)123424xyy = x² − 4x + 3Vertex (2, -1)Root (1, 0)Root (3, 0)y = 0
Graph of y=x2−4x+3 y = x^2 - 4x + 3 . The vertex coordinates are given by x=−b/(2a)=2 x = -b/(2a) = 2 and y=f(2)=−1 y = f(2) = -1 . The x x -intercepts (1,0) (1,0) and (3,0) (3,0) represent the real solutions to x2−4x+3=0 x^2 - 4x + 3 = 0 .
Quadratic formula
−b±b2−4ac2a \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}
Difference of squares
(a−b)(a+b) (a-b)(a+b)
Exponents
aman=am+n a^{m}a^{n}=a^{m+n}
Absolute value
x=±c x=\pm c
⚠️ Classic ACT Traps: Algebra & Quadratics
  • No Formula Sheet Provided: Unlike the SAT, the ACT provides zero reference formulas. You must memorize the Quadratic Formula x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} and the vertex formula x=−b2a x = -\frac{b}{2a} .
  • Flipping Inequality Signs: When multiplying or dividing an inequality by a negative number, always reverse the inequality symbol: −3x<12⇒x>−4 -3x < 12 \Rightarrow x > -4 .
  • Vieta's Sum of Roots Shortcut: For any quadratic ax2+bx+c=0 ax^2 + bx + c = 0 , the sum of the solutions is −b/a -b/a and the product of the solutions is c/a c/a . When the ACT asks for 'the sum of all solutions', never solve with the quadratic formula — simply evaluate −b/a -b/a !
  • Discriminant Rules: The quantity Δ=b2−4ac \Delta = b^2 - 4ac reveals solution count: Δ>0 \Delta > 0 (2 real roots), Δ=0 \Delta = 0 (1 real root), Δ<0 \Delta < 0 (2 complex/imaginary roots).
Worked example 2.A

Solve 3(x−2)=12 3(x-2)=12 .

x−2=4 x-2=4 :

x=6. x=6.

Worked example 2.B

Simplify 50 \sqrt{50} . 25⋅2=52. \sqrt{25\cdot 2}=5\sqrt{2}.

For what values of xx is ∣2x−3∣<7|2x - 3| < 7?
−7<2x−3<7⇒−4<2x<10⇒−2<x<5-7 < 2x-3 < 7 \Rightarrow -4 < 2x < 10 \Rightarrow -2 < x < 5.
What is the sum of the solutions of the quadratic equation 3x2−12x−5=03x^2 - 12x - 5 = 0?
By Vieta's formulas, for any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is −ba-\frac{b}{a}. Here a=3a = 3 and b=−12b = -12, so the sum of the solutions is −−123=4-\frac{-12}{3} = 4.
📝 Chapter Quiz
3

Functions & Graphs

🗻 Functions & Graphs 12–15%  ·  Function notation  ·  Domain & range  ·  Transformations  ·  Composition & inverse

Function notation, evaluating and composing functions, and reading the vertex of a parabola.

▲ Reference — Functions

f(a) f(a) substitutes a a for x x ; a composite f(g(x)) f(g(x)) works inside-out. In vertex form y=a(x−h)2+k y=a(x-h)^{2}+k , the vertex is (h,k) (h,k) and the axis is x=h x=h .

Rational Function with Asymptotes: f(x) = 1/(x - 2) + 1 — Coordinate plane with graph of f(x) = 1/(x-2) + 1; marked points (3, 2) (3, 2), (1, 0) (1, 0)-2-1123456-224xyf(x) = 1/(x-2) + 1(3, 2)(1, 0)H.A. y = 1
Graph of rational function f(x)=1x−2+1 f(x) = \frac{1}{x - 2} + 1 . The vertical asymptote occurs where the denominator equals zero (x=2 x = 2 , restricted from the domain). The horizontal asymptote y=1 y = 1 is the value approached as x→±∞ x \to \pm\infty .

Slide the vertex (h,k) (h,k) of y=(x−h)2+k y=(x-h)^{2}+k . The amber dot is the vertex; the dashed line is the axis of symmetry.

⊙ Explorer · Vertex forminteractive
vertex—
axis of symmetry—

Moving h h slides the curve left/right; k k slides it up/down.

Evaluate
f(a) f(a)
Composite
f(g(x)) f(g(x))
Vertex form
a(x−h)2+k a(x-h)^{2}+k
Rate of change
f(b)−f(a)b−a \dfrac{f(b)-f(a)}{b-a}
▲ ACT Traps & Pitfalls — Functions & Graphs
  • Order of Composition: (f∘g)(x)=f(g(x)) (f \circ g)(x) = f(g(x)) means evaluate g(x) g(x) FIRST, then substitute the result into f f . Evaluating outside-in is one of the most common ACT distractor answer choices.
  • Domain Restrictions: Watch for two key restrictions: denominators cannot equal zero (e.g., 1x−3  ⟹  x≠3 \frac{1}{x-3} \implies x \neq 3 ), and expressions under an even radical cannot be negative (e.g., 2x−8  ⟹  2x−8≥0  ⟹  x≥4 \sqrt{2x-8} \implies 2x-8 \ge 0 \implies x \ge 4 ).
  • Vertical Line Test vs. Horizontal Line Test: A vertical line tests whether a graph is a function (each input has at most one output). A horizontal line tests whether a function is one-to-one (has an invertible function f−1(x) f^{-1}(x) ).
  • Function Transformations: f(x−c) f(x - c) shifts RIGHT by c c , while f(x+c) f(x + c) shifts LEFT by c c (horizontal operations are counter-intuitive). Vertical shifts f(x)+c f(x) + c go UP, and f(x)−c f(x) - c go DOWN.
Worked example 3.A

If f(x)=x2−1 f(x)=x^{2}-1 , find f(3) f(3) . 9−1=8. 9-1=8.

Worked example 3.B

If f(x)=2x+1 f(x)=2x+1 and g(x)=x2 g(x)=x^{2} , find f(g(2)) f(g(2)) .

g(2)=4 g(2)=4 , then f(4)=9 f(4)=9 :

9. 9.

If f(x)=x2−1f(x) = x^2 - 1 and g(x)=2x+3g(x) = 2x + 3, find f(g(1))f(g(1)).
g(1)=5g(1) = 5. f(5)=25−1=24f(5) = 25 - 1 = 24.
What is the domain of the function f(x)=x+5x−3f(x) = \dfrac{\sqrt{x + 5}}{x - 3}?
For the radical in the numerator to be defined, the radicand must be non-negative: x+5≥0⇒x≥−5x + 5 \ge 0 \Rightarrow x \ge -5. For the rational expression to be defined, the denominator cannot equal zero: x−3≠0⇒x≠3x - 3 \neq 0 \Rightarrow x \neq 3. Combining both restrictions gives x≥−5x \ge -5 and x≠3x \neq 3, or in interval notation [−5,3)∪(3,∞)[-5, 3) \cup (3, \infty).
If f(x)=3x−7f(x) = 3x - 7, what is the formula for the inverse function f−1(x)f^{-1}(x)?
Set y=3x−7y = 3x - 7. To find the inverse, interchange xx and yy: x=3y−7x = 3y - 7. Solve for yy: x+7=3y⇒y=x+73x + 7 = 3y \Rightarrow y = \dfrac{x + 7}{3}. Therefore, f−1(x)=x+73f^{-1}(x) = \dfrac{x + 7}{3}.
📝 Chapter Quiz
4

Coordinate Geometry

🗻 Coordinate Geometry 12–15%  ·  Slope & equations of lines  ·  Distance & midpoint  ·  Circles  ·  Parabolas

Lines in the plane: slope, intercepts, distance and midpoint, and the equations of lines.

▲ Reference — The coordinate plane

Slope m=y2−y1x2−x1 m=\dfrac{y_2-y_1}{x_2-x_1} ; line y=mx+b y=mx+b . Distance (x2−x1)2+(y2−y1)2 \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} ; midpoint (x1+x22,y1+y22) \left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right) .

Adjust slope m m and intercept b b of y=mx+b y=mx+b ; the amber dots are the intercepts.

⊙ Explorer · Slope & interceptsinteractive
slope—
y-intercept—
x-intercept—

Parallel lines share a slope; perpendicular slopes multiply to −1 -1 .

Perpendicular Lines: Negative Reciprocal Slopes — Coordinate plane with graph of L₁: y = 2x - 1 (m = 2), L₂: y = -0.5x + 4 (m = -1/2); marked points Intersection (2, 3) (2, 3)123424xyL₁: y = 2x − 1 (m = 2)L₂: y = −0.5x + 4 (m = −1/2)Intersection (2, 3)y = 0
Perpendicular lines L1 L_1 and L2 L_2 intersecting at right angles at (2,3) (2, 3) . The product of their slopes satisfies m1⋅m2=2⋅(−0.5)=−1 m_1 \cdot m_2 = 2 \cdot (-0.5) = -1 (negative reciprocals).
Slope
y2−y1x2−x1 \dfrac{y_2-y_1}{x_2-x_1}
Distance
Δx2+Δy2 \sqrt{\Delta x^{2}+\Delta y^{2}}
Midpoint
(x1+x22,y1+y22) \left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)
Line
y=mx+b y=mx+b
⚠️ Classic ACT Traps: Coordinate Geometry
  • Perpendicular Slopes: Perpendicular lines have slopes that are negative reciprocals: m2=−1m1 m_2 = -\frac{1}{m_1} . A line perpendicular to y=23x−5 y = \frac{2}{3}x - 5 has slope −32 -\frac{3}{2} , NOT 32 \frac{3}{2} and NOT −23 -\frac{2}{3} .
  • Circle Equation Center Signs: The equation (x−h)2+(y−k)2=r2 (x - h)^2 + (y - k)^2 = r^2 subtracts coordinates. A circle with equation (x+3)2+(y−4)2=49 (x + 3)^2 + (y - 4)^2 = 49 has center (−3,4) (-3, 4) and radius r=7 r = 7 , NOT 49 49 !
  • Finding an Missing Endpoint from Midpoint: If M(xm,ym) M(x_m, y_m) is the midpoint of AB‾ \overline{AB} and you know A(xa,ya) A(x_a, y_a) , do NOT average A A and M M ! Use the step method: xb=2xm−xa x_b = 2x_m - x_a and yb=2ym−ya y_b = 2y_m - y_a .
  • Horizontal and Vertical Lines: Horizontal lines have slope m=0 m = 0 and equation y=c y = c ; vertical lines have undefined slope and equation x=c x = c .
Worked example 4.A

Find the distance between (0,0) (0,0) and (3,4) (3,4) . 9+16=5. \sqrt{9+16}=5.

Worked example 4.B

Find the midpoint of (2,4) (2,4) and (6,8) (6,8) . (4, 6). (4,\,6).

What is the equation of the circle with centre (−2,4)(-2, 4) and radius 55?
(x+2)2+(y−4)2=25(x+2)^2 + (y-4)^2 = 25.
Point M(3,−2)M(3, -2) is the midpoint of line segment AB‾\overline{AB}. If point AA has coordinates (−1,4)(-1, 4), what are the coordinates of point BB?
Using the midpoint formula for each coordinate:
xM=xA+xB2⇒3=−1+xB2⇒6=−1+xB⇒xB=7x_M = \frac{x_A + x_B}{2} \Rightarrow 3 = \frac{-1 + x_B}{2} \Rightarrow 6 = -1 + x_B \Rightarrow x_B = 7.
yM=yA+yB2⇒−2=4+yB2⇒−4=4+yB⇒yB=−8y_M = \frac{y_A + y_B}{2} \Rightarrow -2 = \frac{4 + y_B}{2} \Rightarrow -4 = 4 + y_B \Rightarrow y_B = -8.
Coordinates of BB: (7, -8).
📝 Chapter Quiz
5

Plane Geometry

🗻 Plane Geometry 12–15%  ·  Angles & parallel lines  ·  Triangles  ·  Polygons & circles  ·  Area & volume

Angles, triangles and polygons, circles, and area, perimeter and volume.

▲ Reference — Shapes

Triangle angles sum to 180∘ 180^{\circ} ; a polygon of n n sides sums to (n−2)⋅180∘ (n-2)\cdot180^{\circ} . Pythagoras a2+b2=c2 a^{2}+b^{2}=c^{2} . Circle area πr2 \pi r^{2} , circumference 2πr 2\pi r .

Parallel lines intersected by a transversal line120°60°120°60°
Parallel lines cut by a transversal. Alternate interior angles and corresponding angles are equal (120∘ 120^\circ ), while consecutive interior angles are supplementary (120∘+60∘=180∘ 120^\circ + 60^\circ = 180^\circ ).
Triangle area
12bh \tfrac12 bh
Pythagoras
a2+b2=c2 a^{2}+b^{2}=c^{2}
Polygon angles
(n−2)⋅180∘ (n-2)\cdot180^{\circ}
Circle
πr2, 2πr \pi r^{2},\ 2\pi r
⚠️ Classic ACT Traps: Plane Geometry
  • Figures Are Not Necessarily to Scale: Never trust eye measurements on the ACT. An angle that looks like 90° is only 90° if marked with a square or proven with given equations.
  • The Triangle Inequality Theorem: The sum of any two sides must strictly exceed the third: a+b>c a + b > c . If two sides are 7 7 and 11 11 , the third side x x must satisfy 11−7<x<11+7⇒4<x<18 11 - 7 < x < 11 + 7 \Rightarrow 4 < x < 18 .
  • Special Right Triangles (No Formula Sheet!): Memorize 30∘−60∘−90∘ 30^\circ-60^\circ-90^\circ sides x:x3:2x x : x\sqrt{3} : 2x , and 45∘−45∘−90∘ 45^\circ-45^\circ-90^\circ sides x:x:x2 x : x : x\sqrt{2} . Common Pythagorean triples include 3−4−5 3-4-5 , 5−12−13 5-12-13 , and 8−15−17 8-15-17 .
  • Regular Polygon Angles: The sum of interior angles is (n−2)⋅180∘ (n-2)\cdot180^\circ . Each interior angle of a regular n n -gon is (n−2)⋅180∘n \frac{(n-2)\cdot180^\circ}{n} , and each exterior angle is 360∘n \frac{360^\circ}{n} .
Worked example 5.A

A right triangle has legs 9 9 and 12 12 . Find the hypotenuse. 81+144=15. \sqrt{81+144}=15.

Worked example 5.B

Find the interior-angle sum of a pentagon. (5−2)⋅180∘=540∘. (5-2)\cdot180^{\circ}=540^{\circ}.

A cylinder has radius 3 cm and height 8 cm. Find its total surface area.
TSA =2πr2+2πrh=2π(9)+2π(3)(8)=18π+48π=66π≈207.3= 2\pi r^2 + 2\pi r h = 2\pi(9) + 2\pi(3)(8) = 18\pi + 48\pi = 66\pi \approx 207.3 cm2^2.
Two sides of a non-degenerate triangle measure 7 inches and 11 inches. What is the range of possible perimeters for this triangle?
Let xx be the third side. By the Triangle Inequality Theorem, 11−7<x<11+7⇒4<x<1811 - 7 < x < 11 + 7 \Rightarrow 4 < x < 18. The perimeter is P=7+11+x=18+xP = 7 + 11 + x = 18 + x. Substituting the bounds for xx: 18+4<P<18+18⇒22<P<3618 + 4 < P < 18 + 18 \Rightarrow 22 < P < 36 inches.
📝 Chapter Quiz
6

Trigonometry

🗻 Trigonometry 7–10%  ·  SOH-CAH-TOA  ·  Unit circle  ·  Trig identities  ·  Graphs of trig functions

Right-triangle ratios, the values of common angles, and reading the sine curve.

▲ Reference — SOH-CAH-TOA

sin⁡θ=opphyp \sin\theta=\dfrac{\text{opp}}{\text{hyp}} , cos⁡θ=adjhyp \cos\theta=\dfrac{\text{adj}}{\text{hyp}} , tan⁡θ=oppadj \tan\theta=\dfrac{\text{opp}}{\text{adj}} . Common values: sin⁡30∘=12 \sin30^{\circ}=\tfrac12 , cos⁡60∘=12 \cos60^{\circ}=\tfrac12 , tan⁡45∘=1 \tan45^{\circ}=1 .

Unit circle diagram showing terminal angle 150: Unit Circle Reference: θ = 150° (5π/6 radians)Unit Circle Reference: θ = 150° (5π/6 radians)xy150°(-√3/2, 1/2)
Unit circle with terminal angle θ=150∘ \theta = 150^\circ (5π6 \frac{5\pi}{6} rad). The terminal point has coordinates (cos⁡θ,sin⁡θ)=(−32,12) (\cos\theta, \sin\theta) = (-\frac{\sqrt{3}}{2}, \frac{1}{2}) . In Quadrant II, sine is positive while cosine and tangent are negative.

Slide the angle θ \theta along the sine curve. The readouts give sin⁡θ \sin\theta and cos⁡θ \cos\theta .

⊙ Explorer · The sine curveinteractive
sin θ—
cos θ—

Sine peaks at 90∘ 90^{\circ} , is zero at 0∘ 0^{\circ} and 180∘ 180^{\circ} , and bottoms at 270∘ 270^{\circ} .

Sine
opphyp \tfrac{\text{opp}}{\text{hyp}}
Cosine
adjhyp \tfrac{\text{adj}}{\text{hyp}}
Tangent
oppadj \tfrac{\text{opp}}{\text{adj}}
Identity
sin⁡2θ+cos⁡2θ=1 \sin^{2}\theta+\cos^{2}\theta=1
▲ ACT Traps & Pitfalls — Trigonometry
  • Radian vs. Degree Mode: ACT problems frequently alternate between degrees and radians. Always verify your calculator status line before evaluating trig functions, especially when finding angles or amplitudes.
  • Pythagorean Identity Variations: sin⁡2θ+cos⁡2θ=1 \sin^2\theta + \cos^2\theta = 1 is the most tested trig identity on the ACT. Look for algebraic rearrangements such as 1−sin⁡2θ=cos⁡2θ 1 - \sin^2\theta = \cos^2\theta and sin⁡2θ1−cos⁡θ=1+cos⁡θ \frac{\sin^2\theta}{1 - \cos\theta} = 1 + \cos\theta .
  • ASTC Quadrant Signs: Remember 'All Students Take Calculus': Quadrant I (All positive), Quadrant II (Sine positive), Quadrant III (Tangent positive), Quadrant IV (Cosine positive). A common error is missing the negative sign when given θ \theta in Quadrants II or III.
  • Period of a Sinusoid: For y=Asin⁡(Bx) y = A \sin(Bx) or y=Acos⁡(Bx) y = A \cos(Bx) , the period is 2πB \frac{2\pi}{B} (or 360∘B \frac{360^\circ}{B} ), NOT 2πB 2\pi B . Amplitude is ∣A∣ |A| , and the vertical midline is y=D y = D .
Worked example 6.A

A right triangle has opposite 3 3 and hypotenuse 5 5 . Find sin⁡θ \sin\theta . sin⁡θ=35. \sin\theta=\tfrac{3}{5}.

Worked example 6.B

Evaluate tan⁡45∘ \tan45^{\circ} . 1. 1.

In right triangle ABCABC with C=90°C = 90°, a=5a = 5, c=13c = 13. Find sin⁡A\sin A, cos⁡A\cos A, tan⁡A\tan A.
b=169−25=12b = \sqrt{169-25} = 12. sin⁡A=5/13\sin A = 5/13, cos⁡A=12/13\cos A = 12/13, tan⁡A=5/12\tan A = 5/12.
If cos⁡θ=−45\cos\theta = -\frac{4}{5} and θ\theta lies in Quadrant II, what is the value of tan⁡θ\tan\theta?
In Quadrant II, x=−4x = -4 and r=5r = 5. Using x2+y2=r2x^2 + y^2 = r^2, we have (−4)2+y2=52⇒16+y2=25⇒y=3(-4)^2 + y^2 = 5^2 \Rightarrow 16 + y^2 = 25 \Rightarrow y = 3 (since sine is positive in Quadrant II). Therefore, tan⁡θ=yx=−34\tan\theta = \frac{y}{x} = -\frac{3}{4}.
What is the period of the trigonometric function f(x)=4sin⁡(3x)−2f(x) = 4\sin(3x) - 2?
For any sinusoidal function of the form f(x)=Asin⁡(Bx−C)+Df(x) = A\sin(Bx - C) + D, the period is given by 2πB\frac{2\pi}{B}. Here B=3B = 3, so the period is 2π3\frac{2\pi}{3}.
📝 Chapter Quiz
7

Statistics & Probability

🗻 Statistics & Probability 8–12%  ·  Mean, median, mode  ·  Data spread  ·  Probability  ·  Expected value

Averages and spread, basic counting, and the probability of simple and combined events.

▲ Reference — Data & chance

Mean is total over count; the median is the middle value. Probability =favourabletotal =\dfrac{\text{favourable}}{\text{total}} ; independent events multiply: P(A and B)=P(A)⋅P(B) P(A\text{ and }B)=P(A)\cdot P(B) .

Five-Number Summary (Box-and-Whisker Plot) — Box plot summary: Min 14, Q1 22, Median 28, Q3 35, Max 44Five-Number Summary (Box-and-Whisker Plot)810121416182022242628303234363840424446485052
Five-number summary box-and-whisker plot: Minimum = 14, First Quartile Q1=22 Q_1 = 22 , Median =28 = 28 , Third Quartile Q3=35 Q_3 = 35 , Maximum = 44. The Interquartile Range is IQR=Q3−Q1=35−22=13 \text{IQR} = Q_3 - Q_1 = 35 - 22 = 13 , containing the middle 50% of the distribution.
Mean
sumcount \dfrac{\text{sum}}{\text{count}}
Median
middle value
Probability
favtotal \dfrac{\text{fav}}{\text{total}}
Independent
P(A)⋅P(B) P(A)\cdot P(B)
▲ ACT Traps & Pitfalls — Statistics & Probability
  • Outlier Sensitivity (Mean vs. Median): The mean is heavily pulled by extreme outliers, whereas the median is resistant. If an unusually high score is added to a dataset, the mean increases significantly, but the median often changes little or not at all.
  • Expected Value (Net vs. Gross): On ACT raffle/carnival game problems, expected value is E(X)=∑[x⋅P(x)] E(X) = \sum [x \cdot P(x)] . Watch out whether the question asks for net winnings (subtracting the ticket cost) or gross payout!
  • Mutually Exclusive vs. Independent: For mutually exclusive events, P(A or B)=P(A)+P(B) P(A \text{ or } B) = P(A) + P(B) . For non-mutually exclusive events, you must subtract the overlap: P(A∪B)=P(A)+P(B)−P(A∩B) P(A \cup B) = P(A) + P(B) - P(A \cap B) . For independent events, P(A and B)=P(A)×P(B) P(A \text{ and } B) = P(A) \times P(B) .
  • Combinations vs. Permutations: Order matters   ⟹   \implies Permutation nPr=n!(n−r)! _nP_r = \frac{n!}{(n-r)!} (e.g. officer positions: President, VP, Secretary). Order does NOT matter   ⟹   \implies Combination nCr=n!r!(n−r)! _nC_r = \frac{n!}{r!(n-r)!} (e.g. committee, team, selecting toppings).
Worked example 7.A

Find the mean of 4, 8, 10, 18 4,\,8,\,10,\,18 . 404=10. \frac{40}{4}=10.

Worked example 7.B

A fair coin is tossed twice. Find P(two heads) P(\text{two heads}) . 12⋅12=14. \tfrac12\cdot\tfrac12=\tfrac14.

A data set is: 4, 7, 7, 9, 12, 15. Find the mean, median, and range.
Mean =54/6=9= 54/6 = 9. Median =(7+9)/2=8= (7+9)/2 = 8. Range =15−4=11= 15-4 = 11.
A carnival game costs $5 to play. There is a 110\frac{1}{10} probability of winning $30, a 210\frac{2}{10} probability of winning $10, and a 710\frac{7}{10} probability of winning nothing. What is the expected net profit or loss per play?
The expected gross payout is E(gross)=30(110)+10(210)+0(710)=3+2+0=$5E(\text{gross}) = 30\left(\frac{1}{10}\right) + 10\left(\frac{2}{10}\right) + 0\left(\frac{7}{10}\right) = 3 + 2 + 0 = \$5. Since the ticket costs $5, the expected net profit is 5−5=$05 - 5 = \$0 (a fair game).
A committee of 3 students is to be chosen from a group of 8 candidates. How many different committees can be formed?
Because the order of committee members does not matter, this is a combination: 8C3=8!3!(8−3)!=8×7×63×2×1=56_8C_3 = \dfrac{8!}{3!(8-3)!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 different committees.
📝 Chapter Quiz
8

Matrices, Vectors & Complex Numbers

🗻 Advanced Topics (Matrices/Vectors/Complex) ~5%  ·  Matrix operations  ·  Vector components  ·  Complex arithmetic  ·  Polar form

The hardest ACT items touch matrices, vectors, logarithms and the imaginary unit i i .

▲ Reference — Advanced topics

i2=−1 i^{2}=-1 ; add matrices entrywise; a vector (a,b) (a,b) has magnitude a2+b2 \sqrt{a^{2}+b^{2}} ; log⁡bx=y \log_b x=y means by=x b^{y}=x .

Complex Plane (Argand Diagram): z = 3 + 4i — Coordinate plane with graph of |z| = 5; marked points Origin (0,0) (0, 0), z = 3 + 4i (3, 4)12341234xy|z| = 5Origin (0,0)z = 3 + 4iIm = 4
Complex number z=3+4i z = 3 + 4i on the complex plane (real axis horizontal, imaginary axis vertical). The modulus is the distance from the origin ∣z∣=32+42=25=5 |z| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 .
Imaginary
i2=−1 i^{2}=-1
Matrix
add entrywise
Vector
a2+b2 \sqrt{a^{2}+b^{2}}
Logarithm
log⁡bx \log_b x
▲ ACT Traps & Pitfalls — Advanced Topics
  • Matrix Dimensions for Multiplication: To multiply Am×k×Bk×n A_{m \times k} \times B_{k \times n} , the inner dimensions MUST match (number of columns of AA = number of rows of BB). The product has dimensions m×n m \times n . Matrix multiplication is NOT commutative: AB≠BA AB \neq BA in general.
  • Matrix Determinant & Inverse: For a 2×2 2 \times 2 matrix (abcd) \begin{pmatrix} a & b \\ c & d \end{pmatrix} , det⁡(A)=ad−bc \det(A) = ad - bc . If det⁡(A)=0 \det(A) = 0 , the matrix has NO inverse. To form A−1=1ad−bc(d−b−ca) A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , swap the main diagonal and negate the off-diagonal entries!
  • Powers of i i Cycle of 4: i1=i i^1 = i , i2=−1 i^2 = -1 , i3=−i i^3 = -i , i4=1 i^4 = 1 . To evaluate in i^n , divide n n by 4 and look at the remainder: remainder 0   ⟹  1 \implies 1 , remainder 1   ⟹  i \implies i , remainder 2   ⟹  −1 \implies -1 , remainder 3   ⟹  −i \implies -i .
  • Vector Magnitude & Addition: Vector addition is component-wise: ⟨u1,u2⟩+⟨v1,v2⟩=⟨u1+v1,u2+v2⟩ \langle u_1, u_2 \rangle + \langle v_1, v_2 \rangle = \langle u_1+v_1, u_2+v_2 \rangle . Magnitude is ∣v⃗∣=v12+v22 |\vec{v}| = \sqrt{v_1^2 + v_2^2} . Do not simply add magnitudes together!
Worked example 8.A

Simplify i2+i4 i^{2}+i^{4} . −1+1=0. -1+1=0.

● Complex Numbers & Polar Form

A complex number z=a+biz = a+bi has modulus ∣z∣=a2+b2|z| = \sqrt{a^2+b^2} and argument θ=arctan⁡(b/a)\theta = \arctan(b/a). Polar form: z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta). Multiplying in polar: multiply moduli, add arguments.

Worked example — Matrix determinant & inverse

Find A−1A^{-1} for A=(3152)A = \begin{pmatrix}3&1\\5&2\end{pmatrix}.

det⁡A=6−5=1\det A = 6-5 = 1. A−1=11(2−1−53)=(2−1−53)A^{-1} = \dfrac{1}{1}\begin{pmatrix}2&-1\\-5&3\end{pmatrix} = \begin{pmatrix}2&-1\\-5&3\end{pmatrix}.

Simplify (2+3i)(1−i)(2+3i)(1-i) and write in the form a+bia+bi.
(2+3i)(1−i)=2−2i+3i−3i2=2+i+3=5+i(2+3i)(1-i) = 2 - 2i + 3i - 3i^2 = 2 + i + 3 = 5+i.
If A=(2−134)A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} and B=(5−2)B = \begin{pmatrix} 5 \\ -2 \end{pmatrix}, what is the matrix product ABAB?
The product of a 2×22 \times 2 matrix and a 2×12 \times 1 matrix is a 2×12 \times 1 matrix. Compute row by column: the first entry is 2(5)+(−1)(−2)=10+2=122(5) + (-1)(-2) = 10 + 2 = 12; the second entry is 3(5)+4(−2)=15−8=73(5) + 4(-2) = 15 - 8 = 7. Thus AB=(127)AB = \begin{pmatrix} 12 \\ 7 \end{pmatrix}.
What is the value of the complex expression i47−i18i^{47} - i^{18}?
Divide the exponents by 4: 47=4(11)+347 = 4(11) + 3, so i47=i3=−ii^{47} = i^3 = -i. 18=4(4)+218 = 4(4) + 2, so i18=i2=−1i^{18} = i^2 = -1. Therefore, i47−i18=−i−(−1)=1−ii^{47} - i^{18} = -i - (-1) = 1 - i.
📝 Chapter Quiz
§

Practice set

Fifty ACT-style problems spanning numbers, algebra, functions, geometry, trigonometry and statistics, tagged by difficulty. Click any problem for a full worked solution.

1

Evaluate 15% 15\% of 200 200 .

Basic
0.15×200=30. 0.15\times200=30.
2

Simplify 34+18 \dfrac{3}{4}+\dfrac{1}{8} .

Basic

68+18 \tfrac{6}{8}+\tfrac{1}{8} :

78. \tfrac{7}{8}.
3

Split 30 30 in the ratio 2:3 2:3 . Find the smaller part.

Intermediate

Five parts, each 6 6 :

2×6=12. 2\times6=12.
4

12 12 is what percent of 48 48 ?

Intermediate
1248=25%. \frac{12}{48}=25\%.
5

Find the least common multiple of 6 6 and 8 8 .

Intermediate
LCM=24. \text{LCM}=24.
6

A recipe is scaled from 2 to 5 servings. How many cups replace 4 4 cups?

Intermediate

Multiply by 52 \tfrac{5}{2} :

4×52=10. 4\times\tfrac{5}{2}=10.
7

A \$80 item is marked down 20% 20\% . Find the sale price.

Advanced
80×0.80=$64. 80\times0.80=\$64.
8

A car travels 120 120 miles in 3 3 hours. Find the average speed.

Intermediate
1203=40 mph. \frac{120}{3}=40\ \text{mph}.
9

Solve 2x+7=19 2x+7=19 .

Basic

2x=12 2x=12 :

x=6. x=6.
10

Solve 3(x−2)=12 3(x-2)=12 .

Intermediate

x−2=4 x-2=4 :

x=6. x=6.
11

Factor x2−9 x^{2}-9 .

Intermediate
(x−3)(x+3). (x-3)(x+3).
12

Solve x2−7x+10=0 x^{2}-7x+10=0 .

Intermediate

(x−2)(x−5)=0 (x-2)(x-5)=0 :

x=2, 5. x=2,\ 5.
13

Simplify 2x+3x−5 2x+3x-5 .

Intermediate
5x−5. 5x-5.
14

If y=2x+1 y=2x+1 , find x x when y=9 y=9 .

Intermediate

2x=8 2x=8 :

x=4. x=4.
15

Solve ∣x−3∣=5 |x-3|=5 .

Advanced

x−3=±5 x-3=\pm 5 :

x=8, −2. x=8,\ -2.
16

Simplify x2⋅x4 x^{2}\cdot x^{4} .

Intermediate
x6. x^{6}.
17

Solve 5x−3=2x+9 5x-3=2x+9 .

Intermediate

3x=12 3x=12 :

x=4. x=4.
18

Simplify 50 \sqrt{50} .

Advanced
25⋅2=52. \sqrt{25\cdot2}=5\sqrt{2}.
19

Evaluate 23+32 2^{3}+3^{2} .

Intermediate
8+9=17. 8+9=17.
20

Solve the system x+y=7 x+y=7 , 2x−y=2 2x-y=2 .

Advanced

Add: 3x=9⇒x=3 3x=9\Rightarrow x=3 , then y=4 y=4 :

x=3, y=4. x=3,\ y=4.
21

If f(x)=x2−1 f(x)=x^{2}-1 , find f(3) f(3) .

Intermediate
9−1=8. 9-1=8.
22

State the vertex of y=(x−2)2+3 y=(x-2)^{2}+3 .

Intermediate
(2, 3). (2,\,3).
23

If f(x)=2x+1 f(x)=2x+1 and g(x)=x2 g(x)=x^{2} , find f(g(2)) f(g(2)) .

Advanced

g(2)=4 g(2)=4 , then f(4)=9 f(4)=9 :

9. 9.
24

For what value of x x is 1x−2 \dfrac{1}{x-2} undefined?

Intermediate

The denominator is zero:

x=2. x=2.
25

Find the y-intercept of y=3x−6 y=3x-6 .

Intermediate
(0, −6). (0,\,-6).
26

For f(x)=x2 f(x)=x^{2} , find the average rate of change on [1,3] [1,3] .

Advanced
9−13−1=4. \frac{9-1}{3-1}=4.
27

Find the slope of the line through (0,0) (0,0) and (2,6) (2,6) .

Basic
62=3. \frac{6}{2}=3.
28

Find the midpoint of (2,4) (2,4) and (6,8) (6,8) .

Intermediate
(2+62,4+82)=(4,6). \left(\tfrac{2+6}{2},\tfrac{4+8}{2}\right)=(4,6).
29

Find the distance between (0,0) (0,0) and (3,4) (3,4) .

Intermediate
9+16=5. \sqrt{9+16}=5.
30

Find the slope of y=2x+5 y=2x+5 .

Intermediate
m=2. m=2.
31

Find the x-intercept of 2x+3y=12 2x+3y=12 .

Intermediate

Set y=0 y=0 : 2x=12 2x=12 :

x=6. x=6.
32

A line passes through (1,2) (1,2) with slope 3 3 . Find y y when x=4 x=4 .

Advanced

y−2=3(4−1) y-2=3(4-1) :

y=11. y=11.
33

Give the equation of the horizontal line through (0,5) (0,5) .

Intermediate
y=5. y=5.
34

Find the area of a rectangle 7×4 7\times4 .

Basic
7×4=28. 7\times4=28.
35

A triangle has angles 50∘ 50^{\circ} and 60∘ 60^{\circ} . Find the third.

Intermediate
180∘−110∘=70∘. 180^{\circ}-110^{\circ}=70^{\circ}.
36

A right triangle has legs 9 9 and 12 12 . Find the hypotenuse.

Intermediate
81+144=15. \sqrt{81+144}=15.
37

Find the area of a triangle with base 10 10 and height 8 8 .

Intermediate
12×10×8=40. \tfrac12\times10\times8=40.
38

Find the area of a circle with radius 5 5 .

Intermediate
π(5)2=25π. \pi(5)^{2}=25\pi.
39

Find the circumference of a circle with r=7 r=7 . Take π=227 \pi=\tfrac{22}{7} .

Intermediate
2×227×7=44. 2\times\tfrac{22}{7}\times7=44.
40

Find the interior-angle sum of a pentagon.

Advanced
(5−2)⋅180∘=540∘. (5-2)\cdot180^{\circ}=540^{\circ}.
41

Find the volume of a box 2×3×4 2\times3\times4 .

Intermediate
2×3×4=24. 2\times3\times4=24.
42

A square has area 36 36 . Find its perimeter.

Advanced

Side =6 =6 :

4×6=24. 4\times6=24.
43

A right triangle has opposite 3 3 and hypotenuse 5 5 . Find sin⁡θ \sin\theta .

Intermediate
sin⁡θ=35. \sin\theta=\tfrac{3}{5}.
44

Evaluate sin⁡30∘ \sin30^{\circ} .

Intermediate
12. \tfrac12.
45

Evaluate cos⁡60∘ \cos60^{\circ} .

Intermediate
12. \tfrac12.
46

Evaluate tan⁡45∘ \tan45^{\circ} .

Advanced
1. 1.
47

A right triangle has adjacent 4 4 and hypotenuse 5 5 . Find cos⁡θ \cos\theta .

Intermediate
cos⁡θ=45. \cos\theta=\tfrac{4}{5}.
48

A right triangle has opposite 6 6 and adjacent 8 8 . Find tan⁡θ \tan\theta .

Advanced
tan⁡θ=68=34. \tan\theta=\tfrac{6}{8}=\tfrac{3}{4}.
49

Find the mean of 4, 8, 10, 18 4,\,8,\,10,\,18 .

Intermediate
404=10. \frac{40}{4}=10.
50

A fair coin is tossed twice. Find P(two heads) P(\text{two heads}) .

Intermediate
12⋅12=14. \tfrac12\cdot\tfrac12=\tfrac14.
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