ACT Mathematics Tests 2025

$ f(x, y) = 3x^2 - 2y \\[3ex] f(2, 3) \\[3ex] = 3(2)^2 - 2(3) \\[3ex] = 3(4) - 6 \\[3ex] = 12 - 6 \\[3ex] = 6 $

$ 9 - 11x = 13 - 9x \\[3ex] 9 - 13 = -9x + 11x \\[3ex] -4 = 2x \\[3ex] 2x = -4 \\[3ex] x = -\dfrac{4}{2} \\[5ex] x = -2 $

$ \$10 \text{ per day for 2 days } = 10(2) = \$20 \\[3ex] \$0.445 \text{ per mile for 120 miles } = 0.445(120) = \$53.4 \\[3ex] \$0.445 \text{ per mile for 100 miles } = 0.442(100) = \$44.5 \\[3ex] \text{Amount of Jorge's reimbursement} \\[3ex] = 20 + 53.4 + 44.2 \\[3ex] = \$117.90 $

$ a \triangle b = a^2 + b - \dfrac{a}{b} \\[5ex] a \triangle b = a^2 + b - (a \div b) \\[5ex] 5 \triangle \dfrac{1}{2} \\[5ex] = 5^2 + \dfrac{1}{2} - \left(5 \div \dfrac{1}{2}\right) \\[5ex] = 25 + \dfrac{1}{2} - \left(5 * \dfrac{2}{1}\right) \\[5ex] = \dfrac{50}{2} + \dfrac{1}{2} - \dfrac{20}{2} \\[5ex] = \dfrac{50 + 1 - 20}{2} \\[5ex] = \dfrac{31}{2} \\[5ex] = 15\dfrac{1}{2} $

Haley purchased lemonade mix for $2.30 and a package of cups for $1.50.
Cost Price: = 2.30 + 1.50 = $3.80

Haley sells each cup of lemonade for $0.50.
Selling Price: x cups of lemonade @ $0.50 per cup = $0.5x

$ \text{Profit} = \text{Selling Price} - \text{Cost Price} \\[3ex] = 0.5x - 3.8 $

$ \underline{\text{Given}} \\[3ex] \angle B = 90^\circ \\[3ex] \angle A = 4 * \angle C \quad\dots\text{Given} \\[5ex] \angle A + \angle B + \angle C = 180^\circ \quad\dots\text{sum of angles of a triangle} \\[3ex] 4 * \angle C + 90 + \angle C = 180 \\[3ex] 5\angle C = 180 - 90 \\[3ex] \angle C = \dfrac{90}{5} \\[5ex] \angle C = 18^\circ $

$ 11\text{ wins} \\[3ex] 8\text{ losses} \\[3ex] 0\text{ ties} \\[3ex] 14\text{ remaining games} \\[3ex] \text{Total number of games} = 11 + 8 + 14 = 33 \\[3ex] 50\%\text{ of all the team's games} \\[3ex] = \dfrac{50}{100} * 33 \\[5ex] = \dfrac{33}{2} \\[5ex] $ Let the least number of its remaining 14 games the team must win to finish the season winning more than 50% of all the team’s games = x

$ 11 + x \gt \dfrac{33}{2} \\[5ex] 2(11 + x) \gt 33 \\[3ex] 22 + 2x \gt 33 \\[3ex] 2x \gt 33 - 22 \\[3ex] 2x \gt 11 \\[3ex] x \gt \dfrac{11}{2} \\[5ex] x \gt 5.5 \\[3ex] x \ge 6 \\[3ex] \text{Least number} = 6 \\[3ex] $ To finish the season winning more than 50% of all the team’s games, the team must win at least 6 games.

Only one point is insufficient to determine the nature of the slope.
The slope of the line cannot be determined from the given information.

S = Sample Space = all available free throws

$ P(\text{make a free throw}) = \dfrac{n(\text{free throws made})}{n(S)} \\[5ex] 0.8 = \dfrac{20}{n(S)} \\[5ex] 0.8 * n(S) = 20 \\[3ex] n(S) = \dfrac{20}{0.8} \\[5ex] n(S) = 25 $

$ \underline{\text{Cost for 1 day using the Unity Fraction Method}} \\[3ex] \text{Set up units:} \\[3ex] \dfrac{...\text{ gallon}}{...\text{ miles}} * ...\text{ miles} * \dfrac{...\$}{...\text{ gallon}} \\[5ex] \text{Set up measurements and units:} \\[3ex] \dfrac{1\text{ gallon}}{20\text{ miles}} * 6\text{ miles} * \dfrac{\$ 5}{1\text{ gallon}} \\[5ex] = \$ 1.5 \\[5ex] \text{Cost for 10 days @ } \$ 1.5 \text{ per day} \\[3ex] = 10(1.5) \\[3ex] = \$ 15 $

$ 2x^2 + 5x = 3 \\[3ex] 2x^2 + 5x - 3 = 0 \\[3ex] ............................................ \\[3ex] 2x^2(-3) = -6x^2 \\[3ex] \text{Factors are: } 6x \text{ and } -x \\[3ex] ............................................ \\[3ex] 2x^2 + 6x - x - 3 = 0 \\[3ex] 2x(x + 3) - 1(x + 3) = 0 \\[3ex] (x + 3)(2x - 1) = 0 \\[3ex] x + 3 = 0 \hspace{1em} or \hspace{1em} 2x - 1 = 0 \\[3ex] x = -3 \hspace{1em} or \hspace{1em} 2x = 1 \\[3ex] \hspace{4em} or \hspace{1em} x = \dfrac{1}{2} \\[5ex] \text{Because } x \gt 0: \\[3ex] x = \dfrac{1}{2} $

The rate of change of y with respect to x is the slope of the function.
Let us represent the function in slope–intercept form so that we can determine the slope.

$ \sin\theta = \dfrac{opp}{hyp} \quad\dots\text{SOHCAHTOA} \\[5ex] \sin\theta = \dfrac{2}{5} \\[5ex] \theta = \sin^{-1}\left(\dfrac{2}{5}\right) $

$ \text{Mean} = \dfrac{21.2 + 22.01 + 21.33 + 22.13 + 21.2 + 21.5 + 21.48 + 22.19}{8} \\[5ex] = \dfrac{173.04}{8} \\[5ex] = 21.63\text{ seconds} $

Because of time, I suggest you use this approach to check your work (if you have time).

$ \text{Stadium's capacity} = 5000\text{ people} \\[3ex] \text{Price of ticket} = \$8 \\[3ex] \text{Number of tickets} = 85\% \text{ of the stadium's capacity} \\[3ex] = \dfrac{85}{100} * 5000 \\[5ex] = 4250\text{ tickets} \\[5ex] 4250\text{ tickets @ } \$8 \text{ per ticket} = 4250(8) = \$ 34,000 $

$ \underline{\text{SHS track}} \\[3ex] \text{diameter} = d = 240\;m \\[3ex] \text{radius} = r = \dfrac{d}{2} = \dfrac{240}{2} = 120\;m \\[5ex] \text{angle} = \theta = 60^\circ \\[5ex] \text{Area of a sector} = \dfrac{\pi r^2\theta}{360} \\[5ex] = \dfrac{\pi * 120^2 * 60}{360} \\[5ex] = 2400\;pi\;\;m^2 $

Let us review each option to find the correct answer.
Note thet the correct pair must satisfy both inequalities.

$ \underline{\text{Process of Elimination}} \\[3ex] \text{Testing each } (x, y) \text{ pair} \\[3ex] \text{1st Inequality: } x \gt y \\[3ex] (0, 4) \implies 0 \gt 4 \quad\dots\text{Not true, Discard} \\[3ex] (1, 3) \implies 1 \gt 3 \quad\dots\text{Not true, Discard} \\[3ex] (2, 1) \implies 2 \gt 1 \quad\dots\text{okay} \\[3ex] (3, 0) \quad\dots\text{okay} \\[3ex] (4, 1) \quad\dots\text{okay} \\[5ex] \text{2nd Inequality: } x \gt y \\[3ex] x + y \gt 3 \\[3ex] (2, 1) \implies 2 + 1 \gt 3 \quad\dots\text{Not true, Discard} \\[3ex] (3, 0) \implies 3 + 0 \gt 3 \quad\dots\text{Not true, Discard} \\[3ex] $ The correct answer is option E.
But if you want to confirm:

$ (4, 1) \implies 4 + 1 \gt 3 \quad\dots\text{True} $

$ \text{Area of the trapezoid} = \dfrac{h(a + b)}{2} \\[5ex] \text{where} \\[3ex] h = \perp\text{height} = 5\;feet \\[3ex] a = \text{length of one base} = 2\;feet \\[3ex] b = \text{length of the other base} = ? \\[3ex] \text{Area} = 20\;ft^2 \\[3ex] \implies \\[3ex] 20 = \dfrac{5(2 + b)}{2} \\[5ex] 5(2 + b) = 20(2) \\[3ex] 2 + b = \dfrac{20(2)}{5} \\[5ex] 2 + b = 8 \\[3ex] b = 8 - 2 \\[3ex] b = 6\;feet $

$ \underline{\text{Rectangle}} \\[3ex] L = \text{Length} = 196\;cm \\[3ex] W = \text{Width} = 4\;cm \\[3ex] A_R = \text{Area} = L * W \\[3ex] = 196 * 4 \\[3ex] = 784\;cm^2 \\[5ex] \underline{\text{Square}} \\[3ex] S = \text{Side} = ? \\[3ex] A_S = \text{Area} = S^2 \\[5ex] \text{A square and a rectangle have the same area.} \\[3ex] A_S = A_R \\[3ex] S^2 = 784 \\[3ex] S = \sqrt{784} \\[3ex] S = 28\;cm $

$ (3x - 1)(2x - 5) = ax^2 + bx + c \\[3ex] 6x^2 - 15x - 2x + 5 = ax^2 + bx + c \\[3ex] ax^2 + bx + c = 6x^2 - 17x + 5 \\[3ex] \implies \\[3ex] a = 6 \\[3ex] b = -17 \\[3ex] c = 5 \\[5ex] a + b + c \\[3ex] = 6 + (-17) + 5 \\[3ex] = -6 $

$ \dfrac{a^{2n + 1}}{a^n} \\[5ex] = a^{(2n + 1) - n} \quad\dots\text{Law 2 Exp} \\[3ex] = a^{2n + 1 - n} \\[3ex] = a^{n + 1} $

$ (\sqrt{8a})(\sqrt{2a}) = 16 \\[3ex] \sqrt{16a^2} = 16 \\[3ex] 4a = 16 \\[3ex] a = \dfrac{16}{4} \\[5ex] a = 4 $

Let:
red = R
m(R) = 3

green = G
n(G) = 4

yellow = Y
n(Y) = 2

sample space = S = {3R, 4G, 2Y}
n(S) = 3 + 4 + 2 = 9 apples

An apple is drawn at random, eaten, and then another apple is randomly drawn.
That first apple is red.
So, 8 apples remain.
What is the probability of selecting a yellow out of 8 apples that consists of 2 red, 4 green, and 2 yellow apples?

$ P(RY) = \dfrac{2}{8} \\[5ex] = \dfrac{1}{4} $

Irrational numbers cannot be expressed as a fraction/ratio of two numbers.
They are non-terminating non-repeating decimals.

$\pi = 3.141592654...$ is irrational because it is non-terminating and non-repeating.

$3\dfrac{1}{3} = \dfrac{10}{3}$ is rational because it is expressed as a fraction.

$2.193876 = \dfrac{2193876}{1000000} = \dfrac{548469}{250000}$ is expressed as a fraction.

$\sqrt{144} = 12 = \dfrac{12}{1}$ is expressed as a fraction.

$4 = \dfrac{4}{1}$ is expressed as a fraction.

$ 7[3 - 2(x - 4)] \\[3ex] = 7[3 - 2x + 8] \\[3ex] = 7[-2x + 11] \\[3ex] = -14x + 77 $

The rate of change of a function is its slope.
A linear function (straight-line graph) has a constant slope.
Option G. is the correct answer.

Student: SamDom For Peace
Teacher: What's good?
Student: Could you give the type of function graphs?
Teacher: Sure.
The functions whose graphs are represented in options:

F. is a piecewise function.
J. is an exponential function.
G. is a linear function.
K. is a quadratic function.
H. is an absolute value function.

The rational number = x-coordinate of the Midpoint = x
On the number line, the horizontal distance between $\dfrac{3}{8}$ and the rational number is the same as the horizontal distance between the rational number and $\dfrac{7}{16}$.

$ x - \dfrac{3}{8} = \dfrac{7}{16} - x \\[5ex] 2x = \dfrac{7}{16} + \dfrac{3}{8} \\[5ex] x = \left(\dfrac{7}{16} + \dfrac{6}{16}\right) \div 2 \\[5ex] x = \dfrac{13}{16} * \dfrac{1}{2} \\[5ex] x = \dfrac{13}{32} $

The question is asking for the Least Common Multiple (LCM) of 4 and 10

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 4, 6 \\[3ex] 4 = \color{black}{2} * 2 \\[3ex] 6 = \color{black}{2} * 3 \\[5ex] LCM = \color{black}{2} * 2 * 3 \\[3ex] LCM = 12 \\[3ex] $ 12 seconds will elapse until the 2 signs next flash at the same time.

$ \text{Set up Units: } \dfrac{\$...}{...yd^3} * \dfrac{...yd^3}{...ft^3} * ...ft^3 \\[5ex] \text{Set up Measurements and Units: } \dfrac{\$54}{1\;yd^3} * \dfrac{1\;yd^3}{27\;ft^3} * 42\;ft^3 \\[5ex] = \$84 $

The patio is a rectangle.
Let us find the area of the rectangle.

$ \underline{\text{Rectangular Patio}} \\[3ex] \text{diagonal} = 25\;ft \\[3ex] \text{width} = 7\;ft \\[3ex] \text{length} = ? \\[5ex] \text{length}^2 + \text{width}^2 = \text{diagonal}^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] \text{length}^2 + 7^2 = 25^2 \\[3ex] \text{length} = \sqrt{25^2 - 7^2} \\[3ex] \text{length} = 24\;ft \\[5ex] \text{Area} = \text{length} * \text{width} \\[3ex] = 24 * 7 \\[3ex] = 168\;ft^2 $

$ \underline{\text{Rectangular Prism Concrete Step}} \\[3ex] \text{Volume} = 3 * 1.5 * 1 \\[3ex] = 4.5\;ft^3 \\[5ex] \underline{\text{Bags of Concrete Mix}} \\[3ex] \text{Set up Units: } \dfrac{...\text{bag}}{...ft^3} * ...ft^3 \\[5ex] \text{Set up Measurements and Units: } \dfrac{1\text{ bag}}{0.4\;ft^3} * 4.5\;ft^3 \\[5ex] = 11.25 \\[3ex] \approx 11.3\text{ bags} \quad\dots\text{to the nearest } 0.1 \text{ bag} $

Length of single board = 72 inches
Length of 1st piece to be cut from the single board = $35\dfrac{5}{8}$ inches

Length of 2nd piece to be cut from the single board = $27\dfrac{1}{2}$ inches

Length of the 2 cuts (kerf or offcut of the 2 cuts) = $\dfrac{1}{8}$ inch

$ \underline{\text{Total Length of 2 pieces and 2 cuts}} \\[3ex] 35\dfrac{5}{8} + 27\dfrac{1}{2} + \dfrac{1}{8} \\[5ex] = 35 + \dfrac{5}{8} + 27 + \dfrac{1}{2} + \dfrac{1}{8} \\[5ex] = 35 + 27 + \dfrac{5}{8} + \dfrac{4}{8} + \dfrac{1}{8} \\[5ex] = 62 + \dfrac{10}{8} \\[5ex] = 62 + \dfrac{5}{4} \\[5ex] = 62 + 1\dfrac{1}{4} \\[5ex] = 62 + 1 + \dfrac{1}{4} \\[5ex] = 63 + \dfrac{1}{4}\;inches \\[5ex] \underline{\text{Left Over}} \\[3ex] \text{Remaining} = 72 - \left(63 + \dfrac{1}{4}\right) \\[5ex] = 72 - 63 - \dfrac{1}{4} \\[5ex] = 9 - \dfrac{1}{4} \\[5ex] = 8\dfrac{3}{4}\;inches $

$ B - A = difference \\[3ex] B = difference + A \\[5ex] = \begin{bmatrix} 6 & 7 & 4 \\[2ex] 1 & -1 & -4 \end{bmatrix} + \begin{bmatrix} -7 & 2 & 4 \\[2ex] -1 & 0 & -3 \end{bmatrix} \\[7ex] = \begin{bmatrix} 6 + -7 & 7 + 2 & 4 + 4 \\[2ex] 1 + -1 & -1 + 0 & -4 + -3 \end{bmatrix} \\[7ex] = \begin{bmatrix} -1 & 9 & 8 \\[2ex] 0 & -1 & -7 \end{bmatrix} $

$ \text{Proficient} = 16 + 12 + 10 = 38 \\[3ex] \text{Not Proficient} = 6 + 16 = 22 \\[3ex] \text{Ratio of Proficient to Not Proficient} = \dfrac{38}{22} \\[5ex] = \dfrac{19}{11} \\[5ex] = 19 : 11 $

This is a case of the Fundamental Counting Principle.
4 friends to be seated, out of 5 chairs available.

The first friend may sit on any of the 5 seats.
The 2nd friend may sit on any of the remaining 4 (5 − 1) seats
The 3rd friend may sit on any of the remaining 3 (4 − 1) seats
The last friend may sit on any of the remaining 2 (3 − 1) seats

$ \begin{array}{c c c c} \text{1st Friend} & \text{2nd Friend} & \text{3rd Friend} & \text{4th Friend} \\ \rule{3cm}{0.5mm} & \rule{3cm}{0.5mm} & \rule{3cm}{0.5mm} & \rule{3cm}{0.5mm} \\ 5 & 4 & 3 & 2 \\[3ex] \end{array} \\[5ex] \text{Number of different seating arrangements} = 5 * 4 * 3 * 2 = 120 \text{ arrangements.} \\[5ex] $ Alternatively, we can use the Permutation formula.
This is also a case of Permutations because the order of seating matters.

$ \text{Perm 4 from 5} \\[3ex] P(n, r) = \dfrac{n!}{(n - r)!} \\[5ex] P(5, 4) = \dfrac{5!}{(5 - 4)!} \\[5ex] = \dfrac{5!}{1!} \\[5ex] = 120\;\text{ arrangements} \\[3ex] $ Student: SamDom For Peace, I thought this was a case of Combination.
But you did it as a Permutation.
May you explain?
Teacher: Here is something to bear in mind when deciding whether order matters or not.
Let's say Friend 1 sits on Seat 1 and Friend 2 sits on Seat 2
If Friend 1 sits on Seat 2 and Friend 2 sits on Seat 1, would it be considered a different seating arrangement or not?
Student: It is a different seating arrangement.
Teacher: Good.
In that case, the seating arrangement (order in which they are seated) is important.
Hence, it is a Permutation.
Student: Can you reword the same question to illustrate a case of Combination?
Teacher: Sure... here is a reworded question that illustrates a Combination
A group of 4 friends decide to go to a movie.
After buying their tickets, they find 5 empty chairs in the theater.
How many different sets of 4 chairs can be chosen from the 5 available chairs for the friends to sit in?
It can also be worded as:
A group of 4 friends find 5 empty chairs in a theater.
In how many ways can 1 empty chair be left over after they all sit down?
In this case, any one of them can sit in any one chair.
Instead of focusing on who sits where, the question asks which 4 chairs will be occupied, without worrying about specific seating order.

We can solve this question using at least 2 approaches
Use any approach you prefer, however, it is highly recommended that you use the first approach.

$ \text{Let } S = \text{sample space} \\[3ex] n(S) = faces^{tosses} \\[4ex] n(S) = 6^2 = 36 \\[3ex] $ Assume event, E = sum of the numbers on the faces landing up is 3 or more
Complement of the event, E' = sum of the numbers on the faces landing up is less than 3

$ \underline{\text{1st Approach: Complementary Rule}} \\[3ex] \text{1st cube} = \{1, 2, 3, 4, 5, 6\} \\[3ex] \text{2nd cube} = \{1, 2, 3, 4, 5, 6\} \\[3ex] $ Let us find all the possible event spaces where the sum of the numbers on both cubes is less than 3

$ 1 + 1 = 2\quad\dots\text{only one possibility} \\[3ex] n(E') = 1 \\[3ex] n(E) + n(E') = n(S) \\[3ex] P(E) + P(E') = 1 \quad\dots\text{Complementary Rule} \\[3ex] P(E) = 1 - P(E') \\[3ex] = 1 - \dfrac{n(E')}{n(S)} \\[5ex] = 1 - \dfrac{1}{36} \\[5ex] = \dfrac{36}{36} - \dfrac{1}{36} \\[5ex] = \dfrac{35}{36} \\[5ex] $ 2nd Approach: Punnett Square
Sample Space for a Toss of Two Fair Cubes

A Cube in the Column and A Cube in the Row

1 Cube → (+) 1 Cube ↓	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

$ n(E) = 35 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{35}{36} $

We can solve the question using at least two approaches.
Because of the time limit of about 1 minute per question on the ACT, I recommend the first approach.

$ \triangle ECD \\[3ex] \overline{EC}^2 = \overline{ED}^2 + \overline{DC}^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] \overline{EC}^2 = 4^2 + 3^2 \\[3ex] \overline{EC}^2 = 16 + 9 \\[3ex] \overline{EC}^2 = 25 \\[3ex] \overline{EC} = \sqrt{25} \\[3ex] \overline{EC} = 5 \\[5ex] \underline{\text{1st Approach: Similar Triangles}} \\[3ex] \triangle ACB\;\;and\;\;\triangle ECD \\[3ex] \overline{AB} || \overline{DE} \\[3ex] \angle ABC = \angle EDB = 90^\circ \quad\dots\text{alternate angles are congruent} \\[3ex] \angle ACB = \angle ECD \quad\dots\text{vertical angles are congruent} \\[3ex] \angle BAC = \angle DEC \quad\dots\text{alternate angles are congruent} \\[3ex] \triangle ACB \sim \triangle ECD \quad\dots\text{Angle — Angle Similarity} \\[3ex] $ Because the triangles are similar, the corresponding sides will have the same ratio.

$ \underline{\text{Ratio of Corresponding Sides}} \\[3ex] \dfrac{\overline{AC}}{\overline{AB}} = \dfrac{\overline{EC}}{\overline{ED}} \\[5ex] \dfrac{\overline{AC}}{5} = \dfrac{5}{4} \\[5ex] \overline{AC} = \dfrac{5 \cdot 5}{4} \\[5ex] \overline{AC} = \dfrac{25}{4}\;inches \\[5ex] \underline{\text{2nd Approach: Trigonometric Ratios and Laws}} \\[3ex] \angle ABC = \angle EDB = 90^\circ \quad\dots\text{alternate angles are congruent} \\[5ex] \triangle ECD \\[3ex] \sin \angle ECD = \dfrac{opp}{hyp} = \dfrac{4}{5} \quad\dots\text{SOHCAHTOA} \\[5ex] \triangle ACB \\[3ex] \angle ACB = \angle ECD \quad\dots\text{vertical angles are congruent} \\[3ex] \implies \\[3ex] \sin\angle ACB = \sin\angle ECD = \dfrac{4}{5} \\[5ex] \dfrac{\overline{AC}}{\sin\angle ABC} = \dfrac{\overline{AB}}{\sin\angle ACB} \quad\dots\text{Sine Law} \\[5ex] \dfrac{\overline{AC}}{\sin 90^\circ} = \dfrac{5}{\dfrac{4}{5}} \\[8ex] \dfrac{\overline{AC}}{1} = 5 \div \dfrac{4}{5} \\[5ex] \overline{AC} = 5 \cdot \dfrac{5}{4} \\[5ex] \overline{AC} = \dfrac{25}{4}\;inches $

The function is a rational radical function.

Dealing with Rational Function:
1st Condition: The denominator should not be zero

$ \sqrt{x - 3} \ne 0 \\[3ex] x - 3 \ne 0^2 \\[3ex] x - 3 \ne 0 \\[3ex] x \ne 0 + 3 \\[3ex] x \ne 3 \\[3ex] $ Dealing with Radical Function:
2nd Condition: The radicand should must be non-negative (greater than or equal to zero)

$ x - 3 \ge 0 \\[3ex] x \ge 3 \\[3ex] \text{Combining both conditions}: \\[3ex] Domain = \{x: x \gt 3\} $

$ f(x) = 2x + 3 \hspace{3em} g(x) = x^2 - 4 \\[3ex] f(g(x)) \\[3ex] = f(x^2 - 4) \\[3ex] = 2(x^2 - 4) + 3 \\[3ex] = 2x^2 - 8 + 3 \\[3ex] = 2x^2 - 5 $

$ \dfrac{y - 1}{y + 1} - \dfrac{2(y - 1)}{4y + 4} \\[5ex] LCD = (y + 1)(4y + 4) \\[3ex] = \dfrac{(y - 1)(4y + 4)}{(y + 1)(4y + 4)} - \dfrac{2(y - 1)(y + 1)}{(y + 1)(4y + 4)} \\[5ex] (y - 1)(y + 1) = y^2 - 1^2 \quad\dots\text{Difference of Two Squares} \\[3ex] = \dfrac{4y^2 + 4y - 4y - 4 - 2(y^2 - 1^2)}{(y + 1)(4y + 4)} \\[5ex] = \dfrac{4y^2 - 4 - 2(y^2 - 1)}{(y + 1)\cdot 4(y + 1)} \\[5ex] = \dfrac{4y^2 - 4 - 2y^2 + 2}{4(y + 1)(y + 1)} \\[5ex] = \dfrac{2y^2 - 2}{4(y + 1)(y + 1)} \\[5ex] = \dfrac{2(y^2 - 1)}{4(y + 1)(y + 1)} \\[5ex] = \dfrac{2(y^2 - 1^2)}{4(y + 1)(y + 1)} \\[5ex] = \dfrac{2(y + 1)(y - 1)}{4(y + 1)(y + 1)} \\[5ex] = \dfrac{y - 1}{2(y + 1)} $

A wheel with a radius of 1 centimeter rolls ...
How many centimeters has the wheel traveled along the ground?

Assume a circular wheel...of course it should be circular
This implies that 1 revolution of rolling the wheel is the circumference of the circle. So, 2 revolutions is twice the circumference of the circle

$ radius = 1\;cm \\[3ex] Circumference = 2 \cdot \pi \cdot radius \\[3ex] \text{2 revolutions = twice the circumference} \\[3ex] = 2 \cdot 2 \cdot \pi \cdot 1 \\[3ex] = 4\pi $

4 different kinds of pizza
1st, 2nd, 3rd, 4th pizza

1st pizza: 6 slices
Fraction per slice = $\dfrac{1}{6}$

2nd pizza: 7 slices
Fraction per slice = $\dfrac{1}{7}$

3rd pizza: 8 slices
Fraction per slice = $\dfrac{1}{8}$

4th pizza: 9 slices
Fraction per slice = $\dfrac{1}{9}$

5 slices given to each boy
At least 3 different kinds of pizza ⇒ 3 or more (in this case: 3 or 4 different kinds of pizza)

For example:
2 slices of 1st pizza: $\dfrac{2}{6}$, 1 slice of 2nd pizza: $\dfrac{1}{7}$, 1 slice of 3rd pizza: $\dfrac{1}{8}$, 1 slice of 4th pizza: $\dfrac{1}{9}$

Another example: 3 slices of 1st pizza: $\dfrac{3}{6}$, 1 slice of 2nd pizza: $\dfrac{1}{7}$, 1 slice of 3rd pizza: $\dfrac{1}{8}$
... and so on and so forth
We are interested in finding the largest fraction of pizza that any boy ate
1st Approach would be to add the fractions in each option and determine the greatest sum
But this approach may take some time.

So, let us do it this way:
Assume 3 different kinds of pizza
Which is greater: $\dfrac{3}{6}$ OR $\dfrac{3}{9}$?
$\dfrac{3}{6}$ is greater

Assume 4 different kinds of pizza
Which is greater: $\dfrac{2}{6}$ OR $\dfrac{2}{9}$?
$\dfrac{2}{6}$ is greater

Now compare the greater of both options
Which is greater: $\dfrac{3}{6}$ OR $\dfrac{2}{6}$?
$\dfrac{3}{6}$ is greater

So, we shall go with this option because it gives the largest fraction: 3 slices of 1st pizza: $\dfrac{3}{6}$, 1 slice of 2nd pizza: $\dfrac{1}{7}$ and 1 slice of 3rd pizza: $\dfrac{1}{8}$

$ x^0 = 1 ...Law\;3...Exp \\[3ex] y^1 = y ...Law\;4...Exp \\[3ex] z^{-1} = \dfrac{1}{z^1} = \dfrac{1}{z} ...Law\;6...Exp \\[5ex] \implies \\[3ex] x^0 y^1 z^{-1} \\[5ex] = 1 \cdot y \cdot \dfrac{1}{z} \\[5ex] = \dfrac{y}{z} \\[5ex] For\;\; \dfrac{y}{z} \text{ to be } \gt 1 \\[5ex] y \text{ must be} \gt z $

Regression Equation: $y = 0.16x − 0.44$
x = amount of check (in dollars)
y = amount of tip (in dollars)
Slope = $0.16 = 16 cents
This is a positive slope. The interpretation for the slope in this context implies that: On average, for every unit change ($1 increase in) in x (the amount of check), the y (amount of tip) increases by the slope (16 cents).

From her statistical analysis, Sumiko can correctly conclude that every $1 increase in the total amount of the check will result in approximately a 16 cent increase in her tip.

$ \text{square yard} = yard^2 = yard \cdot yard \\[3ex] \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units}: \\[3ex] 12\;feet \cdot 18\;feet \cdot \dfrac{...yard}{...feet} \cdot \dfrac{...yard}{...feet} \cdot \dfrac{...\$}{...yard \cdot yard} \\[5ex] 3\; feet = 1\; yard \\[3ex] \implies \\[3ex] 12\;feet \cdot 18\;feet \cdot \dfrac{1\;yard}{3\;feet} \cdot \dfrac{1\;yard}{3\;feet} \cdot \dfrac{\$10}{1\;yard \cdot yard} \\[5ex] = 4 \cdot 6 \cdot \$10 \\[3ex] = \$240 $

$ f(x) = x^5 \\[3ex] y = x^5 \\[3ex] \text{Interchange x and y} \\[3ex] x = y^5 \\[3ex] y^5 = x \\[3ex] y = \sqrt[5]{x} \\[3ex] \text{This new } y \text{ is the inverse} \\[3ex] f^{-1}(x) = \sqrt[5]{x} $

$ s...t...d \\[3ex] speed \cdot time = distance \\[5ex] \text{60 seconds = 1 minute} \\[3ex] \text{60 minutes = 1 hour} \\[5ex] \underline{\text{1st Part}} \\[3ex] \text{average speed} = 30 \text{ miles per hour} \\[3ex] \text{Convert to miles per second} \\[3ex] \underline{\text{Unity Fraction Method}} \\[3ex] \dfrac{30\;miles}{1\;hour} \cdot \dfrac{...hour}{...minute} \cdot \dfrac{...minute}{...second} \\[5ex] \dfrac{30\;miles}{1\;hour} \cdot \dfrac{1\;hour}{60\;minute} \cdot \dfrac{1\;minute}{60\;second} \\[5ex] \dfrac{1}{120}\;\text{miles per second} \\[5ex] \text{time} = 15 \text{ seconds} \\[3ex] \text{distance} = \dfrac{1}{120}\;\text{miles per second} \cdot 15 \text{ seconds} \\[5ex] \text{distance} = \dfrac{1}{8}\;mile \\[5ex] \underline{\text{2nd Part}} \\[3ex] \text{speed} = 60 \text{ miles per hour} \\[3ex] \text{Convert to miles per second} \\[3ex] \underline{\text{Unity Fraction Method}} \\[3ex] \dfrac{60\;miles}{1\;hour} \cdot \dfrac{...hour}{...minute} \cdot \dfrac{...minute}{...second} \\[5ex] \dfrac{60\;miles}{1\;hour} \cdot \dfrac{1\;hour}{60\;minute} \cdot \dfrac{1\;minute}{60\;second} \\[5ex] \dfrac{1}{60}\;\text{miles per second} \\[5ex] \text{time} = 15 \text{ seconds} \\[3ex] \text{distance} = \dfrac{1}{60}\;\text{miles per second} \cdot 15 \text{ seconds} \\[5ex] \text{distance} = \dfrac{1}{4}\;mile \\[5ex] \underline{\text{Entire Trip}} \\[3ex] \text{distance} = \dfrac{1}{8}\;mile + \dfrac{1}{4}\;mile \\[5ex] = \dfrac{1}{8}\;mile + \dfrac{2}{8}\;mile \\[5ex] = \dfrac{3}{8}\;mile $

$ a - b = 5 ...eqn.(1) \\[3ex] c = 7a - 9 - 7b ...eqn.(2) \\[5ex] From\;\;eqn.(1) \\[3ex] a = 5 + b ...eqn.(3) \\[5ex] \text{Substitute for a in eqn.(2)} \\[3ex] c = 7(5 + b) - 9 - 7b \\[3ex] c = 35 + 7b - 9 - 7b \\[3ex] c = 26 $

$ \tan\theta = \dfrac{\sin\theta}{\cos\theta} \quad\dots\text{SOHCAHTOA} \\[5ex] \cos\theta = \dfrac{\sin\theta}{\tan\theta} \\[5ex] = \dfrac{1}{2} \div -\dfrac{\sqrt{3}}{3} \\[5ex] = \dfrac{1}{2} \cdot -\dfrac{3}{\sqrt{3}} \\[5ex] = -\dfrac{3}{2\sqrt{3}} \\[5ex] \text{Rationalize the denominator} \\[3ex] = -\dfrac{3}{2\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = -\dfrac{3\sqrt{3}}{2 \cdot 3} \\[5ex] = -\dfrac{\sqrt{3}}{2} $

Reflected across the x-axis is Vertical Reflection
In Vertical Reflection: y changes, x does not change
y-coordinate is multiplied by −1
The transformed y-coordinate becomes −y

$ y = a(x - b)^2 \\[3ex] -1 \cdot y = -1 \cdot a(x - b)^2 \\[3ex] = -a(x - b)^2 \\[3ex] \text{Transformed } y = -a(x - b)^2 $

$ \dfrac{2 + 3i}{i} \\[5ex] \text{Rationalize the denominator} \\[3ex] = \dfrac{2 + 3i}{i} \cdot \dfrac{i}{i} \\[5ex] = \dfrac{i(2 + 3i)}{i^2} \\[5ex] = \dfrac{2i + 3i^2}{-1} \\[5ex] = -1(2i + 3(-1)) \\[3ex] = -1(2i - 3) \\[3ex] = -2i + 3 \\[3ex] = 3 - 2i $

Let:
n = number of students
x = score
$\bar{x}$ = average score
Σx = summation of the scores

$ \underline{\text{Initial: Graded Midterm Test}} \\[3ex] n = 30 \\[3ex] \bar{x} = p \\[3ex] \Sigma x = \bar{x} \cdot n \\[3ex] = p \cdot 30 \\[3ex] = 30p \\[5ex] \underline{\text{Corrected Grading}} \\[3ex] n = 30 \\[3ex] \Sigma x = 30p + 5 \quad\dots\text{increase of 5 points} \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] = \dfrac{30p + 5}{30} \\[5ex] \underline{\text{Change in the Average Score}} \\[3ex] \text{Corrected Average} - \text{Initial Average} \\[3ex] = \dfrac{30p + 5}{30} - p \\[5ex] = \dfrac{30p + 5}{30} - \dfrac{30p}{30} \\[5ex] = \dfrac{30p + 5 - 30p}{30} \\[5ex] = \dfrac{5}{30} \\[5ex] = \dfrac{1}{6} $

$ 2x + y - z = 0 ...eqn.(1) \\[3ex] x - 3y + z = 18 ...eqn.(2) \\[3ex] x + y = 1 ...eqn.(3) \\[5ex] \underline{\text{Elimination Method}} \\[3ex] eqn.(2) - eqn.(3) \rightarrow \\[3ex] x - 3y + z - (x + y) = 18 - 1 \\[3ex] x - 3y + z - x - y = 17 \\[3ex] -4y + z = 17 ...eqn.(4) \\[3ex] 2 \cdot eqn.(2) \rightarrow \\[3ex] 2(x - 3y + z) = 2(18) \\[3ex] 2x - 6y + 2z = 36 ...eqn.(5) \\[3ex] eqn.(1) - eqn.(2) \rightarrow \\[3ex] 2x + y - z - (2x - 6y + 2z) = 0 - 36 \\[3ex] 2x + y - z - 2x + 6y - 2z = -36 \\[3ex] 7y - 3z = -36 ...eqn.(6) \\[3ex] 7 \cdot eqn.(4) + 4 \cdot eqn.(6) \rightarrow \\[3ex] 7(-4y + z) + 4(7y - 3z) = 7(17) + 4(-36) \\[3ex] -28y + 7z + 28y - 12z = 119 - 144 \\[3ex] -5z = -25 \\[3ex] z = \dfrac{-25}{-5} \\[5ex] z = 5 $

The maximum value of a quotient is when the numerator is the biggest and the denominator is the smallest.
You may try it with some values to confirm.
This implies that we have to use the greatest value of x and the least value of y in the set.

$ \underline{\text{Both Sets}} \\[3ex] \text{greatest value of } x = 1 \\[3ex] \text{least value of } y = \dfrac{1}{8} \\[5ex] \dfrac{x}{y^2} \\[5ex] = \dfrac{1}{\left(\dfrac{1}{8}\right)^2} \\[7ex] = 1 \div \dfrac{1}{64} \\[5ex] = 1 \cdot 64 \\[3ex] = 64 \\[3ex] $ Student: SamDom For Peace, for questions like this, can I just guess that the answer is 64 because it is the greatest number in the options?
Teacher: Yes, you may...only if you do not have much time left
I recommend that you solve it if you have time, but then guess rather than leave any question blank.
Student: I'm not sure you understand my question.
Okay, let me put it this way: is it possible for ACT to include a number greater than 64 in the options?
Teacher: I think it is possible.
But let's attempt more questions like this and see.
Let's solve more past questions and if we notice a trend (where the maximum answer in the option is always the correct answer), then we can decide what to do.
Until then, I recommend you solve it.

Let us analyze this question.

$ \text{Area of a Triangle} = \dfrac{1}{2} \cdot base \cdot \perp height \\[3ex] $ $\triangle ACB$ has the same area as $\triangle AGH$
This is because they have:
(I.) the same base: $\overline{BC} = \overline{GH}$
(II.) the same perpendicular height: because a square has equal sides: $\overline{AB} = \overline{AH}$

The other triangles: $\triangle ACD$, $\triangle ADE$, $\triangle AEF$, and $\triangle AFG$ have:
(III.) the same base: $\overline{CD} = \overline{DE} = \overline{EF} = \overline{FG}$

But what about their perpendicular heights?
How do we calculate the area of a triangle if the triangle does not have a perpendicular height?
One of the ways to calculate the area is to extend the base to the same level as the opposite vertex, then draw a perpendicular height from that vertex to the extension of the base.
This is what I mean.
$\triangle ACD$ does not a perpendicular height



One of the ways to calculate the area of the triangle is to draw the perpendicular height $\overline{AB}$ (Please see the red lines)



This implies that the perpendicular height is still the same side of the square
(IV.) the same perpendicular height: the side of the square

Therefore all five triangles: $\triangle ACB$, $\triangle ACD$, $\triangle ADE$, $\triangle AEF$, and $\triangle AFG$ have the same area as $\triangle AGH$.

Area of the shaded portion = Area of Square ABCD − Area of Square KLMN
Area of a Square = Side Length²

$ \underline{\text{Square ABCD}} \\[3ex] \text{side length} = 14\;m \\[3ex] \overline{AB} = \overline{BC} = \overline{CD} = \overline{DA} = 14\;m \\[3ex] Area = 14^2 \\[3ex] Area = 196\;m^2 \\[5ex] \underline{\text{Triangle ALK}} \\[3ex] \text{This is a right triangle because a square has 4 right angles} \\[3ex] \overline{AL} = \dfrac{\overline{AB}}{2} \quad\dots\text{Midpoint} \\[5ex] = \dfrac{14}{2} \\[5ex] = 7\;m \\[5ex] \overline{AK} = \dfrac{\overline{DA}}{2} \quad\dots\text{Midpoint} \\[5ex] = \dfrac{14}{2} \\[5ex] = 7\;m \\[5ex] \overline{KL}^2 = \overline{AL}^2 + \overline{AK}^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] \overline{KL}^2 = 7^2 + 7^2 \\[3ex] \overline{KL}^2 = 49 + 49 \\[3ex] \overline{KL}^2 = 98 \\[3ex] \overline{KL} = \sqrt{98} \\[5ex] \underline{\text{Square KLMN}} \\[3ex] \text{side length} = \overline{KL} = \sqrt{98}\;m \\[3ex] Area = \sqrt{98}^2 \\[3ex] Area = 98\;m^2 \\[5ex] \underline{\text{Shaded Portion}} \\[3ex] Area = 196 - 98 \\[3ex] Area = 98\;m^2 $

$ Area = \dfrac{1}{2}ab\sin C\quad\dots\text{one of the formulas} \\[5ex] Area = \dfrac{1}{2} \cdot 4 \cdot 6 \cdot \sin 30^\circ \\[5ex] = 12 \cdot \dfrac{1}{2} \\[5ex] = 6\;m^2 $

1st: Let us determine the asymptotes of the hyperbola

$ Hyperbola:\;\; \dfrac{x^2}{9} - \dfrac{y^2}{4} = 1 \\[5ex] \text{This is a horizontal hyperbola} \\[3ex] \text{Standard Form: center at the origin}: \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \\[5ex] a^2 = 9 \hspace{3em} b^2 = 4 \\[3ex] a = \sqrt{9} \hspace{3em} b = \sqrt{4} \\[3ex] a = 3 \hspace{3em} b = 2 \\[3ex] Asymptotes:\;\; y = \pm \dfrac{b}{a}x \\[5ex] y = \pm \dfrac{2}{3}x \\[5ex] $ Let us analyze each option.

$ F.\;\; \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 \\[5ex] \text{This is a horizontal ellipse} \\[3ex] \text{Incorrect option} \\[5ex] G.\;\; \dfrac{x^2}{4} - \dfrac{y^2}{9} = 1 \\[5ex] \text{This is a horizontal hyperbola} \\[3ex] a^2 = 4 \hspace{3em} b^2 = 9 \\[3ex] a = \sqrt{4} \hspace{3em} b = \sqrt{9} \\[3ex] a = 2 \hspace{3em} b = 3 \\[3ex] Asymptotes:\;\; y = \pm \dfrac{b}{a}x \\[5ex] y = \pm \dfrac{3}{2}x \\[5ex] \pm \dfrac{3}{2}x \ne \pm \dfrac{2}{3}x \\[5ex] \text{Incorrect option} \\[5ex] H.\;\; \dfrac{y^2}{9} + \dfrac{x^2}{4} = 1 \\[5ex] \text{This is a vertical ellipse} \\[3ex] \text{Incorrect option} \\[5ex] J.\;\; \dfrac{y^2}{9} - \dfrac{x^2}{4} = 1 \\[5ex] \text{This is a vertical hyperbola} \\[3ex] b^2 = 9 \hspace{3em} a^2 = 4 \\[3ex] b = \sqrt{9} \hspace{3em} a = \sqrt{4} \\[3ex] b = 3 \hspace{3em} a = 2 \\[3ex] Asymptotes:\;\; y = \pm \dfrac{b}{a}x \\[5ex] y = \pm \dfrac{3}{2}x \\[5ex] \text{Incorrect option} \\[5ex] K.\;\; \dfrac{y^2}{4} - \dfrac{x^2}{9} = 1 \\[5ex] \text{This is a vertical hyperbola} \\[3ex] b^2 = 4 \hspace{3em} a^2 = 9 \\[3ex] b = \sqrt{4} \hspace{3em} a = \sqrt{9} \\[3ex] b = 2 \hspace{3em} a = 3 \\[3ex] Asymptotes:\;\; y = \pm \dfrac{b}{a}x \\[5ex] y = \pm \dfrac{2}{3}x \\[5ex] \text{This is the correct option} $

For a:
Left-skewed distribution (longer tail on the left): Mean < Median < Mode
Symmetric distribution: Mean = Median = Mode
Right-skewed distribution (longer tail on the right): Mean > Median > Mode

The distribution is right-skewed; hence the mean is greater than the median, and the median is greater than the mode.
This implies that:
The mode is less than the median, and the median is less than the mean.

$ Volume = Length \cdot Width \cdot Height \\[3ex] \underline{\text{Solid Rectangular Block: 1}} \\[3ex] Length = L \\[3ex] Width = W \\[3ex] Height = H \\[3ex] Volume = V_1 \\[3ex] Weight = w ...Given \\[3ex] V_1 = L \cdot W \cdot H \\[3ex] V_1 = LWH \\[5ex] \underline{\text{Larger Solid Rectangular Block: 2}} \\[3ex] Length = 4L \\[3ex] Width = 4W \\[3ex] Height = 4H \\[3ex] Volume = V_2 \\[3ex] Weight = ? \\[5ex] V_2 = 4L \cdot 4W \cdot 4H \\[3ex] V_2 = 64LWH \\[5ex] Volume \alpha Weight\quad\dots\text{Direct Proportion...assumed} \\[3ex] Volume = k \cdot Weight \quad\dots\text{k = constant of proportionality} \\[3ex] k = \dfrac{Volume}{Weight} \\[5ex] \implies \\[3ex] k = \dfrac{V_1}{w} = \dfrac{V_2}{Weight} \\[5ex] \dfrac{LWH}{w} = \dfrac{64LWH}{Weight} \\[5ex] \text{Cross Multiply} \\[3ex] LWH \cdot Weight = w \cdot 64LWH \\[3ex] Weight = \dfrac{w \cdot 64LWH}{LWH} \\[5ex] Weight = 64w\;pounds $