Chapter 2 — Number and Quantity

Answer: D

Explanation

Work from the inside out. A power of a power multiplies:

$$(x^3)^4 = x^{3 \times 4} = x^{12}$$

Dividing subtracts:

$$\frac{x^{12}}{x^5} = x^{12 - 5} = x^{7}$$

Why each wrong option is wrong:

• A, $x^{12}$ — did the top correctly and then forgot to divide.
• B, $x^{60}$ — multiplied all three numbers together.
• C, $x^{2}$ — added 3 and 4 instead of multiplying them, giving $x^7$ on top and then $x^2$ after dividing. The rule for a power of a power is multiply; the rule for a product is add. Mixing them up is the most common exponent error there is.

Takeaway: power of a power multiplies; multiplying like bases adds; dividing subtracts. Say which one you are using before you write anything.

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Answer: J

Explanation

$$8^{2/3} = \left(\sqrt[3]{8}\right)^{2}$$

The 3 on the bottom says cube root; the 2 on top says square.

$$\sqrt[3]{8} = 2 \qquad 2^2 = 4$$

Do the root first. Going the other way, $8^2 = 64$ and then $\sqrt[3]{64} = 4\mathclose{,}$ gives the same answer but with much bigger numbers.

Why each wrong option is wrong:

• F, $512$ — used the 3 as the power and ignored the 2.
• G, $16$ — multiplied 8 by the 2 on top of the fraction and ignored the 3 underneath it.
• H, $\frac{16}{3}$ — multiplied 8 by $\frac{2}{3}\mathclose{.}$ An exponent is not a multiplier.

Takeaway: in $x^{m/n}$ the bottom is the root and the top is the power. Take the root first to keep the arithmetic small.

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Answer: B

Explanation

A negative exponent means the reciprocal:

$$4^{-2} = \frac{1}{4^{2}} = \frac{1}{16}$$

The minus sign belongs to the exponent, and its job is to flip the fraction over. It never makes the value negative. A positive base raised to any power whatsoever stays positive.

Why each wrong option is wrong:

• A, $- \frac{1}{16}$ — inverted correctly and kept a minus sign, doing the job twice.
• C, $\frac{1}{8}$ — multiplied 4 by 2.
• D, $16$ — ignored the minus sign on the exponent and squared.

Takeaway: a negative exponent flips, it does not negate. $x^{-a} = \frac{1}{x^a}\mathclose{,}$ and the sign of the answer is unchanged.

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Answer: H

Explanation

Find the largest square number that divides 72. The squares are 4, 9, 16, 25, 36, 49… and 36 divides 72.

$$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$$

Check with a calculator: $6 \times 1.414 = 8.485\mathclose{,}$ and $\sqrt{72} = 8.485\mathclose{.}$ Correct.

Why each wrong option is wrong:

• F, $8\sqrt{2}$ — used 64, which is a square but does not divide 72.
• G, $2\sqrt{18}$ — took out the factor 4 rather than 36, giving $2\sqrt{18}\mathclose{.}$ That equals the right value, but 18 still contains the square factor 9, so it is not in simplest form. This is why "simplest" is stated in the question.
• J, $36$ — halved 72.

Takeaway: pull out the largest square factor, then check what is left has no square factor of its own. If it does, you have not finished.

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Answer: D

Explanation

To undo a fourth root, raise both sides to the fourth power:

$$\left(\sqrt[4]{x}\right)^4 = 3^4$$

$$x = 81$$

Check: $\sqrt[4]{81} = 3\mathclose{,}$ because $3 \times 3 \times 3 \times 3 = 81\mathclose{.}$ Correct.

Why each wrong option is wrong:

• A, $0.75$ — divided by 4.
• B, $12$ — multiplied 3 by 4. A root is not a division and its inverse is not a multiplication.
• C, $64$ — computed $4^3$ instead of $3^4\mathclose{.}$ The base and the exponent are not interchangeable: $4^3 = 64$ but $3^4 = 81\mathclose{.}$

Takeaway: an $n$th root is undone by an $n$th power. And keep the base and the exponent the right way round — they give different answers.

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Answer: G

Explanation

Two separate rules:

$$3^0 = 1 \qquad \text{(anything except 0, to the power 0, is 1)}$$

$$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$$

$$1 + \frac{1}{2} = \frac{3}{2}$$

Why each wrong option is wrong:

• F, $\frac{1}{2}$ — took $3^0$ to be 0. The exponent being zero does not make the term zero; it makes it one.
• H, $3$ — read $2^{-1}$ as 2, ignoring the minus sign. A negative exponent is a reciprocal: $2^{-1} = \frac{1}{2}\mathclose{.}$
• J, $\frac{7}{2}$ — took $3^0$ to be 3, as though the exponent were 1.

Takeaway: $x^0 = 1\mathclose{,}$ always. A zero exponent means "no copies of the base multiplied together", and an empty product is 1, not 0.

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Section 2 — Number Properties and Scientific Notation

A handful of definitions that the ACT uses without explaining:

• a factor of $n$ divides into $n$ exactly; a multiple of $n$ is $n$ times a whole number
• a prime has exactly two factors, itself and 1. 2 is prime; 1 is not
• the greatest common factor is the largest number dividing both
• the lowest common multiple is the smallest number both divide into

"Shared out with nothing left over" is a common factor. "Happens together again" is a common multiple.

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Answer: A

Explanation

Divide by primes until nothing composite is left:

$$84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7$$

So

$$84 = 2^2 \times 3 \times 7$$

Check by multiplying back: $4 \times 3 \times 7 = 84\mathclose{.}$ Correct.

Why each wrong option is wrong:

• B, $2 \times 42$ and C, $4 \times 21$ — correct products, but 42, 4 and 21 are not prime, so neither is a prime factorisation. Multiply them out and both give 84, which is exactly why the word "prime" in the question matters.
• D, $2 \times 3 \times 7$ — dropped one of the two factors of 2, giving 42.

Takeaway: keep dividing until every factor is prime, then check by multiplying back. Getting the right product is not enough; the form is what is being asked for.

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Answer: J

Explanation

Move the decimal point so that exactly one non-zero digit stands in front of it:

$$0.00042 \longrightarrow 4.2$$

The point moved 4 places to the right, and moving right makes the number bigger, so the power of ten must make it smaller again:

$$4.2 \times 10^{-4}$$

Check: $4.2 \times 0.0001 = 0.00042\mathclose{.}$ Correct.

Why each wrong option is wrong:

• F, $4.2 \times 10^{-3}$ — counted one place too few. Expanded, this is 0.0042.
• H, $4.2 \times 10^{-5}$ — counted one place too many. Expanded, this is 0.000042.
• G, $42 \times 10^{-5}$ — the right size, since $42 \times 0.00001 = 0.00042\mathclose{,}$ but not scientific notation: the leading number must be at least 1 and below 10.

Counting the places is most of the question, so count them twice, and always check by expanding the answer back out.

Takeaway: a number smaller than 1 has a negative power of ten. Count the places the point moves, then expand your answer to check.

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Answer: B

Explanation

They coincide at times that are multiples of both 12 and 18, so the first is the lowest common multiple.

$$12 = 2^2 \times 3 \qquad 18 = 2 \times 3^2$$

Take the highest power of each prime that appears:

$$2^2 \times 3^2 = 36$$

Check: $36 = 3 \times 12$ and $36 = 2 \times 18\mathclose{.}$ Both whole. Correct.

Why each wrong option is wrong:

• A, $6$ — gave the greatest common factor, 6. That is the wrong tool: a factor is smaller than both intervals, and they cannot coincide before either has flashed.
• C, $72$ — a common multiple, but not the first: by then the lights have already flashed together once, at 36 seconds.
• D, $216$ — multiplied them. 216 is a common multiple, but it is not the lowest one, and the question asks when they next coincide.

Takeaway: "when do they happen together again" is the lowest common multiple. Multiplying always gives a common multiple, but rarely the lowest.

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Section 3 — Complex Numbers

The ACT tests this and the SAT does not, so it is worth reading carefully even if you have never met it.

There is no real number whose square is negative. So mathematicians defined one:

$$i = \sqrt{-1} \qquad \text{which means} \qquad i^2 = -1$$

That single fact is nearly the whole topic. Everything else follows from it, and the powers of $i$ cycle every four:

$$i^1 = i \qquad i^2 = -1 \qquad i^3 = -i \qquad i^4 = 1 \qquad i^5 = i \dots$$

To find any power of $i\mathclose{,}$ divide the exponent by 4 and use the remainder.

A complex number is written $a + bi\mathclose{:}$ a real part and an imaginary part. Add them like algebra, keeping real with real. Multiply them like brackets, then replace every $i^2$ with $-1\mathclose{.}$

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Answer: H

Explanation

Divide the exponent by 4 and keep the remainder:

$$27 \div 4 = 6 \text{ remainder } 3$$

So $i^{27} = i^{3}\mathclose{,}$ and from the cycle

$$i^{3} = -i$$

Why it works: $i^{27} = (i^4)^6 \times i^3 = 1^6 \times i^3 = i^3\mathclose{,}$ because $i^4 = 1$ and 1 to any power is 1.

Why each wrong option is wrong:

• F, $1\mathclose{,}$ G, $i$ and J, $-1$ — each uses a different remainder. Only the arithmetic of $27 \div 4$ separates them, so do that division carefully; it is the entire question.

Takeaway: powers of $i$ cycle $i, -1, -i, 1$ with period 4. Divide the exponent by 4 and use the remainder: 1, 2, 3, 0 give $i\mathclose{,}$ $-1\mathclose{,}$ $-i\mathclose{,}$ $1\mathclose{.}$

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Answer: D

Explanation

Expand as ordinary brackets:

$$(3 + 2i)(4 - i) = 12 - 3i + 8i - 2i^2$$

Collect the $i$ terms: $-3i + 8i = 5i\mathclose{.}$

Now the key step — replace $i^2$ with $-1\mathclose{:}$

$$-2i^2 = -2(-1) = +2$$

So

$$12 + 2 + 5i = 14 + 5i$$

Why each wrong option is wrong:

• A, $12 + 5 i$ — expanded correctly but left the $-2i^2$ alone, so the real part stayed 12. This is the mistake the question exists to catch.
• B, $10 + 5 i$ — subtracted the 2 rather than adding it, forgetting that $-2 \times -1$ is positive.
• C, $14 - 5 i$ — sign slip when collecting $-3i + 8i\mathclose{.}$

Takeaway: expand normally, then hunt for every $i^2$ and replace it with $-1\mathclose{.}$ A $-i^2$ becomes $+1$ — two negatives, and it is easy to keep only one.

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Answer: F

Explanation

Convert each root separately first:

$$\sqrt{-9} = \sqrt{9}\sqrt{-1} = 3i \qquad \sqrt{-16} = 4i$$

Now add, treating $i$ like a variable:

$$3i + 4i = 7i$$

Why you must convert first. The rule $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ only holds when $a$ and $b$ are not negative. Combining under one root here would give $\sqrt{-25} = 5i\mathclose{,}$ which is wrong — and it is option G.

Why each wrong option is wrong:

• G, $5 i$ — combined the two roots into $\sqrt{-25}\mathclose{.}$
• H, $-7$ — dropped the $i$ and kept a minus sign.
• J, $5$ — ignored the minus signs and worked with $\sqrt{9} + \sqrt{16}\mathclose{.}$

Takeaway: rewrite each $\sqrt{\text{negative}}$ as a multiple of $i$ before doing anything else. The usual root rules do not survive a negative under the sign.

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Section 4 — Matrices

A matrix is a rectangular block of numbers, described by its size: a $2 \times 3$ matrix has 2 rows and 3 columns. Rows first, always.

The ACT asks for very little:

• Adding — add matching entries. Both matrices must be the same size.
• Multiplying by a number (a scalar) — multiply every entry.
• The determinant of a $2 \times 2$ — for $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ it is $ad - bc\mathclose{:}$ the product of the main diagonal minus the product of the other one.

That is essentially the whole syllabus. The commonest error is multiplying only the first entry by the scalar.

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Answer: B

Explanation

Multiply every one of the four entries by 5:

$$5\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 \times 1 & 5 \times 2 \\ 5 \times 3 & 5 \times 4 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix}$$

Why each wrong option is wrong:

• A, $\left[\begin{matrix}5 & 2\\3 & 4\end{matrix}\right]$ — multiplied only the top-left entry.
• C, $\left[\begin{matrix}6 & 7\\8 & 9\end{matrix}\right]$ — added 5 to each entry instead of multiplying.
• D, $\left[\begin{matrix}5 & 10\\3 & 4\end{matrix}\right]$ — multiplied the first row and left the second alone.

Takeaway: a scalar reaches every entry. Count them: a $2 \times 2$ has four, and all four must change.

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Answer: F

Explanation

For $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ the determinant is $ad - bc\mathclose{:}$

$$(6 \times 2) - (3 \times 4) = 12 - 12 = 0$$

A determinant of zero is not a mistake — it is a meaningful answer, telling you the two rows are multiples of one another. Here the first row is exactly $1.5$ times the second.

Why each wrong option is wrong:

• G, $10$ — multiplied down the columns, $6 \times 3 - 4 \times 2\mathclose{,}$ instead of along the diagonals. Note that subtracting the right products in the wrong order, $bc - ad\mathclose{,}$ would give 0 here as well — with this matrix that particular slip is invisible, which is a good reason to fix the order by habit rather than by checking the answer.
• H, $12$ — used only the main diagonal.
• J, $24$ — added the two products instead of subtracting.

Takeaway: $ad - bc\mathclose{,}$ main diagonal first. A determinant of 0 is a real answer, and it means the rows are proportional.

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Answer: A

Explanation

Add each pair in matching positions:

$$\begin{bmatrix} 2+3 & -1+4 \\ 0+(-2) & 5+1 \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ -2 & 6 \end{bmatrix}$$

Why each wrong option is wrong:

• B, $\left[\begin{matrix}5 & 3\\2 & 6\end{matrix}\right]$ — got $+2$ in the bottom left, ignoring the minus sign on $-2\mathclose{.}$
• C, $\left[\begin{matrix}6 & -4\\0 & 5\end{matrix}\right]$ — multiplied matching entries rather than adding them.
• D, $\left[\begin{matrix}-1 & -5\\2 & 4\end{matrix}\right]$ — subtracted the first matrix from the second.

Takeaway: matrix addition is entry-by-entry, and the negative signs are where the marks go. Work position by position, not left to right along the whole block.

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Section 5 — Vectors

A vector has both a size and a direction, written by its components: $\langle 3, 4 \rangle$ means 3 across and 4 up.

• Add two vectors by adding matching components.
• Its magnitude — its length — comes from Pythagoras:

$$\left| \langle a, b \rangle \right| = \sqrt{a^2 + b^2}$$

That is the entire ACT vector syllabus. Notice that magnitude is Pythagoras wearing different notation, so if you can find the hypotenuse of a right triangle you can already do this.

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Answer: G

Explanation

Magnitude is Pythagoras on the two components:

$$\sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

Notice that squaring removes the minus sign, which is why a magnitude can never come out negative. A length has no direction.

Why each wrong option is wrong:

• H, $7$ and F, $17$ — subtracted or added the sizes without squaring. Both would give the right answer only if the two components pointed the same way, and they do not — they are at right angles.
• J, $-7$ — added the components, giving a negative "length".

Takeaway: magnitude is $\sqrt{a^2 + b^2}\mathclose{.}$ Square first, so the signs disappear, and expect a positive answer every time. (5, 12, 13 is a Pythagorean triple worth memorising — see Chapter 8.)

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Answer: D

Explanation

Add the matching components:

$$\langle 2 + (-5), \; -3 + 8 \rangle = \langle -3, \; 5 \rangle$$

Take the two separately and slowly. $2 + (-5) = -3\mathclose{;}$ $-3 + 8 = 5\mathclose{.}$

Why each wrong option is wrong:

• A, $\langle 7, -11 \rangle$ — subtracted the second vector from the first.
• B, $\langle -10, -24 \rangle$ — multiplied the components.
• C, $\langle -3, -5 \rangle$ — got the first component right and slipped a sign on the second.

Takeaway: vector addition is component by component. Write the two calculations on separate lines — running them together in your head is where the sign errors come from.

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Section 6 — Sequences

A sequence is a list of numbers following a rule. The ACT tests two kinds, and the first job is always to work out which one you have.

Arithmetic — you add the same amount each time. That amount is the common difference, $d\mathclose{.}$

$$n\text{th term} = a + (n - 1)d$$

Geometric — you multiply by the same amount each time. That amount is the common ratio, $r\mathclose{.}$

$$n\text{th term} = a \cdot r^{\,n-1}$$

In both, $a$ is the first term. And in both, the exponent and the bracket contain $n - 1\mathclose{,}$ not $n$ — because getting from term 1 to term $n$ takes $n - 1$ steps. That off-by-one is the single most common sequence error.

How to tell them apart: subtract consecutive terms. If you always get the same number, it is arithmetic. If not, divide consecutive terms instead; the same number means geometric.

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Answer: H

Explanation

First check which kind of sequence it is. Subtract consecutive terms:

$$21 - 7 = 14 \qquad 35 - 21 = 14 \qquad 49 - 35 = 14$$

Always 14, so it is arithmetic with $a = 7$ and $d = 14\mathclose{.}$

$$\text{20th term} = a + (n - 1)d = 7 + 19 \times 14 = 7 + 266 = 273$$

Why each wrong option is wrong:

• F, $133$ — used 7 as the common difference.
• G, $140$ — multiplied the first term by 20, which would only work if the sequence started at 0 and had difference 7.
• J, $287$ — used $a + nd$ instead of $a + (n-1)d\mathclose{,}$ adding one difference too many. This is the off-by-one the formula is designed to prevent.

Takeaway: $a + (n-1)d\mathclose{.}$ There are $n - 1$ steps to reach the $n$th term, not $n$ — you are already standing on the first one.

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Answer: C

Explanation

Subtracting gives $12 - 3 = 9$ and $48 - 12 = 36$ — not constant, so it is not arithmetic. Divide instead:

$$\frac{12}{3} = 4 \qquad \frac{48}{12} = 4 \qquad \frac{192}{48} = 4$$

Constant, so it is geometric with $a = 3$ and $r = 4\mathclose{.}$

$$\text{7th term} = a \cdot r^{\,n-1} = 3 \times 4^{6} = 3 \times 4096 = 12288$$

Why each wrong option is wrong:

• A, $27$ — treated the sequence as arithmetic with a difference of 4.
• B, $84$ — multiplied the three numbers together.
• D, $49152$ — used $r^n$ instead of $r^{n-1}\mathclose{,}$ giving a term four times too big.

Takeaway: subtract to test for arithmetic; if that fails, divide to test for geometric. Then use $a \cdot r^{\,n-1}$ — the exponent is one less than the term number.

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Answer: J

Explanation

Substitute $n = 1, 2, 3$ in turn:

$$n = 1: \; 5(1) + 3 = 8$$ $$n = 2: \; 5(2) + 3 = 13$$ $$n = 3: \; 5(3) + 3 = 18$$

$$8 + 13 + 18 = 39$$

Notice the terms go up by 5 each time, which matches the coefficient of $n$ — that is a useful check that the substitution was done correctly.

Why each wrong option is wrong:

• F, $8$ and G, $18$ — gave a single term rather than the sum.
• H, $33$ — added $1 + 2 + 3$ first and applied the rule once, to $n = 6\mathclose{.}$ The rule has to be applied to each term separately, then the results added.

Takeaway: an $n$th-term expression is a machine — feed it each $n$ in turn. Do not put a sum of $n$ values into it.

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Answer: C

Explanation

Multiply both components by 3:

$$3\langle -2, 5 \rangle = \langle -6, 15 \rangle$$

The sign is carried through: $3 \times (-2) = -6\mathclose{.}$

Why each wrong option is wrong:

• A, $\langle -2, 15 \rangle$ and B, $\langle -6, 5 \rangle$ — multiplied one component and left the other.
• D, $\langle 1, 8 \rangle$ — added 3 to each rather than multiplying.

Takeaway: a scalar reaches every component, signs included. The direction is unchanged; only the length scales.

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Answer: F

Explanation

Subtract matching components, keeping the order the question gives:

$$\langle 7 - 3, \; -2 - 4 \rangle = \langle 4, \; -6 \rangle$$

Do the two on separate lines. The second is $-2 - 4\mathclose{,}$ which is $-6\mathclose{,}$ not $2$ — running them together in your head is where the sign goes.

Why each wrong option is wrong:

• G, $\langle -4, 6 \rangle$ — computed $\mathbf{v} - \mathbf{u}\mathclose{,}$ which is the negative of the answer.
• H, $\langle 10, 2 \rangle$ — added the vectors.
• J, $\langle 4, 2 \rangle$ — got the first component right and slipped the sign on the second.

Takeaway: vector subtraction is component by component and is not symmetric: $\mathbf{u} - \mathbf{v}$ and $\mathbf{v} - \mathbf{u}$ point opposite ways.

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Answer: A

Explanation

A vector's components say how far it moves across and how far up: where it ends, minus where it starts.

$$\langle 6 - 1, \; -2 - 4 \rangle = \langle 5, \; -6 \rangle$$

Check by walking it: from $(1, 4)\mathclose{,}$ 5 to the right and 6 down lands on $(6, -2)\mathclose{.}$ Correct.

Why each wrong option is wrong:

• B, $\langle -5, 6 \rangle$ — subtracted the terminal point from the initial one. That is the vector pointing the opposite way.
• C, $\langle 7, 2 \rangle$ — added the coordinates instead of subtracting them.
• D, $\langle 5, 6 \rangle$ — got the first component right and subtracted the second pair backwards.

Takeaway: component form is terminal minus initial, one coordinate at a time. The same vector can start anywhere; only the change matters.

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Answer: H

Explanation

Add the real parts and the imaginary parts separately:

$$\text{real: } 4 + (-1) = 3 \qquad \text{imaginary: } -3 + 7 = 4$$

$$(4 - 3i) + (-1 + 7i) = 3 + 4i$$

There is no $i^2$ here, so nothing needs replacing — that only happens when complex numbers are multiplied.

Why each wrong option is wrong:

• F, $3 - 4 i$ — subtracted the imaginary parts.
• G, $5 + 4 i$ — subtracted the real parts, giving $4 - (-1) = 5\mathclose{.}$
• J, $3 + 10 i$ — computed $7 - (-3) = 10$ rather than $-3 + 7\mathclose{.}$

Takeaway: real with real, imaginary with imaginary. Adding complex numbers never produces an $i^2\mathclose{,}$ so no substitution is needed.

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