ACT practice test

ACT Preparing for Higher Math Practice Test

Equations, exponents, quadratics, functions and sequences, with every step shown. Here are 10 ACT preparing for higher math questions, each with the answer and the reason.

Updated 2026-11-14 by the Tigertest team. Original passages and questions written to the enhanced ACT format (2025): four choices, 50/45/36/40 questions, Science optional. Not affiliated with ACT, Inc. Score estimates are unofficial.

1. 1.

   If $\frac{2x}{3} - 4 = 8$, then $x =$

   • A.

     $6$

   • B.

     $8$

   • C.

     $18$

   • D.

     $36$

   Answer: C.

   Add $4$ to both sides first: $\frac{2x}{3} = 12$. Then multiply both sides by $\frac{3}{2}$ (or multiply by $3$ and divide by $2$): $x = 18$. Check: $\frac{2(18)}{3} - 4 = 12 - 4 = 8$. The choice $36$ comes from multiplying by $3$ but forgetting to divide by $2$, and $6$ comes from subtracting $4$ instead of adding it.

2. 2.

   If $f(x) = 3x - 2$, which of the following is equivalent to $f(a + 1)$?

   • A.

     $3a - 1$

   • B.

     $3a + 1$

   • C.

     $3a - 2$

   • D.

     $3a + 3$

   Answer: B.

   Replace every $x$ in the rule with the entire expression $a + 1$: $f(a + 1) = 3(a + 1) - 2 = 3a + 3 - 2 = 3a + 1$. The choice $3a - 2$ comes from substituting only $a$ and ignoring the $+ 1$, and $3a - 1$ comes from adding $1$ after multiplying instead of distributing the $3$ to it.

3. 3.

   If $y = 2x - 1$ and $3x + 2y = 19$, what is the value of $x + y$?

   • A.

     $3$

   • B.

     $5$

   • C.

     $8$

   • D.

     $15$

   Answer: C.

   Substitute $2x - 1$ for $y$ in the second equation: $3x + 2(2x - 1) = 19$, so $3x + 4x - 2 = 19$, $7x = 21$, and $x = 3$. Then $y = 2(3) - 1 = 5$, so $x + y = 8$. The choices $3$ and $5$ are the values of $x$ and $y$ alone, and $15$ is their product rather than their sum.

4. 4.

   Two shuttle buses leave a parking garage at 8:00 a.m. One bus leaves the garage every 12 minutes, and the other leaves every 18 minutes. At what time will the two buses next leave the garage at the same moment?

   • A.

     8:06 a.m.

   • B.

     8:30 a.m.

   • C.

     8:36 a.m.

   • D.

     9:12 a.m.

   Answer: C.

   The buses leave together at every common multiple of 12 and 18 minutes after 8:00. The least common multiple is 36 ($12 = 2^2 \cdot 3$ and $18 = 2 \cdot 3^2$, so the LCM is $2^2 \cdot 3^2 = 36$), which gives 8:36 a.m. Choice A uses the greatest common factor, 6, and choice D uses 72, a common multiple but not the least one.

5. 5.

   For all $x$ such that $x \neq -2$ and $x \neq -3$, which of the following is equivalent to $\dfrac{x^2 - 9}{x^2 + 5x + 6}$?

   • A.

     $\dfrac{x - 3}{x + 2}$

   • B.

     $\dfrac{x + 3}{x + 2}$

   • C.

     $\dfrac{x - 3}{x + 3}$

   • D.

     $-\dfrac{9}{5x + 6}$

   Answer: A.

   Factor both parts: the numerator is a difference of squares, $(x - 3)(x + 3)$, and the denominator is $(x + 2)(x + 3)$. The common factor $x + 3$ cancels, leaving $\dfrac{x - 3}{x + 2}$. The choice $-\dfrac{9}{5x + 6}$ cancels the $x^2$ terms, which is not allowed because they are terms being added, not factors being multiplied. The choice $\dfrac{x + 3}{x + 2}$ cancels the wrong factor from the numerator.

6. 6.

   What is the value of $\sin\left(\dfrac{7\pi}{6}\right)$?

   • A.

     $-\dfrac{\sqrt{3}}{2}$

   • B.

     $-\dfrac{1}{2}$

   • C.

     $\dfrac{1}{2}$

   • D.

     $\dfrac{\sqrt{3}}{2}$

   Answer: B.

   An angle of $\dfrac{7\pi}{6}$ radians is $210^\circ$, which lies in Quadrant III with a reference angle of $\dfrac{\pi}{6}$ ($30^\circ$). Since $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$ and sine is negative in Quadrant III, $\sin\dfrac{7\pi}{6} = -\dfrac{1}{2}$. Choice C forgets that sine is negative below the $x$-axis, and choice A is the cosine value for this angle.

7. 7.

   The greatest common factor of two positive integers is 6, and their least common multiple is 180. If one of the integers is 36, what is the other integer?

   • A.

     5

   • B.

     30

   • C.

     60

   • D.

     180

   Answer: B.

   For any two positive integers, the product of the integers equals the product of their GCF and LCM. So the other integer is $\dfrac{6 \times 180}{36} = 30$. Check: $36 = 2^2 \cdot 3^2$ and $30 = 2 \cdot 3 \cdot 5$ share $2 \cdot 3 = 6$, and their LCM is $2^2 \cdot 3^2 \cdot 5 = 180$. Choice C, 60, has the right LCM with 36 but a GCF of 12, not 6, and choice A is $180 \div 36$ with the GCF ignored.

8. 8.

   The function $f$ is defined by $f(x) = 2x^2 - 3x + 1$. What is the value of $f(-2)$?

   • A.

     $-1$

   • B.

     $3$

   • C.

     $11$

   • D.

     $15$

   Answer: D.

   Substitute $-2$ for $x$: $f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 15$. The choice $3$ comes from treating $-3(-2)$ as $-6$ instead of $+6$, and $-1$ comes from also treating $(-2)^2$ as $-4$.

9. 9.

   If $4^{x + 1} = 8^{x - 1}$, then $x =$

   • A.

     $-5$

   • B.

     $1$

   • C.

     $4$

   • D.

     $5$

   Answer: D.

   Write both sides with base $2$: $4^{x + 1} = (2^2)^{x + 1} = 2^{2x + 2}$ and $8^{x - 1} = (2^3)^{x - 1} = 2^{3x - 3}$. Equal bases mean equal exponents, so $2x + 2 = 3x - 3$ and $x = 5$. Check: $4^6 = 4096 = 8^4$. The choice $-5$ comes from swapping the bases ($3x + 3 = 2x - 2$), and $4$ comes from distributing the $2$ to only the $x$ ($2x + 1 = 3x - 3$).

10. 10.

    If $\log_2 x + \log_2 (x - 6) = 4$, what is the value of $x$?

    • A.

      2

    • B.

      8

    • C.

      11

    • D.

      22

    Answer: B.

    The sum of two logarithms with the same base is the logarithm of the product, so $\log_2 [x(x - 6)] = 4$ and $x(x - 6) = 2^4 = 16$. Then $x^2 - 6x - 16 = 0$, which factors as $(x - 8)(x + 2) = 0$. The solution $x = -2$ is rejected because $\log_2 (-2)$ is undefined, leaving $x = 8$. Choice C comes from adding the arguments instead of multiplying them: $x + (x - 6) = 16$ gives $x = 11$.