Preparing for the Accuplacer Math test? To succeed on the Accuplacer Math test, you need to practice as many real Accuplacer Math questions as possible. There’s nothing like working on Accuplacer Math sample questions to measure your exam readiness and put you more at ease when taking the Accuplacer Math test. The sample math questions you’ll find here are brief samples designed to give you the insights you need to be as prepared as possible for your Accuplacer Math test.

Check out our sample Accuplacer Math practice questions to find out what areas you need to practice more before taking the Accuplacer Math test!

Start preparing for the 2021 Accuplacer Math test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions you need to practice.

## The Absolute Best Book** to Ace the Accuplacer** **Math** Test

## 10 Sample **Accuplacer** Math Practice Questions

1- If \(f(x) = 5 + x\) and \(g(x) = – x^2 – 1 – 2x\), then find \((g – f)(x)\)?

A. \(x^2 – 3x – 6\)

B. \(x^2 – 3x + 6\)

C. \(–x^2 – 3x + 6\)

D. \(–x^2 – 3x – 6\)

2- \(\frac{|3+x|}{7} ≤ 5\), then \(x =\) ?

A. \(–38 ≤ x ≤ 35\)

B. \(–38 ≤ x ≤ 32\)

C. \(–32 ≤ x ≤ 38\)

D. \(–32 ≤ x ≤ 32\)

3- tan \((– \frac{π}{6})\) = ?

A. \(\frac{\sqrt{3}}{3}\)

B. \(-\frac{\sqrt2}{2}\)

C. \(\frac{\sqrt2}{2}\)

D. \(-\frac{\sqrt3}{3}\)

4-\(\frac{\sqrt{32a^5 b^3}}{\sqrt{2ab^2}}=?\)

A. \(4a^2 \sqrt {b}\)

B. \(2b^2 \sqrt {a}\)

C. \(4b^2 \sqrt {a}\)

D. \(– 4a^2 \sqrt {b}\)

5- The cost, in thousands of dollars, of producing \(x\) thousands of textbooks is \(C (x) = x^2 + 10x + 30\). The revenue, also in thousands of dollars, is \(R(x) = 4x\). Find the profit or loss if 3,000 textbooks are produced. (profit \(=\) revenue\( –\) cost)

A. $21,000 loss

B. $57,000 profit

C. $3,000 profit

D. $57,000 loss

6- Suppose a triangle has the dimensions indicated below:

Then Sin B = ?

\(\img{https://appmanager.effortlessmath.com/public/images/questions/test1.q26.jpg

}\)

A. \(\frac{3}{5}\)

B. \(\frac{4}{5}\)

C. \(\frac{4}{3}\)

D. \(\frac{3}{4}\)

7- Find the slope–intercept form of the graph \(6x – 7y = – 12\)

A. \(y = – \frac{7}{6} x – \frac{12}{7}\)

B. \(y = – \frac{6}{7} x + 12\)

C. \(y = \frac{6}{7} x + \frac{12}{7}\)

D. \(y = \frac{7}{6}x – 12\)

8- Ella (E) is 4 years older than her friend Ava (A) who is 3 years younger than her sister Sofia (S). If E, A and S denote their ages, which one of the following represents the given information?

A. \(E=A+4\)

\(S=A-3\)

B. \(E=A+4\)

\(A=S+3\)

C. \(A=E+4\)

\(S=A-3\)

D. \(E=A+4\)

\(A=S-3\)

9-Which of the following point is the solution of the system of equations?

\(5x + y = 9\)

\(10x-7y= -18\)

A. (2, 4)

B. (2, 2)

C. (1, 4)

D. (0, 4)

10- Find the Center and Radius of the graph \((x – 3)^2 + (y + 6)^2 = 12\)

A. \((3, 6), \sqrt3\)

B. \((3, –6), 2\sqrt3\)

C. \((–3, 6), 2\sqrt3\)

D. \((3, –6), \sqrt3\)

## Best **Accuplacer** Math Prep Resource for 2021

## Answers:

1- **D**

\((g – f)(x) = g(x) – f(x) = (– x^2 – 1 – 2x) – (5 + x)\)

\(– x^2 – 1 – 2x – 5 – x = – x^2 – 3x – 6\)

2- **B**

\(\frac{|3+x|}{7} ≤5⇒|3+x|≤35⇒-35≤3+x≤35⇒-35-3≤x≤35-3⇒

-38≤x≤32 \)

3- **D**

\((– \frac{π}{6}) = -\frac{\sqrt 3}{3}\)

4- **A**

\(\frac{\sqrt{32a^5 b^3}}{\sqrt{2ab^2}}=\frac{4a^2 b\sqrt{2ab}}{b\sqrt{2a}}= 4a^2 \sqrt b\)

5- **D**

\(c(3)=(3)^2+10(3)+30=9+30+30=69\)

\(4×3=12⇒12-69=-57⇒57,000 \space loss\)

6- **B**

sinB\(=\frac{the \space length \space of \space the \space side \space that \space is \space opposite \space that \space angle}{the \space length \space of \space the \space longest \space side \space of \space the \space triangle} = \frac{4}{5}\)

7- **C**

\(-7y=-6x-12⇒y=\frac{-6}{-7}x-\frac{12}{-7}⇒y=\frac{6}{7}x+\frac{12}{7}\)

8- **D**

\(E = 4 + A\)

\(A = S – 3\)

9- **C**

\(5x + y = 9\)

\(10x-7y= -18\)

⇒ Multiplication \((–2)\) in first equation ⇒

\(-10x- 2y = -18 \)

\(10x-7y= -18\)

Add two equations together \(⇒ –9y = –36 ⇒ y = 4\) then: \(x = 1\)

10- **B**

\((x – h)^2 + (y – k)^2 = r^2\) ⇒ center: (h,k) and radius: r

\((x – 3)^2 + (y + 6)^2 = 12 \)⇒ center: \((3,-6)\) and radius: \(2\sqrt {3}\)