ACCUPLACER College Level Math Practice Test

The ACCUPLACER College Level Math test is the most challenging of the three ACCUPLACER math tests. The college math test has 20 problems that cover the following topics: algebraic operations, algebraic applications, trigonometry, functions, coordinate geometry, and solutions of equations and inequalities. Be sure to review all of these topics during your test prep, and make sure you work through plenty of practice questions. Get started right now with our free College Level Math ACCUPLACER practice test.

Directions: For each question, choose the best answer from the five choices. You may use paper and a pencil for computations, but you may not use a calculator.

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Question 1

If $f(x) = 4x^2$ and $g(x) = 5x$, what is $g(f(2x))$?

A
$100x^2$
B
$80x^2$
C
$20x^2$
D
$10x^2$
E
$16x^2$
Question 1 Explanation: 
The correct answer is (B). First evaluate the inner function by substituting $2x$ into the expression for $f(x)$:

$f(2x) = 4(2x)^2$
$f(2x) = 4(4x^2)$
$f(2x) = 16x^2$

Next substitute the expression for $f(2x)$ into the expression for $g(x)$:

$g(f(2x)) = g(16x^2)$
$g(f(2x)) = 5(16x^2)$
$g(f(2x)) = 80x^2$
Question 2

What is the value of:

$\dfrac{5!}{3! \ast 2}$

A
$\dfrac{1}{6}$
B
$\dfrac{25}{15}$
C
$10$
D
$15$
E
$30$
Question 2 Explanation: 
The correct answer is (C). Recall that the factorial operation represents the product of an integer and all of the positive integers less than the integer. For example:

$4! = 4 * 3 * 2 * 1 = 24$

Expand the factorials in the numerator and denominator and simplify where possible:

$\dfrac{5!}{3! \ast 2} =\dfrac{5 \ast 4 \ast 3 \ast 2 \ast 1}{3 \ast 2 \ast 1 \ast 2}$

$= \dfrac{5 \ast 4}{2}$

$= 10$
Question 3

$6y(3y − 2) + (3y − 2) =$

A
$2(3y + 2) − (6y)$
B
$(3y + 2)(6y − 1)$
C
$(3y − 2)(6y + 1)$
D
$(6y − 1)(3y − 2)$
E
$(6y)(3y)(y − 2)(−1)$
Question 3 Explanation: 
The correct answer is (C). Consider the form of the answer choices before evaluating the problem. Notice that $(3y − 2)$ is common to both terms, meaning it can be factored out. After factoring out the $(3y − 2)$, only the terms $6y + 1$ remain:

$(3y − 2)(6y + 1)$
Question 4

A line in the coordinate plane has the equation $5a + b = 7$, where a represents the independent variable, and b represents the dependent variable. Which of the following is the equation of a line that is perpendicular to $5a + b = 7$?

A
$b = 5a + 3$
B
$b = -5a - 5$
C
$b = \left(\frac{1}{5}\right)a - 2$
D
$b = \left(-\frac{1}{5}\right)a + 1$
E
$b = \left(-\frac{7}{5}\right)a - 2$
Question 4 Explanation: 
The correct answer is (C). The independent variable is conventionally written as $x$, so we can make a trivial replacement to rewrite the equation into a more familiar form:

$5a + b = 7$

$→ 5x + y = 7$

$→ y = −5x + 7$

Recall that perpendicular lines have slopes that are negative reciprocals of each other. The slope in the given line is $m = −5$. The negative reciprocal is: $\frac{1}{5}$. Only answer choice (C) has a slope of $m =\frac{1}{5}$.
Question 5

A candy store sells bags of cotton candy for \$0.70 and chocolate bars for \$0.50. Sheila bought several bags of cotton candy and several chocolate bars. She spent \$6.30. How many items did Sheila purchase in total?

A
$4$
B
$7$
C
$11$
D
$13$
E
$17$
Question 5 Explanation: 
The correct answer is (C). This question involves setting up several linear equations:
.70x + .50y = 6.30
Multiply the entire equation by 10 to remove the decimals:
7x + 5y = 63

What numbers will fit? The largest x could be is 9, but 7 * 9 = 63, and we know Sheila also bought chocolate bars. If x = 8, 7 * 8 = 56, but 63 − 56 = 7 and 5 will not divide evenly into 7.

If x = 7, then 7 * 7 = 49
63 − 49 = 14, which is not a multiple of 5.

If x = 6, then 7 * 6 = 42
63 − 42 = 21, which is not a multiple of 5.

If x = 5, then 7 * 5 = 35
63 − 35 = 28, which is not a multiple of 5.

If x = 4, then 7 * 4 = 28
63 − 28 = 35
Finally, we have a multiple of 5! This means x = 4, and y = 7. The total number of bags and bars Sheila bought is 4 + 7 = 11.
Question 6

In the coordinate plane, point $Z$ lies on the positive $y$-axis and point $Q$ has a $y$-coordinate of $0$ and is five units away from the origin, $O$. The area of triangle $OZQ$ is $12$, what is the $y$-coordinate of point $Z$?

A
$3.0$
B
$4.2$
C
$4.4$
D
$4.8$
E
$5.0$
Question 6 Explanation: 
The correct answer is (D). The best way to approach a geometry word problem such as this is to draw a figure to represent the situation:



In this case, the easiest way to solve is using the formula for the area of a triangle where $b$ represents the $x$-coordinate of point $Q$, and the variable $h$ represents the $y$-coordinate of $z$:

$\text{Area} = \frac{1}{2}bh$

$12 = \frac{1}{2}5h$

$24 = 5h$

$h = \frac{24}{5} = 4.8$
Question 7

If $−3$ is one solution of the equation $x^2 + 3x + z = 12$, where $z$ is a constant, what is the other solution?

A
$0$
B
$3$
C
$6$
D
$9$
E
$12$
Question 7 Explanation: 
The correct answer is (A). Given that $x = −3$ is a solution, substitute this value for $x$ in the equation to solve for $z$. Substitute the calculated value of $z$ back into the original equation and solve for $x$:

$(−3)^2 + 3(−3) + z = 12$

$9 − 9 + z = 12$

$z = 12$

Substitute this value into the original equation:

$x^2 + 3x + 12 = 12$

$→ x^2 + 3x = 0$

$→ x(x + 3) = 0$

$x = 0 \;$ and $\; x = −3$
Question 8

An equation of the line that contains the origin and the point (−4, 7) is

A
$y = \left(-\frac{7}{4}\right)x$
B
$y = \left(-\frac{4}{7}\right)x$
C
$y = \left(\frac{7}{4}\right)x$
D
$y = \left(-\frac{7}{4}\right)x + 2$
E
$y = \left(-\frac{7}{4}\right)x - 2$
Question 8 Explanation: 
The correct answer is (A). The slope-intercept form of the equation of a line in the coordinate plane is $y = mx + b$, where $b$ is the $y$-intercept and $m$ is the slope.

Since the line passes through the origin $(0, 0)$, the $y$-intercept will be $0$, so the equation will simply read: $y = mx$. Immediately, we can eliminate all answer choices that do not have $0$ as the $y$-intercept.

Recall that:

$m = \text{Slope} = \frac{\text{Rise}}{\text{Run}}$

(Or the change in the $y$-coordinates divided by the change in the $x$-coordinates.)

Our two coordinates are $(0, 0)$ and $(-4, 7)$. The slope is:

$m = \dfrac{7 − 0}{−4 − 0} = −\dfrac{7}{4}$

And the equation of the line is:

$y = -\dfrac{7}{4}x$
Question 9

If $\;x = p^2\;$ and $\;y = \sqrt{p}\;$ what is the value of $\;x^2y^2 + (xy)^2\;$?

A
$0$
B
$1$
C
$p^5$
D
$2p^5$
E
$p^{10}$
Question 9 Explanation: 
The correct answer is (D). Recall the rules of combining exponents of the same base:

$x^a \ast x^b = x^{a + b}$

$\sqrt{x} = x^{\frac{1}{2}}$

$(x^a)^b = x^{a \ast b}$

Substitute the values of $x$ and $y$ into the expression and evaluate:

$x^2y^2 + (xy)^2$

$= (p^2)^2 (\sqrt{p})^2 + \left((p^2)(\sqrt{p})\right)^2$

$= p^4 \ast p + p^4 \ast p$

$= p^5 + p^5 = 2p^5$
Question 10

In the $xy$ plane, the graph of $y = x^2$ and a circle with the center at $(1, 1)$ and radius $1$ have how many points of intersection?

A
$0$
B
$1$
C
$2$
D
$3$
E
$4$
Question 10 Explanation: 
The correct answer is (C). One possible method for solving this problem is to set the equation of the parabola equal to the equation of the circle and solve for the point(s) of intersection. However, because the specific point(s) of intersection are not required, a better approach is to generate the graphs of the equations and count the number of intersection points.



It can be directly seen that there are 2 points of intersection.
Question 11

If $\left(5^2 \sqrt{5} \right)^y$ $= \left(\sqrt[3]{5} \right)^{y + 1}$
then what is the value of $y$?

A
$\dfrac{1}{5}$
B
$\dfrac{2}{13}$
C
$\dfrac{2}{15}$
D
$\dfrac{5}{3}$
E
$\dfrac{15}{2}$
Question 11 Explanation: 
The correct answer is (B). Use the rules of exponents to simplify and express each side of the equation as a power of 5:

$\left( 5^2 \cdot \sqrt{5} \right)^y$ $= \left(\sqrt[3]{5} \right)^{y + 1}$

$\left( 5^2 \cdot 5^{\frac{1}{2}} \right)^y$ $= \left(5^{\frac{1}{3}} \right)^{y + 1}$

$\left( 5^{\frac{5}{2}} \right)^y$ $= \left(5^{\frac{1}{3}} \right)^{y + 1}$

$(5)^{\frac{5}{2}y}$ $= (5)^{\frac{1}{3}(y + 1)}$

Because the bases on both sides of the equation are equal, we can equate the exponents and solve for $y$:

$\dfrac{5}{2}y = \dfrac{1}{3}(y + 1)$ → $6 \left[\dfrac{5}{2}y = \dfrac{1}{3}(y + 1) \right]$

→ $15y = 2y + 2$ → $13y = 2$ → $y = \dfrac{2}{13}$
Question 12

In the arithmetic sequence $x_1, x_2, x_3,..., x_n$, $x_1 = 24$ and $x_n = x_{n − 1} + 2$ for each $n > 1$. What is the value of $n$ when $x_n = 60?$

A
$16$
B
$17$
C
$18$
D
$19$
E
$20$
Question 12 Explanation: 
The correct answer is (D). For an arithmetic sequence with common difference, $d$, the $n$th term is $x_n = x_1 + (n − 1)d$, where $x_n$ is the $n$th term, $x_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference between the terms.

For the given sequence, $d = 2$ since the problem states that $a_{n + 1} = a_n + 2$. Substituting the given information into the formula for the nth term of an arithmetic sequence, we get:
$x_n = x_1 + (n − 1)d$
$60 = 24 + (n − 1)2$
$60 = 24 + 2n − 2$
$38 = 2n$
$n = 19$
Question 13

For which of the following is $q(v) + q(u) = q(u + v)$ for all positive numbers $u$ and $v$?

A
$q(x) = x^2$
B
$q(x) = x + 2$
C
$q (x) = \sqrt{x} − 2$
D
$q(x) = \dfrac{3}{x}$
E
$q(x) = −3x$
Question 13 Explanation: 
The correct answer is (E). Choose values that satisfy the conditions stated in the problem. Let $u = 1$ and $v = 2$. Substitute these values into the answer choices and disqualify those that do not yield a true statement:

(A): $q(1) + q(2) = q(1 + 2)$

$(1)^2 + (2)^2 = (1 + 2)^2$ → $5 ≠ 9$,

so (A) is incorrect.

(B): $q(1) + q(2) = q(1 +2)$

$(1) + 2 + (2) + 2 = (1 + 2) + 2$ → $7 ≠ 5$,

so (B) is incorrect.

(C): $q(1) + q(2) = q(1 + 2)$

$\sqrt{1} − 2 + \sqrt{2} − 2 = \sqrt{1 + 2} − 3$

→ $\sqrt{2} − 3 ≠ \sqrt{3} − 3$, so (C) is incorrect.

(D): $q(1) + q(2) = q(1 + 2)$

$\dfrac{3}{1} + \dfrac{3}{2} = \dfrac{3}{1 + 2}$ → $\dfrac{9}{2} ≠ 1$, so (D) is incorrect.

(E): $q(1) + q(2) = q(1 + 2)$

$−3(1) + −3(2) = −3(1 + 2)$ → $ −9 = −9$, so (E) is correct.
Question 14

What value of $x$ satisfies the equation: $\log_3 18 − \log_3 2 = \log_4 x$?

A
$8$
B
$16$
C
$20$
D
$24$
E
$32$
Question 14 Explanation: 
The correct answer is (B). Simplify the left side of the equation using the division law of logarithms:

$\log_b x − \log_b y = \log_b \left(\dfrac{x}{y}\right)$
$\log_3 18 − log_3 2 = log_3 \left(\dfrac{18}{2}\right)$
$= \log_3 9$

Recall that logarithms and exponential functions are different ways of writing the same number:
$\log_b x = y ≡ b^y = x$.
So, $\log3 = 9$ can be read as $3$ to what power equals $9$:
$\log3 9 = 2$.

Use this equality to simplify the original problem:
$log_3 9 = \log_4 x$ → $2 = \log4 x$
Which can be rewritten as: $4^2 = x$ → $16 = x$
Question 15

The menu of a coffee store lists 9 beverages of 3 different types:

  • 2 cold brews
  • 3 blended drinks
  • 4 lattes
A customer orders two different beverages from the menu. What is the probability that the customer ordered two different types of beverages?

A
$\dfrac{1}{6}$
B
$\dfrac{1}{9}$
C
$\dfrac{2}{9}$
D
$\dfrac{5}{18}$
E
$\dfrac{13}{18}$
Question 15 Explanation: 
The correct answer is (E). The probability that the customer ordered two different drinks is the total probability, or $1$, minus the probability of ordering the same two drinks.

Find the probability that the customer orders two of the same drinks. $C$ = cold brew, $B$ = blended, $L$ = latte. Total drinks = $CC, BBB, LLLL$. For the first drink, select one drink out of the $9$, and for the second drink, select one of the SAME drinks left out of $8$ (we are only choosing out of $8$ for the second drink because we already picked the first drink).

The probability that two drinks of the same type are ordered:

$P(CC) = \dfrac{2}{9} \cdot \dfrac{1}{8} = \dfrac{1}{36}$

$P(BB) = \dfrac{3}{9} \cdot \dfrac{2}{8} = \dfrac{1}{12}$

$P(LL) = \dfrac{4}{9} \cdot \dfrac{3}{8} = \dfrac{1}{6}$

Combining these probabilities yields the total probability of selecting $2$ drinks of the same type. It is important to recognize that the probability of selecting $2$ drinks of the same type combined with the probability of selecting $2$ drinks of a different type will add to $1$.

Consequently, subtracting the probability of selecting two of the same drink from $1$ will yield the probability of selecting two different drinks:

$1 − \left( \dfrac{1}{36} + \dfrac{1}{12} + \dfrac{1}{6} \right)$

$= 1 − \dfrac{5}{18}$

$\dfrac{13}{18}$
Question 16

m5

The area of the square $ACEG$ is $81$, and the area of square region $IDEF$ is one-fourth of the area of square region $ABIH$. What is the length of segment $CD$?

A
$4$
B
$5$
C
$6$
D
$7$
E
$8$
Question 16 Explanation: 
The correct answer is (C). In problems that describe the dimensions of a figure, begin by labeling the known values and expressing the unknown values in terms of these known values. In this case, $CE = 9$ since the area of square $ACEG$ is $81$. Let $CD = x$ and $DE = 9 − x$ (because $CD + DE = CE$). The ratio of the area of the squares is $1:4$, as described in the question-stem:

$x^2 = 4(9 − x)^2$

$\sqrt{x^2} = \sqrt{4 \cdot (9 − x)^2}$

$x = 2(9 − x)$

$3x = 18$

$x = 6$
Question 17

For which of the following values of $x$ is $1 − \left( \sqrt{(2 − \sqrt{x})} \right)$ NOT a real number?

A
$1$
B
$2$
C
$3$
D
$4$
E
$5$
Question 17 Explanation: 
The correct answer is (E). Real numbers are integers, fractions and irrational numbers. Non-real numbers are roots of negative numbers. Any answer choice that leaves a negative number under the root will be imaginary.

If the parenthetical expression beneath the square root is less than $0$, the entire expression will yield a non real number. Set up an inequality and solve for the variable $x$:

$0 > \sqrt{(2 − \sqrt{x})}$
$0 > 2 − \sqrt{x}$
$\sqrt{x} > 2$
$x > 4$

Only answer choice (E) satisfies this condition.
Question 18

If $a ≠ 0$ and $a > 0$, then $\dfrac{a}{\sqrt{2a} + \sqrt{a}} =$ ?

A
$2a$
B
$2a + \sqrt{2a} + \sqrt{a}$
C
$\sqrt{2a} − \sqrt{a} + a$
D
$\sqrt{2a} − \sqrt{a}$
E
$\dfrac{1}{\sqrt{2a}} − \sqrt{a}$
Question 18 Explanation: 
The correct answer is (D). Begin by rationalizing the denominator by multiplying top and bottom of the expression with the conjugate. Recall that the conjugate is the same terms but the opposite sign of the expression in the denominator:

$\dfrac{a}{\sqrt{2a} + \sqrt{a}} \cdot \dfrac{\sqrt{2a} − \sqrt{a}}{\sqrt{2a} − \sqrt{a}}$

$= \dfrac{a \left(\sqrt{2a} − \sqrt{a}\right)}{2a − a}$

$= \dfrac{a \left(\sqrt{2a} − \sqrt{a}\right)}{a}$

$= \left(\sqrt{2a} − \sqrt{a}\right)$

Apply the distributive property to evaluate:
$\left(\sqrt{2a} + \sqrt{a}\right)\left(\sqrt{2a} − \sqrt{a} \right)$, but also recognize that this is the expanded form of a difference of squares:

$(x^2 − y^2) = (x + y)(x − y)$, so $\left(\sqrt{2a}\right)^2 − \left(\sqrt{a}\right)^2 = 2a − a = a$.
Question 19

$f(x) = \sqrt{x} − 40$, for all positive numbers of $x$. If $a = f(b)$ for some positive numbers $b$ and $a$, then what is $b$ in terms of $a$?

A
$\sqrt{a + 40}$
B
$\left(\sqrt{a + 40}\right)^2$
C
$\sqrt{a^2 + 40}$
D
$\left(a^2 + 40\right)$
E
$(a + 40)^2$
Question 19 Explanation: 
The correct answer is (E). The question provides the condition that:

$a = f(b)$ and that $f(x) = \sqrt{x} − 40$.

These conditions can be combined to yield:

$a = f(b) = \sqrt{b} − 40$.

Solving this resulting equation for the variable $b$:

$a = \sqrt{b} − 40$

$a + 40 = \sqrt{b}$

$(a + 40)^2 = b$
Question 20

If $\sin x = \dfrac{\sqrt{3}}{2}$ and $x > \dfrac{π}{2}$, what is $\cos x$?

A
$−1$
B
$−\dfrac{1}{2}$
C
$0$
D
$\dfrac{1}{2}$
E
$1$
Question 20 Explanation: 
The correct answer is (B). Since $x > \frac{π}{2}$, $x$ must be in quadrant II ($\sin x$ is only positive in quadrants I and II).

Since $\sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$, if $\sin x = \dfrac{\sqrt{3}}{2}$, then the opposite side is $\sqrt{3}$, and the hypotenuse is $2$. This is shown in the figure below.



Since $\cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$, $\cos x = \dfrac{−1}{2} = −\dfrac{1}{2}$

Since the ratios in the figure correspond to a $30-60-90$ triangle, the third side must have magnitude $1$. Notice that the adjacent side must be $−1$ if the angle is in Quadrant II because cosine is negative in Quadrants II and III.
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