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.

Congratulations - you have completed . You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1
If ƒ(x) = 4x² and g(x) = 5x, what is g(ƒ(2x)) ?

A
100x²
B
80x²
C
20x²
D
10x²
E
16x²
Question 1 Explanation: 
The correct answer is (B). First evaluate the inner function by substituting 2x into the expression for ƒ(x):
ƒ(2x) = 4(2x
ƒ(2x) = 4(x²)
ƒ(2x) = 16x²

Next substitute the expression for ƒ(2x) into the expression for g(x):
g(ƒ(2x)) = g(16x²)
g(ƒ(2x)) = 5(16x²)
g(ƒ(2x)) = 80x²
Question 2
What is the value of:

m1

A
m2a
B
m2b
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:
m2
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 certain 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
m40a
B
m40b
C
m40c
D
m40d
E
m40e
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: 1/5. Only answer choice (C) has a slope of m = 1/5.
Question 5
A candy store sells a bag of cotton candy for $0.70 and chocolate bars for $0.50. Sheila bought some cotton candy and several chocolate bars from the store and spent a total of $6.30. How many bags of cotton candy and bars of chocolate did Sheila purchase?

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.
m10
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.
m11
Question 7
If −3 is one solution of the equation + 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 orginal equation and solve for x:
(−3)² + 3(−3) + z = 12
9 − 9 + z = 12
z = 12
Substitute this value into the original equation:
x² + 3x + 12 = 12 → x² + 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
m6a
B
m6b
C
m6c
D
m6d
E
m6e
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:
m7ex
(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:
m7ex2
And the equation of the line is:
m7ex3
Question 9
m7

A
0
B
1
C
p5
D
2p5
E
p10
Question 9 Explanation: 
The correct answer is (D). Recall the rules of combining exponents of the same base:
m10ex1
Substitute the values of x and y into the expression and evaluate:
m10ex2
Question 10
In the xy plane, the graph of yx2 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.
m8
It can be directly seen that there are 2 points of intersection.
Question 11
m9

A
m11a
B
m11b
C
m11c
D
m11d
E
m11e
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:
m11ex1
Because the bases on both sides of the equation are equal, we can equate the exponents and solve for y:
m11ex2
Question 12
In the arithmetic sequence x1, x2, x3,..., xnx1 = 24 and xn = xn − 1 + 2 for each n > 1. What is the value of n when xn = 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 nth term is xn = x1 + (n − 1)d, where xn is the nth term, x1 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 an + 1 = an + 2. Substituting the given information into the formula for the nth term of an arithmetic sequence, we get:
xn = x1 + (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
m50a
B
m50b
C
m50c
D
m50d
E
m50e
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:
m13ex
Question 14
What value of x satisfies the equation: log3 18 − log3 2 = log4 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:
m4ex
log3 18 − log3 2 = log3(18/2)
= log3 9

Recall that logarithms and exponential functions are different ways of writing the same number:
logb x = yby = 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:
log3 9 = log4 x → 2 = log4 x
Which can be rewritten as: 4² = x → 16 = x
Question 15
A coffee store sells 2 cold brews, 3 blended drinks, and 4 lattes. A customer orders two beverages from the menu. What is the probability that the customer ordered two different drinks?

A
m5a
B
m5b
C
m5c
D
m5d
E
m5e
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:
m5ex1
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:
m5ex2
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:
m6
Question 17
m12

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:
m17
Only answer choice (E) satisfies this condition.
Question 18
m13

A
m13a
B
m13b
C
m13c
D
m13d
E
m13e
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:
m18
Question 19
m14

A
m14a
B
m14b
C
m14c
D
m14d
E
m14e
Question 19 Explanation: 
The correct answer is (E). Here is the solution:
m19
Question 20
m15

A
−1
B
−½
C
0
D
½
E
1
Question 20 Explanation: 
The correct answer is (B). Since x > π/2, x must be in quadrant II (sin x is only positive in quadrants I and II).

The figure below illustrates how to calculate the cos of x:
m16
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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