ACCUPLACER Elementary Algebra Practice Test

The ACCUPLACER Elementary Algebra test is the second of the three ACCUPLACER math tests. There are 12 algebra problems to solve on this section of the test. If you haven’t studied algebra for awhile, then you will definitely want to review all the basic concepts. There will be questions about monomials and polynomials, simplifying algebraic fractions, and factoring. You will also need to be able to translate written phrases into algebraic expressions. Continue your test prep and review right now with our free ACCUPLACER Elementary Algebra practice test.

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

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Question 1
Solve for x:

3(x + 1) = 5(x − 2) + 7

A
−2
B
2
C
½
D
3
Question 1 Explanation: 
The correct answer is (D). We can simplify the equation as follows:
3(x + 1) = 5(x − 2) + 7
3x + 3 = 5x − 10 + 7
3x + 3 = 5x − 3
We can add 3 to both sides to get the following:
3x + 6 = 5x
We can subtract 3x from both sides and then divide the resulting equation by 2 to solve for x as follows:
6 = 2x
3 = x
Question 2
Evaluate:

x4 − y   if   x = 3 and y = −20

A
101
B
61
C
47
D
32
Question 2 Explanation: 
The correct answer is (A). Substitute the values of x and y into the given equation and evaluate: = 34 − (−20)
= 81 + 20
= 101
Question 3
alg3

A
alg3a
B
alg3b
C
alg3c
D
alg3d
Question 3 Explanation: 
The correct answer is (B). If you add any number to itself, the result is twice that number. For example: a + a = 2a. Therefore:
alg3ex
Question 4
What is the radius of a circle with an area of 25π in2?

A
5 in.
B
10 in.
C
20 in.
D
25 in.
Question 4 Explanation: 
The correct answer is (A). Plug the given information into the formula for the area of a circle and solve for the radius:
A = πr2
25π = πr2
25 = r2
r = 5 in.
Question 5
m70

A
6 + 6x6
B
4 + 6x6
C
4x3 + 6x9
D
4x3 + 6x3
Question 5 Explanation: 
The correct answer is (B). Begin every problem involving polynomials in the numerator or denominator by considering the factoring options. Here, the numerator can be factored as there is a coefficient and variable common to both terms: 10x3. However, before factoring out the full coefficient, notice that dividing out the full 10 would entail multiplying the resulting 2 back through the numerator. Factor out 5x3 and divide by the denominator to arrive at the answer:
m72
Question 6
Joe’s current age is five times Mary’s age ten years ago. If Mary is currently m years old, what is Joe’s current age in terms of m?

A
5m
B
5m − 10
C
5m − 50
D
5m + (m − 10)
Question 6 Explanation: 
The correct answer is (C). Mary’s age ten years ago is:
m − 10
So Joe’s age is:
5(m − 10)
= 5m − 50
Question 7
alg1

A
alg1a
B
alg1b
C
alg1c
D
alg1d
Question 7 Explanation: 
The correct answer is (C). We can use the following rule of radicals to answer this question:
alg1ex
Question 8
(3x + 4y)(2x + 5y) = ?

A
3x2 + 8xy + 20y2
B
3x2 + 15xy + 20y2
C
6x2 + 15xy + 20y2
D
6x2 + 23xy + 20y2
Question 8 Explanation: 
The correct answer is (D). The “FOIL method” is the easiest way to remember how to multiply two-termed expressions. Multiply the First two terms, then the Outer two terms, then the Inner two terms, and then the Last two terms, then sum all four to arrive at the answer:
(3x + 4y)(2x + 5y)
= 6x2 + 15xy + 8xy + 20y2
= 6x2 + 23xy + 20y2
Question 9
A wheel has a diameter of 6 meters. If a rope can completely wrap around the wheel 3 times, what is the length of the rope?

A
6π meters
B
8π meters
C
12π meters
D
18π meters
Question 9 Explanation: 
The correct answer is (D). The diameter of the wheel and the number of times the rope wraps around the wheel, which is the same as the circumference of the wheel, are given. Use the given diameter to find the circumference, and then multiply the result by 3 to find the length of the rope:
C = d * π = 6π meters
3 * 6π = 18π meters
Question 10
Solve:

alg4

A
alg4a
B
alg4b
C
alg4c
D
alg4d
Question 10 Explanation: 
The correct answer is (D). In order to isolate x, we must first subtract 5 from both sides as follows:
alg4ex1
Next, we multiply both sides by 5, and then divide both sides by 2 to get:
alg4ex2
Question 11
If the area of a square is equal to its perimeter, what is the area of the square?

A
2
B
4
C
8
D
16
Question 11 Explanation: 
The correct answer is (D). If A is the area of a square, S is the length of a side of the square, and P is the perimeter, then:
A = S2
P = 4S
Since we are told that the area is equal to the perimeter, we can set the two expressions equal to each other and solve as follows:
A = P
S2 = 4S
S2 − 4S = 0
S(S − 4) = 0
S = 0,4
Since S cannot be zero, the only possible value of S in this case is 4. The area of the square is:
A = S2 = 42 = 16
Question 12
alg5

A
alg5a
B
alg5b
C
alg5c
D
alg5d
Question 12 Explanation: 
The correct answer is (D). Begin by rewriting the expression as the multiplication of the reciprocal of the second fraction:
alg5ex1
Factor the numerator and denominator of each fraction and reduce where possible:
alg5ex2
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