The Accuplacer math prep problems on this page are up-to-date for the new exam.
Instructions: There are fifty Accuplacer math prep problems on this page. The math problems are grouped into sets of five. You can complete the all of the practice problems in one sitting, or check your answers at the end of each set. You can also try the other materials below.
The questions in this Accuplacer math practice test are ordered from basic to advanced.
So, we recommend that you complete the practice problems in the order provided.
3) Simplify, then state whether the result is a rational or irrational number:
4) Simplify, then state whether the result is a rational or irrational number:
5) 3 + 150 ÷ 6 ÷ 2 × 4 – 8 = ?
Before competing exercises 6 and 7, you may want to view more information on perfect squares.
8) Solve for x and y if 3x + y = 20 and 2x + 2y = 16. Use the substitution method.
9) Group like terms together and then factor: ab 2 y – 3a + 9b 2 y – 27
10) 3 is added to the original value, then 5 is subtracted from this sum to get 68.
What is the original value?
11) Factor: 18a 2 b 3 c + 12a 3 b 2 c 2 – 24ab 4 c 3
If | x + y | = 5 and x = 3, then what are the possible value of y ?
14) Identify the variables, the coefficients, and the constant in the expression below.
4a 2 + 16ab + 24b 4 – 8
15) Factor: a 2 b 3 c 2 + a 3 b 4 c 2 + ab 2 c
16) Find the greatest common factor:
100x 4 y – 300x 3 y 2 + 400x 2 y 2 – 200
17) Factor: 5b + 25c
18) Which of the following are monomials?
14
15b + ab
3xy
43xy + x + y
25x 2
19) Balance the equation and express your answer as a simplified fraction, if possible.
If A + B – C = D, then A = ?
(6x 2 + 9xy + 7y 2 ) + (8x 2 + 2xy + 5y 2 ) = ?
24) Perform the operation:
(5x 2 – 3xy + 4y 2 ) – (2x 2 + 9xy – 7y 2 ) = ?
25) Factor: 9a –3b + 15c 2
26) (16x 3 + 8x 2 + 4x) ÷ (2x) = ?
27) Solve for a and b using the elimination method:
3a + 5b = 4
2a – b = 7
$latex \sqrt \times \sqrt$
10x + 5y – 6x + 12 + 3y – 6
31) Use the FOIL method to simplify: (3x – 2) 2
32) Use the FOIL method to simplify: (5a – 1)(3a + 9)
33) Expand by multiplying polynomials: a 2 b 2 (2a 2 b + 3a 3 b 3 )
34) Expand by multiplying polynomials: x 3 y 3 (4x 3 y 2 + 5x 2 y 2 )
35) Which of the following are rational numbers:
36) Express in scientific notation: 19,875.246
37) Represent the solution to the following on a number line:
38) Find all possible values of x for the inequality:
39) Use the elimination method to solve the system of equations:
Solve for x and y if x = y + 7 and x + y – 1 = 10
40) Factor: a 3 – 27 3
41) Rewrite in the form ax 2 ± bx ± c = 0. Then solve for x.
42) Rewrite in the form ax 2 ± bx ± c = 0. Then solve for x.
43) Find the possible solutions for the quadratic equation:
44) Find the possible solutions for the quadratic equation:
45) Factor the following quadratic equation:
For questions 46 and 47, identify which of the equations are quadratic equations, giving the reason for your answer.
Then re-write those equations into the standard quadratic equation format, using variables b and c where necessary.
48) Use the quadratic formula to solve for x: x 2 + 13x +36 = 0
49) Use the remainder theorem to find the remainder in the following problem:
x 3 + 4x 2 – 5x – 18 ÷ (x + 4)
50) Re-write the following quadratic equation in the completed square form. Then solve for x.
The correct answer is: a 7
For algebra problems like this one, you need to understand the laws of exponents.
a 3 × a 4 = a 3 + 4 = a 7
The correct answer is: y 8
If you have problems with algebra questions on exponents, study these exponent laws.
(y 2 ) 4 = y 2 x 4 = y 8
The correct answer is: irrational
Click on these links for further practice with algebra problems on rational numbers and irrational numbers.
15π is an irrational number.
When π is multiplied by a rational number like 15, the result is irrational.
It is irrational because π has an infinite number of decimal places.
Any number multiplied by π has an infinite number of decimal places.
The correct answer is: rational
When simplified, we get 3, which is a rational number.
$latex \sqrt \times \sqrt =$
The correct answer is: 45
To solve algebra equations like this one, you should study PEMDAS – Order of Operations.
There are no parentheses or exponents.
Do the multiplication and division from left to right.
3 + 150 ÷ 6 ÷ 2 × 4 – 8 =
3 + (150 ÷ 6) ÷ 2 × 4 – 8 =
Then do the addition and subtraction from left to right.
The correct answer is: (x + y)(x – y)
For Accuplacer math prep problems like questions 6 and 7, you may need to review perfect squares.
You should know from memory that x 2 – y 2 = (x + y)(x – y).
The correct answer is: 2(a + 7b)(a – 7b)
The correct answer is: x = 6 and y = 2
To solve algebra questions like this one, you can use the substitution method.
Step 1: Label the equations
Our equations were 3x + y = 20 and 2x + 2y = 16.
Label the equations A and B:
Step 2: Isolate one of the variables
We will isolate variable y in equation A in this solution.
Equation A is 3x + y = 20.
Perform inverse operations to isolate y.
3x – 3x + y = 20 – 3x
3x – 3x + y = 20 – 3x
Step 3: Substitute the value of the isolated variable
We have just determined that y = 20 – 3x.
Equation B is 2x + 2y = 16.
So, we will substitute 20 – 3x for the value of y in equation B.
2x + 2(20 – 3x) = 16
Step 4: Then perform the multiplication
2x + 2(20 – 3x) = 16
2x + (2 × 20) – (2 × 3x) = 16
Step 5: Perform inverse operations to solve for x
2x + 40 – 40 – 6x = 16 – 40
2x + 40 – 40 – 6x = 16 – 40
Step 6: Substitute the value of x to solve for y
We have just determined that x = 6.
Equation B is 2x + 2y = 16.
12 – 12 + 2y = 16 – 12
12 – 12 + 2y = 16 – 12
The correct answer is: (b 2 y – 3)(a + 9)
To find the solution to Accuplacer math prep problems such as this, you need to group like terms together and then factor.
ab 2 y – 3a + 9b 2 y – 27 =
(ab 2 y – 3a) + (9b 2 y – 27) =
[(a × b 2 × y) – (3 × a)] + [(9 × b 2 × y) – (9 × 3)] =
[(a × b 2 × y) – (a × 3)] + [(9 × b 2 × y) – (9 × 3)] =
a[( a × b 2 × y) – ( a × 3)] + 9[( 9 × b 2 × y) – ( 9 × 3)] =
a[(b 2 × y) – (3)] + 9[((b 2 × y) – (3)] =
a(b 2 y – 3) + 9(b 2 y – 3) =
(b 2 y – 3)[a( b 2 y – 3 ) + 9( b 2 y – 3 )] =
The correct answer is: 70
For algebra problems asking for the original value, you need to solve by using inverse operations.
Set up the equation: x + 3 – 5 = 68
Set up the inverse operation: 68 + 5 – 3 = x
Solve to get the original value: 68 + 5 – 3 = 73 – 3 = 70
The correct answer is: 6ab 2 c(3ab + 2a 2 c – 4b 2 c 2 )
For help with algebra questions asking you to factor, you can study factoring quadratics.
18a 2 b 3 c + 12a 3 b 2 c 2 – 24ab 4 c 3 =
ab 2 c(18ab + 12a 2 c – 24b 2 c 2 ) =
ab 2 c[(6 × 3)ab + (6 × 2)a 2 c – (6 × 4)b 2 c 2 ] =
6ab 2 c(3ab + 2a 2 c – 4b 2 c 2 )
The correct answer is: x = 25/3
To learn how to solve Accuplacer math prep problems that ask you to simplify, have a look at simplifying expressions.
Eliminate the fraction by multiplying each side by the denominator.
$ \dfrac \times 2 = 10 \times 2$
Deal with the integer.
3x – 5 + 5 = 20 + 5
Divide by the coefficient to solve.
The correct answer is: –8 = y = 2
To solve similar algebra questions on your exam, learn more about absolute value.
If | x + y | = 5 and x = 3, we can substitute 3 for the value of x to solve.
However, for absolute value problems, you also have to consider other possible values.
If | x + y | = 5 then | x + y | = | –5 |
We know this because the absolute value of –5 is 5.
So again substitute for the value of x to solve.
When dealing with a negative number, we have to reverse the way the inequality sign points.
So, the possible value of y is:
The correct answers are: the variable in the first term is a; the variables in the second term are a and b; the variable in the third term is b.
To answer these types of algebra equations, look at the page on algebraic terms.
The coefficient in the first term is 4. The coefficient in the second term is 16. The coefficient in the third term is 24.
The constant is –8.
The correct answer is: ab 2 c(abc + a 2 b 2 c + 1)
For algebra equations on factoring, you may want to have another look at how to factor quadratics.
a 2 b 3 c 2 + a 3 b 4 c 2 + ab 2 c =
(a × a × b × b × b × c × c) + (a × a × a × b × b × b × b × c × c) + (a × b × b × c) =
ab 2 c( a × a × b × b × b × c × c) + ( a × a × a × b × b × b × b × c × c) + ( a × b × b × c ) =
ab 2 c[(a × b × c) + (a × a × b × b × c) + (1)] =
ab 2 c(abc) + (a 2 × b 2 × c) + (1)] =
ab 2 c(abc + a 2 b 2 c + 1)
The correct answer is: 100
Click on the following link for help with Accuplacer math prep problems on how to find the greatest common factor.
The last term on this expression (200) does not contain any variables.
So, our greatest common factor consists of a number only.
100x 4 y – 300x 3 y 2 + 400x 2 y 2 – 200 =
(100 × 1)x 4 y – (100 × 3)x 3 y 2 + (100 × 4)x 2 y 2 – (100 × 2) =
100[( 100 × 1)x 4 y – ( 100 × 3)x 3 y 2 + ( 100 × 4)x 2 y 2 – ( 100 × 2)] =
100(x 4 y – 3 3 y 2 + 4x 2 y 2 – 2)
So, the greatest common factor is 100.
The correct answer is: 5(b + 5c)
For more algebra questions like this, view our posts on factoring.
5[( 5 × b) + ( 5 × 5 × c)] =
The correct answers are: 4, 3xy and 25x 2
For algebra problems like this one, you need to understand monomials.
The correct answer is: A = D + C – B
To solve algebra equations like the one below, you might like to study balancing equations again.
If A + B – C = D, then A = ?
A + B – C + C = D + C
A + B – C + C = D + C
A + B – B = D + C – B
A + B – B = D + C – B
The correct answer is: 14x 2 + 11xy + 12y 2
To learn how to solve these types of algebra equations, study our post on grouping like terms.
(6x 2 + 9xy + 7y 2 ) + (8x 2 + 2xy + 5y 2 ) =
6x 2 + 9xy + 7y 2 + 8x 2 + 2xy + 5y 2 =
6x 2 + 8x 2 + 9xy + 7y 2 + 8x 2 + 2xy + 5y 2 =
6x 2 + 8x 2 + 9xy + 7y 2 + 2xy + 5y 2 =
6x 2 + 8x 2 + 9xy + 2xy + 7y 2 + 2xy + 5y 2 =
6x 2 + 8x 2 + 9xy + 2xy + 7y 2 + 5y 2 =
(6x 2 + 8x 2 ) + (9xy + 2xy) + (7y 2 + 5y 2 ) =
14x 2 + 11xy + 12y 2
The correct answer is: a = 10
Click on the following link to learn how to solve algebraic equations for an unknown variable.
Deal with the constants.
3a + 8 + 2 = 4a – 2 + 2
3a + 8 + 2 = 4a – 2 + 2
Then get the variable on one side of the equation.
3a – 3a + 10 = 4a – 3a
3a – 3a + 10 = 4a – 3a
The correct answer is: 14
To solve algebra questions like 22 and 23, you need to memorize the perfect squares.
196 is a perfect square.
The correct answer is: 11
121 is a perfect square.
The correct answer is: 3x 2 – 12xy + 11y 2
To find the solutions for algebra equations like this one, learn how to subtract polynomials.
(5x 2 – 3xy + 4y 2 ) – (2x 2 + 9xy – 7y 2 ) =
5x 2 – 3xy + 4y 2 – 2x 2 – 9xy – –7y 2 =
5x 2 – 3xy + 4y 2 – 2x 2 – 9xy + 7y 2 =
5x 2 – 2x 2 – 3xy + 4y 2 – 2x 2 – 9xy + 7y 2 =
5x 2 – 2x 2 – 3xy + 4y 2 – 9xy + 7y 2 =
5x 2 – 2x 2 – 3xy – 9xy + 4y 2 – 9xy + 7y 2 =
5x 2 – 2x 2 – 3xy – 9xy + 4y 2 + 7y 2 =
(5x 2 – 2x 2 ) + (–3xy – 9xy) + (4y 2 + 7y 2 ) =
3x 2 + –12xy + 11y 2 =
3x 2 – 12xy + 11y 2
The correct answer is: 3(3a – b + 5c 2 )
Click on this link for help with Accuplacer math prep questions on factoring.
(3 × 3)a – (3 × 1)b + (3 × 5)c 2 =
The correct answer is: 8x 2 + 4x + 2
Get help with Accuplacer math prep questions like this on dividing polynomials.
(16x 3 + 8x 2 + 4x) ÷ (2x) =
(16x 3 ÷ 2x) + (8x 2 ÷ 2x) + (4x ÷ 2x) =
(16x 3 ÷ 2x) + (8x 2 ÷ 2x) + (2) =
(16x 3 ÷ 2x) + (4x) + 2 =
The correct answer is: a = 3 and b = –1
To solve algebra problems like this, you need to learn the elimination method.
Our algebra equations were:
Multiply the second equation by 5.
(2a × 5) – (b × 5) = (7 × 5)
Now perform the elimination method by adding this to the first equation.
13a ÷ 13 = 39 ÷ 13
Now substitute the value of a to solve for b.