Our Accuplacer Math Course download provides practice test problems in all of the skills that are covered on the test.

This page has free math samples from our download. If you would like further information on our math course download, please navigate to the second half of this page.

## Math Samples – Practice Test Problems

1) 2.37 + .123 + .005 = ?

2) 7.02 × 4.2 = ?

3) Two volunteers have offered their time to a charity for a fund-raising event. Shannon is going to work for half of the time. Terry is going to work for one-sixth of the time. What fraction represents the amount of time for which there are currently no volunteers?

4) If 6x − 3(x + 4) = 0, then x = ?

5) Simplify this algebraic expression:

6) (a + 4b)(a − b) = ?

7) Simplify the following:

8) Determine the amount of three-letter permutations that can be derived from the following set: Q R W X Y Z?

9) Determine the midpoint of the line that connects (−6, 3) and (2, −7).

10) The perimeter of a rectangle is 64 meters. If the width were increased by 2 meters and the length were increased by 3 meters, what is the perimeter of the new rectangle?

Here is an example of the solutions we provide with the math course.

## Math Practice Test Solutions and Explanations

1) The answer is: 2.498

You will be able to use scratch paper on the test. When you are working out the solution, you have to align the decimals by adding zeros where necessary like this:

2.370

0.123

__0.005__

2.498

2) The answer is: 29.484

Remember to position the decimal correctly after performing the multiplication.

The decimal is placed three numbers from the right hand side of the result because 7.02 has 2 decimal places and 4.2 has got only 1 decimal place. So you simply have to add these together to determine where to put the decimal in the final product.

The multiplication is done as follows:

7.02

__x 4.2__

1.404

__28.080__

29.484

3) The answer is: 1/3

The total amount of time that all of the potential volunteers are going to work will be equal to 100%. We can simplify 100% to 1 for purposes of making an equation to solve the practice problem.

In order to make the equation, we will assign S to Shannon and T to Terry. The work to be done by the other volunteers will be represented letter V.

Therefore, the equation needed to solve the problem is:

S + T + V = 1

Now substitute with the fractions that have been provided:

1/2 + 1/6 + V = 1

Then find the lowest common denominator (LCD) of the fractions.

1/2 is equal to 3/6 (1/2 × 3/3 = 3/6), so the LCD is 6.

Now substitute 3/6 for 1/2 in the equation:

1/2 + 1/6 + V = 1

3/6 + 1/6 + V = 1

4/6 + V = 1

U = 1 − 4/6

U = 2/6

Then simplify this as follows: 2/6 ÷ 2/2 = 1/3

4) The answer is: 4

First of all, you need to multiply the items in the parentheses:

6x − 3(x + 4) = 0

6x − (3x + 12) = 0

6x − 3x − 12 = 0

Next, isolate the integers to one side of the equation:

6x − 3x − 12 = 0

6x − 3x − 12 + 12 = 0 + 12

3x = 12

Then solve for x:

3x = 12

3x ÷ 3 = 12 ÷ 3

x = 12 ÷ 3

x = 4

5) The answer is:

You will often see algebraic expressions in this format on the exam:

These kinds of algebra problems are called binomials – “bi” since they have two terms inside each of the pairs of parentheses.

First, expand the equation like this:

Then you have to multiply the terms from each of the pairs of parentheses in the following order:

FIRST: Take the first term from each set of parentheses and multiply them to get a product.

OUTSIDE: Now look at the first set of parentheses and take the first term from there, and then look at the second set of parentheses and take the last term form there. In other words, these terms are positioned on the outer edge of each set of parentheses so they are on the outside.

INSIDE: Now find the product of the second term from the first pair of parentheses and the first term from the second pair of parentheses. This is called the inside because the terms are situated together at the inner part of the equation.

LAST: Now you need to multiply the final or last terms in each of the pairs of parentheses.

Then we add all of the above parts together to get:

The First-Outside-Inside-Last method of multiplying binomials is sometimes referred to with the acronym “FOIL”.

6) The answer is:

You will see several “FOIL” method questions on the test, so we are giving you another chance to practice “FOIL” below.

FIRST:

OUTSIDE:

INSIDE:

LAST:

Then add the four above products together like this to get the answer to this practice test question:

This practice test item covers exponent laws. In order to solve these problems, you need to determine whether the base numbers in each part of the equation are the same. 8 is the base number in this equation. Since the base numbers are the same, we need to subtract the exponents and leave the base number unchanged.

Remember that if you see multiplication of exponents, you need to add the exponents.

8) Determine the amount of three-letter permutations that can be derived from the following set: Q R W X Y Z?

The answer is: 120 permutations

To determine the amount of permutations of size *S* that can be derived from a set of *N* items, you need the permutations formula:

*N*! ÷ (*N* − *S*)! =

In this question *N* = 6 since there are six items in the letter set Q R W X Y Z.

*S* = 3 because the problems is asking you to find three-letter permutations.

The exclamation point means that you have to multiply the number by every integer that precedes it.

*N*! ÷ (*N* − *S*)! =

(6 × 5 × 4 × 3 × 2 × 1) ÷ (6 − 3) =

(6 × 5 × 4 × 3 × 2) ÷ (3 × 2 × 1)=

720 ÷ 6 =

120 permutations

Remember that the formula for permutations is distinct from the formula for combinations.

Permutations consider the order of the items in each set, but combinations do not.

So, A B C and B C A are different permutations.

However, A B C and B C A are the same combination.

9) Determine the midpoint of the line that connects (−6, 3) and (2, −7).

The answer is: (−2, −2)

The midpoint formula is (*x*_{1} + *x*_{2}) ÷ 2 , (*y*_{1} + *y*_{2}) ÷ 2

Note that *x*_{1} means the value of *x* from the first set of coordinates, while *x*_{2} represents the value of *x* from the second set of coordinates.

Now use the formula above to determine midpoint *x*:

(−6 + 2) ÷ 2 =

−4 ÷ 2 =

−2 = midpoint *x*

Then use the other part of the midpoint formula to determine midpoint *y*:

(3 + −7) ÷ 2 =

−4 ÷ 2 =

−2 = midpoint *y*

The answer is: (−2, −2)

10) The answer is 74 meters.

**Step 1** – Set up equations for the areas of the rectangles both before and after the change.

Express the width as the variable W and the length as variable L.

Remember that perimeter is calculated as two times the width plus two times the length, so the formula for perimeter is 2W + 2L = P

So the equation we need *before* the change is:

2L + 2W = 64

**Step 2** – Isolate variable W to one side of the equation:

2L + 2W = 64

2L − 2L + 2W = 64 − 2L

2W = 64 − 2L

2W ÷ 2 = (64 − 2L) ÷ 2

W = 32 − L

**Step 3** – Then set up the equation that will calculate the perimeter of the rectangle *after* the changes to its dimensions.

P = 2(L + 3) + 2(W + 2)

We add 3 to the length and 2 to the width because width is increased by 2 meters and length is increased by 3 meters, according to the facts in the question provided.

**Step 4** – Now substitute 32 − L from the W = 32 − L from the “before” equation above into the W in the equation below to arrive at the final solution:

P = 2(L + 3) + 2(W + 2)

P = 2(L + 3) + 2(32 − L + 2)

P = (2L + 6) + 2(34 − L)

P = 2L + 6 + 68 − 2L

P = 6 + 68

P = 74

The problems above are from our Accuplacer Math Course download, which contains 200 practice test problems.

## Accuplacer Math Course Download

Our self-study math course is a PDF download that includes 200 math practice test problems. The problems in our math course are not the same as those in any of our three practice tests.

So, if you are taking the complete exam in English, reading, and math, you would probably prefer the online version of our practice tests.

If you are taking the complete exam but think you need extra help with math, you would probably prefer to get the online tests plus our math course.

However, if you are testing only in math, you would probably prefer to get just the math course.

The math course includes Accuplacer math test practice problems with detailed explanations so you can learn how to solve each type of math question on the test.

**More information on our Accuplacer Math Practice Test Download**

The Accuplacer math practice test materials in our math course include all of the math formulas covered on the test, with step-by-step illustrations for each type of math problem.

Since it simulates the real test, our math course has problems in arithmetic, algebra, advanced algebra, geometry, and trigonometry.

You will learn how to crack even the toughest math problems and ace your math test with the tips, tricks, equations, formulas, and explanations in our publications.

In order for you to see the method we use in explaining the answers in our math practice test materials, here are some excerpts from our math course practice tests.

You will see all of the problems together at the start of our math practice test materials.

Just like on the examination, the problems will be scaled in order of difficulty from easiest to the most difficult.

So, you will see the arithmetic problems first, followed by algebra, and then finally you will have to solve the college-level math problems in advanced algebra, geometry, and trigonometry.

The solutions and explanations to the math practice test are provided at the end of the test.

To see some different sample problems from our Accuplacer practice tests, which are available in both online and download PDF formats, please visit the following links: