College Math Placement Test

College Math Placement Test – Set 1

Instructions: The college math placement test example quiz below contains 10 questions. You will see your score and the explanations when you have completed the test.

Then try the second set of college math placement test questions in the following section.

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College Math Placement Test Download

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

If you would like further information on our math PDF download, please navigate to the bottom of this page.

College Math Placement Test – Set 2

Instructions: Now try 10 more college math placement test questions. The answers and solutions are given in the next section.

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?

Placement Test Solutions & Explanations

Here is an example of the kinds of solutions we provide with our college math placement test download.

Question 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

Question 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

Question 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

Question 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

Question 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.

TO SOLVE: 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”.

Question 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:

TO SOLVE:

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

Question 7

The answer is:

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.

Question 8

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.

Question 9

The answer is: (−2, −2)

The midpoint formula is (x₁ + x₂) ÷ 2 , (y₁ + y₂) ÷ 2

Note that x₁ means the value of x from the first set of coordinates, while x₂ represents the value of x from the second set of coordinates.

The line connects (−6, 3) and (2, −7).

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)

Question 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 math download, which contains 400 practice test problems.

Math Placement – Get the Download

Our self-study college math placement test course is a PDF download that includes 400 math practice test problems.

Online Exams Plus Math PDF

If you are taking the exam and think you need extra help with math, you should try our free online tests plus our math downloads.

Why Get the Math Download?

If you are testing only in math, you would probably prefer to get our math download.

Detailed solutions help you learn!

The math course includes math test practice problems with detailed explanations.

It has step-by-step illustrations for each type of math problem.

So, you can learn how to solve each type of math question on the test.

More Information on our Math Download

Get all the math formulas on the test!

The math practice test materials in our math course include all of the math formulas covered on the test.

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

Learn how to solve the toughest math problems!

You will learn how to crack even the toughest math problems and ace your math test.

That’s because you get tips, tricks, equations, formulas, and explanations in our publications.

For you to see how we explain the answers in our math material, we have given 20 questions from our math download on this page.

Build your math skills!

You will see all of the 400 problems at the start of our college math placement test download.

Just like on the examination, the problems will be ordered from the easiest to the most difficult.

So, you will see the arithmetic problems first, followed by algebra.

Then finally you will have to solve the problems in advanced algebra, geometry, and trigonometry.

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

More College Math Placement Test Help

To see sample problems like those in our materials (which are available in online and download PDF formats) please visit the following links:

Arithmetic exercises

Algebra questions

Advanced algebra and geometry

Trigonometry