Algebra Placement Test - Free CPT Algebra Problems

Looking for a free algebra placement test? You will see a sample algebra test below.

The college placement math test in quantitative reasoning, algebra, and statistics is one part of the CPT exam.

The quantitative reasoning, algebra, and statistics part of the test contains twenty questions.

Here is a free sample from our CPT math download.

Free Algebra Placement Test Questions

Example 1:

Polynomials are algebraic expressions that contain more than one term.

You will definitely have CPT algebra problems about multiplying polynomials on the exam.

(2a − 3b)² = ?

Example 2:

This is a question on inequalities. Inequalities contain the less than or greater than signs.

Algebra questions on the CPT will include inequalities problems like this one.

35 − ¹/₄X > 31, then X < ?

Example 3:

You will also see exam problems on the division of polynomials.

In order to solve this type of problem, you must do long division of the polynomial:

(a² + a − 12) ÷ (a − 3) = ?

Free Algebra Placement Test – Solutions

Solution 1:

When multiplying polynomials, you should use the F-O-I-L method.

This means that you multiply the variables two at a time from each of the two parts of the equation in the parentheses in the order First – Outside – Inside – Last

(2a − 3b)² = (2a − 3b)(2a − 3b)

So, the first variables in each set of parentheses are 2a and 2a.

FIRST: 2a × 2a = 4a²

The variables on the outside of each set of parentheses are 2a and 3b.

OUTSIDE: 2a × −3b = −6ab

The variables on the inside of each set of parentheses are −3b and 2a.

INSIDE: −3b × 2a = −6ab

The last variables in each set of parentheses are −3b and −3b.

LAST: −3b × −3b = 9b²

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

4a² − 6ab − 6ab + 9b² =

4a² − 12ab + 9b²

Solution 2:

First, deal with the whole numbers on each side of the equation first:

35 − ¹/₄X > 31

(35 − 35) − ¹/₄X > (31 − 35)

−¹/₄X > −4

Then deal with the fraction:

−¹/₄X > −4

4 × −¹/₄X > −4 × 4

−X > −16

Then, deal with the negative number:

−X > −16

−X + 16 > −16 + 16

−X + 16 > 0

Finally, isolate the unknown variable as a positive number:

−X + 16 > 0

−X + X + 16 > 0 + X

16 > X

X < 16

Solution 3:

Remember, you must to long division of the polynomial until you have no remainder.

>>>>>>>a + 4

a − 3)a² + a − 12

>>>>>−(a² − 3a

>>>>>>4a − 12

>>>>>−(4a − 12

>>>>>>>>>>>0