# Advanced Algebra and Geometry for the Accuplacer

Advanced algebra and geometry problems are included in the college-level math section of the Accuplacer.

Geometry questions will cover: midpoints, x and y intercepts, slope, arcs, chords, cones, and triangles.

To see a FREE sample of our Accuplacer practice tests, please click on the "Free Accuplacer Practice Test" button on the main menu.

Example 1:

Find the coordinates (x, y) of the midpoint of the line segment on a graph that connects the points (5, 2) and (1, -10).

Solution 1:

In order to find midpoints on a line, you need to use the following formula:

For two points on a graph (x1, y1) and (x2, y2), the midpoint is:

(x1 + x2) ÷ 2 , (y1 + y2) ÷ 2

Now calculate for x and y:

(5 + 1) ÷ 2 = midpoint x, (2 + -10) ÷ 2 = midpoint y

6 ÷ 2 = midpoint x, -8 ÷ 2 = midpoint y

-3 = midpoint x, -4 = midpoint y

(3, -4)

Example 2:

Consider the vertex of an angle at the center of a circle. The diameter of the circle is 4. If the angle measures 90 degrees, what is the arc length relating to the angle?

Solution 2:

To solve this Accuplacer problem, you need these three principles:

(1) Arc length is the distance on the outside (or circumference) of a circle.

(2) The circumference of a circle is always π times the diameter.

(3) There are 360 degrees in a circle.

The angle in this problem is 90 degrees.

360 ÷ 90 = 4

In other words, we are dealing with the circumference of 1/4 of the circle.

Given that the circumference the circle is 4π, and we are dealing only with 1/4 of the circle, then the arc length for this angle is:

4π ÷ 4 = π

Example 3:

You will have at least one question on combinations or permutations on the Accuplacer Test.

How many 2 letter combinations can be made from the letter set?:
H A N D Y

Solution 3:

Tip 1: Remember that a combination, unlike a permutation, does not take into account the order of the items in the combination.

For example, the combination H A is considered the same as the combination A H.

To determine the number of combinations of S at a time that can be made from a set containing N items, you need this formula:

(N!) ÷ [(N - S)! x S!]

In the example above, S = 2 and N = 5

Remember: S represents how many letters each combination should contain. Each combination will contain 2 letters in this exercise, so S = 2.

N represents total set (H A N D Y in this example). So, N = 5 because there are five letters in the set.

Here is the formula again: (N!) ÷ [(N - S)! x S!]

Tip 2: The exclamation point means that you have to multiply the stated number by every number less than it. For example, 5! = 5 x 4 x 3 x 2 x 1

Now substitute the values for S and N and carry out the operation represented by the exclamation point:

(5 x 4 x 3 x 2 x 1) ÷ [(5 - 2)! x 2!] =

(5 x 4 x 3 x 2) ÷ [(3 x 2 x 1) x (2 x 1)] =

120 ÷ (6 x 2) =

120 ÷ 12 = 10

So 10 two-letter combinations can be made from a five letter set.