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

If you think that you need further help with math, you can study the problems on the math sample page when you have completed the questions here.

You can also check out additional practice questions for Algebra and College Math.

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

## 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 (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}), the midpoint is:

(*x*_{1} + *x*_{2}) ÷ 2 , (*y*_{1} + *y*_{2}) ÷ 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 geometry 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 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*)! × *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*)! × *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 × 4 × 3 × 2 × 1

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

(5 × 4 × 3 × 2 × 1) ÷ [(5 − 2)! × 2!] =

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

120 ÷ (6 × 2) =

120 ÷ 12 = 10

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

The final math exercises in this series can be found at: