Praxis Mathematics: Content Knowledge (5161) Ultimate Guide2019-12-09T20:35:13+00:00

Praxis Mathematics: Content Knowledge (5161) Ultimate Guide and Practice Test

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Praxis Mathematics: Content Knowledge Quick Facts

The Praxis Mathematics: Content Knowledge test is created to assess the mathematical skills of a novice secondary mathematics teacher. Examinees will be assessed on their ability to see patterns, make conjectures, reason mathematically, solve multi-step problems, create mathematical models from real-world problems, and solve problems in a variety of mathematical areas.

Format: 

The Praxis Mathematics: Content Knowledge test consists of a total of 60 questions and lasts 150 minutes. The questions are made up of two categories. Category one covers number and quantity, functions, algebra, and calculus and makes up the majority of the exam with approximately 41 questions. Category two covers geometry, probability and statistics, and discrete mathematics and is tested in about 19 questions. This exam includes different types of test questions, including drag-and-drop, selected response with one answer or more than one answer, numerical entry questions, and text completion as well as other test formats.

An on-screen calculator is provided.

Cost: 

$120

Scoring: 

The score range for this exam is 100-200. As of 2019, the average performance range is 137-169. Different states require different passing scores on this exam. Maryland and Virginia, for example, require a 160, while South Carolina requires a 150. Be sure to check the requirements for your state of certification.

Study time: 

The Praxis Mathematics: Content Knowledge Exam is a comprehensive test that covers topics from discrete mathematics, algebra, and geometry through calculus and statistics. Due to the broad range of topics tested on this exam, it is strongly encouraged that you spend many weeks, or even months, preparing. Taking practice tests and continuing to improve your skills in areas of weakness will help increase your chances of success.

What test takers wish they’d known: 

  • An on-screen calculator is provided.
  • You shouldn’t round answers until the last step of the problem.
  • The test score is based on the number of questions that are correct, so it is better to guess than leave an item blank. 
  • You may skip a question and return to it later in the test time frame.
  • Scratch paper is provided but all answers must be entered into the computer to be graded.
  • You should request any needed disability or health-related accomodations in advance.
  • Food, water, and personal items should not be brought into the test center. You should familiarize yourself with which items are allowed in the test center.
  • You’ll need to bring a valid photo identification.

I: Number and Quantity, Algebra, Functions, and Calculus

Overview

This content category has about 41 questions. These questions account for 68% of the entire exam.

Let’s explore some specific concepts from each of the sections from this content category.

Number and Quantity

Order of Operations

In order to evaluate a mathematical expression, the set of rules to apply is called the order of operations. The order of operations can best be remembered using the acronym “PEMDAS” which describes the order in which to evaluate the expression.

First, evaluate anything in parentheses (or brackets). Next, evaluate any exponents (this includes radicals that can be rewritten as exponents). The third rule is to evaluate using multiplication or division. When working through this step, evaluate the multiplication or division by working left to right. Finally, evaluate any addition or subtraction. Again, work from left to right.

Example 1: Evaluate 3 + 4 x 5 – 6 ÷ 3

First, evaluate the multiplication and division by working left to right. So 4 x 5 = 20 and 6 ÷ 3 = 2. The expression is then 3 + 20 – 2. Now add and subtract the terms from left to right to find the result of 23 – 2 = 21.

Example 2: Evaluate (3 x 7 – 2 )² – 3

First, evaluate what is in the parentheses. Inside the parentheses are both multiplication and subtraction. Using PEMDAS, multiply first and then subtract, so that (3 x 7 – 2) = (21 – 2) = 19. After working in the parentheses, apply the exponent (19)² = 361. Finally, subtract the 3 on the outside: 361 – 3 = 358.

Example 3: Evaluate √(3 x 5 + 3² +1)

Recall that the square root is really an exponent of ½, so evaluate the exponent. There is an expression inside of the radical, so we must evaluate that using order of operations. First, evaluate 3² = 9. Next, evaluate 3 x 5 = 15. The expression now reads √(15 + 9 -1). Evaluate addition and subtraction from left to right so that 15 + 9 = 24, and then 24 + 1 = 25. Now, take the √(25) = 5.

Polynomial Identities Using Complex Numbers

Polynomial Identities can be extended into complex numbers. Many of these involve factoring.

Recall that √(-1) = i and i² = -1.

One identity allows factoring of the sum of two squares: x² + y² = (x + yi)(x – yi).

Example 1: 4x² + 9y² = (2x + 3yi)(2x – 3yi)

Example 2: Factor x⁴ + 5x² + 4

First factor the trinomial into (x² + 4)(x² +1). Now, using the polynomial identity, each binomial can be factored further:

(x² + 4) = (x + 2i)(x – 2i)
(x² + 1) = (x + i)(x – i)
Finally, x⁴ + 5x² + 4 = (x² + 4)(x² + 1) = (x + 2i)(x – 2i)(x + i)(x-i).

Scientific Notation

Scientific notation involves numbers that are written as the product of a number from 1 to 10 and a power of 10. Scientific numbers can be added, subtracted, multiplied, and divided.

Adding or Subtracting Scientific Numbers

Example: (4.3 x 10²) + (3.2 x 10⁵)

To add or subtract scientific numbers, first rewrite all the number(s) in the same power of 10 as the number that has the higher power of 10. If you are converting to a different power of 10, the decimal point will need to be moved.

Since (4.3 x 10²) has the smaller exponent, rewrite it as (0.0043 x 10⁵). Notice that the decimal point moved three places to the left.

Now the problem is (0.0043 x 10⁵) + (3.2 x 10⁵).

Since the powers of 10 are the same, the numbers can be added, so 0.0043 + 3.2 = 3.2043.

The result is (3.2043 x 10⁵).

Multiplying Scientific Numbers

Example: (4.8 x 10⁸)(3.5 x 10⁴)

To multiply scientific numbers, multiply the numbers and then multiply the powers of 10. So 4.8 x 3.5 = 16.8.

When multiplying the powers of 10, apply the rules of exponents by adding the powers: 10⁸ x 10⁴ = 10⁸⁺⁴ = 10¹².

Now (4.8 x 10⁸)(3.5 x 10⁴) = (16.8 x 10¹²).

Notice that 16.8 is not between 1 and 10 and therefore this is not in proper scientific notation. Move the decimal point to the left one place and add one to the exponent in the power of 10.

 (4.8 x 10⁸)(3.5 x 10⁴) = (16.8 x 10¹²) = (1.68 x 10¹³).

Dividing Scientific Numbers

Example: (5.2 x 10⁷) ÷ (2.6 x 10³)

To divide scientific numbers, divide the numbers and then divide the powers of 10: 5.2 ÷ 2.6 = 2.

When dividing the powers of 10, apply the rules of exponents and subtract the powers: (10⁷) ÷ (10³) = 10⁷⁻³ = 10⁴.

(5.2 x 10⁷) ÷ (2.6 x 10³) = 2 x 10⁴

Algebra

Add, Subtract, and Multiply Polynomials

To add or subtract polynomials, be sure to combine like terms.

Example 1: (4x² – 5x + 2) + (3x² + 7x – 3)

Add like terms so that 4x² + 3x² = 7x², -5x + 7x = 2x, and 2 + -3 = -1.
So (4x² – 5x + 2) + (3x² + 7x – 3) = 7x² + 2x – 1.

Example 2: (5xy – 7x + 4y – 2) – (3xy – 2y + 3x + 5)

Subtract like terms so that 5xy – 3xy = 2xy:

-7x – 3x = -10x, 4y – (-2y) = 6y and -2 – 5 = -7.
So (5xy – 7x + 4y – 2) – (3xy – 2y + 3x + 5) = 2xy – 10x + 6y – 7.

To multiply polynomials, be sure to multiply each term in one polynomial by the terms in the other polynomial. For binomials, the acronym FOIL (first, outer, inner, last) is often used to make sure all terms are multiplied.

Example 1: (8x – 3)(2x + 4)

Since this is a binomial multiplied by a binomial, FOIL can be used. 
Multiply (8x) x (2x) = 16x².
Then (8x) x (4) = 32x.
Next, multiply (-3) x (2x) = -6x, then (-3) x (4) = -12.
Putting all terms together yields (8x – 3)(2x + 4) = 16x² + 32x – 6x – 12 = 16x² + 26x – 12.

Example 2: (2x – 3)(3x² + 4x – 2)

Multiply 2x by each term in (3x² + 4x – 2), 2x(3x² + 4x – 2) = 6x³ + 8x² – 4x.
Next, multiply -3 by each term in (3x² + 4x – 2): -3(3x² + 4x – 2) = -9x² – 12x + 6.
Now, combine all terms: 6x³ + 8x² – 4x – 9x² – 12x + 6 = 6x³ – x² – 16x + 6.

Completing the Square

Completing the square is a strategy used to help solve quadratic equations.

A quadratic function is written in the form f(x) = ax² + bx + c.

When completing the square, the “b,” or coefficient of x, is divided by 2 and then squared. This helps create a perfect square trinomial and allows an equation to be written in an equivalent form, (x – p)² = q.

Example 1: Write x² + 6x + 2 = 7 into (x – p)² = q form

Step 1: Bring constants to the right side of the equation and leave a space on the left side:
x² + 6x + ____ = 7 – 2 so x² + 6x + ____ = 5.

Step 2: Take the “b” and divide by 2, then square. 
For this equation, b = 6 so (6 ÷ 2)² = 9. Add this value to both sides of the equation.
Now, x² + 6x + 9 = 5 + 9 so x² + 6x + 9 = 14.

Step 3: Factor the perfect square trinomial to put the equation in (x – p)² = q form.
(x + 3)² = 14.

Sometimes, the coefficient of x² is not equal to 1.

Example 2: 3x² + 4x – 9 = 2

Step 1: Bring constants to the right side of the equation and leave a space on the left side.
3x² + 4x + _____ = 11

Step 2. Factor out the “a” from the left side. This will change the “b” in the parentheses.
3(x² + (4/3)x + ___) = 11

Step 3: Complete the square inside the parentheses. Here, b = 4/3:
(4/3) / 2 = 4/6 = ⅔. After dividing by 2, square the result: (⅔)² = 4/9. 
Place this value in the space on the left side: 3(x² + (4/3)x + (4/9)) = 11

Step 4: To balance the equation, the value needs to be added to the right-hand side. However, since there is a 3 on the outside of the parentheses, 4 x (4 / 9) = 16 / 9 was really added and this is what needs to be added to the right side.
3(x² + (4 / 3)x + (4 / 9)) = 11 + (16 / 9)
3(x² + (4 / 3)x + (4 / 9)) = (99 / 9) + (16 / 9)
3(x² + (4 / 3)x + (4 / 9)) = 115 / 9

Step 5. Divide both sides by the coefficient that was factored out, 3.
x² + (4 / 3) x + (4 / 9) = (115 / 9) ∙ (⅓)
x² + (4 / 3) x + (4 / 9) = 115 / 27

Step 6: Factor the perfect square trinomial to put the equation in (x – p)² = q form.
(x + ⅔)² = 115 / 27

Systems of Linear Equations

A system of two linear equations can be solved graphically or algebraically.

Example: y = 2x + 3 and y = -x

Graphically: Graph each line and find the intersection.
Solution: (-1, 1)

Algebraically:

Step 1: Solve each equation for y if needed.

Step 2: Set the two equations equal.
2x + 3 = -x

Step 3: Combine like terms to solve for x.
2x + 3 + x = -x + x
3x + 3 = 0
3x = -3
x = -3/3
x = -1

Step 4: Substitute the value of x into one of the original functions.
y = 2x +3
y = 2 (-1) +3
y = -2 +3
y = 1

Step 5: Check the solution in the other equation.
y = -x
1 = -(-1)
1 = 1

The solution is (-1, 1)

More complicated linear systems of equations can be solved algebraically using elimination (addition method) or substitution.

Example: Find the solution of the system 2x + y = 5 and 4x – y = 1.

Using elimination:

Step 1: Make sure the same term in each equation has the same coefficient, one positive and one negative, so the terms can cancel. If they are not the same number with opposite signs, multiply one or both equations to achieve this.
2x + y = 5
4x – y = 1

For this system, the y terms have a coefficient of one and opposite signs.

Step 2: Add the two equations vertically to combine like terms.
6x = 6

Step 3: Solve for the variable.
x = 1

Step 4: Substitute into one of the original equations to find the other variable.
2x + y = 5
2(1) + y = 5
y = 5 – 2 = 3

Step 5: Check your solution in the other equation. 
4x – y = 1
4(1) – 3 = 1

Step 6: Write the solution as an ordered pair.
(1, 3)

Now solve the same system using substitution.

Step 1: Solve one equation for one of the variables. 
2x + y = 5
y = -2x +5

Step 2: Substitute this equation into the other equation in the system.
4x – y = 1
4x – (-2x + 5) = 1

Step 3: Combine like terms to solve for the other variable.
6x – 5 = 1
6x = 6
x = 1

Step 4: Substitute the variable found into one of the original equations to solve 

for the other variable.
2x + y = 5
2(1) + y = 5
y = 5 – 2 = 3

Step 5: Check your solution against the other original equation that wasn’t just used.
4x – y = 1
4(1) – 3 = 1

Step 6: Write your solution as an ordered pair.
(1, 3)

Functions

Domain and Range

The domain of a function is the set of all of the x-values of the function. The range of a function is the set of all y-values of the function.

Example: Identify the domain and range of the function f(x) = x² +3.

To find the domain by looking at the graph of the function, look at the values from the left to the right. The domain is all real numbers. The y-values of the range start at y = 3 and extend up to infinity, so the range is y ≥ 3.

Writing a Function

A function relates two quantities, an input value and an output value. A function can be written from a table, graph, or situation.

Table: 
Write a function rule that describes the data in the table below.

First find the slope. Since the y values increase by two as every x value increases by 1, the slope is 2 and this table of values represents a linear function. Find the y-intercept by counting back in the table to find the value where x = 0. The y value there will be -1. The function rule for this table is f(x) = 2x – 1.

From a graph:

First, identify that slope by looking at the slope between points using rise/run. From one point to another, rise 3 then run right 2.

Next, identify the y-intercept. For this graph, the y-intercept is 1.5.

The equation for the graphed function is f(x) = (3/2)x + (1.5).

From a word problem: Sue has $240 in her savings account and earns $6 a week for her allowance, which she adds to her savings. Write a function that can determine the amount of savings she has after x weeks. 

For the function, the y-intercept, or initial value is 240. The slope or change is 6 so the function rule for this situation is f(x) = 240 + 6x.

Inverses

A function has an inverse if each y-value only has one x-value. On the graph, the function should pass the horizontal line test, which says that a horizontal line should not touch the graph of a function more than once if the function has an inverse.

Example 1: Find the inverse of f(x) = 2x – 3

Step 1: Call f(x) = y.
y = 2x – 3

Step 2: Switch the x and the y.
x = 2y – 3

Step 3: Solve for the “new” y.
x + 3 = 2y
(x+3) / 2 = y

Step 4: Call this “new y” the inverse with the notation f⁽⁻¹⁾⁽ˣ⁾.
f⁽⁻¹⁾⁽ˣ⁾ = (x+3) / 2

Example 2: Find the inverse for f(x) = (x – 2) / (x + 1).

Step 1: Call f(x) = y.
y = (x – 2) / (x + 1)

Step 2: Switch the x and y variables.
x = (y – 2) / (y + 1)

Step 3: Solve for the “new” y.
x (y + 1) = y – 2 (get rid of the rational expression by multiplying)
xy + x = y – 2 (distribute)
xy – y = -2 – x (combine terms with the y on one side of the equation)
y (x – 1) = -2 – x (factor out the y)
y = (-2 – x) / (x – 1) (solve for y)

Step 4: Call this “new y” the inverse with the notation f⁽⁻¹⁾⁽ˣ⁾.
f⁽⁻¹⁾⁽ˣ⁾ = (-2 – x) / (x – 1)

Calculus

Approximate a Derivative

A table of values can be used to approximate the derivative over a period of time or at a particular point on a function.

Example 1: Given the table below, approximate the derivative between t = 2 and t = 5.

To find an approximation of the derivative from a table, use two points from the table to find the slope of the secant line.

Using the points (2, 3) and (5, 6), find the slope between the two points to approximate the derivative over that time period. 
m = (6-3) / (5-2) = 3 / 3 = 1

Example 2: Using the same table above, approximate the derivative at t = 2.

Use two points around t = 2 or the point at t = 2 and another point given that is close to t = 2.
From the table, let’s use the points (2,3) and (3, 7).
The slope of the secant line between these points is m = (7 – 3) / (3 – 2) = 4 / 1 = 4
The approximate derivative at t = 2 is f’(2) = 4.

Apply Derivatives

There are a variety of rules to find the derivative of functions.

Power rule: If f(x) = xⁿ, then f’(x) = n(xⁿ⁻¹)

Example: Find the derivative of f(x) = x⁴ – 5x³ + 3x² – 9x + 8

For each term, the exponent comes down to multiply by the term’s coefficient, then subtract one from the exponent. Recall: The derivative of a constant is 0.

f’(x) = 4x⁴⁻¹ – 5(3)x³⁻¹ + 3(2)x²⁻¹ – 9(1)x¹⁻¹
f’(x) = 4x³ – 15x² + 6x – 9

One application of derivatives is related rates. Related rates require that the derivative of each variable is taken with respect to time.

Example: A tanker is leaking oil in a circular pattern whose radius increases at a rate of 10 feet per minute. At the instant when the radius is 45 feet, how fast is the area of the spill increasing?

Step 1: Create a formula. The formula for the area of the circle is A = πr².

Step 2: Substitute in any constant values. This problem does not have any unchanging values.

Step 3: Take the derivative of the formula with respect to time.
(dA / dt) = 2πr(dr / dt)

Step 4: Substitute rates and any “at the given instant” values to solve the problem.
(dA / dt) = 2π(45)(10) = 900π square feet.

Note that in related rates, we represent a decreasing rate with a negative sign.

Infinite Series

A series is a sum of terms. An infinite series converges if the numbers in the sum approach a definite sum. An infinite series diverges if the numbers in the sum approach infinity.

Example 1: Determine if the series 1+1+1+1+1+… converges or diverges.

Since the numbers keep adding together to make a larger sum, this series diverges to infinity.

Example 2: Determine if the series 3 + (3/2) + (¾) + (⅜) + (3/16) + …  converges or diverges.

Since the numbers are getting smaller, the series is approaching a definite sum and converges. Note: This is a geometric series with common ratio of ½.

II: Geometry, Probability and Statistics, and Discrete Mathematics

Overview

This content category has about 19 questions. These questions account for 32% of the entire exam.

Let’s explore some specific concepts from each of the sections from this content category.

Geometry

Pythagorean Theorem

The Pythagorean theorem is used to find the length of a missing side in a right triangle.

a² + b² = c²

Example 1: Solve for x.

In the example, the hypotenuse, or “c,” is missing.
Substitute the known values into the formula: 8² + 6² = x².
Simplify: 64 + 36 = x².
Solve for x by taking the square root of both sides: 100 = x²
x = 10.

Example 2: Solve for x.

In the example, the length of one of the legs is missing.
Substitute known values into the formula: x² + 15² = 17².
Simplify: x² + 225 = 289.
Isolate the term x²: x² = 64.
Solve for x by taking the square root of both sides: x = 8.

Angle Relationships

When two parallel lines are cut by a transversal, a variety of angle relationships are formed.

Vertical angles are congruent and make an “x” shape. In the figure above, the pairs that are vertical angles are 1 and 4, 2 and 3, 5 and 8, and 6 and 7.

Linear pairs share a line and are supplementary, meaning that together they have a sum of 180 degrees. In the figure above, the linear pairs are 1 and 2, 1 and 3, 2 and 4, 3 and 4, 5 and 6, 5 and 7, 6 and 8, and 7 and 8.

Same side interior angles or consecutive interior angles are supplementary, meaning that together they equal 180 degrees. In the figure above, pairs of these angles are 3 and 5 and 4 and 6.

Corresponding angles are congruent. They must be in the same position (above or below) the parallel lines and on the same side of the transversal line. In the figure above, the corresponding angles are 1 and 5, 3 and 7, 2 and 6, and 4 and 8.

Alternate exterior angles are congruent. They are on the outside of the parallel lines and on alternate sides of the transversal. In the figure above they are 1 and 8, and 2 and 7.

Alternate interior angles are congruent. They are on the inside of the parallel lines and on alternate sides of the transversal. In the figure above they are 3 and 6, and 4 and 5.

Example 1: Lines m and n are parallel and cut by a transversal, t. Find the value of x using the figure below.

Since these angles are alternate interior angles, they are congruent. Set the two angles equal to each other: 

2x – 1 = 3x + 4

Solve for x:

-1 = 3x – 2x + 4
-1 = x +4
-1 – 4 = x
x = -5

Probability and Statistics

Correlation and Causation

Correlation measures the relationship between two quantities. 

  • A positive correlation occurs when as one quantity increases, the other also increases.
  • A negative correlation occurs when as one quantity increases, the other decreases.
  • Correlation does not imply causation.

Example: As the temperature decreases, the amount of hot chocolate sold at a stadium increases. While the temperature decreasing is related to the amount of hot chocolate sold, it does not cause the hot chocolate to be sold.

Causation occurs when one event causes another to occur.

Example: The more jumping jacks Jill completes, the more calories she burns. The number of jumping jacks has a direct effect on how many calories are burned.

Determining Probability of a Compound Event

An event that is the result of two simple events taken together is called a compound event.

To find the probability of a compound event, multiply the probabilities of the simple events together.

Example 1: Monte rolls a die and flips a coin. What is the probability that the dice lands on 5 and the coin lands on heads?

The probability of rolling a 5 is ⅙. The probability that a coin lands on heads is ½. 

The probability of both events happening together, the compound probability, is (⅙)(½) = 1/12.

Example 2: Bryce chooses two cards from the same deck of cards. What is the probability he chooses a king, and then, without replacing the first card, chooses an ace?

There are 4 kings and 4 aces in a deck of 52 cards. So the initial probability of choosing a king is 4/52. If that card is not replaced, the probability of choosing an ace is 4/51.

To find the probability of choosing both in order, multiply the two events. The compound probability is (4/52)(4/51) = 16/2652 = 4/663.

Discrete Mathematics

Geometric Sequences

A geometric sequence is a sequence whose terms are found by multiplying the previous term by the same nonzero number, the common ratio.

An example of a geometric sequence is 4, 2, 1, ½, ¼, … where the common ratio between each term is ½.

Example: Find the 8th term in the following geometric sequence: 2, 6, 18, 54, …

Identify the key parts needed for the formula. The first term is 2, the common ratio is 3. Then, substitute into the formula.

A₈ = 2 ∙ 3⁸⁻¹ = 2 ∙ 3⁷ = 2,187

The 8th term of this geometric sequence is 2,187.

Permutations

The number of permutations of n objects by choosing k objects at a time is notated by the permutation formula:

where ! is the symbol for factorial. This formula comes from the fundamental counting principle, which finds the total number of outcomes of two or more events by multiplying the number of outcomes for each event together.

Example: An ice cream shop carries 10 flavors of ice cream. Customers can choose to put the ice cream in a waffle cone, sugar cone, or cup. How many options are there for how the customer can order their ice cream?

There are 10 options for the ice cream and three options for how to hold the ice cream. Using the counting principle, the number of options would be 10 x 3 = 30.

Example: For a three-course meal special at a restaurant, customers can choose from 5 appetizers, 6 main courses, and 3 desserts. How many options do customers have to create their meals?

Using the counting principle, multiply the options of each course:

5 x 6 x 3 = 90 options.

And that’s some basic information about the second content category.

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