Page 1  Mathematics Study Guide for the HiSET Test
General Information
This test requires you to answer 55 questions in 90 minutes. The questions involve math content from the following areas, with approximately the noted level of emphasis:
 Numbers and Operations on Numbers (19%)
 Measurement/Geometry (18%)
 Data Analysis/Probability/Statistics (18%)
 Algebraic Concepts (45%)
Additionally, you’ll need to be fluent in these process skills to answer the questions correctly:
 Understand Mathematical Concepts and Procedures
 Analyze and Interpret Information
 Synthesize Data and Solve Problems
In this study guide, we will focus on the content first. You will not be able to get all the information you need in this guide, but you will have an idea of the topics and concepts you’ll need to know. Following that, we will outline the types of mathematical processes you’ll need to know. Be sure to seek extra help and practice with any content or procedures that give you trouble. Use our practice tests and flashcards for more information about just where you stand with mathematics before taking the test.
When taking the HiSET test, you will have access to a formula sheet. However, you must memorize the following three formulas because you may need them on the test and they are not on the formula sheet.
The Pythagorean Theorem:
\[{a^2} + {b^2} = {c^2}\]The Distance Formula:
\[d = \sqrt {{{\left( {{x_2}  {x_1}} \right)}^2} + {{\left( {{y_2}  {y_1}} \right)}^2}}\]The Quadratic Formula:
\[x = \frac{b \pm \sqrt{b^{2}  4ac}}{2a}\]The content covered by the questions on this test is wide and varied, touching several areas of mathematics. This study guide takes you through the types of questions you will see and explains the procedures necessary to solve problems and find answers. There are also example problems and situations within the text.
Numbers and Operations
In mathematics, we have numbers such as \(4,\;  3,\;  \frac{3}{4},\;\sqrt 7 ,\;16,\;\pi ,\;{\rm{and}}\;0.\) These numbers are real numbers, and we can add them together, subtract them, multiply them, and divide them (as long as we do not divide by \(0\)) In this section, we will explore various properties and concepts of all different types of numbers, and how to perform operations using these properties and concepts.
Properties of Operations
When we are performing operations with numbers, there is a special order in which we must perform the operations. Thus, we must remember to follow the order of operations (PEMDAS):
Do everything in parentheses (P), left to right.
Evaluate any exponents (E), left to right.
Do all multiplication and division MD), in order, left to right.
Then do all addition and subtraction (AS), in order, from left to right.
A good way to remember this order is:
Please Excuse My Dear Aunt Sally
Example:
Simplify this expression.
\(6 + 3(5 + 2)  2 \cdot 2\)
Add \(5 + 2\)
\(6 + 3(7)  2 \cdot 2\)
Multiply \(3(7)\) and \(2 \cdot 2\)
\(6 + 21  4\)
Add or subtract from left to right to get \(23.\)
Numbers belong to sets. The most common set of numbers is called the real numbers. These numbers are actually used in the “real world.” The set includes natural numbers, whole numbers, integers, rational numbers, and irrational numbers. An important note about the set of real numbers is that we can find them, or find where they would be, on the real number line.
The set of natural numbers is sometimes called the counting numbers. These numbers are the set: \(\{ 1,\,2,\,3,\,4,...\}\) This set does not include negative numbers, fractions, or decimals.
The set of whole numbers is the natural numbers plus \(0.\) These numbers are the set: \(\{ 0,\,1,\,2,\,3,\,4,...\}\) This set does not include negative numbers, fractions, or decimals.
The set of integers is the set of all positive numbers, negative numbers, zero, with no fractions or decimals. These numbers are the set: \(\{ ...  3, \, 2,\,  1,\,0,\,1,\,2,\,3,\,4,...\}\)
The set of rational numbers is the set of all numbers that can be written as a fraction. This set includes the previous sets of numbers, all terminating decimals, nonterminating repeating decimals, and radicals that have exact values. An example of this set of numbers is: \(\left\{ {  5,\,\frac{2}{3},\,0,7,\,\sqrt {25},\,2.\bar 4,\,\sqrt {0.04} } \right\}\)
The set of irrational numbers is the set of all numbers that cannot be written as a fraction. This set includes many radicals (that do not have exact values), and the two commonly used irrational numbers \(\pi\) and \(e.\) Examples of these numbers are in the set: \(\left\{ {\sqrt 5 ,\,\sqrt {26} ,\,\pi ,\,e, \, \sqrt 8 ,\,  \sqrt[3]{{25}}} \right\}\)
There are four basic mathematics operations, upon which all other processes are based. These operations are:

Finding a sum is the result of adding numbers, such as \(6 + 15 = 21\)

Finding a difference is the result of subtracting one number from another, such as \(18  7 = 11\)

Finding a product is the result of multiplying numbers, called factors of the product. Multiplication is sometimes shown by a raised dot \(\cdot\) instead of the times sign \(\times.\) Examples are \(5 \cdot 7 = 35\) or \(5 \times 7 = 35\)

Finding a quotient is the result of dividing one number by another number. Division is indicated by the symbol \(\div\), a horizontal bar, or a slanted fraction bar. Examples are \(8 \div 4 = 2\), \(\frac{8}{4} = 2,\) or \(8/4 = 2.\) It is important to note that you cannot divide by \(0\) because the result is undefined.
Properties of Exponents
Exponents are used as a shortcut for writing repeated multiplication. For example: \(4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\) is written as \({4^7}\) using exponents. In this example, \(4\) is the base, and \(7\) is the exponent.
Several points about exponents to remember are:

If an expression does not have a visible exponent, that expression is understood to have an exponent of \(1.\) For example, \(a = {a^1}\).

An exponent applies only to the term before it. For example, in the expression \(3{a^n},\) only \(a\) is raised to the \(n\)th power.

If a negative expression is raised to a power, the negative is not raised to the power unless it is enclosed in parenthesis with the numeral being raised. For example: \( {3^2} =  9\) but \({\left( {  3} \right)^2} = 9.\)
Let’s review the properties of exponents. These rules apply to all exponents, whether they are integers, fractions, or decimals:
 When multiplying terms with exponents, the terms must have the same base. The property, known as the product rule of exponents, tells us to add the exponents.
Example:
\[{a^m} \cdot {a^n} = {a^{m + n}}\]Using numbers:
\[\begin{array}{l}{3^2} \cdot {3^5} &=& {3^{2 + 5}}\\ &=& {3^7}\end{array}\] When dividing terms with exponents, the terms, again, must have the same base. The property, known as the quotient rule of exponents, tells us to subtract the exponent in the denominator from the exponent in the numerator.
Example:
\[\frac{{{a^m}}}{{{a^n}}} = {a^{m  n}}\]Using numbers:
\[\begin{array}{l}\frac{{{7^6}}}{{{7^3}}} &=& {7^{6  3}}\\ &=& {7^3}\end{array}\] When raising a product of two numbers to a power, we use the power of a product rule. This rule tells us to raise all parts of the product to the power.
Example:
\[{\left( {abc} \right)^n} = {a^n}{b^n}{c^n}\]Using numbers:
\[\begin{array}{l}{\left( {3xyz} \right)^2} &=& {3^2}{x^2}{y^2}{z^2}\\ &=& 9{x^2}{y^2}{z^2}\end{array}\] When raising a term with an exponent to a power, we use the power of a power rule. This rule tells us to multiply the two exponents.
Example:
\[{\left( {{a^m}} \right)^n} = {a^{mn}}\]Using numbers:
\[\begin{array}{l}{\left( {{2^3}} \right)^4} &=& {2^{3 \cdot 4}}\\ &=& {2^{12}}\end{array}\] When raising a term to the power of \(0\), we use the power property of 0. This rule tells us that anything raised to the power of \(0\) results in \(1.\) However, keep in mind other exponent rules that must be followed when we use this rule.
Examples:
\[{5^0} = 1,\;  {5^0} =  1,\;2{a^0} = 2 \cdot 1 = 2,\;{\left( {ab} \right)^0} = 1\]Using numbers:
\[{\left( {\frac{{32}}{5}} \right)^0} = 1\] If we have a negative exponent, we can remove the negative exponent using the negative exponent rule and then continue the process using the other exponent rules. This rule tells us that if we have a negative exponent, we can remove that negative on the exponent by moving the term with the exponent from the numerator to the denominator or from the denominator to the numerator.
Example:
\({a^{  m}} = \frac{1}{{{a^m}}}\) or \(\frac{1}{{{a^{  m}}}} = {a^m}\)
 Radicals can be written using exponents. The general radical expression \(\sqrt[m]{a}\) means the \(m\)th root of \(a.\) The \(m\) is called the index of the radical, and the \(a\) is called the radicand of the radical. This radical can be rewritten using the rational exponent rule, which tells us that the radical expression can be written with the radicand as the base with an exponent of \(1\) divided by the index as the exponent.
Example:
\[\sqrt[m]{a} = {a^{\frac{1}{m}}}\]Using numbers:
\[\sqrt[5]{{625}} = {625^{\frac{1}{5}}}\] If the radical has exponents inside it, then the exponent inside the radical is in the numerator of the rational exponent:
Example:
\[\sqrt[m]{{{a^n}}} = {a^{\frac{n}{m}}}\]Using numbers:
\[\sqrt[4]{{{7^3}}} = {7^{\frac{3}{4}}}\] The radical expression \(\sqrt a\) is read as the square root of \(a\). This radical expression has no visible index. It is understood that the index is \(2\) so the denominator of the rational exponent is \(2.\)
Example:
\[\sqrt a = {a^{\frac{1}{2}}}\]Using numbers:
\[\sqrt{17} = {17^{\frac{1}{2}}}\]As mentioned earlier, once a radical expression is converted to a rational expression, we can perform operations using the exponent rules.
Example:
\[\begin{array}{l}\sqrt[3]{6} \cdot \sqrt[4]{6} &=& {6^{\frac{1}{3}}} \cdot {6^{\frac{1}{4}}}\\ &=& {6^{\frac{1}{3} + \frac{1}{4}}}\\ &=& {6^{\frac{7}{{12}}}}\end{array}\]Scientific Notation
Scientific notation is a method of writing numbers that are very large or very small. Such numbers might be very difficult to visualize and perform operations with. The method of writing a number using scientific notation follows this format:
\(a \times {10^n}\) where \(1 \le a < 10\) and \(n\) is an integer
The integer \(n\) represents the number of places the decimal point is moved to create the correct \(a.\) If the number is larger than \(10,\) the decimal place is moved to the left, and \(n\) is positive. If the number is smaller than \(1,\) the decimal place is moved to the right, and \(n\) is negative.
Lastly, before we do some examples, please remember that scientific notation typically estimates very large and very small numbers. Thus, if we wanted to use the number \(9,274,258,752\) using scientific notation, we would probably change it to \(9,270,000,000\) depending on how accurate we want our calculations.
Examples:
\[7,600,000 = 7.6 \times {10^6}\]The requirement for \(a\) \(1 \le a < 10\) so we need \(a=7.6.\) This requires us to move the decimal point \(6\) places to the left, so the exponent of \(10\) is \(6.\)
\[0.000000045 = 4.5 \times {10^{8}}\]The requirement for \(a\) \(1 \le a < 10\) so we need \(a=4.5.\) This requires us to move the decimal point \(8\) places to the left, so the exponent of \(10\) is \(8.\)
To convert from scientific notation to standard notation, we do the opposite.
Example:
\[9.2 \times {10^8} = 920,000,000\]The exponent is positive \(8\), so we move the decimal point \(8\) places to the right.
\[5.3 \times {10^{  6}} = 0.0000053\]The exponent is negative \(6\), so we move the decimal point \(6\) places to the left.
We can multiply numbers in scientific notation. We do this using multiplication and the product rule for exponents.
Example:
\[\begin{array}{l}\left( {6.4 \times {{10}^4}} \right) \cdot \left( {1.2 \cdot {{10}^5}} \right) &=& \left( {6.4 \cdot 1.2} \right) \times {10^{4 + 5}}\\ &=& 7.68 \times {10^9}\end{array}\]When we multiply the \(a\) terms, we need to remember that \(a\) has the restriction \(1 \le a < 10\) so typically, we must adjust the decimal point after multiplying.
Examples:
\[\begin{array}{l}\left( {5.7 \times {{10}^8}} \right) \cdot \left( {6.1 \times {{10}^7}} \right) &=& \left( {5.7 \cdot 6.1} \right) \times {10^{8 + 7}}\\ &=& 34.77 \times {10^{15}}\\ &=& 3.477 \times {10^{16}}\end{array}\]Round to the nearest \(2\) decimal places:
\[\begin{array}{l}\left( {9.47 \times {{10}^6}} \right) \cdot \left( {6.15 \times {{10}^{17}}} \right) &=& \left( {9.47 \cdot 6.15} \right) \times {10^{6 + 17}}\\ &=& 58.2405 \times {10^{23}}\\ &=& 5.82 \times {10^{24}}\end{array}\]We can divide numbers in scientific notation. We do this using division and the quotient rule for exponents.
Example:
\[\begin{array}{*{20}{l}}{\frac{{8.4 \times {{10}^{15}}}}{{4.2 \times {{10}^6}}}}& = &{\frac{{8.4}}{{4.2}} \times {{10}^{15  6}}}\\{}& = &{2 \times {{10}^9}}\end{array}\]When we divide the \(a\) terms, we need to remember that \(a\) has the restriction \(1 \le a < 10\) so typically, we must adjust the decimal point after dividing.
Example:
\[\begin{array}{l}\frac{{1.43 \times {{10}^{13}}}}{{7.15 \times {{10}^{19}}}} &=& \frac{{1.43}}{{7.15}} \times {10^{13  19}}\\ &=& 0.2 \times {10^{  6}}\\ &=& 2 \times {10^{  7}}\end{array}\]When we divided the \(a\) values, the result was less than \(1.\) Therefore, we had to move the decimal to the left.
We can also multiply and divide in the same operation.
Example:
\[\begin{array}{l}\frac{{\left( {5 \times {{10}^7}} \right)\left( {1.8 \times {{10}^5}} \right)}}{{3 \times {{10}^8}}} &=& \left( {\frac{{5 \cdot 1.8}}{3}} \right) \times {10^{7 + 5  8}}\\ &=& 3 \times {10^4}\end{array}\]Problem Solving
We use mathematics to solve problems algebraically. The concepts we learn in mathematics transfer directly to realworld problems. To solve realworld problems, we should always use this strategy:
Step 1—Read and understand the problem. Read the problem carefully a few times. Decide what numbers are asked for and what information is given. Making a sketch may be helpful.
Step 2—Choose a variable and use it with the given facts to represent the number(s) described in the problem. Labeling your sketch or arranging the given information in a chart may help.
Step 3—Reread the problem. Then write an equation that represents the relationships among the numbers in the problem.
Step 4—Solve the equation and find the required number(s).
Step 5—Check your results with the original statement of the problem. Did you answer the correct question? Give the answer using units.
Example 1:
At noon a plane leaves Tampa Airport and heads north at \(180\) miles per hour. Its destination is Chicago, \(1,174\) miles away. To the nearest halfhour, what time does the plane arrive in Chicago. Note that Tampa is in Eastern Time Zone and Chicago is in Central Time Zone.
Answer: \(7:30\) p.m.
Explanation:
To find the length of time the plane takes to fly from Tampa to Chicago, write and solve the equation \(t = 1174 \div 180.\) Divide, resulting in \(t=6.52222.\) This number equates to \(6.5\) hours. Next, determine the time the plane took off in Central Time Zone. Central Time Zone is \(1\) hours behind Eastern Time Zone. The plane took off at \(11:00\) a.m. Last, add \(6.5\) hours to \(11:00\) a.m., giving \(5.5\) p.m. The plane arrived in Chicago at \(5:30\) p.m. Central Time.
Example 2:
A petroleum tanker is fully loaded with a capacity of \(5.3 \times {10^6}\) barrels. The ship begins offloading immediately upon arrival in port and inspection. The offloading time takes \(160\) hours. To the nearest \(1,000\), what is the pumping rate offloading the tanker?
Answer: \(33,000\) barrels per hour
Explanation:
Divide the capacity by the number of hours. \(\frac{{5.3 \times {{10}^6}}}{{160}} = 33,125.\) Next, round the answer to the nearest thousand: \(33,000\). Add units to the answer, resulting in \(33,000\) barrels per hour.
Example 3:
The mass of the Sun, \(2.0 \times {10^{30}}\) kg, is \(3.33 \times {10^5}\) times the mass of the Earth. What is the mass of the Earth? Give your answer using scientific notation rounded to one decimal place.
Answer: and \(6.0 \times {10^{24}}\) kg
Explanation:
Divide the mass of the Sun by the multiplier. \(\frac{{2.0 \times {{10}^{30}}}}{{3.33 \times {{10}^5}}}.\) The result is \(6.0 \times {10^{24}}.\) Add units to the answer, resulting in \(6.0 \times {10^{24}}\) kg.
Level of Accuracy
When solving a problem, we must determine which level of accuracy is appropriate. Different problems require different levels of accuracy. A common mistake many students make is treating every problem the same and trying to get an exact answer every time. Let’s look at a few examples to decide whether an exact answer is needed or if an estimate is okay.
Example 1:
Determine how many pizzas are needed to feed the high school football team.
Answer: Since we don’t know the exact number of slices each player will eat, or how many players will actually be there, an estimate is okay.
Example 2:
Ask each band member to reimburse the bandleader for the airfare on a recent trip.
Answer: We need an exact amount of money so everyone pays their actual cost.
Example 3:
Determine the diameter to manufacture the shaft for hydraulic cylinders.
Answer: Although there might be a tolerance, the hydraulic cylinder must not leak fluid. Therefore, we want an exact measurement of the diameter.
Example 4:
Each Friday morning Chef Bob purchases enough chicken for his restaurant to serve on the weekend. Based on the average from the past three weekends, he expects to serve \(85\) orders of dark meat and \(65\) orders of white meat. He orders chicken in cases that hold \(25\) orders of each type of meat. Is he going to buy the exact amount he expects to serve or an estimate?
Answer: Since he buys the chicken in bulk, he will have to buy the closest estimate of what he thinks he needs.
Example 5:
Jason and Deidra buy some new furniture. The total cost is \(\$ 2,656.88.\) They paid \(\$500\) down and will pay the balance in \(24\) equal payments with no interest. How much is their monthly payment? Is this an exact answer or an estimate?
Answer: \(\$89.87\) This is an exact answer because they must pay the entire amount, and they do not want to pay more than the entire amount. They will calculate the payment as follows:
\[\begin{array}{c}\$ 2656.88  \$ 500 &=& \$ 2156.88\\\$ 2156.88 \div 24 &=& \$ 89.87\end{array}\]Example 5:
Jack wants to add builtin bookshelves along a certain wall in his office. The wall is \(8\) feet tall and he plans to install \(4\) evenly spaced \(1\frac{1}{2}\) thick shelves. How much material should he buy? Does this problem involve exact measurements or approximate measurements?
Answer: Both. He probably doesn’t want to run out of material while he is building the shelves, so he will most likely round up, estimating the amount of material he needs. Then, when he is building the shelves, he will want them to be spaced correctly, so he will make exact measurements both cutting the material and positioning the material on the wall.