Using the TI-30XS Calculator to pass the GED - FINAL

Copyright © 2020 by Advanced Computing, LLC

All rights reserved. No part of this book may be used or reproduced in any manner without the written

permission of the publisher.

This edition published by Advanced Computing, LLC

Written by Jeremy F. Tinsley, M.Ed.

Cover Design: Template used from www.canvas.com

Printed in the United States of America

ISBN-13: 978-0-9829871-1-7

GED® and GED Testing Service® are registered trademarks of the American Council on Education (ACE)-,

which is not affiliated with this book. This material is not endorsed or approved by ACE or GED Testing

Service.

Excerpts from TI-30XS Multiview – A Guide for Teachers used with permission from GED and GED

Testing service

Disclaimer

Although the author and publisher have made every effort to ensure the information in this book was

correct at press time, the author and publisher do not assume and hereby disclaim any liability to any

party for any loss, damage, or disruption caused by errors or omissions, whether such errors or

omissions result from negligence, accident, or any other cause.

The information provided in this book is designed to provide helpful information on the subjects

discussed. This book does not guarantee or imply successful passing of the GED Math exam.

Advanced Computing, LLC

Dedication

I dedicate this book to all the individuals who at one time or another thought they couldn’t do it! This is

your time!

To my children, the world is your oyster. Whatever you choose to do, DO YOUR BEST!

To my loving wife, without you I would not be the man I am today.

To my mom and dad, I thank you most of all for teaching the importance of LOVE in this cold world!

And finally, the over two thousand students which I have instructed, “STOP PLAYING”! Go GET IT!

Jeremy F. Tinsley, M.Ed.

Table of Contents

Dedication .................................................................................................................................................... 3

Table of Contents ......................................................................................................................................... 4

Introduction ................................................................................................................................................. 7

Assumptions................................................................................................................................................................................. 8

Algebraic Expressions................................................................................................................................................................... 8

Algebraic Equations ..................................................................................................................................................................... 9

Algebraic Inequalities ................................................................................................................................................................... 9

Functions.................................................................................................................................................................................... 11

Getting Familiar with the TI-30XS ........................................................................................................... 12

Using the calculator on your exams ........................................................................................................................................ 12

Accessing the Calculator on-screen ......................................................................................................................................... 12

Don’t have the TI-30XS calculator ........................................................................................................................................... 12

The TI-30XS Basics .................................................................................................................................. 13–

TI-30XS BASICS (cont’d).................................................................................. Error! Bookmark not defined.

Calculator Reference Sheet ...................................................................................................................................................... 16

The "First Five" ............................................................................................................................................ 17

Lesson 1 – Number Sense, Operations, and Rational Numbers ............................................................. 20

Basic Operators ........................................................................................................................................................................ 20

Signed Numbers ....................................................................................................................................................................... 21

Rounding ................................................................................................................................................................................... 23

Lesson 2 – Powers and Roots ................................................................................................................... 27

Exponent Rules .......................................................................................................................................................................... 30

Lesson 3 – Order of Operations ................................................................................................................ 32

Lesson 4 – Decimals and Fractions .......................................................................................................... 34

Fractions ................................................................................................................................................................................... 34

Entering Fractions .................................................................................................................................................................... 35

Multiplying/Dividing Fractions ............................................................................................................................................... 38

Converting Fractions to Decimals ........................................................................................................................................... 40

Lesson 5 – Percent ..................................................................................................................................... 43

Entering a Percent ................................................................................................................................................................... 43

Converting Decimals to a Percent ........................................................................................................................................... 43

Converting Fractions to a Percent ........................................................................................................................................... 44

[GED MATH IN 30 DAYS]

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Solving for the Part ................................................................................................................................................................... 45

Solving for the Percent ............................................................................................................................................................. 46

Solving for the Whole ............................................................................................................................................................... 47

Percent Change ......................................................................................................................................................................... 48

Lesson 6 – Ratios/Proportions ................................................................................................................. 50

Solving Proportions.................................................................................................................................................................. 50

Using Unit Rate ......................................................................................................................................................................... 50

Using Equivalent Fractions ...................................................................................................................................................... 51

Using Cross Multiply ................................................................................................................................................................ 51

Lesson 7 – Probability ............................................................................................................................... 53

Solving Simple Probability ....................................................................................................................................................... 53

Solving Compound Probability ................................................................................................................................................ 54

Lesson 8 – Combinations and Permutations ........................................................................................... 58

Solving Combinations .............................................................................................................................................................. 58

Lesson 9 – Statistics .................................................................................................................................. 61

Calculating the Mean (Average) .............................................................................................................................................. 61

Calculating the Median ............................................................................................................................................................. 62

Calculating the Range (Variance)............................................................................................................................................ 62

Calculating the Mode ................................................................................................................................................................ 62

Find Mean, Median, Minimum, and Maximum using the TI-30XS ......................................................................................... 63

Lesson 10 –Scientific Notation ................................................................................................................. 66

Converting to Scientific Notation ............................................................................................................................................ 66

Lesson 11 – Pythagorean Theorem.......................................................................................................... 69

Solving Pythagorean Theorem ................................................................................................................................................ 69

Lesson 12 – Area and Perimeter .............................................................................................................. 72

(Triangles, Rectangles, Squares, Trapezoids, and Parallelograms) ......... Error! Bookmark not defined.

Solving Perimeter problems .................................................................................................................................................... 72

Solving Area problems (Triangle, Rectangle, Square, Trapezoid, and Parallelograms) ...................................................... 73

Lesson 13 – Circles .................................................................................................................................... 75

Solving Area/Circumference problems .................................................................................................................................. 77

Lesson 14 – Volume/Surface Area ........................................................................................................... 78

Solving Volume/Surface Area problems ................................................................................................................................. 78

Practice ..................................................................................................................................................................................... 81

Concept Questions .................................................................................................................................................................... 81

Lesson 15 – Linear Equations and the Coordinate Plane ....................................................................... 82

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Finding the Slope of a Line ....................................................................................................................................................... 82

Solving for the Slope using a table of values ........................................................................................................................... 86

Finding the Equation of a Line using two points .................................................................................................................... 87

Finding the slope of a line given an equation ......................................................................................................................... 88

Finding the Equation of a Line of a Parallel line ..................................................................................................................... 89

Finding the Equation of a Line of a Perpendicular line .......................................................................................................... 91

Lesson 16 – Functions ............................................................................................................................... 94

Storing a number ...................................................................................................................................................................... 95

Solving Functions using Substitution ...................................................................................................................................... 96

Using Ask-x ................................................................................................................................................................................. 96

Lesson 17 – Quadratic Equations ........................................................................................................... 104

Find zeroes .............................................................................................................................................................................. 106

Answers ..................................................................................................................................................... 110

Table of Contents Page 6


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Introduction

The essential aspects of studying for your HSE are:

• Time Management

• Dedication

• TI-30XS Calculator

• Understanding your skill set. Identify your strengths and weaknesses because you can't

study everything

• GED Assessment – 80% weak areas 20% Review/Practice (Overall skill assessment and 2

GED Ready Practice tests)

• Get familiar with GED type questions

• Key tips – Look at the multiple choice answers

This handbook was written to assist students in passing the math section of the GED. Being an adult

educator for the last twenty-two years, I have realized the difficulty many students have with

passing the math portion of the GED.

1. Quality over Quantity

The way you spend your study hours is more important than the number you put into studying.

Make sure you are using your time wisely.

Focus your studying on what YOU need most. It's a good idea to take a diagnostic practice test

when you're first starting to study so that you can identify your strengths and weaknesses.

Targeting your weaknesses as you prepare and not waste time going over things you've already

mastered. I recommend studying 90-120 minutes daily.

2. Make a study schedule and stick to it.

As an adult learner with other responsibilities, it's essential to make a study schedule. Whether

its 30 minutes, an hour, or more, it's crucial STUDY TIME happens EVERY DAY!

3. Perfect using the calculator

Practice, practice, and practice some more. Hopefully, out of the forty questions on the math exam,

you will be able to complete at least 20 problems using the calculator. Each lesson includes stepby-step

instructions to use the TI-30XS calculator to solve the problem.

Introduction Page 7


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Assumptions

There are certain areas that I believe are necessary for success in mathematics. These are basic

foundation skills.

• Place Value

• Multiplication (1-12) (Complete 100 problems in 5 minutes)

• Basic Math (Addition, subtraction, multiplication, and division)

• Ratio/Proportion and Percentages

"Making Sense" of Word Problems

• Read, Understand and Visualize the problem

• Identify given information needed to solve the problem

• Choose an appropriate method to solve the problem (Use formula sheet, when appropriate)

• Solve the problem

• Check the answer

Algebra accounts for over 50% of the GED exam.

Basic algebra can be viewed as an extension of arithmetic. The central concept that distinguishes

algebra from arithmetic is the concept of variables, which are letters that represent a quantity

whose value is unknown. The letters x and y are often used as variables, but any letter can be used.

Variables enable you to present a word problem in terms of unknown quantities by using algebraic

expressions, equations, inequalities, and functions.

Algebraic Expressions

Algebraic Expressions account 1-3 questions on the GED exam.

An algebraic expression has one or more variables and can be written as a single term or as a sum

of terms. Algebraic expressions cannot be solved. Here are examples of algebraic expressions.

Algebraic expression

Algebraic expression (in words)

2n Two times a number n.

3t-5 Five subtracted from three times a number, t.

2x + 5

4x

Two times a number x plus five divided by the

product of 4 and x

3x 2 − 5xy + 8xy − 7x 2 + 9

3x 2 and − 7x 2

−5xy and 8xy

Introduction Page 8


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The examples above contain "like terms" because they have the same variables, and the

corresponding variables have the same exponents. A term that has no variable is called a constant.

A number that is multiplied by variables is called the coefficient of a term. Like terms can be

combined. Therefore, the coefficients 3 and -7 of x 2 can be combined with the result being -4. The

coefficients of xy, -5 and 8 can be combined to 3.

3x 2 and − 7x 2 = −4x 2

−5xy and 8xy = 3xy

Therefore, the expression can be simplified to

−4x 2 + 3xy + 9

Algebraic Equations

The difference between an algebraic expression and an algebraic equation is an equal sign. An

equation can be solved. A statement of equality between two algebraic expressions that are true for

only specific values of the variables involved is called an equation. The values are called the

solutions of the equation.

The following are three basic types of equations.

Type 1: A linear equation in one variable: for example,

2x + 5 = 15

Type 2: A linear equation in two variables: for example,

2x + 3y = 15

Type 3: A quadratic equation in one variable: for example

x 2 + 5x = −6

Algebraic Inequalities

A mathematical statement that uses one of the following four inequality signs is called an

inequality.

Algebraic inequalities account for 1-4 questions on the GED exam.

< the less than sign

> the greater than sign

≤

the less than or equal to sign

≥

the greater than or equal to sign

Introduction Page 9


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To solve an inequality means to find the set of all values of the variable that make the inequality

true. This set of values is also known as the solution set.

e.g.

3x − 5 ≤ 16

3x − 5 ≤ 16 (add 5 to both sides)

3x ≤ 21 (divide both sides by 3)

x ≤ 7

Therefore, all values of x less than or equal to seven are the solution set.

Important rules of inequalities

Rule

Multiplying/Dividing by a negative number

When multiplying or dividing by a negative

number, reverse the inequality

The variable is on the right.

Switch sides and the inequality sign

−3x ≤ 21

(divide both sides by -3 and reverse the inequality)

x ≥ 7

15 ≤ x − 5

20 ≤ x

Therefore, x ≥ 20

Introduction Page 10


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Functions

Functions will probably account for 1-4 questions on the GED exam.

An algebraic expression in one variable can be used to define a function of that variable. Functions

are usually denoted by letters such as f, g, and h. For example, the algebraic expression 3t + 5 can

be used to define a function f by

f(x) = 3x + 5

where f(x) f of x is called the value of f at x and is obtained by substituting the value of x in the

expression above. For example,

f(1) = 3x + 5 denotes the value of the function when x=1

f(1) = 3(1) + 5

f(1) = 3 + 5

f(1) = 8

It might be easier to think of a function f as a machine that takes an input, which is a value of the

variable x and produces a corresponding output.

Important to understanding if an equation is a valid function, for any function, each input x gives

exactly one output.

FOR EVERY X, there is ONLY ONE Y

Introduction Page 11


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Getting Familiar with the TI-30XS

Types of Problems that can be solved using the TI-30XS

• Math Basics

• Working with Powers and Roots

• Order of operations

• Signed Numbers

• Fractions

• Rounding

• Convert Decimal →Fractions→%

• Percent

• Probability (Simple and Compound)

• Mean, Median and Range

• Area

• Surface Area and Volume

• Scientific Notation

• Radicals

• Pythagorean Theorem

• Permutations and combinations

• Evaluate Algebraic Expressions and Functions (Storing a variable),

• Slope, equation of a line

• Quadratic Formula

WOW! Yes, you are probably in disbelief but the majority of the GED Math exam can be solved using the

TI-30XS calculator.

Using the calculator on your exams

The GED Calculator (TI-30XS) will be available on-screen for the majority for the Mathematical

Reasoning test, and some of the items on the Science and Social Studies tests.

Note: The calculator can be used for the entire Mathematical Reasoning exam except for the first

five questions.

Accessing the Calculator on-screen

The calculator is built into the GED test software on the computer. You can also use a physical

calculator distributed by the testing center.

You can access it during the test by clicking on 'Calculator' in the top left-hand side of the test.

This will open the GED Calculator window, within the test window.

Don’t have the TI-30XS calculator

Don’t worry! You can download a 90 day trial for the TI-SmartView emulator for MathPrint

calculators. This should be more than enough time to get ready for your exam.

https://education.ti.com/en/software/details/en/F4D60658D2D947F38F0A4C9C043B12E6/tismartview-for-mathprint-calculators

Getting Familiar with the TI-30XS Page 12


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The TI-30XS Basics

The TI-30XS Basics Page 13


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Calculator Reference Sheet

It is possible to open the Calculator Reference sheet whenever you have access to the calculator. Do

this by clicking on the Calculator Reference button in the upper right-hand corner of the screen.

The Calculator Reference sheet explains the basic operations and can be referred to when you are

confused about how to complete certain operations on the calculator.

Note: Hopefully, after completing this book, the calculator reference sheet will not be needed.



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The "First Five"

The "First Five" are the first five problems on the exam. The TI-30XS calculator cannot be used.

The "First Five" typically come from the following concepts:

• Undefined

• Absolute Value

• Greatest Common factor

• Fractions (Converting from an improper fraction to a mixed number)

• Multiplying/Dividing Decimals

• Ordering Fractions, Decimals and Percents

• Square, Square Roots, Cube & Cube Roots

Examples of the “First Five”

Undefined

A fraction is undefined for any value that where the denominator is equal to 0 or square root of a

negative number.

Find the value for x that makes each function undefined.

12x − 5

5 + x

(x − 7)(x + 3)

x − 4

√

15x

x(x + 3)

x 2 − 12x − 5

(x + 2)(x − 8)

x = −5

x = +4

= any negative number

x = 0, −3

x = −2, +8

Absolute Value

Which expression correctly represents the distance between the two points marked on this number

line?

The distance between the two points is 6 but one must select the appropriate absolute value

A | − 1 + 5|

B | − 1 − 5|

C | + 1 − 5|

D −1 + 5



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To find the distance between two points. Subtract the two values and take the absolute value.

Therefore, B is the correct answer.

Greatest Common Factor

What is the GCF of 21 and 56?

Factors of 21:1, 3, 7, and 21

Factors of 56:1, 2, 4, 7, 8, 14, 28, and 56

The largest number in both lists is 7, this is

the GCF

What is the GCF of 16, 20 and 44?

Factors of 16:1, 2, 4, 8, and 16

Factors of 20:1, 2, 4, 5, 10, and 20

Factors of 44:1, 2, 4, 11, 22, and 44

The largest number in all three lists is 4, this

is the GCF

Converting Improper Fractions↔Mixed Numbers

2 7 8 = 8 × 2 + 7 23

8

52

52 ÷ 9 = 5

5 7 9

Remainder = 7

9

Multiplying/Dividing Decimals

15.05 ÷ .5 = 30.1 2.5 × .3 = .75

Ordering Fractions, Decimals and Percents

Order the following from least to greatest

3

4 , 3 2

, .7, .007, 7.5, 55%

Easiest way to compare values of fractions, decimals and percents is to covert each to a decimal and

add corresponding zeroes, if needed

3

3

. 7 . 007 7.5 55%

4

2

. 75 1.5 . 7 . 007 7.5 .55

. 750 1.500 . 700 . 007 7.500 .550

Then it’s much easier to put them in order



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. 007 55% . 7

3

4

3

2

7.5

Square, Square Root, Cube & Cube Root

√6 2 + 8 2 5 2 − 3 2 √144

3

√125

√36 + 64 25 − 9 12 5

√100 = 10 16



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Lesson 1 – Number Sense, Operations, and

Rational Numbers

The TI-30XS works similar to a standard 4-function calculator with so much more. You must learn

the basics of the TI-30XS calculator.

Basic Operators

Addition

Subtraction

12

6

Multiplication

24

Note: I recommend using parentheses to multiply

Division

3

Note: Recommended method to divide

Lesson 1 – Number Sense, Operations, and Rational Numbers Page 20


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Signed Numbers

One of the significant and important aspects of the TI-30XS calculator is the difference between the

subtraction sign and the negative sign. As the negative sign implies use the negative sign when the

expression starts with a negative sign or if the sign is inside parentheses. Another suggestion when

using the negative sign is when two operators are side by side. In this case, enclose the second value

in parentheses.

Examples: Addition, subtraction, multiplication, and division of signed numbers

Note: If an equation starts with a negative or a negative number is inside of parentheses use the

negative sign.

See examples below:

Addition of signed numbers

Subtraction of signed numbers



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Multiplication of signed numbers



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Division of signed numbers

Practice

Complete with and without the calculator. (Answers on p. 109)

−3(7) = 17 − (−3) = (−7.8)(−2.1) =

−3 − 9 = −(−39) + 4 = 16.53 + 17 =

−12 − (−9) = (9)(−8) = 23.4 − 14.04 =

−108

12 = (−3)(−7) = 3.31(.3) =

144

= 16.78(2.3) = (−24

−12 4 )(3.2) =

Rounding

Rounding is important because, at times, an exact result is not needed. Therefore, we estimate the

result.

e.g. Round 657.739 to the nearest whole number, nearest tenth, and hundredth



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Nearest Whole Number

until FLOAT

until 0

Note: The zero represents no decimal places

Nearest tenth

until FLOAT

until 1

Note: The one represents the first decimal place or

the tenths place



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Nearest Hundredth

until FLOAT

until 2

Note: The two represents the second decimal

place or the hundredths place

Important: Be sure to return the mode back to normal

Rounding on the GED exam is an important concept which needs to be mastered.



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Concept Questions

Answers on page 110

1) Give three (3) places when rounding may be used on the GED.

a) _____________________________________________________________

b) _____________________________________________________________

c) _____________________________________________________________

2) To round using the TI-30XS calculator, what key does one press initially? ________

3) To round to the nearest whole number, using the TI-30XS calculator, under FLOAT, which

number must be selected? ________ nearest tenth? _______ nearest hundredth? ____

Practice

Answers on page 106

Please complete each problem by hand and then check using the calculator

Round the following numbers to the indicated place

652.789

Whole Number ____________

Tenth _______________________

Hundredth _________________

156.805

Whole Number _____________

Tenth _______________________

Hundredth _________________

9,873.327


Tenth _______________________

Hundredth _________________

2.017


199.9505

Whole Number _____________

Tenth _______________________

Hundredth _________________

Tenth _______________________

Hundredth _________________



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Lesson 2 – Powers and Roots

TI-30XS button

Squares a number

Raises a number to a power

Use the negative symbol with negative

exponents

Square Root of a number

Lesson 2 – Powers and Roots Page 27


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Finds the cube root of a number

Note: The 3 is want results in the cube root

It is vital to understand squaring and square root are inverse operations. The same holds for cubing

and cube root.

√ 2 = 6

(√ 2 ) 2 = 6 2

2 = 36

= 6

3

√

= 4

3

(√) 3 = 4 3

= 64



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Square

1 2 = 1

2 2 = 4

3 2 = 9

4 2 = 16

5 2 = 25

6 2 = 36

7 2 = 49

8 2 = 64

9 2 = 81

10 2 = 100

11 2 = 121

12 2 = 144

Square Root

√1 2 = √1 = 1

√2 2 = √4 = 2

√3 2 = √9 = 3

√4 2 = √16 = 4

√5 2 = √25 = 5

√6 2 = √36 = 6

√7 2 = √49 = 7

√8 2 = √64 = 8

√9 2 = √81 = 9

√10 2 = √100 = 10

√11 2 = √121 = 11

√12 2 = √144 = 12

Cube

1 3 = 1

2 3 = 8

3 3 = 27

4 3 = 64

5 3 = 125

Cube Root

3

√1 = 1

3

√8

3

√27

3

√64

3

√125

= 2

= 3

= 4

= 5

Knowing the square and square roots will also assist one in the following areas on the GED exam:

• Distance between two points

• Pythagorean Theorem



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Exponent Rules

Any base to the 0 power is 1.

Any base to the 1 st power is its base

Negative Exponent

x 0 = 1

10 0 = 1

20 0 = 1

x 1 = x

10 1 = 10

10 −3 = 1

10 3

c −3 d 4 = d4

c 3

Multiplying exponents with the same base, add

the exponents

Dividing exponents with the same base, subtract

the exponents

Raising to a power, multiply the exponents

x 2 x 5 = x 7

(x 3 y 5 )(x 4 y 3 ) = x 7 y 8

w 7

w 7

w 4 = w3

w 9 = w−2 = 1 w 2

(c 3 ) 3 = c 9

(xy 2 z 5 ) 4 = x 4 y 8 z 20



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Practice

Answers on p. 111

Complete the following with and without the calculator

8 2 = 2 3 3

= √64 = √64

=

11 2 = 3 3 3

= √25 = √125

=

4 2 = 4 3 3

= √36 = √1,000

=

5 2 = 5 3 3

= √100 = √8

=

10 2 = 10 3 3

= √169 = √27

=

Complete the following with or without the calculator

√7 + 3√7 =

10√3 − 5√3 =

√ 169

4 =

√48 = √96 = 2 4 (2 −3 ) =

4√5 × 3√5 =

3

√27

+ 2 3 = 5 3 + 3 3 =

3 2 + 4 2 = √13 2 − 5 2 = √6 2 + 8 2 =

1. The inverse operation of squaring a number is ______________________________________

2. Any base raised to the zero power is _________________________________________________

3. When multiplying exponents with the same base, __________________________________ the exponents.

4. When dividing exponents with the same base, ______________________________________ the exponents.

5. √ 2 = 12 = ____________________



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Lesson 3 – Order of Operations

A pneumonic device to remember the Order of Operations is "Please Excuse My Dear Aunt Sally".

Luckily, the TI-30XS automatically follows the Order of Operations.

P

E

M

D

A

S

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

Examples

Jolene goes to the grocery store in wake of the Coronavirus. She buys 3 cases of water that cost

$3.97 each. She also buys 4 packages of batteries that cost 6.45 each. Finally, she buys a power pack

for her cell phone that costs 24.99. What is her total before tates?

3(3.97) +4(6.45)+24.99

Jadah's monthly rent is $2,100. She splits the rent with two other roommates. Their utility bills total

$150.00/month, which is also, split three ways. What are her total expenses for the month?

Lesson 3 – Order of Operations Page 32


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2100/3 + 150/3

Lesson 3 – Order of Operations Page 33


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Lesson 4 – Decimals and Fractions

To have a complete conceptual understanding of fractions, one must understand the following:

In mathematics, fractions can be identified in various ways:

Fractions

Part of a whole

e.g. ¼ means a whole has been separated into four (4) equal parts and one of these parts has been

selected

As a Quotient

e.g. ¼ means 1 is being divided by 4

As a Ratio

e.g. the fraction ¼ can be also be viewed as a ratio 1:4 or 1 to 4

Understanding these distinct representations is essential for passing the GED.

Note: All proper fractions and decimals are between 0 and 1

Zero Rule - 0 divided by any number is 0 e.g.

0

5 = 0

Any number divided by 0 is UNDEFINED.

e.g. 8 0 = UNDEFINED

Lesson 4 – Decimals and Fractions Page 34


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Entering Fractions

Proper/Improper

Mixed Number

Note: When you enter a mixed number the calculator automatically converts to an improper

fraction.

Converting Improper Fractions to Mixed Numbers



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Practice

Answers on p. 112

Complete the following table:

Improper Fraction

18

5

Mixed Number

9 1 4

12 5 8

32

9

Proper Fraction

15

20

32

100

20

90

Simplest Form

45

200

Note: You should have noticed the TI-30XS reduces fractions to simplest form AUTOMATICALLY!

Adding/Subtracting Fractions

The TI-30XS will complete all operations regardless of like/unlike denominators



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Improper Fractions

Convert to a mixed number

Mixed Numbers



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Note: Convert to a mixed number

Practice

Answers on p. 109

Add/Subtract the following fractions (if necessary, convert to a mixed number, then convert to a

decimal)

Equation Fraction Decimal

3

4 + 4 1 2 =

4

5 + 2 10 =

8 1 8 − 5 3 4 =

34 3 10 − 23 7 10 =

Multiplying/Dividing Fractions

Enter 2 5

× 7 8 =



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Enter 3 2 5

× 5 1 4 =

The answer is displayed as an improper fraction. Therefore, the result must be converted to a

mixed number.

Note: There are several ways to demonstrate multiplication and division.

5 × 3

5(3)

(5)(3)

(5)3

15 ÷ 3

15

3


[GED MATH]

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Converting Fractions to Decimals

or

Practice

Answers on p. 112

Complete the following table. If necessary, convert to a mixed number.

(5 3 4 )(4 1 2 )=

(10 4 5 )(2 1 10 ) =

2 1 8 ÷ 3 4 =

7 1 3 ÷ 23 7 10 =


8 1 5

4

5


[GED MATH]

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Common Fractions, Decimals, and Percent

Fraction Decimal Percent

1

2

1

3

1

4

1

5

1

8

1

9

1

10

.5 50%

.33 33%

.25 25%

.20 20%

.125 12.5%

.11 11%

.1 10%

Therefore, if you commit the common fractions to memory, calculating multiples becomes easy.

e.g.

3

= 1 4 4

× 3 =.25 × 3=.75 =

75%

4

= 1 9 9

× 4 =.11 × 4=.44 =

44%

5

= 1 8 8

× 5 =.125 × 5=.625 =

62.5%


[GED MATH]

Advanced Computing

Practice

Answers on p. 113

Complete the following table

Hours Studying

Monday Tuesday Wednesday Thursday Friday Total (Mixed

Fraction)

Joe

Jeremy

Journey

Jonah

Jerel

2 ½ ½ 1 1 ¼ 2

2 2 2 1 ½ 2 ½

3 1 ½

4 3.5 2.25 .75 3.75

2.5 2 1.25 1.4 2.6

Total

(Decimal)

Concept Questions

1. What are the three types of fractions?

• _________________________

• _________________________

• _________________________

2. How can 4 be expressed as an improper fraction?

__________________________________

3. What key(s) converts fractions to decimals and decimals to fractions?

__________________________________

12

0 = 0

3 =

What values of x would make this function

"undefined."

x = _______ and ________

(x 2 + 5x − 9)

3(x − 1)(x + 2)


[GED MATH]

Advanced Computing

Lesson 5 – Percent

Just like fractions, think of a percentage as a part of a whole. Per – means "out of" Cent- means 100.

Therefore, percent can be thought of as "out of 100."

85% = 85

100 = .85

7% =

7

100 = .07

165% = 165

100 = 1.65

Entering a Percent

Converting Decimals to a Percent

Lesson 5 – Percent Page 43

[GED MATH]

Advanced Computing

Converting Fractions to a Percent

Practice

Answers on p. 115

Complete the following table: (If necessary, convert improper fractions →Mixed Numbers

Percent Decimal Fraction

65%

.72

.03

3

4

28%

50%

.8

5

8

112%

3.35


[GED MATH]

Advanced Computing

Solving for the Part

Part = Percent × Whole

e.g. 65% of 320

Or

Note: 65% = .65

Practice

Answers on p. 116

Part

50% of 75

100% of 175

40% of 80

15% of 70

8% of 65


[GED MATH]

Advanced Computing

Solving for the Percent

Percent =

Part

Whole

35 is what percent of 140?

Practice

Answers on p. 116

Percent

25 of 75

20 of 80

40 of 80

10.5 of 70

1.3 of 65


[GED MATH]

Advanced Computing

Solving for the Whole

Whole =

Part

Percent

72 students attended graduation which accounted for 90% of the class? How many students are in

the class?


[GED MATH]

Advanced Computing

Practice

Answers on p. 117

Whole

40 is 50% of what number


30 is 25% of what number?



Percent Change

Percent Change =

(New − Old)

Old

Note: The sign indicates an increase or decrease

John was rewarded with a raise from his supervisor. His pay increased from $15 to $15.80. What

was the percent increase, John received from his supervisor?


[GED MATH]

Advanced Computing

Practice

Answers on p. 117

Percent Change (Round to

nearest whole percent)

50 →80

120→100

1,200→1,500

63→80

150→115

Concept Questions

1. To change a decimal to a percent, one must move the decimal ___________ places to the left/right.

2. To change a percent to a decimal, one must move the decimal ___________ places to the left/right.

3. Why is the decimal moved to the left? ________________________________________________________

4. Why is the decimal moved to the right? ______________________________________________________

5. Percent =


[GED MATH]

Advanced Computing

Lesson 6 – Ratios/Proportions

Solving Proportions

On the GED exam, proportions can be used to solve scale, grocery and shopping, recipe and cooking,

and slope problems among others. A ratio is a comparison between two numbers. A proportion is

two ratios set equal to each other.

Betty can bake 9 cakes in 2 hours. How many cakes can she bake in 12 hours?

Set up the proportion,

Known

ratio

9 cakes

2 hours = x

12 hours

Note: You must have "like terms" on both sides of the equal sign. e.g. Hours are on the bottom on

both sides of the equal

Proportions can be solved in various ways:

Using Unit Rate

9 cakes cakes

= 4.5

2 hours hour

Note: Rate per hour

To calculate the number of cakes in 12 hours,

the rate per hour must be multiplied by 12

Lesson 6 – Ratios/Proportions Page 50

[GED MATH]

Advanced Computing

Using Equivalent Fractions

2

9 = 12

?

2

9 × 6 6 = 12

?

2

9 × 6 6 = 12

54

Using Cross Multiply

Note: Set up algebraically

2x = (9)(12)

2x = 108

2x

2 = 108

2

x = 54

Note: To solve for the variable x, 9 and 12 must be multiplied and then divided by 2

Note: Only one pair of parentheses was used to multiply in the example above.


[GED MATH]

Advanced Computing

Practice

Answers on p. 118

Solve for x

x

2

3 = x 9

3

5 = x 60

4

9 = 24

x

1.5

6.0 = x 18

17

4 = 51

x

Concept Questions

1. What are three ways one can solve a proportion?

______________________________________

______________________________________

______________________________________

2. What type of problems can be solved using proportions?

______________________________________

______________________________________

______________________________________


[GED MATH]

Advanced Computing

Lesson 7 – Probability

Probability is a number that reflects the chance or likelihood that a particular event will occur. In

mathematical terms, probability can be thought of as a fraction. Probability will account for 1-2

problems on the GED exam.

Simple probability is the likelihood of one event.

Note: Probability can be expressed as a fraction, decimal, ratio, or percent.

Solving Simple Probability

Using the TI-30XS calculator, probability problems can be solved in the same manner we computed

fractions.

Probability can be shown as a:

• Fraction – ¼

• Decimal .25

• Ratio 1:4

• Percentage 25%

Simple Probability =

Desired Outcome

Total Possible Outcomes

e.g.

In a bag of marbles, there are 5 white, 6 blue, 4 green, 4 black, and 1 yellow. What is the probability

of selecting a blue marble? What is the probability of choosing a blue marble or white marble?

Probability = 6 blue marbles/20 Total marbles

P(Blue or White Marble)

Possible Outcomes = (6 blue and 5 white) = 11

Total Outcomes = (Total Marbles) = 20

P(Blue or White Marble) = 11

20

11

20

= .55 = 55%

Lesson 7 – Probability Page 53

[GED MATH]

Advanced Computing

Convert to

a Decimal

Convert to

a Percent

Solving Compound Probability

Compound probability is the likelihood of more than one event occurring.

To solve a compound probability problem, one needs to multiply each respective simple probability

of each event.

e.g.

P(Event 1) × P(Event 2) × P(Event n)

What is the probability of rolling an even number on a die, and selecting a King from a deck of

cards?

Possible Outcomes = (2,4,6)

Total Out comes = (1,2,3,4,5,6)

P(Even number) × P(King)

P(Even number) = 3

6 = 1 2


[GED MATH]

Advanced Computing

Possible Outcomes = (King of Hearts, Diamonds, Spades, and Clubs) = 4 Kings

Total Out comes = (Entire Deck) = 52 cards

P(King) = 4

52 = 1 13

P(Even number) × P(King)

1

26

( 1

2 ) ( 1 13 ) = 1 26

= .038 = 3.8%

Convert

Fraction

→Decimal

Decimal →

Fraction

Rounded Decimal = .038


[GED MATH]

Advanced Computing

Convert to a

Percent

If we needed to round the result - Percent = 3.8%

Round to the nearest tenth

0 - Round to the whole number

1 – Round to the nearest tenth

2 - Round to the nearest hundredth etc.

To return to normal


[GED MATH]

Advanced Computing

Practice

Answers on p. 118

1. A bag of marbles contains 5 blue, 4 red, 3 green, 7 black, and 1 white marble.

What is the chance a blue marble is selected? __________________________________________________________

What is the percentage of a blue or black marble being selected? ____________________________________

2. Ben selects a card from a deck of cards, without replacement, selects another card. What is the

probability he will select a King then a club? ___________________________________________________________

3. Joan and Heaven each buy raffles from their local basketball team. Raffles cost $1.00. Joan buys

20 raffles and Heaven buys 15 raffles. If 225 raffles were sold, what is the probability, to the

nearest whole percent, of either winning the raffle? ___________________________________________________


[GED MATH]

Advanced Computing

Lesson 8 – Combinations and Permutations

Solving Combinations

The difference between combinations and permutations is ordering. With permutations the order

of the elements is important, whereas with combinations order is not. Expect one question on your

exam from this topic. Combinations and Permutations account 1-2 questions on the GED exam.

At a local all-you-can-eat buffet store, you can choose from 25 types of meat, 10 different

vegetables, and 7 starches. How many different combinations do you have to choose from, if you

can select 1 from each category?

To solve this problem, all these values must be multiplied.

Solving Permutations

There are no supplied formulas for combinations and permutations on the GED formula sheet.

Therefore, so you'll have to memorize the following formula to solve permutations problems:

n items and want to find the number of ways k items can be ordered:

P(n, k) =

n!

(n − k)!

5! read 5 factorial = 5 x 4x 3 x 2 x 1

The number of ways to combine k items from a set of n:

The best way to conceptualize these types of problems n choose k

Lesson 8 – Combinations and Permutations Page 58

[GED MATH]

Advanced Computing

C(n, k) =

n!

(n − k)! k!

e.g.

How many different 4-letter combinations can be formed from the word TASSELS? Therefore, this

is a 7 choose 4 problem.

factorial

choose ! for

choose ! for factorial

OR

7 Choose 4

Choose nPr


[GED MATH]

Advanced Computing

Practice

Answers on p. 119

1. You just got a free ticket for a boat ride, and you can bring along 2 friends! Unfortunately, you

have 5 friends who want to come along. How many different groups of friends could you take with

you? (Hint 5 choose 2)

2. Michael is packing his bags for his vacation. He has 5 unique shirts, but only 4 fit in his bag. How

many different groups of 4 shirts can he take? (Hint 5 choose 4)

3. William is packing his bags for his vacation. He has 8 unique books, but only 5 fit in his bag. How

many different groups of 5 books can he take? (Hint 5 choose 4)


[GED MATH]

Advanced Computing

Lesson 9 – Statistics

The TI-30XS does your central tendency for you. The GED exam will usually contain a few problems

from this concept area. Fortunately, the TI-30XS can calculate the mean, median, minimum, and

maximum. These concepts will account 2-4 questions on the math and science exams.

The mean represents the average, and, in fact, the word "mean" is often interchangeable with the

word "average." To find the mean of a set of numbers you add up all of the numbers and divide that

sum by the number of values you added together.

The median number is the number that is in the middle of a set of numbers. To find the median you

first need to order the numbers from smallest to largest.

The mode is the value that occurs the most often in a set of values. If no number appears the most

there is no mode. If several numbers appear the most, those numbers are the mode.

e.g. 73, 64, 78, 89 ,93 – no mode 73,73, 75, 89, 89, 93 – mode = 73,89

The range is just the difference between the maximum or largest value and the minimum or the

smallest value

Data Set {89, 76, 85, 76, 77, 84}

Calculating the Mean (Average)

This set of calculator entries calculates the total or sum of the data set. Now, the total must be

divided by the number of values, 6.

Therefore, the average or mean is 85.6.

Lesson 9 – Statistics Page 61

[GED MATH]

Advanced Computing

Calculating the Median

First, we must place the data in order from smallest to largest.

Since, there is an even number of values, the sum of the two values in the center must be then

divided by 2. The data set ordered is now

{76, 76, 77, 84, 85, 89}

Visually, one can see two numbers make up the "middle" of the data set. Two numbers now lie

outside of the middle on both sides.

77 + 84 = 159

159 ÷ 2 = 79.5

Median = 79.5

Calculating the Range (Variance)

Using the sorted data set, determining the maximum and minimum is quite evident. Take the

difference and the range is easily calculated.

89 − 76 = 13

Calculating the Mode

The mode must also be done manually. Using the sorted data list, the mode is the value that appears

the most. Since, 76 appears twice, it is the mode.

{76, 76, 77, 84, 85, 89}

Find missing value given mean

At times, the mean is given and the missing value must be determined.

e.g.

Let the TI-30XS do it for you.

e.g. Jubilee wants the average of all her exams to be a 90, so she can receive an A in her Math class.

She has already taken five (5) exams. She received 90, 85, 83, 95, and 91. What is the minimum

score she can receive on her last exam to get an A in the class?


[GED MATH]

Advanced Computing

Add the existing values

To get an A, she must have a total of at least 540

points (6 exams x 90). Subtract the previous

total from 540 to get the missing value.

Find Mean, Median, Minimum, and Maximum using the TI-30XS

Data Set: {76, 76, 77, 84, 85, 89}


[GED MATH]

Advanced Computing

Note:

n = the number of values = 6

x̅ = mean or the average = 81.16666667

MED = median = 80.5

MIN = minimum = 76

MAX = maximum = 89

Note:

Average = 81.16666667

Median = 80.8

Practice

Answers on p. 119

Type of Nuts

Price per lb. of

Peanuts 2.50

Almonds 14.00

Cashews 9.00

Pistachios 2.50

Walnuts 10.00

1. What is the median price for nuts? ____________________

2. What is the mean price of nuts? ______________________

3. What is the mode price of nuts? ______________________


[GED MATH]

Advanced Computing

4. What is the range of the prices of nuts? ________________

Jabes counted the number of students in each class on the 2 nd floor. There were: 20, 31, 20, 18, 16, 20,

16, and 11 respectively.

5. What is the median number of students on the second floor? __________________

6. What is the mean number of students on the second floor? ___________________

7. What is the mode for number of students on the second floor? ________________

8. What is the range of the number of students on the second floor? ______________

9. If Jabes counts one more room of students, the new average is 20. How many students were in this

class? ___________________________________________

10. Steve found a few old books. The books have the following numbers of pages: 28, 35, 15, and 18. He

then found one more book. The new average of pages after finding the new book is 26. How many

pages are in the new book? __________________________

Concept Questions

1. What is the mathematical term for "average? _________________________________________________

2. Before determining the median, the set of values must be __________________________________

3. Another term used for the range is ____________________________________________________________

4. The sum is a term that means __________________________________________________________________


[GED MATH]

Advanced Computing

Lesson 10 –Scientific Notation

Converting to Scientific Notation

Scientific Notation is a way of displaying numbers that are too big or too small to be conveniently

written in decimal form. Commonly, scientific notation problems will appear on the science portion

of the GED.

Scientific notation is the way one can display very large numbers or very small numbers. For

example, instead of writing 0.0000000056, we write 5.6 x 10 -9 . So, how does this work?

We can think of 5.6 x 10-9 as the product of two numbers: 5.6 (the digit term) and 10 -9 (the

exponential term).

Here are some examples of scientific notation.

Common Place Values

Standard and Scientific

Forms

1

10,000 = 1 x 10 4 23,328 = 2.3328 x 10 3

1,000 = 1 x 10 3 8,353 = 8.353 x 10 3

100 = 1 x 10 2 382 = 3.82 x 10 2

10 = 1 x 10 1 89 = 8.9 x 10 1

1 = 1 x 10 0

1

10 = 0.1 = 1 x 10-1 0.32 = 3.2 x 10 -1

100 = 0.01 = 1 x 10-2 0.053 = 5.3 x 10 -2

1

1,000 = 0.001 = 1 x 10-3 0.0088 = 8.8 x 10 -3

Lesson 10 –Scientific Notation Page 66

[GED MATH]

Advanced Computing

To convert standard form to scientific form using the TI-30XS

0.053 = 5.3 x 10 -2

Note:

The number is displayed in Scientific Notation

SCI – Scientific Notation

To return to Normal Mode

The number is displayed in Standard Form

select Norm


[GED MATH]

Advanced Computing

Practice

Answers on p. 121

Complete the table

Standard form

Scientific Notation

.00000734

.00045

5,234,000

678,000

.0000003968

4,153,000

7,800,000,000

3.45 x 10 −5

2.45 x 10 6

7.5 x 10 5

5 x 10 2

1. There are 25 cells each measuring 1.5 x 10 −6 on a microscope slide. What is the total width of

the cells’? __________________

2. The population of the US is 3.282 x 10 8 and New York city has a population of 8.4 x 10 6 . What

percent of the US lives in New York City. _____________________________________________

3. The volume of the Atlantic Ocean is about 3.1 x 10 17 .The Missouri River has an annual flow of 5.8 x

10 11 cubic meters. How many times would the annual flow of the Missouri River fit in the Atlantic

Ocean?

Write your final answer in scientific notation, and round to two decimal places. _________________


[GED MATH]

Advanced Computing

Lesson 11 – Pythagorean Theorem

Solving Pythagorean Theorem

Pythagorean Theorem is one of the most fundamental theorems in mathematics and it defines the

relationship between the three sides of a right-angled triangle. The Pythagorean Theorem can be

manipulated to find any missing side of a right triangle. A keyword to identify Pythagorean

Theorem problems is diagonal. This concepts will account 1-2 questions on the math exam.

The side opposite the right angle has a unique name. It is called the "hypotenuse". It is the longest

side of the right triangle.

a 2 + b 2 = c 2

If the missing side is

c = use √a 2 + b 2

a = use √c 2 − b 2

b = use √c 2 − a 2

Two special right triangles often on standardized tests are the 3-4-5 and 5-12-13 triangles. The

form denotes the sides are 3 and 4, the hypotenuse is 5. The same holds for the other special

triangle. The sides are 5 and 12 and the hypotenuse is 13. The multiples of these triangles also hold

true. Therefore, if the sides of a triangle were 6 and 8 (a multiple of 2, 3-4-5 triangle), the

hypotenuse would be 10.

Lesson 11 – Pythagorean Theorem Page 69

[GED MATH]

Advanced Computing

a = 5

b =?

c = 13

Since one leg is known 5 and the hypotenuse is 13, the Pythagorean theorem can be modified to

find the missing leg.

√13 2 − 5 2 = b

√169 − 25 = b

√144 = b

12 = b


[GED MATH]

Advanced Computing

Practice

Answers on p. 122

Find the missing side (If necessary, round to the nearest tenth)

Sides of Right Triangle

a b c (hypotenuse)

3 4

5 12

9 12

15 20

15 36

40 50

40 41

6 9

8 17

A 15-foot ladder is leaning against a 30-foot wall. The bottom end of the ladder is 9 feet from the

wall. How many feet above the ground does the ladder touch the wall? _________________________________

Using the Pythagorean Theorem can also be used to solve the distance between two points.

Concept Questions

1. What is the Pythagorean Theorem? ___________________________________________________

2. The longest side of a "right" triangle is called the ___________________________________ .

3. Two special "right" triangles discussed above are the ______________ and ___________

4. The two legs of a right triangle are denoted by the variables _______________________ .

5. The hypotenuse is denoted by the variable __________________________________________ .


[GED MATH]

Advanced Computing

Lesson 12 – Area and Perimeter

Solving Perimeter problems

Perimeter is the distance "around". To determine the perimeter of any shape, take the sum of all

exterior sides. At times, they will give you the area or perimeter and one of the sides and the other

unknown side must be determined.

Rectangle

P = 2L + 2W

e.g. What is the width of a rectangular garden that has a perimeter of 40 feet and the length is 8

feet?

P = 2L + 2W

40 = 2(8) + 2W

40 = 16 + 2W

24 = 2W

12 = W

Lesson 12 – Area and Perimeter Page 72

[GED MATH]

Advanced Computing

Practice

Answers on p. 122

Complete the following table for rectangles using the given

Length Width Perimeter

10 in 12 in

8 in 50 in

14.5 in 20 in

15 in 55 in

17 ft. 98 ft.

Solving Area problems (Triangle, Rectangle, Square, Trapezoid, and

Parallelograms)

The area is defined as the size of a surface and is calculated by determining the amount of space

inside the boundaries. The surface area is defined as the sum of the area of each respective side.

Use the appropriate formula from the GED formula sheet and substitute where necessary.

Shape

Rectangle

Formula

A = L × W

L = A W

Triangle

W = A L

A = 1 2 bh

b = 2A h

Parallelogram

h = 2A b

A = bh

b = A h

h = A b

Trapezoid

A = 1 2 (b 1 + b 2 )h

h =

2A

(b 1 + b 2 )


[GED MATH]

Advanced Computing

Practice

Answers on p. 123

Solve the following given:

(b 1 + b 2 ) = 2A h

Sphere SA = 4πr 2

r = √ SA

4π

Shape Base(s) Side(s) Height Perimeter Area

Rectangle 8 in 40 sq. in.

Rectangle 6 ft 36 ft

Triangle

Triangle

b =12 cm

s2 = 5 cm

s3 = 5 cm

b =10 cm

s2 = 7 cm

s3 = 7 cm

5 cm

25 sq. ft

Parallelogram 15 ft 8 ft

Parallelogram 10 ft 60 sq. ft.

Trapezoid

b1 =10 yd

b2 = 12 yd

s3 = 11 yd

s4 = 14 yd

77 sq. yd.

Concept Questions

1. What is the formula to find the area of a triangle? ___________________________________

2. What is the formula to find the area of a parallelogram? ____________________________

3. What is the formula to find the area of a circle? ______________________________________

4. What is the formula to find the area of a trapezoid? _________________________________

5. The radius is ______________ the diameter.

6. The diameter is ___________ the radius.


[GED MATH]

Advanced Computing

Lesson 13 – Circles

Area and Circumference of a circle are almost guaranteed to be 1 or 2 of the problems on your GED

exam.

A = πr 2

C = 2πr or C = πd

Circle A = πr 2

r = √ A π

d = C π

Note: Only the first two formulas are included on the GED formula sheet. The other formulas have

been included to expedite the learning process.

What are the radius, diameter, and circumference of a circle if given its area is 379.94

cm 2 ?

Find the correct formula on your formula sheet.

A = πr 2

Since, the radius, diameter, and circumference

are wanted, the formula needs to be changed to

solve for r

r = √ A π

Once r is obtained, the diameter and

circumference can be easily obtained.

Lesson 13 – Circles Page 75

[GED MATH]

Advanced Computing

Now, the radius is known. Since, r = 11 then

the diameter is 22 cm

Now, the circumference is easily found

C = πd

Radius = 11 cm

Diamter = 22 cm

Circumference = 69.08 cm


[GED MATH]

Advanced Computing

Solving Area/Circumference problems

Practice

Answers on p. 124


Radius Diameter Circumference Area

5 ft.

10 in.

1 1 4 yd

19 cm

153.86 sq. in

706.5 sq. ft.

34.54 in.

665.68 ft.

Concept Questions

1. What is the formula to find the circumference of a circle? ___________________________________

2. What is the formula to find the area of a circle? _______________________________________________

3. What is the formula to find the radius given the area? _______________________________________

4. What is the formula to find the diameter given the circumference? _________________________


[GED MATH]

Advanced Computing

Lesson 14 – Volume/Surface Area

Volume is a measurement of the amount of space that a 3 - dimensional object holds. The difference

between area and volume is the third dimension, height. Due to this, all volume is represented by

cubic units. e.g. How much does a cereal box hold with the dimensions 12 inches x 5 inches x 2

inches? The result would be 120 in 3 . Another common term used synonymously with volume is

capacity.

Given the aforementioned formulas, substitute where necessary using your TI-30XS calculator.

Solving Volume/Surface Area problems

Shape

Rectangular Prism

Cylinder

Formula

V = L × W × H

V

L =

HW

V

W =

HL

V

H =

LW

V = πr 2 h

h =

V

πr 2

r = √ A πh

Cone

Pyramid

Sphere

V =

V = πr2 h

3 = 1 3 πr2 h

r = √ 3V

πh

h = √ 3V

πr 2

L × W × H

== 1 3 3 LWH

h = 3V

LW

l = 3V

HW

w = 3V

HL

V = 4 3 πr3

Lesson 14 – Volume/Surface Area Page 78

[GED MATH]

Advanced Computing

John received a package in the shape of a

rectangular prism as shown below. Its volume is

748 cubic inches. What is the height of the

package?

Note: cubic inches = volume


V = L × W × H

Since the height is wanted, the formula needs to

be changed to solve for h

V

H =

LW


[GED MATH]

Advanced Computing

The Amish Oats company requires that its oatmeal containers have a capacity for 176 cubic

inches to be in the shape of a cylinder, and be made from cardboard. Determine the radius, to

the nearest tenth, if the height of the container is 8 in.?


V = πr 2 h

Since, the radius is wanted, the formula needs to

be changed to solve for r.

r = √ A πh


[GED MATH]

Advanced Computing

Practice

Answers on p. 125

1. Find the height of a cylinder with a volume of 2260.8 ft 3 and a radius of 6 ft. _______________

2. Find the radius of a cylinder with a volume of 150 in3 and a height of 10 in. ________________

3. Find the height of a cone with a volume of 21 ft3 and a radius of 4 ft _____________________

4. Find the radius, diameter and circumference for each given the area is 200.96 sq. ft. of the

circle.

r=________ d= _________ C= __________

5. Find the radius, diameter and area for each given the circumference is 78.5 ft. of the circle.

r=________ d= _________ A= __________

Concept Questions

1. What is the formula to find the volume of a cylinder? ________________________________________

2. What is the formula to find the volume of a rectangular prism? _____________________________

3. What is the formula to find the volume of a pyramid? ________________________________________

4. What is the formula to find the volume of a cone? ____________________________________________

5. What is the formula to find the radius if given the height and the volume of a cylinder? __

_________________________________________________________________________________________________________

6. What is the formula to solve for the radius given the area of a circle? ______________________

7. What is the formula to solve for the radius given the circumference? _______________________


[GED MATH]

Advanced Computing

Lesson 15 – Linear Equations and the

Coordinate Plane

Finding the Slope of a Line

There will be a question on the GED math test that gives you two points or a line on a coordinate

plane and asks you to provide the slope of the line. The slope is just the "steepness" of a line. The

formula for finding the slope is provided on the formula sheet:

m = y 2 − y 1

x 2 − x 1

Looks complicated right?. It's not; m is the variable for slope. Take the vertical difference between

the two points and take the horizontal distance between the two points. Finally, divide the

numerator by the denominator. Yes, the slope can be displayed as a fraction. If the slope is going up

from left to right then the slope is positive, if the line is going down the slope is negative. Each point

is represented by an ordered pair (x,y).

The TI-30XS can be used to determine the slope between two points in several ways.

e.g. Find the slope between the points (3, 7) and (5, 8)

Method #1

(3,7) is the 1 st point. 3 is the x value and 7 is the y value.

(5,8) is the 2 nd point. 5 is the x value and 8 is the y value.

Therefore, x1 =3, y1 =7, x2 = 5 and y2 = 8

m = y 2 − y 1

x 2 − x 1

x1 =3, y1 =7, x2 = 5 and y2 = 8

Slope (m) = ½

Lesson 15 – Linear Equations and the Coordinate Plane Page 82

[GED MATH]

Advanced Computing

Method #2 – Using Data on the calculator

Find the slope between the points (3, 7) and (5, 8)

Enter the two points. Enter the x values in the 1 st

column and the y values in the 2 nd column

Steps to find the slope, and y-intercept of a line.

a = slope

b = y-intercept

Therefore, the slope = .5 and the y-intercept is

5.5.


[GED MATH]

Advanced Computing

The last step is converting the values into fractions

Take note of both the slope and y-intercept (

and b) (a = .5 and b = 5.5) Clear the screen

Convert values into fractions

m = 1 2

b = 11

2

Therefore, the equation of the line is

y = 1 11

x +

2 2


+3

[GED MATH]

Advanced Computing

Method #3 – Finding slope from a graph

+4

Practice

Complete the table with/without the calculator

m = rise

run = +3

+4

What is the slope of the line that passes through

(-3,-2) and (1,6)?

y 2 = y 1 =

x 2 = y 1 = m =


(-7,+5) and (1,9)?

y 2 = y 1 =

x 2 = y 1 =

m =


(-3,-2) and (1,-3)?

y 2 = y 1 =

x 2 = y 1 =

m =


[GED MATH]

Advanced Computing


(-12,2) and (-2,-3)?

y 2 = y 1 =

x 2 = y 1 =

m =


(-3,-2) and (1,6)?

y 2 = y 1 =

x 2 = y 1 =

m =

Solving for the Slope using a table of values

At times, points on a line are given the form of a table.

Month

Cell Phone Charges

January 120

February 170

March 220

April 270

The month in this table is the independent variable or x.

The charges in this table are the dependent variable or y.

Therefore, the table can be written with the following values:

Month

Cell Phone Charges

0 120

1 170

2 220

3 270

Therefore, the points would be (0,120), (1,170), (2,220) and (3,270)

Month 0 could be the cost of the phone or a fee.

Note: Any two points can be used to calculate the slope.


[GED MATH]

Advanced Computing

m =

170 − 120

1 − 0

m = 50

1

m = 50

The y-intercept is the point where a line crosses or intercepts the y-axis of a

graph.

The y-intercept is (0,120)

Therefore, the equation of the line is (y= mx + b)

y = 50x + 120

After the 2 nd month the charges would be:

y = 50(2) + 120

y = 100 + 120

y = 220

Finding the Equation of a Line using two points

See Method #2 from above.


[GED MATH]

Advanced Computing

Practice

Answers on p. 126

Point(s) Slope Equation of line

(1,-3) (-2,3)

(2,3) (1,2)

(-9,-6) (-5,2)

(-8,-3) (-4,0)

(0,4) (-5,3)

Finding the slope of a line given an equation

A linear equation can be displayed in various forms. The first is slope-intercept form y =mx+b and

the other is Standard form Ax+By = C. Be familiar with both types of equations and being able to

determine the slope and y-intercept.

Linear Equation form Example Slope, y-intercept

Slope intercept form

y = mx + b

y = 2x + 3 Slope (m) =2 ; y-intercept (b)= +3

Standard form

Ax + Bx = C

2x + 3y = 9

A = 2

B = 3

C =9

m = − A B ; b = C B ;

m = − 2 3

b = 9 3 = +3

Therefore, y = − 2 3 x+3


[GED MATH]

Advanced Computing

Practice

Answers on p. 126

Determine the slope of each line

Equation

Slope and y-intercept(m,b)

5x + 2y = 8 m= b=

−x + 2y = 8 m= b=

y = 2 3 x − 10 m= b=

y = 6x − 10 m= b=

−8x + 3y = 21 m= b=

Finding the Equation of a Line of a Parallel line

Recently, the GED has asked for the equation of a parallel line. The important thing to know about

parallel lines is parallel lines have the same slope.

What is the equation of the line that passes through ( -5, -5 )

and is parallel to y = 3x + 2

To use the TI-30XS calculator to solve this problem, it needs two points. From this problem, the

point and the slope are given. Therefore, two points can be obtained.

The point (−5, −5) and m = 3 = rise

run = 3 1

Therefore, the rise or y will change by +3 and the run or x will change by +1.

The second point will be (-5(+1), -5(+3)) = (-4,-2).

Now we have two points (-5,-5) and (-4,-2).


[GED MATH]

Advanced Computing

Enter two points

Note: Enter the x values in the 1 st column and

the y values in the 2 nd column

Get slope and Y-intercept

a = 3; b =10

y = mx + b

y = 3x + 10


[GED MATH]

Advanced Computing

Practice

Answers on p. 127

Find the equation of the parallel line

Given

Passes through (3,1)and is parallel to

y = 3x + 10

Passes through (−8,1)and is parallel to

y = 3 2 x − 5

Equation of a parallel line

Passes through(−4 − 1)and parallel to

−3x + 4y = 8


4x − 3y = 9

Finding the Equation of a Line of a Perpendicular line

e.g. What is the equation of the line that passes through ( 1, 2 ) and is perpendicular to y =- 1 6 x - 5

The vital thing to know about perpendicular lines is the slope of a line, which is perpendicular, is

the negative inverse. In simpler terms, change the sign and flip. To use the TI-30XS calculator to

solve this problem, it needs two points. From this problem, a point and the slope are given.

Therefore, two points can be obtained.

m = − 1 6

Therefore, m (perpendicular) = + 6 = rise

= +6

1 run 1

The point (1, 2) and m = +6 = rise

run = +6

+1

Therefore, the y will change by +6, and the x will change by +1. The second point will be

(1(+1), 2(+6)) = (2,8).

Now we have two points (1,2) and (2,8).


[GED MATH]

Advanced Computing

Enter the two points

Get slope and Y-intercept

a = 6

b = -4

y = mx + b

y = 6x − 4


[GED MATH]

Advanced Computing

Practice

Given

Passes through (3,1)and is perpendicular to

y = 3x + 10

Passes through (−12,4)and is perpendicular to

y = 3 2 x − 5

Equation of a perpendicular line


−x − 4y = 8

Complete the following table.

Slope The slope of a parallel line The slope of a Perpendicular

line

4

3

2

− 5 2

1

5

− 1 8

Concept Questions

1. Parallel lines have the ______________________ slope.

2. The slope is the _____________________________ of the slope of a perpendicular line.

3. The slope of the line, which is decreasing from left to right, is _______________________ .

4. _______ is the variable for slope.

5. _______ is the variable for the y-intercept.

6. At + By = C is the___________________________ form of a linear equation.

7. y = mx + b is the ____________________________ form of a linear equation.


[GED MATH]

Advanced Computing

Lesson 16 – Functions

A function is a relationship from a set of inputs to a set of possible outputs where each input is

related to exactly one output. For a function to be valid "every x there must be only one y."

The TI-30XS can store several variables including x, y, z, a, b, c, and x. The GED exam, for the most

part, only utilizes these variables.

A function can be represented by an equation, a set of points, a table or in a graph. Being able to

determine a valid function is an essential part of the GED.

Vertical Line Test - Draw a line parallel to the y-axis for any chosen value of x. If the vertical line you

drew intersects the graph more than once for any value of x then the graph is not a function.

e.g. For every value of x there must be only one corresponding y.

(1,1)

(2,4)

(2,7)

(3,8)

Points Table Graph

x

y

1 1

2 4

2 7

3 8

For the value x = 2; there are two different values of y

These are all examples of invalid functions.

Lesson 16 – Functions Page 94

[GED MATH]

Advanced Computing

Practice

Answers on p. 129

Determine if the following examples are functions by circling VALID or INVALID

x

y

3 17

5 19

5 23

7 32

VALID

INVALID

x

y

-5 4

5 14

10 10

0 9

VALID

INVALID

x

y

-2 1

-2 -3

2 1

2 5

VALID

INVALID

(-7, 23) (0,13) (2,24) (2,32) VALID INVALID

(0,3) (1,5) (2,7) (3,9) VALID INVALID

Storing a number

x= 3


[GED MATH]

Advanced Computing

Solving Functions using Substitution

f(8) = x 2 − 3x

means

f(x) = x 2 − 3x

when x =8

Store the variable x=8

Enter function into calculator using the variable

x

The calculator evaluates the function when x =8

Using Ask-x

Enter equation

Select Ask-x

Click Ok

Enter x=8


[GED MATH]

Advanced Computing

when x=8; y=40

Jethro had $400 to spend on a diet plan. The initial start-up fee was $50.00. The monthly payment was

$75. How many months could Jethro afford this diet plan?

A. 3

B. 4

C. 5

D. 6

Use the multiple-choice answers as x to solve this

problem


[GED MATH]

Advanced Computing

Method #1

Enter multiple choice answers until the $400 limit

is reached.

There is no reason to enter the six since the limit

was reached. Therefore, Jethro could only pay for

four months.

Set up as a linear equation and solve for x 75x + 50 = 400

350 = 75x

350

75 = 75x

75

350

75 = x

x = 4.67

Again, Jethro could only pay for four months.

Note: The ability to use Ask –x and Table can also be used to solve inequalities


[GED MATH]

Advanced Computing

Practice

Answers on p. 129

Two variables

1. The cost of a school banquet is $200 plus

$15.00 for each person attending. Write the

equation that gives the total cost as a

function of the number of people attending.

What is the cost of 75 people?

2. The size of a tree to be planted is 4.2 ft. Each

day the tree grows .002 inches. Write the

equation that gives the total size as a

function of the number of days. What is the

size of the tree in feet after 3 years?

3. Billy plans to paint baskets. The paint costs

$13.25. The baskets cost $6.75 each. Write

an equation that gives the total cost as a

function of the number of baskets made.

Determine the cost of nine baskets.


[GED MATH]

Advanced Computing

System of equations

Solve the system of equations

−4x − 2y = −12

4x + 8y = −24

A. {1,4}

B. {6,-6}

C. {-2,-2}

D. {-1,4}

Enter 1 st Equation

Solve for y

y = −2x + 6

y = − 1 2 x − 3

Select Ask-x and Click OK

Substitute the values of x and see if the result is y

Note: This option is valid for 1 st equation

Note This option is valid for 1 st equation

Note: This option is not valid for 1 st equation


[GED MATH]

Advanced Computing

Note: This option is not valid for 1 st equation

Enter second equation and check the first two

options

Ask-x is already selected, click OK

Clear Table (if necessary)

Enter values of x

Note: This option is not valid for 2 nd equation

Note: This option is valid for 2 nd equation

The solution is {6,-6}


[GED MATH]

Advanced Computing

Solve the system of equations

−4x − 2y = −12

4x + 8y = −24

A. (1,4)

B. (6,-6)

C. (-2,-2)

D. (-1,4)

Since, the multiple choice answers contain

possible solutions; one can substitute the

corresponding values into the two equations.

Store x and y for option A

Enter first equation

Enter 2 nd Equation

The second equation does not hold true.

Repeat for remaining multiple choice

answers until both hold true

Both equations are TRUE for both x any y.

Therefore, are solution is the ordered pair (6,-6).

Note: The intersection of these equations is at the

point (6,-6)


[GED MATH]

Advanced Computing


[GED MATH]

Advanced Computing

Lesson 17 – Quadratic Equations

A quadratic equation is an equation of the second degree, meaning it contains at least one term that

is squared. The standard form is Ax² + Bx + C = 0 with a, b, and c being constants, or numerical

coefficients, and x is an unknown variable. One absolute rule is that the first constant "a" cannot be

a zero. There are several ways to solve a quadratic equation. Two of these ways will be discussed in

this book. One way to solve a quadratic equation is by factoring. Another is using the quadratic

formula.

Quadratic Formula

x = −b ± √b2 − 4ac

2a

The variables used in the quadratic formula are obtained from the Standard form of a quadratic

equation.

Standard Form of a Quadratic Equation

Ax 2 +Bx + C = 0

A quadratic equation can have up to two solutions. To solve by factoring:

• Set equation equal to 0

• Find values that will make each factor 0

Solve by factoring

x 2 − 2x − 15=0

A =1; B=-2; C=-15

If a = 1,

15

Find the factor pairs of c

+1, -15 -1, +15

Find pair equal to b

+3, -5 -3, +5

+1-15=-14 -1+15 = 14

3-5 =-2 -3+5= 2

(3,-5) Factored - (x+3)(x-5)

Find the zeroes or the solutions

x=-3,5

to the quadratic equation

Lesson 17 – Quadratic Equations Page 104

[GED MATH]

Advanced Computing

Solve using the Quadratic formula

Standard form

Solve:

Ax 2 + Bx + C = 0

x = −b ± √b2 − 4ac

2a

Identify A, B, and C. x 2 − 2x − 15

A =1; B=-2; C=-15

Store variables in calculator

Enter the number to be stored,

Press then press

until the respective variable is

visible.

Repeat for B and C

Enter Quadratic Formula into

Calculator.

Note: When two variables are adjacent press

Repeat for the – part of the ±

Change + → -

once.

The solutions are 5 and -3.


[GED MATH]

Advanced Computing

Find zeroes

Once factored, find the value of x would result in the solution evaluating to be zero.

Quadratic equation Factored Solution(s)

x 2 − 8x + 16 = 4

To change to standard form,

6,2

subtract four from both

(6-6)(6-2)=0

(x-6)(x-2)

sides

and

x 2 − 8t + 12 = 0

(2-6)(2-2)=0

( 1 3 , −5)

(3x 2 + 14x − 5) (3x − 1)(x + 5)

(3 ( 1 3 ) − 1), (1 3 − 5)

(1 − 1) ( 1 3 − 5) = 0

(3(−5) − 1)(−5 + 5)

(−16)(−5 + 5) = 0

6x 2 − 5x + 1

(2x-1)(3x-1)

+ 1 2 , 1 3

Solving Quadratic Equations

What is the solution set of

x 2 − 9x = −18

{1,2}

{-3,6}

{3,6)

{-3,-6)

Plug quadratic equation into calculator

Convert to STANDARD FORM

x 2 − 9x + 18 = 0


[GED MATH]

Advanced Computing

Select Ask-x and Click OK

Enter values of x

Note: No reason to enter the second value of x

Note: No reason to enter the second value of x

Note: The solutions of the quadratic equation

are three and six {3,6}.

For both values of x, y=0

Solving Quadratic Equations

Converting solutions of quadratic equations to factored form.

Solutions of quadratic equation:

If solutions are whole numbers

e.g. x= {a, b}

(x-a)(x-b)

If solutions are fractions

x= { a , c } b d

(bx-a)(dx-c)

x= { 5 , −3} a=5, b=2, c=-3 and d=1 (2x-5)(x+3)

2

Multiplying binomials

(2x-5)(x+3)

(2x)(x) = 2x 2

(2x)(+3) = 6x

(-5)(x) = -5x

(-5)(+3) = -15


[GED MATH]

Advanced Computing

Combine like terms

2x 2 +6x-5x-15

2x 2 +x-15

Practice

Answers on p. 130

Multiply Binomials

(x+4)(x+2)

(x-6)(x-3)

(x+8)(x-2)

(x-7)(x+5)

(2t+3)(2t-3)

(3x + 3) 2

(2t+1)(2t-3)


[GED MATH]

Advanced Computing

Complete table

Answers on p. 131


x 2 − 2x − 15

x 2 + 4x − 32

(x − 7)(x + 5)

(2x + 3)(x − 4)

(+3, −8)

(− 5 4 , 3)

Solve the following Quadratic equations using the quadratic formula. (Calculator use permitted)

Note: Be sure the quadratic equation is in standard form . (Ax 2 +Bx + C = 0)

Quadratic Equation

Solution(s)

2x 2 − 3x − 5 = 0

2m 2 + 3m − 20 = 0

p 2 + 4p = −3

4x 2 + 8x + 7 = 4

2f 2 − 7f − 13 = −10


[GED MATH]

Advanced Computing

Answers

Practice (from p. 20)

Complete with and without the calculator

−3(7) =-21 17 − (−3) = 20 (−7.8)(−2.1) =16.38

−3 − 9 =-12 −(−39) + 4 = 43 16.53 + 17 = 33.53

−12 − (−9) = -3 (9)(−8) = -72 23.4 − 14.04 =9.36

−108

12

144

−12

= -9 (−3)(−7) = 21 3.31(.3) =.993

= -12 16.78(2.3) = 38.594 (−24)(3.2) =-19.2

4

Concept Questions

1) Give three (3) places when rounding may be used on the GED.

a) Whole Number

b) Tenths place

c) Hundredths place

2) To round using the TI-30XS calculator, what key does one press initially? Mode

3) To round to the nearest whole number, using the TI-30XS calculator, under FLOAT, which

number must be selected? 0nearest tenth? 1 nearest hundredth? 2

Answers Page 110

[GED MATH]

Advanced Computing

Practice

Please complete each problem by hand and then check using the calculator

Round the following numbers to the indicated place

652.789

Whole Number 653

Tenth 652.8

Hundredth 652.79

156.805

Whole Number 157

Tenth 156.8

Hundredth 156.81

9,873.327

Whole Number 9,873

Tenth 9,873.3

Hundredth 9,873.33

2.017

Whole Number 2

199.9505

Whole Number 200

Tenth 200.0

Hundredth 199.95

Tenth 2.0

Hundredth 2.02

Answers Page 111

[GED MATH]

Advanced Computing

Practice

Complete the following with and without the calculator

8 2 = 64 2 3 3

=8 √64 =8 √64

=4

11 2 =121 3 3 3

=27 √25 =5 √125

=5

4 2 =16 4 3 3

=64 √36 =6 √1,000

=10

5 2 =25 5 3 3

=125 √100 =10 √8

=2

10 2 =100 10 3 3

=1,000 √169 =13 √27

=3

Complete the following with or without the calculator

√7 + 3√7 = 4√7

10√3 − 5√3 = 5√3

√ 169

4 = 13

2

√48 = 4√3 √96 = 4√6 2 4 (2 −3 ) = 2

4√5 × 3√5 = 60

3

√27

+ 2 3 = 11 5 3 + 3 3 = 152

3 2 + 4 2 =25 √13 2 − 5 2 = 12 √6 2 + 8 2 = 10

7. The inverse operation of squaring a number is Square Root

8. Any base raised to the zero power is 1

9. When multiplying exponents with the same base, add the exponents.

10. When dividing exponents with the same base, subtract the exponents.

11. √ 2 = 12 =12

Answers Page 112

[GED MATH]

Advanced Computing

Practice

Add/Subtract the following fractions (if necessary, convert to a mixed number, then convert to a

decimal)


3

+ 4 1 = 5 1 5.25

4 2

4

4

5 + 2 10 = 1 1

8 1 8 − 5 3 4 = 2 3 8

2.375

Practice

34 3 10 − 23 7 10 = 10 3 5

10.6

Complete the following table. If necessary, convert to a mixed number.


(5 3 4 )(4 1 2 )= 25 7 8

25.875

(10 4 5 )(2 1 ) = 27 27

10

2 1 8 ÷ 3 4 = 2 5 6

7 1 3 ÷ 23 7 10 = 220

711

2.8333333

.309423347

8 1 5

4

5

= 10 1 4

10.25

Answers Page 113

[GED MATH]

Advanced Computing

Practice

Complete the following table

Hours Studying

Monday Tuesday Wednesday Thursday Friday Total (Mixed

Fraction)

Joe

Jeremy

Journey

Jonah

Jerel

2 ½ ½ 1 1 ¼ 2 7 1 4

2 2 2 1 ¼ 2 ½ 10 3 4

3 1 ½ 5 1 2

4 3.5 2.25 .75 3.75 14 1 4

2.5 2 1.25 1.4 2.6 9 3 4

Total

(Decimal) 15.0 8 6.5 5.65 11.35 47.5

Concept Questions

What are the three types of fractions?

1. Proper

2. Improper

3. Mixed Number

How can 4 be expressed as an improper fraction?

4

1

What key(s) converts fractions to decimals and decimals to fractions?

12

0 =Undefined 0

3 =0

What values of x would make this function

"undefined."

x =1 and -2

(x 2 + 5x − 9)

3(x − 1)(x + 2)

Answers Page 114

[GED MATH]

Advanced Computing

Practice

Complete the following table: (If necessary, convert improper fractions →Mixed Numbers

Percent Decimal Fraction

65% .65

72% .72

75% .75

3% .03

28% .28

80% .8

62.5% .625

13

20

18

25

3

4

3

100

7

25

4

5

1

50% .5

2

112% 1.12 1 3

25

335% 3.35 3 7

20

5

8

Answers Page 115

[GED MATH]

Advanced Computing

Practice

Part

50% of 75 37.5

100% of 175 175

40% of 80 32

15% of 70 10.5

8% of 65 5.2

Practice

Percent

25 of 75 33%

20 of 80 25%

40 of 80 50%

10.5 of 70 15%

1.3 of 65 2%

Answers Page 116

[GED MATH]

Advanced Computing

Practice

Whole

40 is 50% of what number 80


30 is 25% of what number? 120



116.66667 or 116 2 3

Practice

Percent Change (Round to nearest whole percent)

50 →80 60% Increase

120→100

17% Decrease

1,200→1,500

25% Decrease

63→80

27% Increase

150→115

23% Decrease

Concept Questions

1. To change a decimal to a percent, one must move the decimal 2places to the left/right.

2. To change a percent to a decimal, one must move the decimal 2places to the left/right.

3. Why is the decimal moved to the left? Dividing by 100

4. Why is the decimal moved to the right? Multiplying by 100

5. Percent = Part

Whole

Answers Page 117

[GED MATH]

Advanced Computing

Practice

Concept Questions

Solve for x

2

3 = x 9

3

5 = x 60

4

9 = 24

x

1.5

6.0 = x 18

17

4 = 51

x

x

18

3 = 6

180

5 = 36

116

4 = 29

27

6 = 4.5

204

17 = 12

1. What are three ways one can solve a proportion?

Equivalent Fractions

Unit Rate

Cross Multiply

2. What type of problems can be solved using proportions?

Scale

Recipes

Gas efficiency

Answers Page 118

[GED MATH]

Advanced Computing

Practice

1. A bag of marbles contains 5 blue, 4 red, 3 green, 7 black, and 1 white marble.

What is the chance a blue marble is selected? 5 = 1 =. 25 = 25%

20 4

What is the percentage of a blue or black marble being selected? 12

20 = 3 5

=. 6 = 60%

2. Ben selects a card from a deck of cards, without replacement, selects another card. What is the

probability he will select a King then an club? ( 1 ) 13 (1) = 1 = 1. 92%

4 52

3. Joan and Heaven each buy raffles from their local basketball team. Raffles cost $1.00. Joan buys

20 raffles and Heaven buys 15 raffles. If 225 raffles were sold, what is the probability, to the

nearest whole percent, of either winning the raffle? 35

= 15. 56%

Practice

225 = 7 45

You just got a free ticket for a boat ride, and you can bring along 2 friends! Unfortunately, you

have 5 friends who want to come along. How many different groups of friends could you take with

you? 20

Michael is packing his bags for his vacation. He has 5 unique shirts, but only 4 fit in his bag.

How many different groups of 4 shirts can he take? 120

William is packing his bags for his vacation. He has 8 unique books, but only 5 fit in his bag.

How many different groups of 5 books can he take? 6720

Answers Page 119

[GED MATH]

Advanced Computing

Practice

Type of Nuts

Price per lb. of

Peanuts 2.50

Almonds 14.00

Cashews 9.00

Pistachios 2.50

Walnuts 10.00

1. What is the median price for nuts? $9.00

2. What is the mean price of nuts? $7.60

3. What is the mode price of nuts? $2.50

4. What is the range of the prices of nuts? $11.50

Jabes counted the number of students in each class on the 2 nd floor. There were: 20, 31, 20, 18, 16, 20,

16, and 11 respectively.

5. What is the median number of students on the second floor? 19

6. What is the mean number of students on the second floor?19

7. What is the mode for number of students on the second floor?20 (appears 3t)

8. What is the range of the number of students on the second floor?20 (31-11)

9. If Jabes counts one more room of students, the new average is 20. How many students were in this

class?28

10. Steve found a few old books. The books have the following numbers of pages: 28, 35, 15, and 18. He

then found one more book. The new average of pages after finding the new book is 26. How many

pages are in the new book?34

Concept Questions

1. What is the mathematical term for "average? mean

2. Before determining the median, the set of values must be sorted – put in order.

3. Another term used for the range is variance.

4. The sum is a term that means addition.

Answers Page 120

[GED MATH]

Advanced Computing

Practice

Complete the table

Standard form

Scientific Notation

.00000734 7. 34 × 10 −6

.00045 4. 5 × 10 −4

5,234,000 5. 234 × 10 6

678,000 6. 78 × 10 5

.0000003968 3. 968 × 10 −7

4,153,000 4. 153 × 10 6

7,800,000,000 7. 8 × 10 9

.0000345 3.45 x 10 −5

2,450,000 2.45 x 10 6

7.5 x 10

1,500

5

5 x 10 2

Write your final answer in scientific notation, and round to two decimal places.

1. There are 25 cells each measuring 1.5 x 10 −6 on a microscope slide. What is the total width of

the cells’? 3.75 x 10 -5

2. The population of the US is 3.282 x 10 8 and New York city has a population of 8.4 x 10 6 . What

percent of the US lives in New York city. 2.56%

3. The volume of the Atlantic Ocean is about 3.1 x 10 17 .The Missouri River has an annual flow of 5.8 x

10 11 cubic meters. How many times would the annual flow of the Missouri River fit in the Atlantic

Ocean?5.34 x 10 5

Answers Page 121

[GED MATH]

Advanced Computing

Practice

Find the missing side (If necessary, round to the nearest tenth)

Concept Questions

Sides of Right Triangle

a b c

3 4 5

5 12 13

9 12 15

15 20 25

15 36 39

30 40 50

9 40 41

6 9 3√13 = 10. 82

8 15 17

1. What is the Pythagorean Theorem?a 2 + b 2 = c 2

2. The longest side of a "right" triangle is called the hypotenuse.

3. Two special "right" triangles discussed above are the3-4-5 and 5-12-13

4. The two legs of a right triangle are denoted by the variables a,b.

5. The hypotenuse, the longest side, is denoted by the variable c.

Practice

Complete the following table for rectangles using the given

Length Width Perimeter

10 in 12 in 44 in

8 in 17 in 50 in

12.5 in 15 in 55 in

14.5 in 20 in 69 in

32 in 17 ft. 98 ft.

Answers Page 122

[GED MATH]

Advanced Computing

Practice


Shape

Base(s)

Side(s)

Height Perimeter Area

Rectangle 8 in 5 in 26 in. 40 sq. in.

Rectangle 12 ft 6 ft 36 ft 72 sq. ft.

Triangle

Triangle

Parallelogram

Parallelogram

Trapezoid

Concept Questions

b =12 cm

s2 = 5 cm

s3 = 5 cm

b =10 cm

s2 = 7 cm

s3 = 7 cm

15 ft

8 ft

6 ft

8 ft

b1 =10 yd

b2 = 12 yd

s3 = 11 yd

s4 = 14 yd

1. What is the formula to find the area of a triangle?A = 1 2 bh

2. What is the formula to find the area of a parallelogram? A = bh

3. What is the formula to find the area of a circle? A = πr 2

4. What is the formula to find the area of a trapezoid? A = (b 1+b 2 )

h

2

5. The radius is 1 the diameter.

2

6. The diameter is twice the radius.

5 cm 19 cm 30 sq. cm

5 cm 24 cm 25 sq. ft

8 ft 46 ft 120 sq. ft.

10 ft 28 ft 60 sq. ft.

7 yd 47 yd 77 sq. yd.

Answers Page 123

[GED MATH]

Advanced Computing

Practice


Radius Diameter Circumference Area

5 ft. 10 ft 31.4 ft 78.5 sq. ft.

10 in. 20 in 62.8 in 314 sq. in

1 1 yd 2 1 yd 7.85 yd 4.90625 sq. yd

4 2

9.5 cm 19 cm 59.66 cm 283.385 sq cm

7 in 14 in 43.96 in 153.86 sq. in

15 ft 30 ft 94.2 ft 706.5 sq. ft.

5.5 in 11 in 34.54 in. 94.985 sq in

106 ft 212 ft 665.68 ft. 35,281.04 sq ft

Concept Questions

1. What is the formula to find the circumference of a circle? C = dπ or 2πr

2. What is the formula to find the area of a circle? A = πr 2

3. What is the formula to find the radius given the area? r = √ A π

4. What is the formula to find the diameter given the circumference? d = C π

Answers Page 124

[GED MATH]

Advanced Computing

Practice

1. Find the height of a cylinder with a volume of 2260.8 ft 3 and a radius of 6 ft. 20

2. Find the radius of a cylinder with a volume of 150 in 3 and a height of 10 in. 2.185

3. Find the height of a cone with a volume of 21 ft 3 and a radius of 4 ft 1.25

4. Find the radius, diameter and circumference for each given the area is 200.96 sq. ft. of the

circle.

5. r= 8 ft d= 16 ft C= 50.24 ft.

6. Find the radius, diameter and area for each given the circumference is 78.5 ft. of the circle.

7. r= 12.5 ft d= 25 ft A= 490.625 ft 2

Concept Questions

1. What is the formula to find the volume of a cylinder? V = πr 2 h

2. What is the formula to find the volume of a rectangular prism? V = L × W × H

3. What is the formula to find the volume of a pyramid? V = L×W×H

4. What is the formula to find the volume of a cone? V = πr2 h

3

3

5. What is the formula to find the radius if given the height and the volume of a cylinder r = √ A πh

6. What is the formula to solve for the radius given the area of a circle? r = √ A π

7. What is the formula to solve for the radius given the circumference? r = C

2π

Answers Page 125

[GED MATH]

Advanced Computing

Practice

Complete the table with/without the calculator

What is the slope of the line that passes

through (-3,-2) and (1,6)?

y 2 = 6 y 1 = -2

x 2 = 1

x 1 = −3

m = (6−(−2)

(1−(−3) = 8 4 = 2


through (-7,+5) and (1,9)?

y 2 =9 y 1 =+5

x 2 = 1 x 1 =-7

m = (9−5)

(1−(−7) = 4 8 = 1 2


through (-3,-2) and (1,-3)?

y 2 = −3

x 2 =1

y 1 = −2

x 1 = −3

m = (−3−(−2))

(−3−(+1) = −1

−4 = 1 4


through (-12,2) and (-2,-3)?

y 2 = -3

x 2 = −2

y 1 = 2

x 1 = −12

m = (+2−(−3))

(−12−(−2) = −1

−10 = 1 10


through (-3,-2) and (1,6)?

Practice

y 2 = 6 y 1 = -2

x 2 = 1

y 1 = −3

m = (−2−(6))

(−3−(1) = −8

−4 = 2 1

Point(s) Slope Equation of line

(1,-3) (-2,3) m = −2 a = −2, b = −1

y = −2x − 1

(2,3) (1,2) m = 1 a = +1, b = +1

y = x + 1

(-9,-6) (-5,2) m = +2 a = +2, b = +12

y = 2x + 12

(-8,-3) (-4,0)

(0,4) (-5,3)

m = . 75 = 3 4

m = . 2 = 1 5

a = . 75, b = +3

y = 3 x + 3

4

a = . 2, b = +4

y = 1 x + 4

5

Answers Page 126

[GED MATH]

Advanced Computing

Practice

Determine the slope of each line

Equation

5x + 2y = 8

Slope and y-intercept(m,b)

m = −5

2

b = 8 2 = 4

−x + 2y = 8

m = 1 2

b = 8 2 = 4

y = 2 m

x − 10

3

y = 6x − 10

= 2 3

b = −10

m = 6

b = −10

Practice

−8x + 3y = 21

m = 8 3

b = 21

3 = 7

Find the equation of the parallel line

Given

Passes through (3,1)and is parallel to

y = 3x + 10


y = 3 2 x − 5

Passes through(−4, −1)and parallel to

−3x + 4y = 8


4x − 3y = 9

Equation of a parallel line

y = 3x − 8

y = 3 x + 13

2

y = 3 x + 4

4

y = 4 10

x +

3 3

Answers Page 127

[GED MATH]

Advanced Computing

Practice

Given

Passes through (3,1)and is perpendicular to

y = 3x + 10


y = 3 2 x − 5


−x − 4y = 8

Equation of a perpendicular line

y = − 1 x + 2

3

y = − 2 x − 4

3

y = 4x + 6

Complete the following table.

Slope

The slope of a parallel line

The slope of a Perpendicular

line

4 4 − 1 4

3

2

2

5

−5

− 1 8

Concept Questions

1. Parallel lines have the same slope.

3

2

2

5

−5

− 1 8

− 2 3

− 5 2

1

5

+8

2. The slope is the negative inverse of the slope of a perpendicular line.

3. The slope of the line, which is decreasing from left to right, is negative.

4. m is the variable for slope.

5. b is the variable for the y-intercept.

6. Ax + By = C is the standard form of a linear equation.

7. y = mx + b is the slope intercept form of a linear equation.

Answers Page 128

[GED MATH]

Advanced Computing

Practice

Determine if the following examples are functions by circling VALID or INVALID

x

y

3 17

5 19

5 23

7 32

VALID

INVALID

x

y

-5 4

5 14

10 10

0 9

VALID

INVALID

x

y

-2 1

-2 -3

2 1

2 5

(-7, 23) (0,13) (2,24) (2,32)

VALID

VALID

INVALID

INVALID

Practice

Two variables

1. The cost of a school banquet is $200 plus

$15.00 for each person attending. Write the

equation that gives the total cost as a

function of the number of people attending.

What is the cost of 75 people?

2. The size of a tree to be planted is 4.2 ft. Each

day the tree grows .002 inches. Write the

equation that gives the total size as a

function of the number of days. What is the

size of the tree in feet after 3 years?

200 is constant = b

15 per person = m (the number of people is

variable)

Therefore, y = 15x + 200

y = 15(75) + 200

y = 1,125 + 200

y = 1,325

4.2 is the size of the tree when planted = b

The tree grows .002 feet per day. (the number of

days is variable)

365 = 1 year

Therefore y = .002(365 x 3) + 4.2

y = 2.19 + 4.2

y = 6.39 feet

Answers Page 129

[GED MATH]

Advanced Computing

3. Billy plans to paint baskets. The paint costs

$13.25. The baskets cost $6.75 each. Write

an equation that gives the total cost as a

function of the number of baskets made.

Determine the cost of nine baskets.

Cost of paint, no matter the number of baskets is

13.25

6.75 per basket (the number of baskets are

variable)

Y = 6.75b + 13.25

Cost of 9 baskets

Y = 6.75(9) + 13.25

Y = $74.00

Find zeroes

Once factored, find the value of x would result in the solution evaluating to be zero.


x 2 − 8x + 16 = 4

To change to standard form,

subtract four from both sides

x 2 − 8x + 12 = 0

(x-6)(x-2)

6,2

(6-6)(6-2)=0

and

(2-6)(2-2)=0

( 1 3 , −5)

(3x 2 + 14x − 5) (3x − 1)(x + 5)

(3 ( 1 3 ) − 1), (1 3 − 5)

(1 − 1) ( 1 3 − 5) = 0

(3(−5) − 1)(−5 + 5)

(−16)(−5 + 5) = 0

6x 2 − 5x + 1

(2x-1)(3x-1)

+ 1 2 , 1 3

Answers Page 130

[GED MATH]

Advanced Computing

Practice

Multiply Binomials

Complete table

(x+4)(x+2)=

(x-6)(x-3)=

(x+8)(x-2)=

(x-7)(x+5)=

(2x+3)(2x-3)=

(3x + 3) 2 =

(2x+1)(2x-3)=

x 2 + 6x + 8

x 2 − 9x + 18

x 2 + 6x − 16

x 2 − 2x − 35

4x 2 − 9

9x 2 + 18x + 9

4x 2 − 4x − 3


x 2 − 2x − 15

(x − 5)(x + 3) x = {5, -3}

x 2 + 4x − 32

(x − 4)(x + 8) x = {4, -8}

x 2 − 2x − 35 (x − 7)(x + 5)

x = {7,-5}

2x 2 + 5x − 12 (2x + 3)(x − 4) x = {− 3 2 , 4}

x 2 + 5x − 24 (x − 3)(x + 8) x = {+3, −8}

4x 2 − 7x − 15 (4x + 5)(x − 3) x = {− 5 4 , 3}

Answers Page 131

[GED MATH]

Advanced Computing

Quadratic Equation

2x 2 − 3x − 5 = 0

Solution(s)

x = { 5 2 , −1}

2m 2 + 3m − 20 = 0

x = { 5 2 , −4}

p 2 + 4p = −3

x = {−1, −3}

4x 2 + 8x + 7 = 4 x = {− 1 2 , − 3 2 }

2f 2 − 7f − 13 = −10

x = {

7 ± √73

4

Answers Page 132

[GED MATH]

Advanced Computing

Resources

The following resources are free of charge and ones that I recommend:

GED Services

CK-12

• Free Practice Questions – (https://app2.ged.com/portal/home-practice)

• Practice Test – (https://app2.ged.com/portal/study/practice#gedReadyPracticeTest)

• Mathematics Study Guide - (https://app2.ged.com/portal/study-guide/MATH)

• Most Missed GED Questions (See Below.)

• GED Mathematics Formula Sheet - (https://ged.com/wpcontent/uploads/math_formula_sheet.pdf)

Brain Genie (http://braingenie.ck12.org/) (Highly recommended)

A website that gives practice for any of the aforementioned topics

Khan Academy (http://khanacademy.org/)

A website that gives practice for any of the aforementioned topics

Resources Page 133

[GED MATH]

Advanced Computing

Resources Page 134

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