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CHAPTER 5 – Equations Instructor: Dr.Gehan Shanmuganathan 5-1 Learning Outcomes Solve equations using multiplication or division. Solve equations using addition or subtraction. Solve equations using more than one operation. Solve equations containing multiple unknown terms. Solve equations containing parentheses. Solve equations that are proportions. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations using multiplication or division 5-1-1 Section 5-1 Equations An equation is a mathematical statement in which two quantities are equal. Solving an equation means finding the value of an unknown. Example: 8x = 24 To solve this equation, the value of x must be discovered. Division is used to solve this equation. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations using multiplication or division Section 5-1 Equations Letters, such as (x,y,z) represent unknown amounts and are called unknowns or variables. 4x = 16 The numbers are called known or given amounts. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations using multiplication or division Section 5-1 Equations Any operation performed on one side of the equation must be performed on the other side of the equation as well. – If you “multiply by 2” on one side, you must “multiply by 2” on the other side. – If you “divide by 3” on one side, you must also “divide by 3” on the other side. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations using multiplication or division HOW TO: Section 5-1 Equations 8x = 24 STEP 1 Isolate the unknown value and determine if multiplication or division is needed. STEP 2 Use division to divide both sides by 8. STEP 3 Simplify: x = 3 3 x 8 = 24 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Find the value of an unknown using multiplication HOW TO: Section 5-1 Equations Find the value of a: a =6 3 Multiply both sides by 3 to isolate a. The left side becomes 1a or a. The right side becomes the product of 6 x 3, or 18. a = 18 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-1 Equations 2b = 40 STEP 1 Determine which operation is needed. Division STEP 2 Perform the same operation to both sides. Divide both sides by 2. STEP 3 Isolate the variable and solve. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel 40 b= = 20 2 © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations using addition or subtraction 5-1-2 Section 5-1 Equations Adding or subtracting any number from one side must be carried out on the other side as well. – Subtract “the given amount” from both sides. Would solving 4 + x = 16 require addition or subtraction of “4” from each side? Subtraction Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations using addition or subtraction HOW TO: Section 5-1 Equations 4 + x = 10 STEP 1 Isolate the unknown value and determine if addition or subtraction is needed. STEP 2 Use subtraction to isolate x. STEP 3 Simplify: x = 6 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-1 Equations b - 12 = 8 STEP 1 Determine which operation is needed. Addition STEP 2 Perform the same operation to both sides. Add 12 to both sides STEP 3 Isolate the variable and solve. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel b = 8 + 12 = 20 © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations using more than one operation 5-1-3 Section 5-1 Equations Isolate the unknown value. – Add or subtract as necessary first. – Multiply or divide as necessary second. Identify the solution. – The number on the side opposite the unknown. Check the solution by “plugging in” the number using the original equation. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Order of operations Section 5-1 Equations When two or more calculations are written symbolically, the operations are performed according to a specified order of operations. – First — perform multiplication and division as they appear from left to right. – Second — perform addition and subtraction as they appear from left to right. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Order of operations Section 5-1 Equations To solve an equation, undo the operations, working in reverse order – First — undo the addition or subtraction. – Second — undo multiplication or division. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-1 Equations 7x + 4 = 39 STEP 1 Undo the addition by subtracting 4 from each side. 7x = 35 STEP 2 Divide each side by 7. 35 x= =5 7 STEP 3 Verify by plugging in 5 in place of x . 7 (5) + 4 = 39 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel 35 + 4 = 39 © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations containing multiple unknown terms 5-1-4 Section 5-1 Equations In some equations, the unknown value may occur more than once. The simplest instance is when the unknown value occurs in two addends, such as 3a + 2a = 25 – Add the numbers in each addend (2+3). – Multiply the sum by the unknown (5a = 25). – Solve for a (a = 5). Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-1 Equations Find a if: a + 4a – 5 = 39 STEP 1 Combine the unknown value addends. a + 4a = 5a 5a – 5 = 30 STEP 2 Undo the subtraction. 5a = 35 STEP 3 Undo the multiplication. a=7 STEP 4 Check by replacing a with 7. 7 + 4(7) = 35 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel Correct! © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations containing parentheses 5-1-5 Section 5-1 Equations Eliminate the parentheses. – Multiply the number just outside the parentheses by each addend inside the parentheses. – Show the resulting products as addition or subtraction, as indicated Solve the resulting equation. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-1 Equations Solve: 6A + 2 = 24 STEP 1 Multiply 6 by each addend. 6 multiplied by A + 6 multiplied by 2 STEP 2 Show the resulting products. 6A + 12 = 24 STEP 3 Check by replacing a with 7. 7 + 4(7) = 35 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-1 Equations 5 (x - 2) = 45 TIP: Remove the parentheses first. 5x -10 = 45 5x = 55 x = 11 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations that are proportions 5-1-6 Section 5-1 Equations A proportion is based on two pairs of related quantities. The most common way to write proportions is to use fraction notation—also called a ratio. – When two ratios are equal, they form a proportion. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Solve equations that are proportions 5-1-6 Section 5-1 Equations A cross product is the product of the numerator of one fraction, times the denominator of another fraction. – An important property of proportions is that the cross products are equal. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Verify that two fractions form a proportion HOW TO: Section 5-1 Equations 6 4 Do and form a proportion? 12 18 STEP 1 Multiply the numerator from the first fraction by the denominator of the second fraction. 4 x 18 = 72 STEP 2 Multiply the denominator of the first fraction by the numerator of the second fraction. 6 x 12 = 72 Are they equal? Yes, they form a proportion. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved 5-2 Learning Outcome Use the problem-solving approach to analyze and solve word problems. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Use the problem-solving approach to analyze and solve word problems. 5-2-1 Section 5-2 Using Equations to Solve Problems Five step problem solving approach: – What you know. • Known or given facts. – What you are looking for. • Unknown or missing amounts. – Solution Plan. • Equation or relationship among known/unknown facts. – Solution. • Solve the equation. – Conclusion. • Solution interpreted within context of problem. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Use the problem-solving approach to analyze and solve word problems. 5-2-1 Section 5-2 Using Equations to Solve Problems Key words in Table 5-1 will guide you in using the problem-solving approach. See page 533 These words help you interpret the information and begin to set up the equation to solve the problem. “of” often implies multiplication. Example: “¼ of her salary” means “multiply her salary by ¼” Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Full time employees work more hours than part-time employees. If the difference is four per day, and part-time employees work six hours per day, how many hours per day do full-timers work? What are we looking for? Number of hours that FT employees work. What do we know? PT employees work 6 hours, and the difference between FT and PT is 4 hours. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Full time employees work more hours than part-time employees. If the difference is four per day, and part-time employees work six hours per day, how many hours per day do full-timers work? We also know that Set up a solution plan. “difference” implies What are we looking for? subtraction. FT – PT = 4 Number of hours that FT employees work. FT = N [unknown] PT = 6 hours What do we know? Conclusion: PT employees andtime the employees Solution plan: N = 4 + 6 =work 10 6 hours,Full difference between FT and PT iswork 4 hours. 10 hours. N–6=4 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved HOW TO: Section 5-2 Use the solution plan Using Equations to Solve Problems Jill has three times as many trading cards as Matt. If the total number both have is 200, how many does Jill have? 1. What are you looking for? The number of cards that Jill has. 2. What do you know? The relationship in the number of cards is 3:1; total is 200 3. Set up a solution plan. x(Matt’s) + 3x(Jill’s) = 200 4. Solve it. x + 3x = 200; 4x = 200; x = 50 5. Draw the conclusion. Jill has 3x, or 150 cards Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved HOW TO: Section 5-2 Use the solution plan Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? What are you looking for? How many humorous cards were ordered and how many nature cards were ordered—the total of H + N = 600 or N = 600 – H. If we let H represent the humorous cards, Nature cards will be 600 – H, which will simplify the solution process by using only one unknown: H. MORE Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved HOW TO: Use the solution plan Section 5-2 Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? What do you know? A total of $950 was spent. Two types of cards were ordered. The total number of cards ordered was 600. Humorous cards cost $1.75 each and nature cards cost $1.50 each. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel MORE © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Use the solution plan HOW TO: Section 5-2 Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? Solution Plan Set up the equation by multiplying the unit price of each by the volume, represented by the unknowns equaling the total amount spent. $1.75(H) + $1.50 (600 – H) = $950.00 Unit prices Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel Volume unknowns Total spent MORE © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved HOW TO: Section 5-2 Use the solution plan Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? Solution $1.75H + $1.50(600 - H) = $950.00 $1.75H + $900.00 - $1.50H = $950.00 $0.25H + $900.00 = $950.00 $0.25H = $50.00 MORE H = 200 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved HOW TO: Section 5-2 Use the solution plan Using Equations to Solve Problems Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order? Conclusion The number of humorous cards ordered is 200. Since nature cards are 600 – H, we can conclude that 400 nature cards were ordered. Using “200” and “400” in the original equation proves that the volume amounts are correct. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved HOW TO: Section 5-2 Use the solution plan Using Equations to Solve Problems Denise ordered 75 dinners for the awards banquet. Fish dinners cost $11.75 and chicken dinners cost $9.25 each. If she spent a total of $756.25, how many of each type of dinner did she order? $11.75(F) + $9.25(75 - F) = $756.25 $11.75F + $693.75 - $9.25F = $756.25 $2.50F + $693.75 = $756.25 $2.50F = $62.50 F = 25 Conclusion: 25 fish dinners and 50 chicken dinners were ordered. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Proportions Section 5-2 Using Equations to Solve Problems The relationship between two factors is often described in proportions. – You can use proportions to solve for unknowns. Example: The label on a container of weed killer gives directions to mix three ounces of weed killer with every two gallons of water. For five gallons of water, how many ounces of weed killer should you use? Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Proportions Section 5-2 Using Equations to Solve Problems The relationship between two factors is often described in proportions. – You can use proportions to solve for unknowns. 1. What are you looking for? Example: Number of ounces of weed killer needed for 5 gallons of water. The label on a container of weed killer gives directions to mix 2. What do you know?killer with every two gallons of water. three ounces of weed 2 gallons of water, need 3 oz. ofof weed killer. For For fiveevery gallons of water, how you many ounces weed killer should you use? 2 5 Set up a solution plan. = Conclusion. 3 x Solve it. You need 7.5 oz of weed killer for 5 gal of water. Cross multiply: 2x = 15; x = 7.5 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Proportions Section 5-2 Using Equations to Solve Problems Many business-related problems that involve pairs of numbers that are proportional involve direct proportions. – An increase (or decrease) in one amount causes an increase (or decrease) in the number that pairs with it. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-2 Using Equations to Solve Problems Your car gets 23 miles to the gallon. How far can you go on 16 gallons of gas? 1 gallon 16 gallons = 23 miles x miles Cross multiply: 1x = 368 miles Conclusion: You can travel 368 miles on 16 gallons of gas. In this example, an increase in the amount of gas would directly and proportionately increase the mileage yielded. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved 5-3 Learning Outcomes Evaluate a formula. Find an equivalent formula by rearranging the formula. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved 5-3-1 Evaluate the formula Section 5-3 Formulas Write the formula. Rewrite the formula substituting known values for the letters of the formula. Solve the equation for the unknown letter or perform the indicated operations, applying the order of operations. Interpret the solution within the context of the formula. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-3 Formulas A plasma TV that costs $2,145 is marked up $854. What is the selling price of the TV? Use the formula S = C + M where S is the selling price, C is the cost, and M is Markup. S = $2,145 + $854 S or Selling Price = $2,999 Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved 5-3-2 Section 5-3 Find an Equivalent Formula by Rearranging the Formula Formulas Determine which variable of the formula is to be isolated (solved for). Highlight or mentally locate all instances of the variable to be isolated. Treat all other variables of the formula as you would treat numbers in an equation, and perform normal steps for solving an equation. If the isolated variable is on the right side of the equation, interchange the sides so that it appears on the left side. Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved An Example… Section 5-3 Formulas The formula for Square Footage = Length x Width or S = L x W. Solve the formula for W or width. Isolate W by dividing both sides by L. The new formula is: S =W L Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved EXERCISE SET A Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved EXERCISE SET A Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved EXERCISE SET A Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved EXERCISE SET A Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved EXERCISE SET A Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved EXERCISE SET A Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved PRACTICE TEST Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved Business Math, Ninth Edition Cheryl Cleaves, Margie Hobbs & Jeffrey Nobel © 2012 Pearson Education, Inc. Upper Saddle River, NJ 07458 All Rights Reserved