1. Copyright © 2011 Pearson Education, Inc. Problem Solving CHAPTER 3.1Ratios and Proportions 3.2Percents 3.3Problems with Two or More Unknowns 3.4Rates 3.5Investment and Mixture 3

2. Copyright © 2011 Pearson Education, Inc. Ratios and Proportions 3.1 1.Solve problems involving ratios. 2.Solve for a missing number in a proportion. 3.Solve proportion problems. 4.Use proportions to solve for missing lengths in figures that are similar.

3. Slide 3- 3 Copyright © 2011 Pearson Education, Inc. Ratio: A comparison of two quantities using a quotient. Ratios are usually expressed in fraction form or with a colon. When a ratio is written in English, the word to separates the numerator and denominator quantities. The ratio of 12 to 17 translates to NumeratorDenominator Unit ratio: A ratio with a denominator of 1.

4. Slide 3- 4 Copyright © 2011 Pearson Education, Inc. Example 1 A bin at a hardware store contains 120 washers and 85 bolts. Write the ratio of washers to bolts in simplest form. Solution The ratio of washers to bolts =

5. Slide 3- 5 Copyright © 2011 Pearson Education, Inc. Example 2 A bin at a hardware store contains 120 washers and 85 bolts. Write the ratio of washers to bolts in simplest form. Express the ratio as a unit ratio. Interpret the answer. Solution The unit ratio indicates that there are 1.41 washers for every bolt.

6. Slide 3- 6 Copyright © 2011 Pearson Education, Inc. Proportion: An equation in the form, where and. Proportions and Their Cross Products If where and then ad = bc.

7. Slide 3- 7 Copyright © 2011 Pearson Education, Inc. Solving a Proportion To solve a proportion using cross products: 1. Calculate the cross products. 2. Set the cross products equal to each other. 3. Use the multiplication principle of equality to isolate the variable.

8. Slide 3- 8 Copyright © 2011 Pearson Education, Inc. Example 3a Solve. Solution Calculate the cross products. Set the cross products equal to one another. Divide both sides by 5 to isolate x.

9. Slide 3- 9 Copyright © 2011 Pearson Education, Inc. Example 3b Solve. Solution Calculate the cross products. Set the cross products equal to one another. Multiply both sides by 5/26 to isolate w.

10. Slide 3- 10 Copyright © 2011 Pearson Education, Inc. Example 4 Pedro drives his semi truck 2200 miles in 4 days. At this rate, how many miles will he drive in 15 days? Understand We are to find the amount of miles that Pedro will drive in 15 days if he drives 2200 miles in 4 days. Translating 2200 miles every 4 days to a ratio, we have Plan At this rate indicates stays the same for 15 days, so we can use a proportion.

11. Slide 3- 11 Copyright © 2011 Pearson Education, Inc. continued Execute Set the cross products equal to one another. Divide both sides by 4 to isolate x.

12. Slide 3- 12 Copyright © 2011 Pearson Education, Inc. continued Answer Pedro will drive 8250 miles in 15 days. Check Verify the reasonableness of the answer. Pedro drives about 550 miles per day so is correct.

13. Slide 3- 13 Copyright © 2011 Pearson Education, Inc. Solving Proportion Application Problems To solve proportion problems: 1. Set up a proportion in which the numerators and denominators of the ratios correspond in a logical manner. 2. Solve using cross products.

14. Slide 3- 14 Copyright © 2011 Pearson Education, Inc. Congruent angles: Angles that have the same measure. The symbol for congruent is. Similar figures: Figures with congruent angles and proportional side lengths.

15. The two figures are similar. Find the missing lengths. Slide 3- 15 Copyright © 2011 Pearson Education, Inc. Example 5 4 m 7 m 8 m c 3 m b a Understand Since the figures are similar, the lengths of the corresponding sides are proportional. Ratio of larger to smaller

16. Slide 3- 16 Copyright © 2011 Pearson Education, Inc. continued Plan Write a proportion to find each missing side length. Execute It is not necessary to write a proportion for c. The corresponding sides must be equal so we can conclude that c = 3. To find a: To find b:

17. Slide 3- 17 Copyright © 2011 Pearson Education, Inc. continued Answer The missing lengths are a = 6 m, b = 5.25 m, and c = 3 m. Check Verify the reasonableness of the answers by quickly doing estimates of the missing side lengths. These do check out.

18. Slide 1- 18 Copyright © 2011 Pearson Education, Inc. Solve. a) 12 b) 18 ¾ c) d) 21 ¾ 3.1

19. Slide 1- 19 Copyright © 2011 Pearson Education, Inc. Solve. a) 12 b) 18 ¾ c) d) 21 ¾ 3.1

20. Slide 1- 20 Copyright © 2011 Pearson Education, Inc. On a game preserve 112 deer were caught, tagged, and released. Later, 82 were caught in which 32 had tags. Estimate the number of deer in the preserve. a) 23 b) 44 c) 214 d) 287 3.1

21. Slide 1- 21 Copyright © 2011 Pearson Education, Inc. On a game preserve 112 deer were caught, tagged, and released. Later, 82 were caught in which 32 had tags. Estimate the number of deer in the preserve. a) 23 b) 44 c) 214 d) 287 3.1

22. Copyright © 2011 Pearson Education, Inc. Percents 3.2 1.Write a percent as a fraction or decimal number. 2.Write a fraction or decimal number as a percent. 3.Translate and solve percent sentences. 4.Solve problems involving percents.

23. Slide 3- 23 Copyright © 2011 Pearson Education, Inc. Percent: A ratio representing some part out of 100. The symbol for percent is %. For example, 20 percent, which means 20 out of 100, is written 20%. 20% = 20 out of 100

24. Slide 3- 24 Copyright © 2011 Pearson Education, Inc. Rewriting a Percent To write a percent as a fraction or decimal: 1. Write the percent as a ratio with 100 in the denominator. 2. Simplify to the desired form. Note: When simplifying, remember that dividing a decimal number by 100 moves the decimal point two places to the left.

25. Slide 3- 25 Copyright © 2011 Pearson Education, Inc. Example 1 Write each percent as a decimal number and a fraction. a. 14.4% Solution Write 14.4 over 100, then simplify. 14.4% = = 0.144 Write as a ratio with 100 in the denominator. Write the decimal form. Write the fraction form. Simplify to lowest terms.

26. Slide 3- 26 Copyright © 2011 Pearson Education, Inc. Example 1 b. Solution Write over 100, then simplify. Write as a ratio with 100 in the denominator. Rewrite the division. Write an equivalent multiplication. Multiply.

27. Slide 3- 27 Copyright © 2011 Pearson Education, Inc. Writing a Fraction or Decimal as a Percent To write a fraction or decimal number as a percent: 1. Multiply by 100%. 2. Simplify. Note: Multiplying a decimal number by 100 moves the decimal point two places to the right.

28. Slide 3- 28 Copyright © 2011 Pearson Education, Inc. Example 2 Write as a percent. a. b. 0.348 Solution a. 2 25 b. 0.348 = 0.348100% = 34.8%

29. Slide 3- 29 Copyright © 2011 Pearson Education, Inc. A percent of a whole is a part of the whole. A percent ofa wholeis a part. Unknown part:42% of68is what amount? Unknown whole:76% ofwhat numberis 63.84? Unknown percent:What percent of72is 63?

30. Slide 3- 30 Copyright © 2011 Pearson Education, Inc. Translating Simple Percent Sentences Method 1. Translate the sentence word for word. 1. Select a variable for the unknown. 2. Translate the word is to an equal sign. 3. If of is preceded by the percent, translate it to multiplication. If of is preceded by a whole number, translate it to division. Method 2. Translate to a proportion by writing the following form: where the percent is expressed as a fraction with a denominator of 100.

31. Slide 3- 31 Copyright © 2011 Pearson Education, Inc. Example 3 36% of 72 is what number? Solution Notice that the part is unknown. 36%of 72iswhat number? Percentofthe wholeisthe part

32. Slide 3- 32 Copyright © 2011 Pearson Education, Inc. continued Method 1: Word-for-word translation 36% of 72 is what number? 0.36  72 = n 25.92 = n

33. Slide 3- 33 Copyright © 2011 Pearson Education, Inc. continued Method 2. Proportion:  Whole  Part Equate the cross products. Divide both sides by 100 to isolate n.

34. Slide 3- 34 Copyright © 2011 Pearson Education, Inc. Example 4 42% of what number is 31.92? Solution Note the three pieces: 42%ofwhat numberis31.92? Percentofthe wholeisthe part

35. Slide 3- 35 Copyright © 2011 Pearson Education, Inc. continued Method 1. Word-for-word translation: 0.42 n = 31.92 n = 76 42% of what number is 31.92?

36. Slide 3- 36 Copyright © 2011 Pearson Education, Inc. continued Method 2. Proportion:  Whole  Part Equate the cross products. Divide both sides by 42 to isolate n.

37. Slide 3- 37 Copyright © 2011 Pearson Education, Inc. Example 5 What percent of 63 is 56? Solution Note the pieces: What percentof63is56? Percentofthe wholeisthe part

38. Slide 3- 38 Copyright © 2011 Pearson Education, Inc. continued Method 1. Word-for-word translation: p 63 = 56 63p = 56 p = 0.889 Answer 0.889  100% = 88.9% What percent of 63 is 56?

39. Slide 3- 39 Copyright © 2011 Pearson Education, Inc. continued Method 2. Proportion: Answer 88.9%  Whole  Part Equate the cross products. Divide both sides by 63 to isolate p.

40. Slide 3- 40 Copyright © 2011 Pearson Education, Inc. Solving Percent Application Problems To solve problems involving percent, 1. Determine whether the percent, the whole, or the part is unknown. 2. Write the problem as a simple percent sentence (if needed). 3. Translate to an equation (word for word or proportion). 4. Solve for the unknown.

41. Slide 3- 41 Copyright © 2011 Pearson Education, Inc. Example 6 Mia gets 97 questions correct on a test of 120 questions. What percent correct did she receive on the test? Understand The unknown is the percent of the questions she got correct. There were 120 questions, which is the whole. She got 97 correct, which is the part. Plan Write a simple percent sentence, then solve. We will let P represent the unknown percent, and we will translate to a proportion.

42. Slide 3- 42 Copyright © 2011 Pearson Education, Inc. continued Execute What percent of 120 is 97? Answer Mia got about 80.8% on the test. Check 0.808 120 = 96.96, therefore 80.8% is correct.  Part  Whole

43. Slide 3- 43 Copyright © 2011 Pearson Education, Inc. Example 7 Jamal purchases a new surfboard that costs $570. The sales tax is 7%, what is the total cost of the surfboard? Understand Determine the amount of sales tax and add to the price of the surfboard. Plan Write a simple percent sentence, then solve. We will let P represent the unknown percent, and translate word for word.

44. Slide 3- 44 Copyright © 2011 Pearson Education, Inc. continued Execute The sales tax is 7% of $570. Sales tax is $39.90 Answer The total cost is $570 + $39.90 = $609.90. Check The check is left to the student.

45. Slide 3- 45 Copyright © 2011 Pearson Education, Inc. Example 8 A pediatrician plots the growth of a child as growing from 27.5 inches to 31.25 inches. What is the percent of increase? Understand We are to determine the percent of increase in growth. Plan To determine the percent of increase in growth, we need the amount of the increase, which is found by subtracting the initial amount from the final amount.

46. Slide 3- 46 Copyright © 2011 Pearson Education, Inc. continued Execute Amount of increase = 31.25 – 27.5 = 3.75 What percent of 27.5 is 3.75? Answer The percent of increase in growth is 13.6%. Check 0.136 27.5 = 3.74. Add the increase: 27.5 + 3.74 = 31.24. The answer is reasonable.  Part  Whole

47. Slide 1- 47 Copyright © 2011 Pearson Education, Inc. What is 7/8 as a percent? a) 78% b) 87% c) 87.5% d) 718% 3.2

48. Slide 1- 48 Copyright © 2011 Pearson Education, Inc. What is 7/8 as a percent? a) 78% b) 87% c) 87.5% d) 718% 3.2

49. Slide 1- 49 Copyright © 2011 Pearson Education, Inc. 2.4% of 130 is what number? a) 31.2 b) 3.12 c) 1.83 d) 0.18 3.2

50. Slide 1- 50 Copyright © 2011 Pearson Education, Inc. 2.4% of 130 is what number? a) 31.2 b) 3.12 c) 1.83 d) 0.18 3.2

51. Slide 1- 51 Copyright © 2011 Pearson Education, Inc. What percent of 48 is 51.84? a) 93% b) 98% c) 108% d) 109% 3.2

52. Slide 1- 52 Copyright © 2011 Pearson Education, Inc. What percent of 48 is 51.84? a) 93% b) 98% c) 108% d) 109% 3.2

53. Copyright © 2011 Pearson Education, Inc. Problems with Two or More Unknowns 3.3 1.Solve problems involving two unknowns. 2.Use a table in solving problems with two unknowns.

54. Slide 3- 54 Copyright © 2011 Pearson Education, Inc. Solving Problems with Two or More Unknowns To solve problems with two or more unknowns: 1. Determine which unknown will be represented by a variable. Tip: Let the unknown that is acted on be represented by the variable. 2. Use one of the relationships to describe the other unknown(s) in terms of the variable. 3. Use the other relationship to write an equation. 4. Solve the equation.

55. Slide 3- 55 Copyright © 2011 Pearson Education, Inc. Example 1 At a book sale, Franklin buys one paperback and one hardback book. The hardback book is $3.00 more than the paperback. If his total purchase is $7.80 before tax, how much does each book cost? Understand There are two unknowns in the problem. We must find the price of the paperback book as well as the hardback book. Relationship 1: The cost of the hardback book is $3 more than the paperback book. Relationship 2: The total cost for both books is $7.80. Plan Translate each relationship into an equation and then solve.

56. Slide 3- 56 Copyright © 2011 Pearson Education, Inc. continued Execute Use the first relationship to determine the variable. Relationship 1: The cost of the hardback book is $3 more than the paperback book. Cost of hardback = $3 + cost of paperback Cost of hardback = $3 + n Now use the second relationship to write an equation. Relationship 2: The total cost for both books is $7.80. Cost of paperback + Cost of hardback = $7.80 n + 3 + n = 7.80 2n + 3 = 7.80 2n = 4.80 n = 2.40

57. Slide 3- 57 Copyright © 2011 Pearson Education, Inc. continued Answer Since n represents the cost of the paperback book, the paperback costs $2.40. The hardback book costs $3.00 more than the paperback, so $2.40 +$ 3.00 = $5.40. Check Verify the total cost is $7.80, which it is because $5.40 + $2.40 = $7.80.

58. Slide 3- 58 Copyright © 2011 Pearson Education, Inc. Example 3 At rectangular patio is to be built so that the length is 12 feet more than the width. The perimeter of the patio is to be 72 feet. Find the dimensions of the patio. Understand Draw a picture and list the relationships. Relationship 1: The length is 12 feet more than the width. Relationship 2: The perimeter must be 72 feet. Plan Translate to an equation using the key words and then solve the equation.

59. Slide 3- 59 Copyright © 2011 Pearson Education, Inc. continued Execute Relationship 1: The length is 12 feet more than the width. Length = Width + 12 Length = w + 12 Relationship 2: The perimeter is 72 feet. The formula for perimeter is P = 2l + 2w Perimeter = 72 2l + 2w = 72 2(w + 12) + 2w = 72 2w + 24 + 2w = 72 4w + 24 = 72 4w = 48 w = 12 w w w + 12 Substitute w + 12 for l. Distribute. Combine like terms. Subtract 24 from both sides. Divide both sides by 4 to isolate w.

60. Slide 3- 60 Copyright © 2011 Pearson Education, Inc. continued Answer The width is 12 feet. To determine the length use Relationship 1. If the length = w + 12, then the length = 12 + 12 = 24 feet. Check Verify that both conditions in the problem are satisfied. The length is 12 feet more than the width. The perimeter must be 72 feet. This is true: 2(24) + 2(12) = 72 feet.

61. Slide 3- 61 Copyright © 2011 Pearson Education, Inc. Complementary angles: Two angles are complementary if the sum of their measures is 90 . Supplementary angles: Two angles are supplementary if the sum of their measures is 180 . A D CB 58  32  and are complementary because 32  + 58  = 90  and are supplementary because 20  + 160  = 180  D CB 160  A 20 

62. Slide 3- 62 Copyright © 2011 Pearson Education, Inc. Example 4 Two angles are complementary. If the measure of one angle is 32 degrees less than the other angle, find the measure of the two angles. Understand We must find the measure of the two angles. Relationship 1: One angle is 32 degrees less than the other. Relationship 2: Angle 1 + Angle 2 = 90. Plan Translate the relationships to an equation and then solve.

63. Slide 3- 63 Copyright © 2011 Pearson Education, Inc. continued Execute Relationship 1: One angle is 32 degrees less than the other. a – 32 Relationship 2: The two angles are complementary. Angle 1 + Angle 2 = 90 a + a – 32 = 90 2a – 32 = 90 2a = 122 a = 61 Answer One angle is 61 degrees the other angle is 61 – 32 = 29 degrees. Combine like terms. Add 32 to both sides. Divide both side by 2 to isolate a.

64. Slide 3- 64 Copyright © 2011 Pearson Education, Inc. continued Check Verify that both conditions in the problem are satisfied. One angle is 32 degrees less than the other. This is true: 29 is 32 less than 61. The two angles are complementary: 29 + 61 = 90

65. Slide 3- 65 Copyright © 2011 Pearson Education, Inc. Another group of problems in which the relationships are not obvious involve consecutive integers. The word consecutive means that the numbers are in sequence. With 1 as the First Integer With n as the First Integer First integer:1n Next integer:1 + 1 = 2n + 1 Next integer:1 + 2 = 3n + 2 Patterns of consecutive integers…

66. Slide 3- 66 Copyright © 2011 Pearson Education, Inc. With 1 as the First Odd Integer With n as the First Odd Integer First odd integer:1n Next odd integer:1 + 2 = 3n + 2 Next odd integer:1 + 4 = 5n + 4 Consecutive Odd Integers Consecutive Even Integers With 2 as the First Even Integer With n as the First Even Integer First even integer:2n Next even integer:2 + 2 = 4n + 2 Next even integer:2 + 4 = 6n + 4

67. Slide 3- 67 Copyright © 2011 Pearson Education, Inc. Example 5 The sum of three consecutive even integers is 384. What are the integers? Understand The key word sum means to “add”. If we let n represent the smallest of the even integers, then the pattern for three consecutive even integers is: Smallest unknown even integer: n Next even integer: n + 2 Third even integer: n + 4 Plan Translate to an equation, then solve.

68. Slide 3- 68 Copyright © 2011 Pearson Education, Inc. continued Execute Translate: The sum of three consecutive even integers is 384. Smallest even integer Next even integer ++ Third even integer = 384 n n + 2 n + 4 = 384 3n + 6 = 384 – 6 – 6 3n = 378 n = 126

69. Slide 3- 69 Copyright © 2011 Pearson Education, Inc. continued Answer The smallest of the unknown integers, represented by n, is 126. Next even integer: n + 2 = 126 + 2 = 128 Third even integer: n + 4 = 126 + 4 = 130 Check The numbers 126, 128, and 130 are three consecutive integers and 126 + 128 + 130 = 384.

70. Slide 3- 70 Copyright © 2011 Pearson Education, Inc. Example 6 A day spa sells two sizes of bottles of lotion. A small bottle sells for $10.25 and a large bottle sells for $16.75. One weekend the spa sells twice as many large bottles as small bottles. If the total revenue that weekend for these bottles of lotion was $393.75, how many bottles of each size were sold? Understand We know the total revenue is $393.75. + = $393.75 = Revenue from the small bottles Revenue from the large bottles Price per bottle Number of bottles sold Revenue for that size bottle

71. Slide 3- 71 Copyright © 2011 Pearson Education, Inc. continued With so much information, it is helpful to use a table to list it all. n represents the number of small bottles, then translated “twice as many large bottles as small” CategoriesPrice per BottleNumber of BottlesRevenue Small$10.25n10.25n Large$16.752n2n16.75(2n) Price per bottle Number of bottles Revenue from each size of bottle = Givens

72. Slide 3- 72 Copyright © 2011 Pearson Education, Inc. continued Plan Translate the information to an equation, and then solve. Execute Now we can use our initial relationship: + = 393.75 Revenue from the small bottles Revenue from the large bottles

73. Slide 3- 73 Copyright © 2011 Pearson Education, Inc. continued Answer The number of small bottles sold, n, is 9. The number of large bottles sold, 2n, is 18. Check First, verify that twice the number of small bottles sold, 9, is 18, which is true. Second, verify that the total revenue from the sale of lotion is $393.75. 10.25(9) + 16.75(18)= 92.25 + 301.50 = 393.75. It checks.

74. Slide 3- 74 Copyright © 2011 Pearson Education, Inc. Example 7 A bakery sells muffins and bagels. The muffins sell for $6 per dozen and the bagels sell for $8 per dozen. The number of dozen of each kind was lost. The bakery knows that there were 32 dozen bagels and muffins sold and that the total revenue was $222. How many dozen of each were sold? Understand We know the total revenue is $222. + = $222 $6 times dozens of muffins $8 times dozens of bagels

75. Slide 3- 75 Copyright © 2011 Pearson Education, Inc. continued With so much information, it is helpful to use a table to list it all. Total number of dozen sold was 32. Dozens of muffins + dozens of bagels = 32 Let n represent the number of dozen bagels. muffins = 32 – n CategoriesPrice per DozenNumberRevenue Muffins$6.00 Bagels$8.00

76. Slide 3- 76 Copyright © 2011 Pearson Education, Inc. continued Plan Translate the information to an equation, and then solve. Execute Now we can use our initial relationship: + = 222 Revenue from the muffins Revenue from bagels

77. Slide 3- 77 Copyright © 2011 Pearson Education, Inc. continued Answer The number of dozens of bagels sold, n is 15. The number of dozen of muffins sold is 17 dozen (found by 32 – 15) Check First, verify that the sum is 32 dozen. and the revenue is correct. 15 + 17 = 32 17(6) + 15(8) = 102 + 120 = 222 It checks.

78. Slide 1- 78 Copyright © 2011 Pearson Education, Inc. One number is 4 more than the other. The sum of the numbers is 58. What is the larger of the two numbers? a) 24 b) 27 c) 31 d) 34 3.3

79. Slide 1- 79 Copyright © 2011 Pearson Education, Inc. One number is 4 more than the other. The sum of the numbers is 58. What is the larger of the two numbers? a) 24 b) 27 c) 31 d) 34 3.3

80. Slide 1- 80 Copyright © 2011 Pearson Education, Inc. Adrian is 3 times as old as his younger brother George. In five years, Adrian will only be twice as old as George. How old is Adrian now? a) 5 years old b) 10 years old c) 15 years old d) 20 years old 3.3

81. Slide 1- 81 Copyright © 2011 Pearson Education, Inc. Adrian is 3 times as old as his younger brother George. In five years, Adrian will only be twice as old as George. How old is Adrian now? a) 5 years old b) 10 years old c) 15 years old d) 20 years old 3.3

82. Copyright © 2011 Pearson Education, Inc. Rates 3.4 1.Solve problems involving two objects traveling in opposite directions. 2.Solve problems involving two objects traveling in the same direction.

83. Slide 3- 83 Copyright © 2011 Pearson Education, Inc. Example 1 Two trains are traveling on parallel tracks toward each other from a distance of 533 miles. If the freight train is traveling at 27 miles per hour and the passenger train is moving at 55 miles per hour, how long will it take for them to pass each other? Understand Draw a picture of the situation: 533 miles 55 mph 27 mph

84. Slide 3- 84 Copyright © 2011 Pearson Education, Inc. continued Use a table to find expressions for the individual distances. Distance traveled by passenger train Distance traveled by freight train += 533 CategoriesRateTimeDistance Passenger55 mpht Freight27 mpht 55t 27t Multiplying the rate value by the time value gives the distance value. Both trains start at the same time and meet at the same moment in time, so they traveled the same amount of time, t.

85. Slide 3- 85 Copyright © 2011 Pearson Education, Inc. continued PlanUse the information to write an equation, then solve. Execute 55t + 27t = 533 82t = 533 82 t = 6.5 Answer The trains will meet in 6.5 hours. Check Verify that in 6.5 hours, the trains will travel a combined distance of 533 miles. 55(6.5) + 27(6.5) ? 533 357.5 + 175.5 = 533

86. Slide 3- 86 Copyright © 2011 Pearson Education, Inc. Two Objects Traveling in Opposite Directions To solve for time when two objects are moving in opposite directions: 1. Use a table with columns for categories, rate, time, and distance. Use the fact that rate  time = distance. 2. Write an equation that is the sum of the individual distances equal to the total distance of separation. 3. Solve the equation.

87. Slide 3- 87 Copyright © 2011 Pearson Education, Inc. Example 2 Michael and Renee are traveling north in separate cars on the same highway. Michael is traveling at 55 mph and Renee is traveling at 65 mph. Michael passes Exit 54 at 2:30 p.m. Renee passes the same exit at 2:45 p.m. At what time will Renee catch up with Michael? Understand Calculate the amount of time it will take Renee to catch up to Michael. Add that time to 2:45. 15 minutes = ¼ of an hour or 0.25

88. Slide 3- 88 Copyright © 2011 Pearson Education, Inc. continued The distance Michael and Renee will have traveled when they catch up to each other is the same amount, so their distances are equal. Plan Set the expressions of their individual distances equal, and solve for t. CategoriesRateTimeDistance Michael Renee 55 mph 65 mph t + 0.25 t 55(t + 0.25) 65t

89. Slide 3- 89 Copyright © 2011 Pearson Education, Inc. continued Execute65t = 55(t + 0.25) 65t = 55t + 13.75 –55t –55t 10t = 13.75 10 10 t = 1.375 Answer It will take Renee 1.375 hours, which is 1 hour and 22.5 minutes to catch up to Michael. 2:45 + 1 hour 22.5 minutes = 4:07 p.m.

90. Slide 3- 90 Copyright © 2011 Pearson Education, Inc. continued Check Verify that Renee and Michael are equal distances from the exit after traveling their respective times. Renee: d = 65(1.375) = 89.375 miles Michael: d = 55(1.625) (1.375 + 0.25) = 89.375 miles

91. Slide 3- 91 Copyright © 2011 Pearson Education, Inc. Two Objects Traveling in the Same Direction To solve problems involving two objects traveling in the same direction in which the objective is to determine the time for one object to catch up to the other: 1. Use a table to organize the rates and times. Let t represent the time for the object to catch up, and add the time difference to t to represent the other object’s time. 2. Set the expressions for the individual distances equal. 3. Solve the equation.

92. Slide 1- 92 Copyright © 2011 Pearson Education, Inc. Two trains are traveling toward each other from a distance of 208 miles. One train is traveling at 18 miles per hour and the other at 46 miles per hour. How long will it take for them to pass each other? a) 3.25 hours b) 4.5 hours c) 6.75 hours d) 11.6 hours 3.4

93. Slide 1- 93 Copyright © 2011 Pearson Education, Inc. Two trains are traveling toward each other from a distance of 208 miles. One train is traveling at 18 miles per hour and the other at 46 miles per hour. How long will it take for them to pass each other? a) 3.25 hours b) 4.5 hours c) 6.75 hours d) 11.6 hours 3.4

94. Slide 1- 94 Copyright © 2011 Pearson Education, Inc. Two trains are delivering to the same site. One leaves at 8:00 a.m. and the other leaves at 8:15 a.m. If the first train is traveling at 55 mph and the second at 60 mph, at what time will the second catch up to the first? a) 9:45 a.m. b) 11:00 a.m. c) 10:15 a.m. d) 11:30 a.m. 3.4

95. Slide 1- 95 Copyright © 2011 Pearson Education, Inc. Two trains are delivering to the same site. One leaves at 8:00 a.m. and the other leaves at 8:15 a.m. If the first train is traveling at 55 mph and the second at 60 mph, at what time will the second catch up to the first? a) 9:45 a.m. b) 11:00 a.m. c) 10:15 a.m. d) 11:30 a.m. 3.4

96. Copyright © 2011 Pearson Education, Inc. Investment and Mixture 3.5 1.Use a table to solve problems involving two investments. 2.Use a table to solve problems involving mixtures.

97. Slide 3- 97 Copyright © 2011 Pearson Education, Inc. Example 1 Geraldine invests a total of $4000 in two different accounts. The first account earns 6% while the second account earns 4%. If the total interest earned is $210 after one year, what principle was invested in each account? Understand Since Geraldine invests a total of $4000, we can say Principal invested in first account Principal invested in second account += 4000

98. Slide 3- 98 Copyright © 2011 Pearson Education, Inc. continued We can isolate one of the unknown amounts by writing a related subtraction statement. AccountsRatePrincipalInterest First0.064000 – P Second0.04P 0.06(4000 – P) 0.04P Principal invested in first account Principal invested in second account 4000  = = 4000 – P

99. Slide 3- 99 Copyright © 2011 Pearson Education, Inc. continued PlanWrite an equation, then solve for p. Execute 0.06(4000 – p) + 0.04p = 210 240 – 0.06p + 0.04p = 210 240 – 0.02p = 210 –240 –240 – 0.02p = – 30 – 0.02 –0.02 p = 1500 Answer The principal invested in the second account is $1500. In the first account is $4000  $1500 = $2500.

100. Slide 3- 100 Copyright © 2011 Pearson Education, Inc. continued Check Verify that investing $2500 at 6% and $1500 at 4% result in a total of $210 in interest. 2500(0.06) + 1500(0.04) ? 210 50 + 60 = 210 It checks.

101. Slide 3- 101 Copyright © 2011 Pearson Education, Inc. Solving Problems Involving Investment To solve problems involving two interest rates, in which the total interest is given: 1. Use a table to organize the interest rates and principals. Multiply the individual rates and principals to get expressions for the individual interests. 2. Write an equation that is the sum of the expressions for interest set equal to the given total interest. 3. Solve the equation.

102. Slide 3- 102 Copyright © 2011 Pearson Education, Inc. Chemicals are often mixed to achieve a solution that has a particular concentration. Concentration refers to the portion of a solution that is pure. Concentration  Whole solution = Volume volume particular chemical

103. Slide 3- 103 Copyright © 2011 Pearson Education, Inc. Example 2 Mitchell has a bottle containing 80 milliliters of 40% HCl solution and a bottle of 15% HCl solution. He wants a 25% HCl solution. How much of the 15% solution must be added to the 40% solution so that a 25% concentration is created? Understand Since more than one solution is involved, a table is helpful in organizing the information.

104. Slide 3- 104 Copyright © 2011 Pearson Education, Inc. continued SolutionConcentration of HCl Volume of Solution Volume of HCl 40%0.4080 15%0.15x 25%0.2580 + x 0.4(80) 0.15x 0.25(80 + x) Percent written in decimal form. The 40% and 15% solutions are combined to get the 25% solution, so we add. Multiply straight across to generate the expressions for the volume of HCl in each solution.

105. Slide 3- 105 Copyright © 2011 Pearson Education, Inc. continued PlanWrite an equation, then solve for x. Execute 0.4(80) + 0.15x = 0.25(80 + x) 32 + 0.15x = 20 + 0.25x  20 –20 12 + 0.15x = 0 + 0.25x  0.15x  0.15x 12 = 0.1x 0.1 0.1 120 = x Answer 120 milliliters of the 15% solution must be added to 80 milliliters of 40% solution to create a 25% solution.

106. Slide 3- 106 Copyright © 2011 Pearson Education, Inc. continued Check Verify that the volume of HCl in the two original solutions combined is equal to the volume of HCl in the combined solution. 0.40(80) + 0.15(120) ? 0.25(80 + 120) 32 + 18 = 0.25(200) 50 = 50 It checks.

107. Slide 3- 107 Copyright © 2011 Pearson Education, Inc. Solving Mixture Problems To solve problems involving mixing solutions: 1. Use a table to organize the concentrations and volumes. Multiply the individual concentrations and solution volumes to get expressions for the volume of the particular chemical. 2. Write an equation that the sum of the volumes of the chemical in each solution set is equal to the volume in the combined solution. 3. Solve the equation.

108. Slide 1- 108 Copyright © 2011 Pearson Education, Inc. Lisa invests a total of $6000 in two different accounts. The first account earns 8% while the second account earns 3%. If the total interest earned is $390 after one year, what amount is invested at 8%? a) $1800 b) $2100 c) $4200 d) $4800 3.5

109. Slide 1- 109 Copyright © 2011 Pearson Education, Inc. Lisa invests a total of $6000 in two different accounts. The first account earns 8% while the second account earns 3%. If the total interest earned is $390 after one year, what amount is invested at 8%? a) $1800 b) $2100 c) $4200 d) $4800 3.5

110. Slide 1- 110 Copyright © 2011 Pearson Education, Inc. Martin has a bottle containing 120 milliliters of 30% HCl solution and a bottle of 15% HCl solution. He wants a 25% HCl solution. How much of the 15% solution must be added to the 30% solution so that a 25% concentration is created? a) 30 milliliters b) 45 milliliters c) 60 milliliters d) 75 milliliters 3.5

111. Slide 1- 111 Copyright © 2011 Pearson Education, Inc. Martin has a bottle containing 120 milliliters of 30% HCl solution and a bottle of 15% HCl solution. He wants a 25% HCl solution. How much of the 15% solution must be added to the 30% solution so that a 25% concentration is created? a) 30 milliliters b) 45 milliliters c) 60 milliliters d) 75 milliliters 3.5