New Jersey Center for Teaching and Learning Progressive Mathematics Initiative Click to go to website: Algebra I

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1 New Jersey Center for Teaching and Learning Slide 1 / 121 Progressive Mathematics Initiative This material is made freely available at and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Click to go to website: Algebra I Slide 2 / 121 Open Ended Application Problems Table of Contents Slide 3 / 121 Number Problems Age Problems Geometry Problems Percent Problems Mixture Problems Uniform Motion Problems Work Problems Proportionality Problems Click to go to that topic

2 Slide 4 / 121 Problem Solving Strategies Most problems can be solved by incorporating one or more strategies. Work backwards Make a table, chart or diagram Solve a simpler or similar problem Guess and check Look for a pattern Eliminate possibilities Draw a picture Slide 5 / 121 Plan for Solving a Word Problem Plan: Read the problem several times. What do you know? What do you need to find? Eliminate any unnecessary information. Set Up: Define a variable. Making a chart or drawing a picture may be helpful. Write an open sentence. Solve the open sentence. Slide 6 / 121 Plan for Solving a Word Problem Check: Reread the problem. Did you answer the question? Did you state your answer clearly with the appropriate units? Is your answer consistent with the information given in the problem?

3 Slide 7 / 121 Number Problems Return to Table of Contents Vocabulary Slide 8 / 121 Consecutive integers are obtained when you count by ones from any given integer. Examples 1, 2, 3, 4, 5-6, -5, -4, -3 x, x + 1, x + 2 x - 1, x, x + 1 Vocabulary Slide 9 / 121 Even integers are integers that are multiples of two. Consecutive even integers are obtained when you count by twos from any given even integer. Examples 2, 4, 6-6, -4, -2 x, x + 2, x + 4 x - 2, x, x + 2

4 Vocabulary Slide 10 / 121 Odd integers are integers that are not even. Consecutive odd integers are obtained when you count by twos from any given odd integer. Examples 3, 5, 7-9, -7, -5 x, x + 2, x + 4 x - 2, x, x + 2 Example 1 One number is 10 greater than another. If the lesser number is subtracted from three times the greater number, the difference is 42. Find the numbers. Slide 11 / 121 Set Up 2 numbers - one is 10 greater than the other : n + 10 = greater number n = lesser number Write an open sentence: 3(n + 10) - n = 42 Slide 12 / 121 3(n + 10) - n = 42 3n n = 42 2n + 30 = n = n = 6 So n + 10 = = 16 The two numbers are 6 and 16. Check : Reread the problem. Does your answer make sense?

5 Example 2 One number is 12 greater than another. If the sum of the two numbers is 88, find the numbers. Slide 13 / 121 Set Up 2 numbers - one is 12 greater than the other : n + 12 = one number n = second number (n + 12) + n = 88 Slide 14 / 121 (n + 12) + n = 88 2n + 12 = 88 2n + 12 = n = n = 38 So n + 12 = = 50 The two numbers are 38 and 50. Check : Reread the problem. Does your answer make sense? Example 3 Find three consecutive odd integers whose sum is 183. Slide 15 / 121 Set Up Find three consecutive odd integers : n = 1st consecutive odd integer n + 2 = 2nd consecutive odd integer n + 4 = 3rd consecutive odd integer n + n n + 4 = 183

6 Slide 16 / 121 n + n n + 4 = 183 3n + 6 = n = n = 59 So n + 2 = 61 and n + 4 = 63 The three consecutive odd integers are 59, 61 and 63. Check : Reread the problem. Does your answer make sense? Practice 1 One number is 70 greater than a second number. If the lesser number is subtracted from twice the greater number, the difference is 174. Find the numbers. Slide 17 / 121 2(n + 70) - n = 174 The two numbers are 34 and 104. Check Your Solution : Practice 2 Find three consecutive integers whose sum is Slide 18 / 121 x + x x + 2 = -315 The three numbers are -106, -105 and Check Your Solution :

7 Practice 3 Find three consecutive even integers such that the sum of the least integer and the greatest integer is Slide 19 / 121 The three consecutive odd integers are -92, -90 and -88. Check Your Solution : 1 Find the largest of four consecutive integers whose sum is 130. Slide 20 / The lengths of the sides of a triangle are consecutive odd integers. The perimeter is 27 meters. Find the length of the smallest side. Slide 21 / 121

8 3 Find two consecutive even integers whose sum is 148. Slide 22 / Sam has 6 more than twice as many newspaper customers as when he started selling newspapers. If he currently has 98 customers, how many did he have when he started? Slide 23 / There are fifty coins in a jar that contains only dimes and quarters. The number of dimes in the jar is 2 less than three times the number of quarters. How many dimes are in the jar? Slide 24 / 121

9 Slide 25 / 121 Age Problems Return to Table of Contents Slide 26 / 121 Problem Solving Strategy Sometimes using a chart to organize the information given in a word problem can be helpful. You can use this strategy to solve age problems. Example 1 Jake is 12 years older than his dog. Next year he will be four times as old as his dog will be. How old is Jake now? Slide 27 / 121 Find Jake's age now. Age Now Age Next Year Dog x x + 1 Jake x + 12 (x + 12) + 1 Next year, Jake will be four times as old as his dog. (x + 12) + 1 = 4(x + 1)

10 (x + 12) + 1 = 4(x + 1) x + 13 = 4x + 4 -x -x 13 = 3x = 3x = x Slide 28 / 121 Jake's dog is 3 years old now and = 15. Therefore, Jake is 15 years old now. Check : Reread the problem. Does your answer make sense? Example 2 Erica is now four years older than her sister Alicia. In ten years, Erica will be twice Alicia s present age. Find the age of each girl now. Slide 29 / 121 Find the age of each girl. Age Now Age in 10 years Erica x + 4 (x + 4) + 10 Alicia x x + 10 x + 14 = 2x x + 14 = 2x -x -x 14 = x Slide 30 / 121 Alicia is 14 years old now and Erica is 18 years old now. Check : Reread the problem. Does your answer make sense?

11 Practice 1 Anthony is 9 years older than his sister Marie. Next year, he will be four times as old as his sister. How old is Anthony now? Slide 31 / 121 Find Anthony's age now. Anthony is 11 years old now. Check : Practice 2 Cara is six years older than her brother. In three years, she will be twice as old as her brother will be. How old is Cara now? Slide 32 / 121 Cara is 9 years old now. Check : 6 Bebe is twice as old as Marcus. The sum of their ages is 57. How old is Bebe? Slide 33 / 121

12 7 Each sister is two years older than the next. The oldest sister is twice the age of the youngest sister, with two sisters in between. How old are the sisters? Slide 34 / Deanna's age is eight years greater than half of Metri's age. If the sum of their ages is 17, how old is Deanna? Slide 35 / Zach's age is three less than twice Matt's age. In five years, the sum of their ages will be 19. How old is Zach now? Slide 36 / 121

13 10 The son is 28 years younger than his father. The sum of their ages is 84. How old is dad? Slide 37 / 121 Slide 38 / 121 Geometry Problems Return to Table of Contents Formulas to Remember Slide 39 / 121 Perimeter is the distance around a figure. Area of a rectangle A = l w Area of a triangle A = 1/2 b h

14 Example 1 - Area Find the measure of the area of the shaded region in the figure below. Slide 40 / 121 x + 6 4x 2x x + 4 Given the length and width of large and small rectangles. Set Up : A = area of shaded region Open Sentence : A = 4x(x + 6) - 2x(x + 4) x + 6 Slide 41 / 121 4x 2x x + 4 4x(x + 6) - 2x(x + 4) = A 4x x - 2x 2-8x = A 2x x = A The area of the shaded region is (2x x) units 2 Check : Example 2 - Perimeter/Area The measure of the perimeter of a square is 12a + 16b. Find the measure of the area of the square. Slide 42 / 121 Find the area of the square. A = s 2 s = (12a + 16b) 4 s = 3a + 4b Open Sentence : A = (3a + 4b) 2

15 Slide 43 / 121 (3a + 4b) (3a + 4b) 9a ab +12ab +16b 2 9a ab + 16b 2 The area of the square is (9a 2 +24ab + 16b 2 ) units 2. Check : Example 3 - Construction The length of a rectangular lot is 5 yards less than three times the width. If the length was decreased by 2 yards and the width increased by 5 yards, the area would be increased by 17 square yards. Find the original dimensions of the lot. Slide 44 / 121 Find the area of the original lot. It may help to draw a picture. Set Up : w = original width w + 5 = new width 3w - 5 = original length 3w - 7 = new length Open Sentence : w(3w - 5) + 17 = (w + 5)(3w - 7) Slide 45 / 121 w(3w - 5) + 17 = (w + 5)(3w - 7) 3w 2-5w + 17 = 3w w - 7w w 2-5w + 17 = 3w 2 + 8w w 2-3w 2-5w + 17 = 8w w -8w -13w + 17 = w = w = 4 so, 3w - 5 = 7 The original dimensions were 4 yards by 7 yards. Check :

16 Practice 1 Melissa has a rectangular garden that is 10 feet longer than it is wide. A brick path that is 3 feet wide surrounds the garden. The total area of the path is 396 square feet. What are the dimensions of the garden? Slide 46 / 121 The width is 25 feet and the length is 35 feet. Check Your Solution : Practice 2 A new athletic field is being sodded at Lawrence High School using 2-yard by 2-yard squares of sod. If the width of the field is 70 yards less than its length and its area is 6000 square yards, how many squares of sod will be needed? Slide 47 / squares of sod will be needed. Check Your Solution : 11 The length of a garden is one more foot than twice the width. The area of the garden is 55 feet. What is the width of the garden? Slide 48 / 121

17 12 A 3 x 4 picture sits in a picture frame. The total area of the picture and frame is 56 inches. How wide is the frame? Slide 49 / x Slide 50 / 121 x + 15 The garage door is a square that measures 2x feet on each side. How many square feet of house surrounds the garage? PULL 2x A square has its side doubled in length. How much does the area of the square increase? Slide 51 / 121 x

18 15 An area rug's length is 3 feet less than three times its width. The area of the rug is 90 square feet. What is the length of the rug? Slide 52 / 121 Slide 53 / 121 Percent Problems Return to Table of Contents Important Concepts to Recall Slide 54 / 121 Part Percent Whole = 100 List Price - Discount = Sale Price Simple Interest = Principle x Rate x Time Percent of Change: Amount of Change Original Amount

19 Example 1 Mia bought a sweater at a 15% discount. If she paid $38.25 for the sweater, what was the original price? Slide 55 / 121 Find the original price of the sweater. If the sweater is 15% off, then she paid 85% of the original price. Set Up : x = original price of the sweater. Open Sentence : 85% of x = Slide 56 / % of x = (.85)(x) = x = 45 The original price of the sweater was $45. Check : 15% of 45 = 6.75 and = Slide 57 / 121 Example 2 The Smiths invest part of $8000 in bank accounts that pay 5% simple annual interest and the rest in bonds that pay 12% simple annual interest. How much money is invested in each account if the total annual income from these investments is $610? They invest part of the money at 5% and part of the money at 12%. Remember: part + part = whole x = amount invested at 5% x = amount invested at 12% Open Sentence : 0.05x ( x) = 610

20 Slide 58 / x ( x) = x x = x = x = x = 5000 $5,000 was invested at 5% and $3,000 was invested at 12%. Check : 5000(0.05) (0.12) = = = 610 Practice 1 The O'Connors paid $15,000 in closing costs when purchasing their new home. If this amount represents 6% of the purchase price, how much did they pay for their home? Slide 59 / 121 They paid $250,000 for their house. Check Your Solution : Practice 2 A jacket was on sale for $ If the original selling price was $75, what was the percent of the discount? Slide 60 / 121 The discount percent was 15%. Check Your Solution :

21 Practice 3 Marco invested $10,000, part at an annual interest rate of 5% and the rest at an annual rate of 10.25%. How much money did he invest at each rate if his total income on the investment for one year was $867.50? Slide 61 / 121 He invested $3,000 at 5% and $7,000 at 10.25% Check Your Solution : 16 The sale price of a dress is $ after a 25% discount is taken. Find the regular price of the dress. Slide 62 / The bill at the restaurant is $35. You want to leave a 20% tip. How much should you leave? Slide 63 / 121

22 18 One store has a $200 bicycle on sale for 40% off. Another store has the same bicycle for $200 with a 30% off plus an additional 10% off. Will the bicycles cost the same at both stores? Slide 64 / The price of the CD increased from $12 to $15. What is the percent of increase? Slide 65 / Mark invested $20,000, part at an annual interest rate of 6% and the rest at an annual rate of 2.5%. How much money did he invest at each rate if his total income on the investment for one year was $960.50? Slide 66 / 121

23 Slide 67 / 121 Mixture Problems Return to Table of Contents Slide 68 / 121 Mixture Problems Sometimes a chemist mixes solutions of different strengths to obtain a desired solution. Or a business mixes two or more goods in order to sell a blend at a given price. Mixture problems are problems related to these situations. Example 1 Steven bought some 44 cent stamps and some 28 cent stamps. He bought 35 stamps in all and paid $13.00 for them. How many stamps of each kind did he buy? Slide 69 / 121 He bought 2 kinds of stamps and paid $ Find how many of each kind he bought. 44 cent stamps 28 cent stamps Number of stamps Value per stamp Total Value s $ s 35 - s $ (35 - s) Mixture 35 $13.00

24 Slide 70 / 121 Open Sentence : 0.44s (35 - s) = s s = s = s = s = s = 15 He bought 20 forty-four cent stamps and 15 twenty-eight cent stamps. Check : Example 2 A 40L solution is 15% salt. How much water must be added to make it an 8% salt solution? Slide 71 / 121 How much water must be added to change the solution from 15% salt to 8% salt Original Solutio n Water Added New Solutio n Volume of Solution % Salt Volume of Salt 40 15% 0.15(40) s 0% 0(s) 40 + s 8% 0.08(40 + s) Slide 72 / 121 Open Sentence : 0.15(40) + 0(s) = 0.08(40 + s) 0.15(40) + 0(s) = 0.08(40 + s) 6 = s = 0.08s = s 35 liters of water need to be added to make an 8% salt solution. Check :

25 Practice 1 How many pounds of dried apricots worth $7.50 per pound must be added to 5 pounds of dried bananas worth $5.25 per pound to form a mixture worth $6.00 per pound? Slide 73 / pounds of dried apricots are needed. Check Your Solution : Practice 2 A 15L solution is 70% antifreeze. How much antifreeze must be added to produce a solution that is 80% antifreeze? Slide 74 / L of antifreeze must be added. Check Your Solution : 21 Sofie has twice as many dimes as nickels in her piggy bank. If the dimes and nickels together total $18.00, how many dimes does she have? Slide 75 / 121

26 Slide 76 / Alberto has 48 ml of solution that is 50% acid. How many ml of a 15% acid solution should he add to obtain a solution that is 35% acid? 23 There are 15 lbs. of nuts valued at $7 per lb. The mixture consists of peanuts, valued at $2.20 per pound and cashews valued at $8.50 per pound. Approximately how many pounds of peanuts are in the mixture? Slide 77 / In your chemistry class, you have a bottle of 5% boric acid solution and a bottle of 2% boric acid solution. You need 50 ml of a 4% boric acid solution. How much of the 2% acid solution must be used? Slide 78 / 121

27 25 The register at the carnival has 200 bills in $1 and $5 denominations. The value of the money is $660. How many $5 bills are in the register? Slide 79 / 121 Slide 80 / 121 Uniform Motion Problems Return to Table of Contents Uniform Motion Problems Slide 81 / 121 An object that moves at a constant rate, or speed, is considered to be in uniform motion. A formula that is used in solving uniform motion problems is: rate x time = distance To apply the formula correctly, the units used for the time and the distance measurements must be the same as those used for the rate. Three types of uniform motion problems are shown in the following examples.

28 Example 1 - Motion in the Same Direction Two speedboats leave from the same dock at the same time traveling to Point Pleasant. The faster boat arrives in 6 hours. The slower boat arrives in 9 hours. The slower boat travels at an average speed that is 15 km/h slower than the faster boat. What is the average speed of the faster boat? Slide 82 / 121 Find the rate or speed of each boat. Faster boat Slower boat Rate (km/h) Time (h) Distance (km) x 6 6x x (x - 15) Slide 83 / 121 Open Sentence : 6x = 9(x - 15) 6x = 9x x - 9x -3x = x = 45 x - 15 = 30 The faster boat travels at a rate of 45 km/h and the slower boat travels at 30 km/h. Check : Example 2 - Motion in the Opposite Direction Jazmine and Meghan are 630 meters apart. Jazmine walks toward Meghan at the rate of 2.5 m/sec and Meghan runs toward Jazmine. What is Meghan's rate if she reaches Jazmine in 1.4 min? Slide 84 / 121 Find Meghan's rate. Remember 1.4 min =? sec Rate (m/s) Time (s) Distance (m) Jazmin e (84) Meghan x 84 84x

29 Slide 85 / 121 Open Sentence : 2.5(1.4) + 84x = x = x = x = 5 Meghan runs at a rate of 5 m/s. Check : Example 3 - Round Trip Javone is a member of the cross country team at his school. On Monday he ran from school to Central Park and back again. On the way to the park he ran at a rate of 7.5 m/hr and on the return he ran at a rate of 5 m/hr. If he took 1 hour 15 minutes to run the entire distance, how far is the school from Central Park? Slide 86 / 121 Find the distance from the school to Central Park. Rate (m/h) Time (h) Distance (m) To the park Back to school 7.5 x 7.5x x 5( x) Slide 87 / 121 Open Sentence : 7.5x = 5( x) 7.5x = x + 5x + 5x 12.5x = x = 0.5 The time is 0.5 hours, so the distance from the school to the park is 7.5(0.5) which equals 3.75 miles. Check :

30 Practice 1 Jack and Jessie are cycling in the same direction on the same bike path. Jack's average speed for the trip is 20 miles per and Jesse's is 14 miles per hour. After how many hours will they be 7.5 miles apart? Slide 88 / 121 It will take them 1.25 hours to be 7.5 miles apart. Check Your Solution : Practice 2 Jeff and Brian live 1.5 mile apart. They agree to meet at the library directly between their homes. Jeff needs 12 minutes and Brian needs 18 minutes to get to the library. If they both travel at the same average speed, how far do Jeff and Brian live from the library? Slide 89 / 121 Jeff lives 0.6 miles from the library and Brian lives 0.9 miles from the library. Check Your Solution : Practice 3 Scott delivers newspapers every morning. The trip delivering them takes 30 minutes. The return trip over the same route takes 20 minutes. If his average rate going is 6 km/h slower than returning home, how far does he travel each morning? Slide 90 / 121 Scott's entire trip is 12 km long. Check Your Solution :

31 26 An airplane flies 1500 miles due west in 3 hours and 1000 miles due south in 2 hours. What is the average speed of the airplane? Slide 91 / Fred and Ted leave their home at the same time, traveling in opposite directions. Fred travels 50 miles per hour and Ted travels 54 miles per hour. In how many hours will they be 572 miles apart? Slide 92 / Mark travels at the rate of 32 mph for four hours. His sister, travels at the rate of 40 miles per hour. How long will it take her to travel the same distance as her brother? Slide 93 / 121

32 29 Joey left home on his bicycle for a long distance ride. Marcie left 2 hours later on her motorcycle carrying his lunch. Marcie traveled at 45 miles per hour and she caught up to Joey in 1.75 hours. How fast was Joey traveling? Slide 94 / Bob and Sally are in a race. Bob is running at the rate of 8 mph and Sally is running at the rate of 6 mph. How long will it take them to be 11 miles apart? Slide 95 / 121 Slide 96 / 121 Work Problems Return to Table of Contents

33 Work Problems Slide 97 / 121 When solving problems that involve finding how long it takes to complete a task, a constant rate of work is assumed. Work rate is the fraction of the whole job that can be done per unit of time. work rate x time = work done Another way to do work problems is to think: Time Together + Time Together = 1 (for one job Time Alone Time Alone done) Example 1 Maya can paint a room in four hours. Takira can paint a room in six hours. How long would it take them to paint the room if they worked together? Slide 98 / 121 Find the number of hours it will take them to paint the room together. x = # of hours to paint the room together Since Maya can paint the whole room in 4 hours, her work rate is 1/4 of the job per hour. In x hours, she could paint 1/4 times x of the job, or x/4 of the job. Therefore, Takira could do x/6 of the job. Open Sentence : Maya's work + Takira's work = Job (1 room painted) x/4 + x/6 = 1 Slide 99 / (x/4 + x/6) = 1(12) 3x + 2x = 12 5x = x = 12/5 or 2 or It would take 2.4 hours to paint the room if they worked together. Check :

34 Example 2 Kyle can mow the lawn in three hours. Dean can mow the lawn in two hours. How long would it take them to mow the lawn if they worked together? Slide 100 / 121 Find the number of hours it will take them to mow the lawn together. Set Up : x = # of hours to mow the lawn together Kyle can do 1/3 of the work in 1 hour. Dean can do 1/2 of the work in 1 hour. Open Sentence : x/3 + x/2 = 1 Slide 101 / 121 6(x/3 + x/2) = 1(6) 2x + 3x = 6 5x = x = 6/5 or 1 or It would take 1.2 hours to mow the lawn if they worked together. Check : Practice 1 Jake can service a car in 4 hours. Jared can service a car in 5 hours. How long would it take them to service nine cars if they worked together? Slide 102 / 121 If they work together, they can service 9 cars in 20 hours. Check Your Solution :

35 Practice 2 Daisy can vacuum and dust the house in 2 hours. Jessica can do the same job in 1.2 hours. How long will it take them to vacuum and dust the house if they work together? Slide 103 / 121 It will take them 0.75 hours working together. Check Your Solution : 31 Bob can paint a room in 8 hours. Mark can paint a room in 10 hours. How long will it take them to paint the room if they work together? Slide 104 / Sal and Sam can row across the lake together in 3 hours. Alone, it takes Sal 5 hours to row across the lake. How long does it take Sam to row across the lake alone? Slide 105 / 121

36 33 A 500 gallon pool takes 6 days to fill with one hose. If the neighbors put their hose in the pool to help fill it, it will take only 4 days. If only the neighbor's hose is used, how long will it take to fill the pool? Slide 106 / It takes you 30 minutes to clean your room. It takes your brother 45 minutes to clean his room. How long does it take the two of you to clean your rooms if you work together? Slide 107 / You can knit two squares for an afghan in 3 hours and your friend can knit 4 squares in 5 hours. How long does it take both of you to knit 10 squares when working together? Slide 108 / 121

37 Slide 109 / 121 Proportionality Problems Return to Table of Contents Proportion Problems Slide 110 / 121 Concepts to Remember Part = Percent Whole 100 Complementary Angles are two angles whose measures add up to 90 degrees. Supplementary Angles are two angles whose measures add up to 180 degrees. Example 1 Mr. Jones bought 10 pounds of grass seed to seed an area of 2000 square feet. At this rate, how much seed would he need to seed 3200 square feet? Slide 111 / 121 Set Up Given 10 lbs of seed for 2000 sq ft Find how much for 3200 sq ft : x = pounds of grass seed for 3200 sq ft Open Sentence : 10 x =

38 Slide 112 / = x x = 10(3200) 2000x = x = 16 Mr. Jones needs 16 lbs of grass seed. Check : Example 2 The measures of two complementary angles are in the ratio of 3:7. Find the measure of each angle in degrees. Slide 113 / 121 The sum of the measures of complementary angles is 90 degrees. Their ratio is 3:7. Set Up : x = measure of smaller angle 90 - x = measure of larger angle Open Sentence : 3 x = x 3 = x x 3(90 - x) = 7x 270-3x = 7x + 3x +3x 270 = 10x = x Slide 114 / 121 The measures of the two complementary angles are 27 degrees and 63 degrees. Check :

39 Practice 1 In a survey of a high school with 1272 students, 7 out of every 12 students said that they do not like the school lunches. How many students do like the lunches? Slide 115 / students do like the lunches. Check Your Solution : Practice 2 The measures of two supplementary angles are in the ratio 3:2. Find the measure of each angle. Slide 116 / degrees and 72 degrees. Check Your Solution : Slide 117 / A 96 mile trip requires 8 gallons of gasoline. At that rate, how many gallons would be required for a 156 mile trip?

40 Slide 118 / The scale on the blueprint for a house is 1 inch to 3 feet. If the living room on the blueprint is 5.5 inches by 7 inches, what is the area of the actual room? Slide 119 / The measures of two adjacent angles are in the ratio of 2 to 5. If the sum of the angle measures is 98 degrees, what is the measure of the larger angle? 39 The measure of supplementary angles are in the ratio of 3 to 5. Find the measure of the smaller angle. Slide 120 / 121

41 40 The ratio of boys to girls is 4:5. If there are 2000 students in the auditorium, how many are girls? Slide 121 / 121