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This material is made freely available at www.njctl.org 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: www.njctl.org New Jersey Center for Teaching and Learning Progressive Mathematics Initiative

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Algebra I Open Ended Application Problems www.njctl.org 2012-08-14

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Use Normal View for the Interactive Elements To use the interactive elements in this presentation, do not select the Slide Show view. Instead, select Normal view and follow these steps to set the view as large as possible: On the View menu, select Normal. Close the Slides tab on the left. In the upper right corner next to the Help button, click the ^ to minimize the ribbon at the top of the screen. On the View menu, confirm that Ruler is deselected. On the View tab, click Fit to Window. Use Slide Show View to Administer Assessment Items To administer the numbered assessment items in this presentation, use the Slide Show view. (See Slide 21 for an example.) Setting the PowerPoint View

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Number Problems Proportionality Problems Age Problems Geometry Problems Percent Problems Mixture Problems Uniform Motion Problems Work Problems Table of Contents Click on the topic to go to that section

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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 Problem Solving Strategies

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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.

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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? Plan for Solving a Word Problem

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Number Problems Return to Table of Contents

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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

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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 Vocabulary

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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 Vocabulary

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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. Plan: 2 numbers - one is 10 greater than the other Set Up:n + 10 = greater number n = lesser number Write an open sentence: 3(n + 10) - n = 42

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Solve:3(n + 10) - n = 42 3n + 30 - n = 42 2n + 30 = 42 -30 -30 2n = 12 2 2 n = 6 So n + 10 = 6 + 10 = 16 The two numbers are 6 and 16. Check:Reread the problem. Does your answer make sense?

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Example 2 One number is 12 greater than another. If the sum of the two numbers is 88, find the numbers. Plan:2 numbers - one is 12 greater than the other Set Up:n + 12 = one number n = second number Write an open sentence: (n + 12) + n = 88 X

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Solve:(n + 12) + n = 88 2n + 12 = 88 -12 -12 2n = 76 2 2 n = 38 So n + 12 = 38 + 12 = 50 The two numbers are 38 and 50. Check:Reread the problem. Does your answer make sense? X

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Example 3 Find three consecutive odd integers whose sum is 183. Plan:Find three consecutive odd integers Set Up:n = 1st consecutive odd integer n + 2 = 2nd consecutive odd integer n + 4 = 3rd consecutive odd integer Write an open sentence: n + n + 2 + n + 4 = 183 X

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Solve:n + n + 2 + n + 4 = 183 3n + 6 = 183 - 6 - 6 3n = 177 3 3 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? X

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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. Plan: Set Up: Write an open sentence: 2(n + 70) - n = 174 Solve: The two numbers are 34 and 104. Check Your Solution:

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Practice 2 Find three consecutive integers whose sum is -315. Plan: Set Up: Write an open sentence: x + x + 1 + x + 2 = -315 Solve: The three numbers are -106, -105 and -104. Check Your Solution:

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Practice 3 Find three consecutive even integers such that the sum of the least integer and the greatest integer is -180. Plan: Set Up: Write an open sentence: Solve: The three consecutive odd integers are -92, -90 and -88. Check Your Solution:

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1Find the largest of four consecutive integers whose sum is 130.

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2 The lengths of the sides of a triangle are consecutive odd integers. The perimeter is 27 meters. Find the length of the smallest side.

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3Find two consecutive even integers whose sum is 148.

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4 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?

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5 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?

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Age Problems Return to Table of Contents

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Sometimes using a chart to organize the information given in a word problem can be helpful. Problem Solving Strategy You can use this strategy to solve age problems.

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Plan:Find Jake's age now. Set Up: Age NowAge Next Year Dogxx + 1 Jakex + 12(x + 12) + 1 Write an open sentence: Next year, Jake will be four times as old as his dog. (x + 12) + 1 = 4(x + 1) 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? X

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Solve: (x + 12) + 1 = 4(x + 1) x + 13 = 4x + 4 -x -x 13 = 3x + 4 - 4 - 4 9 = 3x 3 3 3 = x Jake's dog is 3 years old now and 3 + 12 = 15. Therefore, Jake is 15 years old now. Check: Reread the problem. Does your answer make sense? X

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Plan:Find the age of each girl. Set Up: Age NowAge in 10 years Ericax + 4(x + 4) + 10 Aliciaxx + 10 Write an open sentence: x + 14 = 2x 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. X

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Solve: Alicia is 14 years old now and Erica is 18 years old now. Check: Reread the problem. Does your answer make sense? x + 14 = 2x -x 14 = x X

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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? Plan:Find Anthony's age now. Set Up: Write an open sentence: Solve:Anthony is 11 years old now. Check:

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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? Plan: Set Up: Write an open sentence: Solve:Cara is 9 years old now. Check:

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6 Bebe is twice as old as Marcus. The sum of their ages is 57. How old is Bebe?

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7Each 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?

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8 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?

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9Zach'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?

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10 The son is 28 years younger than his father. The sum of their ages is 84. How old is dad?

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Geometry Problems Return to Table of Contents

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Formulas to Remember Perimeter is the distance around a figure. Area of a rectangle A = l w Area of a triangle A = 1/2 b h

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Example 1 - Area Find the measure of the area of the shaded region in the figure below. x + 6 x + 4 2x 4x Plan: 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

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Solve: 4x(x + 6) - 2x(x + 4) = A 4x 2 + 24x - 2x 2 - 8x = A 2x 2 + 16x = A The area of the shaded region is (2x 2 + 16x) units 2 Check: x + 6 x + 4 2x 4x X

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Example 2 - Perimeter/Area The measure of the perimeter of a square is 12a + 16b. Find the measure of the area of the square. Plan: Find the area of the square. Set Up: A = s 2 s = (12a + 16b) ÷ 4 s = 3a + 4b Open Sentence: A = (3a + 4b) 2 X

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Solve: (3a + 4b) (3a + 4b) 9a 2 + 12ab +12ab +16b 2 9a 2 + 24ab + 16b 2 The area of the square is (9a 2 +24ab + 16b 2 ) units 2. Check: X

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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. Plan: 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) X

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Solve: w(3w - 5) + 17 = (w + 5)(3w - 7) 3w 2 - 5w + 17 = 3w 2 + 15w - 7w - 35 3w 2 - 5w + 17 = 3w 2 + 8w - 35 -3w 2 -3w 2 -5w + 17 = 8w - 35 -8w -8w -13w + 17 = -35 -17 -17 -13w = -52 -13 -13 w = 4 so, 3w - 5 = 7 The original dimensions were 4 yards by 7 yards. Check: X

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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? Plan: Set Up: Write an open sentence: Solve: The width is 25 feet and the length is 35 feet. Check Your Solution:

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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? Plan: Set Up: Write an open sentence: Solve: 1500 squares of sod will be needed. Check Your Solution:

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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?

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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?

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-----------------6x-------------------------------- x + 15 The garage door is a square that measures 2x feet on each side. How many square feet of house surrounds the garage? 13

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x 14 A square has its side doubled in length. How much does the area of the square increase?

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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?

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Percent Problems Return to Table of Contents

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Important Concepts to Recall Part Percent Whole 100 List Price - Discount = Sale Price Simple Interest = Principle x Rate x Time Percent of Change: Amount of Change Original Amount =

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Example 1 Mia bought a sweater at a 15% discount. If she paid $38.25 for the sweater, what was the original price? Plan: 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 = 38.25 X

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Solve:85% of x = 38.25 (.85)(x) = 38.25.85.85 x = 45 The original price of the sweater was $45. Check:15% of 45 = 6.75 and 45 - 6.75 = 38.25 X

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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? Plan: They invest part of the money at 5% and part of the money at 12%. Remember: part + part = whole Set Up: x = amount invested at 5% 8000 - x = amount invested at 12% Open Sentence: 0.05x + 0.12(8000 - x) = 610 X

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Solve: 0.05x + 0.12(8000 - x) = 610 0.05x + 960 - 0.12x = 610 -0.07x + 960 = 610 - 960 -960 -0.07x = -350 -0.07 -0.07 x = 5000 $5,000 was invested at 5% and $3,000 was invested at 12%. Check:5000(0.05) + 3000(0.12) = 610 250 + 360 = 610 610 = 610 X

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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? Plan: Set Up: Write an open sentence: Solve: They paid $250,000 for their house. Check Your Solution:

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Practice 2 A jacket was on sale for $63.75. If the original selling price was $75, what was the percent of the discount? Plan: Set Up: Write an open sentence: Solve: The discount percent was 15%. Check Your Solution:

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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? Plan: Set Up: Write an open sentence: Solve: He invested $3,000 at 5% and $7,000 at 10.25% Check Your Solution:

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16 The sale price of a dress is $112.50 after a 25% discount is taken. Find the regular price of the dress.

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17 The bill at the restaurant is $35. You want to leave a 20% tip. How much should you leave?

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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?

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19 The price of the CD increased from $12 to $15. What is the percent of increase?

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20 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?

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Mixture Problems Return to Table of Contents

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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. Mixture Problems

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Plan: He bought 2 kinds of stamps and paid $13.00. Find how many of each kind he bought. Set Up: 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? Number of stamps Value per stamp Total Value 44 cent stamps s$0.440.44s 28 cent stamps 35 - s$0.280.28(35 - s) Mixture 35$13.00 X

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Open Sentence: 0.44s + 0.28(35 - s) = 13 Solve: 0.44s + 9.80 - 0.28s = 13 0.16s + 9.80 = 13 - 9.80 - 9.80 0.16s = 3.20 0.16 0.16 s = 20 35 - s = 15 He bought 20 forty-four cent stamps and 15 twenty-eight cent stamps. Check: X

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Plan:How much water must be added to change the solution from 15% salt to 8% salt Set Up: Example 2 A 40L solution is 15% salt. How much water must be added to make it an 8% salt solution? Volume Of Solution % Salt Volume of Salt Original Solution 4015%0.15(40) Water Added s0%0(s) New Solution 40 + s8%0.08(40 + s) X

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Open Sentence: 0.15(40) + 0(s) = 0.08(40 + s) Solve:0.15(40) + 0(s) = 0.08(40 + s) 6 = 3.2 + 0.08s - 3.2 - 3.2 2.8 = 0.08s 0.08 0.08 35 = s 35 liters of water need to be added to make an 8% salt solution. Check: X

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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? Plan: Set Up: Write an open sentence: Solve: 2.5 pounds of dried apricots are needed. Check Your Solution:

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Practice 2 A 15L solution is 70% antifreeze. How much antifreeze must be added to produce a solution that is 80% antifreeze? Plan: Set Up: Write an open sentence: Solve: 7.5L of antifreeze must be added. Check Your Solution:

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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?

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22 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?

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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?

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24 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?

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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?

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Uniform Motion Problems Return to Table of Contents

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Uniform Motion Problems 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.

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Plan: Find the rate or speed of each boat. Set Up: 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? Rate (km/h) Time (h) Distance (km) Faster boat x66x Slower boat x - 1599(x - 15) X

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Open Sentence:6x = 9(x - 15) Solve 6x = 9x - 135 -9x - 9x -3x = -135 -3 -3 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: X

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Plan: Find Meghan's rate. Remember 1.4 min = ? sec Set Up: 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? Rate (m/s) Time (s) Distance (m) Jazmine2.5842.5(84) Meghanx8484x X

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Open Sentence: 2.5(1.4) + 84x = 630 Solve: 210 + 84x = 630 - 210 -210 84x = 420 84 84 x = 5 Meghan runs at a rate of 5 m/s. Check: X

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Plan:Find the distance from the school to Central Park. Set Up: 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? Rate (m/h) Time (h) Distance (m) To the park 7.5x7.5x Back to school 51.25 - x5(1.25 - x) X

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Open Sentence: 7.5x = 5(1.25 - x) Solve: 7.5x = 6.25 - 5x + 5x + 5x 12.5x = 6.25 12.5 12.5 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: X

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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? Plan: Set Up: Write an open sentence: Solve: It will take them 1.25 hours to be 7.5 miles apart. Check Your Solution:

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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? Plan: Set Up: Write an open sentence: Solve: Jeff lives 0.6 miles from the library and Brian lives 0.9 miles from the library. Check Your Solution:

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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? Plan: Set Up: Write an open sentence: Solve: Scott's entire trip is 12 km long. Check Your Solution:

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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?

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27 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?

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28 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?

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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?

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30 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?

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Work Problems Return to Table of Contents

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Work Problems 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)

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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? Plan: Find the number of hours it will take them to paint the room together. Set Up: 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. X

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Open Sentence: Maya's work + Takira's work = Job (1 room painted) x/4 + x/6 = 1 Solve: 12(x/4 + x/6) = 1(12) 3x + 2x = 12 5x = 12 5 5 x = 12/5 or 2 or 2.4 It would take 2.4 hours to paint the room if they worked together. Check: 2525 X

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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? Plan: 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. X

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Open Sentence: x/3 + x/2 = 1 Solve: 6(x/3 + x/2) = 1(6) 2x + 3x = 6 5x = 6 5 5 x = 6/5 or 1 or 1.2 It would take 1.2 hours to mow the lawn if they worked together. Check: 1515 X

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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? Plan: Set Up: Write an open sentence: Solve: If they work together, they can service 9 cars in 20 hours. Check Your Solution:

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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? Plan: Set Up: Write an open sentence: Solve: It will take them 0.75 hours working together. Check Your Solution:

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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?

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32 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?

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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?

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34 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?

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35 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?

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Proportionality Problems Return to Table of Contents

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Proportion Problems 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. =

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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? Plan: Given 10 lbs of seed for 2000 sq ft Find how much for 3200 sq ft Set Up: x = pounds of grass seed for 3200 sq ft Open Sentence: 10 x 2000 3200 = X

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Solve: 10 x 2000 3200 2000x = 10(3200) 2000x = 32000 2000 2000 x = 16 Mr. Jones needs 16 lbs of grass seed. Check: = X

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Example 2 The measures of two complementary angles are in the ratio of 3:7. Find the measure of each angle in degrees. Plan: 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 7 90 - x = X

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Solve: 3 x 7 90 - x 3(90 - x) = 7x 270 - 3x = 7x + 3x +3x 270 = 10x 10 10 27 = x The measures of the two complementary angles are 27 degrees and 63 degrees. Check: = X

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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? Plan: Set Up: Write an open sentence: Solve: 530 students do like the lunches. Check Your Solution:

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Practice 2 The measures of two supplementary angles are in the ratio 3:2. Find the measure of each angle. Plan: Set Up: Write an open sentence: Solve: 108 degrees and 72 degrees. Check Your Solution:

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36 A 96 mile trip requires 8 gallons of gasoline. At that rate, how many gallons would be required for a 156 mile trip?

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37 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?

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38 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?

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39 The measure of supplementary angles are in the ratio of 3 to 5. Find the measure of the smaller angle.

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

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