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Grade 10 Academic Math Chapter 1 – Linear Systems Modelling Word Problems Days 4 through Days 9

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Day 4 Agenda 1.Warm-up 2.Types of Modelling Problems 3.Mixture Problems 4.Relative Value Problems 5.Practice

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Learning Goal By the end of the lesson… … students will be able to read and interpret a mixture or relative value word problems and create a pair of linear relation equations, resulting in a linear system

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Curriculum Expectations Solve problems that arise from realistic situations described in words… by choosing an appropriate algebraic… method Ontario Catholic School Graduate Expectations: The graduate is expected to be… a self-directed life long learner who CGE4f applies effective… problem solving… skills

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Mathematical Process Expectations Connecting – make connections among mathematical concepts and procedures; and relate mathematical ideas to situations or phenomena drawn from other contexts

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Modelling Types 1.Break-Even Problems 2. Mixture Problems 3. Relative Value Problems 4. Rate Problems

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Mixture Problems 2 things come together to give a total number or amount 2 things come together to form a total cost, weight, points, etc. Equations are usually in form Ax + By = C

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Mixture Problems Ex. 1 Henry sharpens figures skates for $3 a pair and hockey skates for $2.50 per pair. If he earns $240 and sharpens 94 pairs of skates, how many pairs of each type of skate does he sharpen?

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Example 1 Mixture (Cont’d)

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Mixture Problems Let x represent# of figure skates Let y represent# of hockey skates x + y = 94 (# of skates eq’n) 3x + 2.5y = 240 (earnings eq’n)

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Mixture Problems Ex. 2 Joe has 38 loonies and toonies totalling $55. How many of each type of coin does he have?

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Example 2 Mixture (Cont’d)

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Mixture Problems Let l represent# of loonies Let t represent# of toonies l + t = 38 (# of coins equation) l + 2t = 55 (value equation)

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Mixture Problems Ex. 3 (p.44, #11e) Benoit invested some money at 8% and some at 10%. He earned a total of $235 in interest.

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Example 3 Mixture (Cont’d)

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Mixture Problems Let x representAmount of $ invested at 8% Let y representAmount of $ invested at 10% 0.08x + 0.1y = 235 (interest equation) Note: In order to do a $ invested eq’n, we need the amount invested

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Mixture Problems Ex. 4, p.51, #4c The total value of nickels and dimes is 75¢

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Example 4 Mixture (Cont’d)

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Mixture Problems Let n represent# of nickels Let d represent# of dimes 0.05n + 0.10d = 0.75 (value equation) Note: In order to do a # of coins eq’n, we need to know the # coins

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Relative Value Problems Usually 2 unknown numbers, ages, etc. No set form to the equations Must follow the directional words such as more than, less, times, is, twice, sum, difference, etc.

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Relative Value Problems Ex. 1, p.51, #7 The sum of two numbers is 72. Their difference is 48. Find the numbers.

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Example 1 Relative Value (Cont’d)

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Relative Value Problems Let x representthe first number Let y representthe other number x + y = 72 (sum equation) x – y = 48 (difference equation)

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Relative Value Problems Ex. 2, p.51, #8) A number is four times another number. Six times the smaller number plus half of the larger number equals 212. Find the numbers.

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Example 2 Relative Value (Cont’d)

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Relative Value Problems Let x representthe first number Let y representthe other number x = 4y (multiplication eq’n) 0.5x + 6y = 212 (difference equation)

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Relative Value Problems Ex. 3, p.24, #7 At the December concert, 209 tickets were sold. There were 23 more student tickets sold than twice the number of adult tickets. How many of each were sold?

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Example 3 Relative Value (Cont’d)

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Relative Value Problems Let x represent# of student tickets Let y represent# of adult tickets x - 23 = 2y (relative # of tickets) x + y = 209 (# of tickets)

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Relative Value Problems Ex. 4, p.24, #8 A rectangle with a perimeter of 54cm is 3m longer than it is wide. What are its length and width?

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Relative Value Problems Let l representwidth of the rect. Let w representlength of the rect. 2x + 2y = 54 (perimeter eq’n) l – 3 = w (relative length to width eq’n)

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Humour Break

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Break-Even Problems Usually look for the point at which two things cost the same Can refer to the point at which cost and number of things are equal Equations usually take the form of y = mx + b

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Break-Even Problems Ex. 1. Barney’s Banquet Hall charges $500 to rent the room, plus $15 for each meal and Patrick’s Party Palace charges $400 for the hall plus $18 for each meal. When will both places cost the same amount?

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Example 1 Break-Even (Cont’d)

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Break-Even Problems Let x represent# meals Let y representthe cost y = 15x + 500 (Barney’s BH) y = 18x + 400 (Patrick’s PP)

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Break-Even Problems Ex. 2. The Millennium Wheelchair Co. has just started its business. It costs them $125 to make each wheelchair plus $15,000 in start-up costs. They plan to sell the chairs for $500 each. How many chairs do they have to sell in order to break even?

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Example 2 Break-Even (Cont’d)

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Break-Even Problems Let x represent# of wheelchairs Let y representcost or revenue y = 125x + 15000 (Cost eq’n) y = 500x (Revenue eq’n)

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Break-Even Problems Ex. 3. p.44, #11c It costs $135 to rent the car, based on $25 per day, plus $0.15/km

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Example 3 Break-Even (Cont’d)

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Break-Even Problems Let x represent# of days Let y represent# of km driven 25x + 0.15y = 135(Cost eq’n) Note: This is not a usual example. Usually if you are dealing with car rental, you have an eq’n like y = 0.15x + 25

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Humour Break

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Rate (Speed Distance Time) Problems (Copy) Usually looking for time, speed or distance Distance = Speed x Time (from science – can be rearranged for speed and time also) Easiest to use a chart to help develop the equations

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Rate (Speed Distance Time) Problems But first, we have the Distance = Speed x Time (equation) Or... D = S x T

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Rate (Speed Distance Time) Problems We can also rearrange this eq’n to solve for speed... Speed = Distance ------------ Time Or...

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Rate (Speed Distance Time) Problems We can also rearrange this eq’n to solve for Time... Time = Distance ------------ Speed

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Rate (Speed Distance Time) Problems Ex. 1 Fred travelled 95 km by car and train. The car averaged 60 km/h and the train averaged 90 km/hr. If the trip took 1.5 hours, how long did he travel by car? Let’s use a speed distance time chart to organize our information...

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Example 1 Rate (Cont’d)

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Rate (Speed Distance Time) Problems Let x representthe time in the car Let y representthe time on the train Distance (km) Speed (kph) Time (h) Car Train Total

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) Car60 Train Total

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) Car60 Train90 Total

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Rate (Speed Distance Time) Problems Distance (km) (recall D = S x T... Speed (kph) Time (h) Car60x60x Train90 Total

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) Car60x60x Train90y90y Total

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) Car60x60x Train90y90y Total95

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) Car60x60x Train90y90y Total951.5

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Rate (Speed Distance Time) Problems x + y = 1.5 (total travelling time) 60x + 90y = 95 (total distance travelled)

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Rate (Speed Distance Time) Problems Ex. 2 (text p.137, #6) A traffic helicopter pilot finds that with a tailwind, her 120km trip away from the airport takes 30 minutes. On her return trip to the airport, into the wind, she finds that her trip is 10 minutes longer. What is the speed of the helicopter? What is the speed of the wind?

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Example 2 Rate (Cont’d)

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Rate (Speed Distance Time) Problems Let h represent the speed of the helicopter Let w represent the speed of the wind Distance (km) Speed (kph) Time (h) With tail wind With headwind Total

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) With tailwind 120 With headwind 120 Total

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) With tailwind 120h + w With headwind 120 Total

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Rate (Speed Distance Time) Problems Distance (km) Speed (kph) Time (h) With tailwind 120h + w½ With headwind 120h - w2/3 (keep as a fraction) Total

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Rate (Speed Distance Time) Problems Recall that Speed = Distance ------------ Time

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Rate (Speed Distance Time) Problems h + w = 120/0.5 (with tailwind... ) h – w = 120/(2/3) (with headwind) h + w = 240 (simplified) h – w = 120 x (3/2) (flip 3/2 and x’s) h – w = 180 (simplified)

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Humour Break

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