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Key words: rate, ratio, per, proportion, exchange, yield, interest.

Overview. Key words: rate, ratio, per, proportion, exchange, yield, interest.

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Key words: rate, ratio, per, proportion, exchange, yield, interest.

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  1. Overview Key words: rate, ratio, per, proportion, exchange, yield, interest. Purpose: This unit is designed to help tutors who teach courses in which working with rates is important. These courses may involve, rates of exchange, interest rates, yield rates, time and distance rates, rates of coverage Tutor Outcomes: By the end of the unit tutors should be able to: 1. Recognise contexts and problems that involve rates and can be solved using proportional thinking 2. Develop lessons in their teaching context that help learners to solve problems using rates. Resources you will need for this unit: Calculators

  2. Mathematical Background A unit rate is essentially a ratio used to compare a quantity (unit) to the way it is being measured. The two numbers being compared are not the same type, they are different units e.g dollars per kilogram, NZ Dollars per US Dollars A rate describes a relationship between two different measures. (Some rates have more than 2 units, eg., an IV rate might be express mg/kg/hour) In describing the units of a rate, the word "per" is used to separate the units of the two measurements used to calculate the rate (for example a heart rate is expressed "beats per minute"). The two parts are not mixed. Rate example: 50km/hr, comparison of speed to time. Ratio example: 1:2 fertiliser to water (the water and the fertiliser can be mixed)

  3. Mathematical Background • Why do we work with rates? • The basis of calculating an amount or value as a reference point. • A benchmark for calculating more or less, bigger or smaller, faster or slower. • To keep the rates constant we need to keep the amounts in proportion. • Rates have two features • Two measurement units, e.g for the price of coffee the units were dollars and grams • The word “per” which means “for every”, the cost of coffee was $3.50 for every 100 grams ($3.50 per 100g).

  4. Examples Recognising Rates Some contexts are easy to recognise as rates: 60 km/hr, $5 for 50 Mb data, $14 / hr 1 litre per 15 m2

  5. Examples Recognising Rates Some contexts are less easy to recognise as rates: 50 mg tablet 3 m length boards 60% yield It takes 2 sheets of metal to make 3 ducts

  6. Examples Working with rates Rates are often expressed using the symbol / to mean ‘per’ or ‘for every’. The same symbol is used for division in some contexts. The symbols used in maths often get in the way of understanding for learners who struggle with maths. A good teaching tip for working with rates and proportions is to organise what we know, and what we’re looking for in a table or box. For example 80 km/hr involves two units, km and hours:

  7. Examples Working with rates Start with a context most learners can relate to. Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 320 km? Think about what 80 km/hr means… Every 1 hour Sam travels 80 km. So in 2 hours he travels…

  8. Examples Working with rates Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 320 km? Now build up the table: Every 1 hour Sam travels 80 km. So in 2 hours he travels 160 km. How many hours to go 320 km? x 2 x 2 x 2 x 2

  9. Mathematical Background Direct Proportionality To keep the rate constant we need to keep the amounts in direct proportion. If the value of one unit in a rate doubles then the value of the other unit doubles, and so on. Our table so far, gives us three ways of expressing a constant rate: 80 km/hr is the same rate as 160 km/ 2hr or 320 km/ 4hr. Note there are horizontal and vertical relationships between the numbers. x 4 ÷ 80 ÷ 80 x 4

  10. Mathematical Background Proportionality There is also a relationship between diagonal numbers: 80 x 2 = 1 x 160 2 x 320 = 160 x 4 1 x 320 = 80 x 4 This is a very useful check when we’re solving problems using rates.

  11. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? This is a slightly harder problem because 300 is not a whole number multiple of 80. You and your learners may come up with several strategies for working out the answer. But before you even start looking, ask yourself (and your learners): What do we already know about the answer? We already know that the answer is between 2 and 4 hours, (and it’s closer to 4 than 2.)

  12. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? Note that we can build up the table using additive thinking: Sam travels 80 km in 1 hour and 160 km in 2 hours so in 3 hours he travels 160 + 80 = 240 km OR multiplicative thinking: Sam travels 80 km in 1 hour and 3 hours is 3 times longer; so in 3 hours he travels 80 x 3 = 240 km

  13. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? Because 300 km is between 240 and 320, we know it takes between 3 and 4 hours to travel 300 km. Do we need to be more exact than this? It depends on the context. If we need an exact answer, several strategies are available.

  14. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? Possible Strategy 1: Extend the table for km per part hours It takes ¼ hour to travel 20 km and ½ hour to travel 40 km, so ¾ hour to travel 60 km. (Note that the table can be built up in either direction.) 3 hours takes Sam 240 km, ¾ hour will take him another 60 km. So to travel 300 km takes 3 ¾ or 3.75 hours. This strategy will suit learners who are more comfortable thinking additively than multiplicatively.

  15. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? Possible Strategy 2: Work out how many times bigger 300 is than 240, and apply that multiple to 3. 300 ÷ 240 = 1.25 240 x 1.25 = 300 x ? We now know that 300 is 1.25 times bigger than 240 x 1.25 So the travel time is 1.25 times bigger than 3 hours x 1.25

  16. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? Check the answer using the diagonals: 3 x 300 = 900 240 x 3.75 = 900 x 1.25 x 1.25

  17. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? Possible Strategy 3: Use the diagonal pattern to find the missing value 3 x 300 = 240 x ? 900 = 240 x ? ? = 900 ÷ 240 We know that 3 x 300 = 240 x missing value Note that to use this strategy with understanding, your learners must understand that 900 ÷ 240 means 240 x ‘what’ = 900

  18. Examples Finding less obvious values Example: Sam’s average speed for a road trip is 80 km/hr. How long will it take him to travel 300 km? The beauty of the diagonals is that it doesn’t matter which diagonal combination you use: 1 x 300 = 80 x ? 300 = 80 x ? ? = 300 ÷ 80

  19. Examples Board lengths are 1.5 m, how many boards are needed to cover 4.5 m? How might you solve it? Are there any problems like this in the context of your course?

  20. Worked Examples The implied rate in this problem is 1.5 m per board Now build up the table: Every 1 board covers 1.5 m. So 2 boards cover 3 m. How many boards will cover 4.5 m? Board lengths are 1.5 m, how many boards are needed to cover 4.5 m? x 2 x 2

  21. Worked Examples Board lengths are 1.5 m, how many boards are needed to cover 4.5 m? Additive Strategy: 1 board covers 1.5 m and 2 boards cover 3 m. So 3 boards cover 1.5 + 3 = 4.5 m.

  22. Worked Examples Multiplicative strategy 1: Find how many times bigger 4.5 is than 1.5 4.5 ÷ 1.5 = 3 Apply that multiple to both rows: Board lengths are 1.5 m, how many boards are needed to cover 4.5 m? x 3 x 3

  23. Worked Examples Multiplicative strategy 2 (diagonals): 1 x 4.5 = 1.5 x ?? = 1 x 4.5 ÷ 1.5 = 3 Board lengths are 1.5 m, how many boards are needed to cover 4.5 m? OR: 2 x 4.5 = 3 x ?? = 2 x 4.5 ÷ 3 = 3

  24. Worked Examples Suppose 1 NZ dollar buys 76 Australian cents:$NZ1 = $AU0.76How much Australian currency will $NZ1350 buy? • Can you solve this using a table? • Any similar problems in your course?

  25. Worked Examples Now build up a table: 1 NZ dollar buys Aussie $0.76. So 10 NZ dollars buys $7.60 and 100 NZ dollars buys $AU76.00. And so on… $NZ1 = $AU0.76How much Australian currency will $NZ1350 buy? x 10 X 10

  26. Worked Examples Additive strategy: $NZ1000 buys $AU760 Plus $NZ300 buys 3 x $AU76 (3 x 76 = $AU228) Plus $NZ50 buys half of $AU76 (76 ÷ 2 = $AU38) $NZ1350 = $AU (760 + 228 + 38) = $AU 1026 $NZ1 = $AU0.76How much Australian currency will $NZ1350 buy? How would you use this additive strategy to work out how much $NZ1357.50 would buy?

  27. Worked Examples Multiplicative strategy: $NZ1 buys $AU0.76 $NZ1350 is 1350 times bigger than $NZ1 so $AU0.76 x 1350 = $1026 $NZ1 = $AU0.76How much Australian currency will $NZ1350 buy? x 1350 x 1350

  28. Worked Examples A slight twist on the problem:$NZ1 = $AU0.76How much NZ currency is required to buy $AU1000? We can build the table up to estimate that we’ll need a bit less than $NZ1500.

  29. Worked Examples A slight twist on the problem:$NZ1 = $AU0.76How much NZ currency is required to buy $AU1000? To work out the exact answer for this problem, it’s useful to work out how many $NZ are equivalent to $1 Aussie. ÷ 0.76 To change 0.76 to 1, we divide by 0.76. To keep the table proportional, we need to also divide the top line by 0.76 ÷ 0.76

  30. Worked Examples A slight twist on the problem:$NZ1 = $AU0.76How much NZ currency is required to buy $AU1000? 1 ÷ 0.76 = 1.315, so every 1 Aussie $ buys $NZ1.32 ÷ 0.76 ÷ 0.76 x 1000 Now build the table to $AU1000 x 1000

  31. Activity Use the rates in your course or context Think about the calculations in your course that are based on rates. Get your learners to set the problems up using a proportion table (ratio box). Ask them to think about what would be a reasonable answer before they calculate the exact answer.

  32. Practice Problems Following are some problems from various contexts to practice with. It takes two sheets of metal make three ducts. How many ducts can we make from 12 sheets?

  33. Practice Problems You get 4.5 glasses of wine from a bottle, how many bottles are required for 90 glasses of wine? You get 4.5 glasses of wine from a bottle, how many glasses of wine can you get from a case of 12 bottles?

  34. Practice Problems Since percent means ‘for every 100’, percentage problems are also rate problems: In Frank’s business last month, electricity costs were $900 and total costs were $4000. Electricity accounts for what percentage of the total expenses?

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