3.3 Ratios, Rates, Unit Rates & Proportions

3.3 Ratios, Rates, Unit Rates & Proportions

Quick Summary: Proportional reasoning is a unifying mathematical concept connecting geometry, statistics, algebra, and financial literacy. On the Ontario Mathematics Proficiency Test (MPT), candidates must distinguish between part-to-part and part-to-whole ratios, compute unit rates for comparative analysis, solve linear proportions via cross-multiplication, and correctly apply area scale transformations (( k^2 )).

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1. Ratios: Part-to-Part vs. Part-to-Whole

A ratio is a multiplicative comparison of two or more quantities of the same dimension or unit. Ratios can be expressed using colon notation (( a:b )), phrase notation ("( a ) to ( b )"), or fractional notation (( \frac{a}{b} )).

Structural Distinctions

• Part-to-Part Ratio: Compares one subgroup to another subgroup within the same set. Example: In a science class with 12 boys and 16 girls, the ratio of boys to girls is ( 12:16 ), which simplifies to ( 3:4 ).

• Part-to-Whole Ratio: Compares one subgroup to the entire set. Example: In the same class, the ratio of boys to total students is ( 12:(12+16) = 12:28 ), which simplifies to ( 3:7 ).

Pedagogical Note for Teachers

Only part-to-whole ratios can be interpreted directly as standard fractions or percentages of the overall set. A common student error on the MPT is treating a part-to-part ratio (e.g., ( 3:4 )) as if it represented ( \frac{3}{4} ) of the total population instead of ( \frac{3}{7} ).

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2. Rates vs. Unit Rates

• Rate: A comparison of two quantities having different units (e.g., distance and time, cost and mass). Examples: ( 450 \text{ km} ) per ( 5 \text{ hours} ); ( $14.25 ) for ( 3 \text{ kg} ).

• Unit Rate: A rate simplified so that the denominator is exactly 1 unit of the second quantity: [ \text{Unit Rate} = \frac{\text{Quantity 1}}{\text{Quantity 2}} ] Examples: ( \frac{450 \text{ km}}{5 \text{ h}} = 90 \text{ km/h} ); ( \frac{$14.25}{3 \text{ kg}} = $4.75/\text{kg} ).

Best-Buy & Efficiency Calculations

To compare value across products of different sizes, compute the unit rate for each product: [ \text{Unit Price} = \frac{\text{Total Price}}{\text{Quantity (g, kg, L, mL)}} ]

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3. Setting Up and Solving Proportions

A proportion is a mathematical statement asserting that two ratios or rates are equivalent: [ \frac{a}{b} = \frac{c}{d} \quad (b, d \neq 0) ]

Fundamental Property of Proportions (Cross-Multiplication)

By multiplying both sides of ( \frac{a}{b} = \frac{c}{d} ) by the common denominator ( b \cdot d ), we derive the Cross-Product Property: [ \frac{a}{b} = \frac{c}{d} \iff a \cdot d = b \cdot c ]

Algebraic Proportion Solving

When solving algebraic proportions involving binomial expressions, distribute carefully after cross-multiplying: [ \frac{x + 3}{8} = \frac{15}{20} ] Cross-multiply: ( 20(x + 3) = 8 \cdot 15 \implies 20x + 60 = 120 \implies 20x = 60 \implies x = 3 ).

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4. Scale Drawings, Map Scales & Area Scaling Transformations

Linear Scale Factor (( k ))

The linear scale factor ( k ) is the ratio of a length on a drawing/map to the corresponding actual length in reality: [ k = \frac{\text{Map Length}}{\text{Actual Length}} ]

The Area Scaling Law (( k^2 ))

If two geometric figures are similar with linear scale factor ( k ), their corresponding surface areas scale by the square of the linear scale factor (( k^2 )): [ \text{Actual Area} = \text{Map Area} \times \left(\frac{\text{Actual Linear Unit}}{\text{Map Linear Unit}}\right)^2 ] Exam Tip: If linear scale is ( 1 \text{ cm} : 5 \text{ m} ), then ( 1 \text{ cm}^2 ) on the map corresponds to ( 5^2 = 25 \text{ m}^2 ) in reality, not ( 5 \text{ m}^2 ).

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5. Summary Matrix of Proportional Reasoning

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6. Step-by-Step Worked Problems

Worked Problem 1: Multi-Part Ratio Budget Allocation

Problem: A school board allocates a total budget of ( $45,000 ) among three extracurricular programs—Athletics, Music, and Drama—in the ratio ( 4 : 3 : 2 ). Calculate the exact dollar amount allocated to each program and determine what fraction of the total budget Drama receives.

Step-by-Step Solution:

1. Identify Ratio Parts

   • Athletics : Music : Drama = ( 4 : 3 : 2 ).

2. Calculate Total Ratio Parts [ \text{Total Parts} = 4 + 3 + 2 = 9 \text{ parts} ]

3. Calculate Value per Ratio Part [ \text{Value per part} = \frac{$45,000}{9} = $5,000 ]

4. Determine Program Allocations

   • Athletics: ( 4 \times $5,000 = $20,000 ).
   • Music: ( 3 \times $5,000 = $15,000 ).
   • Drama: ( 2 \times $5,000 = $10,000 ).

5. Determine Drama's Fractional Share of Total Budget

   • Part-to-whole ratio for Drama: ( \frac{2}{9} ).
   • Check: ( \frac{$10,000}{$45,000} = \frac{2}{9} ).
   • Final Answer: Athletics = $20,000, Music = $15,000, Drama = $10,000; Drama receives ( \frac{2}{9} ) of the total budget.

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Worked Problem 2: Map Scale and Area Dilation Transformation

Problem: On a municipal plan drawn to a scale of ( 1 \text{ cm} : 500 \text{ m} ), a rectangular school yard measures ( 4 \text{ cm} ) by ( 7 \text{ cm} ). Calculate: (a) the actual perimeter of the school yard in metres, and (b) the actual area of the school yard in square kilometres (( \text{km}^2 )).

Step-by-Step Solution:

1. Determine Linear Scale Factors

   • ( 1 \text{ cm} = 500 \text{ m} = 0.5 \text{ km} ).

2. Calculate Actual Dimensions

   • Actual Length: ( 7 \text{ cm} \times 500 \text{ m/cm} = 3,500 \text{ m} = 3.5 \text{ km} ).
   • Actual Width: ( 4 \text{ cm} \times 500 \text{ m/cm} = 2,000 \text{ m} = 2.0 \text{ km} ).

3. Part (a): Calculate Actual Perimeter [ \text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (3,500 + 2,000) = 2 \times 5,500 = 11,000 \text{ metres} ]

4. Part (b): Calculate Actual Area in Square Kilometres

   • Method 1 (Product of Actual Kilometre Dimensions): [ \text{Area} = 3.5 \text{ km} \times 2.0 \text{ km} = 7.0 \text{ km}^2 ]
   • Method 2 (Area Scale Factor ( k^2 )):
     • Map Area ( = 4 \text{ cm} \times 7 \text{ cm} = 28 \text{ cm}^2 ).
     • Linear scale ( k = 0.5 \text{ km/cm} ).
     • Area scale factor ( k^2 = (0.5)^2 = 0.25 \text{ km}^2/\text{cm}^2 ).
     • Actual Area ( = 28 \text{ cm}^2 \times 0.25 \text{ km}^2/\text{cm}^2 = 7.0 \text{ km}^2 ).
   • Final Answer: (a) ( 11,000 \text{ m} ), (b) ( 7.0 \text{ km}^2 ).

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Worked Problem 3: Multi-Variable Worker-Rate Problem

Problem: Five identical 3D printers in a STEM lab produce 150 prototype parts in 6 hours of continuous operation. How many hours would it take 8 of these 3D printers to produce 300 prototype parts at the same constant rate?

Step-by-Step Solution:

1. State the Compound Rate Formula [ \frac{W_1}{P_1 \cdot T_1} = \frac{W_2}{P_2 \cdot T_2} ] where ( W ) is work (parts), ( P ) is printers, and ( T ) is time (hours).

2. Calculate Unit Rate per Printer per Hour [ \text{Unit Rate} = \frac{150 \text{ parts}}{5 \text{ printers} \times 6 \text{ hours}} = \frac{150}{30} = 5 \text{ parts per printer-hour} ]

3. Set Up Equation for 8 Printers Producing 300 Parts

   • Output rate for 8 printers = ( 8 \text{ printers} \times 5 \text{ parts/printer-hour} = 40 \text{ parts/hour} ).

4. Solve for Required Time ( T_2 ) [ T_2 = \frac{\text{Total Parts Required}}{\text{Combined Rate}} = \frac{300 \text{ parts}}{40 \text{ parts/hour}} = 7.5 \text{ hours} ]

   • Final Answer: It will take 7.5 hours (7 hours 30 minutes).