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 $5 \text{ hours}$; $\text{\textdollar}14.25$ 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: $\frac{450 \text{ km}}{5 \text{ h}} = 90 \text{ km/h}$; $\frac{\text{\textdollar}14.25}{3 \text{ kg}} = \text{\textdollar}4.75/\text{kg}$.

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 $\text{\textdollar}45,000$ 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: $4 \times \text{\textdollar}5,000 = \text{\textdollar}20,000$.
   • Music: $3 \times \text{\textdollar}5,000 = \text{\textdollar}15,000$.
   • Drama: $2 \times \text{\textdollar}5,000 = \text{\textdollar}10,000$.

5. Determine Drama's Fractional Share of Total Budget

   • Part-to-whole ratio for Drama: $\frac{2}{9}$.
   • Check: $\frac{\text{\textdollar}10,000}{\text{\textdollar}45,000} = \frac{2}{9}$.
   • 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).