7.4 Ratios, Proportions, Rates & Weighted Averages

7.4 Ratios, Proportions, Rates & Weighted Averages

Key Exam Fact: Ratios, dimensional rates, and weighted averages represent the fourth pillar of Bocconi Numerical Reasoning. These questions test structural proportionality, unit conversions, and blended averages under time pressure. Candidates who write out complex systems of equations often run out of time; candidates who utilize proportional scaling, work-rate inversion, and the alligation lever rule solve these problems in under 45 seconds.

Proportional reasoning is foundational to economic analysis. Whether modeling labor productivity, asset allocations, currency conversions, or product recipe blends, the mathematical principles governing these phenomena rely on consistent ratio properties.

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1. Ratio Mechanics: Part-to-Part vs. Part-to-Whole

A ratio expresses the relative magnitude of two or more quantities. Ratios can be expressed as $a:b$, as a fraction $\frac{a}{b}$, or verbally as "a to b".

Scaling & Equivalence

Multiplying or dividing all terms of a ratio by a non-zero constant $k$ preserves its proportional value: $a : b = ka : kb$

Part-to-Part vs. Part-to-Whole Conversion

• Part-to-Part: If the ratio of domestic to international students in a Bocconi master's cohort is $3 : 5$, there are 3 domestic students for every 5 international students.
• Total Parts: $3 + 5 = 8$ total parts.
• Part-to-Whole Fractions: $\text{Domestic Fraction} = \frac{3}{3 + 5} = \frac{3}{8} = 37.5\%$ $\text{International Fraction} = \frac{5}{3 + 5} = \frac{5}{8} = 62.5\%$

Chained Ratios (Unifying Three Variables)

A classic Bocconi item provides two disjoint two-variable ratios sharing a common middle variable and asks for the unified ratio $A : B : C$.

The LCM Unification Protocol:

1. Write down the two given ratios: $A : B = a_1 : b_1$ and $B : C = b_2 : c_2$.
2. Determine the Least Common Multiple (LCM) of the common terms $b_1$ and $b_2$: $L = \text{LCM}(b_1, b_2)$.
3. Scale the first ratio by $\frac{L}{b_1}$ and the second ratio by $\frac{L}{b_2}$.
4. Merge into the single unified string $A : B : C$.

Example: If $A : B = 5 : 6$ and $B : C = 4 : 3$:

• Shared variable is $B$, with values $6$ and $4$.
• $\text{LCM}(6, 4) = 12$.
• Scale Ratio 1 ($\times 2$): $A : B = 10 : 12$.
• Scale Ratio 2 ($\times 3$): $B : C = 12 : 9$.
• Unified Ratio: $\mathbf{A : B : C = 10 : 12 : 9}$.

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2. Direct vs. Inverse Proportionality

Direct Proportionality ($y = kx$)

Two quantities $x$ and $y$ are directly proportional if their ratio remains constant: $\frac{y_1}{x_1} = \frac{y_2}{x_2} = k \implies y_1 x_2 = y_2 x_1$

• If $x$ doubles, $y$ doubles.
• Real-world examples: Direct material cost vs. production volume; fuel consumed vs. distance traveled.

Inverse Proportionality ($y = \frac{k}{x}$)

Two quantities $x$ and $y$ are inversely proportional if their product remains constant: $x_1 \cdot y_1 = x_2 \cdot y_2 = k \implies y_2 = y_1 \left(\frac{x_1}{x_2}\right)$

• If $x$ doubles, $y$ is halved ($1/2$).
• Real-world examples: Speed vs. travel time ($v \cdot t = d$); number of workers vs. days required to complete a fixed project.

Compound Proportionality (The Master Work Identity)

When a task involves workers ($W$), days ($D$), hours per day ($H$), and units of work completed ($Q$): $\frac{W_1 \cdot D_1 \cdot H_1}{Q_1} = \frac{W_2 \cdot D_2 \cdot H_2}{Q_2}$

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3. Work-Rate and Pipe/Tank Mechanics

Work-rate problems require determining the time needed for multiple agents (workers, machines, or inlet/drain pipes) acting concurrently to finish a fixed job.

The Reciprocal Rate Principle

If an agent completes $1$ full job in $T$ hours, its work rate $R$ (jobs per hour) is the reciprocal of time: $R = \frac{1}{T}$

Combined Cooperative Work Rate

When independent agents work together, their individual rates add directly: $R_{\text{total}} = R_1 + R_2 + \dots + R_n = \frac{1}{T_1} + \frac{1}{T_2} + \dots + \frac{1}{T_n}$ $T_{\text{together}} = \frac{1}{R_{\text{total}}} = \frac{1}{\sum \frac{1}{T_i}}$

• Two-Agent Shortcut Formula: $T_{\text{together}} = \frac{T_1 \cdot T_2}{T_1 + T_2}$

Opposing Rates (Pipes and Drains)

If Inlet Pipe 1 fills a tank in $T_{\text{in1}}$ hours, Inlet Pipe 2 fills it in $T_{\text{in2}}$ hours, and Drainage Pipe 3 empties the tank in $T_{\text{drain}}$ hours, the net rate is: $R_{\text{net}} = R_{\text{in1}} + R_{\text{in2}} - R_{\text{drain}} = \frac{1}{T_{\text{in1}}} + \frac{1}{T_{\text{in2}}} - \frac{1}{T_{\text{drain}}}$ $T_{\text{net}} = \frac{1}{R_{\text{net}}}$

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4. Currency Exchange & Dimensional Analysis

Currency problems assess transactional conversions, bid-ask spreads, and cross-currency triangulation.

Direct vs. Indirect Quotations

• Direct Quote: Cost of 1 unit of foreign currency expressed in domestic units (e.g., $1\text{ USD} = 0.92\text{ EUR}$ from an EU perspective).
• Indirect Quote: Quantity of foreign currency purchased by 1 unit of domestic currency (e.g., $1\text{ EUR} = 1.087\text{ USD}$).
• Identity: $\text{Direct Quote} = \frac{1}{\text{Indirect Quote}}$.

Units of Measure and Conversion Chains

Bocconi's mathematics syllabus lists a bullet that is easy to skip and cheap to master: "be able to solve simple problems involving various units of measure and conversions between them." These items are pure dimensional analysis — multiply by conversion factors written as fractions equal to 1, arranged so that the unwanted units cancel.

$\text{Target quantity} = \text{Given quantity} \times \frac{\text{target unit}}{\text{source unit}} \times \cdots$

The three conversion factors most likely to appear together:

The Squared-Factor Trap: because $1\text{ m} = 100\text{ cm}$, candidates convert $1\text{ m}^2$ to $100\text{ cm}^2$. It is $100^2 = 10{,}000\text{ cm}^2$. Any conversion of an area scales by the square of the linear factor, and any conversion of a volume by its cube — the same scaling law met in the plane-geometry section.

Worked Example A (compound-unit chain). A plant consumes 45 grams of resin per unit and produces 8,000 units per hour. Express annual consumption in tonnes for 2,000 operating hours per year. $45 \frac{\text{g}}{\text{unit}} \times 8{,}000 \frac{\text{units}}{\text{h}} \times 2{,}000 \frac{\text{h}}{\text{year}} = 720{,}000{,}000 \frac{\text{g}}{\text{year}}$ Dividing by $10^6$ grams per tonne gives 720 tonnes per year. Notice that every unwanted unit cancels diagonally; if it does not, the chain is assembled upside down.

Worked Example B (rate conversion). A conveyor moves at 1.5 m/s. In km/h that is $1.5 \times 3.6 = 5.4\text{ km/h}$, because $\frac{1\text{ m}}{1\text{ s}} \times \frac{3{,}600\text{ s}}{1\text{ h}} \times \frac{1\text{ km}}{1{,}000\text{ m}} = 3.6 \frac{\text{km}}{\text{h}}$.

Worked Example C (unit-price comparison). Supplier X quotes €18 per 750 mL bottle; Supplier Y quotes €22 per litre. Converting X to the same basis: $\frac{18}{0.75} = \text{EUR } 24$ per litre. Supplier Y is cheaper by €2 per litre, a conclusion invisible until both quotations sit on one unit.

Cross-Rate Currency Triangulation

Given two bilateral exchange rates sharing a common reference currency (e.g., the US Dollar), compute the synthetic cross-rate: $\frac{\text{EUR}}{\text{GBP}} = \frac{\text{EUR} / \text{USD}}{\text{GBP} / \text{USD}}$

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5. Weighted Averages & The Alligation (Lever) Rule

A weighted average reflects the relative importance (weight) of each component value in a composite distribution.

Standard Weighted Average Formula

$\bar{x}_w = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i} = \frac{w_1 x_1 + w_2 x_2 + \dots + w_n x_n}{w_1 + w_2 + \dots + w_n}$

The Alligation Alternate (The Lever Rule) Shortcut

When mixing two groups of values $x_1$ (lower) and $x_2$ (higher) to achieve a target blended mean $\bar{x}$, setting up algebraic equations is slow. Instead, use the Alligation Lever Rule.

          LOWER VALUE (x_1)                     HIGHER VALUE (x_2)
                 o----------------------^----------------------o
                                  TARGET (x_bar)
                 | <---- d_1 ----> |    | <----- d_2 ------> |
                 
                 d_1 = x_bar - x_1          d_2 = x_2 - x_bar

By the physical law of moments (torque balance around the fulcrum $\bar{x}$): $w_1 (\bar{x} - x_1) = w_2 (x_2 - \bar{x})$ $\frac{w_1}{w_2} = \frac{x_2 - \bar{x}}{\bar{x} - x_1} = \frac{d_2}{d_1}$

THE ALLIGATION SHORTCUT: The ratio of the weights is the inverse ratio of the distances from the component values to the target mean!

Instant Application: Blend a $15%$ acid solution ($x_1$) with a $40%$ acid solution ($x_2$) to create a $25%$ acid solution ($\bar{x}$):

• Distance $d_1 = 25 - 15 = 10$.
• Distance $d_2 = 40 - 25 = 15$.
• Ratio of quantities: $\frac{w_1}{w_2} = \frac{d_2}{d_1} = \frac{15}{10} = \frac{3}{2}$.
• Exactly $3$ parts of the $15%$ solution are required for every $2$ parts of the $40%$ solution. Solved in 5 seconds.

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6. Worked Step-by-Step Examples

Problem 1: Tank Filling with Inflow & Drainage

An industrial chemical vat has two inlet pipes and one drainage pipe. Inlet Pipe A fills the vat in 6 hours. Inlet Pipe B fills the vat in 8 hours. Drainage Pipe C empties the vat in 12 hours. If the vat is empty and all three pipes are opened simultaneously, how long does it take to fill the vat completely?

1. Establish Individual Hourly Rates: $R_A = \frac{1}{6}, \quad R_B = \frac{1}{8}, \quad R_C = -\frac{1}{12}$
2. Sum Net Hourly Rate (Find Common Denominator = 24): $R_{\text{net}} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{4 + 3 - 2}{24} = \frac{5}{24}\text{ vats per hour}$
3. Invert Rate to Find Time: $T = \frac{1}{R_{\text{net}}} = \frac{24}{5} = 4.8\text{ hours}$ $0.8\text{ hours} = 0.8 \times 60 = 48\text{ minutes}$ The vat will fill in 4 hours and 48 minutes (4.8 hours).

Problem 2: Portfolio Fixed-Income Allocation via Alligation

A wealth manager must allocate €$5,000,000$ across two bond funds to achieve an aggregate annual yield of exactly $6.2%$. Fund A yields $4.6%$ annually, and Fund B yields $7.0%$ annually. How much capital must be invested in Fund B?

1. Apply the Alligation Lever Rule:
   • Lower Yield ($x_1$): $4.6%$
   • Higher Yield ($x_2$): $7.0%$
   • Target Yield ($\bar{x}$): $6.2%$
2. Calculate Distance Delays:
   • Distance from Fund A to Target: $d_1 = 6.2% - 4.6% = 1.6%$.
   • Distance from Fund B to Target: $d_2 = 7.0% - 6.2% = 0.8%$.
3. Determine Ratio of Weights: $\frac{W_A}{W_B} = \frac{d_2}{d_1} = \frac{0.8\%}{1.6\%} = \frac{1}{2}$ Fund A receives $1$ part, and Fund B receives $2$ parts (Total parts = $1 + 2 = 3$).
4. Compute Capital for Fund B: $W_B = \frac{2}{3} \times \text{EUR } 5,000,000 = \frac{\text{EUR } 10,000,000}{3} \approx \mathbf{\text{EUR } 3,333,333}$ The manager must allocate €$3,333,333$ to Fund B (and €$1,666,667$ to Fund A).