4.2 Ratios, Proportions, Direct & Inverse Variations

4.2 Ratios, Proportions, Direct & Inverse Variations

Operational Imperative: Military officers continuously handle proportional reasoning: scaling supply requisitions according to troop strength, calculating convoy fuel burn rates across variable terrain, scheduling combat engineering projects under changing personnel numbers, and converting navigational coordinates between metric, imperial, and maritime nautical standards. Mastery of direct, inverse, and partitive variations provides the algebraic toolkit to resolve these problems in under one minute per item.

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The Architecture of Ratios: Part-to-Part vs. Part-to-Whole

A ratio is a mathematical comparison of two or more quantities having the same unit of measurement, expressed as $a : b$, $\frac{a}{b}$, or "$a$ to $b$". Ratios represent relative scale rather than absolute quantities. A ratio of $3:5$ means that for every 3 units of the first entity, exactly 5 units of the second entity exist.

Part-to-Part vs. Part-to-Whole Ratios

One of the most frequent traps on the AFPSAT involves confusing part-to-part relationships with part-to-whole relationships:

• Part-to-Part Ratio ($a : b$): Compares one subgroup directly to another subgroup. For example, if a specialized scout platoon contains 6 officers and 24 enlisted personnel, the part-to-part ratio of officers to enlisted personnel is $6 : 24$, which simplifies to $1 : 4$.
• Part-to-Whole Ratio ($a : [a + b]$): Compares one subgroup to the aggregate total. In the same platoon, the total personnel count is $6 + 24 = 30$. The part-to-whole ratio of officers to the entire platoon is $\frac{6}{30} = \frac{1}{5}$ (or $20%$). The part-to-whole ratio of enlisted soldiers is $\frac{24}{30} = \frac{4}{5}$ (or $80%$).

Rule for Ratio Conversion: If a part-to-part ratio between two groups is $a : b$, the fractional share of group $a$ relative to the whole is $\frac{a}{a + b}$, and the fractional share of group $b$ is $\frac{b}{a + b}$. Misidentifying a part-to-part ratio as a part-to-whole fraction (e.g., assuming $1:4$ means $1/4 = 25%$ rather than $1/5 = 20%$) is a major source of examinee error.

Scaling and Simplifying Ratios

Ratios behave identically to fractions regarding multiplication and division: multiplying or dividing all terms in a ratio by the same non-zero constant preserves the proportional relationship: $3 : 5 = (3 \times 4) : (5 \times 4) = 12 : 20$ $36 : 48 = (36 \div 12) : (48 \div 12) = 3 : 4$

However, adding or subtracting a constant from ratio terms destroys the ratio. If two platoons have personnel in the ratio $3:5$, transferring or adding 10 soldiers to both platoons completely alters the proportion.

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Partitive Proportion & Multi-Part Distribution

A partitive proportion involves partitioning an aggregate quantity into multiple unequal portions according to a designated multi-part ratio $a : b : c$.

The "Total Parts" Scaling Algorithm

To divide total quantity $T$ according to the ratio $a : b : c$:

1. Sum the ratio parts: $S = a + b + c$.
2. Derive the scaling multiplier ($k$): Determine the physical value of a single ratio part by dividing the total quantity by the sum of the parts: $k = \frac{T}{S}$.
3. Compute each individual share: Multiply each ratio term by $k$: $\text{Share } A = a \times k, \quad \text{Share } B = b \times k, \quad \text{Share } C = c \times k$

Step-by-Step Worked Example

Problem: A joint task force allocates a logistics shipment of $9,600$ crates of field rations among the Army, Navy, and Air Force components in the ratio $5 : 2 : 1$. How many crates does the Army component receive?

1. Sum the parts: $S = 5 + 2 + 1 = 8$ parts.
2. Find the value of 1 part ($k$): $k = \frac{9,600}{8} = 1,200 \text{ crates per part}$
3. Compute allocations:
   • Army ($5$ parts): $5 \times 1,200 = 6,000$ crates.
   • Navy ($2$ parts): $2 \times 1,200 = 2,400$ crates.
   • Air Force ($1$ part): $1 \times 1,200 = 1,200$ crates.
4. Verification: $6,000 + 2,400 + 1,200 = 9,600$ crates. The Army component receives $6,000$ crates.

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Direct Variation: Concepts, Formulations & Logistical Scenarios

A direct proportion (or direct variation) describes a mathematical relationship between two variables where their ratio remains constant. When one quantity increases, the other increases at the same proportional rate; when one decreases, the other decreases identically.

Mathematical Formulation

Variable $y$ varies directly as $x$ if: $y = kx \quad \iff \quad \frac{y}{x} = k$ where $k$ is the non-zero constant of proportionality (the rate). For any two states $(x_1, y_1)$ and $(x_2, y_2)$: $\frac{y_1}{x_1} = \frac{y_2}{x_2} \quad \iff \quad \frac{x_1}{x_2} = \frac{y_1}{y_2}$ Cross-multiplying yields: $x_1 y_2 = x_2 y_1$

Classic Military Direct Variation Scenarios

• Convoy Fuel Consumption: Fuel burned ($F$) varies directly with travel distance ($D$): $F = kD$.
• Ammunition Expenditure: Rounds fired ($R$) vary directly with firing duration ($t$): $R = kt$.
• Subsistence Rations: Rations required ($R$) vary directly with troop count ($P$): $R = kP$.

Worked Scenario: A military transport convoy of six-wheeled utility trucks consumes $180$ liters of diesel fuel while traversing $75$ kilometers of mountain terrain. Under identical driving conditions, how many liters of diesel will the convoy consume over a distance of $225$ kilometers?

$\frac{F_1}{D_1} = \frac{F_2}{D_2} \implies \frac{180}{75} = \frac{F_2}{225}$ Instead of long division, notice the horizontal scaling factor: $225 = 75 \times 3$. Because the distance increases by a factor of 3, fuel consumption must scale by the exact same factor of 3: $F_2 = 180 \times 3 = 540 \text{ liters}$ Solving via cross-multiplication: $75 F_2 = 180 \times 225 \implies F_2 = \frac{180 \times 225}{75} = 180 \times 3 = 540$ liters.

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Inverse (Indirect) Variation: Worker-Time & Speed-Time Trade-Offs

An inverse proportion (or indirect variation) describes a relationship between two variables where their product remains constant. As one variable increases, the other variable decreases proportionally.

Mathematical Formulation

Variable $y$ varies inversely as $x$ if: $y = \frac{k}{x} \quad \iff \quad x \times y = k$ where $k$ is the invariant product. For any two corresponding states $(x_1, y_1)$ and $(x_2, y_2)$: $x_1 y_1 = x_2 y_2 \quad \iff \quad \frac{x_1}{x_2} = \frac{y_2}{y_1}$

Notice that in an inverse proportion, the ratio of the second variable is inverted (flipped) relative to the first.

The Worker-Time Principle (The "Man-Days" Model)

The most pervasive inverse proportion on military exams involves workforce size and project completion time. Total required labor is a fixed constant measured in man-days or man-hours: $\text{Work} = \text{Workers} \times \text{Time} = M_1 D_1 = M_2 D_2$

• Doubling the workforce halves the required completion time.
• Reducing the workforce to one-third triples the required completion time.

Exam Trap Alert: Test writers frequently ask "How many additional workers must be assigned?" rather than "What is the total new worker count?" If 10 men take 6 days ($60$ man-days), finishing in 4 days requires $60 / 4 = 15$ men total. If the question asks for additional men, the correct answer is $15 - 10 = 5$ additional men, not $15$.

The Speed-Time Distance Model

When transit distance $D$ is constant, speed ($S$) and travel time ($T$) are inversely proportional: $D = S_1 T_1 = S_2 T_2$ Increasing convoy cruising speed reduces transit time proportionally: $\frac{S_1}{S_2} = \frac{T_2}{T_1}$.

Garrison Provision Depletion Model

A fixed quantity of rations in an isolated military outpost will feed a certain number of soldiers for a specific number of days ($P_1 D_1 = P_2 D_2$). If reinforcements arrive (increasing $P$), the days of food remaining ($D$) decreases inversely.

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Joint & Combined Variation: The Master Compound Proportion

Many real-world operational problems involve multiple variables fluctuating concurrently:

• Joint Variation: A variable varies directly as the product of two or more other variables: $z = kxy$.
• Combined Variation: A variable varies directly with one variable and inversely with another: $z = \frac{kx}{y}$.

The Master Compound Work Formula

When multiple workers ($M$) operate for a given number of days ($D$) and daily hours ($H$) to complete a measurable quantity of work ($W$), the overarching relationship is: $\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}$

Worked Scenario: If $12$ combat engineers working $8$ hours per day can excavate $480$ meters of defensive antitank ditch in $5$ days, how many meters of ditch can $18$ engineers excavate in $4$ days working $6$ hours per day?

1. Set up the master compound proportion: $\frac{12 \times 5 \times 8}{480} = \frac{18 \times 4 \times 6}{W_2}$
2. Simplify the left-hand side numerator: $12 \times 5 \times 8 = 480$ $\frac{480}{480} = 1$
3. Evaluate the right-hand side numerator: $18 \times 4 \times 6 = 18 \times 24 = 432$
4. Solve for $W_2$: $1 = \frac{432}{W_2} \implies W_2 = 432 \text{ meters}$

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Dimensional Analysis & Military Unit Conversion Chains

In military operations, calculations frequently span different measurement systems: kilometers to nautical miles, meters per second to kilometers per hour, and map millimeters to field meters. Dimensional Analysis (the Factor-Label Method) guarantees precision by treating measurement units as algebraic factors that cancel diagonally.

The Unit Multiplier Rule

Multiply the given quantity by conversion fractions that are mathematically equivalent to $1$ (where numerator and denominator represent identical physical magnitudes in different units), arranging units so that unwanted units cancel: $\text{Given Unit} \times \left( \frac{\text{Desired Unit}}{\text{Given Unit}} \right) = \text{Desired Unit}$

Key Conversion Chains for AFPSAT

1. Speed Conversions: km/h to m/s

To convert kilometers per hour ($\text{km/h}$) to meters per second ($\text{m/s}$): $1 \text{ km/h} = \frac{1,000 \text{ m}}{3,600 \text{ s}} = \frac{5}{18} \text{ m/s} \approx 0.2778 \text{ m/s}$

• To convert $\text{km/h} \implies \text{m/s}$: Multiply by $\frac{5}{18}$ (or divide by $3.6$).
  • Example: Convert $72 \text{ km/h}$ to $\text{m/s}$: $72 \times \frac{5}{18} = 4 \times 5 = 20 \text{ m/s}$.
• To convert $\text{m/s} \implies \text{km/h}$: Multiply by $\frac{18}{5}$ (or multiply by $3.6$).
  • Example: Convert $25 \text{ m/s}$ to $\text{km/h}$: $25 \times \frac{18}{5} = 5 \times 18 = 90 \text{ km/h}$.

2. Maritime Navigation: Knots, Nautical Miles, and Kilometers

In naval operations (Philippine Navy and Philippine Coast Guard) and tactical aviation (Philippine Air Force):

• 1 International Nautical Mile (NM) $= 1.852 \text{ kilometers} = 1,852 \text{ meters} \approx 1.151 \text{ statute miles}$.
• 1 Knot (kt) $= 1 \text{ Nautical Mile per hour} = 1.852 \text{ km/h} \approx 0.5144 \text{ m/s}$.

3. Topographic Map Scale Conversions

Standard military topographic maps use standard representative fractions:

• Scale 1 : 50,000: $1 \text{ cm}$ on the map represents $50,000 \text{ cm} = 500 \text{ meters} = 0.5 \text{ km}$ on the ground. Therefore, $2 \text{ cm} = 1.0 \text{ km}$.
• Scale 1 : 25,000: $1 \text{ cm}$ on the map represents $25,000 \text{ cm} = 250 \text{ meters} = 0.25 \text{ km}$ on the ground. Therefore, $4 \text{ cm} = 1.0 \text{ km}$.

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Comparative Framework: Direct, Inverse, and Partitive Proportions

The following table synthesizes the three primary forms of proportion tested on the AFPSAT: