Study guide
4.2 Ratios, Proportions, Direct & Inverse Variations
Armed Forces of the Philippines AFP
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1 Chapter 1: Exam Overview, Eligibility, Written Battery & The Officer Selection Pipeline
1.1 AFPSAT Format, Scoring Cutoffs & Testing Logistics 1.2 OCC Eligibility Criteria, Educational Standards & Document Requirements 1.3 The Complete Officer Selection Pipeline: PFT, Medical, Neuro-Psychiatric & Board Interview 1.4 After the AFPSAT: The Army Qualifying Exam (AQE) & Special Written Exam (SWE)
2 Chapter 2: Verbal Reasoning — Vocabulary, Context & Analogies
2.1 Vocabulary Expansion, Context Clues & Synonym Identification 2.2 Antonyms, Word Nuance & Paired Contrasts 2.3 Single-Word & Paired Word Analogies: Classification of Logical Relationships
3 Chapter 3: Verbal Reasoning — Grammar, Sentence Mechanics & Comprehension
3.1 English Grammar Mastery: Subject-Verb Agreement, Pronouns & Verb Tenses 3.2 Sentence Correction, Modifiers, Parallelism & Idiomatic Prepositions 3.3 Reading Comprehension: Main Idea, Critical Inference & Passage Analysis
4 Chapter 4: Numerical Reasoning — Core Arithmetic & Algebraic Foundations
4.1 Arithmetic Fluency: Fractions, Decimals, Percentages & Mental Math Shortcuts 4.2 Ratios, Proportions, Direct & Inverse Variations 4.3 Basic Algebra: Linear Equations, Systems of Equations & Inequalities
5 Chapter 5: Numerical Reasoning — Word Problems, Geometry, Series & Data
5.1 Motion, Work Rate & Fluid Flow Word Problems 5.2 Practical Arithmetic: Simple Interest, Percentage Change, Profit/Loss & Age Problems 5.3 Geometry Fundamentals: Perimeter, Area, Volume, Pythagorean Theorem & Angles 5.4 Number Series, Letter Series, Averages & Data Interpretation
6 Chapter 6: Abstract Reasoning — Pattern Sequences & Matrix Progression
6.1 Pattern Progression: Step-by-Step Movement, Alternating Cycles & Position Shifts 6.2 Visual Element Arithmetic: Shape Addition, Subtraction, Overlap & Shading Logic 6.3 Spatial Rotation, Reflections & 2D-to-3D Mental Cube Folding
7 Chapter 7: Abstract Reasoning — Advanced Logical & Diagrammatic Analysis
7.1 Odd-One-Out Classification & Rule Identification 7.2 Diagrammatic Analogies & Topological Relationships 7.3 Time Management & Distractor Elimination Strategies for Fast Abstract Testing
8 Chapter 8: General Information, Philippine Constitution & Military Leadership
8.1 The 1987 Philippine Constitution: Preamble, State Policies & Bill of Rights 8.2 AFP Organization, Branch Missions, Rank Structure & Military History 8.3 Military Core Values, Code of Ethics, Situational Leadership & Officer Judgment 8.4 General Information: Philippine Government, National Security Issues & General Science
4.2 Ratios, Proportions, Direct & Inverse Variations
Key Takeaways Explain This Topic AI
Ratios express relative quantitative magnitude and must be distinguished between part-to-part comparisons (a:b) and part-to-whole proportions (a / [a+b]), which define total fractional shares.
Partitive proportions divide a total quantity across three or more entities using the total-parts method, where a common scaling multiplier k is derived by dividing the total quantity by the sum of ratio terms.
Direct variation (y = kx) governs linear relationships where quantities scale in the same direction, while inverse variation (xy = k) governs reciprocal trade-offs, prominently seen in personnel-time (man-days) and speed-time problems.
Combined and joint variations unite multiple rate factors into a single master compound proportion equation: (Workers₁ × Time₁) / Work₁ = (Workers₂ × Time₂) / Work₂.
Dimensional analysis (the Factor-Label Method) systematically eliminates unit conversion errors across military metrics, such as converting knots to km/h (factor of 1.852) and km/h to m/s (factor of 5/18 or divide by 3.6).
Last updated: September 2026
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.
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 a : b a : b , a b \frac{a}{b} b a , or " a a a to b b b ". Ratios represent relative scale rather than absolute quantities. A ratio of 3 : 5 3:5 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 a : b 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 6 : 24 6 : 24 , which simplifies to 1 : 4 1 : 4 1 : 4 .
Part-to-Whole Ratio ( a : [ a + b ] a : [a + b] a : [ a + b ] ): Compares one subgroup to the aggregate total. In the same platoon, the total personnel count is 6 + 24 = 30 6 + 24 = 30 6 + 24 = 30 . The part-to-whole ratio of officers to the entire platoon is 6 30 = 1 5 \frac{6}{30} = \frac{1}{5} 30 6 = 5 1 (or 20 % 20\% 20% ). The part-to-whole ratio of enlisted soldiers is 24 30 = 4 5 \frac{24}{30} = \frac{4}{5} 30 24 = 5 4 (or 80 % 80\% 80% ).
Rule for Ratio Conversion: If a part-to-part ratio between two groups is a : b a : b a : b , the fractional share of group a a a relative to the whole is a a + b \frac{a}{a + b} a + b a , and the fractional share of group b b b is b a + b \frac{b}{a + b} a + b b . Misidentifying a part-to-part ratio as a part-to-whole fraction (e.g., assuming 1 : 4 1:4 1 : 4 means 1 / 4 = 25 % 1/4 = 25\% 1/4 = 25% rather than 1 / 5 = 20 % 1/5 = 20\% 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 × 4 ) : ( 5 × 4 ) = 12 : 20 3 : 5 = (3 \times 4) : (5 \times 4) = 12 : 20 3 : 5 = ( 3 × 4 ) : ( 5 × 4 ) = 12 : 20 36 : 48 = ( 36 ÷ 12 ) : ( 48 ÷ 12 ) = 3 : 4 36 : 48 = (36 \div 12) : (48 \div 12) = 3 : 4 36 : 48 = ( 36 ÷ 12 ) : ( 48 ÷ 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 3:5 3 : 5 , transferring or adding 10 soldiers to both platoons completely alters the proportion.
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 a : b : c a : b : c .
The "Total Parts" Scaling Algorithm
To divide total quantity T T T according to the ratio a : b : c a : b : c a : b : c :
Sum the ratio parts: S = a + b + c S = a + b + c S = a + b + c .
Derive the scaling multiplier ( k k k ): Determine the physical value of a single ratio part by dividing the total quantity by the sum of the parts: k = T S k = \frac{T}{S} k = S T .
Compute each individual share: Multiply each ratio term by k k k : Share A = a × k , Share B = b × k , Share C = c × k \text{Share } A = a \times k, \quad \text{Share } B = b \times k, \quad \text{Share } C = c \times k Share A = a × k , Share B = b × k , Share C = c × k
Step-by-Step Worked Example
Problem: A joint task force allocates a logistics shipment of 9 , 600 9,600 9 , 600 crates of field rations among the Army, Navy, and Air Force components in the ratio 5 : 2 : 1 5 : 2 : 1 5 : 2 : 1 . How many crates does the Army component receive?
Sum the parts: S = 5 + 2 + 1 = 8 S = 5 + 2 + 1 = 8 S = 5 + 2 + 1 = 8 parts.
Find the value of 1 part ( k k k ): k = 9 , 600 8 = 1 , 200 crates per part k = \frac{9,600}{8} = 1,200 \text{ crates per part} k = 8 9 , 600 = 1 , 200 crates per part
Compute allocations:
Army ( 5 5 5 parts): 5 × 1 , 200 = 6 , 000 5 \times 1,200 = 6,000 5 × 1 , 200 = 6 , 000 crates.
Navy ( 2 2 2 parts): 2 × 1 , 200 = 2 , 400 2 \times 1,200 = 2,400 2 × 1 , 200 = 2 , 400 crates.
Air Force ( 1 1 1 part): 1 × 1 , 200 = 1 , 200 1 \times 1,200 = 1,200 1 × 1 , 200 = 1 , 200 crates.
Verification: 6 , 000 + 2 , 400 + 1 , 200 = 9 , 600 6,000 + 2,400 + 1,200 = 9,600 6 , 000 + 2 , 400 + 1 , 200 = 9 , 600 crates. The Army component receives 6 , 000 6,000 6 , 000 crates.
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 y y varies directly as x x x if: y = k x ⟺ y x = k y = kx \quad \iff \quad \frac{y}{x} = k y = k x ⟺ x y = k
where k k k is the non-zero constant of proportionality (the rate). For any two states ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) : y 1 x 1 = y 2 x 2 ⟺ x 1 x 2 = y 1 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} x 1 y 1 = x 2 y 2 ⟺ x 2 x 1 = y 2 y 1
- Cross-multiplying yields: x 1 y 2 = x 2 y 1 x_1 y_2 = x_2 y_1 x 1 y 2 = x 2 y 1
- Classic Military Direct Variation Scenarios
Convoy Fuel Consumption: Fuel burned ( F F F ) varies directly with travel distance ( D D D ): F = k D F = kD F = k D .
Ammunition Expenditure: Rounds fired ( R R R ) vary directly with firing duration ( t t t ): R = k t R = kt R = k t .
Subsistence Rations: Rations required ( R R R ) vary directly with troop count ( P P P ): R = k P R = kP R = k P .
Worked Scenario: A military transport convoy of six-wheeled utility trucks consumes 180 180 180 liters of diesel fuel while traversing 75 75 75 kilometers of mountain terrain. Under identical driving conditions, how many liters of diesel will the convoy consume over a distance of 225 225 225 kilometers? F 1 D 1 = F 2 D 2 ⟹ 180 75 = F 2 225 \frac{F_1}{D_1} = \frac{F_2}{D_2} \implies \frac{180}{75} = \frac{F_2}{225} D 1 F 1 = D 2 F 2 ⟹ 75 180 = 225 F 2
Instead of long division, notice the horizontal scaling factor: 225 = 75 × 3 225 = 75 \times 3 225 = 75 × 3 . Because the distance increases by a factor of 3, fuel consumption must scale by the exact same factor of 3: F 2 = 180 × 3 = 540 liters F_2 = 180 \times 3 = 540 \text{ liters} F 2 = 180 × 3 = 540 liters
Solving via cross-multiplication: 75 F 2 = 180 × 225 ⟹ F 2 = 180 × 225 75 = 180 × 3 = 540 75 F_2 = 180 \times 225 \implies F_2 = \frac{180 \times 225}{75} = 180 \times 3 = 540 75 F 2 = 180 × 225 ⟹ F 2 = 75 180 × 225 = 180 × 3 = 540 liters.
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 y y varies inversely as x x x if: y = k x ⟺ x × y = k y = \frac{k}{x} \quad \iff \quad x \times y = k y = x k ⟺ x × y = k
where k k k is the invariant product. For any two corresponding states ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) : x 1 y 1 = x 2 y 2 ⟺ x 1 x 2 = y 2 y 1 x_1 y_1 = x_2 y_2 \quad \iff \quad \frac{x_1}{x_2} = \frac{y_2}{y_1} x 1 y 1 = x 2 y 2 ⟺ x 2 x 1 = y 1 y 2
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 : Work = Workers × Time = M 1 D 1 = M 2 D 2 \text{Work} = \text{Workers} \times \text{Time} = M_1 D_1 = M_2 D_2 Work = Workers × 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 60 60 man-days), finishing in 4 days requires 60 / 4 = 15 60 / 4 = 15 60/4 = 15 men total. If the question asks for additional men, the correct answer is 15 − 10 = 5 15 - 10 = 5 15 − 10 = 5 additional men, not 15 15 15 .
The Speed-Time Distance Model
When transit distance D D D is constant, speed ( S S S ) and travel time ( T T T ) are inversely proportional: D = S 1 T 1 = S 2 T 2 D = S_1 T_1 = S_2 T_2 D = S 1 T 1 = S 2 T 2
Increasing convoy cruising speed reduces transit time proportionally: S 1 S 2 = T 2 T 1 \frac{S_1}{S_2} = \frac{T_2}{T_1} S 2 S 1 = T 1 T 2 .
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 P_1 D_1 = P_2 D_2 P 1 D 1 = P 2 D 2 ). If reinforcements arrive (increasing P P P ), the days of food remaining ( D D D ) decreases inversely.
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 = k x y z = kxy z = k x y .
Combined Variation: A variable varies directly with one variable and inversely with another: z = k x y z = \frac{kx}{y} z = y k x .
The Master Compound Work Formula
When multiple workers ( M M M ) operate for a given number of days ( D D D ) and daily hours ( H H H ) to complete a measurable quantity of work ( W W W ), the overarching relationship is: M 1 × D 1 × H 1 W 1 = M 2 × D 2 × H 2 W 2 \frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2} W 1 M 1 × D 1 × H 1 = W 2 M 2 × D 2 × H 2
Worked Scenario: If 12 12 12 combat engineers working 8 8 8 hours per day can excavate 480 480 480 meters of defensive antitank ditch in 5 5 5 days, how many meters of ditch can 18 18 18 engineers excavate in 4 4 4 days working 6 6 6 hours per day?
Set up the master compound proportion: 12 × 5 × 8 480 = 18 × 4 × 6 W 2 \frac{12 \times 5 \times 8}{480} = \frac{18 \times 4 \times 6}{W_2} 480 12 × 5 × 8 = W 2 18 × 4 × 6
Simplify the left-hand side numerator: 12 × 5 × 8 = 480 12 \times 5 \times 8 = 480 12 × 5 × 8 = 480 480 480 = 1 \frac{480}{480} = 1 480 480 = 1
Evaluate the right-hand side numerator: 18 × 4 × 6 = 18 × 24 = 432 18 \times 4 \times 6 = 18 \times 24 = 432 18 × 4 × 6 = 18 × 24 = 432
Solve for W 2 W_2 W 2 : 1 = 432 W 2 ⟹ W 2 = 432 meters 1 = \frac{432}{W_2} \implies W_2 = 432 \text{ meters} 1 = W 2 432 ⟹ W 2 = 432 meters
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 1 1 (where numerator and denominator represent identical physical magnitudes in different units), arranging units so that unwanted units cancel: Given Unit × ( Desired Unit Given Unit ) = Desired Unit \text{Given Unit} \times \left( \frac{\text{Desired Unit}}{\text{Given Unit}} \right) = \text{Desired Unit} Given Unit × ( Given Unit Desired Unit ) = Desired Unit
- Key Conversion Chains for AFPSAT
- 1. Speed Conversions: km/h to m/s
To convert kilometers per hour ( km/h \text{km/h} km/h ) to meters per second ( m/s \text{m/s} m/s ): 1 km/h = 1 , 000 m 3 , 600 s = 5 18 m/s ≈ 0.2778 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} 1 km/h = 3 , 600 s 1 , 000 m = 18 5 m/s ≈ 0.2778 m/s
To convert km/h ⟹ m/s \text{km/h} \implies \text{m/s} km/h ⟹ m/s : Multiply by 5 18 \frac{5}{18} 18 5 (or divide by 3.6 3.6 3.6 ).
Example: Convert 72 km/h 72 \text{ km/h} 72 km/h to m/s \text{m/s} m/s : 72 × 5 18 = 4 × 5 = 20 m/s 72 \times \frac{5}{18} = 4 \times 5 = 20 \text{ m/s} 72 × 18 5 = 4 × 5 = 20 m/s .
To convert m/s ⟹ km/h \text{m/s} \implies \text{km/h} m/s ⟹ km/h : Multiply by 18 5 \frac{18}{5} 5 18 (or multiply by 3.6 3.6 3.6 ).
Example: Convert 25 m/s 25 \text{ m/s} 25 m/s to km/h \text{km/h} km/h : 25 × 18 5 = 5 × 18 = 90 km/h 25 \times \frac{18}{5} = 5 \times 18 = 90 \text{ km/h} 25 × 5 18 = 5 × 18 = 90 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 kilometers = 1 , 852 meters ≈ 1.151 statute miles = 1.852 \text{ kilometers} = 1,852 \text{ meters} \approx 1.151 \text{ statute miles} = 1.852 kilometers = 1 , 852 meters ≈ 1.151 statute miles .
1 Knot (kt) = 1 Nautical Mile per hour = 1.852 km/h ≈ 0.5144 m/s = 1 \text{ Nautical Mile per hour} = 1.852 \text{ km/h} \approx 0.5144 \text{ m/s} = 1 Nautical Mile per hour = 1.852 km/h ≈ 0.5144 m/s .
3. Topographic Map Scale Conversions
Standard military topographic maps use standard representative fractions:
Scale 1 : 50,000: 1 cm 1 \text{ cm} 1 cm on the map represents 50 , 000 cm = 500 meters = 0.5 km 50,000 \text{ cm} = 500 \text{ meters} = 0.5 \text{ km} 50 , 000 cm = 500 meters = 0.5 km on the ground. Therefore, 2 cm = 1.0 km 2 \text{ cm} = 1.0 \text{ km} 2 cm = 1.0 km .
Scale 1 : 25,000: 1 cm 1 \text{ cm} 1 cm on the map represents 25 , 000 cm = 250 meters = 0.25 km 25,000 \text{ cm} = 250 \text{ meters} = 0.25 \text{ km} 25 , 000 cm = 250 meters = 0.25 km on the ground. Therefore, 4 cm = 1.0 km 4 \text{ cm} = 1.0 \text{ km} 4 cm = 1.0 km .
Comparative Framework: Direct, Inverse, and Partitive Proportions
The following table synthesizes the three primary forms of proportion tested on the AFPSAT:
Proportion Type Core Mathematical Model Invariant Constant ( k k k ) Defining Characteristics & Behavior Operational Keyword Indicators Worked Military Scenario
Direct Proportion y 1 x 1 = y 2 x 2 \frac{y_1}{x_1} = \frac{y_2}{x_2} x 1 y 1 = x 2 y 2 or y = k x y = kx y = k x k = y x k = \frac{y}{x} k = x y (Ratio is constant) As x x x increases, y y y increases proportionally; graphs as a straight ray through the origin "At a constant rate," "per unit," "consumed proportionally," "burn rate" A patrol vessel burns 160 L of fuel in 4 hrs; burning at the same rate, 10 hrs consumes 160 4 × 10 = 400 \frac{160}{4} \times 10 = 400 4 160 × 10 = 400 L.
Inverse Proportion x 1 y 1 = x 2 y 2 x_1 y_1 = x_2 y_2 x 1 y 1 = x 2 y 2 or y = k x y = \frac{k}{x} y = x k k = x × y k = x \times y k = x × y (Product is constant) As x x x increases, y y y decreases proportionally; graphs as a rectangular hyperbola "Shared equally," "working together," "man-days," "speed vs travel time" 12 engineers build a fortification in 6 days ( 12 × 6 = 72 12 \times 6 = 72 12 × 6 = 72 man-days). 8 engineers require 72 ÷ 8 = 9 72 \div 8 = 9 72 ÷ 8 = 9 days.
Partitive Proportion T = k ( a + b + c ) T = k(a + b + c) T = k ( a + b + c ) where x = a k , y = b k , z = c k x = ak, y = bk, z = ck x = ak , y = bk , z = c k k = T ∑ parts k = \frac{T}{\sum \text{parts}} k = ∑ parts T (Value per single part) A single aggregate whole is divided into distinct fractional portions according to multi-term ratio "Divided in the ratio of," "apportioned among," "split respectively into" 4,500 rounds partitioned 2:3:4 ( 9 9 9 parts). Value per part = 4 , 500 / 9 = 500 4,500 / 9 = 500 4 , 500/9 = 500 . Shares: 1,000, 1,500, and 2,000 rounds.