4.3 Ratios, Proportions, Rates, Work and Time Problems

Ratios, Proportions, Rates, Work and Time Problems

Tactical planning is fundamentally an exercise in rate balancing and proportional allocation. Military engineers must determine how many combat pioneer personnel and operational hours are required to construct defensive berms or bridge crossings before an advancing adversary arrives. Staff officers must calculate fuel flow rates through logistical pipeline bladders, pace mechanized convoys across differing road surfaces, and schedule rendezvous timings between dispersed reconnaissance patrols.

On the Ghana Armed Forces Officer Cadet Written Examination, questions on ratios, work-time mechanics, and kinematic motion test a candidate's structured problem-solving discipline and mental speed. Deploying standardized rate templates and algebraic identities prevents mathematical confusion under exam conditions.

Ratio Theory, Compound Ratios & Proportional Division

A ratio is an ordered comparison of two or more quantities of identical dimension, expressed as $a:b$ or $\frac{a}{b}$. The first term $a$ is the antecedent, and the second term $b$ is the consequent.

Properties and Classification of Ratios

• Scale Invariance: Multiplying or dividing both antecedent and consequent by a non-zero constant $k$ preserves the ratio: $a:b = ka:kb$.
• Compound Ratio: The compound ratio of two ratios $a:b$ and $c:d$ is the product of their terms: $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$, written as $ac:bd$.
• Power and Root Ratios:
  • Duplicate Ratio: $a^2:b^2$
  • Sub-duplicate Ratio: $\sqrt{a}:\sqrt{b}$
  • Triplicate Ratio: $a^3:b^3$
  • Sub-triplicate Ratio: $\sqrt[3]{a}:\sqrt[3]{b}$

Continued Ratio Unification

A standard test item provides pairwise ratios (e.g., $A:B$ and $B:C$) and requires unifying them into a continuous multi-term ratio $A:B:C$.

$\begin{matrix} A & : & B \\ & & B & : & C \end{matrix} \implies A:B:C = (A \times B_2) : (B_1 \times B_2) : (B_1 \times C)$

Worked Example: In a combined-arms brigade, the ratio of infantry to armor personnel is $5:3$, and the ratio of armor to artillery personnel is $4:7$. What is the continuous ratio of Infantry to Armor to Artillery?

• Align on common term (Armor): Multiply the first ratio by $4$ and the second ratio by $3$:
  • $\text{Infantry} : \text{Armor} = 5 \times 4 : 3 \times 4 = 20 : 12$
  • $\text{Armor} : \text{Artillery} = 4 \times 3 : 7 \times 3 = 12 : 21$
• Combined Ratio: $\text{Infantry} : \text{Armor} : \text{Artillery} = 20 : 12 : 21$.

Proportional Division of Logistical Assets

To divide a total quantity $Q$ into parts proportional to $a : b : c$: $\text{Sum of Ratio Parts } S = a + b + c$ $\text{Part } A = \frac{a}{S} \times Q, \quad \text{Part } B = \frac{b}{S} \times Q, \quad \text{Part } C = \frac{c}{S} \times Q$

Operational Scenario: A logistical consignment of $1,800$ mortar shells is distributed among forward outposts Alpha, Bravo, and Charlie in the ratio $4:5:9$.

• Total ratio parts: $S = 4 + 5 + 9 = 18$.
• Outpost Alpha: $\frac{4}{18} \times 1800 = 400$ shells.
• Outpost Bravo: $\frac{5}{18} \times 1800 = 500$ shells.
• Outpost Charlie: $\frac{9}{18} \times 1800 = 900$ shells.

Variation & The Multi-Variable Chain Rule

Variation describes how a dependent variable changes in response to alterations in independent driving variables.

• Direct Variation ($y \propto x$): $y = kx \implies \frac{y_1}{x_1} = \frac{y_2}{x_2}$.
• Inverse Variation ($y \propto \frac{1}{x}$): $y = \frac{k}{x} \implies x_1 y_1 = x_2 y_2$.
• Joint Variation: When an outcome depends directly on some factors and inversely on others.

The Master Chain Rule for Labor and Construction

In field engineering problems, the total work output $W$ is directly proportional to the number of personnel ($M$), the duration in days ($D$), daily operating hours ($H$), and worker efficiency ($E$): $W \propto M \times D \times H \times E$

This yields the universal balanced chain rule invariant: $\frac{M_1 \times D_1 \times H_1 \times E_1}{W_1} = \frac{M_2 \times D_2 \times H_2 \times E_2}{W_2}$

If worker efficiencies are equal, the efficiency term $E$ cancels out: $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$

Step-by-Step Chain Rule Execution

Problem: A field engineering detachment of $36$ combat engineers working $8$ hours per day constructs an anti-tank trench measuring $720\text{ meters}$ in $12$ days. How many days will it take a reinforced detachment of $48$ engineers working $9$ hours per day to construct a similar trench measuring $1,080\text{ meters}$?

1. Identify Known Parameters:
   • $M_1 = 36$, $D_1 = 12$, $H_1 = 8$, $W_1 = 720$
   • $M_2 = 48$, $D_2 = ?$, $H_2 = 9$, $W_2 = 1080$
2. Set up the Equation: $\frac{36 \times 12 \times 8}{720} = \frac{48 \times D_2 \times 9}{1080}$
3. Simplify the Left Side: $\frac{36 \times 12 \times 8}{720} = \frac{12 \times 8}{20} = \frac{96}{20} = 4.8$
4. Simplify the Right Side: $\frac{48 \times 9 \times D_2}{1080} = \frac{48 \times D_2}{120} = \frac{2 \times D_2}{5} = 0.4 \times D_2$
5. Solve for $D_2$: $4.8 = 0.4 \times D_2 \implies D_2 = \frac{4.8}{0.4} = 12\text{ days}$

The reinforced detachment requires exactly $12$ days.

Work and Time Mathematics & Negative Drainage Rates

Work and time problems are solved by converting total project deliverables into unitary daily rates.

The Reciprocal Work Rate Principle

If an operative or unit can complete an entire project in $T$ days, their daily work rate is $\frac{1}{T}$ of the project per day. When multiple units cooperate independently, their individual work rates sum linearly: $\text{Combined Work Rate } R_{\text{comb}} = \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}$ $\text{Total Days to Complete } T_{\text{comb}} = \frac{1}{R_{\text{comb}}}$

• Two Workers ($A$ and $B$): $T_{\text{comb}} = \frac{t_A t_B}{t_A + t_B}$
• Three Workers ($A$, $B$, and $C$): $\frac{1}{T_{\text{comb}}} = \frac{1}{t_A} + \frac{1}{t_B} + \frac{1}{t_C} = \frac{bc + ac + ab}{abc} \implies T_{\text{comb}} = \frac{abc}{ab + bc + ca}$

Worker Efficiency Ratios

When problem statements specify that Worker $A$ is twice as efficient as Worker $B$ ($E_A = 2 E_B$), then for any fixed task, the time required by $A$ is half the time required by $B$ ($t_A = \frac{t_B}{2}$).

Negative Work: Pipes, Cisterns & Fuel Bladders

In liquid fuel supply and drainage problems, filling pipes represent positive work, while consumption drainage outlets or puncture leaks represent negative work: $\text{Net Rate } R_{\text{net}} = \sum \frac{1}{T_{\text{fill}}} - \sum \frac{1}{T_{\text{drain}}}$ $\text{Net Time to Fill } T = \frac{1}{R_{\text{net}}}$

Worked Logistics Example: A forward aviation fuel bladder can be filled by Pipe A in $6$ hours and by Pipe B in $8$ hours. A gravity distribution pipe drains the full bladder to supply refueling helicopters in $12$ hours. If both inflow pipes and the drainage pipe are opened simultaneously, how long does it take to fill an empty bladder?

1. Establish Individual Hourly Rates:
   • Rate of Pipe A: $+\frac{1}{6}$
   • Rate of Pipe B: $+\frac{1}{8}$
   • Rate of Drainage Pipe: $-\frac{1}{12}$
2. Sum to Find Net Hourly Inflow Rate: $R_{\text{net}} = \frac{1}{6} + \frac{1}{8} - \frac{1}{12}$ Common denominator of $\{6, 8, 12\}$ is $24$: $R_{\text{net}} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{4 + 3 - 2}{24} = \frac{5}{24}\text{ bladder per hour}$
3. Invert to Find Net Fill Time: $T = \frac{24}{5} = 4.8\text{ hours} = 4\text{ hours and } 48\text{ minutes}$

The bladder reaches full operational capacity in $4$ hours and $48$ minutes.

Time, Speed, and Distance Kinetics

The kinematics of troop movements and vehicular transport rest upon the foundational relationship: $\text{Distance } = \text{Speed} \times \text{Time} \iff D = v \times t$

Essential Unit Conversion Factors

Velocity in military operations is frequently expressed interchangeably in kilometers per hour ($\text{km/h}$) and meters per second ($\text{m/s}$): $1\text{ km/h} = \frac{1,000\text{ m}}{3,600\text{ s}} = \frac{5}{18}\text{ m/s}$ $1\text{ m/s} = \frac{18}{5}\text{ km/h} = 3.6\text{ km/h}$

• To convert from $\text{km/h}$ to $\text{m/s}$, multiply by $\frac{5}{18}$.
  • Example: $72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$.
• To convert from $\text{m/s}$ to $\text{km/h}$, multiply by $\frac{18}{5}$.
  • Example: $25\text{ m/s} = 25 \times \frac{18}{5} = 90\text{ km/h}$.

Average Speed: The Harmonic Mean Principle

Average speed is defined as the total distance traversed divided by the total elapsed time: $v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$

• Special Case: Equal Distance Segments (Harmonic Mean): When a vehicle travels a distance $d$ at speed $v_1$ and returns along the same distance $d$ at speed $v_2$, the average speed is NOT the arithmetic mean $\frac{v_1 + v_2}{2}$. Because more time is spent at the slower speed, the average speed is given by the harmonic mean: $v_{\text{avg}} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}$ Example: A patrol vehicle travels to an outpost at $60\text{ km/h}$ and returns at $40\text{ km/h}$: $v_{\text{avg}} = \frac{2 \times 60 \times 40}{60 + 40} = \frac{4,800}{100} = 48\text{ km/h}$
• Special Case: Equal Travel Time Segments (Arithmetic Mean): When a vehicle travels for time $t$ at speed $v_1$ and for an identical time $t$ at speed $v_2$, the average speed is the arithmetic mean: $v_{\text{avg}} = \frac{v_1 t + v_2 t}{2t} = \frac{v_1 + v_2}{2}$

Convoy and Obstacle Crossing Mechanics

When a moving body with non-negligible physical length $L_1$ (e.g., a motorized column or military freight train) crosses a landmark:

• Crossing a Stationary Point Object (Tree, Sentry Post): $\text{Distance Traversed} = L_1, \qquad t = \frac{L_1}{v}$
• Crossing an Extended Stationary Platform / Bridge / Defile ($L_2$): $\text{Distance Traversed} = L_1 + L_2, \qquad t = \frac{L_1 + L_2}{v}$
• Crossing Another Moving Unit ($L_2$): $\text{Distance Traversed} = L_1 + L_2, \qquad t = \frac{L_1 + L_2}{v_{\text{rel}}}$

Relative Speed, Tactical Pacing & Rendezvous Problems

Relative velocity defines the rate at which the separation distance between two mobile bodies changes.

OPPOSITE DIRECTIONS (Closing In / Head-on Approach):
    [Unit A: v1]  ====>>              <<====  [Unit B: v2]
    Relative Speed = v1 + v2

SAME DIRECTION (Pursuit / Overtaking):
    [Unit A: v1]  ====================>>
           [Unit B: v2]  ============>>
    Relative Speed = |v1 - v2|

1. Relative Velocity Formulas

• Moving in Opposite Directions (Approaching or Receding): The distance between the units closes or widens at the sum of their speeds: $v_{\text{rel}} = v_1 + v_2$ $\text{Time to Rendezvous / Collision} = \frac{\text{Initial Separation Distance}}{v_1 + v_2}$
• Moving in the Same Direction (Pursuit / Interception): The distance between the units changes at the difference of their speeds: $v_{\text{rel}} = v_{\text{faster}} - v_{\text{slower}} = |v_1 - v_2|$ $\text{Time to Overtake} = \frac{\text{Lead Distance}}{v_{\text{faster}} - v_{\text{slower}}}$

2. Worked Interception Problem: Convoy Pursuit

A motorized logistics convoy departs a supply base heading north along a highway at a steady speed of $54\text{ km/h}$. Exactly $30\text{ minutes}$ later, a fast armored scout car departs the same base in pursuit along the identical route at a speed of $90\text{ km/h}$. At what distance from the supply base will the scout car intercept the logistics convoy?

1. Determine the Convoy's Lead Distance:
   • Lead time: $30\text{ minutes} = 0.5\text{ hours}$.
   • Distance covered by convoy prior to scout car departure: $D_{\text{lead}} = 54\text{ km/h} \times 0.5\text{ h} = 27\text{ km}$
2. Calculate Relative Velocity:
   • Both units move in the same direction: $v_{\text{rel}} = v_{\text{scout}} - v_{\text{convoy}} = 90 - 54 = 36\text{ km/h}$
3. Compute Time Required to Close the Separation Gap: $t = \frac{D_{\text{lead}}}{v_{\text{rel}}} = \frac{27\text{ km}}{36\text{ km/h}} = 0.75\text{ hours} = 45\text{ minutes}$
4. Calculate Total Distance Traveled by the Scout Car: $D = v_{\text{scout}} \times t = 90\text{ km/h} \times 0.75\text{ h} = 67.5\text{ km}$

Verification: In the total $1.25\text{ hours}$ of convoy motion ($0.5\text{ h lead} + 0.75\text{ h pursuit}$), the convoy travels $54 \times 1.25 = 67.5\text{ km}$. The interception occurs exactly $67.5\text{ km}$ from the base camp.

Comprehensive Kinematic Reference Matrix