3.3 Ratios, Proportions & Practical Rates

3.3 Ratios, Proportions & Practical Rates

In military operations and engineering logistics, calculations rarely involve isolated quantities. Instead, commanders and logisticians must constantly balance resources proportionally—allocating ammunition based on squad strength, projecting deployment timelines when manpower fluctuates, computing transit intervals for road convoys, and evaluating joint maintenance throughput. Section 3.3 explores the quantitative relationships governing ratios, direct and inverse proportion, speed-distance-time mechanics, and cooperative work rates. Mastering these principles enables candidates to systematically deconstruct complex real-world aptitude problems without algebraic stagnation.

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Understanding Ratios and Proportional Division

A ratio expresses the relative magnitude of two or more quantities of the same kind, separated by colons (e.g., $a:b$ or $a:b:c$). Ratios must always be simplified to lowest integer terms by dividing all terms by their greatest common divisor.

The Part-to-Whole Unitary Method

When a total quantity is partitioned according to a given ratio $a:b:c$:

1. Calculate the total number of ratio parts: $\text{Total Parts} = a + b + c$
2. Determine the value of one unitary part: $\text{Unitary Value} = \frac{\text{Total Quantity}}{\text{Total Parts}}$
3. Multiply each ratio coefficient by the unitary value to find each share: $\text{Share of } A = a \times \text{Unitary Value}$

Practical Example: Supply Distribution

Suppose $1,200\text{ rounds}$ of ammunition are apportioned among three forward observation posts in the ratio $2:3:5$.

• Total ratio parts: $2 + 3 + 5 = 10\text{ parts}$
• Value per part: $\frac{1,200}{10} = 120\text{ rounds}$
• Distribution:
  • Post 1: $2 \times 120 = 240\text{ rounds}$
  • Post 2: $3 \times 120 = 360\text{ rounds}$
  • Post 3: $5 \times 120 = 600\text{ rounds}$
• Verification: $240 + 360 + 600 = 1,200\text{ rounds}$.

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Direct Proportion vs. Inverse Proportion

Recognizing whether two related variables change in direct or inverse proportion is essential for setting up correct equations.

The "Worker-Days" Concept in Inverse Proportion

In workforce and construction scenarios, the total volume of work is constant and measured in compound units like person-days or man-hours:

$$\text{Total Work} = \text{Personnel} \times \text{Days} \times \text{Hours per Day}$$

If the workforce changes, the total required work remains constant: $P_1 \times D_1 = P_2 \times D_2$.

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Speed, Distance, and Time Relationships

Motion problems evaluate a candidate's grasp of rates and dimensional units. The fundamental relationship is expressed by the speed triangle:

$$\text{Distance } (d) = \text{Speed } (s) \times \text{Time } (t)$$ $$\text{Speed } (s) = \frac{\text{Distance } (d)}{\text{Time } (t)}, \quad \text{Time } (t) = \frac{\text{Distance } (d)}{\text{Speed } (s)}$$

Converting Between $\text{km/h}$ and $\text{m/s}$

Converting between kilometers per hour ($\text{km/h}$) and meters per second ($\text{m/s}$) relies on the conversion factor $1\text{ km/h} = \frac{1,000\text{ m}}{3,600\text{ s}} = \frac{5}{18}\text{ m/s}$:

• To convert $\text{km/h}$ to $\text{m/s}$: Multiply by $\frac{5}{18}$ (or divide by $3.6$).
  • Example: $72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$.
  • Example: $90\text{ km/h} = 90 \times \frac{5}{18} = 25\text{ m/s}$.
• To convert $\text{m/s}$ to $\text{km/h}$: Multiply by $\frac{18}{5}$ (or multiply by $3.6$).
  • Example: $15\text{ m/s} = 15 \times \frac{18}{5} = 54\text{ km/h}$.

Average Speed on Multi-Stage Journeys

Average speed is strictly defined as total distance divided by total time, not the simple average of speeds:

$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$

For equal round-trip distances $d$ at speeds $s_1$ and $s_2$, use the harmonic mean: $\text{Average Speed} = \frac{2 \times s_1 \times s_2}{s_1 + s_2}$.

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Collaborative Rates of Work

When two or more teams or machines work simultaneously, their individual production rates combine additively.

Work Rate Formulas

If Team A finishes in $t_A$ hours (rate $\frac{1}{t_A}$) and Team B finishes in $t_B$ hours (rate $\frac{1}{t_B}$):

$$\text{Combined Rate} = \frac{1}{t_A} + \frac{1}{t_B}$$ $$\text{Time Together } (T) = \frac{1}{\text{Combined Rate}} = \frac{t_A \times t_B}{t_A + t_B}$$

Opposing Rates (Inflow and Outflow)

If an inlet fills a bladder in $6\text{ hours}$ and an outlet drains it in $10\text{ hours}$, subtract the rates:

$$\text{Net Rate} = \frac{1}{6} - \frac{1}{10} = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15}\text{ per hour}$$

The bladder fills in $15\text{ hours}$ with both taps open.

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Worked Practical Examples

Example 1: Map Scale and Transit Duration

Problem: On a map with scale $1:50,000$, the distance to an objective is $16\text{ cm}$. A patrol travels at $32\text{ km/h}$. How many minutes does the trip take?

Step-by-step Solution:

1. Ground distance in cm: $16\text{ cm} \times 50,000 = 800,000\text{ cm}$.
2. Convert to km: $800,000\text{ cm} \div 100,000 = 8\text{ km}$.
3. Time in hours: $t = \frac{8\text{ km}}{32\text{ km/h}} = \frac{1}{4}\text{ hour}$.
4. Convert to minutes: $\frac{1}{4} \times 60 = 15\text{ minutes}$. Final Answer: $15\text{ minutes}$.

Example 2: Inverse Proportion Manpower Adjustment

Problem: A squad of $24$ soldiers takes $15\text{ days}$ to complete a field trench. To finish in $10\text{ days}$, how many additional soldiers are needed?

Step-by-step Solution:

1. Total work: $24\text{ soldiers} \times 15\text{ days} = 360\text{ person-days}$.
2. Required personnel: $\frac{360\text{ person-days}}{10\text{ days}} = 36\text{ soldiers}$.
3. Additional personnel: $36 - 24 = 12\text{ soldiers}$. Final Answer: $12\text{ additional soldiers}$.