6.3 Ratios, Proportions & Variations

Ratios, Proportions & Variations

Ratios, proportions, and mathematical variations form the operational backbone of resource management, staffing allocations, logistical planning, and rate calculations in public safety administration. The NAPOLCOM PNP Entrance Examination evaluates a candidate's ability to recognize whether two changing quantities vary directly, inversely, or partitively, and to solve multi-variable word problems rapidly without relying on electronic calculating aids.

---

1. The Structure of Ratios: Part-to-Part vs. Part-to-Whole

A ratio is a mathematical comparison of two numbers or quantities measured in the same unit. It indicates how many times one value contains or is contained within another.

Ratio Notations and Simplification

A ratio comparing quantity $a$ to quantity $b$ can be expressed in three equivalent notations:

1. Word Form: $a$ to $b$
2. Colon Form: $a : b$
3. Fraction Form: $\frac{a}{b}$ (where $b \neq 0$)

Like fractions, ratios must always be reduced to lowest terms by dividing both terms by their Greatest Common Factor (GCF). For example, a police unit with $48$ patrol officers and $18$ investigative detectives has an officer-to-detective ratio of $48:18$. Dividing both numbers by $\text{GCF}(48,18) = 6$ yields the simplified ratio $8 : 3$.

The Critical Distinction: Part-to-Part vs. Part-to-Whole

A frequent trap in competitive civil service exams arises from confusing a part-to-part comparison with a part-to-whole comparison:

• Part-to-Part Comparison: Compares one subset directly against another subset within the same population.
  • Example: In a station with 25 male officers and 15 female officers, the part-to-part ratio of male officers to female officers is $25 : 15 = 5 : 3$.
• Part-to-Whole Comparison: Compares one subset against the combined aggregate total.
  • Example: The ratio of female officers to total station personnel is $15 : (25 + 15) = 15 : 40 = 3 : 8$.
  • Expressed as a fraction of the total force, female officers represent $\frac{3}{8}$ (or $37.5%$) of the station, whereas the part-to-part ratio was $5:3$.

Exam Trap Alert: If a question states that the ratio of traffic officers to patrol officers is $2:5$, what fraction of the total force do traffic officers represent? The answer is not $2/5$; it is $\frac{2}{2 + 5} = \frac{2}{7}$.

Unit Consistency in Ratios

Before forming a valid ratio, both quantities must be converted into the same unit of measurement:

• Problem: What is the ratio of 45 minutes to 2 hours?
• Convert 2 hours to minutes: $2 \times 60 = 120\text{ minutes}$.
• Compute simplified ratio: $\frac{45}{120} = \frac{45 \div 15}{120 \div 15} = \frac{3}{8}$ (or $3:8$). Setting the ratio as $45:2$ is a fatal error.

---

2. Proportions & The Fundamental Cross-Product Rule

A proportion is a formal mathematical statement declaring that two ratios are strictly equal:

$\frac{a}{b} = \frac{c}{d} \quad \text{or} \quad a : b = c : d$

Extremes and Means

In the proportion $a : b = c : d$:

• The outer terms, $a$ and $d$, are called the extremes.
• The inner terms, $b$ and $c$, are called the means.

The Fundamental Property of Proportions

In any valid proportion, the product of the extremes is strictly equal to the product of the means:

$a \times d = b \times c \quad (ad = bc)$

This cross-product rule allows instant solution for any single unknown term ($x$):

$\frac{14}{x} = \frac{35}{60} \implies 35 \times x = 14 \times 60$ $35x = 840 \implies x = \frac{840}{35} = 24$

Shortcut: Simplify the known ratio first. Since $\frac{35}{60} = \frac{7}{12}$, the equation becomes $\frac{14}{x} = \frac{7}{12}$. Because the numerator doubles from $7$ to $14$, the denominator must double from $12$ to $24$.

---

3. Direct Proportion: Constant Quotients & Linear Scaling

Two variables $x$ and $y$ are in direct proportion when an increase in one variable produces a proportional increase in the other, or a decrease in one produces a proportional decrease in the other. Their ratio remains constant.

Mathematical Formulation

$\frac{y}{x} = k \quad \implies \quad y = kx \quad \text{or} \quad \frac{y_1}{x_1} = \frac{y_2}{x_2}$

where $k$ represents the constant of proportionality (or constant of variation).

Graphical Representation

The graph of a direct proportion is always a straight line passing through the coordinate origin $(0, 0)$, with a slope equal to $k$.

Common Public Safety Scenarios

• Fuel Consumption vs. Patrol Distance: Distance traveled varies directly with fuel consumed at a steady speed.
  • Scenario: A police patrol utility vehicle consumes 32 liters of diesel fuel over a 280-kilometer highway patrol sector. How many liters of diesel are required to patrol a 420-kilometer sector under identical driving conditions?
  • Set up direct proportion: $\frac{\text{Fuel}_1}{\text{Distance}_1} = \frac{\text{Fuel}_2}{\text{Distance}_2}$ $\frac{32}{280} = \frac{x}{420} \implies \frac{4}{35} = \frac{x}{420}$ $35x = 4 \times 420 = 1,680 \implies x = \frac{1,680}{35} = 48\text{ liters}$
• Logistical Scaling: Scaling ammunition requisitions, uniform tailoring orders, or ration distribution in direct proportion to the number of deployed personnel.

---

4. Inverse (Indirect) Proportion: Constant Products & Workforce Scheduling

Two variables $x$ and $y$ are in inverse proportion when an increase in one variable causes a proportional decrease in the other, such that their mathematical product remains constant.

Mathematical Formulation

$x \times y = k \quad \implies \quad y = \frac{k}{x} \quad \text{or} \quad x_1 y_1 = x_2 y_2$

where $k$ represents the constant product.

Graphical Representation

The graph of an inverse proportion is a rectangular hyperbola located in the first quadrant. As $x$ approaches infinity, $y$ approaches zero asymptotically, and vice versa.

Workforce and Task Completion Problems

Workforce problems are the most common application of inverse proportions on the PNPE. Assuming all personnel work at an identical productivity rate, more workers require fewer days to complete a fixed task:

$\text{Workers}_1 \times \text{Time}_1 = \text{Workers}_2 \times \text{Time}_2 = \text{Total Workload (person-hours or person-days)}$

• Problem: If 8 crime scene technicians take 15 hours to process and catalog physical evidence from a large complex crime scene, how many hours would it take 12 technicians working at the same pace?
• Determine total labor requirement: $k = 8\text{ technicians} \times 15\text{ hours} = 120\text{ technician-hours}$.
• Calculate time for 12 technicians: $12 \times t_2 = 120 \implies t_2 = \frac{120}{12} = 10\text{ hours}$

Speed and Travel Time over a Fixed Distance

Travel time varies inversely with speed over a fixed distance ($d = v \times t = k$):

• Problem: A pursuit vehicle traveling at an average speed of 80 km/h reaches a regional security checkpoint in 45 minutes. If road congestion reduces its speed to 60 km/h, how long will the trip take?
• Total travel constant: $80\text{ km/h} \times 45\text{ min} = 3,600\text{ km}\cdot\text{min}$.
• Time at 60 km/h: $60 \times t_2 = 3,600 \implies t_2 = \frac{3,600}{60} = 60\text{ minutes}$ (1 hour).

---

5. Partitive Proportion: Distributing Fixed Totals Across Unequal Shares

Partitive proportion involves dividing a known whole quantity into three or more unequal shares according to a specified continuous ratio ($a : b : c$).

The Standard 3-Step Algorithm

1. Step 1: Calculate Total Ratio Parts ($S$): $S = a + b + c + \dots$
2. Step 2: Determine the Base Multiplier / Unit Value ($u$): $u = \frac{\text{Total Aggregate Quantity}}{S}$
3. Step 3: Calculate Each Component Share: $\text{Share } A = a \times u; \quad \text{Share } B = b \times u; \quad \text{Share } C = c \times u$

Comprehensive Deployment Scenario

A provincial directorate receives a requisition shipment of 240 new patrol communication radios to be distributed among three municipal police stations—Station Alpha, Station Bravo, and Station Charlie—in the ratio of $3 : 4 : 5$. How many radios will each station receive?

1. Sum of Parts: $3 + 4 + 5 = 12\text{ total parts}$.
2. Unit Value per Part: $u = \frac{240}{12} = 20\text{ radios per part}$.
3. Individual Distribution:
   • Station Alpha: $3 \times 20 = 60\text{ radios}$
   • Station Bravo: $4 \times 20 = 80\text{ radios}$
   • Station Charlie: $5 \times 20 = 100\text{ radios}$
4. Verification: $60 + 80 + 100 = 240\text{ radios}$.

Partitive Proportion with Known Differences

Some advanced items state the difference between two parts rather than the total aggregate:

• Problem: The budget allocation between administrative operations and field patrols is in the ratio of $4 : 7$. If field patrols receive PHP 450,000 more than administrative operations, what is the total operational budget?
• The difference in ratio parts is $7 - 4 = 3\text{ parts}$.
• Set 3 parts equal to the difference: $3\text{ parts} = \text{PHP } 450,000 \implies 1\text{ part} = \text{PHP } 150,000$.
• Total ratio parts $= 4 + 7 = 11\text{ parts}$.
• Total budget $= 11 \times 150,000 = \text{PHP } 1,650,000$.

---

6. Compound & Joint Variations

When a quantitative scenario involves three or more variables changing simultaneously, it constitutes a compound or joint variation.

Types of Multi-Variable Variations

• Joint Variation: A variable varies directly as the product of two or more other variables ($z = kxy$).
• Combined / Compound Variation: A variable varies directly with some variables and inversely with others: $z = \frac{k \times x}{y} \quad \implies \quad \frac{z_1 y_1}{x_1} = \frac{z_2 y_2}{x_2} = k$

The Unified Workforce Formula

In complex workforce word problems, the relationship between workers ($M$), daily working hours ($H$), working days ($D$), and total units of work produced ($W$) is expressed as:

$\frac{W}{M \times H \times D} = \text{Constant Individual Productivity Rate } (k)$ $\frac{W_1}{M_1 \times H_1 \times D_1} = \frac{W_2}{M_2 \times H_2 \times D_2}$

• Problem: If 6 police clerks working 7 hours per day can process 420 administrative clearances in 5 days, how many clearances can 9 clerks working 8 hours per day process in 4 days at the same individual pace?
• Baseline Scenario: $\text{Total Clerk-Hours} = 6\text{ clerks} \times 7\text{ hours/day} \times 5\text{ days} = 210\text{ clerk-hours}$ $\text{Rate } (k) = \frac{420\text{ clearances}}{210\text{ clerk-hours}} = 2\text{ clearances per clerk-hour}$
• Second Scenario: $\text{Total Clerk-Hours} = 9\text{ clerks} \times 8\text{ hours/day} \times 4\text{ days} = 288\text{ clerk-hours}$ $\text{Clearances Produced } (W_2) = 288\text{ clerk-hours} \times 2\text{ clearances/clerk-hour} = 576\text{ clearances}$

---

7. Unit Rates & Dimensional Analysis

A unit rate is a specialized ratio comparing a quantity to exactly one unit of another quantity (e.g., kilometers per liter, pesos per cartridge, citations per hour).

Dimensional Analysis (The Factor-Label Method)

Dimensional analysis converts an expression from one unit of measurement to another by multiplying sequentially by unit fractions (conversion factors that equal 1), canceling identical units in numerators and denominators.

Rapid Speed Conversion Shortcut: km/h to m/s

Traffic reconstruction and speed enforcement items frequently require converting speed from kilometers per hour (km/h) to meters per second (m/s):

$1\text{ km/h} = \frac{1,000\text{ meters}}{3,600\text{ seconds}} = \frac{5}{18}\text{ m/s}$

• To convert from km/h to m/s: Multiply by $\frac{5}{18}$ (or divide by $3.6$).
  • Example: A fleeing suspect vehicle clocked at 90 km/h travels at: $90 \times \frac{5}{18} = 5 \times 5 = 25\text{ meters per second}$
• To convert from m/s to km/h: Multiply by $\frac{18}{5}$ (or multiply by $3.6$).
  • Example: An officer sprinting at 8 m/s runs at: $8 \times \frac{18}{5} = \frac{144}{5} = 28.8\text{ km/h}$

---

8. Summary Comparison of Proportional Relationships