Section 3.3: Rates, Ratios, and Proportions

Section 3.3: Rates, Ratios, and Proportions

In daily policing and emergency response, mathematical relationships involving rates, ratios, and proportions are used constantly. Whether calculating the speed of a fleeing vehicle, determining travel time to a high-priority incident, scaling a map to locate a missing person, or calculating the fuel consumption of a patrol fleet, these concepts are fundamental. The Victoria Police Entrance Examination evaluates these skills extensively under the ACER numeracy framework. Candidates must be comfortable with setting up and solving rate equations, dividing resources using ratios, and calculating direct and inverse proportions to pass the subtest with a score of 111 or higher.

Rates: Speed, Distance, and Time

A rate is a comparison of two quantities with different units of measurement. In transportation and emergency response, the most common rate is speed, which measures distance traveled per unit of time (typically kilometres per hour, or km/h).

The formulas relating speed, distance, and time are:

$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$ $\text{Distance} = \text{Speed} \times \text{Time}$ $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

The Challenge of Decimal Time

The most common mistake candidates make on the exam is failing to convert time into decimal form before performing calculations. Time is measured in base 60 (minutes and seconds), whereas our decimal system is base 10.

• Converting minutes to decimal hours: Divide the number of minutes by 60.
  • $15 \text{ minutes} = 15/60 = 0.25 \text{ hours}$
  • $30 \text{ minutes} = 30/60 = 0.5 \text{ hours}$
  • $45 \text{ minutes} = 45/60 = 0.75 \text{ hours}$
  • $40 \text{ minutes} = 40/60 = 0.667 \text{ hours}$
• Converting decimal hours back to minutes: Multiply the decimal portion by 60.
  • $2.35 \text{ hours} = 2 \text{ hours} + (0.35 \times 60) \text{ minutes} = 2 \text{ hours and } 21 \text{ minutes}$

Worked Example: A police response unit in Melbourne’s outer suburbs needs to travel 15 kilometres to respond to an urgent call for assistance. If traffic conditions allow the vehicle to maintain an average speed of 90 km/h, how long will it take to arrive at the scene?

$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{15}{90} = \frac{1}{6} \text{ hours}$

To convert $1/6$ of an hour to minutes:

$\frac{1}{6} \times 60 = 10 \text{ minutes}$

Fuel Consumption Formula

For logistics and fleet management, police vehicles use a fuel consumption rate measured in litres per 100 kilometres (L/100km). The formula to calculate total fuel used is:

$\text{Fuel used (L)} = \left(\frac{\text{Distance (km)}}{100}\right) \times \text{Rate (L/100km)}$

Worked Example: A regional divisional van based in Mildura must transport a custody suspect to a court in Melbourne, covering a distance of 550 kilometres. If the van's fuel consumption rate is 10.4 litres per 100 kilometres, how many litres of fuel are required?

$\text{Fuel used} = \left(\frac{550}{100}\right) \times 10.4 = 5.5 \times 10.4 = 57.2 \text{ litres}$

Ratios: Resource Allocation and Demographics

A ratio compares two or more quantities of the same unit, showing their relative sizes. Ratios are written with a colon (e.g., 3:2) and can be simplified just like fractions by dividing each term by their greatest common divisor.

Dividing a Quantity into a Ratio

In police administration, ratios are used to distribute resources—such as staffing or equipment—proportionally. To divide a total quantity into a ratio:

1. Add the parts of the ratio together to find the total number of parts.
2. Divide the total quantity by the total number of parts to find the value of "one part."
3. Multiply the value of one part by each component of the ratio.

Worked Example: A police command receives a shipment of 54 new hand-held radios to be distributed between the Ballarat station and the Wendouree station in a ratio of 5:4. How many radios does Ballarat receive?

1. Total parts: $5 + 4 = 9 \text{ parts}$.
2. Value of one part: $54 \div 9 = 6 \text{ radios}$.
3. Ballarat's allocation (5 parts): $5 imes 6 = 30 \text{ radios}$.
4. Wendouree's allocation (4 parts): $4 imes 6 = 24 \text{ radios}$. We can verify this by checking that $30 + 24 = 54$.

Proportions and Scaling on Police Maps

A proportion states that two ratios or rates are equal. Proportions can be direct or inverse:

• Direct proportion: As one quantity increases, the other increases at the same rate (e.g., the more hours you patrol, the more kilometres you cover).
• Inverse proportion: As one quantity increases, the other decreases at the same rate. For example, if you increase the number of officers working on a search grid, the time required to complete the search decreases.

Map Scaling

Police operations often require reading maps to coordinate search and rescue missions or document crime scenes. A map scale is a ratio representing the relationship between a distance on a map and the corresponding distance on the ground (e.g., 1:50,000).

• A scale of 1:50,000 means that 1 unit on the map represents 50,000 units in reality.
• Therefore, 1 centimetre on the map represents 50,000 centimetres on the ground.
• To convert centimetres to metres, divide by 100 (50,000 cm = 500 m).
• To convert metres to kilometres, divide by 1,000 (500 m = 0.5 km).

Thus, on a 1:50,000 scale map, 1 cm represents 0.5 km. If a search boundary is measured on a map as 8 cm, the actual distance is:

$8 \times 0.5 \text{ km} = 4 \text{ km}$