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Difficulty: MediumPopulation Distribution and Density Factors

An agricultural province in West Africa covers a total land area of 80,000 km280,000\text{ km}^2. Arable land suitable for crop cultivation accounts for 25%25\% of this total land area. If the physiological population density of the province is 200 persons per km2200\text{ persons per km}^2 of arable land, what is the total population of the province?

Answer: 4000000 people

Answer

The total population of the province is 4,000,0004,000,000 people.
Physiological density measures the population relative to the amount of arable land available (200 persons per km2200\text{ persons per km}^2). First, finding 25%25\% of the total 80,000 km280,000\text{ km}^2 gives 20,000 km220,000\text{ km}^2 of arable land. Multiplying this arable area by the physiological density of 200 persons/km2200\text{ persons/km}^2 yields a total population of 4,000,0004,000,000 people.

Step-by-Step Solution

1
Calculate the total area of arable land in the province
Arable land area = 20,000 km220,000\text{ km}^2
Physiological density is calculated strictly based on arable (cultivable) land area, not total surface area. 25%25\% of 80,000 km280,000\text{ km}^2 equals 20,000 km220,000\text{ km}^2.
2
Apply the physiological density equation to calculate the total population
Total population = 4,000,0004,000,000 people
Since Physiological Density=Total PopulationArable Land Area\text{Physiological Density} = \frac{\text{Total Population}}{\text{Arable Land Area}}, multiplying the physiological density (200 persons/km2200\text{ persons/km}^2) by the arable land area (20,000 km220,000\text{ km}^2) gives the total population (200×20,000=4,000,000200 \times 20,000 = 4,000,000).

Key Concept

Physiological Population Density
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Topics

Population Distribution and Density Factors

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