A administrative district in West Africa covers a total land area of 50,000\text{ km}^2, o… — JAMB UTME

A administrative district in West Africa covers a total land area of $50,000\text{ km}^2$ , of which $35\%$ is classified as arable land suitable for farming. The district has a total population of $7,000,000$ inhabitants, and farmers with their immediate dependents (the agricultural population) make up $60\%$ of this total. What is the agricultural population density of this district in persons per $\text{km}^2$ of arable land?

Answer: 240 persons/km²

Answer

The agricultural population density of the district is

240\text{ persons/km}^2

Agricultural population density measures the number of farmers per unit of arable land. First, calculate arable land area (

50,000 \times 0.35 = 17,500\text{ km}^2

). Next, find the agricultural population (

7,000,000 \times 0.60 = 4,200,000

). Dividing

4,200,000

17,500

yields

240\text{ persons/km}^2

Step-by-Step Solution

Determine the total arable land area in

\text{km}^2

Arable land area =

50,000 \times 0.35 = 17,500\text{ km}^2

Agricultural population density evaluates population relative specifically to cultivated or arable land, not total land surface.

Calculate the total agricultural population.

Agricultural population =

7,000,000 \times 0.60 = 4,200,000\text{ persons}

Agricultural density isolates the farming population rather than using total population.

Compute agricultural population density by dividing agricultural population by arable land area.

\frac{4,200,000}{17,500} = 240\text{ persons/km}^2

Agricultural density formula =

\frac{\text{Agricultural Population}}{\text{Arable Land Area}}

Key Concept

Distinction between arithmetic density, physiological density, and agricultural density

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