Population Density

Answer: B, 20%. Since population is unchanged and density equals population divided by area, density and area vary inversely. Let the original area be A. With fixed population, 320A = 400A₂, so A₂/A = 320/400 = 0.8. The new area is therefore 80% of the old area. The decrease is 100% − 80% = 20%. A is too small, C confuses the 25% rise in density with the area decrease, and D is too large. B is correct. Notice that a 25% density increase does not mean a 25% area decrease; inverse ratios must be calculated. Memory cue: with fixed population, new area/old area = old density/new density.

When area remains unchanged, density and population have a direct proportional relationship: \(D=P/A\). Thus, the percentage increase in density is also the percentage increase in population. First calculate the numerical increase in density, then compare it with the original density.

The increase is \(585-450=135\) persons per square kilometre. Its percentage is \((135/450)\times100=30\%\). Since area is constant, population also increases by 30 percent. Therefore option C is correct. The value 135 is the absolute increase, while 30 percent expresses that increase relative to the original 450.

When population remains fixed and area becomes larger, population density must decrease. Density is population divided by area. A 35 percent increase means the new area is 135 percent of the old area, or \(1.35\) times the original area. Therefore, the old density is divided by 1.35.

The new density is \(900\div1.35=666.67\) persons per square kilometre, which is approximately 667. The same result follows from finding the old area: \(1080000\div900=1200\) square kilometres. The new area is \(1200\times1.35=1620\), and \(1080000\div1620\approx667\). Thus option B is correct.

The average density of the whole region is based on all of its population and all of its area. The selected part contains 36% of the population but occupies only 12% of the area. Its density relative to the regional average can therefore be found by dividing its population share by its area share. This compares concentration without needing actual population or area values.

The relative density is \(0.36/0.12=3\). Hence the selected area has three times the average density of the entire region. Option C is correct. It is not enough to look only at the 36% population share; the fact that this population occupies just 12% of the area makes the density much higher.

The selected area contains 80% of the country's population but only 40% of its land. Its density compared with the national average is therefore \(\frac{0.80}{0.40}=2\) times the average. The national density is 260 persons per square kilometre, so the density of the selected area is \(260\times2=520\) persons per square kilometre. Hence, option C is correct.

Another way to see this is to imagine the whole country divided into equal area units. This part has twice as much population per unit of area as the countrywide average because its population share is double its area share. The calculation concerns the specified 40% area and does not require knowing the country's total population or total area.

Population density gives the average number of people per unit area, but actual resource pressure is not determined by concentration alone. Water availability, soil quality, technology, infrastructure, consumption levels and internal settlement patterns also matter. Therefore option A is correct. A high-density technologically efficient region may manage resource stress, while a low-density arid region may still face severe scarcity. Density should be treated as an indicator rather than a complete measure.

Answer: B, 20%. Let population be P, total area be A, and land area be L. Gross density is P/A = 600, while land-only density is P/L = 750. Dividing these equations gives (P/A)/(P/L) = L/A = 600/750 = 0.8. Therefore, land occupies 80% of the total area. The remaining 20% must be water bodies, assuming total area consists of land and water bodies as stated. A, C and D do not follow from the density ratio. Notice that land-only density is higher because the same population is divided by a smaller land area. Memory cue: gross density divided by land density gives the land fraction; subtract it from 100% for the water fraction.