Population Density

Density equals population divided by area. If population remains constant and density increases by 40 percent, the new density becomes 1.40 times the original density. To keep the same population, area must change inversely. Thus the new area is \(1/1.40=0.714285...\) of the original area, or about 71.4 percent. Therefore option B is correct.

Let the original population be P and area be A, so density is \(P/A\). The new density is \(1.4P/A\). Solving \(P/A_{new}=1.4P/A\) gives \(A_{new}=A/1.4\). Hence about 71.4 percent of the old area remains. It is not 60 percent because a 40 percent density rise is not the same as a 40 percent area reduction.

Population density is population divided by area, so area can be represented as population divided by density. Percentage changes should be handled with multipliers rather than by simply adding or subtracting percentages. Population becomes \(1.08P\), while density becomes \(0.90D\), because it falls by 10 percent.

Since \(D=P/A\), the new-to-old area ratio is \((1.08P/0.90D)\div(P/D)=1.08/0.90=1.20\). Thus the area is 120 percent of its original size, meaning it increased by 20 percent. Therefore, option C is correct. Calling the increase 18 percent would incorrectly ignore the density decrease in the denominator.

If density remains unchanged, population and area must change by the same proportional factor, because density equals population divided by area. The population rises from 240,000 to 324,000. Its absolute increase is 84,000, but the required answer is the percentage increase relative to the original population.

Calculate \(324000-240000=84000\), then \(\frac{84000}{240000}\times100=35\%\). Since density is unchanged, area must also become 1.35 times its original size, which means a 35 percent increase. Therefore option C is correct. The 84,000 figure is a numerical increase, not a percentage.

After the boundary expansion, the city’s new population is the original population plus the residents who were added. The new area is already given as 1,400 square kilometres. New density must be calculated using these new values, not the original population or original area.

New population is \(840000+70000=910000\). Therefore, new density is \(\frac{910000}{1400}=650\) persons per square kilometre. A reverse check gives \(650\times1400=910000\), confirming the calculation. Thus option B is correct. The original density would be \(840000\div1200=700\), but that is not the requested post-expansion density.

Saying that A is 36 percent lower than B means A is the remaining 64 percent of B, not that B is 36 percent more than A. In algebraic form, \(A=(1-0.36)B=0.64B\). This distinction is important because percentages are measured from the stated base, which here is B.

Given \(A=320\), solve \(320=0.64B\). Therefore, \(B=320\div0.64=500\) persons per square kilometre. A quick check confirms this: 36 percent of 500 is 180, and \(500-180=320\). Hence option C is correct. Adding 36 percent directly to 320 would use the wrong base.

Population equals density multiplied by area. Therefore, percentage changes in density and area combine through multiplication, not addition. If density becomes 84 percent of its original value, its factor is \(0.84\). If area becomes 125 percent of its original value, its factor is \(1.25\). The population factor is the product of these two factors.

Thus, the new population relative to the original is \(0.84\times1.25=1.05\). A factor of 1.05 means 105 percent of the original population, or a 5 percent increase. Hence option C is correct. Although density falls, the larger area more than offsets that fall in the population calculation.

Population density is calculated as \(P \div A\), where population is the numerator and area is the denominator. If the population stays unchanged, the density can increase only when the area used in the calculation becomes smaller. A reduced boundary, a change in the officially counted area, or exclusion of some water area could produce this mathematical result without any change in residents.

Thus, option A is correct: the calculated area has decreased. If area increased, density would fall, not rise, and changing the formula is not a normal explanation. Option D is inconsistent with the condition that total population remains unchanged. The key is to examine the denominator: holding the numerator constant while reducing the denominator raises the quotient.

Density is population divided by area: \(D=P/A\). Both population and area are decreasing, so the population change cannot be inferred merely from the fact that density decreases. However, the direction of the proportional changes can be inferred. If area becomes smaller while density also becomes smaller, population must fall by a greater proportion than area.

Suppose area becomes a fraction \(r\) of its original value, where \(r<1\), and density becomes a smaller fraction \(s\), where \(s<1\) and \(s<r\). Then population becomes \(sr\) of its original value, so its proportional fall is larger than the area’s fall. Therefore option A is correct. The other choices are not necessary conclusions.

Both daytime and resident density are calculated using the same municipal area. Since that area does not change, it cancels when their ratio is formed. The effect can therefore be found by comparing the population multipliers: daytime population becomes \(1.30\) times its old value, while resident population becomes \(0.90\) times its old value.

The new ratio compared with the old ratio is \(1.30\div0.90=1.444...\), or approximately 1.44 times. Thus option B is correct. The ratio does not remain unchanged because the two populations change in different directions. The unchanged area is why no separate area calculation is needed.

The regional average density is based on 100 percent of the population living in 100 percent of the area. The remaining portion contains 20 percent of the population because 80 percent is in the first portion. It still covers 50 percent of the total area. Its density relative to the regional average is therefore its population share divided by its area share.

The calculation is \(0.20/0.50=0.4\). Thus the remaining portion has 0.4 times the regional average density, so option B is correct. It is not 0.2 times, because the remaining population must be compared with the remaining area, not with the whole area alone.

Arithmetic density uses the total population divided by the total area. Cultivated-land density uses the same population divided only by the cultivated portion of the area. If total area is A, cultivated land is 20 percent of it, or \(0.20A\). With arithmetic density 250, the population is \(250A\). Hence cultivated-land density is \(\frac{250A}{0.20A}=\frac{250}{0.20}=1250\) persons per square kilometre of cultivated land.

Therefore, option C is correct. The value is five times the arithmetic density because cultivated land is one-fifth of the total area. This does not mean the population has changed; only the reference area used in the denominator has become smaller. The answer must specify cultivated land to distinguish it from ordinary arithmetic density.

Arithmetic density uses total population divided by total area, whereas cultivated-land density uses the same population divided by cultivated land. If cultivated-land density is five times arithmetic density, the denominator for cultivated-land density must be one-fifth of the total area, assuming the population is the same in both measures. The relation can be solved algebraically.

Let total area be \(A\), cultivated land be \(C\), and population be \(P\). Then \(P/C=5(P/A)\). Cancelling the non-zero population gives \(1/C=5/A\), or \(C=A/5\). Thus cultivated land is \(1/5=20\) percent of total area. Therefore, option B is correct.

Gross density uses the entire city area, while net residential density uses only the land occupied by residential areas. Let total population be P, total area be A, and residential land be R. The given relationships are P/A = 450 and P/R = 1125. We need the fraction of total area that is residential, which is R/A.

Divide the two density expressions: (P/A) ÷ (P/R) = R/A. Hence R/A = 450 ÷ 1125 = 0.4. Converting the fraction to a percentage gives 0.4 × 100 = 40 percent. Therefore, option B is correct. The higher net residential density is expected because the same population is concentrated in only part of the total city area. A percentage such as 50 would not match the given density ratio.

Arithmetic density uses the entire area, whereas habitable-area density uses only the part where people can live. Let population be \(P\), total area be \(A\), and habitable area be \(H\). The given relationships are \(P/A=210\) and \(P/H=525\). Since the same population is used in both ratios, dividing the first density by the second gives the fraction of area that is habitable.

Thus, \(H/A=210\div525=0.4\). Converting the fraction into a percentage gives \(0.4\times100=40\) percent. Therefore, option B is correct. The remaining 60 percent may be non-habitable, but that figure is not the requested percentage.

Official or gross density uses the entire area, including inland water, whereas land-only density uses only the land area. If water covers 10 percent of the total area, land occupies 90 percent, or \(0.9A\), where \(A\) is the total area. The population is unchanged in both calculations.

Given gross density \(360 = P/A\), land-only density is \(P/(0.9A)\). This equals \((P/A) \div 0.9 = 360 \div 0.9 = 400\) persons per square kilometre. Therefore, option B is correct. The land-only value is higher because the same population is divided by a smaller area.

Arithmetic density depends on the official population and the official area included in a boundary. If one city boundary contains only a compact urban core, its area may be small and its density may appear high. If another boundary includes a large rural fringe, the same calculation divides the population by a much larger area. That can lower the reported average even when the built-up part is crowded.

Option B is correct because boundary design can distort a direct comparison. A higher official density does not automatically prove that every part of that city is more crowded than every part of the other city. The rural fringe is not irrelevant: it is part of the denominator and changes the average. Using the same formula does not remove differences in what the boundaries include; comparable boundaries and definitions are needed for a fair interpretation.

Population density is calculated by dividing population by land area: \(D=\frac{P}{A}\). If the population remains almost unchanged but reported density suddenly rises, the first suspicion should be that the denominator has become smaller. This can happen when an administrative boundary is redrawn, a district is divided, land area is revised, or the statistical definition of the region changes. A smaller area with the same population automatically produces a higher density.

Option A is therefore the best first check. A fall in rainfall might influence settlement in the long term, but it does not directly explain a sudden statistical jump when population is unchanged. A higher birth rate would normally increase population, so it is not the immediate explanation. The formula itself usually remains the same; the important issue is whether the area being used has changed. Boundary and area records should be checked before proposing demographic causes.

Population density is not merely a number obtained by dividing population by area. That basic ratio is the starting point, but expert analysis must ask what population and what area are being measured. Administrative boundaries may include mountains, forests, water, or protected land, so the effective inhabited area can differ from the total area. Gross density, net density, spatial concentration, and comparability between regions may produce different interpretations.

Option C is correct because it combines the ratio with boundary definitions, effective area, percentage changes, gross and net density, and spatial distribution. Option A ignores area, while option B treats a simple calculation as a complete analysis. Option D also overstates what a high average value means, because average density does not prove that every locality is crowded. A broad analytical framework is therefore required.