Population Density

The correct answer is B. Let the original population be \(P\) and the original area be \(A\). The original density is \(P/A\). After a 30 percent population increase, the population becomes \(1.30P\). After a 12 percent area decrease, the area becomes \(0.88A\). Therefore, the new density is \(\frac{1.30P}{0.88A}=1.47727\frac{P}{A}\).

The density is therefore about 1.477 times the original density. The increase is \(1.47727-1=0.47727\), or approximately 47.7 percent. Hence option B is correct. Simply adding 30 percent and 12 percent would give 42 percent, but that ignores the fact that density is a ratio and the reduced denominator increases the result further. The calculation must apply both percentage changes multiplicatively.

Population density is defined as population divided by area, so the relationship can be rearranged as \(Area=\frac{Population}{Density}\). The new population is 90 percent of the original because it decreases by 10 percent. The new density is 125 percent of the original because it increases by 25 percent.

Therefore, the area multiplier is \(\frac{0.90}{1.25}=0.72\). The new area is consequently 72 percent of the original area. Option A is correct. The calculation also shows why the answer is not 90 percent: area depends on both population and density, and the increase in density makes the required area smaller.

Population density is population divided by area. To compare two regions using ratios, divide the population ratio by the area ratio. If the populations are in the ratio 7:5 and the areas are in the ratio 14:15, the first region has a smaller area relative to its population than the second one.

The density ratio is \(\frac{7/14}{5/15}=\frac{1/2}{1/3}=\frac{3}{2}\). Therefore, the densities are in the ratio 3:2, so option B is correct. Simply using 7:5 would compare populations only and would ignore the different areas. The area ratio must be included in the calculation.

Answer: C, 3:2. Population is found by multiplying density by area: P = D × A. For the first and second regions, the population ratio is therefore (5 × 6):(4 × 5) = 30:20. Dividing both terms by 10 gives 3:2. A is obtained by multiplying the corresponding terms incorrectly, while B merely repeats the area ratio and D merely repeats the density ratio. Neither alone can give population because population depends on both density and area. C is correct. Memory cue: for ratio questions involving population, write P = D × A first, then multiply the density ratio and area ratio term by term.

Density equals population divided by area. Let the original population be P and area be A, so the original density is \(P/A=720\). After the changes, population becomes \(1.16P\), while area becomes \(1.20A\). The new density is therefore \((1.16P)/(1.20A)=720\times(1.16/1.20)\). Since \(1.16/1.20=0.9666\ldots\), the result is 696 persons per square kilometre. Thus option B is correct.

It is important not to apply only the population increase to density, because area also changes. Population rises by 16 percent, but area rises by a larger 20 percent, so the average number of people per square kilometre falls slightly. Calculating \(720\times1.16\div1.20\) gives exactly 696. Option C would be possible only if population and area grew by the same percentage; option D ignores the area increase.

Population equals density multiplied by area, so the population change can be found by multiplying the density factor by the area factor. Density rises from 400 to 520, giving a factor of \(\frac{520}{400}=1.30\). A 10 percent area decrease gives an area factor of \(0.90\). Hence the population factor is \(1.30\times0.90=1.17\).

A factor of 1.17 means the new population is 117 percent of the original, so it increased by 17 percent. Option A is correct. The area decrease offsets part of the density increase, which is why the population does not rise by 30 percent. This reasoning assumes the density and area values refer to the same region and time comparison.

Arithmetic density uses the entire city area, so \(P/A=480\). If only 32 percent of that area is residential, the residential land equals \(0.32A\). When all residents live within this smaller part, the same population is concentrated over less land. Consequently, the density calculated for residential land must be higher than the gross city density.

The net residential density is \(P/(0.32A)=(P/A)/0.32=480/0.32=1500\) persons per square kilometre. Thus option B is correct. The calculation does not multiply 480 by 0.32, because that would describe a different quantity and would make the density lower even though the same people occupy a smaller area.

Arithmetic density uses the total population divided by the total area, including land that may not be habitable. If 35 percent of the region is uninhabitable, the habitable part is 65 percent of the total area, or 0.65A. The total population remains the same, but all of it is assumed to live in this smaller habitable area. Consequently, the density over that area must be higher than the national arithmetic average.

Let total area be A. The population is 260A. The new density is therefore 260A/(0.65A)=260/0.65=400 persons per square kilometre. Option B is correct. Option C would incorrectly use the total area, while option A does not result from the stated 35 percent unusable share. The calculation assumes residents are concentrated entirely in the habitable portion, as the question specifies.

Density is calculated by dividing population by area. Resident density is \(900000\div700\approx1285.7\) persons per square kilometre. Daytime density is \(1260000\div700=1800\) persons per square kilometre. To find how much higher the daytime density is, subtract the resident value from the daytime value: \(1800-1285.7\approx514.3\) persons per square kilometre.

Option D is correct. The difference is not found by subtracting the two populations first and then ignoring area; the population difference is 360,000, and dividing it by 700 also gives approximately 514.3. Option A and option B are too small, while option C rounds the result incorrectly to 500. Daytime population may include commuters, visitors, and workers, so it can exceed the number of registered residents without changing the stated area.

Answer: C, 1,200 persons per square kilometre. A value 50% higher means the original value plus half of the original value. Half of 800 is 400, so daytime density is 800 + 400 = 1,200. Equivalently, multiply by 1 + 50/100 = 1.5: 800 × 1.5 = 1,200. A represents only a 25% increase, and B represents a 37.5% increase, so both are too low. D is double the original value, meaning a 100% increase, not 50%. C is correct. Memory cue: “x% higher” means multiply the original by 1 + x/100; do not add x as a number.

The average density of the whole region is based on total population divided by total area. If 55% of the population lives in 20% of the area, that portion contains population disproportionately concentrated there. Its density relative to the regional average is the population share divided by the area share: \(0.55\div0.20=2.75\).

Multiplying the average density by this factor gives \(150\times2.75=412.5\) persons per square kilometre. Therefore, option C is correct. The result is greater than the regional average because the portion has a much larger share of people than of land. It is not 300 or 375, which would use an incorrect concentration factor.

Density compares the share of population with the share of area. Let total population and area be 1 for easy comparison. The first portion contains 0.30 of the people in 0.10 of the area, so its relative density is \(0.30/0.10=3\). The remaining portion contains 0.70 of the people in 0.90 of the area, giving relative density \(0.70/0.90=0.777\ldots\).

To compare the two densities, divide the first by the second: \(3/(0.70/0.90)=3.857\ldots\), approximately 3.86. Thus the first portion is about 3.86 times denser than the remaining area, so option C is correct. Simply dividing 30 by 10 gives only the first portion’s relative density, not the comparison between both portions.

Let the total population be \(P\) and the total area be \(A\). The regional average density is \(P/A\). The specified portion contains 65 percent of the population and 25 percent of the area, so its density is \(0.65P/0.25A\). Dividing this by the regional average gives \(\frac{0.65P/0.25A}{P/A}=\frac{0.65}{0.25}=2.6\).

Therefore, the density of that portion is 2.6 times the regional average, making option C correct. Exact values for total population and total area cancel out and are not needed. The result is greater than one because the portion holds a much larger share of people than of land. This comparison uses shares, not absolute population figures.

Population is calculated by multiplying density by area: \(P=D\times A\). The density is 625 persons per square kilometre and the area is 1,152 square kilometres. Therefore, \(P=625\times1152\). A useful check is that 625 equals \(5/8\) of 1,000, so \(1152\times625=1152\times5/8\times1000=144\times5\times1000\).

This gives \(720\times1000=720000\) people. Hence option C is correct. The units also confirm the operation: persons per square kilometre multiplied by square kilometres leaves persons. The other choices do not result from the stated multiplication. No percentage adjustment or division is needed because both density and area are already supplied directly.

Population density is found by dividing total population by total area. The required units are persons per square kilometre, so the calculation is \(918000\div1350\). To divide easily, note that \(1350\times680 = 918000\). Therefore the quotient is exactly 680 persons per square kilometre.

Hence option C is correct. The answer can be verified by multiplying the density by the area: \(680\times1350=918000\), which reproduces the stated population. The other choices do not satisfy this check. It is important not to divide by 680 or confuse population with density; population is the numerator and area is the denominator in the density formula.

The original population is obtained by multiplying the original density by the area. With density 540 persons per square kilometre and area 1,600 square kilometres, the original population is \(540\times1600=864000\). An increase of 108,000 makes the new population 972,000. Because the area remains unchanged, the new density is calculated using the same area.

The new density is \(972000\div1600=607.5\) persons per square kilometre. Therefore, option C is correct. Another quick check is that the population increase alone contributes \(108000\div1600=67.5\) persons per square kilometre; adding this to the original 540 gives 607.5. The unchanged area is essential to this result.

When area is unchanged, a change in population produces the same proportional change in density. First recover the original population from density multiplied by area. Then subtract the people who migrated out and divide the remaining population by the unchanged area. This separates the population change from the area change.

The original population is \(450\times2000=900000\). After 180,000 people leave, the population is \(900000-180000=720000\). New density is \(720000\div2000=360\) persons per square kilometre. Therefore, option C is correct. The area remains 2,000, so it must not be reduced during the second division.

Answer: C, 50%. Because P = D × A, the population multiplier equals the density multiplier multiplied by the area multiplier. The density multiplier is 375/300 = 1.25, meaning a 25% density increase. The area multiplier is 1.20. Thus the population multiplier is 1.25 × 1.20 = 1.50. A multiplier of 1.50 means the new population is 150% of the old population, so the increase is 50%. A is wrong because it ignores the area increase. B and D do not follow from the product of the two factors. C is correct. Memory cue: population change factor = density change factor × area change factor.

Answer: C, approximately 46.7%. From P = D × A, the area multiplier equals the population multiplier divided by the density multiplier. Population increases by 10%, so its multiplier is 1.10. The density multiplier is 450/600 = 0.75. Therefore the area multiplier is 1.10/0.75 = 1.4666..., or about 1.467. This means the new area is about 146.7% of the old area, so the increase is about 46.7%. A and B are too low, while D is too high. C is correct. The density can fall even when population rises because the area expands by a still larger proportion. Memory cue: area factor = population factor ÷ density factor.

Population, density and area are related by \(D=P\div A\), so area is proportional to population divided by density. A 35% population decrease leaves 65% of the original population, while a 10% density decrease leaves 90% of the original density. Comparing these two multipliers gives the change in area.

The new area multiplier is \(0.65\div0.90=0.7222\ldots\). Thus, the new area is about 72.2% of the original area. The decrease is \(100-72.2=27.8\)% approximately. Therefore, option A is correct. The area does not decrease by exactly 35% because density also changes.

The stated density of 350 persons per square kilometre is calculated using the entire municipal area. If only 70% of that area is land and the population is associated with the land, the same population is spread over a smaller area. Dividing by a smaller area makes the land-based density higher.

Let total municipal area be \(A\). Population is \(350A\), while land area is \(0.70A\). Therefore, land-based density is \(350A\div0.70A=350\div0.70=500\) persons per square kilometre. Hence, option C is correct. The water area is excluded from the new denominator.

Official density usually divides population by the total administrative area, including water if that area is part of the boundary. If two cities have the same official density, they have the same population per unit of total area. However, the city with 30% water has less actual land available for residents than the city with only 5% water.

For the same total-area density, the population is distributed over a smaller land area in the first city. Consequently, its land-based density can be higher. It is not necessarily double, because the exact comparison depends on the total areas and population definitions. Therefore option A is the most appropriate inference. Water share does affect how official density should be interpreted.

Answer: C, 600 persons per square kilometre of cultivated land. Arithmetic density is total population divided by total area: P/A = 180. Cultivated land equals 30% of total area, or 0.30A. The population per unit of cultivated land is therefore P/(0.30A) = (P/A)/0.30 = 180/0.30 = 600. A, B and D do not use the 30% land share correctly. A is only twice the arithmetic density, B is 2.5 times it, and D is five times it; none matches the required factor 1/0.30 = 3.333. C is correct. This measure indicates pressure on cultivated land, not density over the entire area. Memory cue: density on a land category = total arithmetic density ÷ that category’s area fraction.

Arithmetic density divides total population by total land area. Cultivated-land density, however, relates population to the land that is actually cultivated or usable for farming. A region may have a large total area, including deserts, mountains, forests, or other sparsely used land, and therefore show low arithmetic density. If only a small part is cultivated, many people may depend on that limited land, producing high pressure there.

Thus the most suitable conclusion is B: population pressure on usable agricultural land may be high. Low arithmetic density does not prove that people are evenly spread, nor does it prove that total population is necessarily very low. It also does not make the arithmetic calculation wrong. The two measures answer different questions because they use different area bases.

Population density is population divided by area. If population does not change, density and area move in opposite directions. A 30 percent rise in density means the new density is 1.30 times the original density. To keep population unchanged, the new area must be the reciprocal factor, or 1 divided by 1.30, times the original area.

The area ratio is new area ÷ original area = 1 ÷ 1.30 = 0.76923 approximately. Converting this ratio to a percentage gives about 76.9 percent. Therefore, option B is correct. The area has not fallen by exactly 30 percent; percentage changes in a reciprocal relationship are calculated using the new multiplier. Option 130 percent describes density, not the remaining area.