Population Density

Density is population divided by area. Therefore, when both population and area change, the percentage change in density is found from the ratio of their new multipliers, not by simply subtracting the two percentages. A 44% population increase gives a population multiplier of 1.44, and a 20% area increase gives an area multiplier of 1.20.

The new density multiplier is \(1.44\div1.20=1.20\). A multiplier of 1.20 means the density is 120% of its original value, so it has increased by 20%. Therefore, option B is correct. The population increase is larger, but the simultaneous area increase reduces its effect on density.

Density is the ratio of population to area. A 28 percent population decrease leaves 72 percent of the original population, so its multiplier is 0.72. A 20 percent area decrease leaves 80 percent of the original area, giving a multiplier of 0.80. The new density multiplier is the population multiplier divided by the area multiplier.

Thus, the density multiplier is \(0.72/0.80=0.90\). The new density is 90 percent of the old density, which means it has decreased by 10 percent. Therefore, option A is correct. The decrease in population is proportionally larger than the decrease in area, causing density to fall.

Population density equals population divided by area, so area can be written as population divided by density. To compare the new area with the old area, use multipliers rather than subtracting the percentage changes directly. A 12 percent population increase gives a multiplier of 1.12, while a 4 percent density decrease gives a multiplier of 0.96.

Thus the area multiplier is \(1.12 \div 0.96 = 1.1666\ldots\). This means the new area is about 116.67 percent of the old area, so the increase is about 16.67 percent, or 16.7 percent. Therefore option C is correct. Simply calculating 12 + 4 = 16 gives only a rough approximation and does not use the exact ratio.

Answer: C, 30 percent. Population density is calculated as population divided by area: D = P/A. The population increases from 150,000 to 195,000, so the increase is 45,000. Compared with the original population, the percentage increase is (45,000 ÷ 150,000) × 100 = 30%. Because density remains unchanged, population and area must change in the same proportion. Therefore, the area also increased by 30%. Option A is incorrect because 20% is not the calculated increase. Option B is incorrect because 25% does not match the population change. Option C is correct because it equals the required area increase. Option D is incorrect because 45,000 is the absolute increase, not the percentage. Memory cue: with constant density, equal percentage changes occur in population and area.

Answer: C, 550 persons per square kilometre. First find the new population: 540,000 + 54,000 = 594,000. The new area is given as 1,080 square kilometres. Apply the density formula, D = P/A: 594,000 ÷ 1,080 = 550 persons per square kilometre. Therefore option C is correct. Option A would result from using an incorrect population or denominator. Option B is not the quotient of the new population by the new area. Option C correctly uses both changed values. Option D is the original density, 540,000 ÷ 900 = 600, and therefore ignores the boundary expansion and added residents. Notice that although population increased, area increased proportionally more, so density fell from 600 to 550. Memory cue: after any boundary change, calculate density with the new population and the new area, not with an old value.

The question asks what percentage the lower density represents of the higher density. This is not asking how much lower it is, so the correct operation is lower value divided by higher value, multiplied by 100. The lower density is 250 persons per square kilometre and the higher density is 400 persons per square kilometre.

Required percentage = \((250 \div 400) \times 100 = 62.5\%\). Therefore the lower density is 62.5 percent of the higher density, making option B correct. The percentage by which it is lower would be 37.5 percent, calculated from \((400-250)\div400\times100\); that is a different question and should not be confused with the requested percentage of the higher value.

A percentage increase is calculated from the original quantity. Here, Region B is 60 percent higher than Region A, whose density is 250 persons per square kilometre. A 60 percent increase means the original amount plus 60 percent of that amount, or 160 percent of the original. Therefore, the correct choice is C, 400 persons per square kilometre.

First find 60 percent of 250: \(250\times\frac{60}{100}=150\). Add this increase to the original density: \(250+150=400\). Equivalently, multiply by \(1.60\): \(250\times1.60=400\). Simply adding 60 to 250 would confuse a percentage with an absolute number and would give 310, which is not correct.

Answer: B, 400 persons per square kilometre. The statement says that A is 20% lower than B, so A represents 80% of B. In symbols, 320 = 0.80 × B. Hence B = 320 ÷ 0.80 = 400. Option B is correct. Option A, 384, comes from adding 20% to 320; that is wrong because the 20% decrease was measured from B, not from A. Option C is not obtained by the stated relationship. Option D is also incorrect because it implies a different percentage difference. The important idea is the percentage base: when a value is 20% less than another, it equals 80% of the original reference value. A useful check is that 20% of 400 is 80, and 400 − 80 = 320. Memory cue: “20% less than B” means multiply B by 0.8, then reverse the operation by dividing by 0.8.

Population depends on both density and area: population = density × area. When both quantities change, their percentage multipliers must be multiplied. A density of 125 percent of the original means a multiplier of 1.25, while an area that is 80 percent of the original has a multiplier of 0.80.

The new population compared with the old one is \(1.25 \times 0.80 = 1.00\). A multiplier of 1.00 means the population is exactly equal to its original value. In percentage form, this is 100 percent of the original population. Therefore, option C is correct. The increase in density exactly balances the reduction in area.

Population is obtained by multiplying density by area. If the new density is 90% of the original, its multiplier is 0.90. If the new area is 120% of the original, its multiplier is 1.20. The population multiplier is therefore the product of these two changes, because both changes act together.

Population multiplier = 0.90 × 1.20 = 1.08. This means the new population is 108% of the original population, or 8% higher than before. A lower density does not necessarily mean a lower total population when the area becomes sufficiently larger. Hence option C is correct; 100% would incorrectly ignore the combined effect.

Answer: A. Arithmetic density is D = P/A, where P is population and A is area. A ratio can increase even when its numerator decreases if its denominator decreases by a greater proportion. For example, suppose population falls by 10%, becoming 0.90P, while the included city area falls by 20%, becoming 0.80A. The new density is (0.90P)/(0.80A) = 1.125(P/A), so density rises by 12.5%. Option A correctly describes this boundary contraction. Option B is wrong because it ignores the denominator. Option C is wrong because area does not always change in the same direction or proportion as population. Option D is wrong because population is the numerator in the density formula. The result depends on both population and area, not on population alone. Memory cue: for a ratio, always inspect both the numerator and denominator before deciding whether it rises or falls.

Population density is the ratio \(P/A\), where \(P\) is population and \(A\) is area. A ratio can decrease even when its numerator increases if its denominator increases by a still larger proportion. Thus, rising population alone does not guarantee rising density.

For density to fall while population rises, area must increase faster, proportionally, than population. For example, if population rises by 10% but area rises by 20%, the ratio becomes smaller. Therefore, option A is the mathematically correct explanation. Option B describes a decreasing area, which would normally push density upward, and the remaining options contradict the formula.

Daytime density is calculated using the number of people present during the day divided by the municipal area. The area is unchanged, so density changes in the same proportion as daytime population. If daytime population rises by 25%, the numerator becomes 125% of its former value, while the denominator stays constant. Thus daytime density also rises by 25%, making option A correct.

Resident density does not automatically change because resident population remains unchanged and the municipal area is unchanged. The question distinguishes daytime population from resident population. A 12.5% increase would not follow from the stated information. The conclusion assumes that daytime density is based on the population present during daytime, including people who commute into the city.

To find how much higher one value is than another, compare the difference with the original, or reference, value. The resident density is the reference because the question asks how much higher daytime density is than resident density. The difference is \(1350-1000=350\) persons per square kilometre. Divide this difference by 1,000 and multiply by 100.

The percentage increase is \((350/1000)\times100=35\%\). Therefore, option C is correct. The number 350 is the absolute difference, not the percentage. Using 1,350 as the denominator would answer a different question, such as the difference as a share of daytime density. Percentage increase is always measured from the original or comparison base.

Answer: A, one-third. Density is proportional to population share divided by area share. For the first portion, relative density is 75 ÷ 50 = 1.5. For the second portion, it is 25 ÷ 50 = 0.5. Now compare the second with the first: 0.5 ÷ 1.5 = 1/3. Therefore, the remaining portion has one-third of the density of the first portion. Option A is correct. Option B would mean the second density is half the first, which does not follow from the shares. Option C is also too high because the first portion contains three times as much population in the same area. Option D is wrong because equal areas do not create equal densities when population shares differ. A simple check is to imagine a total population of 100 and total area of 100 units: the first density is 75/50 and the second is 25/50. Memory cue: density comparison = population share ratio ÷ area share ratio.

The remaining portion has the population not included in the first portion. If 70 percent of the population lives in 40 percent of the area, the remaining 30 percent lives in the remaining 60 percent of the area. To compare its density with the regional average, divide its population share by its area share.

The relative density is \(0.30\div0.60=0.50\). Therefore, the remaining portion has one-half, or 0.5 times, the regional average density. The correct choice is C. It is not 0.3 because population share alone does not measure density; the area share must also be included.

Arithmetic density uses the total population divided by the total area. Cultivated-land density uses the same population divided only by cultivated land. If total area is represented by A, arithmetic density of 220 means population is 220A. Cultivated land is 44% of A, or 0.44A.

Therefore cultivated-land density = 220A ÷ 0.44A = 220 ÷ 0.44 = 500 persons per square kilometre of cultivated land. The A cancels because it is present in both population and cultivated area. Thus option C is correct. The density is higher than arithmetic density because the same population is measured against only part of the total area.

Arithmetic density uses the whole area, while cultivated-land density uses only cultivated land. Let total area be \(A\), cultivated area be \(C\), and population be \(P\). Then arithmetic density is \(P/A\), and cultivated-land density is \(P/C\).

The condition \(P/C=4(P/A)\) means the same population is concentrated on an area four times smaller. Cancelling population gives \(A/C=4\), so \(C=A/4\). Therefore cultivated land is \(1/4\times100=25\%\) of the total area. Option B is correct. The answer is not 40%, because the density multiplier must be converted into the reciprocal area fraction.

Gross density uses the entire municipal area, whereas net residential density uses only residential land. Let total area be \(A\), residential area be \(R\), and population be \(P\). The given values mean \(P/A=360\) and \(P/R=900\).

Dividing the first equation by the second gives \((P/A)/(P/R)=360/900\). Population cancels, leaving \(R/A=0.4\). Converting this fraction to a percentage gives 40%. Therefore option B is correct. The higher net density is expected because the same population is measured over the smaller residential portion rather than over all municipal land.

Answer: B, 20 percent. Let P be the population, A the total municipal area, and L the land area. Gross density is P/A = 500, while land-only density is P/L = 625. Dividing the first relationship by the second gives L/A = 500/625 = 0.8. Thus land occupies 80% of the total municipal area. Water bodies occupy the remaining part: 100% − 80% = 20%. Option B is correct. Option A is too small and does not match the density ratio. Option C incorrectly treats the difference between 625 and 500 as the area percentage. Option D is also inconsistent with the ratio. The same population is used in both densities, so the ratio of densities reveals the ratio of land area to total area. Memory cue: when the numerator is unchanged, density is inversely related to the area used as denominator.

Arithmetic density uses the entire area, while habitable-area density uses only the part where people can live. Let total area be \(A\), habitable area be \(H\), and population be \(P\). The two densities are \(P\div A=160\) and \(P\div H=400\). Dividing the first relationship by the second removes the population.

Thus, \((P/A)\div(P/H)=H/A=160\div400=0.4\). Converting 0.4 into a percentage gives \(0.4\times100=40\) percent. Therefore, habitable land is 40 percent of the total area, so option B is correct. The remaining 60 percent is not included in the habitable-area denominator.

Answer: B, approximately 426.7 persons per square kilometre. If water covers 25% of the total area, land covers 75%, or 0.75A. The given gross density is P/A = 320. Land-only density uses the smaller denominator, so it is P/(0.75A) = (P/A) ÷ 0.75 = 320 ÷ 0.75 = 426.666..., approximately 426.7. Option B is correct. Option A, 400, does not result from the correct division. Option C and option D are also higher than the calculated value and use incorrect adjustments. Removing water area does not remove the population from the numerator; it only reduces the area in the denominator. Therefore the land-only density must be greater than the gross density of 320. Memory cue: divide by the remaining land fraction, not by the water fraction.

An official density value is meaningful only when the population and area refer to comparable units. One city may mean a small municipal boundary, while another figure may refer to a much larger urban agglomeration or metropolitan region. Including surrounding rural land can lower density, while using only a built-up core can raise it. Area units must also be consistent, such as square kilometres or square miles.

Option A is correct because boundary definitions and measurement units directly affect the calculation: \(\text{density}=\frac{\text{population}}{\text{area}}\). The length of a city’s name, number of roads, or rainfall does not determine whether the figures are comparable. A higher reported value should therefore not be interpreted without checking what territory each figure covers.

Population density is calculated as population divided by area: \(D=P/A\). If the population remains almost unchanged but reported density suddenly falls by 15 percent, the first suspicion should be a change in the denominator, the area used for calculation. An administrative boundary may have expanded, a district may have been merged, or the recorded land area may have been revised. Checking these metadata and geographic definitions is therefore the best first analytical step, as stated in option A.

Rainfall and birth rate do not directly explain an abrupt density change when the census population is nearly stable. The formula itself normally remains unchanged, although data-processing errors should be examined after checking boundaries and area definitions. A sudden statistical break often reflects a change in geography, measurement, or reporting rather than a sudden demographic transformation. Thus option A is the most appropriate initial check.

Population density is not understood well by looking at population size alone. Its basic measure is population divided by area, written as \(D=P/A\), but interpretation requires more information. The analyst must define the denominator, check whether the area is total, habitable, or cultivated, and compare boundaries consistently. Spatial concentration and the difference between gross and net density may also change the conclusion.

Option C includes this complete framework. A large population can be spread over a huge area and have low density, while a small population in a tiny area can have high density. Simple division without clear spatial definitions may produce a misleading comparison. Therefore C is the only option that combines calculation with proportional change, concentration, and boundary awareness.