Population Density

Population density is population divided by area, so a change in both quantities must be considered together. An increase in population raises density, while a decrease in area raises it further because the same, or a larger, population is distributed over less land. The correct choice is therefore option C, about 28.3 percent.

Let the original density be \(P/A\). After the changes, it becomes \(1.18P/(0.92A)=(1.18/0.92)(P/A)\approx1.2826(P/A)\). This is about 1.2826 times the original density, so the increase is \(0.2826\times100\approx28.3\%\). Simply adding 18% and 8% gives 26%, but division makes the exact result slightly larger.

Use the relationship \(Density=Population/Area\), which can be rearranged as \(Area=Population/Density\). If population rises by 12%, its multiplier is \(1.12\). If density falls by 20%, its multiplier is \(0.80\). Therefore the area multiplier is \(\frac{1.12}{0.80}=1.40\).

A multiplier of 1.40 means the new area is 140% of the original area. The increase is therefore 40%, so option C is correct. Simply subtracting 20% from 12% would not work because population and density affect area through division, not subtraction. The result is exact under the stated percentage changes, although the question uses “approximately” in its wording.

Density is population divided by area. To compare the two densities, divide the population ratio by the area ratio: \(\frac{P_1/A_1}{P_2/A_2}=\frac{8/4}{5/3}\). This equals \(2\times\frac{3}{5}=\frac{6}{5}\), so the density ratio is 6:5. Therefore option A is correct.

The population ratio alone cannot be used because the regions have different areas. The first region has a population factor of 8 and area factor of 4, while the second has corresponding factors 5 and 3. After allowing for area, their density values are proportional to 2 and 5/3, whose ratio is 6:5. Option D, 24:20, simplifies to 6:5 mathematically, so it represents the same ratio but is not the stated simplest choice.

Population, density, and area are related by population = density × area. Therefore, for two regions, the area ratio is found by dividing each region's population ratio term by its corresponding density ratio term. The population ratio alone cannot be used because two regions with equal populations may have different areas if their densities differ.

Using the given ratios, area ratio = (14 ÷ 7) : (9 ÷ 6) = 2 : 1.5. Multiplying both parts by 2 gives 4 : 3. Hence, option A is correct. A direct check is also possible: if the first population and density are represented by 14 and 7, its relative area is 2; the second is 9 ÷ 6 = 1.5, giving 2 : 1.5 = 4 : 3.

Density changes according to the ratio of the population multiplier to the area multiplier. A 25 percent population increase gives a multiplier of 1.25, while a 16 percent area increase gives a multiplier of 1.16. Therefore, the new density is \(840\times\frac{1.25}{1.16}\approx905.17\) persons per square kilometre.

Rounded to the nearest listed value, this is about 905 persons per square kilometre, so option B is correct. The population increase is proportionally greater than the area increase, which is why density rises, but it does not rise by the full 25 percent. Comparing the two percentage changes alone is not enough; the exact ratio of the new population to the new area must be used.

Population can be calculated as density multiplied by area. Therefore, compare the new and old values using multiplication factors. Density changes from 600 to 750, giving a factor of \(750\div600=1.25\). A 12 percent area decrease leaves 88 percent of the original area, giving a factor of 0.88.

The population factor is \(1.25\times0.88=1.10\). Thus the new population is 110 percent of the old population, which means it has increased by 10 percent. Option A is correct. The 25 percent density increase is partly offset by the 12 percent area decrease, so it should not be used alone.

Gross density uses the entire municipal area, whereas net residential density uses only the land where residents live. Let the total area be \(A\). A gross density of 540 means the population is \(P=540A\). Residential land is 45 percent of the total, so its area is \(0.45A\). If every resident lives on this land, the same population is concentrated in the smaller residential area.

The net residential density is therefore \(P/(0.45A)=540A/(0.45A)=540/0.45=1200\) persons per square kilometre. Hence option C is correct. The net figure is higher than the gross figure because only 45 percent of the total land is used in its denominator.

Arithmetic density uses the entire area of a region, even when some parts cannot support settlement. If 28 percent is uninhabitable, the habitable part is 100 minus 28, or 72 percent of the total area. The same population is then distributed over only this smaller usable area, so habitable-area density is higher than arithmetic density.

The habitable-area density is \(288\div0.72=400\) persons per square kilometre. Therefore, option B is correct. The calculation assumes that the population lives in the habitable portion. Dividing by the full area would give the original arithmetic density, not the density over land where people can actually live.

Daytime density counts people present during the day, while resident density counts the usual resident population. Both densities use the same city area, 625 square kilometres, so the difference can be calculated by finding each density separately. Daytime density is \(1,500,000/625=2,400\), and resident density is \(1,000,000/625=1,600\) persons per square kilometre.

The excess is \(2,400-1,600=800\) persons per square kilometre. Therefore option C is correct. The additional 500,000 people present during the day create the difference; dividing those additional people by 625 also gives \(500,000/625=800\).

Answer: C, 40%. The daytime density is being compared with the residential density, so the residential density is the reference value and must be used as the denominator. The absolute difference is 1,260 − 900 = 360 persons per square kilometre. The percentage increase is 360/900 × 100 = 40%. Therefore, daytime density is 40% higher. A and B are smaller than the calculated increase, while D is larger. The value 360 is an absolute difference, not a percentage. A frequent mistake is dividing by the newer value, 1,260, but percentage increase is measured from the original or reference value. The key rule is: percentage higher = difference ÷ reference value × 100.

The portion contains 48 percent of the population but only 16 percent of the area. Relative to the regional average, its density factor is (0.48 divided by 0.16 = 3). Therefore its density is three times the average density. With an average of 210 persons per square kilometre, the portion's density is (210 times 3 = 630) persons per square kilometre.

Option B is correct. The calculation works because population share divided by area share gives the portion's density relative to the whole region. It is not necessary to know the region's total population or total area. A value of 525 would represent a different ratio and does not reflect that 48 percent of people are concentrated in only 16 percent of the land.

Answer: B, 4.125 times. Density is population share divided by area share when both are expressed relative to the same total. For the first portion, relative density = 0.36/0.12 = 3. The remainder contains 64% of the population and 88% of the area, so its relative density is 0.64/0.88 = 0.72727 approximately. Now compare the first portion with the remainder: 3/0.72727 = 4.125. Equivalently, use (0.36/0.12) ÷ (0.64/0.88). Thus the first portion is 4.125 times as dense. A, C and D result from incorrect comparisons or rounding. The key warning is that the comparison is with the remainder, not with the overall regional average.

Answer: C, 2.4 times. Let the total population be P and the total area be A. The regional average density is P/A. The selected area contains 72% of the population, or 0.72P, and 30% of the area, or 0.30A. Its density is 0.72P/0.30A. Compare this with the regional average: (0.72P/0.30A)/(P/A) = 0.72/0.30 = 2.4. Therefore, the selected area has a density 2.4 times the regional average. A, B and D do not equal the correct population-share-to-area-share ratio. Exact totals are unnecessary because the common P and A cancel. Memory cue: density relative to the average equals population share divided by area share.

Population density tells us how many people live in each square kilometre. When density and area are known, total population is found by multiplying them: \(P=D\times A\). The units also confirm the operation, because persons per square kilometre multiplied by square kilometres leaves persons. This is the inverse of calculating density from population and area.

Using the given values, \(P=768\times1250\). Since \(1250=1000+250\), the product is \(768000+192000=960000\) people. Equivalently, multiplying by 1250 gives the same result directly. Thus the total population is 960,000, so option C is correct. The other values do not result from the stated density-area multiplication.

Population density is calculated by dividing population by area. The answer must therefore be expressed as persons per square kilometre. A useful check is to multiply the proposed density by the area and see whether it returns the given population. This also helps prevent errors caused by zeros or by confusing division with multiplication.

Calculate \(1170000\div1560\). Since \(1560\times750=1170000\), the quotient is 750 persons per square kilometre. Equivalently, multiplying 750 by the stated area reproduces the population exactly. Therefore, option C is correct. The nearby values 700 and 780 do not satisfy this check.

When area remains fixed, any change in population changes density in direct proportion. First calculate the original population from the original density and area. Add the stated increase, then divide by the unchanged area to obtain the new density.

The original population is \(625\times1440=900000\). After an increase of 90000, it becomes \(990000\). Since the area remains 1440 square kilometres, the new density is \(990000\div1440=687.5\) persons per square kilometre. Therefore, option C is correct. The unchanged area means no adjustment to the denominator is needed.

To find the new density, first find the original population because density equals population divided by area. The original population is obtained by multiplying the original density by the area. After people migrate out, subtract the migrants from this population. Since the area does not change, divide the remaining population by the same area.

Original population is \(520\times2500=1300000\). After 260000 people leave, the population is \(1300000-260000=1040000\). The new density is \(1040000\div2500=416\) persons per square kilometre. Therefore, option C is correct. The unchanged area is important because only the population changes.

Population is found by multiplying density by area. Therefore, when both density and area change, their percentage effects must be combined through multiplication rather than simple addition. Density changes from 480 to 600, so its multiplier is \(600/480=1.25\). An area increase of 15% gives an area multiplier of 1.15.

The new population multiplier is \(1.25\times1.15=1.4375\). This means the new population is 143.75% of the original population, so the increase is 43.75%. Therefore option C is correct. Adding 25% and 15% would give 40%, but that misses the interaction: the 15% area increase also applies to the already increased density.

Answer: C, 31.25%. Use density = population/area, so area = population/density. The population multiplier is 1.05 because population rises by 5%. The density multiplier is 560/700 = 0.8. Therefore, the area multiplier is 1.05/0.8 = 1.3125. This means the new area is 131.25% of the old area, so the increase is 131.25% − 100% = 31.25%. A, B and D do not satisfy the density equation. The result makes sense: population increased, yet density fell, so the area had to expand by an even larger proportion. A useful memory cue is to rearrange the formula first: area multiplier = population multiplier ÷ density multiplier.

Population, density, and area are connected by \(\text{Density}=\frac{\text{Population}}{\text{Area}}\), or \(\text{Area}=\frac{\text{Population}}{\text{Density}}\). Percentage changes must be handled with multipliers, not by simply subtracting the two percentages. A 22 percent fall leaves 78 percent of the original population, while a 4 percent rise makes density 104 percent of its original value.

Therefore the area multiplier is \(0.78\div1.04=0.75\). The new area is 75 percent of the old area, so it has decreased by 25 percent. Hence option A is correct. Treating the area change as exactly 22 percent would ignore the separate change in density and would not preserve the population-density-area relationship.

Answer: B, 25%. Let P be the population, A the total municipal area, and L the land area. Gross density is P/A = 420, while land-only density is P/L = 560. Dividing these equations gives (P/A)/(P/L) = L/A = 420/560 = 0.75. Thus land occupies 75% of the total municipal area. Water bodies occupy the remainder: 100% − 75% = 25%. A, C and D do not follow from the ratio of the two densities. The same population is used in both densities, which is why it cancels. A useful warning is not to subtract 420 from 560; the difference between densities is not directly the percentage of water. The correct approach is to compare the reciprocal area components through the shared population.

Arithmetic density uses the whole area, while habitable-area density uses only the part of the area where people can live. Let population be P, total area be A, and habitable area be H. The given relationships are \(P/A=225\) and \(P/H=375\). Dividing the first density by the second removes population and compares the areas.

Thus, \(H/A=225/375=0.60\). Converting this decimal to a percentage gives 60 percent. Therefore, option B is correct. The higher density based on habitable land is expected because the same population is concentrated in only part of the total area.

Arithmetic density uses total population divided by total area, written as \(P/A\). Cultivated-land density uses the same population divided by cultivated land, written as \(P/C\). If cultivated-land density is greater, the cultivated land must be a smaller part of the total area, because the same population is being divided by a smaller denominator.

Given \(P/C=2.5(P/A)\), canceling the common population gives \(A/C=2.5\). Therefore, \(C/A=1/2.5=0.4\). Converting 0.4 into a percentage gives 40 percent. Thus cultivated land makes up 40 percent of the total area, so option B is correct.

Arithmetic density uses the whole area, whereas cultivated-land density uses only the land that is cultivated. If cultivated land is 24 percent of the total area, it is represented by \(0.24A\), where A is total area. The same population is therefore concentrated over a smaller denominator, so the cultivated-land density is higher.

Starting with arithmetic density 192, the required density is \(192\div0.24=800\) persons per square kilometre of cultivated land. Hence option C is correct. This does not mean the total population changed; only the reference area changed from all land to cultivated land.

Arithmetic density is population divided by total area, while cultivated-land density is the same population divided by cultivated area. Let population be \(P\), total area be \(A\), and cultivated area be \(C\). The two given relationships are \(P/A=140\) and \(P/C=700\). Dividing the first density by the second removes population and gives \(C/A\), the cultivated share of total area.

Thus \(C/A=140/700=0.20\). Converting 0.20 to a percentage gives 20 percent, so option B is correct. Cultivated-land density is higher because the same population is measured over only the cultivated portion, which is smaller than total area. The result does not mean that 20 percent of people are cultivated; it refers to the area share.