Population Density

When area remains unchanged, a change in density directly represents the change in population per unit area. The fall in density is \(900-750=150\) persons per square kilometre. Since every square kilometre now has 150 fewer people and the area is fixed, multiplying this fall by the area gives the total number of people who left.

Let the area be \(A\). Then \(150A=60000\), so \(A=\frac{60000}{150}=400\) square kilometres. Therefore, option B is correct. Checking confirms that the population loss across 400 square kilometres is \(150\times400=60000\).

Density is \(P/A\), where \(P\) is population and \(A\) is area. If both population and area are reduced by the same percentage, both parts of the ratio are multiplied by the same remaining fraction. A reduction of 32 percent leaves 68 percent, or 0.68, of each original value. Equal proportional changes in numerator and denominator cancel in the ratio.

The new density is \(\frac{0.68P}{0.68A}=\frac{P}{A}\). Therefore the density remains unchanged, and option C is correct. It does not decrease by 32 percent because the area also decreases by 32 percent; fewer people are being distributed over proportionally less land. The conclusion assumes both changes apply to the same region and occur in the stated equal proportion.

Density compares population with area, rather than looking at either quantity separately. If both population and area are multiplied by exactly the same factor, their ratio does not change. An increase of 75% means each quantity becomes 175% of its original value, or 1.75 times the original.

The new density is \(\frac{1.75P}{1.75A}=\frac{P}{A}\). Thus, the absolute number of people and the total area both increase, but the average number of people per square kilometre remains unchanged. Therefore, option C is correct. A 75% density increase would occur only if population rose while area did not rise equally.

A 125% population increase means the new population is 225% of the original, so its multiplier is 2.25. An 80% area increase means the new area is 180% of the original, giving a multiplier of 1.80. Since density is population divided by area, the density multiplier is \(\frac{2.25}{1.80}=1.25\). Thus density rises by 25%, so option B is correct.

It is important to use increase factors rather than compare 125 and 80 directly. The new density is 1.25 times the old density, and the extra 0.25 represents a 25% increase. The population grows proportionally more than the area, so density increases. The answer is not 45%, which would incorrectly subtract the two percentage increases.

Density is proportional to population divided by area. A population decrease of 45 percent leaves 55 percent of the original population, giving a factor of 0.55. An area decrease of 50 percent leaves half the original area, giving a factor of 0.50. The new density relative to the old density is therefore 0.55/0.50=1.10.

A factor of 1.10 means the new density is 110 percent of the original, so it has increased by 10 percent. The population falls, but the area falls more sharply, causing the remaining people to occupy a proportionally smaller area. Hence option B is correct. It would not be correct to subtract 45 percent directly from density, because the area also changes. The density would remain unchanged only if population and area fell by the same percentage.

To update density after percentage changes, convert each percentage into a multiplier. A 14% population increase makes population \(1.14P\), while a 20% area increase makes area \(1.20A\). Since density is population divided by area, the new density is the old density multiplied by \(1.14/1.20\).

The calculation is \(900\times\frac{1.14}{1.20}=900\times0.95=855\) persons per square kilometre. Therefore, option B is correct. Density decreases because area grows by a larger percentage than population. The result is not 900, since equal percentage growth did not occur.

When the area stays fixed, any percentage change in population produces the same percentage change in density. A decrease of 12.5 percent means that 87.5 percent of the original population remains. Since the area has not changed, density also becomes 87.5 percent of its original value. The given area is not needed for this percentage calculation, though it could be used to find the original population.

Twelve and a half percent is one-eighth. One-eighth of 1120 is \(1120 \div 8 = 140\). Subtracting this decrease gives \(1120 - 140 = 980\) persons per square kilometre. Therefore option C is correct. The area of 750 square kilometres remains relevant to the setting but does not alter the density percentage when it is unchanged.

Answer: C, 30 percent. The area remains unchanged, so the percentage change in density is the same as the percentage change in population. The population increase is 144,000. Relative to the original population, the increase is (144,000 ÷ 480,000) × 100 = 30%. Therefore density increases by 30%. To verify numerically, the original density is 480,000 ÷ 800 = 600 persons per square kilometre. New population is 624,000, so new density is 624,000 ÷ 800 = 780. The increase is 180, and 180 ÷ 600 × 100 = 30%. Option A, B, and D use incorrect percentages. Option C is correct because area is fixed and the population rises by 30%. Memory cue: if the denominator stays constant, the ratio changes by the same percentage as the numerator.

First find the original area because only population and density are given. From \(\text{Area}=\frac{\text{Population}}{\text{Density}}\), the original area is \(875000/700=1250\) square kilometres. The new population is 875,000 plus 105,000, and the new area is 1,250 plus 250.

Thus, new population is 980,000 and new area is 1,500 square kilometres. New density is \(980000/1500=653.33\) persons per square kilometre, approximately. Therefore, option C is correct. Rounding to one decimal place gives 653.3; the exact quotient is repeating, not exactly 653.3.

First determine the original area because the question gives population and density but not area. Using \(A=P\div D\), the original area is obtained. Then increase the population by 40% and the area by the stated 180 square kilometres, and divide the new population by the new area.

The original area is \(315000\div525=600\) square kilometres. The new population is \(315000\times1.40=441000\), and the new area is \(600+180=780\) square kilometres. Thus, new density is \(441000\div780\approx565.38\) persons per square kilometre, or about 565.4. Option C is correct.

Let the country's total population be \(P\) and total area be \(A\). Its average density is \(P/A=280\). The selected area contains 35% of the population and 14% of the area, so its density is \(0.35P/0.14A\). Relative to the national density, this is \(0.35/0.14=2.5\) times as large. Therefore the selected area's density is \(280\times2.5=700\) persons per square kilometre.

Option C is correct. The calculation compares the population share with the area share: because the population share is proportionally much larger than the area share, the local density exceeds the national average. The answer is not 560 or 650 because those do not result from the given ratio, and 840 uses an incorrect multiplier. The national average and percentages are sufficient; the actual totals are unnecessary.

Density equals population divided by area. Let the country's total population be \(P\) and total land area be \(A\). The national average density is \(P/A\). In the specified part, population is \(0.54P\) and land area is \(0.18A\), so its density is \(\frac{0.54P}{0.18A}=\frac{0.54}{0.18}\times\frac{P}{A}=3\frac{P}{A}\). Thus it is three times the national average.

Option C is correct. The useful shortcut is to divide the part's population share by its area share: \(54\div18=3\). The result is not 54 times or 18 times because both percentages must be compared as shares of the national totals. A larger population share concentrated in a smaller area produces above-average density.

To compare the two portions fairly, divide each population share by its area share. The first portion has 65 percent of the population in 25 percent of the area, while the second has 35 percent in 75 percent. A higher population share concentrated in a smaller area produces greater density.

The first portion has relative density \(0.65/0.25=2.6\). The remaining portion has relative density \(0.35/0.75\approx0.467\). Since 2.6 is much greater than 0.467, the first 25 percent of the area is denser. Therefore, option A is correct. The comparison is possible because both population and area shares are supplied.

The relationship between density, population, and area is \(P=D\times A\). Here the density is 937.5 persons per square kilometre and the area is 768 square kilometres. Thus total population is \(937.5\times768\). To calculate conveniently, write \(937.5=1875/2\), so the product becomes \(1875\times384\).

Now \(1875\times384=720000\). Therefore, the total population is 720,000 people and option C is correct. A quick check is that 937.5 is close to 1,000 and 768 is close to 800, so a result near 800,000 is reasonable; 720,000 is consistent with the exact multiplication. The other options do not equal the product.

Area can be found by rearranging the density formula. Since density equals population divided by area, area equals population divided by density. The decimal in the density does not change the method; it can be handled directly or removed by multiplying both numerator and denominator by the same number.

Here, \(A=506250\div562.5\). Multiplying numerator and denominator by 2 gives \(1012500\div1125=900\) square kilometres. A quick check confirms this because \(562.5\times900=506250\). Therefore, option C is correct; 800, 850 and 950 do not reproduce the given population.

Density changes according to the change in population relative to the change in area. A 24 percent population increase multiplies population by \(1.24\). A 7 percent area decrease leaves \(93\) percent of the original area, so area is multiplied by \(0.93\). The new density is the old density multiplied by the first factor divided by the second.

The calculation is \(600\times(1.24\div0.93)\). Since \(1.24\div0.93\approx1.3333\), the result is approximately \(600\times1.3333=800\). Therefore, option C is correct. The area decrease raises density because the same general amount of population is spread over a smaller area.

Answer: B, 80%. Population density means population divided by area: D = P/A. A decrease of 8% leaves 92% of the original population, so the population multiplier is 0.92. An increase of 15% makes the new area 115% of the old area, so the area multiplier is 1.15. Therefore, the new density compared with the old density is 0.92/1.15 = 0.80, or 80%. A is incorrect because it does not correctly combine both percentage changes. C ignores most of the effect of the area increase. D is also too high because population falls while area rises. The important memory cue is: for density, divide the population multiplier by the area multiplier; do not simply subtract or add the percentages.

Answer: B, 625 persons per square kilometre. Density is the ratio of population to area. Let the original population be P and the original area be A. The original density is P/A = 625. After a 44% increase, the new population is 1.44P and the new area is 1.44A. Thus the new density is 1.44P/1.44A = P/A = 625. The common factor 1.44 cancels because numerator and denominator increase in exactly the same proportion. A is wrong because density does not fall when both quantities grow equally. C and D are wrong because they assume that the population increase alone determines density and ignore the equal area increase. Memory cue: equal percentage changes in population and area leave density unchanged.

Answer: B, 20%. The area is unchanged, so population and density change by the same percentage. First find the fall in density: 2,250 − 1,800 = 450 persons per square kilometre. Now compare this fall with the original density, not the final density: 450/2,250 × 100 = 20%. Therefore, the population also decreased by 20%. A is too small, while C and D are too large. For example, a 25% decrease from 2,250 would give 1,687.5, not 1,800. The common confusion is using the final value as the denominator; percentage decrease is always measured from the original value. Fixed area is the key condition that allows the density percentage to represent the population percentage.

For a fixed population, density and area change in opposite directions. Since \(D=P/A\), constant population gives \(D_1A_1=D_2A_2\). Therefore, the ratio of the new area to the old area is \(A_2/A_1=450/600=0.75\). The new area is 75 percent of the original area.

A fall from 100 percent to 75 percent means a decrease of 25 percent. Equivalently, the percentage decrease is \((1-0.75)\times100=25\%\). Thus option B is correct. The 33.33 percent figure would describe the increase in density from 450 to 600, not the decrease in area, so it is not the required answer.

Answer: C, 30%. Since the area remains unchanged, any percentage increase in density represents the same percentage increase in population. The increase in density is 728 − 560 = 168 persons per square kilometre. Compare this increase with the original density: 168/560 × 100 = 30%. Hence the population increased by 30%. A and B underestimate the increase. D overestimates it; a 35% increase would produce 756 persons per square kilometre, not 728. The denominator must be the original value, 560, because the question asks how much the quantity increased from its starting level. Memory cue: with fixed area, density and population move together in the same proportion.

Answer: B, 720 persons per square kilometre. Density equals population divided by area. The population does not change, but a 50% area increase makes the new area 1.5 times the old area. Therefore the new density is 1,080/1.5 = 720 persons per square kilometre. The given population is not actually needed for the shortcut. To verify by the long method, original area = 1,296,000/1,080 = 1,200 square kilometres. New area = 1,200 × 1.5 = 1,800 square kilometres. New density = 1,296,000/1,800 = 720. A, C and D do not correctly account for the inverse effect of area. Memory cue: when population is fixed, density changes inversely with area.

The average density of the whole region is based on all its people and all its area. To compare a selected part with that average, use the fraction of the total population in that part divided by the fraction of the total area occupied by it. This comparison does not require the actual population or area values.

The selected area contains 42 percent of the population but only 14 percent of the area. Its density relative to the regional average is therefore \(0.42/0.14=3\). Thus people are concentrated there at three times the region’s average density, making option C correct. The result follows because the population share is three times the area share.

Let the country’s total population be \(P\) and total area be \(A\). National density is \(P/A=320\). The selected region contains 75 percent of the population and 30 percent of the area, so its density is \(0.75P\div0.30A\). This equals \((0.75/0.30)(P/A)=2.5\times320=800\) persons per square kilometre.

Therefore option C is correct. The selected area holds a larger share of people than of land, so its density must be higher than the national average; 800 is 2.5 times 320. The calculation uses percentages as proportions, not as numbers 75 and 30 without their common percent factor. Options A, B, and D do not equal the required relative-density factor of 2.5.

Population density is only an average: it tells us how many people occupy one unit of area. It does not show how much water, food, energy or usable land is available, how resources are distributed inside the region, or how much each person consumes. Technology and infrastructure can also allow a densely settled area to manage resources differently from another area with the same density.

Therefore option A is correct. Two regions may have equal people per square kilometre but very different soils, water supplies, incomes, consumption patterns and resource management systems. Density is useful as a starting indicator, but it cannot by itself measure complete resource pressure. The other options incorrectly deny that density includes population or area.