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IGCSE Environmental Management Paper 2 — Topic 2 Land — Challenge

Environmental Management in Context — Topic 2 Challenge — source led
50 minutes
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This paper covers the whole of Topic 2. The questions are not in syllabus order and they are not one question per sub-topic. Material from 2.1 to 2.3 is mixed across every question, exactly as Cambridge mixes it, so do not expect Question 1 to be about 2.1.

Instructions

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Question 1 — Source A, soil loss and slope angle
Total: 10 marks
SOURCE A — Soil loss from a maize field, measured on plots of different slope angle
0 10 20 30 40 50 60 70 4 1 2 11 3 5 26 7 10 45 12 15 68 18 20 Soil loss / tonnes per hectare per year Slope angle / degrees bare soil, ploughed up and down the slope contour ploughed, cover crop kept
Two plots were laid out at each of five slope angles on the same farm and the soil lost from each was weighed over one year. One plot of each pair was left bare and ploughed up and down the slope; the other was contour ploughed and a cover crop was kept on it between harvests. Everything else — the soil, the rainfall and the crop — was the same for both plots.
(a)(i)[1]
Using Source A, identify the slope angle at which soil loss from the bare plot is greatest, and state that soil loss.
Model Answer — (a)(i)
20 degrees, 68 tonnes per hectare per year
both the angle and the figure with its unit are needed for the one mark
⚠ If you missed marks here: an identify-from-a-source question wants the figure as well as the label, and the unit as well as the number. "20 degrees" alone, or "68" alone, is half an answer. Read the key too: the question asks for the bare plot, which is the first bar of each pair.
(a)(ii)[3]
Using Source A, describe how soil loss from the bare plots changes as the slope angle increases. Support your answer with figures from the chart.
Model Answer — (a)(ii)
soil loss increases as slope angle increases (a positive relationship)
the increase is not in proportion to the angle: it gets steeper, so each extra degree of slope costs more soil than the one before
figures quoted correctly with units, for example it rises from 4 to 68 tonnes per hectare per year between 2 and 20 degrees, which is 17 times as much for 10 times the angle
accept any correctly read pair of figures for the third mark
⚠ If you missed marks here: a describe-the-trend question is marked in three layers and most candidates stop after the first. Layer one is the direction (it goes up). Layer two is the shape — here the rise gets steeper, so saying "it increases steadily" or "it increases in proportion" actually loses that mark, because the data say otherwise: ten times the angle gives seventeen times the soil loss. Layer three is a quoted figure with its unit. Get into the habit of writing "from X to Y" every time you describe a trend.
(b)[3]
Using Source A, calculate the percentage reduction in soil loss achieved on the 15 degree slope by contour ploughing and keeping a cover crop. Show your working.
Model Answer — (b)
M1 reads both values correctly and finds the difference: 45 − 12 = 33 tonnes per hectare per year
M1 divides the difference by the original value and multiplies by 100: (33 / 45) × 100
A1 73.3% (accept 73% or 73.3%)
a correct method with one misread figure still earns the two method marks
⚠ If you missed marks here: the mark most often thrown away is the second: the denominator must be the original value, the 45 from the bare plot, not the 12 it fell to and not the difference. Dividing by 12 gives 275%%, which should look wrong immediately — a reduction cannot exceed 100%%. Show the subtraction and the division on separate lines, because both are worth a mark even if the arithmetic at the end slips, and keep the unit attached to the figures you read off the chart.
(c)[3]
Source A shows that farming on steep slopes is a cause of soil erosion. Explain why, referring to the data.
Model Answer — (c)
water runs faster downhill on a steeper slope, because more of gravity acts along the slope
faster moving water has more energy, so it dislodges and carries more particles and larger particles away
gravity also moves loosened soil directly down the slope, and the data support this: 68 tonnes per hectare per year at 20 degrees against 4 at 2 degrees, 17 times as much
accept "run-off has less time to infiltrate on a steep slope, so more of the rain runs over the surface"
⚠ If you missed marks here: the question says explain why and referring to the data, so it needs both a mechanism and a figure. Answers that only restate the chart ("the steeper it is the more soil is lost") describe rather than explain and cap at one mark. The mechanism is about the speed of the water: faster water carries more, and larger, particles. Notice too that Source A is a fair test — soil, rainfall and crop were held the same — which is what lets you attribute the difference to the slope at all.
Question 2 — Source B, climate and the growing season
Total: 10 marks
SOURCE B — Mean monthly rainfall and temperature at two farming locations
MonthLocation XLocation Y
rainfall / mmmean temperature / °Crainfall / mmmean temperature / °C
January522703
February824554
March3527607
April60295510
May150306014
June240286517
July260267019
August210267519
September110276516
October30268012
November1024857
December422804
Total for the year1122–820–
A crop grown at both locations needs a mean monthly temperature of at least 10 °C and at least 25 mm of rainfall in a month before it will grow. A month counts as part of the growing season only if both conditions are met.
(a)[1]
Using Source B, identify the wettest month at Location X and state its rainfall.
Model Answer — (a)
July, 260 mm
both the month and the figure with its unit are needed for the mark
⚠ If you missed marks here: read down the correct column. Location X and Location Y each have two columns and the rainfall column comes first for both, so it is easy to answer with a temperature by mistake. Give the unit: 260 without "mm" is an incomplete answer in a data question.
(b)[3]
Using Source B and the condition given beneath it, determine the length of the growing season at each location. State which months are included.
Model Answer — (b)
Location X: 8 months, March to October
Location Y: 7 months, April to October
the method is stated or shown: a month counts only if the temperature is at least 10 °C and the rainfall is at least 25 mm
at X every month is warm enough, so rainfall is the limiting condition; at Y every month from April onwards is wet enough, so temperature is the limiting condition
⚠ If you missed marks here: the word that carries the marks is and. At X the temperature never falls below 22 °C, so it is the rainfall that decides; at Y the rainfall never falls below 55 mm, so it is the temperature that decides. Candidates who apply only one condition get 12 months for X and 7 for Y, and lose two of the three marks. Work down the two columns together, month by month, and mark the months that pass both.
(c)[2]
Describe the pattern of rainfall through the year at Location X. Support your answer with figures.
Model Answer — (c)
it is strongly seasonal: a distinct wet season in the middle of the year and a distinct dry season at each end of it
figures quoted correctly, for example a peak of 260 mm in July falling to only 4 mm in December, and a year total of 1122 mm
accept "almost all the year’s rain falls between May and September" as the first mark
⚠ If you missed marks here: the syllabus objective here is only that some areas have wet and dry seasons, but the command word is describe with figures, so a bare statement that "it rains a lot in summer" scores nothing. Quote the extremes and quote the unit. It is worth noticing that Location X receives more rain in the year than Location Y (1122 mm against 820 mm) yet has months with almost none — the annual total hides the pattern, and the pattern is what limits farming.
(d)[4]
Explain two ways in which the rainfall pattern at Location X limits crop growth. For each one, name a strategy from Topic 2.2.3 that would help, and say how it works.
Model Answer — (d)
The dry season. From November to February there is almost no rain, so there is too little water for photosynthesis and to keep cells turgid; the crop wilts or cannot be grown at all, and the growing season is cut to eight months.
A matched strategy. Rainwater harvesting collects and stores rain from roofs and surfaces during the wet season for use in the dry one; or trickle irrigation delivers stored water straight to the base of each plant so almost none is lost to evaporation; or mulch spread on the surface shades the soil and holds moisture in.
The wet season. In June and July over 240 mm falls in a month, so the soil waterlogs and the pore spaces fill with water, excluding air so roots cannot respire; heavy rain also leaches nitrate and other soluble ions out of the soil and erodes any ground left bare.
A matched strategy. Adding organic matter binds particles into crumbs and leaves larger pores so water soaks in and drains rather than standing; or maintaining vegetation cover means leaves intercept the rain and roots take up water so less runs off; or contour ploughing and bunds hold water back so it infiltrates instead of running off and carrying soil.
one mark for each limitation with its mechanism, one mark for each matched strategy with how it works; the strategy must fit the limitation it is paired with
⚠ If you missed marks here: Naming a strategy on its own scores one mark and no more, and only if it is matched to the right problem. "Irrigation" answers the dry season and does nothing for a waterlogged field, so pairing it with the wet season loses the mark even though irrigation is a genuine syllabus strategy. The other frequent loss is treating both halves of the year as the same problem: a strongly seasonal climate causes trouble at both extremes, drought at one end and waterlogging, leaching and erosion at the other, and the question asks for two different ways.
Question 3 — Source C, two ways of growing the same crop
Total: 10 marks
SOURCE C — Two neighbouring maize farms on the same soil, measured in 2026
MeasurementFarm P
maize every year,
inorganic fertiliser only		Farm Q
four-year rotation
with a legume, manure used
Maize yield in 2020 / tonnes per hectare8.48.1
Maize yield in 2026 / tonnes per hectare6.38.6
Nitrogen fertiliser applied / kg per hectare per year210130
Manure applied / tonnes per hectare per year012
Insecticide applications per year41
Nitrate in the stream at the field edge / mg per dm³11.24.1
Depth of soil lost each year / mm1.80.3
Farm P has grown maize in the same fields every year since 2020 and uses inorganic NPK fertiliser only. Farm Q grows maize as part of a four-year rotation that includes a legume, and spreads manure. The two farms lie side by side on the same soil and receive the same rainfall.
(a)[3]
Using Source C, calculate the percentage change in maize yield on Farm P between 2020 and 2026. Show your working.
Model Answer — (a)
M1 finds the change: 6.3 − 8.4 = −2.1 tonnes per hectare
M1 divides by the 2020 value and multiplies by 100: (2.1 / 8.4) × 100
A1 -25.0%, that is a decrease of 25%
the answer must be identified as a fall; 25% written without a sign or the word "decrease" loses the accuracy mark
⚠ If you missed marks here: two traps in one question. First, the denominator is the starting value, 8.4, not the value it fell to. Second, a percentage change has a direction, so say "a decrease of 25%%" or write it as −25%%. It is worth doing the same sum for Farm Q as a check on your method: 8.1 to 8.6 is an increase of about 6%%, and the contrast between the two is the point of the source.
(b)[3]
Farm Q applies less nitrogen fertiliser than Farm P, yet its yield is higher. Using Source C, explain how the practices used on Farm Q make this possible.
Model Answer — (b)
the legume in the rotation restores nitrogen to the soil naturally, so less has to be supplied as fertiliser
the manure adds organic matter, which improves soil structure and water retention and releases nitrate, phosphate and potassium ions slowly as it decomposes, so less is leached away and more reaches the crop
the rotation itself means different crops remove different ions in different proportions so no single nutrient is stripped out, and crop-specific pests cannot build up — which shows in the source as 1 insecticide application a year against 4
accept the lower soil loss (0.3 mm against 1.8 mm a year) as evidence that Farm Q is keeping its topsoil and therefore its nutrients
⚠ If you missed marks here: the question says using Source C, so an answer built entirely from memory with no figure in it is capped. Each of the three marks is a different mechanism, so writing "rotation is better for the soil" three times in three ways scores once. The manure mark is the one most often missed: inorganic fertiliser supplies ions but adds no organic matter, so it does nothing for structure, water retention or the soil organisms — which is exactly what Farm P has been missing for six years.
(c)[2]
Explain why the stream beside Farm P contains more nitrate than the stream beside Farm Q, and state one consequence for life in that stream.
Model Answer — (c)
Farm P applies 210 kg of nitrogen per hectare a year against Farm Q’s 130, and more than the crop can absorb; the excess dissolves in rainwater and is leached out of the soil into the stream, giving 11.2 mg per dm³ against 4.1
consequence: nutrient enrichment feeds an algal bloom that blocks the light so the plants beneath die, and the decomposers that break them down use up the dissolved oxygen, so fish and other aquatic animals suffocate (eutrophication)
accept a shortened chain for the second mark provided it ends at the dissolved oxygen
⚠ If you missed marks here: the first mark needs the word leaching and a figure; the second needs the chain to finish in the right place. Fish do not die because the water turns green and they do not die because the algae are poisonous — they die because decomposers respiring aerobically have stripped the dissolved oxygen out of the water. Note also why the manure on Farm Q leaches less: it releases its ions slowly as it decomposes, so there is less soluble nitrate sitting in the soil waiting for the next storm.
(d)[2]
The owner of Farm P says that growing one crop every year is still the right choice because it produces cheap food. Using Source C, evaluate that claim.
Model Answer — (d)
in support: a monoculture lets the whole farm be sown, sprayed and harvested with one set of specialised machinery at one time, which cuts the cost per tonne, and cheap food matters most to the people with the least money; Farm P did yield 8.4 t per hectare in 2020, slightly more than Farm Q’s 8.1
against, with a judgement: the source shows the advantage has gone. Farm P’s yield has fallen 25% while Farm Q’s has risen, and Farm P is buying 210 kg of nitrogen and 4 insecticide applications a year to get it, so its cost per tonne is now rising, not falling; it is also losing soil 6 times as fast (1.8 mm a year against 0.3), which makes the fall permanent. A reasonable judgement is that the claim was true in 2020 and is no longer true in 2026.
one mark for a point on each side, or one mark for one supported point and one for a judgement that follows from it; a one-sided answer caps at one mark
⚠ If you missed marks here: Evaluate means weigh, so an answer that only attacks the monoculture is marked as incomplete, not as principled. "Intensive" and "monoculture" describe how a farm operates; they are not verdicts. The honest answer here concedes the real advantage — one crop, one set of machines, lower cost per tonne, and cheaper food — and then shows from the source that six years of it have raised the inputs, cut the yield and taken 1.8 mm of topsoil a year, so the economics have turned. Topsoil forms at a few millimetres per century, which is why the last part of that is not recoverable.
Question 4 — Source D, reading a farm for signs of erosion
Total: 10 marks
SOURCE D — A hillside farm, showing four sites W, X, Y and Z
contour interval 25 m 250 225 200 175 150 125 100 Y W direction of ploughing X Z river flow village prevailing wind KEY contour line, height in metres forest still standing cleared and ploughed land field left bare after harvest river bar of deposited silt
The land rises to a summit in the north-east of the map. Site W is a slope that has been cleared of trees and ploughed in the direction shown. Site X is a flat field that has been left bare since the harvest. Site Y is the forest that is still standing. Site Z is a bar of silt that has built up in the river. The village lies downstream of Z.
(a)[2]
Using Source D, identify the site at which soil is most likely to be eroded by wind, and give one reason from the map.
Model Answer — (a)
site X
reason: it is bare (left with no vegetation cover after harvest) and flat and dry, and it lies open to the prevailing wind with no trees or hedge between the wind and the field
do not accept "it is next to the river" or "it is far from the forest" without the link to shelter
⚠ If you missed marks here: wind erosion needs three things and the map shows all three at X: dry, loose, fine particles, no vegetation to bind them, and nothing to slow the air. Site W is losing more soil overall, but it is losing it to water running down the slope, so it is the wrong answer to this question. Read the key and the wind arrow before choosing, and quote what you read.
(b)[3]
Explain how the direction of ploughing at site W increases soil erosion, and state the change the farmer should make.
Model Answer — (b)
the furrows run up and down the slope, that is straight down the line of steepest descent, so each furrow becomes a channel
water is guided straight downhill instead of being held back; it gathers speed all the way down and faster water carries more and larger particles away, and less water has time to infiltrate
the change: contour ploughing — plough along the contours, across the slope, so every furrow lies horizontally and acts as a small barrier holding water back and letting it soak in
accept terracing or bunds as an additional strategy, but the direct correction of the ploughing direction is contour ploughing
⚠ If you missed marks here: the strategy alone is one mark: writing "contour ploughing" and stopping scores one out of three, because the command word is explain. The two mechanism marks are what the current furrows do (channel the water downhill so it gathers speed) and what the new ones would do (lie across the slope and hold the water so it infiltrates). It is also worth saying that this is the cheapest strategy on the whole syllabus — it costs nothing beyond driving in a different direction — because that is an evaluation point examiners reward.
(c)[3]
The silt bar at Z is growing and the village downstream floods more often than it used to. Explain how the farming at W and X has caused this.
Model Answer — (c)
soil eroded from the cleared slope at W and the bare field at X is washed (or blown) into the river and carried downstream
where the current slows, the sediment settles out on the bed — the silting seen at Z — so the river bed rises and the channel holds less water than before
a given rainfall therefore fills the channel sooner and the river overtops its banks, so the village floods more often; suspended sediment also blocks light and smothers the habitats of river organisms
the three marks are the three linked stages: soil in, sediment settles and the bed rises, channel capacity falls so flooding increases
⚠ If you missed marks here: this is marked as a chain, and the middle link is the one that goes missing. "The soil goes in the river so the village floods" jumps straight from the first stage to the third and scores one. The step that earns the second mark is that the deposited sediment raises the bed and reduces the capacity of the channel. Use "so" between each stage — each "so" is a link the mark scheme is looking for. Notice this is also an impact felt by people who did not cause it, which is a useful evaluation point in a longer answer.
(d)[2]
Suggest one strategy for site W and a different one for site X, and in each case say how it reduces erosion.
Model Answer — (d)
Site W (water on a steep slope): terracing cuts the slope into level steps so water stops and soaks in and cannot build up speed, and soil that does move is caught on the step below; or maintaining vegetation cover, so roots bind the soil, leaves intercept the rain and the plants absorb water so less runs off; or bunds built across the slope to trap water and moving soil.
Site X (wind on flat bare ground): a wind break of trees or hedges planted across the prevailing wind slows the air before it reaches the field so it can no longer lift dry particles, and shelters the crop; or bunds, which stand above the surface and break the wind at ground level; or a cover crop kept on the field between harvests so the soil is never bare; or adding organic matter, which binds fine particles into larger crumbs that are heavier and harder to blow away.
one mark for each site, and the mechanism must be given; the two strategies must be different from each other and from the answer given in (b)
⚠ If you missed marks here: the two sites have different agents — water on the slope at W, wind on the flat at X — so the same strategy will not do for both, and offering terracing for a flat field wastes the mark. Work back from the sentence that unlocks the whole sub-topic: soil is lost when it is left bare and held when it is covered, and every strategy either puts a cover back on it, slows the water moving over it, or slows the wind across it. And again, the name alone is one half of the mark at best: the syllabus gives each strategy together with its reason, which means the mechanism is examinable.

Self-Assessment

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IGCSE Environmental Management 0680 — Paper 2 — Topic 2 Land — Challenge
For examination preparation only. Not an official Cambridge Assessment paper.