IGCSE Environmental Management Paper 2 — Environmental Management in Context
Environmental Management in Context — Mock 1
105 minutes
80
6
105:00
0680
Instructions
Answer all questions in the spaces provided.
Show all working for calculations.
Give non-exact answers to 3 significant figures unless the question says otherwise.
Your answers will be automatically graded when you submit.
Question Navigation
Question 1
Total: 12 marks
A student investigated water quality along a 7 km stretch of the River Kelu. She took samples at five sites, shown in Fig. 1.1. At each site she measured the nitrate concentration and the dissolved oxygen concentration of the water, and counted the mayfly nymphs collected in a standard three-minute kick sample. Mayfly nymphs are river invertebrates that need water with a high dissolved oxygen concentration. All five sites were sampled on the same day.
Site
Nitrate concentration / mg per dm³
Dissolved oxygen / mg per dm³
Number of mayfly nymphs
A
2.0
10.4
24
B
5.5
7.8
11
C
to be calculated
5.2
3
D
9.0
6.5
8
E
4.5
9.1
19
Table 1.1 Results of the river survey
(a)[2]
At site C the student took three separate nitrate readings from the same stretch of water: 11.8, 12.3 and 11.9 mg per dm³. Complete Table 1.1 for site C. Show your working.
Mark Scheme — 1(a)
M1 adds the three readings and divides by three: (11.8 + 12.3 + 11.9) ÷ 3 = 36.0 ÷ 3
A1= 12.0 mg per dm³
accept 12 or 12.0; the unit is already given in the table so it is not needed again
⚠ If you missed marks here: three readings from the same place are repeats, so they are averaged into one value — they are not three separate results to enter in the table. The middle value (12.3) is the median, not the mean; the mean here happens to be 12.0.
(b)[2]
Calculate the percentage change in dissolved oxygen concentration between site A and site C.
Mark Scheme — 1(b)
M1 divides the change by the starting value: (5.2 − 10.4) ÷ 10.4 × 100
A1= −50%, that is a decrease of 50%
the direction must be clear: −50%, or 50% decrease. A bare 50 is not enough
⚠ If you missed marks here: percentage change is always divided by the value you started from, which is site A. Dividing by 5.2 gives 100% and answers a different question. The word change also means you must say whether it went up or down.
(c)[3]
Describe the pattern in the number of mayfly nymphs along the river, and use the data in Table 1.1 to explain it.
Mark Scheme — 1(c)
1 pattern described with figures: numbers fall from 24 at A to a minimum of 3 at C, then rise again to 19 at E
2 links the mayfly numbers to the dissolved oxygen: dissolved oxygen is also lowest at C (5.2 mg per dm³) and rises again downstream, and mayfly nymphs need a high dissolved oxygen concentration
3 explains why the oxygen is low: sewage and nitrate from the farm drain cause nutrient enrichment / eutrophication OR decomposers break down the organic matter and use up the dissolved oxygen
a mark is also available for noting that nitrate is highest where mayfly numbers are lowest (12.0 at C against 3 nymphs)
⚠ If you missed marks here: "describe" wants the shape of the pattern with numbers from the table, and "explain" wants a because. Saying the river got more polluted describes nothing and explains nothing — the mayflies fell because the dissolved oxygen fell, and the oxygen fell because decomposers used it up.
(d)[2]
At site D the student recorded three dissolved oxygen readings: 6.4, 6.6 and 9.8 mg per dm³. Identify the anomalous reading and state what the student should do about it.
Mark Scheme — 1(d)
1 the anomalous reading is 9.8 mg per dm³ — it is far from the other two, which agree closely
2 any 1 from: repeat the measurement at site D to see whether 9.8 can be reproduced; exclude it from the mean and record why it was excluded; keep it in the record but do not use it; check the meter was rinsed and fully submerged and take further readings
the table value of 6.5 is the mean of 6.4 and 6.6 — the anomaly has already been left out
do not credit "delete it" on its own: results are never simply deleted without a reason being recorded
⚠ If you missed marks here: an anomaly is a reading that does not fit the repeats taken at the same place, not simply the largest number in the whole table. Site A reads 10.4 and that is not anomalous, because it is a different site. The action mark needs a scientific response — repeat and investigate — not just crossing the number out.
(e)[3]
The student wrote this conclusion. "The village sewage outfall is the cause of the fall in the mayfly population of the River Kelu."Explain why the data collected do not fully support this conclusion.
Mark Scheme — 1(e)
any 3 from:
there are two possible sources, not one — the farm drain enters between B and C, so its effect cannot be separated from the outfall
the data support this: below the outfall alone (A to B) mayfly numbers fell from 24 to 11, but the largest fall (11 to 3) happens after the farm drain
the nitrate rise from 5.5 to 12.0 mg per dm³ between B and C points to fertiliser leaching as well as sewage
all sites were sampled on one day only, so the result may not be typical of other days, seasons or flow conditions
only one kick sample was taken per site, with no repeats, so the counts may not be representative
other factors that were not measured could affect mayfly numbers, for example temperature, pH or flow rate
the data show a correlation between low oxygen and low mayfly numbers; they do not by themselves establish which input caused it
⚠ If you missed marks here: the conclusion is not wrong, it is unsupported — and those are different criticisms. The examiner wants you to show that the method cannot tell the two inputs apart, and to quote the numbers that show it. Writing that sewage is bad for rivers adds nothing, because that is not what is being questioned.
Question 2
Total: 13 marks
A class investigated whether walkers using a footpath across an area of grassland affect the plants growing beside it. They laid a 20 m tape at right angles to the edge of the path and placed a 0.25 m² grid quadrat at every 2 m mark, as shown in Fig. 2.1. In each quadrat they estimated the percentage cover of grass and the percentage of bare ground. Their results are in Table 2.1.
Distance from the footpath / m
0
2
4
6
8
10
12
14
16
18
20
Grass cover / %
8
21
34
46
58
67
74
79
81
82
82
Bare ground / %
76
58
44
32
22
15
9
5
4
3
3
Table 2.1 Percentage cover along the transect
(a)(i)[2]
State the independent variable and the dependent variable in this investigation.
Mark Scheme — 2(a)(i)
1 independent variable: distance from the edge of the footpath (in metres)
2 dependent variable: percentage cover of grass OR percentage of bare ground
do not credit "trampling" for the independent variable — trampling was never measured, only distance was
⚠ If you missed marks here: the independent variable is the one you choose and change, and here that is where you put the quadrat. The dependent variable is the one you measure as a result. A quick check: the independent variable is almost always the heading of the first row or column of the results table.
(a)(ii)[3]
Identify one variable that should be kept constant in this investigation, and explain why it must be kept constant.
Mark Scheme — 2(a)(ii)
1 any 1 from: the size of the quadrat (0.25 m²); the same person estimating cover, or the same counting rule; the time of day and the weather; the slope and aspect of the ground; the soil type; the same day
2 explains what would otherwise happen, for example: a larger quadrat further from the path would cover more ground, so the percentage figures could not be compared between positions
3 states the consequence for the conclusion: a change in cover could then have been caused by the change in quadrat rather than by the distance from the path, so it would be impossible to tell which caused it
full marks need the chain: what is kept constant → what would go wrong → why the conclusion would be unsafe
⚠ If you missed marks here: naming a control variable is worth one mark out of three. The other two are for the reasoning, and "so it is a fair test" is not reasoning — it is the phrase that replaces it. Say what would change, and then say why that would make it impossible to know what caused the difference.
(b)[3]
The class used systematic sampling rather than random sampling. Justify this choice for this investigation, and state one limitation of systematic sampling.
Mark Scheme — 2(b)
1 identifies that the class is looking for a change with distance, so quadrats must be spread evenly along a line (a belt or line transect) from the path outwards
2 any 1 from: fixed 2 m intervals guarantee a reading at every distance, including close to the path; random positions could by chance cluster in one zone and miss the gradient altogether; the pattern in Table 2.1 (8% rising to 82%) is only visible because the positions are evenly spaced
3 limitation, any 1 from: fixed intervals can coincide with a repeating feature in the ground and give a biased picture; a systematic sample is not a random sample, so it should not be used to estimate the average cover of the whole field; the person choosing the line still chooses where the transect starts
accept the reverse argument for mark 3: random sampling would be the better choice if the aim were the overall abundance of grass in the field rather than the change with distance
⚠ If you missed marks here: "justify" means say why this method suits this aim, not recite the definition of systematic sampling. The aim here is a gradient, and a gradient needs evenly spaced points. The third mark is for admitting a genuine weakness — a one-sided answer is capped.
(c)[3]
The class also wants to compare the number of ground-dwelling beetles at 0 m and at 20 m from the footpath. Name the most appropriate sampling technique and justify your choice.
Mark Scheme — 2(c)
1 pitfall trap
2 justification, any 1 from: beetles walking on the soil surface fall into the sunken container and cannot climb out; the trap collects animals continuously so it does not depend on the students being present; it catches animals that are active at night
3 rejects an alternative with a reason, any 1 from: a sweep net samples invertebrates living in the vegetation, not on the ground; a pooter can only take an animal you have already found by hand-searching, so the number caught depends on the searcher; aerial photography and drones cannot resolve an animal of this size
for a valid comparison the traps must be the same size, use the same bait or none, and be left for the same length of time at both distances — credit this as an alternative to mark 3
⚠ If you missed marks here: two of the three marks are for the justification, so naming the trap and stopping loses most of the question. Match the technique to how the animal moves: on the ground means a pitfall trap; in the vegetation means a sweep net; already visible and needs collecting gently means a pooter.
(d)[2]
Identify one risk to the students during this fieldwork and describe a precaution that reduces it.
Mark Scheme — 2(d)
1 a specific risk, any 1 from: sunburn or heat exhaustion in open grassland; a tick or insect bite; contact with a stinging or irritant plant; tripping on uneven ground; contamination of hands with soil bacteria; being struck by a cyclist or vehicle using the path
2 a precaution that matches the risk, any 1 from: sun hat, sunscreen and drinking water; cover arms and legs and check for ticks afterwards; wear gloves when handling vegetation and soil; wear closed walking shoes and watch the ground; wash hands before eating; work to one side of the path and keep the tape flat
the precaution must reduce the risk that was named; a mismatched pair scores one mark only
⚠ If you missed marks here: "be careful" and "wear suitable clothing" are not precautions, because they do not say what you would actually do. Name the hazard specifically, then name the action that reduces it — and make sure the two match each other.
Question 3
Total: 13 marks
Marenda is a country that generates all of its electricity within its own borders. Table 3.1 shows how much electricity Marenda generated from each energy resource in 2015 and in 2025. Fig. 3.1 shows the carbon dioxide emitted by its electricity generation over the same period.
Energy resource
Electricity generated in 2015 / TWh
Electricity generated in 2025 / TWh
coal
120
60
natural gas
40
50
nuclear
10
6
hydro-electric
20
24
wind
8
40
solar
2
20
total
200
200
Table 3.1 Electricity generation in Marenda
(a)[2]
Calculate the percentage of Marenda’s electricity that was generated from renewable energy resources in 2025.
Mark Scheme — 3(a)
M1 selects the renewable resources only — hydro-electric, wind and solar: 24 + 40 + 20 = 84 TWh
A184 ÷ 200 × 100 = 42%
nuclear is not renewable — it uses uranium, which is a finite resource. Including it gives 45% and scores zero for A1
⚠ If you missed marks here: the trap in this question is nuclear. It emits very little carbon dioxide, but low-carbon and renewable are not the same thing — uranium is mined and is finite, so nuclear is classified as non-renewable. Natural gas is a fossil fuel and is non-renewable too.
(b)[2]
Use Fig. 3.1 to state: (i) the carbon dioxide emissions in 2020; (ii) the first year in which emissions fell below 120 million tonnes.
Mark Scheme — 3(b)
1110 million tonnes (accept 108–112)
22019 — the 2018 point is above the 120 gridline and the 2019 point is below it
read across from the point to the axis using the minor gridlines, which are every 10 million tonnes
⚠ If you missed marks here: the labelled gridlines are 20 apart but there is a fainter line every 10, so 110 sits exactly on a line rather than between two. For part (ii), "first year below 120" means the first point underneath that line, not the last one above it.
(c)[2]
Calculate the percentage change in carbon dioxide emissions between 2015 and 2025.
Mark Scheme — 3(c)
M1 reads 150 for 2015 and 90 for 2025 and divides the change by the 2015 value: (90 − 150) ÷ 150 × 100
A1= −40%, a decrease of 40%
⚠ If you missed marks here: the fall is 60 million tonnes, but the question asks for a percentage, and 60 out of 150 is 40%. Dividing by the 2025 value instead gives 66.7%, which measures the change against where you finished rather than where you started.
(d)[3]
Use both Table 3.1 and Fig. 3.1 to explain the change in Marenda’s emissions between 2015 and 2025.
Mark Scheme — 3(d)
1 total generation is unchanged at 200 TWh, so the fall in emissions is not caused by generating less electricity — the mix has changed
2 coal generation halved, from 120 to 60 TWh, and combustion of coal releases carbon dioxide
3 renewable generation rose from 30 to 84 TWh (15% to 42%), and hydro, wind and solar release no carbon dioxide as they generate
a mark is also available for noting that natural gas rose from 40 to 50 TWh, which offsets part of the fall because gas is also a fossil fuel
⚠ If you missed marks here: the strongest point in this answer is the one candidates skip: the total is identical in both years. That is what rules out "they used less electricity" and forces the explanation onto the change in mix. Always check the totals row before explaining a change.
(e)[4]
A newspaper printed this claim. "Marenda cut its emissions by an average of 6 million tonnes each year from 2015 to 2025. At that rate the country will reach zero emissions from electricity by 2040."Using Fig. 3.1, evaluate this claim.
Mark Scheme — 3(e)
1 confirms the average is correct: (150 − 90) ÷ 10 = 6 million tonnes per year, and 90 ÷ 6 = 15 more years, which is 2040
2 identifies that the line is not straight — the fall was about 8 million tonnes per year in the first five years but only about 3 million tonnes per year in the last three (100 to 90 between 2022 and 2025)
3 uses the recent rate instead: at about 3 million tonnes per year, 90 ÷ 3 would take a further 30 years, well beyond 2040 — so the average across the whole period flatters the trend
4 any 1 from: extrapolating 15 years beyond the last data point assumes the trend continues, which the graph cannot show; the remaining emissions come from coal and gas plants that may be the hardest to replace; demand could rise; a new policy could speed the fall up; reaching exactly zero is a stronger claim than the data support
an answer that argues the claim could be met, provided it gives a reason from the data (for example that the 2021 to 2022 drop shows a faster rate is achievable), can take mark 4
⚠ If you missed marks here: the arithmetic in the claim is correct, so simply saying it is wrong scores nothing. What is wrong is the assumption that a mean rate taken over ten years describes the next fifteen, when the graph clearly flattens. Whenever a claim extrapolates, check whether the recent gradient matches the average gradient.
Question 4
Total: 14 marks
A research station tested four ways of managing a sloping field. Twelve plots, each 5 m by 2 m and each on the same 10° slope, were prepared. Three plots were given each treatment: A bare soil · B contour ploughing · C a mulch of crop residue · D strips of grass planted across the slope. Soil washed off each plot during one rainy season was collected in a trough at the bottom of the plot, dried and weighed. Table 4.1 shows the results.
Treatment
Plot 1 soil loss / kg
Plot 2 soil loss / kg
Plot 3 soil loss / kg
Mean soil loss / kg
A
12.6
11.7
12.3
12.2
B
7.2
6.9
7.5
7.2
C
4.2
3.8
11.6
to be calculated
D
2.1
1.8
2.4
2.1
Table 4.1 Soil collected from each plot over one rainy season
(a)[3]
Complete Table 4.1 for treatment C. Justify the way you have treated the results, and show your working.
Mark Scheme — 4(a)
B1 identifies plot 3 (11.6 kg) as an anomalous result — it is nearly three times the other two plots of the same treatment, and higher than every plot given treatment B
M1 excludes it and averages the remaining replicates: (4.2 + 3.8) ÷ 2
A1= 4.0 kg
a mean of all three values (6.5 kg) scores B0 M0 A0, because the anomaly has not been identified
a candidate who keeps the anomaly but states clearly why it is being kept, and calculates 6.5 kg correctly, can take M1 only
⚠ If you missed marks here: an anomaly is judged against the replicates of its own treatment. 11.6 kg would be unremarkable in treatment A, but among 4.2 and 3.8 it stands out, which is exactly why replicates are taken. Note also that the question said "justify", so the mark for spotting it is only earned if you say so in words.
(b)(i)[1]
The bar for treatment C has not been drawn on Fig. 4.1. State the value on the vertical axis at which the top of this bar should be drawn.
Mark Scheme — 4(b)(i)
14.0 kg — on the labelled 4 gridline, between the bars for B and D
allow the value carried forward from part (a) if part (a) was wrong but the reasoning here is right
⚠ If you missed marks here: a bar chart plots the mean, not an individual plot, so this bar goes at the value you calculated in (a) and not at 4.2 or 11.6. Checking the shape helps: C should sit between B (7.2) and D (2.1), which 4.0 does and 6.5 does not.
(b)(ii)[2]
Calculate the percentage reduction in soil loss achieved by treatment D compared with treatment A. Give your answer to three significant figures.
Mark Scheme — 4(b)(ii)
M1(12.2 − 2.1) ÷ 12.2 × 100
A1= 82.8% (82.786... to 3 significant figures)
an answer of 82.79% or 83% loses A1 — the question specified three significant figures
⚠ If you missed marks here: a percentage reduction is measured against what you started with, which is bare soil at 12.2 kg. Dividing 2.1 by 12.2 gives 17.2%, which is the soil loss remaining, not the reduction — and the two always add to 100.
(c)[2]
Suggest a testable hypothesis for this investigation.
Mark Scheme — 4(c)
1 a statement linking the independent variable to the dependent variable, for example: plots that keep a cover of vegetation or crop residue will lose less soil than bare plots
2 it is testable, that is it predicts a direction and names something that was actually measured (mass of soil collected), so the results can support it or contradict it
accept the null form: there will be no difference in soil loss between the four treatments
do not credit a question ("does mulch reduce soil erosion?") or an aim ("to investigate soil erosion") — a hypothesis is a statement that can be shown to be false
⚠ If you missed marks here: an aim says what you are going to look at; a hypothesis predicts what you will find. "To investigate the effect of ground cover on soil erosion" is an aim. Turn it into a prediction, name the measured quantity, and it becomes testable.
(d)[3]
Two variables the researchers kept constant were the slope angle and the plot area. Explain why the slope angle had to be kept constant, and identify one further variable they should have controlled.
Mark Scheme — 4(d)
1 a steeper slope makes water run off faster OR gives the run-off more energy, so more soil particles are carried away
2 so if the plots had different slopes, a difference in soil loss could have been caused by the slope rather than by the treatment, and the two effects could not be separated
3 one further control variable, any 1 from: soil type and texture; rainfall received by each plot; the crop grown; the initial organic content of the soil; the drying method and time before weighing; the length of the season monitored
mark 2 needs the consequence for the conclusion, not just the words "fair test"
⚠ If you missed marks here: notice that this question separates describing how from explaining why. The how is that steeper slopes give faster run-off; the why is that an uncontrolled slope would make the treatments impossible to compare. Both are needed, and the second is the one usually left out.
(e)[3]
Evaluate the method used in this investigation. In your answer, state one limitation, suggest an improvement, and say how much confidence the results give in the ranking of the four treatments.
Mark Scheme — 4(e)
1 one limitation, any 1 from: only one rainy season was monitored, and a wetter or drier season could give a different ranking; only three replicates per treatment, so one odd plot has a large effect on the mean, as treatment C showed; one site and one soil type, so the results may not apply elsewhere; rainfall on each plot was not measured, so it is assumed rather than known to be equal; soil that moved but was not caught by the trough was not counted
2 an improvement that matches the limitation, any 1 from: repeat over several seasons; increase to five or more replicate plots per treatment; repeat at sites with different soil types; place a rain gauge on each plot and record the rainfall; extend the trough across the full width of the plot
3 a judgement supported by the spread of the data, for example: the difference between A (12.2 kg) and D (2.1 kg) is far larger than the spread within any treatment (11.7 to 12.6 kg in A), so the conclusion that ground cover reduces soil loss is well supported; but the ranking is less secure for treatment C, where one of only three plots had to be set aside
⚠ If you missed marks here: "evaluate" is not a list of everything that could be better. The third mark is the one that separates grades: compare the size of the difference between treatments with the size of the spread within a treatment. If the difference is much bigger than the spread, the result is trustworthy even from a small experiment.
Question 5
Total: 14 marks
A tropical cyclone crossed a coastline and passed over two towns of similar height above sea level. Town P received a warning from the national weather service 36 hours before the cyclone arrived. A fault in the local radio transmitter meant Town Q was warned only 6 hours before it arrived. Table 5.1 shows what happened in each town.
Town P
Town Q
population
24 000
30 000
warning received before the cyclone arrived / hours
36
6
percentage of the population evacuated
90%
45%
number of deaths
12
210
buildings damaged
1 800
2 100
cost of damage / million US$
40
52
Table 5.1 The effect of one cyclone on two towns
(a)[2]
Calculate the number of deaths per 10 000 people in each town.
Mark Scheme — 5(a)
1 Town P: 12 ÷ 24 000 × 10 000 = 5.0 per 10 000
2 Town Q: 210 ÷ 30 000 × 10 000 = 70 per 10 000
both values are needed; one correct value scores one mark
⚠ If you missed marks here: a rate per 10 000 lets you compare towns of different sizes. Town Q had 17.5 times as many deaths but only 1.25 times the population, which is why the raw totals understate the difference — the rate is 14 times higher, not 17.5.
(b)[3]
Use the data in Table 5.1 to explain the difference in the death rate between the two towns.
Mark Scheme — 5(b)
1 Town P had 36 hours of warning against 6 hours in Town Q, and evacuated 90% of its people against 45%
2 explains the link: a longer warning gives time to move people inland or to shelters before the flooding and storm surge arrive, so far fewer people are in the damaged buildings when the cyclone strikes
3 shows the storm itself was of similar strength in both towns, so the warning is the likely cause: buildings damaged per 1 000 people is 75 in P and 70 in Q, and the cost per person is similar
accept for mark 3 any use of the damage figures to argue the hazard was comparable, or a comment that other factors such as building quality were not recorded
⚠ If you missed marks here: the third mark is the analytical one. Damage to buildings is almost the same per person in both towns, which is what shows the cyclone hit them about equally hard — so the enormous difference in deaths has to come from whether people were still there. Look for the figure that rules out the alternative explanation.
(c)[3]
Table 5.2 gives conditions at four places in the ocean in the same month.
Place
Latitude
Sea surface temperature / °C
Ocean depth / m
W
2° N
28.5
120
X
15° N
28.0
90
Y
18° S
24.0
150
Z
40° N
27.5
200
Table 5.2 Ocean conditions at four places Identify the place at which a tropical cyclone could form, and justify your answer.
Mark Scheme — 5(c)
1 place X
2 X meets all three conditions: latitude 15° N lies between 5° and 30°, the sea surface is 28.0 °C which is at least 27 °C, and the ocean is 90 m deep which is at least 60 m
3 any 2 of the rejections: W is only 2° from the Equator, inside the 5° limit; Y is 24.0 °C, below the 27 °C needed; Z is at 40° N, outside the 5° to 30° band
note that W, Y and Z each satisfy two of the three conditions — all three must be met
⚠ If you missed marks here: every distractor here is deliberately close. W is warm enough and deep enough but too near the Equator; Z is warm enough and deep enough but too far from it. Check each of the three conditions against each place rather than stopping at the first one that looks right.
(d)[6]
A government has limited money to spend on reducing the impacts of tropical cyclones. One adviser recommends spending it all on monitoring and warning systems. Discuss the benefits and limitations of this recommendation.
Mark Scheme — 5(d)
Level 3 (5–6 marks) both benefits and limitations are developed, at least one point is supported by the data in Table 5.1, and a clear judgement is reached that follows from the points made.
Level 2 (3–4 marks) both sides are given but one is thin, or the points are correct but undeveloped, or there is no judgement.
Level 1 (1–2 marks) only one side is given, or general statements with no reference to cyclones.
Benefits — any 3 from: warning allows evacuation, which is what separated Town P from Town Q (5 against 70 deaths per 10 000); it protects people wherever they are, not only those in newly built houses; it is cheap compared with rebuilding a coastline; it works for the buildings that already exist; the same monitoring network also serves flooding and drought warnings; satellite and computer forecasting keeps improving.
Limitations — any 3 from: a warning does not reduce damage to property at all — Town P still lost 1 800 buildings and US$40 million; it depends on the message reaching people, and the transmitter fault in Town Q shows how a single failure removes the whole benefit; there must be somewhere to evacuate to, so shelters and roads are also needed; some people will not leave livestock or property; forecasts of the exact track are uncertain, and a false alarm reduces the response next time; spending everything on one strategy leaves nothing for building standards or land use zoning, which reduce damage rather than only deaths.
Judgement a supported conclusion either way, for example: warning is the most effective single way to reduce deaths per dollar and should be funded first, but funding it alone is not justified because the damage figures show that lives and property need different measures. An answer arguing that a poorer country should still prioritise warning, because saving lives matters most and the buildings can be rebuilt, is equally creditworthy if argued from the data.
⚠ If you missed marks here: "discuss" caps a one-sided answer at Level 1, no matter how good the points are. Plan two columns before you write. Then note that the judgement mark is not for choosing the "right" answer — there isn’t one — it is for a conclusion that follows from the specific points you made rather than a general opinion tacked on at the end.
Question 6
Total: 14 marks
Fig. 6.1 and Fig. 6.2 show the population structures of two countries in the same year. Country J is classified by the World Bank as a low-income country (LIC) and Country K as a high-income country (HIC). Each bar gives the percentage of the total population in that age group. In this question the young dependent population is aged under 20, the working population is aged 20 to 59, and the elderly dependent population is aged 60 and over.
(a)[2]
Use Fig. 6.1 and Fig. 6.2 to state: (i) the percentage of Country J’s total population that is male and aged 0 to 9; (ii) the percentage of Country K’s total population that is male and aged 40 to 49.
Mark Scheme — 6(a)
114%
27%
both bars end exactly on a gridline; the minor gridlines are every 1% and the labelled ones every 2%
⚠ If you missed marks here: the male axis runs outwards to the left, so the numbers get larger as you move away from the centre, not towards it. Read the value at the end of the bar, and remember each bar is already a percentage of the whole population, not of that sex.
(b)[3]
Calculate the total dependent population, as a percentage of the total population, for Country J and for Country K.
Mark Scheme — 6(b)
M1 adds males and females in the 0–9 and 10–19 bands and in the 60–69 and 70+ bands
a useful check: each pyramid must total 100%, so the working population is 43% in J and 53% in K
⚠ If you missed marks here: the two commonest errors are forgetting the females and forgetting the elderly. Country J looks like the dependent country because of its wide base, but K is not far behind (47% against 57%) — its dependants are simply at the other end of the pyramid.
(c)[3]
Compare the population structures of Country J and Country K, and suggest one problem that each structure creates for its government.
Mark Scheme — 6(c)
1 comparison with figures: J has a wide base and narrow top — 52% are under 20 and only 5% are 60 or over; K is far more even, with 21% under 20 and 26% aged 60 or over
2 a problem for J, any 1 from: 43% of the population supports the other 57%; demand for school places, healthcare for children and, in ten to twenty years, for jobs and housing as the wide base reaches working age
3 a problem for K, any 1 from: 26% are aged 60 or over, so demand for healthcare and pensions is high and rising; only 21% are under 20, so the working population will shrink as the large 40–59 group retires
for mark 1 a comparison must mention both countries; describing only one scores zero for that mark
⚠ If you missed marks here: a comparison needs a comparative word — wider, higher, more than — and a figure from each country. Two separate descriptions placed side by side is not a comparison. Note also that both structures create problems, just different ones; K is not simply the "better" pyramid.
(d)[6]
Country J is considering improving access to education, particularly for girls, as a strategy for managing its population size. Discuss the benefits and limitations of this strategy.
Mark Scheme — 6(d)
Level 3 (5–6 marks) both benefits and limitations are developed, the answer is set in Country J’s situation rather than being general, and a clear judgement is reached that follows from the points made.
Level 2 (3–4 marks) both sides are given but one is thin, or the points are correct but undeveloped, or there is no judgement.
Level 1 (1–2 marks) only one side is given, or general statements about education with no link to population size.
Benefits — any 3 from: women who stay in education longer tend to have their first child later and to have fewer children, which lowers the birth rate; education explains how contraception works and where to obtain it; educated women are more likely to enter paid work, which raises household income and reduces the economic reason for having many children; it also lowers infant mortality, so families do not need extra births to be confident of surviving children; unlike an anti-natalist law it is voluntary and does not restrict anyone’s choices; the same schooling raises the skills of the workforce, which J needs as its large young population reaches working age.
Limitations — any 3 from: it is slow — the effect on the birth rate appears a generation later, and J’s 52% under-20s will reach childbearing age long before then; building schools and training teachers is expensive for a low-income country; families who need children to work may keep them out of school whatever the government provides; where the number of children per family is a strong social or religious expectation, education alone may not change it; it does nothing about the immediate demand for school places, jobs and housing; it may need to be combined with access to healthcare and contraception to have any effect.
Judgement a supported conclusion either way, for example: education is the most durable strategy because it is voluntary and brings benefits beyond population size, but it cannot be J’s only measure because it acts too slowly for a population already this young. An answer arguing that the slowness is acceptable because faster strategies are coercive, or that limited funds should go to healthcare first, is equally creditworthy if it is argued.
⚠ If you missed marks here: the mark that is nearly always lost is the timescale one. Education changes the birth rate a generation later, and Fig. 6.1 shows J cannot wait a generation. Whenever you evaluate a population strategy, ask three things: how fast does it work, who bears the cost, and does it restrict anyone’s choices.
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