G3 SEC Mathematics • Learner’s Guide Vol 0091. Quick route: build a timeline, measure all moving and waiting intervals, solve stage by stage, then check the whole journey.
A journey question may look like a story about buses, bicycles or a walk home. In G3 SEC Mathematics, the real task is to decide which interval each number belongs to. A bus travels for forty minutes, stops for ten minutes and travels again. Its average speed for the whole journey is not found by averaging the two speeds or forgetting the stop. We need the total distance and the total elapsed time, both measured over the same journey.
This Learner’s Guide Vol 0091 turns that simple principle into a complete workshop. You will learn how to build a journey timeline, calculate distances and durations, read distance–time and speed–time graphs, deal with rest periods, solve backward from a known arrival time and check answers against the story. There are original worked examples, independent problems and explained answers. Start with a two-stage journey before attempting several vehicles or unknown times.
The skill connects Primary School word problems to secondary algebra and real-world modelling. If you have recently practised Volume 0087 on successive transfers, you already know the benefit of updating a quantity after an event. Here the quantities that change are distance, speed and clock time. Use the G3 SEC Learner’s Guide route whenever you need a broader study map.
Chapter index
- 1. The quick answer: one journey, three clocks
- 2. Know exactly what K310 examines
- 3. Build the timeline before touching the calculator
- 4. Why averaging the speeds often gives the wrong answer
- 5. Stops, delays and what the question actually asks
- 6. Read distance–time graphs without guessing motion
- 7. Reverse journeys: solve for an unknown stage
- 8. Two travellers: the same clock, different starting lines
- 9. Mixed units and precision without hidden traps
- 10. Original practice clinic: try before reading the answers
- 11. Answer clinic: inspect the route, not only the final number
- 12. A seven-day route from foundation to exam performance
- 13. Where this workshop leads next
1. The quick answer: one journey, three clocks
Every journey has at least three useful measures. Distance describes how far an object moves. Moving time measures the intervals in which that movement takes place. Elapsed time measures the entire interval from departure to arrival and may include stops. Speed is distance divided by the relevant time. The phrase “relevant time” is the part a careful learner protects.
Write a three-column timeline before solving a complicated question: event or stage, duration, and distance. If speeds differ, add a fourth column for speed. Mark every rest period as an event with zero distance. A ten-minute stop is not a missing row: it is a real part of the total elapsed time. A table makes a story inspectable, so an answer can be checked stage by stage rather than trusted because a calculator displayed a plausible number.
For a journey of 18 km in 30 minutes, then a stop of 15 minutes, then 12 km in 20 minutes, the distance travelled is 30 km. The moving time is 50 minutes. The elapsed time is 65 minutes. Average speed while moving is 30 ÷ (50/60) = 36 km/h. Average speed for the entire journey is 30 ÷ (65/60) = 27.6923… km/h, or 27.7 km/h to three significant figures. Both values are mathematically meaningful, but they answer different questions.
Notice that the whole-journey average is lower because the 15-minute stop contributes time without distance. That does not mean either travelling speed changed. It means we chose a wider interval for the denominator. A reliable solution states the interval before giving the result. This small habit prevents a surprisingly large number of errors.
2. Know exactly what K310 examines
The official 2027 G3 Mathematics syllabus K310 includes rate and speed, unit conversion, interpretation of distance–time and speed–time graphs, and applications in real-world settings such as travel schedules. It examines mathematical techniques, problem solving and reasoned communication. Those are connected skills: the right formula with the wrong interval is still the wrong model.
K310 Paper 1 lasts 2 hours 15 minutes and has about 26 compulsory short-answer questions worth 90 marks, weighted at 50%. Paper 2 also lasts 2 hours 15 minutes; it has 9 to 10 compulsory questions worth 90 marks, weighted at 50%. The last question in Paper 2 applies mathematics to a real-world scenario. The syllabus states that omission of essential working leads to loss of marks. An approved calculator may be used in both papers. Do not turn these facts into a promise that a particular journey type will appear: they describe the assessment framework, not a prediction of a specific question.
Unless a question specifies another accuracy, the syllabus requires non-exact numerical answers to three significant figures, and angles in degrees to one decimal place. Work with full calculator precision until the final line. If a question asks for an exact value or a specified accuracy, obey that instruction instead. A journey result rounded halfway through can produce a different final arrival time or threshold decision.
On the Mathematics Hub you can revisit the underlying ideas of ratios, algebra, graphs and measurement. The Secondary 1 percentages and reverse-percentages classroom helps if percentages in a mixed real-world question are a separate weak link. Use these existing owners for foundations; this volume owns the narrower skill of reconstructing multi-stage travel.
3. Build the timeline before touching the calculator
A clear timeline begins at a specified departure, not wherever a question first mentions a speed. Suppose a delivery rider leaves at 08:10, travels for 24 minutes at 20 km/h, rests for 6 minutes, then rides for 18 minutes at 15 km/h. Write four times: 08:10 departure, 08:34 end of stage one, 08:40 end of rest, 08:58 arrival. Only after this clock is secure should you calculate distances.
Stage one covers 20 × (24/60) = 8 km. The rest covers 0 km. Stage two covers 15 × (18/60) = 4.5 km. Total distance is 12.5 km, travelling time 42 minutes and entire journey time 48 minutes. Average speed for the whole journey is 12.5 ÷ (48/60) = 15.625 km/h, or 15.6 km/h to three significant figures. Average speed while moving is 12.5 ÷ (42/60) = 17.857…, or 17.9 km/h.
Why write clock times first? Because clock arithmetic carries a separate danger. One hour has 60 minutes, not 100. From 08:58, another 7 minutes brings you to 09:05, not 08:65. When a problem spans an hour boundary, use durations measured in minutes and convert back to hours and minutes only when you state an arrival. A neat timetable can stop a careless decimal-time mistake before it becomes an algebra mistake.
Do not confuse a stage’s speed with a journey average. The rider’s second-stage speed is 15 km/h. The whole-journey average is approximately 15.6 km/h, even though the rider sometimes moves faster, sometimes more slowly and sometimes not at all. The average is a summary of total distance over total elapsed time, not a claim that the rider travelled at the same speed throughout.
An efficient working layout might read: “Stage 1: 24/60 h, 8 km. Stop: 6/60 h, 0 km. Stage 2: 18/60 h, 4.5 km. Total: 12.5 km in 48/60 h. Whole-trip average = 15.625 km/h ≈ 15.6 km/h.” This gives a marker a visible chain from information to answer. It also gives you somewhere to look when an independent check detects a mismatch.
4. Why averaging the speeds often gives the wrong answer
Imagine a cyclist travels 12 km at 12 km/h and returns along the same road at 24 km/h. It is tempting to calculate (12 + 24)/2 = 18 km/h. That is wrong for the round trip. The outward journey takes 1 hour. The return takes 0.5 hour. Total distance is 24 km, total time is 1.5 hours, and average speed is 24 ÷ 1.5 = 16 km/h.
The difference matters because speed is not an isolated number to average without thinking about exposure. The cyclist spends longer at the slower speed. An ordinary arithmetic mean of the two rates gives equal weight to the rates, even though they apply for unequal amounts of time. Equal distances do not imply equal times. If the question provides equal distances, find the time for each distance, then recombine the totals.
Now reverse the structure. Suppose the cyclist travels for 30 minutes at 12 km/h, then 30 minutes at 24 km/h with no rest. The distances are 6 km and 12 km. Total distance is 18 km and total time is 1 hour, so average speed is 18 km/h. The simple mean happens to work here because the rates apply over equal durations. It is not a universal rule; it is a consequence of equal weighting by time.
One useful diagnostic question is, “What is equal?” The question may give two equal distances, two equal times, or neither. Equal distances suggest calculating separate times. Equal times make the arithmetic mean of the rates appropriate when the entire period is moving and the unit is consistent. If there is a waiting period, include it in a whole-journey average. If the speed changes continuously, use the graph or information given rather than assuming a convenient two-stage pattern.
Practise explaining the wrong method, not merely crossing it out. “The two speed values are not weighted by equal travel times; the slower segment takes longer.” This is a mathematical reason. It transfers to weighted averages, grouped statistics and many other questions in the Mathematics Hub. It is also a good reminder from Volume 0079 on weighted comparisons: always identify what contributes the weight.
5. Stops, delays and what the question actually asks
Questions often contain a sentence such as “the bus stopped at a station for eight minutes.” A stop may be part of the elapsed time, or it may be irrelevant to the speed during a particular moving stage. Highlight the phrase “for the whole journey” or “while the vehicle was moving.” Those words decide the denominator; the word “average” alone does not.
Example: a bus covers 21 km in 35 minutes, waits 10 minutes, and covers another 9 km in 15 minutes. The first moving speed is 21 ÷ (35/60) = 36 km/h. The second is 9 ÷ (15/60) = 36 km/h. The bus travels at the same stage speed while moving. Yet its whole-trip average is 30 ÷ (60/60) = 30 km/h because the journey lasted 60 minutes including the stop. There is no contradiction. We have measured two different things.
Suppose the question instead says the schedule allows 60 minutes including the stop, and asks how much later the bus arrives if the stop becomes 16 minutes while moving speeds and travel distances remain unchanged. The new elapsed time is 35 + 16 + 15 = 66 minutes. Arrival is 6 minutes later. Its whole-trip average becomes 30 ÷ (66/60) = 27.2727… km/h, about 27.3 km/h. The moving speeds remain 36 km/h. A changed rest period does not rewrite the moving distances.
Now add a timetable. The bus leaves at 14:45 and originally arrives at 15:45. With the longer stop, it arrives at 15:51. Write these times separately from the speed calculation. One row explains duration; another explains clock time. The question might ask whether a 15:50 connection can be made. The answer is no, even though the speed is still a respectable 27.3 km/h. The real decision depends on the arrival threshold, not how large the average number looks.
A powerful check is to set every speed temporarily to zero for a rest row. If your table shows distance increasing during a stop, revisit the labels. A second check is to ask whether total distance remains unchanged when only waiting time changes. If not, you may have accidentally used elapsed time as travel time inside a distance calculation.
6. Read distance–time graphs without guessing motion
A distance–time graph records distance from a stated reference against time. In the simplest travel story where distance travelled never decreases, an upward straight line indicates constant speed, a horizontal line indicates a stop, and a steeper upward segment indicates a higher speed. Speed is the change in distance divided by the change in time for that segment. The graph tells you a rate through its gradient, not through the height of a single point.
Suppose a cumulative-distance graph has points (0 min, 0 km), (20 min, 6 km), (30 min, 6 km), and (50 min, 16 km), joined by straight segments. The first segment’s speed is 6 ÷ (20/60) = 18 km/h. The horizontal interval from 20 to 30 minutes is a 10-minute stop. The final segment’s speed is 10 ÷ (20/60) = 30 km/h. The whole-trip average is 16 ÷ (50/60) = 19.2 km/h. If a learner reads 16 km as the speed at 50 minutes, the unit reveals the mistake immediately: kilometres measure distance, not kilometres per hour.
Notice the distinction between distance travelled and displacement or distance from a starting point. A graph showing distance from home can slope downward when a traveller returns home. A cumulative distance-travelled graph does not decrease. Before interpreting a downward line, read the vertical-axis label carefully. Do not infer direction from an unlabeled sketch. This is a graph-language skill as much as a calculation skill.
A speed–time graph requires a different reading habit. Speed sits on the vertical axis. A horizontal segment shows constant speed, even though the vehicle is moving. Under an appropriate speed–time graph, area represents distance travelled: speed multiplied by time. A rectangle 12 km/h high and 0.5 hour wide represents 6 km. When time is shown in minutes, convert the width to hours before multiplying. A triangle with increasing speed needs the area of a triangle, provided the straight-line change shown is part of the given model.
Here is the interpretation test: What does a flat line mean? On cumulative distance–time, it means no additional distance. On speed–time, a flat line above zero means constant positive speed. One graph’s gradient and the other’s area answer different questions. Draw a small axis sketch in your notes and describe it in words before you begin the calculation. Graph literacy saves more time than memorising isolated pictures.
7. Reverse journeys: solve for an unknown stage
Advanced word problems may tell you when a traveller must arrive and ask how fast a stage must be. The reliable method is to work backwards from the total time available, subtract the known durations, then divide the remaining distance by the remaining moving time. If you cannot first say how many minutes are available for the unknown stage, do not invent its speed.
A student leaves home at 06:55 and must reach school by 07:35, a 40-minute window. She walks 1.2 km at 4 km/h, waits at a crossing for 4 minutes, then rides a bicycle for 2.4 km. Walking time is 1.2 ÷ 4 = 0.3 hour = 18 minutes. The remaining cycling time is 40 − 18 − 4 = 18 minutes, or 0.3 hour. Required cycling speed is 2.4 ÷ 0.3 = 8 km/h. Check: 18 minutes walking + 4 waiting + 18 cycling = 40 minutes.
If the student leaves five minutes later while all distances and the wait remain unchanged, she has 35 minutes. The cycling time becomes 35 − 18 − 4 = 13 minutes. Required average cycling speed is 2.4 ÷ (13/60) ≈ 11.1 km/h to three significant figures. The required speed rises because the available time shrank. This is a sensitivity statement grounded in the original model. It does not mean she will actually be able to cycle that quickly; feasibility belongs to the real context.
To make the algebra visible, let t be cycling time in hours. We know 0.3 + (4/60) + t = 40/60. Therefore t = 0.3. If v is the unknown cycling speed, 2.4/v = t, so v = 8. A more complicated question may combine two unknown stages into an equation. Name the unknown with its unit, express every duration consistently, and check that a negative or impossible duration is rejected.
Some learners put 40 in the equation as though it were 40 hours. Others write 07:35 − 06:55 as 0.80 hour by reading clock digits as decimals. Use minutes to subtract clock times, then convert once. The earlier skill in Volume 0075 on plausibility and reverse checking helps you notice answers that are mechanically calculated but physically unreasonable.
8. Two travellers: the same clock, different starting lines
Meeting and catch-up problems demand one common time axis. Imagine Person A starts cycling at 09:00 at 12 km/h. Person B starts at 09:20 along the same route at 18 km/h. If both travel steadily and B starts from the same point, by 09:20 A has a 4 km head start: 12 × (20/60) = 4. The difference in speed is 18 − 12 = 6 km/h. B gains 6 km each hour. Time from B’s departure to catching A is 4 ÷ 6 = 2/3 hour = 40 minutes. They meet at 10:00.
There is an algebra version. Let t be hours after 09:20. A’s total riding time is t + 1/3 hour, so A’s distance is 12(t + 1/3). B’s distance is 18t. At the catch-up point, they have covered the same distance from the same origin, giving 12t + 4 = 18t. Thus t = 2/3 hour. Substituting back yields 12 km for each person’s distance by 10:00. This second method checks the first.
But now change the story: Person B starts 3 km ahead along the route, instead of at A’s origin. A’s and B’s distances from their own starting points will not be equal at the meeting. We must locate both people relative to the same reference point. If B is already ahead and moves faster in the same direction, A will not catch B under constant-speed assumptions. A solver who writes “equal distances travelled” without checking starting positions can produce an equation that is tidy and false.
On a distance-from-origin graph, the two straight lines meet at an intersection if a meeting occurs. The slopes represent speeds and the vertical intercepts represent starting positions at the chosen zero time. A delayed start shifts the domain of a line: B’s journey does not exist in the model before departure. Drawing even a rough sketch reveals whether the solution is after both journeys begin.
Ask four questions before setting up a chase equation: Are the travellers moving in the same direction? What are their locations at one common moment? Which person is ahead then? Can the faster person close the gap under the stated conditions? These questions are more useful than memorising a single relative-speed formula. The formula works after the situation is understood.
9. Mixed units and precision without hidden traps
Before calculating, identify every unit in the problem. Speed may be in km/h or m/s; time may be minutes, hours or seconds. A kilometre contains 1,000 metres and an hour contains 3,600 seconds. Therefore 1 m/s = 3.6 km/h, and 1 km/h = 1/3.6 m/s. Derive this from units if you forget it; do not try to recall a direction from memory under stress.
A runner covers 750 m in 2 minutes 30 seconds. The duration is 150 seconds, so average speed is 750 ÷ 150 = 5 m/s. In km/h, that is 5 × 3.6 = 18 km/h. If you divide 750 by 2.5 and write m/s, you have treated 2.5 minutes as seconds. The numerical answer 300 might look impressive, but its unit and scale show why it cannot describe an ordinary runner.
Consider two bus legs reported in kilometres and a walking leg reported in metres. Convert before adding. A walk of 800 m plus a ride of 3.2 km covers 0.8 + 3.2 = 4.0 km. Avoid combining the raw numbers into 803.2. In a word problem, the arithmetic is often easier than identifying which measures are compatible.
Precision is another kind of compatibility. If your exact intermediate result is 19.047619…, keep it on the calculator until the final answer. If a timetable threshold depends on seconds or whole minutes, attend to the context before rounding. A speed of 11.1 km/h may be a good final rounded answer, but using 11.1 instead of 144/13 in an earlier step may move a calculated arrival across a borderline threshold.
A final answer needs three parts when appropriate: the number, its unit and its meaning. “Average speed = 19.2 km/h for the 50-minute journey, including the 10-minute stop” is harder to misread than “19.2”. Train yourself to give units during practice, not as a decoration you add only when a teacher reminds you.
10. Original practice clinic: try before reading the answers
Question A, foundations. A jogger covers 3 km in 20 minutes, rests for 5 minutes and covers 2 km in 15 minutes. Find the whole-journey average speed and the moving-only average speed. State which duration belongs in each denominator.
Question B, equal distances. A car covers 40 km at 40 km/h, then another 40 km at 80 km/h without stopping. Find the average speed. Explain why 60 km/h is not the correct whole-trip average.
Question C, clock time. A shuttle leaves at 16:35, travels 22 minutes, waits 8 minutes and travels 19 more minutes. At what time does it arrive? By how many minutes does it miss a 17:20 connection, if at all?
Question D, mixed units. A cyclist rides 900 m in 3 minutes, waits 1 minute and rides 600 m in 2 minutes. Find the whole-journey average speed in m/s and km/h.
Question E, reversed timetable. A learner leaves at 07:10 and must arrive at 07:55. The first stage covers 2 km at 8 km/h, followed by a 5-minute wait. The final stage covers 3 km. Find the required speed during the final stage.
Question F, graph reading. A cumulative-distance graph joins (0 min, 0 km), (15 min, 5 km), (25 min, 5 km), and (45 min, 13 km) by straight lines. Describe the journey and find its whole-trip average speed. What part of the graph tells you the rest interval?
Question G, delayed start. One cyclist starts at 10:00 at 10 km/h. A second starts from the same point at 10:30 at 15 km/h in the same direction. When does the second cyclist catch the first? State the distance from the start at the meeting.
Question H, reasoning. Two vehicles each travel at 30 km/h for 20 minutes, but one has a separate 10-minute stop between two moving stages and the other does not. If their total moving distances are equal, must their moving-only average speeds be equal? Must their whole-journey averages be equal? Explain the difference.
Do not read the next chapter until you have written your stages and units. If a question feels too hard, begin by identifying only the timeline. That is still useful practice. The goal is to become able to build the correct model, not to prove that you can do eight questions rapidly on your first attempt.
11. Answer clinic: inspect the route, not only the final number
A. Total distance is 5 km. Moving time is 20 + 15 = 35 minutes; elapsed time is 40 minutes. Whole-journey average is 5 ÷ (40/60) = 7.5 km/h. Moving-only average is 5 ÷ (35/60) ≈ 8.57 km/h. If you obtained 7.5 for both, you probably used total elapsed time in the second calculation.
B. First time is 40 ÷ 40 = 1 hour. Second time is 40 ÷ 80 = 0.5 hour. Total distance is 80 km, total time 1.5 hours, average speed ≈ 53.3 km/h. The arithmetic mean of 40 and 80 is 60 km/h, but the car spends twice as long at 40 km/h. Equal distance is not equal duration.
C. Total time is 22 + 8 + 19 = 49 minutes. From 16:35, 25 minutes reaches 17:00 and 24 more minutes reaches 17:24. Arrival is 17:24, four minutes after the 17:20 connection. Writing a timetable row makes the hour change easy to audit.
D. Total distance is 1,500 m. Total elapsed time is 3 + 1 + 2 = 6 minutes = 360 seconds. Average speed is 1,500 ÷ 360 ≈ 4.17 m/s. Multiplying by 3.6 gives 15.0 km/h to three significant figures. Moving-only speed would be 1,500 ÷ 300 = 5 m/s, but the question asked for the whole journey.
E. Available time is 45 minutes. The first stage takes 2/8 hour = 15 minutes; waiting takes 5. That leaves 25 minutes for the final 3 km. Required speed is 3 ÷ (25/60) = 7.2 km/h. Check the arithmetic: 15 + 5 + 25 = 45. The result is an average needed on the final stage, not a guarantee that conditions allow that rate.
F. From 0 to 15 minutes the traveller covers 5 km. From 15 to 25 minutes the graph is horizontal, so the traveller stops for 10 minutes. From 25 to 45 minutes another 8 km is covered. Total distance is 13 km in 45 minutes; average speed is 13 ÷ (45/60) ≈ 17.3 km/h. The level section, not a low or high y-coordinate by itself, identifies the rest.
G. The first cyclist covers 10 × 0.5 = 5 km before the second starts. The second gains 15 − 10 = 5 km each hour, so catching up takes 1 hour after 10:30. They meet at 11:30. The first cyclist has travelled 10 × 1.5 = 15 km, and the second cyclist 15 × 1 = 15 km. Same meeting distance, same origin, two different travel durations.
H. Yes, under the stated conditions both moving-only averages are 30 km/h: equal total moving distance divided by equal moving time when each moving interval is at 30 km/h. Their whole-journey averages are not equal if the separate stop makes one elapsed time longer. It is possible to have the same moving performance and different whole-trip averages. The difference comes from the denominator.
For every error, classify it before repeating the problem: wrong interval, missing rest, mixed units, inaccurate clock subtraction, unlabelled origin, inappropriate average or final rounding. A category is more useful than “careless mistake”, because it tells you exactly what to practise next.
12. A seven-day route from foundation to exam performance
Day 1: language of journeys. Take six short word problems. Underline the words that indicate departure, stage change, rest and arrival. Without calculating anything, make timelines with units. If the problem states only a distance and a speed, say which time must be found. A strong foundation is knowing what belongs in each row.
Day 2: complete one-stage arithmetic. Solve distance = speed × time, speed = distance ÷ time and time = distance ÷ speed with straightforward units. Convert 15 minutes to 0.25 hour, 36 minutes to 0.6 hour and 90 seconds to 1.5 minutes. Write the units in every line. Arithmetic accuracy matters, but there is little value in speed until the unit model is stable.
Day 3: two stages and pauses. Solve journeys with a rest row, then compare moving-only and whole-trip averages. Do not look at a worked answer until you have named the denominator. Use one question that changes the rest time but not the travel distances. This helps you recognise the difference between a changed schedule and a changed route.
Day 4: graphical translation. Read points, intervals, gradients and areas on simple supplied graphs. Explain why a horizontal line on a distance–time graph differs from a horizontal line on a speed–time graph. Translate one table into words, then back into a table. Correct interpretation should survive a change of representation.
Day 5: reverse planning. Work from an arrival deadline to an unknown travel time or speed. Include a problem with an hour boundary and a problem with mixed units. Check by rebuilding the journey forwards. A correct backward solution must still satisfy the departure time, every stage duration and the final arrival.
Day 6: meeting and catching up. Use a common origin and a common clock. Solve a delayed-start problem both by head start and by algebra. Explain why equal distances only apply when both measured distances begin at the same reference point. Draw a rough graph as an independent check.
Day 7: mixed-question rehearsal. Take three original problems: one numerical, one graphical and one decision task involving a connection or deadline. Work without help, show essential steps and label units. Mark not only the final answers but also whether every interval and assumption is traceable. Keep a short correction log and revisit the weakest category three days later.
For parents, the best check is not “Did you finish a full paper?” Ask the learner to explain where the rest period appears, why the speed average is weighted by time and what makes the answer reasonable. These questions show whether learning is transferring. A student who can explain the denominator is usually closer to solving the unfamiliar version than one who has memorised a single template.
13. Where this workshop leads next
Multi-stage journey problems teach a broader habit: before combining numbers, make their reference periods compatible. That same habit supports Volume 0089 on changed conditions, where a revised input changes only the part of a solution that depends on it. It also helps with scientific time-series data, where a reading taken before an intervention should not be treated as though it describes the system afterwards.
If this work exposes a Primary School gap in fractions, ratios or representing word problems, use the PSLE to SEC subject-by-subject study route or the Mathematics Hub. You do not need to restart every topic. Find the exact missing skill, repair it with a smaller example and then return to the journey.
Official references for exam-specific statements: SEAB 2027 G3 school-candidate syllabus index, 2027 Mathematics K310 syllabus and SEAB’s SEC overview. Always use the syllabus for the actual examination year; the examples in this article are original teaching exercises, not official examination questions.
