The mathematics of moving a harvest
The habitat harvest has been counted. A trolley must carry cartons from a packing bench through a corridor to a dispatch door, then return for the next load. The question is no longer how a force changes the trolley’s motion. It is whether a movement plan works when distances, times, loading limits and the available dispatch window are stated together. An answer which calculates speed correctly but promises an impossible number of trips is not a completed solution.
This volume continues the Physics stopping-distance investigation in Vol 1276. That guide owns forces, acceleration and experimental limitations. Here the focus changes to quantities, elapsed time, unit rates, feasible integer journeys and honest decisions. The earlier harvest allocation mathematics guide owns carton allocation and reverse percentages. Keep the boundaries separate so that the new problem teaches a new skill.
Start with a simple principle: a journey is a sequence of activities, not merely a distance. Moving, stopping, loading, unloading and returning all consume time. A school examination may hide that principle in a dense paragraph. The strongest first move is to write each stage in order before calculating a grand total.
Official 2027 examination frame
The Singapore-Cambridge SEC examinations begin in 2027 under Full Subject-Based Banding. The published SEAB G3 Mathematics syllabus is K310, replacing the earlier reference code 4052. The official 2027 G3 subject index and K310 syllabus are the source of examination rules; the constructed dispatch case below is an original teaching example, not an official question or a prediction of future paper content.
K310 has two written papers. Paper 1 lasts 2 hours 15 minutes, has 90 marks and contributes 50%. Paper 2 also lasts 2 hours 15 minutes, has 90 marks and contributes 50%. In Paper 1, candidates answer roughly 26 short-answer questions. Paper 2 has about nine or ten questions, with the final question devoted to applying mathematics to a real-world situation. Essential working matters; the syllabus specifically warns that its omission loses marks. Unless the question says otherwise, non-exact numerical answers normally require three significant figures, and angles in degrees one decimal place.
This guide concentrates on useful mathematics for an extended real-world task: reading data and schedules, using distance-time relationships, deciding what belongs in a total, dealing with whole numbers of journeys and testing the meaning of the answer. It is not a substitute for the wider algebra, geometry, statistics and probability syllabus. Use it to strengthen the type of reasoning that must transfer between topics.
A complete constructed route: know the numbers before working
Assume the packing bench is P, the corridor checkpoint is C, and the dispatch door is D. The distance P–C is 48 metres; C–D is 72 metres. A loaded outward trip takes 30 seconds on the first section and 45 seconds on the second. The trolley waits 15 seconds at C. Loading at P takes 45 seconds, unloading at D takes 60 seconds, the empty return to P takes 60 seconds, and the trolley needs a further 20 seconds of resetting before the next cycle.
Each journey carries at most 12 cartons. Each carton has a mass of 2.4 kg; the empty trolley has a mass of 30 kg. The dispatch crew has a 42-minute window, but a two-minute close-out buffer must remain unused. All these data are fictional and intended for mathematical reasoning; they do not certify how any real trolley should be operated or how many loads an actual facility can safely transport.
Before choosing a formula, draw a row of activities: load 45, travel 30, checkpoint 15, travel 45, unload 60, return 60, reset 20. The cycle lasts 275 seconds, or 4 minutes 35 seconds. The route length for the loaded movement is 120 metres. The full round-trip movement covers 240 metres if the return uses the same physical route. A full cycle, however, is not a single continuous 240-metre motion. It includes work and waiting. Different questions ask for different quantities.
Average speed is tied to a specific interval
The outward movement covers 120 metres in 75 seconds of actual travelling time, giving an average moving speed of 1.6 metres per second. Add the 15-second checkpoint stop and the outward elapsed time becomes 90 seconds. The average speed over that outward elapsed interval is 120 divided by 90, about 1.33 metres per second. Both calculations are correct when their time intervals are stated honestly. Neither is interchangeable with a peak speed read from a separate instantaneous measurement.
If the question asks for the average speed from the start of loading to the end of unloading, the denominator changes again. The outward process takes 45 + 30 + 15 + 45 + 60 = 195 seconds. The outward distance remains 120 metres and average speed for the whole process is about 0.615 metres per second, if we deliberately define speed over that complete elapsed interval. This process average measures throughput rather than trolley motion alone. A clear answer must define the interval, or the numerical result can mislead.
At Primary level, learners meet the relation speed = distance divided by time. In the SEC setting, they must use it while reading schedules, distinguishing stationary stages from moving stages and checking whether the question asks for movement, elapsed time or a full production cycle. The Primary 6 Mathematics Learning Hub is a useful bridge when the basic units or rate relationship remains unstable.
The dispatch decision: a correct calculation can still be infeasible
The available time for complete cycles is 42 − 2 = 40 minutes, or 2400 seconds. Divide 2400 by 275 to obtain approximately 8.73 cycles. That quotient is not the number of completed journeys. Only eight whole journeys fit, because the ninth would exceed the reserved time. Eight complete journeys move at most 96 cartons. Nine could carry 108 but require 2475 seconds, which exceeds the usable 2400 seconds.
Suppose the order contains 100 cartons. A response of ‘8.73 trips, therefore 104.7 cartons’ would ignore the physical meaning of a trip and the 12-carton maximum. Eight trips cannot deliver 100. A valid proposal must change a genuine constraint—such as adding a separate approved vehicle, extending the available window or revising the order—not invent a fractional trolley journey. This is the role of integer reasoning in a real-world context.
Do not claim that an increase in speed is safe merely because it improves the arithmetic. Stopping distances, corridor traffic and the operator’s controls belong to the supervised physical safety decision discussed in Vol 1276. Mathematics can show feasibility under a model. It cannot by itself prove safe operation. A responsible conclusion names both the mathematical result and the decision outside the evidence.
Build a dependable examination route
Read the final sentence first. It may ask for a speed, a duration, a maximum number of whole loads, a missing deadline or a recommendation. Underline the requested quantity, then map the known facts by units. Convert minutes and seconds only when necessary, keeping intermediate values exact. Work forward through the activity sequence and backward from the deadline to test the same result.
Mark what is fixed and what is optional. In the example, the 15-second checkpoint wait and the two-minute buffer cannot disappear simply because the candidate wants more trips. If an alternative plan legitimately removes a wait, state the changed assumption explicitly and recompute the total. A changed assumption should never be smuggled into the answer as though it were part of the question.
Use an equation or table that another person can audit. Record time in seconds; convert the final answer back to minutes and seconds where helpful. Keep a separate note for carton capacity, because cartons per journey and journeys per window are different multipliers. Lastly, return to the language of the question: eight journeys, 96 cartons, and four cartons left undelivered in the stated scenario.
How graphs help and how they mislead
A distance–time graph shows an accumulated distance from a stated start. Its gradient across a straight segment is average speed for that segment. A flat section means the recorded distance did not change during that interval. An upward line does not mean acceleration by itself; acceleration requires a change in gradient or direct speed information. Read the axes before naming the physical process.
For cumulative distance along a route, the plotted value need not fall during a return trip; it depends on whether the graph tracks total path travelled or displacement from P. One graph can increase throughout, while another returns toward zero. Never infer that a graph is wrong just because it differs from the mental picture. The label defines the quantity.
In a timetable problem, a horizontal gap may describe waiting; a jump in a table could describe a measurement at a new checkpoint, not an impossible instantaneous movement. When a line graph is supplied, convert each section into a clear statement about the state of the journey. When data alone are given, use a table before drawing a line.
Rounding and checking without losing meaning
K310 asks for appropriate numerical precision, but feasibility questions need an additional form of rounding: whole units. You may report an exact rate as a fraction or a three-significant-figure decimal. You may not round 8.73 journeys up to nine when the ninth cannot finish. Rounding of measured values and rounding of a count answer different questions.
Check an average speed by multiplying it by the relevant time. A result of 1.6 metres per second over 75 seconds should recover 120 metres. Check a cycle total by adding the stage durations a second way, perhaps grouping moving and stationary stages: 30 + 45 + 60 = 135 seconds moving; 45 + 15 + 60 + 20 = 140 seconds stopped or handling; total 275 seconds. A good check uses a different route where possible.
If an answer is nearly at a constraint, leave enough precision in intermediate steps to decide correctly. Do not round 4 minutes 35 seconds to five minutes before calculating whether eight or nine trips fit. A convenient-looking approximation may change the decision.
A worked alternative that truly changes the answer
Imagine an authorised process improvement cuts the reset from 20 seconds to 10, with every other stage unchanged. The cycle becomes 265 seconds. In 2400 seconds, nine complete cycles require 2385 seconds, so the same trolley could mathematically carry at most 108 cartons within the usable window. Only 15 seconds would remain. The computation demonstrates the difference a genuine change makes.
Would the proposal be sensible in practice? The arithmetic does not tell us whether a ten-second reset leaves enough time for checks, inspection or operator recovery. That question requires measured evidence and a safety procedure. An exam solution earns clarity by separating ‘mathematically feasible under the stated assumptions’ from ‘approved and safe in reality’.
One tiny number—ten seconds—has changed the result from eight journeys to nine. This is why an extended question may ask candidates to evaluate options rather than merely calculate a speed. The final decision depends on the constraint that actually binds.
Transfer from PSLE to G3 without changing the subject
The PSLE Learning Guide introduces the habit of identifying the first broken relationship. That habit remains useful in Secondary Mathematics. If a learner can compute 120 ÷ 75 but cannot identify which 75 seconds the question refers to, more division exercises will not solve the problem. The first weak link is interpretation.
Use the Mathematics Learning Hub for the wider structure of number, algebra, measurement and reasoning. For Science evidence and fair testing, PSLE Science Learning Guide connects observations and claims, but upper-secondary Physics uses a more formal set of models. These are different owner routes with complementary purposes.
Move from a trusted problem to a changed one. Change the return time, load limit or deadline. Ask which earlier calculations remain valid and which must be rebuilt. The goal is to make the method independent of the story rather than memorise a fixed number sequence.
Practice bank: how to use each question
The 65 checks below are original examples built around the constructed dispatch system. They are not SEAB past-paper questions or official marking guidance. Try one without reading its answer, show each quantity with units, and then use the explanation to find the first incorrect step. The problems move from direct rates to multiple constraints, graph interpretation and judgement. Several deliberately ask for what cannot be concluded.
To get more value from each check, close the answer, change one number and solve again. If the new result changes an operational claim, explain why. A learner is ready to progress when the reasoning works on a modified task without relying on remembered numerical answers.
Sixty-five progressively harder worked checks
Check 01: Find the distance P–D.
Worked answer. Add 48 m and 72 m to obtain 120 m. These two sections form one continuous outward route.
Why this matters. Do not double the distance unless the question includes the return journey. A route sketch makes the ownership of each length visible.
Check 02: Find the outward moving time.
Worked answer. Add the two travel periods, 30 s and 45 s, to obtain 75 s.
Why this matters. A checkpoint stop is not travel time. Read the phrase moving time carefully before building a denominator.
Check 03: Find the average moving speed on P–C.
Worked answer. 48 ÷ 30 = 1.6 m/s.
Why this matters. State metres per second. The result is an average over the section, not proof that the speed was constant throughout.
Check 04: Find the average moving speed on C–D.
Worked answer. 72 ÷ 45 = 1.6 m/s.
Why this matters. Equal sectional averages need not mean the same acceleration or the same momentary speed pattern. The timetable does not describe those details.
Check 05: Find the combined outward average while moving.
Worked answer. The total distance is 120 m and the travel time is 75 s, so the mean is 1.6 m/s.
Why this matters. Using the arithmetic mean of sectional speeds is unreliable when intervals differ. Total distance divided by total travel time is the dependable route.
Check 06: Include the checkpoint wait in outward elapsed time.
Worked answer. 75 s moving plus 15 s stationary gives 90 s.
Why this matters. An elapsed-time question counts time when the position does not change. A student who leaves out the wait has answered a different question.
Check 07: Find average outward speed including the wait.
Worked answer. 120 ÷ 90 = 1.333…, so about 1.33 m/s.
Why this matters. Write the interval in the answer. Average moving speed and average speed across the full elapsed interval are not synonymous.
Check 08: Calculate the outward process time from load to unload.
Worked answer. 45 + 30 + 15 + 45 + 60 = 195 s, or 3 min 15 s.
Why this matters. This includes handling and waiting but excludes the return and reset. Draw the activity chain first so one stage cannot be lost.
Check 09: Find the full cycle duration.
Worked answer. 195 s plus 60 s return plus 20 s reset equals 275 s.
Why this matters. A cycle returns the trolley to a state ready for another load. Omitting reset means the next cycle starts before the prior one is finished.
Check 10: Convert 275 seconds to minutes and seconds.
Worked answer. 275 = 4 × 60 + 35, so 4 min 35 s.
Why this matters. Minutes-and-seconds notation is not a decimal. Writing 4.35 minutes would wrongly represent 261 seconds.
Check 11: Convert 42 minutes to seconds.
Worked answer. 42 × 60 = 2520 s.
Why this matters. Keep the same units when comparing a time limit and per-cycle times. Mixed-unit division is a common source of impossible answers.
Check 12: Allow for the two-minute buffer.
Worked answer. Two minutes is 120 s, leaving 2520 − 120 = 2400 s for cycles.
Why this matters. The buffer is a hard condition. Removing it without authority can make a plan look feasible when it is not.
Check 13: Compute a quotient for the number of trips.
Worked answer. 2400 ÷ 275 ≈ 8.727.
Why this matters. This quotient is a mathematical upper ratio, not a permitted fractional journey. The physical question asks how many full cycles finish.
Check 14: Report the maximum whole journeys.
Worked answer. Eight journeys are possible because 8 × 275 = 2200 s, while nine require 2475 s.
Why this matters. For a maximum count that must fit, round down after checking the next integer. Never apply ordinary nearest-integer rounding blindly.
Check 15: Calculate cartons transported in eight journeys.
Worked answer. 8 × 12 = 96 cartons.
Why this matters. A per-journey capacity becomes a total only after multiplying by completed journeys. The model assumes each journey is fully loaded.
Check 16: Identify the shortfall for 100 cartons.
Worked answer. 100 − 96 = 4 cartons still not transported.
Why this matters. The shortfall is in cartons, not trips or seconds. Put the final quantity next to the number to avoid an ambiguous conclusion.
Check 17: Find the exact time of nine original cycles.
Worked answer. 9 × 275 = 2475 s, or 41 min 15 s.
Why this matters. Nine cycles fit the headline 42-minute period, but not the 40 minutes left after reserving the buffer. Both restrictions matter.
Check 18: Calculate unused cycle time after eight trips.
Worked answer. 2400 − 2200 = 200 s, or 3 min 20 s.
Why this matters. Unused time cannot be assumed sufficient for another whole trip. It is shorter than the 275-second cycle.
Check 19: Compare two proposed shipment totals.
Worked answer. Eight full loads give 96 cartons; nine give 108.
Why this matters. A request for 100 cartons needs nine journeys under a 12-carton capacity. The number of required journeys is the ceiling of 100 ÷ 12.
Check 20: Find the least whole journeys needed for 85 cartons.
Worked answer. 85 ÷ 12 = 7.083…; eight journeys are required.
Why this matters. For a minimum count to meet demand, round up, not down. State why the seventh journey carries only 84 cartons.
Check 21: Calculate the mass of one full load.
Worked answer. 12 × 2.4 = 28.8 kg of cartons.
Why this matters. Carton mass and trolley mass are different quantities. The load mass alone is not the total mass being moved.
Check 22: Find the total mass of loaded trolley and cartons.
Worked answer. 28.8 + 30 = 58.8 kg.
Why this matters. This is mass, not weight; a Physics question may ask for a force in newtons. Here the mathematical job ends with kilograms.
Check 23: Find the mass for a half load.
Worked answer. Six cartons have a combined mass 6 × 2.4 = 14.4 kg.
Why this matters. Do not halve the trolley’s 30 kg empty mass. The fixed part stays fixed when the number of cartons changes.
Check 24: Compute the loaded total for six cartons.
Worked answer. 30 + 14.4 = 44.4 kg.
Why this matters. An expression such as 30 + 2.4n shows the structure. It makes the answer easy to adapt when n changes.
Check 25: Form an algebraic expression for n cartons.
Worked answer. The load-plus-trolley mass is M = 30 + 2.4n kilograms.
Why this matters. State the domain: n is a non-negative whole number, normally at most 12 in this constructed case.
Check 26: Find n if the total mass is 54 kg.
Worked answer. 30 + 2.4n = 54; subtract 30 and divide by 2.4 to get n = 10.
Why this matters. Checking 10 × 2.4 + 30 = 54 proves the equation fits the story. A fractional carton count would need interpretation.
Check 27: Find the required cycle time for nine trips in 40 minutes.
Worked answer. 2400 ÷ 9 = 266.67 s per cycle or less on average.
Why this matters. The original 275-second cycle is too long. This calculation identifies how much change is needed rather than arbitrarily increasing speed.
Check 28: Calculate the minimum reduction per cycle for nine trips.
Worked answer. 275 − 2400/9 ≈ 8.33 s reduction per journey.
Why this matters. A practical whole-second revised cycle needs to be at most 266 s, not 267, because 9 × 267 = 2403 s.
Check 29: Test a ten-second reset instead of twenty.
Worked answer. The new cycle is 275 − 10 = 265 s; nine cycles take 2385 s.
Why this matters. One actual change can reverse a feasibility conclusion. The mathematical result still does not certify adequate safety checks.
Check 30: Find slack with the revised cycle.
Worked answer. 2400 − 2385 = 15 seconds remain.
Why this matters. Fifteen seconds is a narrow tolerance. A model with no allowance for delays should not be described as operationally robust.
Check 31: Test a 5-second saving per trip.
Worked answer. Each cycle becomes 270 s. Eight require 2160 s, nine need 2430 s and do not fit.
Why this matters. A saving that feels large can still miss a hard threshold. Verify the integer boundary with multiplication.
Check 32: Test a 12-second saving per trip.
Worked answer. Each cycle becomes 263 s; nine require 2367 s, leaving 33 s.
Why this matters. The revised model changes the count to nine. State the changed assumption and whether the remaining margin is substantial.
Check 33: Convert the loaded distance for eight trips to kilometres.
Worked answer. 8 × 120 m = 960 m = 0.960 km.
Why this matters. This is outward loaded distance only. Adding returns would create a different total path length.
Check 34: Find all moving distance in eight complete round trips.
Worked answer. Each round trip travels 240 m, so eight cover 1920 m = 1.92 km.
Why this matters. Travel distance is not equal to net displacement. The trolley ends where it began after each full return.
Check 35: Find net displacement after a complete cycle.
Worked answer. The trolley starts and ends at P, so its displacement for the cycle is zero.
Why this matters. Zero displacement does not mean no distance was travelled. Identify which physical quantity the problem asks for.
Check 36: Find the portion of outward route on the second leg.
Worked answer. 72 ÷ 120 = 0.6, or 60%.
Why this matters. The reference whole is the outward route. Do not compare 72 m with 240 m unless the question expressly asks about the round trip.
Check 37: Find the route ratio P–C to C–D.
Worked answer. 48:72 simplifies by dividing both terms by 24 to 2:3.
Why this matters. The ratio is of distances, not necessarily durations or number of stops. A clearly named ratio can be checked against actual values.
Check 38: Find the fraction of outward moving time spent on P–C.
Worked answer. 30 ÷ 75 = 2/5 or 40%.
Why this matters. Use moving time as the denominator. If the question includes the checkpoint stop, the fraction changes.
Check 39: Find the percentage of each cycle spent travelling.
Worked answer. Moving time is 30 + 45 + 60 = 135 s; 135 ÷ 275 × 100 ≈ 49.1%.
Why this matters. This classification assumes the 60-second return is travel. Group stages by action before dividing.
Check 40: Find the handling-and-waiting portion of a cycle.
Worked answer. Non-moving stages total 45 + 15 + 60 + 20 = 140 s, about 50.9% of 275 s.
Why this matters. The two portions must add to 100%. This gives an independent consistency check on a category table.
Check 41: If load time rises by 10 seconds, find the new cycle.
Worked answer. The cycle becomes 285 s; eight cycles take 2280 s; nine take 2565 s.
Why this matters. A slower stage may not change the count immediately. The right answer reports the new capacity and remaining time.
Check 42: If return becomes 75 seconds, find the new maximum.
Worked answer. The cycle becomes 290 s. Eight take 2320 s, while nine need 2610 s.
Why this matters. The time constraint remains eight whole trips. Explain why a longer return does not necessarily reduce capacity below eight.
Check 43: If the waiting stage disappears by authorised redesign, recompute.
Worked answer. The cycle is 260 s; nine take 2340 s and leave 60 s.
Why this matters. Do not remove the checkpoint wait unless the problem explicitly grants the alternative. An option is not an instruction.
Check 44: Find the departure time of trip two from an 08:00 start.
Worked answer. One complete cycle ends at 08:04:35, so trip two begins at 08:04:35 if there is no extra delay.
Why this matters. A timeline makes clock arithmetic clearer. Use 24-hour notation and carry seconds correctly across minute boundaries.
Check 45: Find when the eighth original cycle ends.
Worked answer. Eight cycles use 36 min 40 s, so an 08:00 start ends at 08:36:40.
Why this matters. This is inside the 08:40 usable deadline if the 42-minute window starts at 08:00 with a two-minute buffer.
Check 46: Compute the loaded outward arrival time after loading.
Worked answer. Load 45 s, then travel 30 s, wait 15 s, travel 45 s: 135 s, so D is reached at 08:02:15.
Why this matters. Arrival at the dispatch door is earlier than completion of unloading. Distinguish an arrival-time task from a completed-transfer task.
Check 47: Compute unloading completion time for the first load.
Worked answer. Add another 60 s, giving 195 s after 08:00, or 08:03:15.
Why this matters. A labelled timetable should keep arrival and unloading completion in separate rows; they answer different questions.
Check 48: Find empty return completion after the first unload.
Worked answer. 195 + 60 = 255 s, or 08:04:15.
Why this matters. Reset still remains. The trolley has returned but is not yet ready for the next loaded cycle.
Check 49: Find when reset completes.
Worked answer. 255 + 20 = 275 seconds after 08:00, or 08:04:35.
Why this matters. A stage can be complete physically without the overall cycle being complete. Check the exact completion definition.
Check 50: Interpret a horizontal line on a cumulative-distance graph.
Worked answer. It marks an interval in which the recorded cumulative distance did not increase, for example a waiting period.
Why this matters. A horizontal segment does not prove zero effort or no work elsewhere in the process. It states only what the plotted variable records.
Check 51: Interpret a steeper line on a distance–time graph.
Worked answer. For the same axes and units, a larger positive gradient means a larger speed over that segment.
Why this matters. Do not compare visual steepness across graphs with different scales. Read numbers before judging slopes.
Check 52: Estimate speed from a 0-to-48 m straight segment over 30 s.
Worked answer. The gradient is (48 − 0)/(30 − 0) = 1.6 m/s.
Why this matters. A graph-gradient solution should show change in distance divided by change in time, not simply quote the height of a point.
Check 53: Distinguish a time graph from a speed graph.
Worked answer. Distance–time gradient tells speed; the area under a speed–time curve represents distance when speed stays non-negative.
Why this matters. The same shape can mean different things with different vertical-axis labels. Checking the axis is a prerequisite, not a finishing touch.
Check 54: Evaluate a claim that 1.6 m/s was the trolley’s maximum speed.
Worked answer. The section averages equal 1.6 m/s, but the data do not reveal the maximum instantaneous speed.
Why this matters. Average data alone do not settle a peak claim. An answer should say what is unknown rather than invent a motion profile.
Check 55: Evaluate a claim that nine journeys can deliver within 42 minutes.
Worked answer. Nine cycles take 41 min 15 s, so they fit the raw 42-minute window but violate the two-minute buffer.
Why this matters. Every qualifier matters. The correct decision comes from the full constraints, not a selectively chosen headline.
Check 56: Find required journeys for 120 cartons.
Worked answer. 120 ÷ 12 = 10 complete journeys at maximum loading.
Why this matters. Even though the load count is exact, the 40-minute window permits only eight cycles. Demand and capacity are separate checks.
Check 57: Find how many extra cartons nine journeys carry over eight.
Worked answer. 108 − 96 = 12 cartons.
Why this matters. A change of one whole journey adds one full-load capacity only if the ninth is fully loaded. Do not apply this to partly filled journeys blindly.
Check 58: Test whether 95 cartons fit in eight cycles.
Worked answer. Eight trips can carry 96 cartons, so 95 is within maximum load capacity if other conditions remain unchanged.
Why this matters. The last journey may be partly loaded. Capacity is a ceiling, not a requirement that every trip contains 12 cartons.
Check 59: Find the final load after seven full journeys for 95 cartons.
Worked answer. Seven full loads move 84 cartons, leaving 11 for an eighth journey.
Why this matters. Do not split 95 evenly across eight trips unless the question permits and requires it. Whole-carton arithmetic is appropriate.
Check 60: Find extra time needed for nine original trips while keeping buffer.
Worked answer. Nine cycles take 2475 s but only 2400 s are usable, so the shortfall is 75 s.
Why this matters. The required extension is one minute fifteen seconds, not the full ninth-cycle duration. Compare total demand with usable time.
Check 61: Check how much speed alone must change if moving time falls by 10 s.
Worked answer. With the route distance unchanged, moving time becomes 65 s and average moving speed is 120/65 ≈ 1.85 m/s.
Why this matters. This is an arithmetic what-if, not an instruction to increase trolley speed. Any physical change needs a separate safety assessment.
Check 62: Find the average load throughput for eight complete cycles.
Worked answer. 96 cartons in 2200 s gives 0.0436 carton/s, or about 2.62 cartons/min.
Why this matters. A rate of completed cartons per time is not the same as driving speed. State both the numerator and the chosen interval.
Check 63: Calculate a day’s transport if there are two identical 40-minute slots.
Worked answer. At eight full journeys per slot, the maximum is 16 journeys and 192 cartons under the same model.
Why this matters. This assumes resets, crew and inventory are independently available for both windows. State repeated-condition assumptions.
Check 64: Identify a misleading claim based on 8.73 trips.
Worked answer. The decimal quotient gives no permission for a partially completed trip to count as a delivered load.
Why this matters. The correct answer separates a real-valued calculation from a whole-number completion condition, then checks the boundary.
Check 65: Explain why a safety clearance cannot be established here.
Worked answer. This schedule gives travel distances and times but does not measure stopping capability, surface conditions or operator response.
Why this matters. Mathematical feasibility does not guarantee safe passage. Refer to the Physics investigation before making a claim about clearance.
Check 66: Write a defensible final recommendation for 100 cartons.
Worked answer. With the given cycle and 40-minute usable window, eight trips move at most 96 cartons. Four remain, so the plan does not meet demand.
Why this matters. A strong conclusion contains the binding constraint, computed capacity, shortfall and a realistic next option, without claiming more than the data show.
An independent 40-minute practice sequence
First, solve Checks 1–12 without a calculator so that the units, cycle definition and conversion steps become automatic. Next, solve Checks 13–32 and mark each answer that depends on a whole-number restriction. Then turn to Checks 33–49 to practise proportional reasoning, time arithmetic and comparisons. Finish with Checks 50–65 as a small decision exercise. Explain the reason for each limit in words, not only with a number.
On a second day, change the checkpoint delay to 18 seconds and the cycle capacity to 10 cartons. Rebuild the model before doing any mental substitution. On a third day, remove the two-minute buffer only as a deliberately stated alternative and compare the new answer with the original. These changed cases test transfer. A student who only remembers ‘eight journeys’ has not yet learnt the underlying method.
Keep a correction record with four columns: what I was trying to find, the first place my working became wrong, the corrected representation and the smallest different question I can now answer. A useful record names a specific failure such as ‘I used moving time instead of elapsed time’. Merely writing ‘careless’ rarely suggests an effective next lesson.
The final examination habit
At the end of a long contextual question, ask four questions. Have I answered the requested quantity? Are the units and the time interval correct? Have I honoured the whole-number and deadline restrictions? Does the conclusion stay within what the evidence supports? Those checks are short enough to use under time pressure and strong enough to catch many plausible-looking errors.
The mathematics of this habitat scenario is not the formula speed = distance divided by time in isolation. It is the discipline of using a formula in the right place, keeping the activity sequence intact, and treating a real-world restriction as part of the problem rather than an inconvenience to be rounded away. Continue from the Physics cause-and-effect guide toward the next English reading task, where the same dispatch incident will be presented through a person’s recollection rather than a clean timetable.
Official examination details remain those on SEAB’s 2027 G3 index and K310. The workbook numbers, schedules and solutions above are illustrative educational models. They are not real logistics guidance or a substitute for any workplace safety procedure.
