The point of a real-world question is to make the arithmetic answer a real question
A school club wants to explain how a fictional water-recovery model works. Its science notes describe a separation method, but its Mathematics problem is different. The club needs to estimate how much liquid it could collect, how many sample containers it could fill, and whether a storage limit makes its plan impossible. A pupil can calculate a percentage correctly and still give the wrong answer if the final question asks for whole containers rather than litres. Another pupil can use the right capacity and still fail if the quantity is measured before rather than after a recovery loss. Such errors arise from confusing stages of a plan, not from forgetting multiplication.
This chapter is a Mathematics learning exercise, not a water-treatment guide. All quantities and results are invented for students. A recovered liquid from a classroom demonstration must not be treated as drinking water. The preceding Volume 0589 explains why a separation method must be chosen for its target component. Here we accept a recovery fraction as given and focus on the mathematical consequences. The fraction is a problem assumption, not a universal claim about a particular apparatus.
The question that organizes everything is simple: what is counted at each stage? The starting feed is not the output. Output liquid is not necessarily usable output. Usable output is not necessarily the number of completely filled containers. The total number of containers that could be filled is not necessarily the number that can be stored in the available space. Strong students learn to keep these as separate named quantities before combining them.
The official G3 SEC Mathematics boundary
The first Singapore–Cambridge Secondary Education Certificate examination takes place in 2027. The official G3 Mathematics subject is K310. According to the 2027 G3 syllabus index and K310 Mathematics syllabus, the assessment has two written papers, each lasting two hours fifteen minutes, each carrying 90 marks and 50% of the overall result. Paper 1 contains approximately 26 short-answer questions. Paper 2 contains nine or ten questions, with the final question focused on a real-world context.
This guide teaches calculations, representation, mathematical reasoning and evaluation useful in those contexts. It is not an official SEAB task and it does not predict which real-world setting will appear. The English and Chemistry links are there to support understanding, not because a G3 Mathematics candidate takes a single combined English–Mathematics–Science examination. A useful numerical model must also distinguish what its assumptions can prove and what they cannot.
The syllabus expects clear mathematical working and suitable accuracy. Unless the question specifies otherwise, non-exact numerical answers should generally be given to three significant figures, and angles in degrees to one decimal place. For exact bottle counts and batch counts, however, a count is an integer and the chosen rounding direction depends on the wording of the question. Never use three significant figures to turn a possible 27 complete bottles into 27.4 bottles; the latter is not a possible number of fully filled physical containers.
Build a quantity map before operating a calculator
Begin with a labelled chain: feed volume, recovered volume, volume available for filling, number of complete containers, number of batches, and amount requiring storage. If the task supplies more than one loss or constraint, add a new box for each step. The arrow between boxes must say which operation changes the quantity. For example, 18 L of feed multiplied by an assumed 75% recovery gives 13.5 L recovered. The next conversion, dividing by 0.5 L per bottle, gives 27 complete half-litre bottles. These are two distinct operations with different units.
The quantity map is not a decorative diagram. It tells you whether the denominator of a percentage refers to feed volume, collected volume or the filled portion. Imagine a club that says, “Three quarters of the feed was collected, so three quarters of the bottles are usable.” That sentence does not follow. We know a volume recovery fraction, not the fraction of bottles passing another quality check. To calculate both, the question must give a relationship or separate acceptance rate.
Label every result at the moment it is found. Write “13.5 L recovered per batch,” not just “13.5.” Write “27 whole 500 mL containers per batch,” not simply “27”. Your own labels become a final internal check. If an answer labelled “litres” is used where a question asks “bottles,” the missing division becomes visible before you submit.
Percentages: identify the original quantity before multiplying
A percentage is a relationship to a base. A recovery rate of 75% means 75 parts recovered from 100 comparable parts of feed in the stipulated model. Seventy-five per cent of 18 L is 0.75 × 18 L = 13.5 L. The remaining 4.5 L has not been collected as the defined recovery output. This does not by itself tell us where the remaining fluid has gone physically; its fate depends on the scientific setup, which this Mathematics task does not specify.
A common mistake reverses the operation. A learner sees the words “75% recovered” and divides 18 L by 0.75, obtaining 24 L. That calculation answers a different question: what feed would be required to recover 18 L at 75% efficiency? Both operations can be correct under different questions. The safeguard is to identify which quantity is given and which is unknown before choosing multiply or divide.
Another mistake treats a later percentage as if it were measured against the original 18 L. Suppose the question says 80% of the recovered 13.5 L is allocated to demonstration containers, while 20% is reserved for a separate non-consumption experiment. The allocated amount is 0.80 × 13.5 L = 10.8 L. It is not 80% of 18 L, because the instruction explicitly uses the recovered amount. The combined fraction of original feed allocated is 0.75 × 0.80 = 0.60, or 60%, which is why 10.8 L is six tenths of 18 L.
Units: a decimal error can become a whole-batch mistake
One litre equals 1,000 millilitres. A container labelled 500 mL holds 0.5 L; a 750 mL vessel holds 0.75 L; a 300 mL specimen container holds 0.3 L. When the recovered amount is 13.5 L, filling 500 mL bottles uses 13,500 ÷ 500 = 27 bottles. The same calculation in litres is 13.5 ÷ 0.5 = 27. Both are valid because the units are consistent.
If a pupil calculates 13.5 ÷ 500 and writes 0.027 bottles, the arithmetic itself may be correct for those two bare numbers, but the quantities were not expressed in compatible units. This is why a dimensional check is often faster than redoing all the working. Litres divided by litres per bottle gives bottles. Litres divided by millilitres per bottle does not form a consistent comparison until the units are converted.
For larger storage questions, capacity may be in cubic centimetres or cubic metres. One cubic centimetre corresponds to one millilitre; 1,000 cubic centimetres equal one litre; and one cubic metre equals 1,000 litres. Do not confuse linear scaling with volume scaling. Doubling the side length of a cubical container multiplies its capacity by eight, not two. If a question supplies dimensions rather than a capacity label, calculate the geometric volume first, convert to the same unit as the water volume, then consider any stated fill limit.
The baseline worked example: count what one batch can actually produce
The club’s fictional plan uses 18 L of feed per batch and a stipulated 75% recovered-volume fraction. The recovery output is 13.5 L. For 500 mL containers, 13.5 L is 13,500 mL, so 13,500 ÷ 500 = 27 whole containers. No rounding is necessary in this first case because the division is exact.
Now change the container to 800 mL. The same 13,500 mL divided by 800 mL is 16.875. If the question asks for completely filled 800 mL containers, the answer is 16, with 700 mL left over. Rounding to 17 would promise a complete 17th container without sufficient liquid. If the question instead asks how many containers are needed to hold all the liquid, including one partially filled container, the answer is 17. The numerical expression is identical, but the required interpretation changes the correct rounding rule.
This is a powerful general lesson. “Number of complete items possible” requires rounding down when a fraction remains. “Number of items required to hold or serve the full quantity” requires rounding up if partial items are permitted. Neither is the same as the calculator’s ordinary nearest-integer rounding. An examination task rewards the student who reads the physical constraint as carefully as the calculation.
The inverse problem: determine the least number of batches for a target
Suppose each batch supplies 27 fully filled half-litre display containers. The coordinator asks for at least 100 complete containers. Three batches produce 3 × 27 = 81, insufficient. Four produce 4 × 27 = 108, enough. Therefore the minimum is four batches. Dividing 100 by 27 gives approximately 3.7037; in the physical plan, 3.7037 batches cannot be carried out if every batch is indivisible. The minimum feasible whole number is four.
To write an exam-quality answer, show the comparison to the threshold rather than merely “round up.” State that 3 batches yield fewer than 100 containers and 4 batches reach the requirement. This proves minimality. Without the comparison, an answer could have been guessed by rounding a decimal without understanding the constraint.
The number of batches could change if the task permits partial batches or includes a reserve. If 10 bottles must remain unused for a later demonstration, meeting a requirement of 100 distributed bottles actually needs at least 110 complete bottles available. Four batches produce 108, so five batches would then be necessary. One apparently minor phrase—“plus ten spare”—changes the mathematical target. Always rewrite the target with every added condition before applying a ceiling.
Storage is a separate constraint, not a footnote
Assume the club has one labelled storage vessel with a maximum capacity of 50 L. If each batch recovers 13.5 L, three full batches yield 40.5 L and fit. Four yield 54 L, exceeding capacity. A student who calculates that four batches give 108 complete half-litre containers has answered the production question; the student who checks that 54 L will not fit has answered a different and equally important feasibility question.
If all recovered liquid must be stored before filling, the tank allows at most three batches at a time. That conflicts with the requirement for at least 100 complete containers if the bottles have to come from one storage period. The answer is not to ignore the tank and announce that four batches meet the target. It is to identify the contradiction and propose a change justified by the question, such as an additional approved storage vessel or a different permitted collection schedule.
Do not invent a permitted workaround. If the task allows bottles to be filled immediately after each batch and removed from storage, a 50 L tank might no longer be the relevant limiting container. If the task states all output must be stored together for testing before filling, it definitely is. The mathematics tells us what is feasible under the stated rules. It cannot decide that a prohibited change of process is allowed.
Two-tank planning and the meaning of unused capacity
The club obtains two separate vessels, each holding no more than 30 L. Their combined nominal capacity is 60 L, so four full batches of 13.5 L produce 54 L and fit in total. However, there is another condition: the liquid must be distributed in a way that respects each vessel’s individual limit. Two batches into the first vessel give 27 L and two into the second give 27 L. Both are below 30 L, so the distribution is feasible.
A calculation of “combined capacity 60 L” alone would not prove that any distribution works when vessels have different permitted fills or batches cannot be split. For example, if three indivisible batches were allocated to a 30 L tank, they would total 40.5 L and exceed its limit. The total capacity can be sufficient while a particular allocation is impossible. Show the allocation, not only the combined litres.
The spare capacity is 6 L overall after collecting four batches. That does not mean 6 L of recovered liquid is left unfilled if bottles are filled later. Spare tank capacity is unused storage space; leftover liquid is output remaining after bottling. They have the same unit but different meanings. A word problem deliberately may place both values close together to check that the student maintains the correct quantity label.
Sequential losses: an efficient fraction does not cancel a later loss
Suppose a modified plan collects 80% of a 20 L feed, then loses 5% of the recovered liquid in a separate transfer. The collection output is 0.80 × 20 = 16 L. The usable volume after transfer is 0.95 × 16 = 15.2 L. It is not 20 × (0.80 − 0.05) = 15 L, because the 5% loss is calculated on the 16 L that exists after collection, not on the original 20 L.
The correct combined usable fraction of feed is 0.80 × 0.95 = 0.76, or 76%. The two percentages act on different bases in sequence. This is why a quantity map is more dependable than adding or subtracting percentage points automatically. A student who gets 15 L has mixed a percentage with a percentage-point adjustment without checking what the base is.
Now compare with a different statement: “The collection efficiency fell from 80% to 75%.” That is a drop of five percentage points, not a subsequent 5% transfer loss. With a 20 L feed, the first model would collect 16 L and the second would collect 15 L. The same printed numbers can produce different answers because the language describes different operations.
Multiple constraints: prove a plan meets them all
A fictional exhibition brief demands at least 100 complete 500 mL samples, no more than four runs, and storage of no more than 60 L at one time. At 13.5 L per batch, four runs yield 108 complete samples and 54 L. The plan meets all three conditions: the sample target, the run limit and the capacity limit. The answer should list each check. An expression such as “108 is more than 100” verifies only one requirement.
Change the storage limit to 50 L, and the same plan fails. Change the run limit to three, and the same plan fails. Change the output requirement to at least 120 bottles, and it fails again. A mathematical recommendation is valid only inside the intersection of the constraints, not whenever one attractive number is reached.
When the problem offers two or three feasible options, compare the stated objective as well. If the task asks for minimum runs, choose the fewest-run feasible plan. If it asks for maximum number of full bottles under a storage limit, evaluate that quantity instead. “Best plan” without a declared objective is an incomplete answer. Real-world questions often test the distinction between a constraint that must be satisfied and an objective to be improved.
Rates and unequal intervals: average the correct quantities
Suppose a prototype collects 3.0 L in its first 20 minutes and 5.0 L in the next 40 minutes. The first interval rate is 3.0 ÷ 20 = 0.15 L/min, and the second is 5.0 ÷ 40 = 0.125 L/min. The total is 8.0 L over 60 minutes, so the overall average is 8 ÷ 60 = 0.1333… L/min, or approximately 0.133 L/min to three significant figures.
The unweighted mean of 0.15 and 0.125 is 0.1375 L/min, which is not the correct overall rate because the intervals have different durations. This mistake is closely related to averaging class averages without considering group sizes. Always reconstruct the combined numerator and denominator if an overall rate is requested.
Be cautious about using an observed average rate to predict later output. A model that produced 8 L during one hour has an observed average, not necessarily a constant production rate for every hour. If the question explicitly assumes constant rate for a forecast, state that assumption and calculate. Otherwise use language such as “at the observed average rate, the estimated time would be…” rather than a guarantee.
Tables, graphs and deliberate reasonableness checks
A good working table keeps one row per batch or interval and columns for feed volume, recovery fraction, recovered volume, transferred usable volume and number of full containers. Never put “80%” in a litre column without indicating that it is a ratio. Likewise, a graph of cumulative output against time must use litres on the vertical axis and time on the horizontal axis. The gradient over an interval has units of litres per unit time; cumulative height has units of litres. These are not interchangeable.
A reasonableness check can be performed without repeating the exact calculation. Since 75% is below 100%, the recovered volume should be below 18 L. Since each half-litre container holds less than one litre, 13.5 L can fill more than thirteen containers. Since four batches require more feed and create more output than three, a fixed-capacity storage limit might first become binding at four. Quick relationships like these expose reversed division and unit errors.
For a graph with a non-zero starting reading, distinguish total cumulative output from a change in reading. If a tank already holds 12 L and ends with 39 L, the added quantity is 27 L, not 39 L. The initial level is part of the system; forgetting it inflates the calculated recovery. A graph question may test this with a vertical-axis intercept or a short sentence beneath the graph.
A fully worked decision: production, storage, bottling and distribution
A final fictional case says: Each run begins with 24 L of feed. The model recovers 75% of the input. Every recovered batch undergoes a 10% transfer loss. The club requires at least 90 completely filled 500 mL display bottles. A single storage container is limited to 50 L before bottling. The coordinator may perform whole runs only, and all usable output must be held in the storage container at once. Decide whether the club can meet the brief with the current container.
First find recovered liquid: 24 × 0.75 = 18 L per run. Then find usable liquid: 18 × 0.90 = 16.2 L per run. Each run could fill 16.2 ÷ 0.5 = 32 full half-litre bottles, with 0.2 L remaining if bottles are filled separately after that run. It is important not to round the usable liquid before finding the required number of runs. Three runs yield 48.6 L usable and can fill 97 complete 500 mL bottles if the liquid from runs can be pooled before filling, leaving 0.1 L. Two runs yield 32.4 L and therefore at most 64 complete bottles if pooled, which is below 90.
The minimum number of runs is three. The 48.6 L accumulated after three runs fits the 50 L container. Thus the plan satisfies the two key conditions under the stated assumptions. The result is different from counting 32 full bottles from each individual run and multiplying by three to get 96: pooling the 0.2 L surplus from each run creates enough additional liquid to fill one more bottle. The brief specifically says all usable output is collected before bottling, so the pooled count of 97 applies. If leftover liquid from each run had to be discarded before the next run, the outcome would be 96.
The last step is a complete sentence: “Three whole runs produce 48.6 L of usable liquid, enough for 97 complete 500 mL display bottles after pooling, and the liquid fits the 50 L tank. The plan meets the 90-bottle requirement under the stated assumptions.” This answer connects calculations, whole-number interpretation and a storage check. It is not a recommendation to use any classroom-produced water for drinking.
Answer quality in Paper 1 and Paper 2
For a short-answer calculation, the most useful structure is quantity, formula, substitution and an answer with units. For a multiple-condition real-world question, show an additional line for each constraint and finish with a comparison. You do not need a long essay to explain that 48.6 L is less than 50 L. You do need to identify that 48.6 L is the usable volume that the brief requires storing, rather than 54 L of pre-transfer recovered liquid.
Use calculator results sensibly. Keep exact values in the working where feasible, then round a decimal answer as instructed. Do not round a fractional number of required batches to the nearest integer when the task requires a minimum. Do not round a fractional number of complete containers up. Do not hide a contradictory limit merely because the target was reached.
The earlier Volume 0586 teaches concentration, units and denominators; Volume 0582 treats integer feasibility in stock allocation. The present topic has a different owner: sequential volume recovery and whole-container logistics. For students rebuilding earlier habits, use the PSLE Mathematics word-problem route and the post-PSLE transition guide.
Twenty-four exam-style practice decisions with complete reasoning
Work through each case before reading its solution. They are deliberately short enough to practise in mixed revision, but each tests a different structural decision: which base to use, what must be counted, how to round, whether pooling is allowed or why a plan fails. Invented classroom figures are used solely for mathematical reasoning.
Practice 1: Reverse the recovery percentage
Task. The output is 15 L at a 60% recovery fraction. What feed volume was required?
Worked solution. Let the feed be F litres. Since 0.60F = 15, F = 15 ÷ 0.60 = 25 L. Dividing is appropriate because the output is supplied and the feed is the unknown.
Transfer check. Check: 60% of 25 L equals the stated 15 L. The forward check prevents confusing the inverse and direct problems.
Practice 2: Convert a mixed unit
Task. A display uses 12.6 L and each container holds 350 mL. How many complete containers?
Worked solution. Convert 12.6 L to 12,600 mL. Divide 12,600 by 350 to get 36 complete containers, exactly.
Transfer check. An answer in containers only makes sense after the liquid volume and container size share the same unit.
Practice 3: Reject a tempting rounded bottle
Task. A run produces 7.3 L for 600 mL bottles. Count complete bottles.
Worked solution. 7,300 ÷ 600 = 12.1666…, so only 12 bottles can be fully filled. Exactly 100 mL is left after the twelfth bottle.
Transfer check. Round down for completely filled items; nearest-integer rounding happens to agree here but is not the governing rule.
Practice 4: Count containers required to hold everything
Task. The same 7.3 L must be stored in containers of at most 600 mL each, including a partial last container.
Worked solution. Twelve containers hold 7.2 L, so a thirteenth is needed for the remaining 0.1 L. The answer is 13 containers, with one partially filled.
Transfer check. Round up when the target is capacity to contain all liquid, not the count of fully filled units.
Practice 5: Find the minimum whole runs
Task. A batch yields 14 complete samples and the club needs at least 100.
Worked solution. Seven batches yield 98, which is insufficient; eight yield 112, which meets the target. Eight whole batches are required.
Transfer check. Comparing the result just below the answer proves minimality.
Practice 6: Apply two losses on their proper bases
Task. From 25 L, 80% is recovered and 10% of the recovered volume is lost during transfer.
Worked solution. The first step gives 20 L. The second retains 90% of 20 L, giving 18 L usable. Multiplying 0.8 × 0.9 gives 0.72 of the original.
Transfer check. Never subtract ten percentage points from the first percentage unless the problem explicitly defines that relationship.
Practice 7: Compare percentage change and points
Task. Recovery efficiency changes from 70% to 77%. What are the point and relative changes?
Worked solution. The difference is 7 percentage points. Relative to the original 70%, the increase is 7 ÷ 70 × 100% = 10%. Both descriptions can be true.
Transfer check. The base for relative change is the original percentage, not the whole 100% automatically.
Practice 8: Check a single tank
Task. Four runs produce 13.5 L each. The tank holds 50 L.
Worked solution. Total production is 54 L, exceeding the limit by 4 L. Therefore all four runs cannot be stored together in that tank.
Transfer check. A target can be achieved mathematically while the plan fails a separate capacity condition.
Practice 9: Allocate indivisible batches to two tanks
Task. Four batches of 13.5 L must fit in two 30 L tanks without splitting any batch.
Worked solution. Put two batches in each tank. Each holds 27 L and stays under its 30 L maximum. The allocation is feasible.
Transfer check. State the actual distribution; summed capacities alone need not show a valid allocation.
Practice 10: Find what is left after complete bottles
Task. A run yields 11 L for 750 mL bottles.
Worked solution. 11,000 ÷ 750 = 14 full bottles with 500 mL left. Fifteen full bottles would require 11.25 L, which is unavailable.
Transfer check. The remainder is a liquid volume, not a fraction of a bottle to be rounded away.
Practice 11: Pool leftovers before filling
Task. Three runs each yield 2.8 L; the bottles hold 500 mL. Bottling is done only after all runs.
Worked solution. The pooled 8.4 L produces 16 complete bottles and leaves 0.4 L. Bottling separately would yield 5 each, or 15 in total.
Transfer check. The time at which pooling occurs changes the whole-number result even when total output is fixed.
Practice 12: Use the correct weighted mean
Task. One stage collects 4 L in 20 min and another 9 L in 60 min. What is the overall rate?
Worked solution. Total volume is 13 L, total time 80 min, so average = 13/80 = 0.1625 L/min, or 0.163 L/min to 3 significant figures.
Transfer check. Do not average stage rates directly when the stage durations differ.
Practice 13: Convert a rate to an hourly figure
Task. A stipulated constant rate is 0.18 L per minute. What volume in 2.5 hours?
Worked solution. 2.5 h = 150 min. At 0.18 L/min, volume = 0.18 × 150 = 27 L.
Transfer check. Rate times duration is meaningful only if the time unit matches the denominator of the rate.
Practice 14: Recover the starting tank quantity
Task. A tank read 8 L before collection and 35.5 L afterward.
Worked solution. The increase is 35.5 − 8 = 27.5 L. The final reading is not itself the recovered output.
Transfer check. Check the initial level before treating a gauge reading as a newly produced volume.
Practice 15: A cubic storage container
Task. A cuboid tank measures 40 cm by 30 cm by 50 cm. Find its full volume in litres.
Worked solution. Volume = 40 × 30 × 50 = 60,000 cm³ = 60 L. A separate fill-limit percentage may reduce the usable capacity.
Transfer check. Length, width and height are multiplied; the cubic-centimetre to litre conversion is 1,000 to one.
Practice 16: Apply a 90% fill limit
Task. A nominal 60 L container must be filled to at most 90% of its capacity.
Worked solution. Maximum permitted liquid = 0.90 × 60 = 54 L. The 60 L physical volume is not the relevant working limit.
Transfer check. A capacity constraint can have both a geometric limit and an operational limit.
Practice 17: Test two alternative plans
Task. Plan A yields 15 L per run for 3 runs. Plan B yields 11 L per run for 4 runs. Which produces more?
Worked solution. A produces 45 L; B produces 44 L. A produces 1 L more, but without a stated cost or capacity goal this does not establish that A is always the best plan.
Transfer check. Compare against the given objective and keep unknown criteria out of the conclusion.
Practice 18: Include a reserve
Task. The club must distribute 90 full bottles and keep 12 spare bottles. Each run supplies 25 complete bottles.
Worked solution. At least 102 bottles are needed. Four runs supply 100, insufficient; five supply 125. Five runs are required.
Transfer check. Translate words such as spare and reserve into the total requirement before dividing.
Practice 19: Interpret a fraction without a calculator shortcut
Task. A 12 L batch yields 7/8 of its initial volume.
Worked solution. Recovered volume = 12 × 7/8 = 10.5 L. The unrecovered amount is 1.5 L.
Transfer check. An exact fraction can be preferable to a rounded decimal in intermediate work.
Practice 20: Handle a transfer-limited target
Task. Four runs produce 20 L each but a transfer process retains only 85%. How much can be bottled?
Worked solution. Recovered total before transfer = 80 L. Usable amount = 0.85 × 80 = 68 L. For 500 mL bottles this is 136 complete bottles.
Transfer check. A full batch’s output is not automatically the amount available for the final product.
Practice 21: Identify a false forecast
Task. A one-hour test produced 6 L. A notice promises that a four-hour run always produces 24 L.
Worked solution. The 24 L forecast requires assuming the same average rate across four hours. The one-hour observation by itself does not establish that constant rate.
Transfer check. Distinguish calculation under an assumption from a guarantee proved by observed data.
Practice 22: Resolve a threshold conflict
Task. The plan needs 120 bottles at 500 mL each, but the usable-volume cap before bottling is 55 L.
Worked solution. The target represents 60 L of usable liquid, exceeding the 55 L cap. The plan as stated is infeasible if all volume must be stored together.
Transfer check. Prove impossibility by converting both the desired quantity and the constraint to the same unit.
Practice 23: Choose objective and feasibility separately
Task. Plan A makes 102 bottles in 5 runs; plan B makes 98 in 3 runs; at least 100 are required.
Worked solution. Only A satisfies the target, so B’s smaller run count cannot make it preferable under the mandatory condition. A is the feasible option among these two.
Transfer check. An infeasible plan cannot win simply because it optimises the wrong quantity.
Practice 24: Write an exact final decision
Task. Three runs give 48.6 L usable, storage is 50 L, and 90 half-litre display bottles are required.
Worked solution. 48.6 L fits the tank; it holds enough for 97 complete half-litre bottles after pooling. Three runs are feasible for this display goal.
Transfer check. A final recommendation must explicitly verify both production and storage constraints.
A reusable final-examination checklist
Read the last sentence of the question first to identify the target output. Read the data to identify the units. Draw the stage chain, calculate percentages on the named base, and convert capacity units before counting containers. If a fraction of a physical item appears, decide whether the word ‘full’, ‘required’, ‘minimum’ or ‘maximum’ determines the rounding direction. Once an attractive number is reached, do not stop: test every storage, timing, stock and minimum-demand constraint independently.
If the calculation cannot give a feasible plan under the stated rules, say so plainly. A correct mathematical conclusion may be that the requirement is impossible with the current equipment or assumptions. That is not a failed attempt; it is the point of a mathematical feasibility problem. Do not quietly enlarge a storage vessel, alter a recovery rate or permit half a batch unless the question authorises it.
Where to continue
The next English guide will show how to write a careful approval email from the same broad type of school exhibition scenario, without treating the calculation as proof of consumer safety. Revisit Volume 0589 for what filtration and distillation actually demonstrate, Volume 0586 for denominators, Volume 0582 for whole-number allocation and the official G3 syllabus index to verify the current examination contract. All practice tasks are original, fictional and not official past-paper material.
