A stock-allocation table becomes a G3 Mathematics problem when quantities must remain whole, groups have different permitted routes, and a grand total alone is not enough to prove that a plan can work. The useful method is to identify each stock count, list the permitted paths, satisfy the most restricted requests first, and check every row, column, total and unit. A tempting arithmetic answer can still be impossible if it sends stock along an excluded path.
This learner’s guide uses fictional ABO-only red-cell labels as a mathematical setting. The biological rule is explained in Vol 0581, and no worked allocation here is a real blood-bank, hospital or patient-care instruction. Our task is mathematical: construct valid integer allocations, understand conditional percentages, test shortages and explain what a table proves and what it cannot prove.
The short answer
A feasible allocation uses only permitted donor-to-recipient routes, never uses more of a stock group than is available, gives every stated request its required whole-number amount, and respects any additional constraints. Check the most restricted recipient first. In the simplified ABO red-cell model, O recipients can receive O only; A recipients may receive A or O; B recipients may receive B or O; AB recipients may receive any ABO group. Total inventory greater than total demand is necessary but not sufficient for feasibility. A group can be short even when the whole cupboard contains spare units.
The 2027 SEC Mathematics examination
The 2027 SEC G3 Mathematics syllabus is K310, replacing the earlier 4052 code in its new SEC numbering. The official paper arrangement is two written papers, each lasting two hours 15 minutes, each worth 90 marks and 50 per cent of the subject. Paper 1 has about 26 short-answer questions. Paper 2 contains nine to ten questions of varying length, with the final question focused on application in a real-world scenario. Approved calculators may be used in both papers, and essential working matters. These details come from SEAB’s 2027 G3 Mathematics syllabus.
The syllabus assesses standard techniques, problems in varied contexts, and mathematical reasoning and communication. Those are exactly the three layers of a constrained table: calculate totals correctly; translate narrative rules into relationships; justify a final verdict. In a real-world-context question, the story might involve transport, household finance or another everyday application. This fictional health-education table is practice for the underlying reasoning, not a prediction of the examination setting.
An answer that gives ’24’ where the examiner asked whether an allocation is possible has not yet answered the question. It has found a total. The missing mathematics is the proof of compatibility, capacity and the absence of an unmet request. Separate these jobs in your working.
Start with an ordinary arithmetic inventory
Imagine a school museum uses counters representing four fictional red-cell stock groups. It labels 12 counters O, 8 counters A, 7 counters B and 3 counters AB. The total is 12 + 8 + 7 + 3 = 30 counters. A basic percentage question could ask what proportion is group O. The answer is 12/30, or 40 per cent. The denominator is 30 because the question says ‘of all stocked counters’.
Now imagine the requested counters are 5 for O, 9 for A, 6 for B and 4 for AB. Total requested is 5 + 9 + 6 + 4 = 24. A learner might write, ’30 exceeds 24, so the requests can all be met.’ That conclusion does not follow. The stock is not interchangeable without conditions. We still need to apply the permitted routes. The 6 spare counters across the entire display may be in a category that cannot cover the particular shortage.
An excellent first habit is therefore to write two separate statements: ‘Total stock = 30; total demand = 24’ and ‘Route feasibility not yet established.’ When the two are kept apart, a later shortage is not a contradiction. It shows why a whole-system sum cannot settle a route-constrained question.
Translate the Biology rule into a directed mathematical relation
The allowed arrows in the simplified ABO-only red-cell classroom model are O to O, A, B or AB; A to A or AB; B to B or AB; and AB to AB. This is a directed relationship. A to O is not allowed just because O to A is allowed. On a grid with stock groups along the rows and request groups across the columns, mark only those permitted cells before entering any number.
Define x(O,A) as the number of group O counters assigned to group A requests. It must be a non-negative integer because we cannot send half a counter. Define x(A,O) similarly, but since this route is excluded in the model, x(A,O) = 0. Your notation can use simpler subscripts or descriptive labels; clarity is more important than elaborate symbols. The crucial idea is that a route that does not exist has a fixed allocation of zero.
For each stock row, the sum of its allocations cannot exceed that row’s starting stock. For each request column, the allocated total must equal that group’s stated demand if full fulfilment is required. These are different mathematical conditions. Row sums control available resources; column sums control what people have asked the fictional display to supply.
Solve the central table completely
Return to stocks O = 12, A = 8, B = 7, AB = 3 and requests O = 5, A = 9, B = 6, AB = 4. Begin with the O request of 5 because only O stock can meet it in the simplified model. Assign 5 O counters to O requests, leaving 7 O counters. For A requests, use all 8 A counters and 1 O counter to reach 9. O stock now has 6 remaining. For B requests, assign 6 B counters, leaving 1 B counter. For AB requests, assign all 3 AB counters plus that remaining 1 B counter. Each assignment follows an allowed route.
The allocation can be written as O to O = 5, O to A = 1, A to A = 8, B to B = 6, B to AB = 1 and AB to AB = 3. All other allocations are zero. Check the demand columns: O receives 5; A receives 1 + 8 = 9; B receives 6; AB receives 1 + 3 = 4. The request total is 24. Check the stock rows: O uses 6 of 12, A uses 8 of 8, B uses 7 of 7 and AB uses 3 of 3. The used total is 6 + 8 + 7 + 3 = 24. The unused balance is six O counters.
This example is deliberately richer than a simple subtraction question. It shows a complete feasible allocation, not necessarily a unique one or a clinically preferred one. Several different assignments might satisfy the mathematics. If the exam asks ‘give one possible allocation’, provide any valid set with enough working. If it asks ‘prove that every allocation fails’, a single unsuccessful attempt is not sufficient.
Why solving the tightest constraint first helps
Group O requests have only one incoming route in the classroom model. If you freely use O counters for every other group before checking O requests, you may leave yourself with an impossible O shortage even though another arrangement could have worked. This is a sequencing heuristic, not a substitute for a proof. It means protect the resource that has the fewest alternatives while exploring a plan.
Group AB requests, in contrast, have all four ABO donor routes in the simple model. They offer the greatest flexibility. When building a feasible allocation by hand, it is often sensible to leave flexible requests until after the more restricted requests have been considered. But do not treat that order as a universal theorem for every possible set of additional conditions. A problem may introduce transport costs, required minimums, expiry dates or restrictions that change the appropriate choice.
Mathematics becomes powerful here because it separates the rules from the search method. The rules determine validity; the order in which you try assignments determines convenience. A quick trial is useful, but the row and column checks are what justify the final answer.
A counterexample to the total-stock shortcut
Consider a new display with O = 4, A = 10, B = 10 and AB = 10 counters. Total stock is 34. Requests are O = 6, A = 5, B = 5 and AB = 5, giving total demand 21. There are 13 more counters than requests overall. Yet the simplified model says O recipients need O counters. Only four O counters exist, so the O request of six cannot be satisfied. The shortage is at least two O counters regardless of how the other 30 counters are distributed.
This is a short proof of infeasibility. State the limiting resource, the one permitted route and the numerical comparison. ‘Although 34 exceeds 21, O demand 6 exceeds O stock 4, and no other group can supply O requests in this model.’ It is stronger than ‘I tried a table and it did not work.’ The latter reports one failed arrangement; the former shows why all arrangements fail.
This style transfers to many K310 problems: a transport schedule may have plenty of seats in total but not enough seats on one departure; a budget may have enough money overall but a rule may cap spending in a particular category; a recipe may contain plenty of total mass but not enough of a necessary ingredient. One scarce input can govern feasibility.
Use inequalities to describe limits
Suppose O requests require at least six counters and stock contains exactly five O counters. If x is the number of O counters assigned to O requests, then x must be at least 6 to meet demand, while x must be at most 5 to respect stock. There is no x that satisfies both inequalities. Writing the contradiction is an elegant exam explanation because it avoids checking many impossible tables.
For an A request, suppose 9 counters are needed and only 7 A counters are available. Let y be the number of O counters diverted to A requests. If no other route or rule changes, then y must be at least 2. It cannot exceed the O counters left after satisfying O requests. If 4 remain, values y = 2, 3 or 4 might satisfy this particular A requirement, subject to all other requests. The valid set is constrained, not automatically determined.
When additional conditions require that an allocation be a multiple of a pack size, include those integer conditions explicitly. If a fictional pack contains four counters and 10 are required, at least three full packs are needed because two provide eight. Never round a needed pack count down simply because the arithmetic quotient is 2.5. Explain why the context forces rounding upward.
Percentages require the correct denominator
Using the first inventory, O stock is 12 of 30 = 40 per cent of all stock. O demand is 5 of 24, approximately 20.8 per cent of all requests. These are percentages of different totals. They should not be subtracted or compared as though both described the same population or event. The denominators belong beside the results, especially when a table contains several categories.
The percentage of the O stock used in our allocation is 6/12 = 50 per cent. That is different again from O’s share of the 24 fulfilled requests, which is 6/24 = 25 per cent. Both are correct answers to different questions. The words ‘of O stock’ and ‘of all fulfilled requests’ are not decorative; they identify the base.
Vol 0578 explains two-way health survey tables and conditional bases. Its central lesson applies here: a fraction is not complete until the denominator is named. In Mathematics, the same 6 can mean 50 per cent, 25 per cent or another percentage depending on the defined whole.
Keep percentage points separate from per cent change
Suppose the fraction of O counters in the display changes from 40 per cent to 45 per cent after a restock. The change is five percentage points. Relative to the original 40 per cent share, the percentage increase is (45 − 40)/40 × 100 = 12.5 per cent. The two descriptions answer different questions. When a question says ‘percentage-point increase’, subtract the percentages. When it says ‘percentage change’, divide by the starting value unless a different base is stated.
Likewise, the number of O counters changing from 12 to 15 is an increase of 3/12 × 100 = 25 per cent. That is not the same as the change in O’s share of a possibly changing total inventory. You must know whether total inventory stayed fixed. If other groups were also restocked, the proportion of O counters might rise less than 25 per cent, remain steady or even fall.
Good exam working therefore preserves the base of comparison at every line. A calculator can compute 3/12; it cannot decide whether the question meant counts or shares. That decision belongs to the reader.
Read a stock table as a mathematical model, not a promise
A labelled table may be an illustrative model, not a verified real-world dataset. In this chapter all counts are fictional counters. They demonstrate constrained allocation and arithmetic. They do not describe actual blood-bank stocks, donation rates, clinical prioritisation, shelf lives or patient needs. A responsible answer stays inside the given model rather than importing medical assumptions or inventing statistics.
When the exam gives extra information, use only what is relevant. If a line says ‘the display closes at 5 pm’, that might be essential to a timing question and irrelevant to a group-allocation question. If it says ‘the display’s O counters cannot be used for two simultaneous demonstrations’, it is a capacity constraint. The skill is to convert meaningful sentences into mathematics and leave irrelevant scenery alone.
A reliable annotation system is simple: circle each available amount, underline each required amount, and box every condition that excludes a route or imposes a whole-number rule. This is an upgraded form of the PSLE representation-handoff method. It protects meaning as sentences become variables and equations.
A second worked model with an unknown
A museum has 6 O counters, 9 A counters, 5 B counters and 4 AB counters. It must fill O = 5, A = 8, B = 4 and AB = 4 requests. Let x denote O counters used to meet A requests. If the museum allocates five O counters to O requests, only one O counter remains. If it uses all nine A counters for A and AB requests, can it fulfil the plan? Yes, one valid plan is O to O = 5; A to A = 8; B to B = 4; AB to AB = 4. No O diversion is needed, and one A counter plus one O counter plus one B counter remain. The unknown x can be zero.
Now change A demand from eight to ten, keeping everything else the same. A stock has nine and O after O requests has one. A demand ten can be met by A nine plus O one, so x must be one if no other route is available. This is a small but useful distinction: the value of a variable depends on the demand and allocation constraints, not on an instinct that O must always be used broadly.
If a further restriction says two O counters must be held as a reserve, this modified plan fails because only one O counter remains after O requests. A solution must be recalculated when a rule changes. Do not reuse the old numerical answer merely because the same table is visible.
The final question of Paper 2 asks for interpretation
In K310, the final Paper 2 question focuses on mathematics in a real-world scenario. That does not mean the topic must be health, or that a candidate should expect a compatibility matrix. It means the candidate should be ready to read a longer situation, identify several relevant ideas, form a plan, use calculations and explain the conclusion. A constrained inventory is a useful practice setting because an answer requires more than one calculation.
Present the interpretation in a full sentence: ‘The fictional request plan is feasible because every assigned route is allowed, no stock group is overused, and all four request totals are met.’ If the plan is infeasible, write the exact reason. For example, ‘Group O demand exceeds group O supply by two, and the given rules provide no alternative incoming route.’ Include the specific numbers rather than an unexplained ‘no’.
When a long question provides many marks, label intermediate working. An examiner cannot award full reasoning credit for an isolated final number if essential working is missing. The official syllabus explicitly warns that omission of essential working costs marks, so make the evidence visible.
Practice problems with worked reasoning
Problem 1 — Inventory total and share
A display contains O = 9, A = 6, B = 3, AB = 2 counters. Total stock is 20. The O share is 9/20 × 100 = 45 per cent. The AB share is 2/20 × 100 = 10 per cent. The denominators are all stocked counters; using only the O and AB subtotal of 11 would answer a different conditional question.
Problem 2 — Which group is limiting?
Stock is O = 3, A = 10, B = 10, AB = 10. Requests include O = 4. No full solution exists under the simplified ABO route rules, because O demand exceeds O stock. A high grand total is irrelevant to this necessary condition. State the O shortage of one and the restricted incoming route.
Problem 3 — An A-only shortage
Stock is O = 7, A = 5, B = 8, AB = 2. Requests are O = 4, A = 7, B = 5 and AB = 3. Allocate O4 to O, A5 plus O2 to A, B5 to B, and AB2 plus B1 to AB. Used O6, A5, B6, AB2; all row limits and column demands hold. This is one feasible assignment.
Problem 4 — The direction mistake
A proposed solution assigns A stock to O requests. The route is excluded in the ABO-only red-cell classroom model. Even if all numerical totals balance, the plan is invalid. A correct mathematical check must inspect both numerical conservation and permitted connections. Arithmetic equality does not override a structural rule.
Problem 5 — An AB demand with alternatives
Stock AB = 1, B = 4 and O = 5, with AB requests = 3. In isolation, the AB requests can be met using AB1 and B2 because AB recipients have both routes in the simple model. But full-system feasibility cannot yet be guaranteed. Other requests may need those B and O counters; missing demands prevent a final global verdict.
Problem 6 — Stock-use percentage
A group B shelf begins with 12 fictional counters and contributes 9 to a feasible allocation. The proportion used is 9/12 = 75 per cent. Three remain. If someone reports 9 divided by all 40 counters in the display, they have calculated B’s contribution as a share of the total inventory, not the share of B stock used.
Problem 7 — Change in percentage points
Group A’s share of all inventory rises from 25 per cent to 31 per cent. This is a rise of six percentage points. The relative percentage increase is 6/25 × 100 = 24 per cent. A question asking for ‘percentage points’ should not receive ’24 per cent’ because it describes a different measure.
Problem 8 — Whole-number packs
A fictional teaching kit holds four counters per sealed pack. Seventeen counters are needed for a school demonstration. Four packs provide 16, which is insufficient; five packs provide 20. The minimum is five packs. The numerical quotient 17/4 = 4.25 must be rounded upward because packs cannot be divided under the given rule.
Problem 9 — Reserve requirement
A display has ten O counters and O requests of seven. Another activity requires a reserve of four O counters. The combined requirement is eleven, exceeding ten by one. No assignment can satisfy both conditions. Write the two demands as 7 + 4 ≤ 10, observe the inequality is false, and conclude infeasibility.
Problem 10 — Two choices for the same request
A recipient-group A demonstration requires six counters. Stock A contains four and O contains five after all O requests are covered. At least two O counters must be diverted to A. The statement ‘use all five O counters’ is not necessary. If the question asks for the minimum diversion, the answer is two, with a reason.
Problem 11 — Sum of stock rows
An allocation uses O = 5 of 8, A = 4 of 4, B = 6 of 7 and AB = 2 of 3 counters. The total used is 17 and the total stock is 22, leaving 5. Add the leftovers separately: 3 + 0 + 1 + 1 = 5. Agreement is a useful check but does not alone prove the assignment used permitted routes.
Problem 12 — Sum of demand columns
Requests are O3, A5, B7 and AB2. Their total is 17. A purported solution allocates O3, A5, B6 and AB3. It still totals 17 but fails the requested categories: B is short by one and AB receives one extra. A matched grand total can conceal wrong column sums, so check every category.
Problem 13 — One impossible cell
A student places x(A,B) = 2 in a table, meaning two A counters go to B requests. That is an excluded route in the classroom model. The correct constraint is x(A,B) = 0. This is not a small subtraction error. It is a misinterpretation of the directed compatibility relation, and recalculating the grand total cannot repair it.
Problem 14 — Different percentage bases
A table contains 8 O counters and 12 others, making O 40 per cent of stock. After an allocation, 6 O counters are used among 15 total counters used. O is also 40 per cent of the used set, but the equal percentages are accidental. If three more O counters were used without other changes, the used-set fraction would change. Always name the base.
Problem 15 — Upper and lower bounds
Let x be the number of O counters assigned to A requests. Demand requires at least 3 and the available O balance after O requests is at most 5. Therefore 3 ≤ x ≤ 5, with x an integer. If a second condition restricts x to even values, only x = 4 qualifies. A final answer should show both the interval and the parity restriction.
Problem 16 — A missing number in a row
Total O stock is 14. An allocation shows 5 sent to O requests, 3 sent to A and 2 sent to B. If the remaining O balance is 1, then O sent to AB must be 14 − (5 + 3 + 2 + 1) = 3. Check that the route to AB is allowed. The arithmetic identifies the missing entry; the route rule validates it.
Problem 17 — Conditional demand percentage
A display serves 5 O, 8 A, 3 B and 4 AB requests. Of the 20 requests, A accounts for 8/20 = 40 per cent. Of the non-O requests, A accounts for 8/(8 + 3 + 4) = 8/15, approximately 53.3 per cent. Both are valid but answer differently worded questions. Read ‘of the non-O requests’ literally.
Problem 18 — A counterexample proof
A plan has 100 units of total stock and 80 total requests, but O stock is 2 and O requests are 3. It is impossible under the simplified routes. The large numbers do not change the local shortage. A clear proof names the necessary O condition and uses 2 < 3 as the decisive comparison, not the irrelevant 100 > 80.
Problem 19 — Additional cost does not alter feasibility
Suppose two valid allocations have costs $10 and $12 in a fictional exercise. Both may be feasible even if one is cheaper. Feasibility asks whether rules and quantities are satisfied; optimisation asks which valid choice minimises a stated quantity. If the question asks only for ‘one feasible plan’, do not invent a cost-minimising requirement.
Problem 20 — Wrong units in a recommendation
A school notice says ‘Six per cent of O stock remains’, but the table actually shows six O counters remaining out of twelve starting O counters. The correct proportion is 6/12 = 50 per cent, not six per cent. The numeral 6 had a count unit, not a percentage unit. Restore the denominator and unit before publishing an explanation.
How to practise this without memorising a solution
Use one page for a table with stock rows and demand columns. Before entering numbers, mark every forbidden route as zero. Then fulfil the restricted columns first and check the remaining flexible columns. When you have one solution, verify each stock row, each request column and the whole-system total. Finally, rewrite the conclusion as a sentence with the correct conditions. This routine trains both calculation and communication.
For a stronger challenge, modify one number at a time. Increase an O request by two. Impose a reserve. Change a pack size. Ask for the minimum O diversion rather than any allocation. These are different mathematical questions, and the student should explain exactly which constraint changed. That is closer to examination transfer than solving the same unchanged table ten times.
An error log can use four labels: total error, route error, denominator error and inference error. Each identifies a different repair. If you are accurate at arithmetic but repeatedly send stock to forbidden cells, more speed drills will not solve the problem. You need to reread the allowed relationship and practise its direction.
Questions families ask
Is this an official hospital-stock topic?
No. It is an original fictional application of the K310 mathematical skills of tables, ratios, percentages, inequalities, whole-number constraints and reasoning. The ABO-only route rule is borrowed from G3 Biology for learning purposes. It is not a description of medical inventory practice, blood supply or patient care.
Why is a large total not enough?
A system with constrained routes cannot freely move every resource to every request. The total provides a necessary arithmetic bound, but a special subgroup may still have insufficient eligible supply. An explicit shortage such as O supply 4 against O demand 6 proves infeasibility in the simplified model.
How many decimals should the student use?
The official K310 directions generally specify non-exact numerical answers to three significant figures, or angles to one decimal place, unless a question says otherwise. Keep exact integer allocations and fractions while working, and only round a non-exact result at the appropriate final stage. A whole-counter requirement cannot be answered as a fractional counter.
How can a student know whether a result is justified?
Check permitted routes, row capacity, column fulfilment, integer restrictions and the final sentence. A valid number without the right constraint is not a valid plan. If the problem provides incomplete information, say which further quantity is required rather than inventing it.
Source and continuity
Official: 2027 G3 Mathematics syllabus K310 and SEAB 2027 SEC G3 syllabuses. For the biological source of the permitted arrows see Vol 0581; for earlier Mathematics on bases and conditional fractions see Vol 0578. The next English chapter will use a fictional community-information briefing, not medical advice, to distinguish a clear public message from a mathematically or biologically overconfident claim.
