PSLE Mathematics word problems become more demanding when a sensible calculation does not yet give a valid decision. A low price per card may look like the best offer, but whole-pack purchasing can change the cheapest order. This school-fair problem-solving workshop teaches you to compare quantities, costs and restrictions together, show why a plan works and justify why a cheaper permitted plan does not exist.
You will begin with one proposed order, then build a systematic comparison. Later, the requirement, stock, delivery charge and discount rule will change. These are separate variations, not hidden additions to the original question. The aim is to learn when a previous solution can be reused and when a changed condition requires a new comparison.
All prices, orders and events below are invented classroom examples. Beatrice and the other learners are fictional. The tasks are not copied PSLE questions or real purchasing advice. Use the Primary 6 Mathematics Learning Hub for the wider syllabus route and the PSLE Learning Guide for English, Mathematics and Science preparation.
The original school-fair problem
The activity team expects 96 participants. Each participant needs one activity card. The team also needs at least 15 additional cards. Cards are sold in packs of 12 for $9 and packs of 20 for $16. The two pack sizes may be mixed. Only whole packs can be bought, stock is sufficient, all prices are final, and there are no delivery charges. The budget is $90.
Find the least expensive permitted order that meets the card requirement. State how many packs of each size should be bought, the total number of cards, the total cost and the amount left from the budget. Explain why your comparison covers the possibilities that could produce a cheaper order.
Before calculating, notice what the task does not ask you to minimise. It does not ask for the fewest packs, the fewest extra cards or the lowest price per card considered alone. Those quantities may help your reasoning, but the stated objective is minimum total cost subject to the requirement and purchasing rules.
The SEAB Mathematics assessment objectives for 2026 include interpreting information, applying mathematical methods and reasoning through problems. This invented case practises those abilities. The labels, comparison method and worked solution here are teaching tools, not an official marking scheme or a claim that one particular method is compulsory.
Translate the requirement into a target quantity
The minimum number of cards is 96 + 15 = 111. Write “minimum cards required” beside that result. The word “at least” permits an order containing more than 111 cards, but an order containing 110 or 108 is not sufficient. Extra cards cannot compensate for a shortage somewhere else unless the question explicitly allows such an exchange.
Do not calculate 15% of 96. The brief asks for 15 additional cards, not a 15% reserve. Do not include the reserve inside the 96 participants either. Each participant still needs one card, and the additional cards are required on top of that allocation. Reading the quantity correctly is the first mathematical step.
There are three useful counts to keep separate: cards purchased, cards allocated to participants and cards beyond the minimum requirement. If an order contains 112 cards, 16 remain after 96 are allocated. Only one card is beyond the minimum requirement of 111. Both differences are correct, but they answer different questions.
Compare unit prices, then stop short of a conclusion
A smaller pack costs $9 for 12 cards, so its unit price is $0.75 per card. A larger pack costs $16 for 20 cards, so its unit price is $0.80 per card. The smaller pack is cheaper per card. That is a useful fact about the offers, not yet a proof about the cheapest permitted order.
Whole packs create unused capacity. If you buy only smaller packs, nine packs provide 108 cards and are insufficient. Ten packs provide 120 cards and cost $90. You cannot buy 9.25 packs just because 111 divided by 12 is 9.25. The supplier’s whole-pack condition remains part of the problem after division.
An order containing some larger packs may provide a closer fit to the required quantity. It can therefore cost less overall despite including cards with a higher unit price. This is not a contradiction. Unit price measures the cost of each card within an offer; total order cost also depends on how many cards you must actually purchase.
Check Beatrice’s proposed mixed order
Beatrice proposes six smaller packs and two larger packs. The smaller packs provide 6 × 12 = 72 cards. The larger packs provide 2 × 20 = 40 cards. Together they provide 112 cards, enough for the 111-card minimum.
The smaller packs cost 6 × $9 = $54. The larger packs cost 2 × $16 = $32. Total cost is $86. The budget remainder is $90 − $86 = $4. The reserve after participant allocation is 112 − 96 = 16 cards.
This proves the proposal is feasible: it obeys the whole-pack rule, meets the quantity requirement and stays within budget. Feasible does not mean automatically cheapest. To prove minimum cost, compare the other relevant permitted orders or use another valid argument that rules out every cheaper possibility.
Keep the proposed plan as a useful benchmark. Once you have a feasible $86 order, any plan already costing more than $86 cannot improve the objective. You do not need to finish calculating unnecessary details for an option that has already lost the cost comparison, although a complete worked table may still help you learn the structure.
Build a comparison that cannot silently miss a plan
Choose one pack type to count systematically. Here, begin with zero larger packs, then one, two, three and so on. For each number of larger packs, buy only the smallest number of smaller packs needed to reach at least 111 cards. Buying an extra smaller pack at the same larger-pack count would add cost without being necessary to meet the requirement.
That last observation explains why the table can be short. For a fixed number of larger packs, you do not have to list every possible smaller-pack count. You need the least feasible count because all greater counts cost more. A systematic comparison is not merely a list of attractive guesses; each row represents a reasoned choice.
Before working through the rows, prepare headings in your notebook: larger packs, cards from larger packs, smaller packs needed, total cards and total cost. The headings keep 20-card quantities separate from $16 costs. A row containing five unrelated numbers is much harder to check than a row with stable quantity names.
Work through the possible larger-pack counts
With zero larger packs, the team needs ten smaller packs. The order contains 120 cards and costs $90. Nine smaller packs would provide only 108, so ten is the smallest feasible smaller-pack count for this row.
With one larger pack, 20 cards are supplied and at least 91 more are needed. Seven smaller packs provide 84, which is not enough. Eight provide 96. The combined order contains 116 cards and costs $16 + $72 = $88.
With two larger packs, 40 cards are supplied and at least 71 more are needed. Six smaller packs provide 72. The combined order contains 112 cards and costs $32 + $54 = $86. This matches Beatrice’s proposal and improves on the first two rows.
With three larger packs, 60 cards are supplied and at least 51 more are needed. Four smaller packs provide only 48, so five are necessary. The order contains 120 cards and costs $48 + $45 = $93. A common mistake is to accept four smaller packs here, producing only 108 cards in total.
With four larger packs, 80 cards are supplied and at least 31 more are needed. Three smaller packs provide 36. The combined order contains 116 cards and costs $64 + $27 = $91. It meets the quantity condition but exceeds the budget and does not beat $86.
With five larger packs, 100 cards are supplied. One smaller pack brings the total to 112. The cost is $80 + $9 = $89. This is feasible within the $90 budget, but it is $3 more expensive than six smaller packs and two larger packs.
With six larger packs, the larger packs alone provide 120 cards and cost $96. Adding smaller packs would only increase that cost. Seven or more larger packs would cost even more before any smaller packs were added. Therefore no order with six or more larger packs can beat the known $86 plan.
The comparison is now complete for the purpose of finding a cheaper order. Every relevant larger-pack count has either been checked or ruled out by cost. For each checked count, the smallest feasible smaller-pack count was used. The minimum is six smaller packs and two larger packs, providing 112 cards for $86.
Write the final answer as a decision
A complete final statement is: “Buy six packs of 12 cards and two packs of 20 cards. The order provides 112 cards and costs $86, leaving $4 from the $90 budget. After allocating 96 cards to participants, 16 additional cards remain. Comparing the least-cost feasible order for each relevant larger-pack count shows that $86 is the minimum.”
Your exact wording need not match this statement. It should identify the pack counts, quantity, cost and requested budget remainder. A final answer of “86” leaves the reader to guess whether you mean cards, packs or dollars. Correct arithmetic needs a correctly identified quantity at the end.
Do not claim a universal buying rule from this one example. “Always mix packs” would be just as unreliable as “always choose the lower unit price”. The result belongs to this requirement and these offers. A changed requirement may make an all-small-pack order cheapest, as a later example will show.
Understand why rounding works differently here
Division tells you how many pack capacities are needed, but the purchase rule decides what to do with a fractional result. For smaller packs alone, 111 ÷ 12 = 9.25. Rounding to the nearest whole number would give nine, but nine packs are insufficient. You need the next whole pack because the requirement is a minimum quantity.
The same reasoning applies within each comparison row. If 51 cards remain to be supplied in 12-card packs, four packs are short because they provide 48. Five packs are needed. You are not following an unexplained rule to “round up in word problems”. You are choosing the smallest permitted whole-pack count that satisfies this particular inequality.
A different question could require a maximum quantity or ask how many complete packs can be assembled from existing cards. In those cases, the direction may differ. Explain the meaning of the quotient and remainder before deciding how to use them. Context determines the permitted result, not the mere presence of a decimal.
Use a quantity check to catch impossible totals
Both pack sizes are multiples of four. Any whole-pack order therefore contains a multiple of four cards. Since 111 is not a multiple of four, no permitted order contains exactly 111 cards. The smallest possible total at or above 111 is at least 112.
This does not by itself prove that 112 can be obtained or that an order containing 112 is cheapest. You still need a permitted combination. Six smaller packs and two larger packs produce 112, so that quantity is achievable. One smaller pack and five larger packs also produce 112, but cost $89 instead of $86.
The example shows why each check has a limited role. The multiple-of-four check can reject an impossible claim of exactly 111 cards. It cannot choose between two possible 112-card orders with different prices. Use it as a consistency check rather than asking it to solve a different question.
Change the objective: the fewest packs
Now consider a separate variation. The group still needs at least 111 cards and has the same $90 budget, but the teacher asks for the fewest packs to carry. This changes the objective. The earlier eight-pack order is cheapest, but it does not automatically minimise pack count.
Five packs cannot provide enough cards, even if all five are the larger size: 5 × 20 = 100. Therefore at least six packs are needed. One smaller pack and five larger packs provide 112 cards in six packs and cost $89, so a six-pack solution is achievable within budget.
Six larger packs also use six packs and provide enough cards, but cost $96 and exceed budget. With six total packs, replacing a larger pack with a smaller one reduces the card total by eight. Two smaller and four larger packs provide only 104 cards, so they are insufficient. The one-small, five-large order is the feasible six-pack choice under these conditions.
Notice that the answer changed because the question changed, not because the first solution was mistaken. Minimum cost and minimum pack count measure different things. In a multi-part examination problem, underline the objective in each part rather than assuming that the previous objective remains in force.
Change the requirement: 108 participants
Return to the minimum-cost objective and change only the participant count to 108. The reserve remains at least 15 cards. The new minimum requirement is 123 cards. The original 112-card order is short by 11 cards and cannot be accepted simply because it was previously optimal.
A quick lower bound reveals a budget problem. Every card costs at least $0.75 under these offers, so 123 cards would cost at least $92.25 even before considering whole-pack fit. Therefore no permitted order can meet the new requirement within the unchanged $90 budget. This argument is enough to answer a question asking only whether the budget is sufficient.
If you also need the new minimum cost, compare again. With zero, one or two larger packs, the smallest feasible orders cost $99, $97 and $95 respectively. Those rows use eleven, nine and seven smaller packs, providing 132, 128 and 124 cards.
With three, four or five larger packs, the smallest feasible orders cost $102, $100 and $98. With six larger packs, one smaller pack is still needed, costing $105 altogether. Seven larger packs cost $112 without smaller packs. Higher counts cannot improve on $95. Thus seven smaller packs and two larger packs provide 124 cards for $95, requiring $5 more than the original budget.
Change the purchasing rule: mixing is not allowed
In another separate variation, the requirement returns to 111 cards, but the group must use only one pack size. The six-small, two-large order is now prohibited. Its arithmetic remains correct, yet it is not a valid answer under the changed rule.
The smallest all-small order is ten packs, providing 120 cards for $90. The smallest all-large order is six packs, providing 120 cards for $96. The all-small order is cheaper and is the only one of these two minimum orders within the $90 budget.
Do not interpret “must use only one pack size” as “must buy one pack”. The restriction concerns the type of pack, not the quantity of packs. Read the noun phrase carefully. Many word-problem errors begin when a condition is attached to the wrong quantity.
This variation also shows why an answer cannot be judged from the final number alone. A student who writes $86 has a correct result for the original problem but an invalid result for this version. Always connect the result to the current set of conditions before carrying it into the next part.
Change the charges: a fee for using the larger-pack seller
Return to the original requirement of 111 cards and allow mixed orders again. In this variation, smaller packs can be collected without a fee. Any order containing one or more larger packs incurs a single $5 delivery charge. The charge is per order using that seller, not per larger pack. All other conditions remain the same.
The earlier $86 order now costs $91. It is no longer within the $90 budget. Do not stop there and conclude that the whole problem is impossible. The all-small order still costs $90 because it uses no larger packs and incurs no delivery charge.
Compare the relevant mixed alternatives by adding $5 once to each earlier row containing larger packs. Their totals become $93, $91, $98, $96, $94 and $101 for one through six larger packs at their minimum feasible smaller-pack counts. None beats the $90 all-small order. More larger packs cannot improve the total.
The new optimum is ten smaller packs for $90. A delivery rule can change which order is best even when pack prices do not change. The mathematical lesson is to attach a charge to the condition that triggers it. Adding $5 to every pack or to an order containing no larger packs would answer a different problem.
Change the stock: only four smaller packs are available
Consider a fresh variation with no delivery charge. The original requirement and $90 budget return, but the supplier has at most four smaller packs available. Larger-pack stock remains sufficient. The old optimum needs six smaller packs, so it is not permitted under this stock limit.
With zero, one or two larger packs, four smaller packs cannot bring the total to 111. With three larger packs and four smaller packs, the total is only 60 + 48 = 108. Four larger packs need three smaller packs to reach 116 cards, costing $91. Five larger packs need one smaller pack to reach 112 cards, costing $89.
Six larger packs cost $96, and more cannot be cheaper. Therefore the least-cost permitted order under the stock restriction is one smaller pack and five larger packs for $89. It meets the quantity condition and stays within budget.
A stock restriction is not a small note to check after selecting your favourite answer. It changes the allowed combinations from the beginning. You can still reuse the earlier comparison as a starting record, but you must remove rows that violate stock and reconsider the remaining choices.
Change the discount: a threshold can reward a different order
In this separate extension, stock is unlimited and there is no delivery charge. The supplier gives a 10% discount on an order whose listed pack cost is at least $90. Orders below $90 receive no discount. The minimum card requirement is 111 and the budget is $90 after any discount.
The earlier $86 order does not qualify, so it still costs $86. Ten smaller packs have a listed cost of $90 and qualify. The discount is $9, making the amount paid $81. That is cheaper than the original mixed order despite supplying more cards.
Why can you be confident that $81 is the minimum under this extension? Every qualifying order has a listed cost of at least $90, so after a 10% discount it costs at least $81. The ten-small-pack order reaches that lower bound and is feasible. Every non-qualifying feasible order costs at least $86, as established by the original exhaustive comparison.
This proof separates two groups of orders: those receiving the discount and those not receiving it. It does not assume that buying more always saves money. The conclusion follows from this particular threshold and discount. In a new problem, check exactly which amount the percentage applies to and which conditions make an order eligible.
Learn when a short proof is better than a long list
An exhaustive table is useful when there are only a few relevant combinations and the conditions are clear. A lower bound is useful when it can rule out a budget or establish a minimum without checking every order. A constructed feasible order is useful for showing that a proposed bound can actually be reached.
These methods can cooperate. In the original problem, systematic enumeration proves the $86 minimum. In the 123-card variation, the unit-price lower bound quickly proves that $90 is insufficient. In the discount variation, a threshold lower bound plus one matching feasible order proves the $81 minimum.
Do not use “I checked several examples” as a substitute for explaining completeness. A list can miss an untested combination. State why the list stops and why each row uses the least necessary number of the other pack. That reasoning turns a collection of calculations into a proof that answers the minimum-cost question.
Reverse the question: how many cards can $90 buy?
A further variation keeps the original pack prices and removes fees, discounts and stock limits. Instead of asking for the cheapest order meeting a requirement, it asks for the greatest number of cards that can be bought with $90. State the changed objective before reaching for the earlier winning combination.
Every card costs at least $0.75 under the two offers. Therefore $90 cannot buy more than $90 ÷ $0.75 = 120 cards. Ten smaller packs provide exactly 120 cards for $90, so the bound is reachable. That order maximises the card quantity under this version of the rules.
The original $86 mixed order remains feasible within the same budget, but it contains only 112 cards. It is the answer to a different question: spend as little as possible while obtaining at least 111 cards. Maximising quantity within a budget and minimising cost above a quantity threshold are related but not interchangeable decisions.
This pair of tasks is useful for checking genuine understanding. A learner who memorised “six small and two large” may keep that answer even after the objective changes. A learner who understands the conditions can explain why ten smaller packs are now appropriate. The offers did not change; the decision did.
Audit a comparison row before trusting the whole table
Choose the row with three larger packs. Name each part: three larger packs supply 60 cards, leaving a shortfall of 51 against the 111-card requirement. Four smaller packs supply 48 cards and are insufficient. Five smaller packs supply 60 cards, giving 120 cards altogether. The cost is three lots of $16 plus five lots of $9, or $93.
Now check the row in reverse. Remove one smaller pack from the proposed five. The total falls to 108, which misses the requirement. That verifies that five is the smallest feasible smaller-pack count for this fixed larger-pack count. It does not prove the row is globally best; the cost comparison with other rows still matters.
Repeating this short audit on one risky row can catch a copied pack count, a confused price or an incorrect stopping decision. It is more purposeful than rereading the table without a question in mind. Check the relationship that could be wrong, then use the corrected row in the wider proof.
Diagnose common errors using their first wrong step
If a learner chooses nine smaller packs, inspect the requirement and the use of the quotient. The multiplication 9 × 12 = 108 may be correct. The error is accepting a total below 111 or rounding without considering the minimum condition. More multiplication practice would not directly repair that decision.
If a learner selects six smaller packs and two larger packs but writes a cost of $112, the card total has been confused with a monetary amount. Label each product before adding. Twelve and 20 are cards per pack; nine and 16 are dollars per pack. A clean two-column quantity-and-cost layout can expose the mismatch.
If a learner says $86 is cheapest after testing only that order and the all-small order, the gap is in justification. The learner found a better example but did not cover all possible competitors. Ask which larger-pack counts remain untested, then complete the systematic comparison or another valid argument.
If a learner keeps $86 after mixing is prohibited, the error is condition transfer. A correct earlier answer has been carried into a changed problem without review. Mark the changed rule first, then identify which earlier conclusions still depend on it. This is a reasoning repair, not a request to erase every calculation and begin blindly again.
Independent practice one: a requirement that fits the smaller packs
The team now needs 76 cards for participants and eight additional cards. The pack offers are unchanged, with unlimited stock, mixed sizes allowed and no fees or discounts. Find the least expensive order. Attempt it before reading the explanation.
The minimum requirement is 84 cards. Seven smaller packs provide exactly 84 cards for $63. Since every card under either offer costs at least $0.75, any order containing at least 84 cards must cost at least $63. The seven-small-pack order reaches that bound, so it is optimal.
This example challenges the incorrect generalisation that mixed orders are always best. Mixing helped in the original problem because it improved the fit between pack sizes and the requirement. Here the lower-unit-price packs fit exactly. The correct purchasing method is stable, but the winning combination changes with the numbers.
Independent practice two: equal costs, different surpluses
A reading festival needs 83 bookmarks and at least nine additional bookmarks. Packs of eight cost $6 and packs of 15 cost $12. Whole packs may be mixed, stock is sufficient and there are no other charges. Find a minimum-cost order that leaves the fewest bookmarks beyond the minimum requirement.
The minimum requirement is 92. Every bookmark costs at least $0.75, so the cost cannot be below $69. Every order cost is a multiple of $6. Therefore a feasible order must cost at least $72. Four smaller packs and four larger packs provide 32 + 60 = 92 bookmarks for $24 + $48 = $72.
That order reaches the cost bound and leaves no surplus beyond the required 92. Twelve smaller packs also cost $72, but provide 96 bookmarks. They tie on cost and lose on the second objective of minimum surplus. Read multiple objectives in their stated order rather than blending them into an undefined idea of the “best” order.
Independent practice three: decide what can be concluded
A pupil tests two feasible orders and finds that Order X costs $86 while Order Y costs $90. What can the pupil conclude? Order X is cheaper than Order Y. The comparison alone does not establish that Order X is cheaper than every permitted order.
Now suppose the pupil has a table containing the least-cost feasible order for every possible larger-pack count that could cost below $86, and none is cheaper. Together with the feasible $86 order, that supports the minimum-cost conclusion. The difference is not confidence or the number of calculation lines. It is whether the reasoning covers the alternatives required by the claim.
Explain this distinction in ordinary language to a partner. If you can say why checking two orders proves less than checking all relevant competitors, you understand the role of the comparison. You are less likely to mistake a good guess for a completed solution on an unfamiliar problem.
A practice routine for dependable multi-step work
First solve a small whole-pack problem with one pack size. Explain why the answer must be a whole number and why the quantity meets the requirement. Next compare two single-size orders. Only then introduce mixed packs, where systematic comparison becomes necessary.
Keep the original school-fair table as a record, but do not memorise its winning row as a method. On a later day, change the participant count or one purchasing condition and work again. Before calculating, state which earlier conclusion is no longer guaranteed. This tests whether you are reading the current problem rather than retrieving an obsolete answer.
When timing practice, keep a short final check: enough items, permitted pack counts, correct total cost, current objective and a justified stopping point for the comparison. Do not sacrifice the condition check to make arithmetic appear faster. A quick invalid order is not a successful solution.
Parent and tutor guidance
Ask the learner what each number means before asking for the next calculation. “What does 111 represent?” can reveal whether the reserve was understood. “What does 16 represent here?” can reveal whether participant allocation and surplus have become confused. These questions inspect meaning without supplying the method.
For a learner who guesses combinations, provide headings for a comparison table but leave the entries blank. For a learner who can build the table, ask why the rows cover the possibilities. For a learner who is already secure, introduce one changed condition and ask which part of the proof must change. Increase the decision demand rather than simply increasing the numbers.
Treat a different valid method fairly. A learner may reason with labelled arithmetic, a table or a carefully defined equation. The important questions are whether the method respects the conditions and whether the conclusion follows. Do not reject understandable primary-level reasoning merely because it differs from the worked presentation.
Questions about pack-size problems
Should I always calculate the unit price first?
It is often useful, but it does not always finish the problem. Unit price can provide a comparison or a lower bound. Whole-pack requirements, fees, stock and discounts may still change the cheapest permitted order. State what the unit price establishes before using it for a broader claim.
Why not just round the number of packs normally?
The task requires enough items and only whole packs are allowed. The smallest permitted pack count must satisfy that requirement. Ordinary rounding to the nearest whole number can produce a shortage. Explain the condition rather than memorising a universal rule about rounding all word-problem answers upward.
Is the order with the least surplus always cheapest?
No. The original problem has two orders containing 112 cards, costing $86 and $89. Equal surplus does not imply equal cost. A different problem may even have a cheaper order with more surplus. Keep the objective and each condition separate.
When can I stop the comparison?
Stop when you can explain why no untested order can improve the current best feasible cost. In the original case, six larger packs already cost $96, and more cannot beat $86. A justified bound is a stopping reason; boredom or a pattern that merely looks convincing is not.
Finish with a valid order and a reason
The original answer is six packs of 12 and two packs of 20: 112 cards for $86, with $4 remaining from budget. The valuable learning is not that particular pair of pack counts. It is the way you connected the minimum requirement, permitted purchases, cost comparison and proof of completeness.
Return to the Primary 6 Mathematics Learning Hub for further number and problem-solving work. Check the current official PSLE examination formats for examination requirements. In your next word problem, let the conditions determine what counts as an answer before you decide which calculation looks familiar.
Connect your calculations to the complete school-fair transfer case. To turn a checked purchase plan into an appropriate request for approval, use the school-fair English writing workshop.
