A classroom shop becomes a mathematics investigation when the learner must explain the money, not just hand over a pile of tokens. How much is in the wallet? Can the same amount be made with different coins? Is the wallet large enough but unsuitable for exact payment? What changes when the buyer receives change? Each question gives numbers a role that the child can inspect.
This investigation is a practical companion to the money-sense and simple-transactions guide within the Primary 1 Mathematics Learning Hub. It supplies an original price list, three specified wallets, teaching conversations, complete combination tasks and twenty questions with explained answers.
All prices and transactions on this page are invented for the activity. They are not current retail prices, tuition fees, offers or recommendations to purchase anything. Large paper value cards can replace money. No real payment, account, personal budget disclosure or collection of children’s financial information is needed.
Separate the core task from optional extensions
The Primary 1 money syllabus includes counting amounts in cents up to one dollar and in whole dollars up to one hundred dollars. See the official MOE primary mathematics syllabus. This investigation begins with cents within one dollar. Change, budget choices and exhaustive combinations are clearly marked teaching extensions, not presented as compulsory Primary 1 money outcomes.
Start with counting and making stated amounts. Introduce transactions only when the learner can keep the value of a token separate from the number of tokens. A child who counts six cards as six cents needs denomination work before a multi-stage purchase ledger.
The extension tasks use familiar whole-number reasoning in a money context. They can be discussed with objects, drawings or spoken explanations. There is no need to introduce decimal notation, formal algebra, percentage discounts or a written optimisation formula.
Prepare the play shop
Make large cards labelled 5 cents, 10 cents, 20 cents and 50 cents. A separate card labelled 100 cents or one dollar can represent a payment of one dollar. These are teaching value cards; their size, colour and shape do not establish their value. The printed label does.
Give every play-shop item one clear price. Keep the price list visible. Decide whether the task permits buying more than one of the same item and whether exact payment is required. Those conditions change which solutions are valid, so they must be stated before the learner starts.
Use three areas on the table: buyer’s wallet, payment area and shop’s change area. Moving a card into the payment area should remove it from the buyer’s available wallet. If change is returned, move those cards back into the wallet before finding the final balance.
For initial counting practice, reset the wallet after each task. Later questions can use a continuous transaction only when the question says that purchases occur in sequence. An unstated carry-over from the previous task creates an avoidable ambiguity.
The invented price list
| Item | Price in cents |
|---|---|
| Bookmark | 15 |
| Eraser | 20 |
| Pencil | 30 |
| Sticker sheet | 35 |
| Notebook | 50 |
Read the table in both directions. Given a pencil, find its price of thirty cents. Given a price of twenty cents, find the eraser. Then ask whether the most expensive item is necessarily the physically largest. Nothing in the table establishes physical size, and price should not be inferred from the drawing of an item.
The price list is a record of stipulated values. In a real shop, a price would need to be read from the actual seller’s information. In this lesson, the printed activity price is the rule so that everyone works with the same mathematical conditions.
Three wallets with equal totals but different contents
| Wallet | Contents in cents | Total value |
|---|---|---|
| P | 50, 20, 20, 10 | 100 cents |
| Q | 50, 20, 10, 10, 5, 5 | 100 cents |
| R | 20, 20, 20, 20, 20 | 100 cents |
Wallet P contains four cards, Q contains six and R contains five. Each has the same total value of one hundred cents. The number of pieces and the value of the collection are different quantities.
Ask the learner to count the cards first, then count their value. Keep both answers. “Six cards worth one hundred cents” is a coherent description of wallet Q. “Six cards, so six cents” confuses the objects being counted with the value each represents.
Session 1: Count value in a traceable order
Count wallet P by moving one value card at a time: fifty, seventy, ninety, one hundred cents. The intermediate totals are visible in the moved cards. The learner can check the last total by adding the two twenty-cent cards first, giving forty, then combining fifty and ten to make sixty; forty and sixty make one hundred.
Both routes are valid. The aim is not to force a single order but to prevent missed or repeated cards. If the learner loses track, keep a counted area separate from an uncounted area, just as in a counting-collections task.
For wallet Q, combine the two five-cent cards into a ten-cent value. Together with the two original ten-cent cards, these contribute thirty cents. Fifty plus twenty plus thirty gives one hundred. The cards remain six physical pieces even when their values are mentally grouped.
A deliberate check asks whether exchanging one twenty-cent card for two ten-cent cards changes the total. It changes the number of pieces but not the value. That distinction is the foundation for making equivalent payments.
Session 2: Make one amount in more than one way
Ask the learner to make sixty cents from an available supply of value cards. Fifty plus ten is one solution. Twenty plus twenty plus twenty is another. The first uses two cards, the second uses three, but each represents sixty cents.
State whether the supply is unlimited for the task. If only wallet P is available, three twenty-cent cards cannot be used because P contains only two. A mathematically correct combination may still violate the inventory condition.
This is an important modelling boundary. An answer must fit both the requested amount and the resources supplied. Do not silently lend a missing token and then describe the original wallet as having made the payment.
Ask the learner to explain equivalence by value rather than appearance. Two differently sized piles can be equal in money value. Moving the cards closer together, spreading them out or changing their order also leaves the total unchanged.
Worked investigation: every way to make forty with tens and twenties
For this task only, allow as many ten-cent and twenty-cent cards as needed. Order does not matter: twenty plus ten plus ten is the same combination as ten plus twenty plus ten. The question is about combinations of values, not arrangements in a row.
Use the number of twenty-cent cards to organise the search. With none, four ten-cent cards are needed. With one twenty-cent card, two ten-cent cards are needed. With two twenty-cent cards, no ten-cent cards are needed.
These are the three combinations: 10 + 10 + 10 + 10; 20 + 10 + 10; and 20 + 20. A third twenty-cent card would already exceed forty cents, so there is no additional case to inspect. This explains why the list is complete.
A child may find the combinations with objects before explaining the organised search. The adult can then draw three rows labelled zero twenties, one twenty and two twenties. The purpose is to make completeness visible, not to demand formal combinatorial language.
Session 3: Enough value is not always exact payment
A sticker sheet costs thirty-five cents. Wallet P holds one hundred cents, so its total value is more than the price. Nevertheless, its cards cannot make an exact payment of thirty-five cents: every available value is a multiple of ten cents.
Keep the explanation concrete. Any sum of P’s ten-, twenty- and fifty-cent values ends at a whole ten-cent step. Thirty-five cents lies five cents beyond thirty and five cents before forty. No available five-cent card can supply that final five.
This is not a lack-of-money problem. It is an exact-combination problem. The purchase might still be possible if the shop accepts a larger payment and can return the correct change, but those are additional conditions that the task must explicitly allow.
Wallet Q can pay thirty-five cents exactly using twenty, ten and five. After moving those three cards out, fifty, ten and five remain. The remaining value is sixty-five cents. The physical wallet provides a direct check on the subtraction 100 − 35 = 65.
Session 4: Change as an optional extension
Use a sticker sheet priced at thirty-five cents and a payment of fifty cents. The change is fifteen cents because thirty-five plus fifteen rebuilds the fifty paid. Ten and five are one available way to return that change.
Read three roles carefully: price, amount paid and change. The amount paid is not automatically the cost of the item. A buyer handing over fifty cents for a thirty-five-cent item has not bought a fifty-cent item; fifteen cents must return under the stated transaction.
Check the equation price plus change equals payment: 35 + 15 = 50. A change answer of eighty-five cents is incompatible with this simple purchase. The error should lead to reviewing the transaction roles, not just rehearsing subtraction facts.
A second example uses a sixty-five-cent purchase paid with one hundred cents. The change is thirty-five cents. Count up from sixty-five to seventy with five cents, then from seventy to one hundred with thirty more cents. The combined difference is thirty-five.
Track a transaction without losing the starting wallet
Begin with wallet Q worth one hundred cents. Pay its fifty-cent card for the thirty-five-cent sticker sheet. Fifty cents remain in the wallet immediately after the payment. Receive fifteen cents in change, producing a final wallet value of sixty-five cents.
Write separate statements: 100 − 50 = 50; then 50 + 15 = 65. The net change is a reduction of thirty-five cents, equal to the price. That agrees with the direct calculation 100 − 35 = 65.
Do not write 100 − 50 = 50 + 15 = 65. The first expression equals fifty and the second equals sixty-five, so the equality chain is false even though the intended sequence of actions is understandable.
This is a practical reason for accurate mathematical recording. The ledger should distinguish successive states, not falsely claim that all the expressions have the same value.
Session 5: Choose a basket within a stated budget
This optional extension uses a budget of one hundred cents. A notebook costs fifty, a pencil thirty and an eraser twenty. Their total is exactly one hundred cents. The basket fits the budget and leaves zero cents.
A notebook and sticker sheet cost eighty-five cents, leaving fifteen cents. That remaining amount can buy a bookmark at its stipulated price of fifteen cents. The full basket then uses the entire budget.
The instruction matters. “Spend at most one hundred cents” allows a cheaper basket. “Spend exactly one hundred cents” does not. A learner who chooses a valid basket costing eighty-five cents should not be marked wrong for an at-most question merely because the adult expected all the money to be spent.
Likewise, “buy two different items” excludes two bookmarks, while “buy any two items” may allow them. State these conditions rather than adding them after the learner has found a solution.
Worked investigation: two different items for at most fifty cents
Use the price list and buy exactly two different item types. The total must be fifty cents or less. To search systematically, pair the bookmark with each more expensive item, then the eraser with each remaining more expensive item, and continue. This avoids listing the same pair twice in reverse order.
The bookmark and eraser cost thirty-five cents. The bookmark and pencil cost forty-five. The bookmark and sticker sheet cost fifty. The bookmark and notebook cost sixty-five, so that pair is too expensive.
The eraser and pencil cost fifty. The eraser with the sticker sheet costs fifty-five, and with the notebook costs seventy. The remaining pairs—pencil with sticker sheet, pencil with notebook, and sticker sheet with notebook—also exceed fifty.
Exactly four pairs satisfy the conditions: bookmark with eraser; bookmark with pencil; bookmark with sticker sheet; eraser with pencil. The organised comparison explains why no further pair is missing.
An optional maximum-item challenge
Allow repeated purchases of the same item, with enough stock available, and ask for the greatest number of items that can be bought with one hundred cents. The cheapest item is the fifteen-cent bookmark. Six bookmarks cost ninety cents and leave ten cents, which is not enough for another item.
No choice of more expensive items can increase the number beyond six because every item costs at least fifteen cents. The explanation is about the minimum possible price per item and the remaining budget, not about a memorised shopping trick.
Keep this challenge optional. A child who can count money accurately does not need to solve optimisation problems to demonstrate that core skill. The task extends reasoning for a learner who is ready to organise cases and justify why a better result is impossible.
Twenty practice questions
Each question starts from the stated wallet or amount afresh unless it explicitly describes a sequence. Prices come from the invented list above. Questions involving change or choosing baskets are optional extensions.
1. How many value cards are in wallet Q, and what is their total value?
2. Wallet P has four cards and Q has six. Which wallet has more money?
3. Find the value of a separate collection containing fifty, twenty and five cents.
4. Give two different ways to make sixty cents using cards labelled ten, twenty or fifty cents, with enough cards supplied.
5. List every combination making forty cents using only ten- and twenty-cent cards. Ignore the order of the cards.
6. A buyer has one hundred cents and buys a notebook and a pencil. What is the total cost and the remaining value?
7. What is the total cost of a sticker sheet and a bookmark?
8. How much more does a notebook cost than a bookmark?
9. A sticker sheet costs thirty-five cents. A buyer has thirty cents. How much more is needed?
10. A buyer pays fifty cents for the thirty-five-cent sticker sheet. Find the change.
11. A buyer pays one hundred cents for a purchase costing sixty-five cents. Find the change.
12. Can wallet P pay thirty-five cents exactly without receiving change or exchanging cards first?
13. Choose an exact thirty-five-cent payment from wallet Q.
14. After the payment in Question 13, how much remains in wallet Q?
15. A buyer spends eighty-five cents on a notebook and sticker sheet from a one-hundred-cent budget. Can the remaining amount buy one bookmark exactly?
16. Find every pair of different item types costing at most fifty cents.
17. Which uses fewer cards to make sixty cents: fifty plus ten, or twenty plus twenty plus twenty?
18. An item costs twenty cents. The question asks for change but gives no payment amount. Can the change be determined uniquely?
19. With repeated items allowed and enough stock, what is the greatest number of items that can be bought with one hundred cents? Explain.
20. A learner records a transaction as 100 − 50 = 50 + 15 = 65. The buyer paid fifty cents and received fifteen cents in change. Repair the working and state the net cost.
Explained answers
1. Six cards worth one hundred cents. The contents are fifty, twenty, ten, ten, five and five. Count the pieces and the values separately. Six is the number of cards, not their value in cents.
2. They have equal value. Each wallet totals one hundred cents. Q has more physical cards, but that does not give it more money than P.
3. Seventy-five cents. Fifty plus twenty is seventy; five more gives seventy-five. Include cents in the answer.
4. Examples include fifty plus ten and twenty plus twenty plus twenty. Both total sixty cents. The question asks for two examples, so other valid combinations may also be accepted.
5. Three combinations. Four tens; one twenty and two tens; or two twenties. Organising cases by zero, one or two twenty-cent cards shows completeness. A third twenty would exceed the target.
6. Eighty cents spent and twenty cents remaining. Fifty plus thirty gives eighty. One hundred minus eighty leaves twenty. Cost and balance are different answers to different parts of the question.
7. Fifty cents. Thirty-five plus fifteen equals fifty. The two item prices are the parts of the combined cost.
8. Thirty-five cents more. Compare the notebook’s fifty-cent price with the bookmark’s fifteen-cent price. Fifty minus fifteen equals thirty-five.
9. Five cents. Thirty cents plus five cents reaches the thirty-five-cent price. This asks for a missing amount, not the combined total of price and available money.
10. Fifteen cents change. Fifty minus thirty-five equals fifteen. Check that the thirty-five-cent price and fifteen-cent change add back to the fifty paid.
11. Thirty-five cents change. One hundred minus sixty-five equals thirty-five. A return of twenty, ten and five cents is one representation of that amount.
12. No exact payment is possible from P. Its available values are fifty, twenty, twenty and ten, all whole ten-cent steps. Their sums cannot make thirty-five. P has enough total value, but not the required exact combination.
13. Twenty, ten and five cents. Wallet Q contains these cards. Together they make thirty-five cents, and each selected card is used only once.
14. Sixty-five cents. Removing twenty, one ten and one five leaves fifty, ten and five. Their total agrees with 100 − 35 = 65.
15. Yes. One hundred minus eighty-five leaves fifteen cents. That equals the stipulated bookmark price exactly, so the remaining amount is sufficient with no value left over.
16. Four pairs. Bookmark and eraser cost thirty-five; bookmark and pencil forty-five; bookmark and sticker sheet fifty; eraser and pencil fifty. Every other pair of different item types costs more than fifty cents.
17. Fifty plus ten uses fewer cards. It uses two cards, whereas three twenties use three. Both make sixty cents, so the difference concerns the number of pieces rather than total value.
18. No. Change depends on the payment as well as the price. Paying twenty cents gives zero change; paying fifty gives thirty cents. Without the payment amount, the answer is not unique.
19. Six items. Six bookmarks cost ninety cents and leave ten cents. Every item costs at least fifteen cents, so the remaining ten cannot buy a seventh. Choosing more expensive items cannot allow more items than choosing the cheapest.
20. Use separate equations. Write 100 − 50 = 50, then 50 + 15 = 65. The net cost is thirty-five cents because fifty was paid and fifteen returned. The original chain incorrectly connects expressions with different values.
What to observe during the activity
If Questions 1 and 2 are difficult, return to coin count versus value. Ask the learner to count the cards first and then add their printed values. An exchange that preserves value while changing the number of pieces is a useful contrast.
If equivalent amounts are understood but Questions 12 and 13 are difficult, inspect the inventory condition. The learner may know a combination that makes the amount but use a card not supplied. That is a constraint error rather than a calculation error.
If change answers are inconsistent, keep the price, payment and change in separate labelled areas. Rebuild payment from price plus change. A learner who cannot say what the subtraction operands represent needs the transaction model made visible before more difficult prices are introduced.
If Question 16 produces a partly correct list, ask how the cases were organised. Finding three valid pairs is not evidence that there are only three. Search by the first item and exclude reversed duplicates. Completeness needs an explanation of where unlisted cases could have occurred.
If Question 19 is too demanding, omit it without treating the core money work as failed. It is an extension designed to connect a price floor, repeated choices and a budget constraint. The child may be ready for counting and exact-payment tasks while still needing substantial support for that reasoning.
Small-group roles and a home version
Rotate buyer, shopkeeper and checker. The buyer states the intended basket and counts the payment. The shopkeeper compares payment with cost. The checker verifies the relevant equation and the final wallet. Give each learner a turn in each role rather than allowing the quickest child to make every mathematical decision.
At home, three item cards and a small supply of value cards are sufficient. Begin with one exact amount, then two different ways to make it. Add change only when the child can explain which amount was paid and which amount represents the price.
Keep the tone investigative rather than competitive. The useful record is the child’s selected method, the condition they noticed or missed, and what changed after a prompt. These original tasks are not a financial-literacy score, a developmental diagnosis or a guarantee of future performance.
For more on the underlying money relationships, return to the money-sense guide. For organised searches and conditions, use the non-routine problem-solving guide. The full sequence remains at the Primary 1 Mathematics Learning Hub.
