BTT Mathematics / Primary Mathematics Learning Hub / Assumption Method
The assumption method begins with an easy trial arrangement, finds how far its total is from the required total and corrects that difference through equal replacements. In a common two-type problem, the number of items stays fixed while replacing one cheaper or smaller item with one more expensive or larger item changes the total by a predictable amount.
Imagine ten tokens, each worth either two points or five points, with a combined value of thirty-eight points. Suppose all ten are worth two points. That trial gives twenty points, eighteen below the required total. Every replacement of a two-point token by a five-point token adds three points. Six replacements add eighteen. The collection therefore has six five-point tokens and four two-point tokens.
The assumption was not a claim about the actual collection. It was a deliberately chosen starting arrangement whose error could be measured. That distinction matters: a mathematical trial may temporarily fail one condition, but the final answer must satisfy every condition in the original problem.
This guide connects organised guess-and-check with the assumption method. It develops item counts, totals, per-item differences, fixed fees, coins, scoring systems and impossible-data checks. It does not replace the general word-problem guide; it supplies a deeper route for one particular structure.
The MOE Primary Mathematics syllabus, pages 13–14, discusses heuristics, representation and the selection of problem-solving strategies. This article’s method names, worked sequence and 24 questions are original teaching choices, not an official list of examination techniques. Select later money and scoring examples only after their component operations are familiar.
Recognise the structure · Organise a trial · Understand replacements · Worked examples · Check feasibility · 24 questions · Worked solutions · Teaching and transfer
1. Two kinds of total must stay separate
A two-type problem commonly gives a total number of objects and a total value contributed by those objects. Twelve coins is an object count. Four hundred twenty cents is a money total. Twenty containers is a container count. Three thousand millilitres is a capacity total. These quantities have different units and must not be added or divided indiscriminately.
Write what one item of each type contributes. A small packet may contain four cards and a large packet seven. Replacing a small packet with a large one preserves the packet count but increases the card total by three. That replacement relationship is the central piece of information.
A question suitable for the method
There are fourteen packets, each containing either four or seven cards. Altogether there are seventy-four cards. The number of packets is known, each packet has one of two known sizes, and the card total is known. Once one packet count is found, the other follows by subtracting from fourteen.
Assume fourteen small packets: fourteen times four gives fifty-six cards. Eighteen more are needed. Every replacement adds three, so six packets must be large. The remaining eight are small. Check eight times four plus six times seven equals thirty-two plus forty-two, or seventy-four cards.
A question not yet suitable for the same shortcut
Suppose there are small, medium and large packets, with three different card counts. Knowing only the packet total and card total may allow several mixtures. One replacement no longer has a single fixed size unless the medium count is known or otherwise constrained.
The method can still help after another quantity is fixed, but the two-type shortcut cannot simply be copied unchanged. A useful strategy must fit the number of unknowns and the relationships given. More unfamiliar words do not necessarily make a problem structurally harder; an extra independent type can.
Keep actual objects separate from model objects
The trial arrangement might contain only small packets even though the problem says the actual collection contains both sizes. This is acceptable as a calculation model. It is not a proposed final solution. Label it “assume all small” so that a reader knows it is temporary.
At the end, return to the actual conditions. If the question requires at least one of each type, a computed answer with zero of one type fails that condition even when the numerical totals match. A baseline may ignore a condition temporarily; the final answer may not.
2. Guess-and-check becomes useful when each trial preserves a condition
Return to the ten tokens worth two or five points, with total value thirty-eight. A random guess might choose seven low-value tokens and seven high-value tokens. Their value could be calculated correctly, but the collection would contain fourteen tokens, contradicting the stated ten before the value check even begins.
Instead, choose a high-value count and determine the low-value count from the fixed total. If there are three high-value tokens, there must be seven low-value tokens. The total value is three times five plus seven times two, or twenty-nine. The item-count condition is now automatically preserved in every trial.
Keep a short trial record
For three high-value tokens, the total is twenty-nine. For four high-value tokens, it is thirty-two. For five, it is thirty-five. For six, it is thirty-eight. Each move replaces one low token by one high token and adds three points.
The trials do more than eventually find the answer. They reveal a fixed change that can be used to jump directly. From the first trial’s twenty-nine, nine more points are needed. Nine divided by three gives three additional high-value tokens, moving from three to six.
Explain why the next guess is different
A productive next trial follows the information from the current one. When the total is too small and the expensive-item count is increased, explain that the item total is held fixed by reducing the cheaper count at the same time. Increasing both counts would not be a valid adjustment.
A learner who writes several unrelated guesses may not yet be using feedback. Ask what the previous trial revealed and how the next trial will move towards the target. The purpose is not to forbid trial work; it is to turn trials into evidence about the structure.
Guess-and-check and assumption are connected
The all-small assumption is a particularly easy first trial. Every item has the same contribution, so its total is simple to calculate. The difference between trial and target is then divided by the change caused by one replacement. The method compresses a series of equally spaced trials into one calculation.
Some learners understand the shortcut more clearly after seeing two or three neighbouring trials. Others can reason directly from the baseline. Use whichever representation allows the learner to explain both the fixed item count and the per-replacement change.
3. The replacement changes two contributions at once
Replacing a two-point token by a five-point token does not add five points to the total. The old two-point contribution is removed and a five-point contribution takes its place. The net change is five minus two, or three points.
This is why dividing the eighteen-point shortfall by five would be wrong in the opening example. The trial already counted two points for every token, including the ones eventually identified as high-value. Only the additional three points per replacement are missing.
Give the difference a unit
Write “three extra points per replacement” rather than only “three.” A shortfall of eighteen points divided by three points per replacement gives six replacements. Those replacements identify six high-value tokens because each replacement changes exactly one item.
The unit check is a useful protection against dividing the shortfall by the number of objects. Eighteen points divided by ten tokens gives an average shortfall per token. That may be meaningful in another analysis, but it is not directly the number of high-value tokens.
Starting with all large items also works
Assume all ten tokens are worth five points. The trial total is fifty, twelve above the target thirty-eight. Replacing one high-value token by one low-value token removes three points. Four replacements remove twelve, so there are four low-value tokens and six high-value tokens.
The two baselines find different unknowns first. An all-small baseline counts upgrades into large items. An all-large baseline counts downgrades into small items. Name what the quotient represents before subtracting it from the total number of items.
The calculation behind the explanation
With N items, a lower contribution s, a higher contribution h and k high-type items, the total is N × s + k × (h − s). The first term counts the all-small arrangement. The second adds the extra contribution carried by the k high-type items.
Formal letters are optional at Primary level. The sentence “all-small total plus extra amount from each replacement” expresses the same relationship. The algebra is included as a bridge for a learner ready to see why one explanation applies across coins, tickets, containers and scores.
What must remain fixed for the shortcut to work?
The total item count must be fixed. Every item must belong to one of the two stated types. The contribution of each type must be known and consistent. There must be no unaccounted third contribution, such as a delivery fee included in the money total or empty containers included in a supposed full-capacity total.
Check these conditions before applying a remembered formula. When a condition changes, adjust the model instead of forcing the printed numbers into the familiar pattern.
4. Worked examples across different contexts
Example A: Tickets with two prices
Sixteen tickets cost either three dollars or seven dollars each. Their total cost is eighty-eight dollars. Assume every ticket costs three dollars: sixteen times three gives forty-eight dollars. The shortfall is forty dollars. Each change to a seven-dollar ticket adds four dollars, so ten tickets cost seven dollars.
The other six tickets cost three dollars. Check both totals: ten plus six equals sixteen tickets; ten times seven plus six times three equals seventy plus eighteen, or eighty-eight dollars. Checking only the money could allow an arrangement with the wrong number of tickets to pass unnoticed.
Example B: Coins in cents
Fourteen coins are either twenty-cent or fifty-cent coins and total four hundred sixty cents. Assume all twenty-cent coins: fourteen times twenty is two hundred eighty cents. The required increase is one hundred eighty cents. Each replacement adds thirty cents, giving six fifty-cent coins and eight twenty-cent coins.
Working entirely in cents avoids mixing fourteen objects, dollars and cents in the same arithmetic line. A check gives six times fifty plus eight times twenty equals three hundred plus one hundred sixty, or four hundred sixty cents. Convert to dollars only when the requested answer requires it.
Example C: Packets with two sizes
Twenty-one packets each contain either five or nine stickers. Altogether there are one hundred thirty-seven stickers. An all-five baseline gives one hundred five. The extra thirty-two stickers come from replacements that each add four. Eight packets are therefore nine-sticker packets and thirteen are five-sticker packets.
The arithmetic is the same as the ticket example, but the quantity being totalled is stickers rather than money. This transfer matters: the method belongs to the relationship between two contributions, not to a keyword such as “cost” or “coins.”
Example D: A fixed charge outside the mixture
Twelve items cost either four dollars or six dollars each. A five-dollar delivery fee brings the total payment to sixty-seven dollars. The items themselves cost sixty-two dollars after the fixed fee is removed. An all-four baseline is forty-eight dollars, giving fourteen dollars to explain.
Each upgrade adds two dollars, so seven items cost six dollars and five cost four dollars. Check the full bill: seven times six plus five times four equals sixty-two; add the fee to obtain sixty-seven. Using the full payment as the mixture total would incorrectly ask replacements to explain the delivery charge.
Example E: Correct answers and zero for incorrect answers
A quiz has eighteen questions. Every correct answer earns five points and every incorrect answer earns zero. All questions are answered, and the score is sixty-five. An all-correct trial gives ninety points. Every incorrect answer reduces that trial by five points. The loss of twenty-five points therefore represents five incorrect answers.
There are thirteen correct answers. A direct route, sixty-five divided by five, also finds thirteen because incorrect answers contribute zero. Use this comparison to show that a general method is not always the shortest method. Strategy choice should follow the structure, not loyalty to a named technique.
Example F: A penalty changes the replacement amount
In a different illustrative quiz, twenty questions are answered. A correct answer earns four points, and an incorrect answer loses one point. The final score is fifty-five. An all-correct trial gives eighty points. Replacing a correct answer with an incorrect one removes the four points that would have been earned and introduces a one-point loss.
Each replacement therefore lowers the total by five points, not by one and not by four. The twenty-five-point deficit represents five incorrect answers. Fifteen are correct: fifteen times four minus five gives fifty-five. This optional example requires understanding subtraction as a penalty; it need not be used before the learner is ready.
Example G: A full-container condition
Thirteen containers hold either two hundred millilitres or five hundred millilitres, and all are full. Together they hold four thousand one hundred millilitres. An all-small trial gives two thousand six hundred. The extra one thousand five hundred is five increments of three hundred, so there are five large containers and eight small ones.
The word full makes the listed capacities equal to the actual quantities contributed. If containers can be partly filled, their capacities alone do not determine the liquid total. The simple replacement calculation would then lack a necessary condition. Reading that word is part of the mathematics.
Example H: Two baselines as a checking pair
Twelve tokens worth either four or nine points total seventy-eight. The all-four baseline gives forty-eight, leaving thirty to add in steps of five. Six tokens are worth nine. The all-nine baseline gives one hundred eight, requiring thirty to be removed in steps of five. Six tokens are worth four.
The two calculations agree on six of each type. They do not constitute unrelated proofs, because both use the same model, but they help detect a reversed label or an incorrect final subtraction. Always finish by calculating the actual total from the recovered counts.
5. Some totals cannot come from the stated objects
The assumption method should not manufacture fractional people, fractional tickets or a negative number of packets merely because a division was possible. A proposed answer has to fit the kind of objects being counted.
Check the smallest and largest possible totals
Ten objects worth either three or eight points must total at least thirty and at most eighty. A stated total of eighty-four is impossible under those conditions. No detailed search is needed: even the all-large arrangement cannot reach it.
This bound also helps catch copied data. It does not justify silently changing eighty-four to another number. State the inconsistency and verify the question. The job of a mathematical check is to expose a problem, not hide it by repairing the data without permission.
Check the allowed increments
With ten objects worth either three or eight points, the possible totals rise from thirty in steps of five: thirty, thirty-five, forty and so on to eighty. A total of thirty-two lies inside the broad range but cannot be obtained, because the two-point excess is not a whole number of five-point replacements.
A range check and an increment check have different jobs. The first asks whether the total is large enough or too large. The second asks whether it falls on one of the totals the available objects can actually produce.
Check the number of replacements
When the quotient represents a number of high-type items, it must lie between zero and the total number of objects. If both types are required, the endpoints are excluded. If at least six high-type items are required, a result of four is not acceptable even when the two headline totals match.
Do not stop reading after the first two numbers needed by the method. Later conditions such as “at least,” “exactly” and “both kinds” remain part of the problem.
Three types can produce more than one answer
Six tokens worth two, three or five points total twenty. Two tokens of each type give four plus six plus ten, or twenty. Five three-point tokens and one five-point token also give twenty. The same item count and point total permit different mixtures.
This does not make the problem meaningless. It means that finding a unique count for every type requires another condition. A systematic list can describe all valid mixtures. A statement that one mixture works must not be presented as proof that it is the only one.
Why the two-type solution is unique when it exists
For fixed item count and two distinct contributions, each extra high-type item changes the total by the same positive difference. The possible totals therefore increase without repeating as the high-type count increases. A target can match at most one of those counts.
The existence checks still matter. A target outside the allowed range or between two allowed increments matches no count. Uniqueness means “at most one,” not “there must be one.” This distinction is a useful extension for stronger learners who want to justify the method rather than merely perform it.
6. Practice: 24 questions
All prices, scores and collections below are hypothetical. Unless a question says otherwise, items are whole and belong only to the two listed types. Show the trial total, the difference from the required total, the change per replacement and a final check of both totals. Keep the solutions covered during the first attempt.
Questions 1–8: Build the replacement model
1. Twelve toy vehicles have either two or three wheels each. There are thirty-one wheels altogether. Find the number of each type.
2. Twenty toy vehicles have either two or four wheels each. The wheel total is sixty-two. Find the two counts.
3. Fifteen coins are either 20-cent or 50-cent coins. Their total value is 540 cents. How many of each coin are there?
4. Fourteen tickets cost either $3 or $8 each. The total is $67. Find each ticket count.
5. Eighteen boxes contain either four or seven counters each. Altogether they contain ninety-six counters. Find the numbers of small and large boxes.
6. A twenty-five-question quiz gives four points per correct answer and zero per incorrect answer. All questions are answered and the score is seventy-six. Find the correct and incorrect counts.
7. Twelve items cost either $4 or $9 each and total $78. Solve by assuming all items cost $9.
8. Ten full containers hold either 3 L or 5 L each. Together they hold 44 L. Find the number of each size.
Questions 9–16: Keep units and extra charges separate
9. Sixteen items cost either $2 or $5 each. A $4 fixed delivery fee makes the payment $60. Find the number of each item type.
10. Twenty full bags contain either 100 g or 250 g of rice each. Their total rice mass is 3,500 g. Find each bag count.
11. Twenty-four tickets cost either $2.50 or $4 each. They total $81. Find the two counts.
12. Thirty questions are answered. Each correct answer earns three points and each incorrect answer loses one point. The score is sixty-two. Find the correct and incorrect counts.
13. Eighteen toy animals represent chickens with two legs or rabbits with four legs. Counting the represented legs gives fifty. Find the number of each type.
14. Seventeen gift bags hold either six or ten counters each. There are 138 counters altogether. Find the two bag counts.
15. Twenty-two coins are either 50-cent or $1 coins. Their total value is $16. Find each coin count.
16. Twenty-eight notebooks cost either $3 or $6 each, totalling $132. Find the counts at each price.
Questions 17–24: Test feasibility and explain the method
17. Ten boxes each contain four or seven counters. Could their total be fifty-six counters? Explain.
18. Eight items each cost $2 or $5. Could their total be $44? Explain.
19. Six tokens, each worth two, three or five points, total twenty points. Give two different valid mixtures. A type may have count zero.
20. Thirty full containers hold either 200 ml or 350 ml each. Together they hold 8,250 ml. Find each count.
21. Two collections each contain twelve items priced at $2 or $5. The second collection costs $12 more than the first. How many more $5 items does the second collection contain?
22. Fourteen tables seat either six or ten people each. Their total seating capacity is one hundred. Use an all-ten-seat assumption to find each count.
23. Eighteen items cost $4 or $7 each and total $90. The question also says there are at least eight $7 items. Can all the conditions hold?
24. Sixteen workshop tickets cost $5 or $9 each. A $12 voucher reduces the final payment to $100. Find the two ticket counts and check the voucher stage.
7. Worked solutions
Solutions 1–8
1. Five two-wheel and seven three-wheel vehicles. Assuming all twelve have two wheels gives twenty-four wheels. Seven more are needed. Each replacement adds one wheel, so seven vehicles have three wheels. Check the count, 5 + 7 = 12, and the wheels, 5 × 2 + 7 × 3 = 31.
2. Nine two-wheel and eleven four-wheel vehicles. The all-two-wheel total is forty. The excess twenty-two is eleven replacements of two wheels each. The other nine vehicles have two wheels. Check: eighteen wheels plus forty-four wheels gives sixty-two.
3. Seven 20-cent and eight 50-cent coins. Fifteen 20-cent coins would total three hundred cents. The shortfall of two hundred forty cents is filled by eight increments of thirty cents. Check: seven times twenty plus eight times fifty equals five hundred forty cents.
4. Nine $3 and five $8 tickets. The baseline is fourteen times three, or forty-two dollars. Twenty-five extra dollars require five replacements at five dollars each. Check: twenty-seven plus forty equals sixty-seven, and the ticket counts total fourteen.
5. Ten small and eight large boxes. An all-four baseline contains seventy-two counters. The twenty-four-counter excess is eight replacements of three counters. Check: ten times four plus eight times seven equals forty plus fifty-six, or ninety-six.
6. Nineteen correct and six incorrect. The all-correct score is one hundred. A loss of twenty-four points represents six incorrect answers because each removes four points. Nineteen times four gives the stated seventy-six. Direct division by four is also valid here.
7. Six of each price. The all-nine-dollar trial costs one hundred eight dollars. It must fall by thirty. Each replacement with a four-dollar item lowers it by five, so six items are the cheaper type. The remaining six cost nine dollars. Check: twenty-four plus fifty-four is seventy-eight.
8. Three 3 L and seven 5 L containers. Ten small containers hold thirty litres. The additional fourteen litres require seven replacements of two litres each. Check: nine plus thirty-five equals forty-four litres. The full-container condition makes capacity equal to the stated contents.
Solutions 9–16
9. Eight $2 and eight $5 items. Remove the four-dollar fee from sixty to obtain fifty-six dollars for items. Sixteen two-dollar items would cost thirty-two, leaving twenty-four to explain in three-dollar increments. Eight replacements are required. Restoring the fee gives the final sixty-dollar payment.
10. Ten bags of each size. Twenty 100 g bags contain 2,000 g. The extra 1,500 g comes in increments of 150 g per replacement, giving ten larger bags. Check: ten times 100 plus ten times 250 equals 3,500 g.
11. Ten $2.50 and fourteen $4 tickets. The cheaper baseline is sixty dollars. The twenty-one-dollar shortfall divided by $1.50 per upgrade gives fourteen. Check: ten cheaper tickets cost twenty-five dollars and fourteen dearer tickets cost fifty-six, totalling eighty-one.
12. Twenty-three correct and seven incorrect. The all-correct score is ninety. Each wrong answer replaces a three-point gain with a one-point loss, lowering the trial by four. The twenty-eight-point deficit represents seven wrong answers. Check: 23 × 3 − 7 = 62.
13. Eleven chickens and seven rabbits. Eighteen two-legged models account for thirty-six legs. Fourteen more legs require seven replacements, each adding two. Check: eleven times two plus seven times four equals fifty legs across eighteen models.
14. Eight six-counter and nine ten-counter bags. The all-six baseline is one hundred two counters. The extra thirty-six comes from nine upgrades of four counters. Check: forty-eight plus ninety equals one hundred thirty-eight, with seventeen bags altogether.
15. Twelve 50-cent and ten $1 coins. In cents, the all-small total is 1,100 and the required total is 1,600. The five hundred extra cents correspond to ten upgrades of fifty cents. Check: six dollars in half-dollar coins plus ten dollars gives sixteen dollars.
16. Twelve $3 and sixteen $6 notebooks. The all-three-dollar total is eighty-four. The extra forty-eight requires sixteen upgrades at three dollars each. Check: twelve times three plus sixteen times six equals thirty-six plus ninety-six, or one hundred thirty-two.
Solutions 17–24
17. No. The all-four baseline is forty. The required excess is sixteen, but each replacement adds three. Sixteen is not divisible by three, so no whole-number count of large boxes produces fifty-six. The nearest trial totals around it are fifty-five and fifty-eight.
18. No. Even eight five-dollar items cost only forty dollars. Forty-four exceeds the maximum possible total. This range check is sufficient to reject the conditions without calculating a replacement count.
19. Two of each type; or five three-point tokens and one five-point token. The first gives six tokens and 4 + 6 + 10 = 20 points. The second also gives six tokens and 15 + 5 = 20. Two valid answers show the mixture is not uniquely determined.
20. Fifteen of each size. Thirty 200 ml containers hold 6,000 ml. The extra 2,250 ml consists of fifteen increments of 150 ml. Check: fifteen small containers contribute 3,000 ml and fifteen large ones 5,250 ml, totalling 8,250 ml.
21. Four more $5 items. Both collections have the same twelve-item baseline. Each additional dearer item replaces a cheaper item and increases total cost by three dollars. A twelve-dollar difference therefore corresponds to four such replacements. The individual counts in each collection are not uniquely fixed by this information alone.
22. Ten six-seat and four ten-seat tables. Fourteen ten-seat tables would seat one hundred forty. The trial must fall by forty seats. Each replacement with a six-seat table removes four seats, so ten tables are smaller. Check: sixty plus forty equals one hundred seats.
23. No. The two headline totals force six seven-dollar items: an all-four baseline is seventy-two, and the eighteen-dollar excess is six upgrades of three. Six fails the additional requirement of at least eight. Matching two conditions does not justify ignoring the third.
24. Eight tickets at each price. Restore the twelve-dollar voucher to obtain an item total of one hundred twelve. An all-five baseline is eighty, leaving thirty-two in four-dollar increments. Eight upgrades give eight nine-dollar tickets and eight five-dollar tickets. Their costs total one hundred twelve; the voucher reduces payment to one hundred.
8. Teaching the assumption method without turning it into a slogan
The phrase “assume all small, subtract and divide” can produce a correct answer without showing that the learner understands what is being replaced. Ask the learner to explain one replacement with actual counters, labelled tokens or a sketch before relying on the compressed calculation.
When the learner divides by the larger value
Return to a single object in the trial arrangement. It already contributes the smaller amount. Cross out that contribution and write the larger contribution beside it. Ask how much the combined total changes. The answer is the difference between the two contributions, not the full larger amount.
A two-row trial record can reinforce the point: five large items and the required complementary small count; then six large items and one fewer small item. Compare the totals. The net change is visible without any general formula.
When the labels reverse at the end
Ask which type was present in the original assumption and which type each replacement introduces. In an all-small model, the quotient counts large items. In an all-large model, it counts small items. Write this meaning beside the quotient before finding the complementary count.
A correct number attached to the wrong type can still produce a plausible-looking answer. The final value check will expose it unless the two counts happen to be equal. Use an example with unequal counts when testing whether this distinction is secure.
When the item count changes during trials
Provide the total number of empty item spaces. Let the learner relabel a space from small to large instead of adding a new space. This models replacement rather than addition. Every trial should have the same number of occupied spaces.
Later, remove the physical spaces and keep a two-column record whose counts add to the fixed total. The representation can become shorter while preserving the relationship it was designed to show.
When an impossible result is accepted
Ask whether a result such as five and one-third tickets belongs to the objects in the question. Then inspect the target range and the possible increments. A fractional quotient may be evidence that the conditions cannot hold, not an instruction to round the object count.
Do not round a replacement count to make the numbers fit. Rounding changes the total. A correct response to inconsistent conditions should explain the contradiction or identify which information needs checking.
A progression from concrete replacement to independent choice
Begin with tokens carrying two printed values. Make an all-small trial and perform replacements physically. Move to a short trial table. Then replace the repeated trials with a difference calculation. Finally present a mixed set containing a two-type question, a reverse-process question and an incomplete-information question.
The last step tests selection, not just execution. Ask the learner why an assumption baseline is useful in one problem and why another needs a stage chain or more information. This proposed progression can be adapted to the learner’s working; it is not a guarantee or a prescribed timetable.
Create a problem forwards before solving it backwards
Choose seven small packets containing three counters each and five large packets containing eight each. The total is twelve packets and sixty-one counters. Hide the two packet counts and write a question from the two totals. The all-small trial has thirty-six counters; twenty-five extra counters require five upgrades of five.
This construction makes the conditions internally consistent. It also gives a useful exercise in explaining uniqueness: increasing the large count by one always adds five, so no different large count can give the same sixty-one-counter total while twelve packets remain fixed.
Continue through the problem-solving collection
For chronological recovery of an earlier amount, use Working Backwards and Reverse Processes. For more than two types or a request for every valid arrangement, use Systematic Listing and Logical Counting. For a fixed total or difference across changing quantities, use Invariants, Before-and-After and Unchanged Quantities.
Return to Equal Groups, Division and Remainders for the multiplication foundation, or to the BTT Primary Mathematics Learning Hub for the wider collection.
Original explanations and 24 original practice questions with separate worked solutions. Prices, scores and quantities are hypothetical. Curriculum reference accessed 6 September 2026. The collection does not reproduce school questions or predict an examination.