BTT Mathematics / Primary Mathematics Learning Hub / Working Backwards
Working backwards means starting with a known final result and undoing the recorded actions in the opposite order. It is useful when a problem gives what remains, what a number becomes or when something must finish, but asks for an earlier quantity. The method succeeds only when each reverse step restores the correct earlier stage.
Suppose a number is increased by eleven and the result is multiplied by three to give sixty-three. The last action was multiplication by three. Undo that first: sixty-three divided by three gives twenty-one. Then undo the addition of eleven: twenty-one minus eleven gives ten. Check forwards: ten plus eleven is twenty-one, and three times twenty-one is sixty-three.
The important lesson is not simply to exchange addition with subtraction and multiplication with division. The order of the actions matters. The quantity at each stage matters. When a fraction of the remainder is taken, the remainder becomes a new reference whole. When information is rounded or discarded, an earlier value may no longer be uniquely recoverable.
This guide is a detailed practice companion to Word Problems, Bar Models and Checking, not a replacement for that wider introduction. It concentrates on chronological reconstruction: what happened first, what happened last and which earlier amount the next calculation recovers.
The MOE Primary Mathematics syllabus, pages 13–14, places problem solving, strategy selection and contextual checking within its framework. The explanations, sequences and questions below are original teaching material. Choose the operations already familiar to the learner; this is not a year-level checklist or an official examination paper.
Choose a starting point · Record the stages · Worked examples · Fractions and remainders · When reversal is not unique · 24 questions · Worked solutions · Teaching and transfer
1. Decide whether the problem really asks you to reverse a process
A reverse-process question contains an earlier unknown, a sequence of actions and a later known result. For example, a child gives away some stickers and has a stated number left. Or an input passes through two arithmetic operations and produces a known output. The task is to reconstruct an earlier state, not merely to calculate something from every number in the paragraph.
The words “at first” can suggest this structure, but they are not a complete method-selection rule. A question may ask for an original amount while giving a ratio, a total or two simultaneous comparisons instead of a reversible chain. Read the relationship before selecting working backwards. When two quantities change together, the unchanged-quantities guide provides a complementary route.
A small readiness check
Ask the learner to complete three statements: an amount plus six becomes nineteen; four equal groups contain twenty-eight altogether; three-fifths of an amount remains after two-fifths is removed. The first two test inverse operations. The third tests whether the learner distinguishes the removed part from the remaining part. Uncertainty in one of these statements identifies a smaller prerequisite to revisit.
Do not make a long story the first test of a relationship the learner has not yet met. A child can practise undoing an addition with counters, then a two-step number machine, then a story using the same structure. Increasing the language and the arithmetic at the same time makes it harder to see which change created the difficulty.
Know what your answer is supposed to describe
Write a label for the unknown: original money, starting number of counters, amount before the second sale or latest starting time. These are not interchangeable. A reverse calculation often recovers a useful intermediate amount before it reaches the quantity actually requested.
If a child spends money, then receives a gift, and the question asks how much remained immediately after spending, there is no need to reconstruct the original balance. Stop at the requested stage. Working backwards is a direction of reasoning, not a rule to undo every action regardless of the question.
2. Record the forward story before reversing it
Consider this process: start with an unknown number, subtract nine, multiply the result by four and finish with fifty-two. A useful record is a chain of quantities connected by labelled actions. The labels on the arrows matter just as much as the numbers in the boxes.
Start → subtract 9 → intermediate amount → multiply by 4 → 52.
Read the same chain backwards. Fifty-two divided by four gives thirteen, the amount after subtracting nine. Add nine to recover twenty-two, the original number. The complete forward check is twenty-two minus nine equals thirteen; thirteen multiplied by four equals fifty-two.
Why the last action must be undone first
The final fifty-two already includes the effect of multiplication by four. Adding nine directly to fifty-two would attach the earlier subtraction to the wrong stage. The nine was removed before multiplication, not afterwards. Undoing in the wrong order changes the process being solved.
A familiar physical comparison helps: a jumper is put on before a jacket. To remove the layers, take off the jacket first. The analogy explains order, but the arithmetic still needs its own labelled relationships. Not every physical action is mathematically reversible, and not every arithmetic operation preserves enough information for a unique reverse.
Use one line for each recovered quantity
A compact solution could read “52 ÷ 4 = 13; 13 + 9 = 22.” Add labels during learning: thirteen is the amount after the first action, while twenty-two is the original amount. These labels prevent a correct intermediate result from being reported as the final answer.
Avoid a false equality chain such as “52 ÷ 4 = 13 + 9 = 22.” The expression fifty-two divided by four equals thirteen, not twenty-two. Separate equations with sentences, semicolons or line breaks. The equals sign must continue to mean that both sides have the same value.
A stage table for longer stories
Use three columns when a chain becomes crowded: stage, what happened and amount at that stage. List the stages in chronological order, even when you fill the amounts from the bottom upwards. This separates the time order of the story from the order in which the solution becomes known.
For an unknown starting stock that is reduced, replenished and then shared, a table also makes it easier to ask whether every stage remains possible. A mathematically calculated starting amount is not acceptable if an earlier step would require giving away more whole objects than were available.
3. Worked examples: reverse the action and identify the recovered stage
Example A: A removed quantity
A class gives twenty-three cards to another class and has forty-one cards left. The original collection contains the cards given away and the cards remaining. Add them: forty-one plus twenty-three equals sixty-four cards.
The operation is addition because it restores a removed part. It is not addition merely because the answer needs to be larger. Check forwards: sixty-four minus twenty-three equals forty-one. This verifies the original relationship and accounts for every card.
Example B: An added quantity
A box receives seventeen more counters and then contains seventy-four. The earlier count is seventy-four minus seventeen, or fifty-seven. Adding another seventeen would describe a future state after another addition, not the state before the recorded addition.
Ask the learner to distinguish the sentences “seventeen were added” and “seventeen were removed.” Keeping the final count seventy-four in both versions creates two different original amounts. The action determines the inverse, even when the numbers and the question wording otherwise look similar.
Example C: Two actions, with division last
An unknown number is increased by eight. The result is divided by six to give seven. Undo the final division by multiplying seven by six, obtaining forty-two. Undo the earlier addition by subtracting eight, obtaining thirty-four.
Check forwards: thirty-four plus eight gives forty-two, and forty-two divided by six gives seven. A proposed answer of fifty would fail because fifty plus eight is fifty-eight, which does not divide by six to give seven. The forward check tests the complete chain rather than one isolated operation.
Example D: Three actions
A number is multiplied by four, reduced by thirteen and then divided by three to give seventeen. Begin with seventeen. Multiply by three to recover fifty-one. Add thirteen to recover sixty-four. Divide by four to recover sixteen.
The recovered quantities have distinct jobs: fifty-one is the value before division, sixty-four is the value before subtraction, and sixteen is the starting number. Check sixteen times four equals sixty-four; subtracting thirteen gives fifty-one; dividing by three gives seventeen.
Example E: Reconstruct a unit amount
Six identical bags contain the same number of counters each. Nine counters are removed from the combined collection, and the remaining counters are shared equally among three children. Each receives eleven counters. How many counters were in each original bag?
Three shares of eleven reconstruct thirty-three counters after the removal. Add back nine to obtain forty-two counters in all six bags. Divide by six to obtain seven counters per bag. The answer is not forty-two, because the question asks for one bag rather than the combined original collection.
A useful check reverses the units as well as the numbers: six bags at seven counters per bag give forty-two counters; removing nine leaves thirty-three; three equal shares contain eleven each. The unit labels show why multiplication, addition and division occur in that order.
Example F: Reverse a timetable
A fictional school activity starts at 10:15 a.m. A pupil needs twelve minutes to walk from the drop-off point and should arrive at the activity five minutes early. The journey to the drop-off point takes twenty-eight minutes. What is the latest departure time under these exact-duration assumptions?
Work backwards from the required arrival of 10:10 a.m. Remove the twelve-minute walk to obtain 9:58 a.m. at the drop-off point. Remove the twenty-eight-minute journey to obtain 9:30 a.m. departure. Check forwards: 9:30 plus twenty-eight minutes is 9:58, plus twelve is 10:10.
This is an arithmetic schedule, not a prediction of traffic or a live journey recommendation. Real travel planning may need extra allowance. In the mathematical problem, every interval must be included once, and the five-minute early-arrival condition belongs before the activity’s start.
4. Fractions of a remainder require a new whole at each stage
Suppose three-sevenths of a collection is removed and twenty-eight counters remain. The remaining twenty-eight is four-sevenths of the original, not three-sevenths. Divide twenty-eight by four to obtain seven counters per seventh. Multiply by seven to obtain forty-nine counters originally.
The method uses a known part to reconstruct its whole. Before calculating, write the fraction that remains. Removing three-sevenths leaves four-sevenths because the original whole contains seven equal seventh-parts. This complement step is the point at which many plausible-looking reverse calculations depart from the story.
Example G: A fixed amount, then a fractional spending
A child spends eighteen dollars, then spends two-thirds of the money remaining. Twenty-two dollars are left. Immediately before the fractional spending, the money contained three equal thirds. The twenty-two left is one third, so that stage contained sixty-six dollars.
Restore the earlier eighteen-dollar spending: sixty-six plus eighteen equals eighty-four dollars at first. Check: eighty-four minus eighteen is sixty-six; spending two-thirds, or forty-four, leaves twenty-two. The first reverse calculation recovers an intermediate whole, not the original whole.
Example H: Two successive fractions
One-quarter of a ribbon is used. Then two-thirds of the remaining ribbon is used. Twelve centimetres remain. Undo the second action first: twelve is the one-third left from that intermediate ribbon, so the intermediate length was thirty-six centimetres.
Thirty-six represents the three-quarters that survived the first action. One quarter is twelve centimetres and four quarters are forty-eight centimetres. The original ribbon was forty-eight centimetres long. Check forward: use twelve, leaving thirty-six; use twenty-four more, leaving twelve.
Do not add one-quarter and two-thirds and subtract that sum from one. The first fraction refers to the original ribbon; the second refers to a shorter ribbon. They are fractions of different wholes. Label the whole after every action before combining any fractions.
Example I: A replenishment between removals
A box loses one-third of its counters, receives ten new counters and then loses half of its contents. Nineteen counters remain. Restore the final half-removal: thirty-eight counters were present just before it. Undo the addition of ten: twenty-eight counters remained after the first removal.
Those twenty-eight represent two-thirds of the original box. One third is fourteen and the original total is forty-two. Check: forty-two minus fourteen leaves twenty-eight; add ten to reach thirty-eight; remove half to leave nineteen. The replenishment interrupts any attempt to multiply only the two remaining fractions.
Example J: Reverse a percentage, then a fixed reduction
An illustrative item is reduced by twenty-five per cent and then by a five-dollar voucher. Its final price is forty-nine dollars. Restore the voucher first: fifty-four dollars was the price after the percentage reduction. Fifty-four is seventy-five per cent, or three-quarters, of the original price.
One quarter is eighteen dollars, so the original price was seventy-two dollars. Check the original condition: a quarter of seventy-two is eighteen, leaving fifty-four; the voucher then leaves forty-nine. Adding twenty-five per cent of forty-nine would use both the wrong stage and the wrong base.
Prices here are invented arithmetic values, not current shop offers. Percentage problems should state whether each reduction applies to the original amount or the amount after an earlier reduction. Different instructions can produce different correct calculations with the same printed percentages.
Keep fractions exact until the task asks for rounding
Suppose two-thirds of a quantity remains. Recovering the whole means dividing by two-thirds, or finding one third first and then three thirds. Replacing two-thirds with 0.67 changes the relationship. That approximation can introduce a discrepancy even if the later decimal arithmetic is flawless.
For whole-object problems, use exact fractions and check that the original quantities are whole numbers when required. For measured quantities, fractions or decimals can be appropriate, but a rounding instruction should be followed deliberately rather than introduced silently.
5. Not every forward process has one recoverable starting value
Adding a known amount can be undone uniquely. Multiplying by a known non-zero number can also be undone within the appropriate number system. But some actions erase information. Working backwards cannot recover information that the question never preserves.
Rounding produces a range
A whole number rounded to the nearest ten gives sixty. Under the usual school halfway-up convention, the original whole number could be any integer from fifty-five through sixty-four. Sixty is one possible original, not the uniquely determined original.
A correct reverse solution states the range and checks its boundaries. Fifty-four rounds to fifty, so it is excluded. Sixty-five rounds to seventy, so it is excluded. The ten integers in between all produce the stated rounded value.
Multiplication by zero loses the input
A number multiplied by zero becomes zero whether the starting number is three, seven or a thousand. Adding nine afterwards gives nine for all those inputs. Knowing the final nine therefore does not recover a unique starting number.
There is no valid step of dividing by zero to retrieve the lost value. The useful answer is that the information is insufficient to determine the original uniquely, together with two different inputs that satisfy the condition.
A quotient without its remainder can hide several originals
A positive whole number divided by five has whole-number quotient eight. Without the remainder, the number might be forty, forty-one, forty-two, forty-three or forty-four. Each gives eight complete groups of five.
If the remainder is specified as three, the starting number is uniquely forty-three. This illustrates how one extra condition changes a set of possible answers into a single answer. Do not invent the missing remainder as zero unless exact division is stated.
A total does not restore the individual parts
Two children combine their counters to make thirty. The total alone does not reveal how many each had. Ten and twenty, twelve and eighteen, or fifteen and fifteen all fit. To reconstruct the two parts, another relationship is needed, such as an equality, difference or ratio.
That is why “reverse the addition” is not enough when both addends are unknown. The systematic-listing guide shows how to describe a finite set of possibilities without pretending one candidate is uniquely forced.
Checking forwards establishes validity, not always uniqueness
A candidate that reproduces the final result is valid for the stated conditions. However, another candidate might do so as well. A complete answer to “find the original number” must either establish that the reversal is unique or explain why several originals remain possible.
This distinction is useful even in ordinary practice. It prevents a learner from treating the first successful guess as a complete solution when the question asks for all possibilities. It also prevents an adult from marking one valid answer wrong merely because a different valid answer appeared in a sample solution.
6. Practice: 24 questions with three entry points
Keep the worked solutions covered until an attempt is complete. For every process, write the stages in forward order before filling them backwards. Questions 1–8 use whole-number operations; 9–16 introduce parts and intermediate totals; 17–24 add longer chains and information limits. These groups are teaching choices, not official achievement bands.
Questions 1–8: Recover one or two earlier stages
1. A number is increased by 17 to give 53. Find the original number.
2. A collection loses 26 counters and has 45 left. How many counters were there at first?
3. A number is multiplied by 7 to give 84. Find the number.
4. An amount is shared equally into 6 groups of 14. Find the original amount.
5. A number is increased by 9 and then multiplied by 4 to give 100. Find the number.
6. A number is multiplied by 3 and then reduced by 8 to give 55. Find the number.
7. Twelve is subtracted from a number. The result is divided by 5 to give 9. Find the original number.
8. A number is divided by 4 and then increased by 7 to give 20. Find the number.
Questions 9–16: Reconstruct the correct whole
9. Two-fifths of a set of counters is removed. There are 36 counters left. Find the original count.
10. Three-eighths of a ribbon is used and 45 cm remains. Find the original length.
11. A child spends $14 and then spends half of the money remaining. There is $23 left. Find the original money.
12. After a 20% reduction, an illustrative price is $96. Find the price before the reduction.
13. A quantity increases by 25% to become 90. Find its original value.
14. One-quarter of a collection is removed. Then one-third of the remainder is removed. Forty counters are left. Find the starting count.
15. Eighteen counters are added to a box. Two-fifths of the resulting contents are removed, leaving 42. Find the count before the addition.
16. Nine identical packets contain equal numbers of cards. Eight cards are removed from the combined contents. The remainder is shared among four children, who receive sixteen cards each. Find the number in each original packet.
Questions 17–24: Longer chains and limits
17. One-third of a collection is removed, then six more counters are removed. Half of what remains is given away, leaving fifteen. Find the starting count.
18. A price is reduced by 10%, then a $5 voucher is applied. The final price is $103. Find the original price.
19. A number is divided by 3, increased by 7 and then multiplied by 2 to give 54. Find the number.
20. A fictional activity starts at 2:10 p.m. Preparation takes 15 minutes, the journey 35 minutes and the final walk 10 minutes, all consecutively. Find the latest time to begin preparation to arrive exactly at the start, assuming the stated durations.
21. A whole number rounded to the nearest ten is 80. State all possible original whole numbers using the halfway-up convention.
22. A non-negative whole number is multiplied by zero and then increased by seven. The final result is seven. Is the original uniquely determined?
23. A positive whole number divided by four has whole-number quotient six. The remainder is not given. State every possible original number.
24. Three-fifths of some money is spent. One-quarter of the remaining money is then spent. There is $18 left. Find the original amount and check both spending stages.
7. Worked solutions: identify what each reverse step recovers
Solutions 1–8
1. 36. The final fifty-three includes the seventeen added. Remove that addition: 53 − 17 = 36. Check forwards: 36 + 17 = 53. The recovered value belongs to the stage before the addition, which is the quantity requested.
2. 71 counters. Reunite the removed and remaining parts: 45 + 26 = 71. Subtract twenty-six from seventy-one to check that forty-five remains. Subtracting again from forty-five would describe a second removal, not undo the first one.
3. 12. Seven equal copies of the original number total eighty-four. One copy is 84 ÷ 7 = 12. Multiplying twelve by seven reconstructs eighty-four, so the inverse operation restores the original factor.
4. 84. The final arrangement contains six groups, each with fourteen. Recombine them: 6 × 14 = 84. The division that created the shares is undone by multiplying the number of shares by the amount in each share.
5. 16. Undo multiplication first: 100 ÷ 4 = 25. This is the amount after adding nine. Now 25 − 9 = 16. Check: (16 + 9) × 4 = 100. The calculation must remove the nine from twenty-five, not from the final hundred.
6. 21. Undo the last subtraction: 55 + 8 = 63. Then undo multiplication by three: 63 ÷ 3 = 21. Check that three times twenty-one gives sixty-three and removing eight gives fifty-five.
7. 57. Reverse the division to recover 9 × 5 = 45. That is the amount after twelve was removed. Restore the twelve: 45 + 12 = 57. Forward checking gives (57 − 12) ÷ 5 = 9.
8. 52. The amount before the addition is 20 − 7 = 13. It is one quarter of the starting number, so the start is 13 × 4 = 52. Check: 52 ÷ 4 = 13, then 13 + 7 = 20.
Solutions 9–16
9. 60 counters. Removing two-fifths leaves three-fifths. Three fifth-parts contain thirty-six counters, so one fifth contains twelve. Five fifths contain sixty. Forward check: two-fifths of sixty is twenty-four, and sixty minus twenty-four leaves thirty-six.
10. 72 cm. Five-eighths remains. Divide forty-five by five to find nine centimetres per eighth, then multiply by eight to obtain seventy-two centimetres. The three-eighths used is twenty-seven centimetres; adding it to forty-five reconstructs the original ribbon.
11. $60. The final twenty-three dollars is half of the amount after the first spending. Double it to recover forty-six dollars at that intermediate stage. Restore the fourteen spent earlier to obtain sixty. Check: $60 − $14 = $46; spending half leaves $23.
12. $120. A twenty per cent reduction leaves eighty per cent. Eight equal ten-per-cent parts total ninety-six dollars, so one such part is twelve dollars. Ten parts give one hundred twenty. The reduction is twenty-four dollars, leaving the stated ninety-six.
13. 72. The final ninety represents one hundred twenty-five per cent, or five quarters, of the original. One quarter is 90 ÷ 5 = 18. Four quarters give seventy-two. A quarter of seventy-two is eighteen, and adding it produces ninety.
14. 80 counters. Forty is two-thirds of the amount after the first removal. One third is twenty, so that intermediate amount is sixty. Sixty is three-quarters of the original, giving twenty per quarter and eighty originally. Check: remove twenty, then twenty, leaving forty.
15. 52 counters. Forty-two is three-fifths of the contents after the addition. One fifth is fourteen, so the intermediate total is seventy. Undo the addition of eighteen: 70 − 18 = 52. Forward checking must include the addition before finding two-fifths.
16. 8 cards per packet. Four shares of sixteen contain sixty-four cards after the removal. Restore eight cards to obtain seventy-two in nine packets. One packet therefore contained 72 ÷ 9 = 8 cards. The answer counts cards per packet, not packets or the combined total.
Solutions 17–24
17. 54 counters. Double the final fifteen to recover thirty before the half was given away. Restore the six removed to obtain thirty-six. Thirty-six is two-thirds of the original count, so one third is eighteen and the original is fifty-four. Check: 54 → 36 → 30 → 15.
18. $120. Restore the voucher first: $103 + $5 = $108. This is ninety per cent of the original price. Ten per cent is twelve dollars and one hundred per cent is one hundred twenty dollars. Check: a twelve-dollar percentage reduction leaves $108, then the voucher leaves $103.
19. 60. Reverse multiplication by two: 54 ÷ 2 = 27. Reverse addition of seven: 27 − 7 = 20. Reverse division by three: 20 × 3 = 60. Forward check: 60 ÷ 3 + 7 = 27, and twice twenty-seven is fifty-four.
20. 1:10 p.m. The three consecutive intervals total fifteen plus thirty-five plus ten, or sixty minutes. Subtract one hour from 2:10 p.m. A stage check gives preparation ending at 1:25, journey ending at 2:00 and the final walk ending at 2:10.
21. 75 through 84 inclusive. Each integer in this range rounds to eighty. Seventy-four rounds to seventy and eighty-five rounds to ninety, so neither belongs. The answer is a set of possibilities; rounding has not preserved the units digit uniquely.
22. No. Every permitted non-negative whole number produces zero after multiplication by zero and seven after the addition. For example, both three and twelve fit. The process erases the starting value, so there is no unique original to reconstruct.
23. 24, 25, 26 or 27. Six complete groups of four account for twenty-four. The remainder may be zero, one, two or three, giving the four possibilities. Twenty-eight is excluded because it would contain seven complete groups.
24. $60. The final eighteen dollars is three-quarters of the amount after the first spending. One quarter is six dollars, so that intermediate amount was twenty-four. Twenty-four is two-fifths of the original; one fifth is twelve, and five fifths are sixty. Check: spend thirty-six, leaving twenty-four; spend six, leaving eighteen.
8. Use the working to choose a smaller, more useful repair
A wrong final answer does not tell you which part failed. Look first at the order of actions, then at the labels attached to the recovered quantities. A learner who writes the correct reverse chain but makes one subtraction error needs different support from a learner who applies a fraction to the wrong stage.
When the inverse operation is correct but the order is wrong
Give two short processes using the same numbers: add five then double; double then add five. Start both with ten. The results are thirty and twenty-five. Since the processes give different outputs, their reverse instructions must also differ. Ask the learner to undo each one separately.
This comparison makes order visible without requiring a complicated story. Only after the difference is explained should the same structure return in a money or sharing context. Keep the initial quantity and arithmetic manageable so that attention can stay on the sequence.
When the complement fraction is missing
Represent a whole as five equal parts. Remove two and ask which parts are left. Label the remaining three parts with the known final amount. The diagram’s job is to show what the given number represents; it should not become an unlabelled decoration beside a remembered calculation.
Ask the learner to say “thirty-six is three-fifths” before finding one fifth. That sentence is more informative than a correct division written without meaning. In a later example, remove a different number of parts and check whether the learner can choose the complement without being prompted.
When a forward check repeats the same mistake
A learner can check against a wrongly copied version of the story and still obtain agreement. Read the original question again before checking. Verify the sequence, the amount removed or added and the precise phrase “of the remainder.” The check should challenge the interpretation, not merely repeat it.
For a long process, a second person can read the actions while the learner follows their reconstructed amounts. That provides a simple separation between reading and calculation. It does not prove independence, so use a fresh unprompted question later to see what the learner can now manage alone.
A practical teaching sequence
Model one example and explain the last action. Complete a second example together, leaving the inverse operation for the learner to choose. On a third, offer only a blank stage table. Finally, change the context and ask the learner whether working backwards is still appropriate. This is a suggested sequence to adapt, not a fixed programme or a promised outcome.
Record the specific improvement: “I now restore the voucher before the percentage” is more useful than “reverse percentages revised.” Keep accuracy, explanation and independence as separate observations. An answer obtained after every operation is supplied is valuable practice, but it is not evidence that the method has been selected independently.
Build a question and test it in both directions
Choose a starting amount of forty-eight counters. Remove one quarter, add six and share the result equally between two children. Each receives twenty-one. Hide the forty-eight and write a reverse question from the final twenty-one. Another learner can reconstruct forty-two, then thirty-six, then forty-eight.
Writing a question reveals whether the process is valid at every stage. It also shows how to keep whole-object shares exact. Change one action and check the whole chain again rather than assuming the previous answer remains suitable.
Continue through the problem-solving collection
When the unknown is a mixture of two types rather than an earlier stage, use The Assumption Method and Guess-and-Check. When the task asks for every possible arrangement, use Systematic Listing and Logical Counting. When a transfer or equal change preserves a useful quantity, use Invariants, Before-and-After and Unchanged Quantities.
Return to Fractions, Decimals and the Same Whole for fraction foundations, or to the BTT Primary Mathematics Learning Hub to choose another route.
Original examples and 24 original practice questions with separate worked solutions. Monetary values and schedules are hypothetical. Curriculum reference accessed 6 September 2026. No examination prediction or performance guarantee is implied.