PRIMARY 4 MATHEMATICS LEARNING GUIDE · BATCH 5 · GUIDE 18
Working backwards means recovering an earlier quantity from a later one without inventing any missing step. When a problem tells us the final result and every action that led to it, we can often undo the actions one at a time. The important detail is that we must reverse their order as well as their operations.
Suppose a number is doubled and then 14 is added, giving 86. The last action was adding 14, so that is the first action to undo. Subtract 14 to recover 72, then halve 72 to recover 36. Starting by halving 86 would act on the wrong stage of the journey.
This extended Primary 4 lesson develops reverse reasoning through number stories, stock, money, equal sharing, lengths, fractions of sets and clock times. It also explains when backwards reasoning cannot recover one exact answer, such as when a remainder or an exact starting measurement has been lost. The twenty-question practice laboratory includes full explanations and forward checks.
Series route: return to the Primary 4 Mathematics Learning Hub. Start with Before-and-After Problems when the main difficulty is identifying which quantity belongs to which moment.
Scope: this is independent problem-solving practice built around familiar arithmetic and the Primary 4 foundations in the MOE Primary Mathematics Syllabus. It is not an official list of examination question types. The deeper challenges are optional, and all situations and numerical inputs are original teaching examples.
Navigate: reverse order · three actions · applications · limits of reversal · practice · answers · teaching routine.
1. An inverse operation restores a previous amount
If adding 17 to a starting number gives 59, subtracting 17 from 59 restores the start: 42. If subtracting 26 gives 81, adding 26 restores the start: 107.
The inverse is not chosen because a teacher says to swap signs. It is chosen because it reverses the effect of a known operation. Adding a fixed amount and then removing that same amount brings us back to the previous value.
Multiplication and exact division form another pair. If four equal groups together contain 156 counters, one group contains 156 ÷ 4 = 39. If a collection is shared equally among six groups and each receives 27, the original collection contains 27 × 6 = 162.
The quantities must still make sense. Dividing by four undoes multiplication by four, but dividing by zero is not permitted. Remainder division also requires extra attention because the quotient alone may not preserve the entire original number. We will return to these limits later.
Begin with these one-step relationships before introducing a long chain. Reversing four steps does not repair uncertainty about what it means to undo one.
2. Undo the last action first
Return to the opening problem: a number is doubled, then 14 is added, giving 86.
Forward: unknown → double → unknown → add 14 → 86 Backward: 36 ← halve ← 72 ← minus 14 ← 86
The final 86 includes the extra 14. Removing that final addition exposes the doubled number. Only then can the doubling be undone.
A common incorrect route is 86 ÷ 2 − 14 = 29. Test 29 forwards: doubling gives 58 and adding 14 gives 72, not 86. The reverse route used inverse operations but applied them in the wrong order.
This is the central discipline of working backwards: start from a known final state, identify the last action, undo it and name the recovered state. Repeat until the requested earlier quantity is reached.
The forward check is not an optional decorative line. It is the test that the recovered number actually belongs to the stated story. A reverse calculation can look tidy while undoing a different sequence.
3. A small change in order creates a different problem
Compare two instructions that use the same numbers. In Problem A, double the starting number and add 14 to obtain 86. In Problem B, add 14 to the starting number and then double the result to obtain 86.
Problem A gives 86 − 14 = 72, followed by 72 ÷ 2 = 36. Problem B gives 86 ÷ 2 = 43, followed by 43 − 14 = 29.
Both answers are correct for their own instructions. They differ because the 14 is doubled in Problem B but not in Problem A. Order is part of the mathematics, not simply a choice of sentence style.
To make this visible with objects, start with six counters. Doubling and adding four gives sixteen. Adding four and then doubling gives twenty. The same two action words appear, but they create different results.
Ask the learner to explain which quantity is doubled. “The starting amount” and “the amount after the addition” are not interchangeable. This one contrast can expose a hidden sequencing weakness before the arithmetic becomes more demanding.
4. Keep a forward record and a reverse record
Write the events in their original order first. A two-column table is often clearer than crossing out words or changing every symbol in the question.
| Action in the original story | Action that restores the previous state |
|---|---|
| Add a known amount | Subtract that same amount |
| Remove a known amount | Add that same amount |
| Multiply by a known non-zero number | Divide by that number |
| Divide exactly into a known number of equal groups | Multiply one group’s amount by the number of groups |
Read the left-hand column in chronological order to understand the story. Read the appropriate reverse actions from the last event to the first to recover earlier states.
Do not erase the forward story once the backwards calculation starts. It is the reference against which the final answer must be checked. Keeping both records prevents a mistaken inverse from quietly rewriting what the question said.
For longer questions, name the states: before sharing, after receiving, before removal. These names stop a number from being used in the wrong place merely because it is the most recent answer on the page.
5. A complete three-action number problem
A starting number has 18 subtracted from it. The result is multiplied by three. Then twelve is added, giving 150. Find the starting number.
Undo the final addition: 150 − 12 = 138. This is the quantity after multiplication by three. Undo that multiplication: 138 ÷ 3 = 46. This is the quantity after subtracting 18. Restore the removed 18: 46 + 18 = 64.
Now check forwards using the original order. Starting at 64, subtracting 18 gives 46. Multiplying 46 by three gives 138. Adding twelve gives 150. The recovered start satisfies every instruction.
A student might perform all three inverse operations mentally, but the written stages are valuable during learning. They show exactly what each intermediate answer represents and where an error would first appear.
The question does not require formal algebra. We can use a labelled timeline and ordinary arithmetic. More advanced notation may eventually compress the route, but it is not needed to understand why the backwards method works.
6. A chain that includes exact sharing
A collection is made four times as large. Twelve counters are then removed. The remaining counters are shared equally among three trays, with twenty counters in each tray and none left over. Find the original collection.
The final twenty is an amount per tray, not the whole remaining collection. Undo the sharing first: 20 × 3 = 60 counters. Restore the twelve removed: 60 + 12 = 72 counters. Undo making the collection four times as large: 72 ÷ 4 = 18 counters.
Check forwards: 18 × 4 = 72; removing twelve leaves 60; sharing 60 among three trays gives twenty per tray.
The unit labels matter. Multiplying twenty counters per tray by three trays reconstructs a total counter count. A reverse route that begins with 20 + 12 would restore the removed counters to only one tray rather than to the whole collection.
Whenever a final answer is expressed as “each”, “per group” or “in every box”, identify how many groups it describes before undoing earlier actions.
7. Work backwards through a stock record
After a shop sells 83 notebooks, it packs all the remaining notebooks into seven boxes with 29 notebooks in each box. None are left unpacked. How many notebooks were there before the sale?
First recover the total after the sale: 7 × 29 = 203 notebooks. Then restore the notebooks sold: 203 + 83 = 286 notebooks.
This is only a two-stage reverse route, but the first stage is essential. The final 29 is not closing stock; it is closing stock divided among seven boxes. Reading the noun after a number is part of reading the mathematics.
The condition “None are left unpacked” tells us that the packed boxes account for the entire remainder. Without it, an unknown number of loose notebooks could make the initial stock larger.
For a changed question, suppose five loose notebooks remain as well as the seven full boxes. The closing quantity becomes 203 + 5 = 208, and the initial quantity becomes 291. We have not invented a new method; we have included one newly stated part of the final state.
8. Recover money without reversing the story’s language
In an invented money problem, Zain spends $7.65 and later receives $5.20. He ends with $21.40. How much did he have before spending?
Undo the final receipt: $21.40 − $5.20 = $16.20. This is the amount after spending. Restore the spent money: $16.20 + $7.65 = $23.85.
Check: $23.85 − $7.65 = $16.20; $16.20 + $5.20 = $21.40.
The original story still says that Zain spent first and received later. Working backwards does not claim that the actual spending was a receipt. It reconstructs the amount before the spending by adding back the known reduction.
Money is useful for checking whether a result is plausible, but do not rely solely on intuition. Since spending exceeded the later receipt by $2.45, the final amount should be $2.45 less than the initial amount. The recovered start of $23.85 and finish of $21.40 satisfy that difference.
All amounts here are teaching inputs, not statements about current prices or financial recommendations.
9. Recover an original measurement
A rope has 2.70 m cut from it. The remainder is divided into four equal pieces, each 3.45 m long. Assume no material is lost in cutting. Find the original length.
The four equal pieces together measure 3.45 × 4 = 13.80 m. Adding the removed 2.70 m gives an original length of 16.50 m.
Check forwards: 16.50 − 2.70 = 13.80 m; dividing that remainder into four equal lengths gives 3.45 m each.
The no-loss condition matters. Real cutting can involve waste, but this mathematical example explicitly excludes it. Do not invent a waste allowance, and do not omit one when a question supplies it.
A useful labelled sketch has one long original strip divided into the removed piece and four equal remaining pieces. That model and the backwards calculation express the same relationship. When a student cannot remember the reverse sequence, the strip can help reconstruct it from the quantities rather than from a memorised procedure.
10. Recover a whole from a remaining fraction
Deeper challenge. One quarter of a set of counters is removed, leaving 36 counters. Find the original number of counters.
The remaining 36 counters represent three quarters, not one quarter. Draw four equal units for the original set and cross out one. The three unremoved units total 36, so one unit contains 36 ÷ 3 = 12 counters. All four original units contain 12 × 4 = 48 counters.
Check: one quarter of 48 is twelve, and removing twelve from 48 leaves 36. This route uses equal-unit reasoning and whole-number arithmetic; it does not require a memorised rule for division by a fraction.
A common error is to multiply 36 by four immediately. That would treat the remainder as one quarter when it is actually three quarters. The fraction named in the event may not be the fraction represented by the given final amount.
Before any calculation, finish this sentence: “The known amount represents ___ equal units of the original whole.” That statement chooses the correct first division.
11. Follow an exact instruction rather than a familiar fraction pattern
Deeper challenge. Two fifths of a collection are removed, and the remaining amount is 42. The remainder represents three fifths. Three units total 42, so one unit is fourteen and the original five units total 70.
Now change the wording: “Two fifths of a collection are 42.” In that case the known amount represents two units, so the whole would be 42 ÷ 2 × 5 = 105. The numbers and fraction look familiar, but the known part has changed.
These are different questions, not alternative answers to one question. The mathematical operation follows the reference quantity described by the words.
For another change, add a fixed amount after the fractional removal. Undo that final addition before matching the remainder to its number of units. For example, if the final amount is 50 after eight were added to a three-fifths remainder, first recover 42; then reconstruct the original 70.
The important habit is to recover one complete state at a time, not to apply every visible fraction to the final number.
12. Work backwards through a transfer between two people
Deeper challenge. Noor gives Kai 28 cards. Afterwards Noor has 77 cards and Kai has 95. Find how many each had before the transfer.
Restore Noor’s transferred cards: 77 + 28 = 105. Remove the receipt from Kai’s final amount: 95 − 28 = 67.
The starting combined total is 105 + 67 = 172. The final combined total is 77 + 95 = 172. An internal transfer leaves the combined number unchanged, which gives an additional check.
Two reverse actions are required because two people’s amounts changed. Undoing only Noor’s reduction would count the transferred cards both in Noor’s restored pile and in Kai’s final pile.
Use a two-row table when several people are involved. Keep one person’s state in each row and one time in each column. A number belonging to Kai after the transfer cannot replace a number belonging to Noor before it merely because both count cards.
13. Recover a starting time across an hour boundary
An activity finishes at 11:10 a.m. after lasting 45 minutes. Find the starting time.
Move back ten minutes to 11:00 a.m. There are 35 minutes still to undo. Moving back another 35 minutes reaches 10:25 a.m.
Check forwards: from 10:25 to 11:00 is 35 minutes, and from 11:00 to 11:10 is ten minutes. Together they make the given 45 minutes.
This uses the same reverse structure as a stock problem, but it uses a clock scale rather than ordinary decimal notation. Do not calculate 11.10 − 0.45 as though the numbers were decimal hours.
When a problem crosses noon, midnight or a date boundary, record that boundary explicitly. For the simple example here, all times fall in the same morning, so no date change is needed. The representation should include every boundary the actual question requires, not additional complications that were never present.
14. A quotient alone may not reveal the original number
A whole number divided by seven gives quotient 24. Is the original number necessarily 168?
Only when there is no remainder. If a remainder is allowed but not reported, the original number could be 168, 169, 170, 171, 172, 173 or 174. These all have 24 complete groups of seven, with remainders from zero to six.
If the question states quotient 24 and remainder five, the original is exactly 7 × 24 + 5 = 173. The reconstruction uses the complete relationship: original = divisor × quotient + remainder.
The missing remainder is not a small detail. It is information needed for a unique reverse answer. Multiplication by the divisor restores only the complete groups, not any unreported remainder.
Revisit Division, Remainders and Quotient Interpretation when this distinction is uncertain. A reliable backwards method begins with knowing whether the forward operation preserved all the information we need.
15. Rounding cannot usually be undone to one exact value
Boundary challenge. A whole-number count is reported as 340 to the nearest ten. Using the usual positive halfway-up school convention, the original whole-number count could be any integer from 335 to 344 inclusive.
We can recover a set of possibilities, not one exact original. Rounding deliberately replaces several nearby values with the same reported value. No opposite operation can identify which of those inputs was used without more information.
This is different from subtracting a known fixed amount. If 17 was added to produce exactly 59, the start is uniquely 42. If a value was rounded to 340, many starts remain possible.
The same caution applies to words such as “about”, “roughly” and “approximately”. Do not quietly convert an approximate final statement into an exact reverse calculation. State the possible range or explain what further information is required.
The lesson is not that working backwards is unreliable. It is that its conclusion must match the information preserved by the forward process.
16. Other situations where one reverse answer is not justified
If a number is multiplied by zero and the result is zero, the original number is not determined. Seven, twenty and many other numbers all produce zero. Dividing zero by zero is not a valid way to recover the start.
If the final statement is “There are more than fifty counters”, it is a condition rather than one exact amount. A backwards calculation must preserve that uncertainty. It cannot begin by pretending the final number is exactly fifty.
If the story says “some counters were removed” but does not identify how many, that unknown action also prevents a direct unique reversal unless another relationship supplies the missing information.
These examples establish an important boundary: working backwards needs an adequate final state and adequate information about every step being undone. A child should not be rewarded for producing a neat number when the question does not support one.
Conversely, do not stop at “not enough information” when a given total, difference, fraction or remainder actually supplies the missing relationship. Read the complete problem before deciding whether reversal is possible.
17. Compare working backwards with drawing a model
Working backwards is especially clear when a single quantity passes through a stated sequence of actions. A bar model can be clearer when several quantities are compared at the same moment.
For “A number is tripled and 14 is added to give 110,” backwards arithmetic gives 110 − 14 = 96, then 96 ÷ 3 = 32. The sequence is explicit.
For “A has three times as many counters as B, and together they have 128,” a unit model is more direct. B is one unit and A is three, so four units make 128 and one unit is 32. There is no historical event in which someone tripled a pile and then added another pile; the model describes a simultaneous comparison.
Both examples produce 32, but they reach it through different relationships. A shared answer is not evidence that the questions are the same.
Choose the representation that exposes the unknown most clearly. The objective is a justified solution, not forcing every problem into the newest technique learnt.
18. What a complete verification looks like
A complete forward check begins with the proposed original number and performs the exact original actions in the exact original order. It does not repeat the reverse calculations and declare that they agree with themselves.
For the three-action example, the recovered 64 is checked by subtracting eighteen, multiplying by three and adding twelve. The final 150 matches the stated output. Every intermediate amount remains meaningful.
Also check context. If the unknown counts indivisible objects, a fractional answer may signal incompatible conditions or a mistake. If the original event removes a positive amount, the quantity immediately before that removal must be at least as large as the amount removed.
Estimation can support this check but cannot replace it. Knowing that an answer should be “around sixty” does not distinguish 61 from 64 when only one satisfies the exact sequence.
A useful final sentence identifies the recovered moment: “There were 64 counters before any were removed.” That is more precise than “There were 64 counters,” which may leave the time reference unclear.
19. Practice laboratory: record the first step to undo
Before solving each question, write the final action in the original story. Then state how you will undo it. Use separate lines for the recovered intermediate quantities. Questions 12–13 and 17–20 are deeper reasoning checks rather than a demand to rush ahead.
- A number increases by 17 and becomes 59. Find the starting number.
- A collection loses 26 counters and has 81 left. Find its original size.
- A number is multiplied by four to give 156. Find the number.
- A collection is shared equally among six boxes, with 27 items in each and none left over. Find the original total.
- A number is tripled and then 14 is added, giving 110. Find the number.
- A number has 14 added to it and the result is doubled, giving 110. Find the number.
- Fifteen is subtracted from a number, the result is multiplied by four, and then 24 is added. The final result is 168. Find the starting number.
- A number is multiplied by four, then eighteen is subtracted. The result is divided exactly by three, giving 22. Find the number.
- A child receives $4.20, spends $5.65 and ends with $17.80. Find the starting amount.
- A rope has 2.70 m cut off. The remainder makes four equal pieces of 3.45 m each, with no waste. Find the original length.
- After 47 notebooks are sold, the remaining notebooks fill eight boxes with 23 in each and none loose. Find the stock before the sale.
- One quarter of a set is removed and 36 counters remain. Find the original set.
- Two fifths of a set are removed and 42 remain. Find the original set.
- A whole number divided by seven gives quotient 24 and remainder five. Find the number.
- An activity finishes at 11:10 a.m. after 45 minutes. When did it begin?
- Noor gives Kai 28 cards. Afterwards they have 77 and 95 respectively. Find their initial amounts.
- A whole-number division by seven gives quotient 24, but no remainder is supplied. List all possible original numbers when a remainder from zero to six is allowed.
- A whole-number count rounds to 340 to the nearest ten under the usual positive halfway-up convention. State the smallest and largest possible original counts.
- A number is multiplied by zero and the result is zero. Can the original number be found uniquely? Explain.
- A student claims that the starting number in Question 5 is 41. Check the claim by replaying the stated operations.
When marking, inspect the first reverse step as well as the final answer. Two students can reach the same wrong answer for different reasons: one may reverse the wrong event, while another may choose the right event and calculate it inaccurately.
20. Explained answers and independent forward checks
1. Undo the addition: 59 − 17 = 42. Forward check: 42 + 17 = 59. The start is smaller than the finish because the event increased it.
2. Restore the removed counters: 81 + 26 = 107. Forward check: 107 − 26 = 81.
3. Undo multiplication by four: 156 ÷ 4 = 39. Check 39 × 4 = 156.
4. Undo the sharing by reconstructing all six groups: 27 × 6 = 162 items. Check that 162 ÷ 6 = 27 with no remainder.
5. The last action is adding fourteen. Undo it first: 110 − 14 = 96. Then undo tripling: 96 ÷ 3 = 32. Check 32 × 3 = 96, then 96 + 14 = 110.
6. The last action is doubling. Undo it first: 110 ÷ 2 = 55. Then undo adding fourteen: 55 − 14 = 41. Check 41 + 14 = 55, then 55 × 2 = 110.
7. Reverse in order: 168 − 24 = 144; 144 ÷ 4 = 36; 36 + 15 = 51. Forward: 51 − 15 = 36; 36 × 4 = 144; 144 + 24 = 168.
8. Undo division first: 22 × 3 = 66. Restore eighteen: 66 + 18 = 84. Undo multiplication: 84 ÷ 4 = 21. Forward: 21 × 4 = 84; 84 − 18 = 66; 66 ÷ 3 = 22.
9. Restore the spending: $17.80 + $5.65 = $23.45. Remove the earlier receipt: $23.45 − $4.20 = $19.25. Check $19.25 + $4.20 − $5.65 = $17.80.
10. The remaining length is 3.45 × 4 = 13.80 m. Restore 2.70 m to obtain 16.50 m. The four pieces and the removed length together account for the entire original rope.
11. After the sale there are 8 × 23 = 184 notebooks. Before the sale there were 184 + 47 = 231 notebooks. Check that selling 47 leaves 184, exactly enough for eight boxes.
12. The remainder is three quarters. Three units equal 36, so one unit is twelve and four units give 48 counters. Removing one quarter, twelve, leaves 36.
13. The remainder is three fifths. Three units equal 42, so one is fourteen and the whole five units total 70. Two fifths is 28; 70 − 28 = 42.
14. Reconstruct complete groups and add the remainder: 7 × 24 + 5 = 173. Since five is smaller than seven, it is a valid remainder.
15. Move back ten minutes to 11:00 a.m., then 35 more minutes to 10:25 a.m. The forward duration is 35 + 10 = 45 minutes.
16. Noor initially had 77 + 28 = 105 cards. Kai initially had 95 − 28 = 67 cards. The combined total is 172 before and after.
17. The possibilities are 168, 169, 170, 171, 172, 173 and 174. Each consists of 24 groups of seven plus a possible remainder from zero to six.
18. The smallest is 335 and the largest is 344. At 335 the stated halfway convention rounds upward to 340; 345 would round to 350 instead.
19. No unique original number is determined. Several different inputs, such as seven and twenty, become zero when multiplied by zero. That operation has removed the information needed for unique reversal.
20. Starting at 41, tripling gives 123 and adding fourteen gives 137, not 110. The claim is false. Forty-one is the answer to Question 6, whose actions occur in a different order.
21. Teach the reverse route without turning it into another trick
Begin with one reversible event and ask the learner to describe why the inverse restores the earlier amount. Then contrast two sequences containing the same operations in different orders. That comparison forces attention to the timeline rather than to a memorised list of signs.
For a learner who is stuck, provide the final state and ask only, “What happened last?” Once that action is named, ask, “What would undo it?” Resist announcing the full sequence. The learner needs practice choosing the next step, not only carrying it out.
Use a small table with a column for the number and a column for its meaning. For example, 138 means “after tripling”; 46 means “before tripling”; 64 means “before removing eighteen”. Remove the table only when the learner can supply these meanings independently.
A suggested practice cycle is one worked example, one close example, one changed-order contrast and one contextual question. Later, revisit a question without the earlier working in view. This is an instructional routine to test and adjust, not a promise that a fixed number of exercises will produce mastery.
Useful correction language
“I reversed the first action first” identifies an order error. “I used the amount in one box as the total” identifies a quantity error. “I forgot the remainder” identifies missing information. “My subtraction was wrong after choosing the correct step” identifies a calculation error. Each statement leads to a different repair.
Ask for a forward check at the end of the correction. A copied reverse solution without replaying the original process does not demonstrate that the learner has understood the relationship.
22. From recovering a quantity to choosing a representation
Working backwards is one route in a larger toolkit. It is strongest when the sequence of transformations is known. A bar model is often clearer when a total or difference links several unknown quantities. Two independent solution routes can then be compared to test both the arithmetic and the interpretation.
Continue to Part-Whole, Difference and Equal-Group Models. For checking and comparing methods, use Multiple Solution Routes.
Final checkpoint: can you identify the last action, reverse it correctly, keep the recovered states in order and replay the original story? Can you also recognise when the given information does not support a unique reverse answer? Both abilities belong to reliable backwards reasoning.
Source and editorial note
The curriculum reference is the MOE Primary Mathematics Syllabus, updated October 2025. The named problem-solving route, examples, boundary tests and practice sequence are independently written. They do not imply MOE endorsement, identical school pacing or a guaranteed assessment outcome.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Recover the missing quantity, preserve the original conditions, test a changed case and verify the forward return.
