PRIMARY 4 MATHEMATICS LEARNING GUIDE · BATCH 5 · GUIDE 17
A number can be correct and still belong to the wrong moment. A shop has one quantity before a delivery, another after the delivery and a third after customers buy some stock. A student who adds every number in the story may carry out accurate arithmetic while describing a quantity that never existed.
Before-and-after problems ask us to keep the story in order. What was present at the beginning? What happened to it? What was present at the end? Which of these quantities is missing? Once those questions are answered, the calculation often becomes much simpler.
This extended guide teaches that distinction through whole numbers, money, lengths, equal groups and carefully bounded transfer problems. It includes complete worked solutions, incorrect routes to inspect, a twenty-question practice laboratory and explanations that return each answer to its original situation. The examples are invented teaching scenarios, not reports about actual pupils, prices or school inventories.
Return route: Primary 4 Mathematics Learning Hub. For written calculation, revisit Whole Numbers and the Four Operations.
Learning boundary: the arithmetic foundations draw on the Primary 4 sections of the MOE Primary Mathematics Syllabus. Before-and-after reasoning is the organising approach of this independent lesson, not a claim that every school teaches a separate chapter under this name. Sections marked deeper challenge are optional; follow your teacher’s sequence.
Navigate: the three quantities · unknown starts · several events · transfers · error diagnosis · practice · explained answers · teaching and return.
1. Separate an amount from an action
“There were 364 books” describes an amount at a particular moment. “128 books arrived” describes an action that changes that amount. “There are 492 books now” describes the resulting amount. These numbers have different jobs even though all are measured in books.
For an increase, the relationship is start + amount added = finish. For a decrease, it is start − amount removed = finish. These statements describe the story. They do not yet tell us which direction to calculate, because the unknown might be the start, the change or the finish.
Use ordinary words before abbreviations. Say, “This is the number of books before the delivery,” rather than merely circling 364. Say, “This is the delivery, not the new total,” beside 128. A useful annotation protects meaning; a page full of circles without labels may not.
With an unfamiliar problem, first retell the event without numbers. For example: “There were some books. More arrived. The question asks how many there are now.” Only then restore the numbers. This reveals whether the structure was understood or whether the learner was guessing an operation from the digits.
A complete first example
A cupboard contains 364 books. A delivery adds 128 books. No books are removed. The final quantity is 364 + 128 = 492 books. To check, remove the delivery from the final quantity: 492 − 128 = 364. We have returned to the stated starting point.
2. A timeline keeps the moments apart
Draw three positions: before the event, the action and after the event. The arrow records what happens in the story, not necessarily the direction in which you will solve it.
Before Action After 364 books add 128 books 492 books
Now replace the starting value with a question mark. The arrow still says “add 128 books”. The story has not changed into a removal. Instead, the calculation must reconstruct an earlier amount from a later amount.
This distinction matters. Some children change the arrow to subtraction simply because they are solving backwards, and then become confused about what actually happened. Keep the forward event on the timeline. Write the reverse calculation on a separate line.
A timeline is particularly useful when a question uses “at first”, “then”, “after that”, “eventually” or “left”. It also helps when the sentences are not written in time order. “There were 492 books after a delivery of 128” still describes an earlier starting amount, even though the final state is mentioned first.
The timeline need not be a beautiful drawing. Its job is to show which amount belongs to which moment. A few words and arrows are enough when they make that relationship unambiguous.
3. Find a final amount after something leaves
A storeroom has 725 notebooks. Teachers take 268 notebooks for lessons. Nothing else enters or leaves. The final amount is 725 − 268 = 457 notebooks.
The subtraction is justified because the 268 notebooks were part of the starting 725 and are no longer in the storeroom. The two numbers are not separate piles that should be combined.
Check both direction and size. Removing a positive number of notebooks must leave fewer than 725. An answer of 993 would fail this check even before we inspect the column subtraction. The inverse check is 457 + 268 = 725.
Keep “taken” and “remaining” distinct. If the question asks how many notebooks were taken, 268 is already given. If it asks how many remain, the answer is 457. If it asks for the original quantity, that is 725. The same story supports several questions, but only one is being asked at a time.
A useful exercise is to cover the final question and ask the learner to invent three possible questions about the story. Then uncover the actual question and identify which quantity it requests. This tests target selection without adding harder arithmetic.
4. Find an unknown start after an increase
After receiving 149 stickers, Hana has 580 stickers. How many did she have at first?
The story relationship is starting stickers + 149 = 580. The starting amount is therefore 580 − 149 = 431 stickers.
The word “receiving” describes an increase, but the solution uses subtraction. This is why “receives means add” is not a reliable complete method. Addition explains what happened; subtraction reconstructs what existed before it happened.
Check forwards, using the actual event: 431 + 149 = 580. Also ask whether the start should be smaller or larger than the finish. Since stickers were received, the start must be smaller. Both the calculation and the meaning agree.
Inspect an incorrect route
A student writes 580 + 149 = 729. Ask what 729 would represent. It would be the amount after another delivery of 149 to the final 580, not the amount before the original delivery. The arithmetic is accurate, but it has created an extra event.
Naming the invented event is more useful than simply replacing a plus sign with a minus sign. It shows the learner why the route does not fit the question.
5. Find an unknown start after a decrease
A teacher gives away 176 counters and has 439 counters left. How many counters were there at first?
The initial collection contains both the counters that remain and the counters that were given away. Recombine these parts: 439 + 176 = 615 counters.
The story is a subtraction story, but recovering the starting collection requires addition. Check forwards: 615 − 176 = 439. Since counters were given away, the starting amount should exceed the remaining amount.
A bar can make this visible. Draw one long bar for the initial collection, divided into a section labelled 176 given away and another section labelled 439 left. The whole bar represents 615. Notice that the two parts describe destinations of the original counters; they are not additional counters created later.
Compare this with the preceding section. After an increase, remove the increase to find the start. After a decrease, restore the decrease to find the start. The reliable question is “What would take this final amount back to the original amount?” rather than “Which word appeared in the story?”
6. Find the change when both states are known
A shelf holds 328 books before a delivery and 507 books afterwards. No books are removed. The delivery contains 507 − 328 = 179 books.
Now consider a shelf with 640 books before borrowing and 458 books afterwards. No books arrive. The borrowed amount is 640 − 458 = 182 books.
Both calculations find a difference between states. Their interpretations differ: 179 arrived in the first problem; 182 left in the second. Write the action in the answer sentence.
Be careful when more than one event occurred. If stock starts at 328 and finishes at 507 after both deliveries and sales, the difference 179 is only the net increase. It is not necessarily the delivery amount. For example, a delivery of 200 followed by sales of 21 would produce the same net increase.
Therefore the sentence “No books are removed” is not decorative. It permits the total change to be identified with the delivery. Without that condition or equivalent information, the two states alone may not determine every action.
7. Keep an intermediate state after every event
A library begins with 1,260 books available. Pupils return 185 books. Later, 247 books are borrowed. How many books are available at the end?
| Moment | Calculation | Books available |
|---|---|---|
| At the start | Given | 1,260 |
| After returns | 1,260 + 185 | 1,445 |
| After borrowing | 1,445 − 247 | 1,198 |
The answer is 1,198 books available. The intermediate 1,445 matters because the borrowing acts on the quantity after the returns, not directly on an unrelated number.
A second route uses the overall change. Borrowing exceeds returns by 247 − 185 = 62, so the library ends with 62 fewer available books than it began with: 1,260 − 62 = 1,198. This works because both changes are known fixed amounts of the same quantity.
Keep the staged table as the first method when the learner is not yet secure. Compressing the route is useful only when the compressed calculation still preserves the story. Fast working that conceals an uncertain intermediate state is not an improvement.
8. Read a story that is presented out of order
At closing time, a shop has 391 exercise books. During the day, it sold 148 books after receiving eight boxes containing 27 books each. How many books were there before the delivery?
The first sentence gives the final amount. The later sentence supplies two earlier events. Put them in order: opening stock → delivery → sales → closing stock.
Delivery: 8 × 27 = 216 books. Before the sales, the shop had 391 + 148 = 539 books. Before the delivery, it had 539 − 216 = 323 books.
Check the complete forward journey: 323 + 216 = 539; 539 − 148 = 391. Every stated event has been used once.
There are two kinds of calculation here. Multiplication converts boxes into books. The later addition and subtraction reconstruct stock levels. Do not add eight directly to 391, because eight counts boxes while 391 counts books. The shared unit must be established before the amounts can interact.
This example shows why chronological order and unit conversion can matter before the main arithmetic begins.
9. Find a missing action inside a sequence
A shop begins with 416 bottles. A delivery brings 89 more. After selling some bottles, 367 remain. How many bottles were sold?
First reconstruct the stock immediately before sales: 416 + 89 = 505 bottles. Sales then reduce 505 to 367, so the number sold is 505 − 367 = 138 bottles.
The missing action sits between two states, but one of those states must first be calculated. This is a useful general pattern: find the state just before the missing action and the state just after it; then connect those states.
The tempting subtraction 416 − 367 = 49 finds the overall decrease across the day. It ignores the delivery and therefore does not give sales. Sales can exceed the net decrease because new bottles arrived during the same period.
Check by replaying the day: 416 + 89 − 138 = 367. A precise answer sentence is “The shop sold 138 bottles,” not merely “138 remained.” The meaning of the output is part of the solution.
10. Money and measurement follow the same event structure
In an invented shopping example, Mira has $18.50. She spends $6.75 and later receives $4.20. After spending, she has $11.75. After receiving the additional amount, she has $15.95.
The decimal calculation must preserve place value, but the before-and-after structure is unchanged: start → spend → receive → finish. Align the decimal points because dollars, tenths of dollars and hundredths of dollars must remain in their correct positions.
For a length example, a roll contains 9.60 m of ribbon and 2.35 m is removed. It now contains 7.25 m. Adding 2.35 m back gives 9.60 m, which checks the calculation.
Do not mix units silently. A question using metres and centimetres requires a consistent representation before combining values. When the lesson is about the timeline rather than unit conversion, rewrite the given amounts into one familiar unit first. Follow the conversions taught in the current school sequence rather than treating every decimal conversion as assumed knowledge.
The central test remains: can the learner identify the quantity after each event and keep its unit attached?
11. A transfer has two effects, not one
Deeper challenge. Aisha has 92 counters and Ben has 56. Aisha gives Ben 18 counters. Find both final amounts and their combined total.
Aisha loses 18: 92 − 18 = 74. Ben gains 18: 56 + 18 = 74. Together they still have 74 + 74 = 148 counters, equal to the starting total 92 + 56.
No counter enters or leaves the pair. The transfer changes ownership, not the total number of counters owned by the two children together. Recording only Aisha’s loss would make the combined total appear to fall incorrectly.
| Person | Before | Change | After |
|---|---|---|---|
| Aisha | 92 | Gives 18 | 74 |
| Ben | 56 | Receives 18 | 74 |
| Combined | 148 | No external change | 148 |
Draw separate rows for separate owners. A transfer arrow between them then makes both effects visible. This table is not required for every simple question, but it is valuable when a learner repeatedly changes only one side.
12. Why equalising a difference can require half the difference
Deeper challenge. Before the transfer, Aisha has 92 and Ben has 56, a difference of 36. Giving one counter from Aisha to Ben reduces Aisha’s lead by two: Aisha has one fewer and Ben has one more.
To remove a difference of 36 in this exact transfer situation, transfer 36 ÷ 2 = 18 counters. The preceding calculation confirms that both finish with 74.
Do not turn this into a universal instruction to halve every difference. If Aisha throws away counters rather than gives them to Ben, Ben’s amount does not rise. Aisha would need to remove all 36 excess counters to match Ben’s unchanged 56.
Likewise, if an adult gives extra counters only to Ben, Ben needs 36 more to catch up. The same starting difference therefore requires different actions depending on where the counters go.
The rule is not “equal means divide by two”. The reasoning is that an internal transfer changes both sides of the comparison in opposite directions. Test the actual action before using the shortcut.
13. Equal additions preserve a difference, not a total
Suppose two children have 67 and 43 cards. Their difference is 24. Each receives eight cards from an adult. They now have 75 and 51, and their difference is still 24.
The combined total changes from 110 to 126 because sixteen new cards entered the pair. Equal additions preserve the difference but do not preserve the total. This contrasts with an internal transfer, which preserves the combined total but usually changes the difference.
A before-and-after table keeps these cases apart. Ask what entered the collection, what left it and what simply moved within it. The boundary of the collection matters. “Both children together” is a different collection from “one child alone”.
This reasoning is useful even without formal algebra. Try the same actions with small counters: start with 7 and 3, add two to each, then reset and transfer two from the larger pile to the smaller. The visible outcomes are different because the actions are different, not because the arithmetic became harder.
14. Fractions must refer to a particular state
A box contains 72 counters. One quarter of them are removed. Then eleven counters are added. How many are in the box now?
The fraction refers to the initial 72 counters. One quarter is 72 ÷ 4 = 18. After removal, 72 − 18 = 54 remain. After the addition, 54 + 11 = 65 counters.
The intermediate 54 is a new state. A later instruction referring to “the remaining counters” would use 54 as its whole, not 72. Keep the exact reference quantity beside each fraction.
For a contrast, imagine that the eleven counters arrive first and then one quarter of the new total is removed. The new total would be 83. A quarter of 83 cannot be removed as a whole number of individual counters. That changed problem needs a different condition or a divisible quantity; it is not the same problem in a different sentence order.
Use this contrast to understand why event order can matter. A fixed addition and a fractional removal do not generally interchange without changing the outcome.
15. Time is another before-and-after quantity
A lesson begins at 9:45 a.m. and lasts 35 minutes. Add fifteen minutes to reach 10:00 a.m., then another twenty minutes. The finish is 10:20 a.m.
If the finish and duration are given instead, travel backwards along the timeline. From 10:20 a.m., subtract twenty minutes to reach 10:00 a.m., then fifteen more to reach 9:45 a.m.
Do not treat clock notation as an ordinary decimal. Forty-five minutes is not 0.45 of an hour. The timeline uses sixty minutes in one hour. Splitting the journey at a whole hour is often clearer than trying to imitate decimal column addition.
This example revisits an earlier measurement foundation. It belongs here because the before, change and after structure is the same, even though the way the quantity is written differs.
A useful check is to replay the duration forwards. Arriving at the given finish after the stated duration confirms both the arithmetic and the interpretation of the clock.
16. Some before-and-after stories do not determine one answer
“A tank contains 40 litres after some water was used. How much was there at first?” does not give enough information. It could have started at 45 litres with 5 used, or 60 litres with 20 used. Both stories fit the final amount.
The missing information could be the amount used, a fraction used, or another relation that determines the earlier state. Do not insert a convenient number simply to create a calculation.
Inconsistency is different from missing information. If a question says two children share all 83 indivisible counters equally, with no counter left and no splitting, those conditions cannot all be satisfied. An equal whole-number share would require an even total. The correct response is to identify the conflict rather than report 41.5 counters as though splitting had been allowed.
A responsible solver distinguishes three outcomes: a unique answer, several possible answers or incompatible conditions. The presence of numbers in a question does not guarantee that one exact answer can be found.
17. Diagnose the first point where the story was lost
| Visible error | What to inspect | Repair question |
|---|---|---|
| Adds a receipt to a known final amount | The unknown start was treated as a finish | Would this calculation undo the receipt or create another one? |
| Uses the difference between opening and closing stock as sales | A delivery was omitted | What was the stock just before sales? |
| Subtracts a transfer from only one child | The receiving side disappeared | Who now owns those counters? |
| Combines boxes with individual items | The quantities use different units | How many items does each box contain? |
| Uses an original total for a fraction of the remainder | The reference state was not updated | Fraction of which amount, at which moment? |
| Gives a plausible number without checking | Only the arithmetic, not the conditions, was tested | Can the answer replay every stated event? |
One wrong answer can contain several later errors caused by the first one. Repair the earliest mismatch between story and representation. There is little value in correcting three later calculations while leaving the initial timeline wrong.
A useful correction has three parts: identify the first wrong state, state the correct relationship and solve a changed question without copying the earlier working. Keep the correction specific enough that the learner knows what to do differently next time.
18. Practice laboratory: identify the state before calculating
Use a short timeline or table when it helps. For each question, label the requested quantity as start, change, intermediate amount or finish. Questions 13–16 and 20 ask for deeper reasoning; use them after the direct examples are secure.
- A cupboard has 364 books. Another 128 arrive. None leave. How many books are there now?
- A storeroom has 725 notebooks. Teachers take 268. How many remain?
- After receiving 149 stickers, Hana has 580. How many did she have before receiving them?
- After giving away 176 counters, a teacher has 439. How many were there initially?
- A shelf has 328 books before a delivery and 507 afterwards. None are removed. How many arrived?
- A box has 640 cards before some are removed and 458 afterwards. None are added. How many were removed?
- A shop begins with 480 items, sells 137 and receives 92. Find its closing stock.
- A shop receives 128 items and then sells 75. Its closing stock is 416. Find its opening stock.
- Seven trays each contain 35 counters. Another 28 counters are added to the collection. Find the final total.
- A shop starts with 416 bottles, receives 89 and sells some. It finishes with 367. How many bottles were sold?
- Mira has $18.50, spends $6.75 and receives $4.20. Find her final amount.
- A 9.60 m ribbon has 2.35 m removed. Find the remaining length.
- A has 84 cards and B has 52. A gives ten cards to B. Find both final amounts and the combined total.
- Starting again with 84 and 52, how many cards must A give B so that both have the same number?
- A has 67 counters and B has 43. Both receive eight counters from an adult. Find the final difference and combined total.
- A box has 90 counters. Two fifths are removed, then twelve are added. Find the final quantity.
- There are 31 pupils in a room. Six leave and four enter. How many are there afterwards?
- A tank has 40 L after some water was used. Can its initial amount be found uniquely? Explain.
- A shop starts with 180 pencils, receives four packets of 25 and sells 63. A pupil writes 180 + 4 − 63. Explain the error and find the correct closing stock.
- Two children are told to share all 83 indivisible counters equally, with none left over. Can all these conditions be met?
Do not read the answer section after every line. Complete a manageable group first, then compare the structures as well as the answers. A correct number reached by an unexplained guess is a reason to ask for a second example, not a reason to assume the concept is secure.
19. Explained answers and forward checks
1. The unknown is the final quantity after an increase. 364 + 128 = 492 books. Removing the arrival gives 492 − 128 = 364, the original amount.
2. The unknown is the final quantity after a decrease. 725 − 268 = 457 notebooks. The used and remaining parts recombine: 268 + 457 = 725.
3. The unknown is the start. Undo the receipt: 580 − 149 = 431 stickers. Check the event in its original direction: 431 + 149 = 580.
4. Restore what was given away: 439 + 176 = 615 counters. The original collection includes both the 176 given away and the 439 left.
5. Find the increase between the two states: 507 − 328 = 179 books. This equals the delivery because no books were removed.
6. Find the decrease: 640 − 458 = 182 cards. Check 458 + 182 = 640. The answer names removed cards, not remaining cards.
7. After sales, 480 − 137 = 343 remain. After the delivery, 343 + 92 = 435 items. The net change is a decrease of 45, also giving 480 − 45 = 435.
8. Before sales, 416 + 75 = 491 items were present. Before the delivery, 491 − 128 = 363 items. Check 363 + 128 − 75 = 416.
9. The seven trays initially contain 7 × 35 = 245 counters. Adding 28 gives 273 counters. Seven is a tray count and must first be converted into a counter total.
10. Before sales, 416 + 89 = 505 bottles were available. Sales are 505 − 367 = 138 bottles. The overall decrease of 49 would not account for the delivery.
11. After spending, $18.50 − $6.75 = $11.75. After receiving, $11.75 + $4.20 = $15.95. Reversing those events gives $15.95 − $4.20 + $6.75 = $18.50.
12. 9.60 − 2.35 = 7.25 m. The removed length and remaining length add to the original roll length.
13. A finishes with 84 − 10 = 74 cards. B finishes with 52 + 10 = 62 cards. Their total is 136, unchanged from 84 + 52.
14. The initial difference is 84 − 52 = 32. Every transferred card reduces that difference by two. Transfer 32 ÷ 2 = 16 cards. Both then have 68, confirming equality.
15. The final amounts are 75 and 51. The difference is 24 counters and the total is 126 counters. Equal additions preserved the difference while increasing the total by sixteen.
16. One fifth of 90 is 18, so two fifths is 36. After removal, 54 remain; after adding twelve, there are 66 counters. The fraction referred to the initial 90.
17. 31 − 6 = 25 after the departure. Then 25 + 4 = 29 pupils. The answer is two fewer than the start because departures exceed arrivals by two.
18. No unique initial amount can be determined. An initial 45 L with 5 L used and an initial 60 L with 20 L used both produce 40 L. The amount used or another determining condition is needed.
19. Four counts packets, not pencils. The delivery contains 4 × 25 = 100 pencils. Closing stock is 180 + 100 − 63 = 217 pencils.
20. No. Eighty-three is odd, so equal whole-number shares would leave one counter over. Reporting 41.5 counters violates the condition that counters cannot be divided.
20. A lesson routine that leaves the decisions with the learner
Begin with one direct final-amount problem. Ask the student to retell the event, identify the unknown and solve it. Then keep the same event but give the final amount instead of the start. The contrast reveals whether the child is thinking about states or merely associating event words with operations.
When the student hesitates, reduce the numbers while keeping the relationship unchanged. “Some counters plus three gives eight” is a clearer starting point than a five-digit calculation if the uncertainty concerns the unknown start. Return to larger numbers after the relationship is explained.
Next introduce one intermediate state. Do not immediately add several transfers, fractions and hidden quantities. Ask what existed immediately before the final action. The learner should supply the state rather than wait for an adult to announce the operation.
Finish with a changed version and a forward check. A suggested later return is to revisit one unknown-start question without the earlier example beside it. This is a practice design, not a guarantee of mastery. Judge the result from the learner’s independent explanation and working.
What a parent can ask without giving the method away
Ask, “Which moment does that number describe?” “What happened between these two amounts?” “Did anything enter or leave the pair?” “What would your answer produce if we replayed the story?” These questions expose the structure while leaving the actual decision to the child.
A useful written correction might say: “I used closing stock as opening stock. Next time I will label the moments before calculating.” That is more actionable than “I must be more careful.”
21. The handover to working backwards and model choice
Before-and-after reasoning establishes what each state means. Working backwards adds a method for recovering earlier states systematically. Model selection helps when several quantities or comparisons interact. Strategy comparison tests whether another route can check the result without losing the same information.
Continue to Working Backwards: Inverse Operations and Missing Quantities. For choosing a representation, use Part-Whole, Difference and Equal-Group Models. For checking through another route, use Multiple Solution Routes.
Final checkpoint: can the learner name the start, the action and the finish, identify the unknown, preserve every intermediate state and replay the result forwards? That complete explanation is the goal of this lesson.
Source and editorial note
The official curriculum reference is the Ministry of Education Primary Mathematics Syllabus, updated October 2025, especially the Primary 4 content on printed pages 37–40. The scenarios, worked examples, diagnostic questions and lesson order are original independent teaching material. No school examination schedule, official endorsement or guaranteed result is implied.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Identify the quantity, preserve the relationship, test a changed case and return the answer to the situation.
