PRIMARY 5 MATHEMATICS LEARNING GUIDE · BATCH 6 · GUIDE 22
Many difficult word problems are not really about complicated arithmetic. They are about changing states. A quantity starts somewhere, changes one or more times, and ends somewhere else. Before–after models make those states explicit. Working backwards then becomes the controlled reversal of those changes.
This guide develops change problems involving addition, subtraction, multiplication, division, fractions, percentages, money and rate. The focus is on state tracking: what existed before, what changed, what remains invariant and which operation reverses each step.
For the official curriculum framework, see the MOE Primary Mathematics Syllabus.
Series route: return to the Primary 5 Mathematics Learning Hub. Earlier: Comparison Bar Models, Difference & Multiplicative Comparison. Continue to Assumption Method, Fixed Totals, Equalisation & Difference Reasoning and Systematic Listing, Case Organisation, Tables & Invariant Thinking.
1. A before–after model separates states
Suppose a shop begins with some pens, sells 120, receives 80, and ends with 350.
The timeline is:
Start → −120 → +80 → 350.
Working backward:
350 − 80 = 270, then 270 + 120 = 390 pens at the start.
2. Reverse steps in reverse order
If a number is multiplied by 4, then 30 is added to give 270, undo +30 first, then undo ×4:
270 − 30 = 240; 240 ÷ 4 = 60.
Reversing the operations in the original order would not reconstruct the state correctly.
3. Addition and subtraction are inverse changes
If 45 is added to a number, subtract 45 to return to the earlier state. If 45 is removed, add 45 to reverse it.
The inverse relationship is exact when no other condition changes the meaning.
4. Multiplication and division are inverse changes
If a quantity is tripled, divide by 3 to recover the previous amount. If it is divided equally among 5 groups and one group amount is known, multiply by 5 to recover the total.
Working backward is inverse thinking across a sequence.
5. Unknown change problems
A tank contains 480 ℓ and later contains 325 ℓ. The change is 325 − 480 = −155 ℓ, so 155 ℓ was removed.
Do not only calculate the magnitude. Interpret direction.
6. Unknown start problems
After 75 books are added, a shelf contains 260 books. Before the addition:
260 − 75 = 185 books.
The final state and change determine the initial state.
7. Unknown final problems
A shelf begins with 185 books and receives 75. Final = 185 + 75 = 260.
The same relationship supports all three unknown positions: start, change or final.
8. Fraction change problems
A tank begins with 360 ℓ. One third is drained.
Drained = 120. Final = 240 ℓ.
Reverse version: if 240 remains after one third is drained, the remaining 2/3 equals 240, so the original is 360 ℓ.
9. Percentage change problems
A price is reduced by 20% and becomes $72. The final state is 80% of the original.
Original = 72 ÷ 0.8 = $90.
Working backward must preserve the reference whole.
10. Two-stage percentage problems
A price is reduced by 10%, then by another 20%, ending at $144.
Backward: before the second discount, 80% = 144, so state = 180. Before the first discount, 90% = 180, so original = $200.
The reference whole changes at each stage.
11. Before–after with equal transfer
Box A has 80 counters and Box B has 40. Some counters move from A to B until both are equal.
Total remains 120. Equal final amounts are 60 each. Therefore 20 counters move.
The invariant total is more useful than tracking every possible transfer.
12. Difference changes under transfer
If x counters move from the larger group to the smaller, the difference shrinks by 2x: one side loses x while the other gains x.
Starting difference 40 and ending difference 0 means 2x = 40, so x = 20.
13. Same addition to both groups preserves difference
If A exceeds B by 35 and both receive 12, the difference remains 35.
This invariant can simplify many before–after comparison problems.
14. Unequal changes alter the difference predictably
If A exceeds B by 35, then A gains 10 while B gains 4. New difference = 35 + 10 − 4 = 41.
Track how each change affects the gap.
15. Before–after model with money
Sara spends 25% of her money, then $18, and has $72 left.
Backward: before spending $18 she had $90. That was 75% of the earlier amount. Original = 90 ÷ 0.75 = $120.
16. Before–after model with rate
A printer works at 30 pages/min for some time, then 120 pages are discarded. Final usable pages = 480.
Before discard = 600 pages. Time = 600 ÷ 30 = 20 min.
The final state reveals the earlier total, which reveals duration.
17. Before–after with repeated changes
A quantity doubles, decreases by 15, then is halved, ending at 42.
Backward: before halving = 84. Before subtracting 15 = 99. Before doubling = 49.5.
If the context requires a whole number, this result may signal that the stated conditions are inconsistent with that constraint.
18. Working backward can expose impossible conditions
If a whole-number count would have to be 49.5, the process cannot have started from a whole-number count under the stated exact operations.
Reverse reasoning is not only a solving method; it is a consistency test.
19. Use a state table for complex changes
| Stage | Quantity |
|---|---|
| Start | ? |
| After 20% removed | ? |
| After +50 | 370 |
Backward: before +50 = 320. That is 80% of start. Start = 320 ÷ 0.8 = 400.
Tables are useful when many states risk being confused.
20. Mark the state before applying a fraction
“One third of the remainder” requires the remainder state to exist first.
Write the remainder explicitly before applying the next fraction. This prevents the learner from applying all fractions to the original quantity.
21. Work forward to verify
If backward reasoning gives a starting value of 120, replay the original sequence. If the final state returns exactly, the solution is verified.
This return path is stronger than repeating the backward arithmetic.
22. Change can be represented on a number line
A number line is useful when direction matters. Moving from 35 to 52 is +17. Moving from 52 to 35 is −17.
Distance and directed change should remain distinct.
23. Before–after geometry
A tank’s water depth rises from 12 cm to 18 cm. Change = 6 cm.
If base area is 500 cm², added volume = 500 × 6 = 3000 cm³.
The state change in one dimension generates a volume change.
24. Error map
| Visible error | Likely cause | Repair question |
|---|---|---|
| Reverses operations in original order | Sequence not inverted | What happened last? |
| Adds 20% of final price to reverse discount | Reference whole wrong | What percentage of original is the final state? |
| Transfer problem tracks guesses | Total invariant missed | What quantity stays constant? |
| Second fraction applied to starting amount | Intermediate state lost | What remains immediately before this step? |
| Backward answer not checked forward | No return path | Does replaying the process reach the stated final value? |
25. Practice laboratory
- A number is multiplied by 5 then 18 is subtracted to give 102. Find the number.
- A shop ends with 350 pens after selling 120 and receiving 80. Find the starting number.
- After 1/4 of a quantity is removed, 180 remains. Find original.
- After a 20% discount, a bag costs $72. Find original price.
- A price is discounted 10% then 20%, ending at $144. Find original.
- Box A has 80 counters, B has 40. How many move from A to B to equalise?
- A exceeds B by 50. A gains 7 and B gains 13. Find new difference.
- Sara spends 25% of her money then $18 and has $72 left. Find original.
- A printer makes 30 pages/min. After 120 are discarded, 480 remain. Find running time.
- A tank depth rises from 9 cm to 15 cm over base area 600 cm². Find added volume.
26. Explained answers
1. (102 + 18) ÷ 5 = 24.
2. 350 − 80 + 120 = 390.
3. 3/4 = 180; whole = 240.
4. 80% = 72; original = $90.
5. 144 ÷ 0.8 ÷ 0.9 = $200.
6. 20 counters.
7. 50 + 7 − 13 = 44.
8. $120.
9. Before discard 600; 600 ÷ 30 = 20 min.
10. Rise 6 cm; 600 × 6 = 3600 cm³.
27. Full mixed problem
A tank loses 20% of its water. Then 30 ℓ is added. It finally contains 270 ℓ. Find the original amount.
Backward: before adding 30, it contained 240 ℓ. That is 80% of original. Original = 240 ÷ 0.8 = 300 ℓ.
Check forward: 20% of 300 = 60; remain 240; add 30 = 270.
28. Final checkpoint
A strong Primary 5 change-problem solver can identify start, change and final states; reverse operations in reverse order; preserve changing percentage and fraction wholes; exploit total and difference invariants; use state tables when needed; and verify by replaying the process forward.
Continue to Primary 5 Mathematics Learning Guide | Assumption Method, Fixed Totals, Equalisation & Difference Reasoning.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Separate the states, preserve what remains invariant, reverse the last transformation first, and prove the recovered starting state by running the process forward again.
