PRIMARY 5 MATHEMATICS LEARNING GUIDE · BATCH 9 · GUIDE 35
Some hard word problems become easy only after the learner notices what did not change. An item may keep the same value while the number of items changes. One person’s amount may remain untouched while another person gives away money. Two different fractions may refer to the same unchanged quantity at different times. A repeated identity can become the bridge between two apparently unrelated states.
The labels in this guide—one item constant, external unchanged, repeated identity and equal-fraction reasoning—are problem-solving descriptions, not official MOE syllabus topic names. Their purpose is to make invariant-based reasoning teachable and searchable within the Primary 5 Mathematics library.
For the official curriculum framework, see the MOE Primary Mathematics Syllabus.
Series route: return to the Primary 5 Mathematics Learning Hub. Earlier: Remainder Concept, Fraction of the Remainder & Branching · Excess & Shortage, Gap & Difference, Distribution Plans. Continue to the Primary 5 Mathematics Examination Guide.
1. Invariant thinking begins with one question
When a problem describes several changes, ask:
What quantity stayed the same?
The unchanged quantity can act as an anchor connecting the before and after states.
2. One item constant: same value per item
Suppose 8 identical pens cost $24. The number of pens may change, but if the unit price remains constant, each pen costs $3.
If 15 pens are bought under the same condition, total cost = 15 × 3 = $45.
The invariant is the value of one item.
3. A one-item constant turns two totals into a common unit
Six notebooks cost $21 and ten notebooks cost $35. Rather than compare totals directly, reduce each to one notebook:
$21 ÷ 6 = $3.50; $35 ÷ 10 = $3.50.
The identical unit value proves the two offers use the same rate.
4. One-item constant is a rate idea
“$3 per notebook”, “18 litres per minute” and “24 pages per booklet” are all per-unit relationships.
The heuristic is useful because a constant per-unit value allows quantities to scale multiplicatively.
5. External unchanged: one group may stay fixed while another changes
Alice has 80 stickers and Ben has 50. Alice gives 20 stickers to Cara, while Ben’s amount does not change.
Before: Alice − Ben = 30.
After: Alice = 60, Ben = 50, difference = 10.
Ben’s unchanged 50 provides the stable comparison anchor.
6. Identify the untouched quantity explicitly
Write:
Ben unchanged = 50.
This prevents the learner from applying every described change to every quantity in the problem.
7. Same total can be an external invariant
If counters move from Box A to Box B, the combined total remains unchanged.
A and B both change, but A + B is invariant.
Different problems preserve different quantities. Do not assume the invariant before checking the action.
8. Repeated identity: the same quantity appears in two descriptions
Suppose 3/5 of a quantity equals 45, and that same 45 later represents 3/8 of another quantity.
The repeated identity is 45.
First whole = 45 ÷ 3/5 = 75.
Second whole = 45 ÷ 3/8 = 120.
The same numerical quantity links two reference wholes.
9. Equal-fraction reasoning begins when two fractional expressions are equal
If 2/3 of A equals 3/5 of B, the common quantity can be represented by a convenient shared number of units.
For example, make both fractional amounts 6 units:
2/3 of A = 6 units → A = 9 units.
3/5 of B = 6 units → B = 10 units.
So A:B follows a 9-to-10 comparison.
This is a reasoning bridge; formal ratio notation belongs to Primary 6.
10. Why common units work
The equal fractional parts represent the same quantity. Choosing a common unit count gives both relationships one shared scale.
The method is not a trick with denominators. It is equal-quantity matching.
11. Example: 3/4 of A equals 2/5 of B
Use a common fractional amount of 6 units.
3/4 of A = 6 → A = 8 units.
2/5 of B = 6 → B = 15 units.
Therefore the whole quantities correspond to 8 and 15 equal units respectively.
12. Same amount before and after can reveal a changed fraction
A student spends money so that the amount remaining changes from 3/5 of the original to a later quantity described as 3/4 of a smaller new reference.
If the actual remaining dollars are unchanged between those two statements, that repeated quantity can connect the two wholes.
Always verify that the problem actually says the physical amount is unchanged.
13. Do not match fractions merely because numerators look alike
3/5 of A and 3/8 of B are not automatically equal. Equality must be stated or deduced from the story.
Visual similarity in fraction notation is not evidence of equal quantity.
14. External unchanged in before–after comparison
A has some money and B has $120. A spends $40 while B remains unchanged. Afterward A has twice as much as B.
After A = 2 × 120 = 240.
Before A = 240 + 40 = $280.
B’s unchanged amount anchors both states.
15. Repeated identity can be a difference
Two quantities differ by 30 before and after an equal addition to both. The repeated identity is the difference, not either quantity.
If both increase by 12, the difference remains 30.
Identify exactly what is being preserved.
16. Repeated identity can be a total
If counters are transferred between two boxes, the total stays the same. If the combined total is 140 before transfer, it remains 140 afterward.
The same total can connect two different comparison states.
17. Repeated identity can be a unit value
If two shopping situations use the same price per item, one unit price links both situations even if quantities and totals differ.
Example: 6 items cost $18 and 14 items cost $42. Both use $3 per item.
18. Equal-fraction reasoning with actual values
Two thirds of Box A equals three fifths of Box B. The common quantity is 60.
A = 60 ÷ 2/3 = 90.
B = 60 ÷ 3/5 = 100.
The repeated 60 is the bridge.
19. Equal-fraction reasoning with unknown common quantity
If no actual common amount is given, use common units instead.
2/3 of A = 3/5 of B. Make the equal parts 6 units, giving A = 9 units and B = 10 units.
If A + B later equals 190, 19 units = 190, one unit = 10. Therefore A = 90 and B = 100.
20. One-item constant with discount must preserve the correct stage
If 5 items cost $50 before a discount, the pre-discount unit price is $10. After a 20% discount on the total, the effective price becomes $8 per item only if the discount applies uniformly to all five items.
The per-item constant can change when the pricing rule changes.
21. An invariant is conditional
A rate stays constant only if the problem says or implies constant rate. A person’s amount stays fixed only if no action changes it. A combined total stays fixed only if nothing enters or leaves the system.
Invariant thinking requires evidence.
22. System boundary matters
If 20 counters move from A to B, A+B is unchanged. If 20 counters leave A and are given to someone outside the two-box system, A+B decreases by 20.
The phrase “external unchanged” is useful because it forces the learner to define what is inside the system.
23. Match unchanged quantities before solving unknowns
In a complex problem, first mark every statement that describes the same quantity across states.
Then build equations or bar models around those anchors. This often reduces the number of unknown relationships that must be held simultaneously.
24. Use equality as the bridge
If “3/5 of A” and “2/3 of B” refer to the same amount, write:
3/5 of A = 2/3 of B.
Once equality is explicit, a common-unit model or later algebraic method becomes natural.
25. Difference between identity and resemblance
Two quantities can have the same numerical value without representing the same object. Conversely, the same physical quantity can appear with different fractional descriptions under changing wholes.
Always ask what the number means, not only whether the digits match.
26. Error map
| Visible error | Likely cause | Repair question |
|---|---|---|
| Assumes a quantity stayed constant | Invariant not justified | What action could have changed it? |
| Matches 3/5 and 3/8 automatically | Notation resemblance mistaken for equality | Does the story say these actual quantities are equal? |
| Uses total as invariant after items enter system | System boundary ignored | Did anything enter or leave? |
| Unit price treated constant after pricing rule changes | Condition changed | Does the same rate still apply? |
| Common-unit method cannot be explained | Heuristic memorised | Which two expressions represent the same quantity? |
27. Practice laboratory
- 8 identical pens cost $24. Find cost of 15 at same unit price.
- A has some money; B has $120 unchanged. A spends $40 and then has twice B. Find A before spending.
- 2/3 of A equals 3/5 of B. If the common amount is 60, find A and B.
- 2/3 of A equals 3/5 of B and A+B=190. Find A and B using common units.
- A and B differ by 30. Both increase by 12. What stays invariant?
- Two boxes total 140. Twenty counters transfer from A to B. What stays invariant?
- Explain why 3/5 of A and 3/8 of B cannot be assumed equal without more information.
28. Answers
1. Unit price $3; total $45.
2. After A = $240; before = $280.
3. A = 90; B = 100.
4. Common 6 units → A 9 units, B 10 units; 19 units = 190; A = 90, B = 100.
5. Difference remains 30.
6. Combined total remains 140.
7. The fractions have different reference wholes; equality must be stated or deduced.
29. Full invariant problem
Three fifths of A equals two thirds of B. Together A and B total 270. Find A and B.
Make the equal fractional amounts 6 units.
3/5 of A = 6 units → A = 10 units.
2/3 of B = 6 units → B = 9 units.
Total = 19 units = 270, so one unit = 270/19. This does not give whole-number quantities.
That is not automatically wrong; it tells us the given total is not compatible with whole-number A and B under this exact relationship. If the context requires whole objects, the data would need reconsideration. Invariant reasoning can test feasibility as well as solve.
30. Final checkpoint
A strong Primary 5 learner can identify what remains constant, distinguish a one-item rate from changing totals, use untouched quantities as anchors, recognise when two fractional expressions represent the same physical amount, construct common units from equality and reject invariants that are not supported by the system boundary or problem conditions.
Continue to the Primary 5 Mathematics Examination Guide | Paper 1, Paper 2, Calculator Discipline & Long-Answer Strategy.
Wintour House V1.0 · CivDJ · eduKate Publishing: identify the quantity that survives the transformation, use it as the common reference between states, and never declare an invariant without checking the system boundary that protects it.
