PRIMARY 4 MATHEMATICS LEARNING GUIDE · BATCH 11 · GUIDE 43
A Constant Total problem changes where a quantity is located without changing how much exists altogether. Counters move from one box to another. Money is transferred from one person to another. Water is poured between containers without loss. The parts change, but the combined amount stays fixed.
This invariant gives the learner a powerful control point. Instead of chasing every before-and-after number separately, we first identify the closed system, calculate or preserve its total, then use the final relationship to reconstruct the parts. The method is especially useful when a transfer creates equality, a fractional relationship, or a new comparison.
“Constant Total” is used here as a Singapore problem-solving heuristic label, not as a separate official syllabus chapter. The official curriculum boundary remains the MOE Primary Mathematics Syllabus, updated October 2025. The examples, models and practice questions are independently written by eduKate Publishing.
Series route: return to the Primary 4 Mathematics Learning Hub. For event sequencing where the total itself changes, use Before-and-After Problems.
Navigate: the invariant · equalisation · fractional final states · three groups · when total is not constant · practice · answers.
1. Internal transfer changes the parts but not the whole
Box A contains 95 counters and Box B contains 55 counters.
Total = 95 + 55 = 150 counters.
If A gives 20 counters to B, the new amounts are 75 and 75.
The combined total remains 150.
No counters entered or left the two-box system. They only changed location.
This is the defining feature of Constant Total reasoning.
2. Draw a boundary around the closed system
Before deciding the total is constant, ask what is inside the system.
If the problem considers only Box A and Box B, a transfer between them is internal.
If A gives counters to a third box outside the system, the A+B total decreases.
If new counters are added from outside, the A+B total increases.
The invariant is therefore relative to the system boundary.
A useful question is: Did anything cross the boundary?
3. Use a before-and-after table
| A | B | Total | |
|---|---|---|---|
| Before | 95 | 55 | 150 |
| Transfer | −20 | +20 | 0 net change |
| After | 75 | 75 | 150 |
The transfer column shows equal and opposite changes.
−20 + 20 = 0, so the total changes by zero.
This representation makes conservation visible without requiring formal algebra.
4. Equal final amounts make the total especially useful
A and B have 150 counters altogether. After A gives 20 counters to B, they have equal amounts. Find their original amounts.
Because the transfer is internal, the total after transfer is still 150.
Equal final amounts mean 150 ÷ 2 = 75 each.
Undo the transfer:
A originally had 75 + 20 = 95.
B originally had 75 − 20 = 55.
Check: 95 + 55 = 150, and 95 − 20 = 55 + 20 = 75.
5. Work backwards from the final distribution
The original story runs forward: A gives 20 to B.
The solution may run backward: recover the final amounts from the conserved total, then undo the transfer.
This is not a contradiction. The event direction and solution direction can differ.
Keep signs attached to the event:
Forward: A −20, B +20.
Backward: A +20, B −20.
The total stays constant in both descriptions.
6. If the original amounts are known, equalisation needs only half the difference
A has 80 counters and B has 70. A gives some counters to B until they are equal.
Total = 150, so each ends with 75.
A gives 5 counters.
Another view: original difference = 10. Every transferred counter reduces the difference by 2 because A loses one while B gains one.
Transfer required = 10 ÷ 2 = 5.
The total-invariant route and difference-change route agree.
7. Constant Total does not mean each part stays constant
Before transfer: A=80, B=70.
After transfer of 5: A=75, B=75.
Both parts changed.
The invariant is their sum, not either individual amount.
This distinction matters because students sometimes see “constant” and freeze every number in the problem.
8. A final fraction relationship can split the conserved total
A and B have 180 counters altogether. After a transfer, A has 2/5 of the total and B has the rest.
Total stays 180.
A after transfer = 2/5 of 180 = 72.
B after transfer = 3/5 of 180 = 108.
If the problem also says A gave 18 counters to B, then before transfer:
A = 72 + 18 = 90.
B = 108 − 18 = 90.
9. The fraction must refer to the conserved whole
“After the transfer, A has 2/5 of their combined counters.”
The whole is the A+B total.
Do not interpret 2/5 as a fraction of A’s original amount unless the wording says so.
Constant Total reasoning is powerful only when the reference whole is identified correctly.
Label the whole before applying the fraction.
10. Final “times as many” relationships can also split the total
A and B have 210 counters altogether. After an internal transfer, A has twice as many as B.
The final state contains three equal units: A=2 units, B=1 unit.
One unit = 210 ÷ 3 = 70.
After transfer, A=140 and B=70.
If A received 15 from B, then before transfer A=125 and B=85.
Check original total: 125+85=210.
11. Final difference plus total can recover final parts
A and B have 174 counters altogether after an internal transfer. A has 38 more than B.
Remove the excess: 174 − 38 = 136.
Split the equalised remainder: 136 ÷ 2 = 68.
B=68, A=106.
If the transfer amount is known, undo it to recover the original parts.
The conserved total supplies one condition; the final difference supplies the second.
12. Transfer between containers preserves liquid volume only if no liquid is lost
Container A has 1.8 L and B has 1.2 L. Some water is poured from A to B without spilling.
Combined volume remains 3.0 L.
If the containers end equal, each has 1.5 L.
A transferred 0.3 L.
The phrase “without spilling” is mathematically important. If 0.1 L spills, the total is no longer constant.
13. Money transfers can preserve combined money
Amir and Bea have $72 altogether. Amir gives Bea $9. No money is spent or added.
The combined amount remains $72.
If they then have equal amounts, each has $36.
Before the transfer Amir had $45 and Bea had $27.
Check: $45−$9=$36 and $27+$9=$36.
Again, the money changes owner but not total quantity within the pair.
14. Constant Total can extend to three groups
Three boxes contain 240 counters altogether.
Counters are moved among the boxes, but none enter or leave.
The total remains 240 regardless of the internal redistribution.
If after redistribution the boxes contain equal amounts, each has 240 ÷ 3 = 80 counters.
If the final relationship is 1:1:2 in unit language, there are four equal units and the boxes contain 60,60,120.
Use only relationships within the learner’s taught scope; the invariant itself remains the same.
15. Track multiple transfers with net change
A gives 12 to B, then B gives 5 to C.
Within A+B+C, total remains constant.
Net changes:
A: −12.
B: +12−5 = +7.
C: +5.
Total net change = −12+7+5 = 0.
This zero-sum check is useful when several internal transfers occur.
16. Do not confuse Constant Total with Constant Difference
When A gives 5 to B, total stays constant.
But the difference between A and B changes by 10.
Example: 80 and70 differ by10. After transfer5, both are75 and difference becomes0.
So one invariant can hold while another changes.
The next guide will study the opposite situation: equal changes to both parts preserve the difference but usually change the total.
17. External addition breaks Constant Total
A has 50 counters and B has 40. A receives 10 new counters from outside the pair, then gives 5 to B.
Original total =90.
After external addition, total=100.
The later internal transfer keeps the new total at100.
The total is constant only within each interval where nothing crosses the system boundary.
Break the story into stages rather than declaring one total constant for the entire question.
18. External removal also breaks Constant Total
A and B have 120 counters. Ten are discarded, then the remaining counters are redistributed.
After removal, the conserved redistribution total is110, not120.
Use the correct total for the correct stage.
Conservation begins after the external loss, not before it.
A timeline can help mark where the system total changes.
19. Growth over time is not a Constant Total situation
Two children’s ages do not have a constant total over time. Each child becomes one year older every year, so the combined age increases by2 each year.
Their age difference, however, remains constant.
This is a useful contrast between Constant Total and Constant Difference.
Ask which quantity is actually invariant before selecting the heuristic.
20. The total can be unknown initially and still be recoverable
After A gives 10 to B, they have 45 each.
Final total = 90.
Because the transfer was internal, original total was also90.
Before transfer A=55 and B=35.
The invariant can be discovered from the final state rather than given directly.
21. A transfer amount alone does not determine original parts
“A gives 10 counters to B. How many did they have originally?”
There is not enough information.
Examples 50 and20, or70 and40, can both support a transfer of10.
We need a total, a final relationship, an original relationship or another independent condition.
Constant Total preserves information; it does not create missing information.
22. Representation choices for Constant Total
| Situation | Useful representation |
|---|---|
| One transfer, final equality | Before/after bars or table |
| Known combined total, final fraction | Equal-unit bar |
| Several transfers | Net-change table |
| External addition then redistribution | Timeline + conserved stage |
| Liquid between containers | Labelled volumes + total check |
The representation should make the invariant visible.
23. Diagnostic error table
| Error | Likely cause | Repair question |
|---|---|---|
| Adds transferred amount to combined total | Internal transfer treated as new quantity | Did anything enter the system? |
| Uses original total after items were discarded | System stage not updated | When did the total actually change? |
| Applies final fraction to one original part | Reference whole misidentified | What does the fraction describe? |
| Assumes difference stays constant during transfer | Invariants confused | What happens to each part when one gives to the other? |
| Cannot recover original after final relationship known | Reverse transfer not understood | How do we undo the internal movement? |
24. Practice laboratory
- A has95 counters and B55. A gives20 to B. Find final amounts and confirm the total.
- A and B have150 counters. After A gives20 to B, they are equal. Find original amounts.
- A has80 and B70. A gives some to B until equal. Find the transfer.
- A and B have180. After transfer, A has2/5 of the total. Find final amounts.
- In Question4, A gave18 to B. Find original amounts.
- A and B have210. After transfer, A has twice B. Find final amounts.
- In Question6, A received15 from B. Find original amounts.
- A and B have174 and after transfer A has38 more than B. Find final amounts.
- Container A has1.8 L and B1.2 L. Water is poured without spilling until equal. Find each final amount and transfer.
- Amir and Bea have$72. Amir gives Bea$9 and they become equal. Find original amounts.
- Three boxes contain240 counters. They are redistributed equally. Find each.
- Three boxes contain240 and finish in1:1:2 unit structure. Find the final amounts.
- A gives12 to B, then B gives5 to C. State each net change and verify the combined net change.
- A has50 and B40. A receives10 from outside, then gives5 to B. What total is constant during the transfer stage?
- A and B have120. Ten are discarded, then the rest are redistributed. What is the conserved total during redistribution?
- After A gives10 to B they have45 each. Find original amounts.
- A gives10 to B but no other information is given. Can the original amounts be found uniquely?
- A and B total100. A gives8 to B. Does the total change? Does the difference necessarily stay constant?
- Two children age one year. Is their combined age constant? Is their age difference constant?
- Create a valid Constant Total problem in which the final amounts are60 and90 after a transfer of15.
25. Explained answers
1. Final A=75, B=75. Total remains150.
2. Final each=75. Undo transfer: A95, B55.
3. Total150 → final75 each. A gives5.
4. A=2/5×180=72; B=108.
5. Undo A−18,B+18: original A90,B90.
6. Three units=210; one70. Final A140,B70.
7. If A received15 from B, before transfer A125,B85.
8. Remove difference:174−38=136; half68. Final A106,B68.
9. Total3.0 L. Final 1.5 L each. A transfers0.3 L.
10. Final each$36. Undo$9: Amir$45, Bea$27.
11. 240÷3=80 each.
12. Four units=240; one60. Final 60,60,120.
13. A−12; B+7; C+5. Sum=0.
14. External addition raises total to100; that total remains constant during the later internal transfer.
15. 120−10=110.
16. Final total90. Undo transfer: A55,B35.
17. No. Many starting pairs can support a transfer of10.
18. Total remains100. The difference generally changes; if A gives8 to B, the gap changes by16.
19. Combined age is not constant; it increases by2. Age difference is constant.
20. Many valid versions. Example: after A gives15 to B, final60 and90. Original A75,B75. Total150 throughout.
26. Teaching routine: ask “what crossed the boundary?”
Draw a box around the people, containers or groups included in the problem.
Mark every movement. If an arrow stays inside the box, it is redistribution. If an arrow crosses the boundary, the total changes.
Once the conserved stage is identified, calculate the total once and keep it visible.
Use the final relationship to split the total, then undo transfers if the original state is required.
This routine makes the invariant a visible object rather than an invisible trick.
27. Handover to Constant Difference
Constant Total preserves the sum under internal transfer. Constant Difference preserves the gap when equal changes happen to both quantities.
Continue to Constant Difference: Equal Changes, Comparison and the Gap That Stays the Same.
Final checkpoint: can the learner define the closed system, identify internal versus external change, preserve the correct total, use the final relationship to recover the parts, and undo the transfer without confusing total with difference?
Source and editorial note
The curriculum boundary is referenced to the MOE Primary Mathematics Syllabus, updated October 2025. “Constant Total” is used here as an instructional heuristic label; all examples and routines are independently written by eduKate Publishing.
Editorial control: Wintour House V1.0 · CivDJ · eduKate Publishing.
