PRIMARY 4 MATHEMATICS LEARNING GUIDE · BATCH 12 · GUIDE 47
Single Unchanged Quantity problems become easier when the learner notices that one subject stays fixed while another changes around it. A child may have the same number of marbles before and after while a friend gains some. A container may keep the same volume while another container is emptied. A fixed amount can act as an anchor that links two different comparison states.
This heuristic is useful because it gives the learner one quantity that does not need to be recalculated. Instead of treating both before-and-after states as entirely new, we hold the constant subject in place and use the changing relationship to recover the missing amount.
This guide uses Single Unchanged Quantity and Constant Single Subject as instructional heuristic labels. They are not presented as separate official syllabus chapters. The official curriculum boundary remains the MOE Primary Mathematics Syllabus, updated October 2025.
Series route: return to the Primary 4 Mathematics Learning Hub. For equal-stage problems, use Equal Concept. For conserved combined amounts, use Constant Total.
Navigate: the anchor · changing comparisons · before and after · fractional states · boundaries · practice · answers.
1. The unchanged quantity is the anchor between two states
Amir has 40 counters throughout the problem.
At first, Bea has 10 fewer counters than Amir.
Later, Bea receives 18 counters.
Amir does not gain or lose any counters.
Because Amir stays at40, Bea’s original amount is30 and her later amount is48.
The unchanged subject gives us a fixed reference point.
2. Mark the unchanged subject explicitly
In a before-and-after table, write Amir as40 in both columns.
| Before | After | |
|---|---|---|
| Amir | 40 | 40 |
| Bea | 30 | 48 |
The repeated40 is not a coincidence. It is the structural bridge.
If the learner mistakenly changes Amir in the second column, the entire comparison becomes unstable.
3. A changing comparison can be measured against the fixed anchor
At first, Bea has10 fewer than Amir.
Later, Bea has8 more than Amir.
Amir remains unchanged.
Bea must have increased by10+8=18.
Why add? Bea moves from10 below the fixed anchor to8 above it, crossing the anchor.
This is the same geometry as moving between two points on opposite sides of a fixed reference.
4. If both comparisons are on the same side, subtract them
At first, Bea has25 fewer than Amir.
Later, Bea has7 fewer than Amir.
Amir remains unchanged.
Bea moved18 closer to Amir.
Increase =25−7=18.
Both states lie below the same fixed anchor, so only the gap reduction matters.
5. Moving farther from the anchor increases the gap
At first, Bea has6 fewer than Amir.
Later, Bea has19 fewer.
Amir remains unchanged.
Bea must have decreased by19−6=13.
The comparison became more negative relative to the same anchor because the changing subject moved farther away.
6. Known change plus one comparison gives the other comparison
Amir stays unchanged.
Bea starts12 fewer than Amir and then gains20.
After gaining12, she would become equal to Amir.
The remaining8 places her8 above Amir.
Final relationship: Bea has 8 more.
The fixed subject lets us transform a known change directly into a new comparison.
7. Work backwards when the final comparison is given
Amir stays unchanged.
After Bea receives15 counters, she has4 more than Amir.
Before receiving15, she must have been11 fewer than Amir.
Why? Moving backward subtracts15 from Bea while Amir stays fixed.
Starting from4 above and moving15 down lands11 below.
This reverse movement is easier when the anchor is kept visible.
8. An actual anchor value can recover exact amounts
Amir has52 counters throughout.
After Bea receives15, she has4 more than Amir.
Bea after=56.
Bea before=56−15=41.
Check original comparison:41 is11 fewer than52.
Once the fixed anchor has an actual value, every relative state can be converted into an exact amount.
9. An unknown anchor can still answer a change question
We are told only:
Bea is15 fewer than Amir at first.
Later Bea is9 more than Amir.
Amir does not change.
Bea’s increase must be15+9=24.
We do not need Amir’s actual amount because the question asks only how far Bea moved relative to the fixed anchor.
10. One subject can remain fixed while the other changes in several stages
Amir remains unchanged.
Bea starts20 below Amir.
She gains8, then gains7, then loses3.
Net change=+12.
Final gap=20−12=8 fewer.
Several changes can be combined before comparing against the fixed anchor.
11. A single unchanged amount can support a multiplicative comparison
Amir stays at40 counters.
At first, Bea has half as many as Amir:20.
Later, Bea has three quarters as many as Amir:30.
Bea increased by10.
The fixed whole40 lets us evaluate both fractional relationships using the same reference.
Do not use this shortcut if Amir’s amount changes between stages.
12. Fixed subject + changing fraction creates a clean before-and-after model
Amir’s collection remains60.
Bea initially has1/3 of Amir’s amount.
Later Bea has2/3 of Amir’s amount.
Before:20.
After:40.
Increase=20.
The denominator and whole are stable because Amir is unchanged.
13. If the reference subject changes, the fraction values must be recomputed
If Amir starts60 but later becomes90, then1/3 of Amir-before and1/3 of Amir-after are20 and30.
The fraction notation stayed the same; the actual value did not.
Single Unchanged Quantity reasoning depends on the anchor actually remaining unchanged.
This is a strong contrast with changing-reference-whole fraction problems.
14. Money comparisons can use a fixed anchor
Bea has $18 less than Amir.
Amir spends nothing.
Bea receives $30.
Bea now has $12 more than Amir.
No actual amount is needed.
The unchanged person turns a money story into movement across a fixed comparison point.
15. Length and mass comparisons use the same structure
Rod A remains120 cm.
Rod B begins35 cm shorter and is extended50 cm.
Rod B ends15 cm longer.
The same logic works for mass, capacity and other quantities as long as the fixed anchor and units remain consistent.
16. The unchanged quantity may be the smaller subject
Bea remains unchanged.
Amir starts12 more than Bea, then loses20.
Amir ends8 less than Bea.
The fixed anchor does not need to be the larger quantity.
Identify which subject is unchanged, not which one looks more prominent in the wording.
17. If both subjects change equally, use Constant Difference instead
A starts20 more than B.
Both gain15.
No subject is unchanged.
The better invariant is Constant Difference: the gap remains20.
Single Unchanged Quantity is not the right owner because neither individual amount stayed fixed.
18. If one gives to the other, use Constant Total or transfer reasoning
A gives10 to B.
Both subjects change.
The combined total may remain constant, but neither individual quantity is unchanged.
This is not a Constant Single Subject problem.
Choose the invariant that actually survives the event.
19. If the anchor changes from outside, split the story into stages
Amir stays unchanged for the first part of a problem, then receives20 from outside.
Do not carry the original anchor value into the later stage.
Use one fixed-anchor interval before the external change and a new anchor afterwards.
Mathematical control often means knowing when an invariant stops being valid.
20. Same name does not guarantee unchanged quantity
“Amir” may appear in every sentence, but if Amir receives, spends, gives or loses something, his amount changed.
The heuristic is about an unchanged quantity, not a repeated name.
Always inspect the actions attached to the anchor subject.
21. A fixed anchor does not determine exact values without enough information
Bea moves from15 fewer to9 more than Amir, while Amir stays fixed.
We know Bea increased24.
But we still do not know Amir’s actual amount.
To determine exact values, we need Amir’s amount, Bea’s amount at one stage, a total or another independent condition.
22. Use a vertical comparison diagram
Draw Amir as a fixed horizontal bar.
Draw Bea-before ending15 units before Amir’s endpoint.
Draw Bea-after extending9 units beyond Amir’s endpoint.
The distance from Bea-before endpoint to Bea-after endpoint is24.
This visual model makes same-side subtraction and opposite-side addition intuitive.
23. Diagnostic error table
| Error | Likely cause | Repair question |
|---|---|---|
| Changes the anchor amount | Unchanged subject not identified | Who or what stays fixed throughout this stage? |
| Adds gaps that are on the same side | Anchor geometry unclear | Are both states below/above the anchor or on opposite sides? |
| Uses Constant Difference when only one subject is fixed | Equal-change condition not checked | Did both quantities change by the same amount? |
| Applies fraction to wrong anchor stage | Reference quantity changed unnoticed | Is the reference subject truly unchanged? |
| Finds movement but invents exact values | Insufficient information ignored | What actual amount fixes the anchor? |
24. Practice laboratory
- Amir stays40. Bea starts10 fewer and gains18. Find Bea’s final amount and final comparison.
- Bea starts10 fewer than unchanged Amir and later8 more. Find Bea’s change.
- Bea starts25 fewer and later7 fewer than unchanged Amir. Find her change.
- Bea starts6 fewer and later19 fewer than unchanged Amir. Find her change.
- Bea starts12 fewer than unchanged Amir and gains20. Find final comparison.
- After gaining15, Bea is4 more than unchanged Amir. Find her original comparison.
- Amir stays52. After gaining15, Bea is4 more. Find Bea before and after.
- Bea goes from15 fewer to9 more than unchanged Amir. Find her increase.
- Bea starts20 below unchanged Amir, then gains8,gains7,loses3. Find final comparison.
- Amir stays40. Bea changes from half Amir to three quarters Amir. Find her increase.
- Amir stays60. Bea changes from1/3 Amir to2/3 Amir. Find her increase.
- Bea is$18 less than unchanged Amir, then receives$30. Find final comparison.
- Rod A stays120 cm. Rod B starts35 cm shorter and extends50 cm. Find final comparison.
- Bea stays unchanged. Amir starts12 more and loses20. Find final comparison.
- A and B both gain15. Does Single Unchanged Quantity apply?
- A gives10 to B. Does Single Unchanged Quantity apply?
- Amir stays fixed at first, then receives20. Can one anchor value be used for the whole story?
- Bea moves from15 fewer to9 more than unchanged Amir. Can Amir’s exact amount be found from this alone?
- Create a valid problem where the changing subject moves from below the anchor to above it by22.
- State one difference between Single Unchanged Quantity and Constant Difference.
25. Explained answers
1. Bea starts30 and ends48, so she has8 more than Amir.
2. 10+8=18.
3. 25−7=18 increase.
4. 19−6=13 decrease.
5. 12 below +20 → 8 above.
6. 4 above−15 → 11 below originally.
7. Amir52. Bea after56, before41.
8. 15+9=24.
9. Net+12, so final8 fewer.
10. 20→30, increase10.
11. 20→40, increase20.
12. Moves from18 below to12 above: final$12 more.
13. Starts85, ends135, so 15 cm longer than A.
14. 12 above−20→8 below.
15. No. Both change; Constant Difference is the cleaner owner.
16. No. Both change; Constant Total/transfer reasoning is more appropriate.
17. No. The anchor changes after the external receipt; split the story into stages.
18. No. Only the movement24 is determined.
19. Many answers. Example start14 below, finish8 above→change22.
20. Single Unchanged Quantity fixes one subject while the other changes; Constant Difference changes both by the same amount while preserving the gap.
26. Teaching routine: freeze one bar
Draw the unchanged subject once.
Do not redraw or resize it across stages.
Place the changing subject relative to that fixed bar before and after.
Ask whether the changing endpoints lie on the same side or opposite sides of the anchor.
This makes the required addition or subtraction visible before numbers are manipulated.
27. Handover to Grouping Concept
The fixed-anchor heuristic controls change around one subject. The final guide in Batch 12 shifts to composite grouping: organise different quantities into repeatable equal sets, find the value of one group, and scale the group structure.
Continue to Grouping Concept: Quantity × Value, Equal Sets and Regrouping.
Final checkpoint: can the learner identify the unchanged anchor, compare before-and-after positions around it, know when to add or subtract gaps, reject the heuristic when the anchor changes, and distinguish movement from exact values?
Source and editorial note
The curriculum boundary is referenced to the MOE Primary Mathematics Syllabus, updated October 2025. “Single Unchanged Quantity” and “Constant Single Subject” are used here as instructional heuristic labels; all examples are independently written by eduKate Publishing.
Editorial control: Wintour House V1.0 · CivDJ · eduKate Publishing.
