PRIMARY 5 MATHEMATICS LEARNING GUIDE · BATCH 6 · GUIDE 23
The assumption method works because an extreme starting case turns an unknown mixture into a measurable difference. Instead of guessing randomly, the learner assumes all objects belong to one category, computes the baseline total, compares it with the actual total, then uses the difference created by one replacement to determine how many replacements are needed.
This guide develops assumption, fixed-total, equalisation and difference reasoning as non-routine problem-solving tools. The methods are especially useful when two categories have different per-unit values but share one total count.
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 and Before–After Models, Change Unknowns & Working Backwards. Continue to Systematic Listing, Case Organisation, Tables & Invariant Thinking.
1. Begin with a baseline assumption
A farm has 12 animals, all chickens or goats, with 32 legs altogether.
Assume all 12 are chickens: 12 × 2 = 24 legs.
The actual total has 8 extra legs. Replacing one chicken with one goat adds 2 legs. Therefore 8 ÷ 2 = 4 goats.
The assumption is temporary, but the difference it creates is useful.
2. The method depends on a constant replacement difference
Each chicken-to-goat replacement adds exactly 2 legs. Because the change per replacement is constant, total difference ÷ difference per replacement gives the number of replacements.
If the per-replacement difference were not constant, the shortcut would fail.
3. Ticket-price problems use the same structure
Twenty tickets are sold. Adult tickets cost $12 and child tickets $8. Total revenue is $208.
Assume all 20 are child tickets: $160.
Actual revenue is $48 higher. Each replacement by an adult ticket adds $4.
48 ÷ 4 = 12 adult tickets.
4. Choose the simpler baseline
You can assume all adults instead. Then total would be 20 × 12 = $240, which is $32 too high. Replacing an adult with a child lowers revenue by $4, so 8 replacements are needed, leaving 12 adults.
Choose the baseline that makes the correction easiest to interpret.
5. Fixed totals are powerful invariants
In a chicken-and-goat problem, the total number of animals stays fixed while category membership changes. In a ticket problem, total tickets stay fixed while ticket types change.
The invariant total lets the learner focus on only one changing quantity: the total legs, money or another per-unit measure.
6. Equalisation removes an imbalance
Two groups total 96 students. One group has 14 more than the other.
Remove the extra 14 from the larger group: 96 − 14 = 82. Now the two equal parts are 41 each.
So the original groups are 55 and 41.
Equalisation transforms an unequal comparison into equal groups.
7. Equalisation can be visualised with bars
Draw the larger bar as the smaller bar plus an extra segment. Remove the extra segment conceptually. The remaining total divides equally between the two aligned bars.
This is why total-and-difference problems can be solved with “subtract difference, then halve”.
8. Difference per replacement is the key quantity
If one category contributes 6 units and another contributes 10, replacing one 6-unit item with a 10-unit item changes the total by 4.
That 4-unit difference is the conversion rate between baseline error and number of replacements.
9. Mixed coin-value problems
A box contains 30 coins, all 50-cent or $1 coins, worth $22 altogether.
Assume all 50-cent coins: value = $15.
Actual value is $7 higher. Replacing one 50-cent coin with a $1 coin adds $0.50.
$7 ÷ $0.50 = 14 one-dollar coins.
10. Mixed item-weight problems
Twenty packages each weigh either 2 kg or 5 kg. Total mass is 70 kg.
Assume all 2 kg: baseline 40 kg.
Extra mass = 30 kg. Each 5 kg replacement adds 3 kg. So 30 ÷ 3 = 10 heavy packages.
11. Assumption is not random guessing
A random guess produces one trial total but no rule for the next guess. The assumption method chooses an extreme that makes every correction identical.
The method narrows the possibilities mathematically.
12. Use assumption only when category contributions are stable
If each adult ticket has one fixed price and each child ticket another fixed price, the method works. If ticket prices vary within categories, the constant replacement difference disappears.
Check the conditions before applying the method.
13. Fixed total with transfers
Box A has 90 counters and Box B has 50. Some counters move from A to B until A has 10 more than B.
Total = 140. Final groups with difference 10 are found by equalisation: 140 − 10 = 130; half = 65. Final amounts are 75 and 65.
A moved from 90 to 75, so 15 counters moved.
14. Transfers change difference twice as fast
When x moves from larger A to smaller B, A decreases by x and B increases by x. The gap shrinks by 2x.
Starting difference 40 and ending difference 10 means gap shrank by 30, so x = 15.
This is a compact invariant-based route.
15. Equalisation in money comparisons
Two accounts total $500. Account A has $80 more than B.
Remove the extra $80: $420 remains. Half = $210. So B = $210 and A = $290.
16. Equalisation in quantities with fractions
A has twice as many as B and together they have 150. This is not a difference problem. Equal-unit reasoning gives 3 units = 150, so values are 100 and 50.
Use equalisation only when the problem describes a difference. Use unit models when it describes a multiplicative scale.
17. Baseline reasoning can use the higher category
Fifteen items cost either $6 or $10 and total $114.
Assume all are $10: baseline $150, which is $36 too high. Each replacement by a $6 item lowers total by $4. Nine replacements are needed.
So there are 6 ten-dollar items and 9 six-dollar items.
18. Reverse-check the category counts
For the previous example: 6 × 10 + 9 × 6 = 60 + 54 = 114 and 6 + 9 = 15.
Both the item-count total and value total must be satisfied.
19. Difference reasoning with two distribution plans
A school has some classes and exercise books. If each class gets 24 books, 36 remain. If each class gets 27, 9 are short.
The plan difference is 3 books per class. Moving from first plan to second consumes the 36 spare and needs 9 more: 45 books.
Classes = 45 ÷ 3 = 15.
Total books = 15 × 24 + 36 = 396.
20. Compare conditions directly when the hidden total cancels
In the book-distribution problem, the total number of books is unknown but identical in both conditions. Subtracting the conditions removes that unknown.
This is a powerful non-routine strategy: compare two states to eliminate what does not change.
21. Assumption with capacity types
Ten vehicles are cars or vans. Cars carry 4 people, vans carry 7. Total capacity is 52.
Assume all cars: 40 places. Extra = 12. Each van replacement adds 3. Vans = 4.
22. Assumption with scores
A quiz has 20 questions. A correct answer earns 5 points and an incorrect answer earns 2 points. A student answers every question and scores 76.
Assume all incorrect: 40 points. Extra = 36. Each correct replacement adds 3. Correct answers = 12.
This is a constructed teaching example; real marking schemes may differ.
23. Assumption can fail if categories exceed two
With three unknown categories, one total count and one value total may not determine a unique solution. More information is needed.
Do not force a two-category method onto an underdetermined problem.
24. Assumption can fail if totals are inconsistent
If 12 animals each have either 2 or 4 legs, the total number of legs must be even and lie between 24 and 48. A stated total of 31 is impossible.
Extreme bounds can test consistency before solving.
25. Error map
| Visible error | Likely cause | Repair question |
|---|---|---|
| Random guesses with no narrowing rule | Baseline not chosen | What if every item were the same type? |
| Difference divided by wrong number | Replacement difference miscalculated | How much does one replacement change the total? |
| Total-and-difference split directly in half | Extra amount not removed | What must be equalised first? |
| Transfer gap reduced by x instead of 2x | Both groups’ changes not tracked | What happens to each side when x moves? |
| Assumption used with varying category values | Constant replacement condition absent | Is each category contribution fixed? |
26. Practice laboratory
- There are 15 chickens/goats and 42 legs. Find goats.
- Twenty-five tickets cost either $6 or $10. Total revenue is $198. Find $10 tickets.
- 30 coins are 50-cent or $1 coins and total $22. Find $1 coins.
- 20 packages weigh 2 kg or 5 kg and total 70 kg. Find 5 kg packages.
- Two groups total 96 and differ by 14. Find both.
- Box A has 90, B has 50. Some move from A to B until A has 10 more. Find moved amount.
- 15 items cost either $6 or $10 and total $114. Find each type.
- If each class gets 24 books, 36 remain; if 27 each, 9 short. Find classes.
- 10 vehicles carry either 4 or 7 people and total capacity is 52. Find vans.
- Explain why assumption may fail with three unknown item types and only two totals.
27. Explained answers
1. All chickens = 30 legs; extra 12; /2 = 6 goats.
2. All $6 = 150; extra 48; /4 = 12 ten-dollar tickets.
3. All 50-cent = $15; extra $7; /$0.50 = 14.
4. All 2 kg = 40; extra 30; /3 = 10.
5. (96 − 14) ÷ 2 = 41; values 55 and 41.
6. Start gap 40, final 10, shrink 30 = 2x, so 15 moved.
7. All $10 = 150, excess 36, /4 = 9 six-dollar; so 6 ten-dollar and 9 six-dollar.
8. Difference in plans 45, /3 = 15 classes.
9. All cars = 40; extra 12; /3 = 4 vans.
10. There can be multiple combinations satisfying the same count and total; another independent condition is needed.
28. Full mixed problem
A theatre sold 40 tickets. Standard tickets cost $18 and premium tickets $25. Total revenue was $790. How many premium tickets were sold?
Assume all standard: 40 × 18 = $720.
Actual revenue is $70 more. One premium replacement adds $7.
Premium tickets = 70 ÷ 7 = 10.
Check: 30 × 18 + 10 × 25 = 540 + 250 = 790.
29. Final checkpoint
A strong Primary 5 non-routine solver can choose a baseline assumption, identify the constant change per replacement, exploit fixed totals, equalise total-and-difference problems, track how transfers alter differences, compare two conditions directly and recognise when the method’s assumptions do not hold.
Continue to Primary 5 Mathematics Learning Guide | Systematic Listing, Case Organisation, Tables & Invariant Thinking.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Choose an extreme baseline, quantify the error it creates, divide by the invariant change per replacement, and verify that both the count total and value total return exactly.
