PRIMARY 5 MATHEMATICS LEARNING GUIDE · BATCH 2 · GUIDE 6
A non-routine problem is difficult because the method is not announced. The learner may know all the necessary arithmetic and still not see how to begin. Heuristics are not magic tricks. They are controlled ways to change the problem until its structure becomes visible.
Primary 5 is an important stage for this shift because students are expected to combine whole numbers, fractions, decimals, percentages, rate, measurement and geometry in unfamiliar ways. The goal is not to memorise a long list of named techniques. The goal is to know when a representation, pattern, reverse step or smaller case will reduce uncertainty.
Series route: return to the Primary 5 Mathematics Learning Hub. Earlier: Word Problems, Bar Models & Multi-Step Reasoning. Continue to Estimation, Calculator Control & Answer Verification and Mixed Practice, Error Analysis & PSLE Runway.
Curriculum boundary: the problem-solving emphasis is consistent with the Singapore Primary Mathematics framework, but the worked problems and heuristic sequence here are independently written. See the MOE Primary Mathematics Syllabus for the official framework.
1. A heuristic is a move that reveals structure
When a problem is unfamiliar, a student needs an action that is cheaper than random calculation. Useful actions include drawing a model, making a table, listing cases systematically, working backward, looking for a pattern, simplifying the numbers, solving a smaller related problem, making a justified guess and checking it, or separating the problem into stages.
The important word is justified. A heuristic should reduce the space of possibilities. If a student guesses repeatedly without using the results to narrow the next guess, the process is not yet mathematical.
Heuristics are temporary scaffolds. Once the underlying relationship is visible, the final solution may become short.
2. Strategy choice begins with the obstacle
Ask why the problem feels difficult. Is the unknown hidden behind several reversible steps? Work backward may help. Are there many possible cases? A table or systematic list may help. Does the problem involve repeated growth? Look for a pattern. Is the diagram crowded? Redraw a simpler version. Are the numbers distracting? Solve a smaller equivalent case first.
Choosing a heuristic by obstacle is more reliable than choosing by chapter title.
3. Work backward when the final state is known
Example: A number is multiplied by 5, then 18 is subtracted. The result is 102. Find the original number.
Reverse the last operation first: 102 + 18 = 120. Then reverse the multiplication: 120 ÷ 5 = 24.
Check forward: 24 × 5 − 18 = 120 − 18 = 102.
Working backward is effective because each forward step has a clear inverse. The order of reversal matters.
4. Working backward in a money problem
Example: After spending 25% of her money and then another $18, Sara has $72 left. How much did she have at first?
Before the $18 spending, she had 72 + 18 = $90. That $90 represents 75% of the amount before the percentage spending. Therefore 100% = 90 ÷ 75 × 100 = $120.
Check: 25% of 120 is 30. 120 − 30 − 18 = 72.
The reverse step must preserve the reference whole. Adding 25% of $72 would use the wrong whole.
5. Make a table when two quantities change together
Example: A staircase pattern uses 3 blocks in Step 1, 5 blocks in Step 2, 7 blocks in Step 3 and 9 blocks in Step 4. How many blocks are used in Step 20?
| Step | Blocks |
|---|---|
| 1 | 3 |
| 2 | 5 |
| 3 | 7 |
| 4 | 9 |
The block count increases by 2 each step. Step 1 begins at 3. By Step 20 there have been 19 increases of 2:
3 + 19 × 2 = 41 blocks.
The table exposes the constant difference and prevents the pattern from remaining a vague visual impression.
6. Look for a pattern, then state what remains constant
A pattern is not just “the numbers are going up”. Name the rule. Does the amount increase by a constant difference? Multiply by a constant factor? Alternate? Repeat in a cycle?
Example: 2, 6, 18, 54, … Each term is three times the previous term. The next term is 162.
But do not assume the first visible pattern is the only possible rule from a short list. In school problems, use the structural information provided by the diagram or situation, not only a few isolated numbers.
7. Solve a smaller related problem
Large numbers can hide a simple structure. Reduce the size while preserving the relationship.
Example: How many handshakes occur if every person in a group of 8 shakes hands with every other person exactly once?
Start smaller. With 2 people: 1 handshake. With 3: 3. With 4: 6. With 5: 10. Each new person shakes hands with everyone already present, adding 1, then 2, then 3, then 4, and so on.
For 8 people: 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 handshakes.
The smaller cases reveal the accumulation rule without requiring a formula first.
8. Simplify the numbers but keep the relationship
Suppose a percentage problem uses awkward numbers and the learner is unsure about the structure. Replace the values with friendly numbers while keeping the wording.
If “35% of 480 is sold, then 20% of the remainder” is confusing, first test the same sequence with 50% of 100, then 20% of the remainder. The simpler case makes the changing reference whole visible.
After the structure is understood, return to the original numbers. The simplified problem is a thinking tool, not the final answer.
9. Guess and check should narrow possibilities
Example: A farm has chickens and goats. There are 12 animals and 32 legs altogether. How many goats are there?
A random sequence of guesses is inefficient. Use structure. If all 12 were chickens, there would be 24 legs. The actual total has 8 extra legs. Replacing one chicken with one goat adds 2 legs, so 8 ÷ 2 = 4 replacements are needed. Therefore there are 4 goats and 8 chickens.
This is sometimes called the assumption method. It turns a guess into a controlled baseline and correction.
10. Use extremes to find boundaries
When a problem involves possible values, test the smallest and largest plausible cases.
Example: A whole number rounds to 3,500 to the nearest hundred. What whole numbers could it be?
The lower boundary is 3,450 and the upper boundary is just below 3,550. For whole numbers, possible values are 3,450 through 3,549 under the usual school rounding convention.
Boundary thinking turns a vague “around 3,500” statement into an interval.
11. Make a systematic list when cases are few
Example: Using digits 2, 4 and 7 exactly once, how many different three-digit numbers can be formed?
Start with 2: 247, 274. Start with 4: 427, 472. Start with 7: 724, 742. There are 6 numbers.
The list is systematic because each possible first digit is handled once, then the remaining two digits are arranged. A random list risks omission or duplication.
12. Organise information in a tree
When choices happen in stages, a tree diagram can organise combinations.
Example: A student chooses one shirt from red or blue and one pair of shorts from black, white or grey. There are 2 × 3 = 6 outfits.
A tree shows each shirt branching to three short choices. The multiplication principle is visible as repeated branching.
For Primary 5, the key is not formal combinatorics. It is learning to count cases without missing or repeating them.
13. Draw an auxiliary line when geometry is blocked
Sometimes a geometry problem becomes easier when an extra line exposes a familiar shape or angle relationship.
For example, a composite figure may be split into a rectangle and triangle by drawing a perpendicular line. A crowded angle diagram may be clarified by extending a straight line.
The auxiliary line must be justified by the geometry. It is not permission to invent a length or angle that was never given.
14. Rearrange a figure when area is easier than direct subtraction
A composite shape may be cut and rearranged conceptually into a familiar rectangle or parallelogram. The area is preserved if pieces are moved without overlap or gaps.
This heuristic is useful when a zigzag boundary makes direct decomposition messy but the missing and protruding parts match.
Always record why the area is preserved. A visual rearrangement is mathematical only when the pieces and dimensions are accounted for.
15. Use before–after conservation
Many non-routine problems become simpler when one quantity stays constant.
Example: A container has red and blue beads. After 20 red beads are removed, the number of red beads equals the number of blue beads. Before removal, there were 20 more red beads than blue beads.
The blue quantity never changed. The removal eliminated exactly the original difference. Identifying the invariant avoids unnecessary equations.
Ask: what changes, and what stays the same?
16. Use total conservation when items move between groups
Example: Box A has 80 counters and Box B has 40 counters. Some counters are moved from A to B until both boxes contain the same number. How many are moved?
Total counters remain 120. Equal final groups must each have 60. Therefore 80 − 60 = 20 counters move from A to B.
The total is invariant even though each box changes. This conservation viewpoint is often faster than trial-and-error transfers.
17. Use difference conservation carefully
If the same amount is added to two quantities, their difference stays the same. If the same amount is subtracted from both, the difference also stays the same.
For example, 70 and 45 differ by 25. Adding 10 to both gives 80 and 55, still a difference of 25.
But multiplying both by the same factor changes the difference while preserving the ratio. Distinguish additive invariants from multiplicative invariants.
18. Assume one extreme, then correct
The chickens-and-goats problem is one example. Another is tickets at two prices.
Example: Twenty tickets are sold. Adult tickets cost $12 and child tickets cost $8. Total revenue is $208. How many adult tickets were sold?
Assume all 20 are child tickets: revenue = 20 × 8 = $160. Actual revenue is $48 higher. Replacing one child ticket with one adult ticket increases revenue by $4. Therefore 48 ÷ 4 = 12 replacements. So 12 adult tickets and 8 child tickets were sold.
The method works because each replacement changes the total by a constant amount.
19. Use a fixed-total model for two groups
If two groups have a known total and known difference, a bar model can solve the problem efficiently.
Example: Two classes have 74 students altogether. Class A has 8 more students than Class B. Find both class sizes.
Remove the extra 8 from the total: 74 − 8 = 66. Split equally: 66 ÷ 2 = 33. Class B has 33; Class A has 41.
This is a heuristic because the “remove the difference, then split” transformation makes the equal shared part visible.
20. Use equalisation when rates or prices differ
When comparing two offers, reducing both to a common unit can reveal the better value.
Pack A: 6 notebooks for $15. Pack B: 10 notebooks for $24. Instead of comparing totals, compare unit prices: $2.50 versus $2.40 per notebook. Or scale both to 30 notebooks: A costs $75; B costs $72.
Equalisation creates a fair comparison basis.
21. Non-routine percentage problem
Problem: A number is increased by 25% and becomes 180. What was the original number?
The new number is 125% of the original. If 125% = 180, then 25% = 36, and 100% = 144.
A direct division route is 180 ÷ 1.25 = 144.
The heuristic is to reinterpret the final state in percentage units rather than subtracting 25% of 180.
22. Non-routine fraction problem
Problem: After 1/3 of a tank is drained, 240 litres remain. Find the original volume.
If one third was drained, two thirds remain. So 2 parts = 240, one part = 120, and three parts = 360 litres.
The problem becomes routine once the remaining fraction is recognised.
23. Non-routine rate problem
Problem: Machine A makes 30 items per minute. Machine B starts 4 minutes later but makes 42 items per minute. How many minutes after B starts will B have made as many items as A?
In A’s 4-minute head start, it makes 120 items. After B starts, B gains on A at 42 − 30 = 12 items per minute. Time to close the gap = 120 ÷ 12 = 10 minutes.
The heuristic is to focus on the difference in rates, not the two totals separately minute by minute.
24. Non-routine geometry problem
Problem: A rectangle has area 96 cm² and length 12 cm. A triangle with the same base as the rectangle and height half the rectangle’s width is drawn. Find the triangle’s area.
Rectangle width = 96 ÷ 12 = 8 cm. Triangle height = 4 cm. Triangle area = 1/2 × 12 × 4 = 24 cm².
The problem hides a needed dimension inside another area relationship. The first task is to recover the width.
25. A heuristic ladder
When stuck, try the following order before random experimentation:
- Restate what is known and unknown.
- Mark units and reference wholes.
- Draw a simple representation.
- Ask what stays constant.
- Try a smaller or friendlier case.
- Make a systematic table or list.
- Work backward if the final state is known.
- Use an extreme or baseline assumption.
- Check whether the result recreates the conditions.
This is not a compulsory nine-step ritual. It is a menu arranged from low-cost clarifications to more specialised transformations.
26. Strategy choice must be inspected after success
A solution can be correct but expensive. After solving, ask whether a shorter invariant, rate comparison or model could have reduced the work.
This reflection matters because examination pressure rewards methods that are both reliable and economical. The goal is not to find the cleverest possible trick. It is to find a method the learner can explain and reproduce.
27. Error map
| Visible behaviour | Likely issue | Repair move |
|---|---|---|
| Random guessing | No narrowing rule | Choose a baseline or organise guesses in a table. |
| Pattern continued from only visual appearance | Rule not stated | Name what changes and what stays constant. |
| Working backward reverses operations in wrong order | Process sequence lost | Write the forward arrows before reversing them. |
| Systematic list misses cases | No organising variable | Fix the first choice, then enumerate the remaining choices. |
| Heuristic used even after structure is obvious | Scaffold became ritual | Switch to the shortest reliable calculation. |
28. Practice laboratory
- A number is multiplied by 6 and then 14 is added. The result is 92. Find the number.
- After spending 20% of her money and then $24, Lina has $72 left. How much did she have at first?
- A pattern has 4, 7, 10, 13, … Find the 25th term.
- Every person in a group of 7 shakes hands with every other person once. How many handshakes occur?
- A farm has 15 animals, all chickens or goats, and 42 legs. Find the number of goats.
- Twenty-five tickets are sold at $6 and $10. Total revenue is $198. How many $10 tickets were sold?
- Two groups have 96 students altogether. One group has 14 more students than the other. Find both sizes.
- After 2/5 of a tank is drained, 270 litres remain. Find the original volume.
- A value is increased by 20% to 360. Find the original value.
- Machine A makes 28 items/min. Machine B starts 5 minutes later at 38 items/min. How long after B starts will their totals be equal?
- Using digits 1, 5 and 8 exactly once, list all three-digit numbers and state how many there are.
- A rectangle has area 150 cm² and length 15 cm. A triangle uses the same base and a height 4 cm shorter than the rectangle’s width. Find the triangle area.
29. Explained answers
1. 92 − 14 = 78; 78 ÷ 6 = 13.
2. Before spending $24 she had 96. That is 80% of the original, so original = 96 ÷ 0.8 = $120.
3. Constant increase 3. Term 25 = 4 + 24 × 3 = 76.
4. 1 + 2 + 3 + 4 + 5 + 6 = 21.
5. If all chickens: 30 legs. Extra legs = 12. Each goat replacement adds 2. Goats = 6.
6. Assume all $6 tickets: $150. Extra revenue = $48. Each $10 ticket adds $4. Number of $10 tickets = 12.
7. Remove difference: 96 − 14 = 82. Half = 41. Groups are 41 and 55.
8. Three fifths remain = 270. One fifth = 90. Original = 450 ℓ.
9. 120% = 360, so 100% = 300.
10. A’s head start = 28 × 5 = 140. Gain rate = 10/min. Time = 14 minutes.
11. 158, 185, 518, 581, 815, 851. 6 numbers.
12. Rectangle width = 150 ÷ 15 = 10. Triangle height = 6. Area = 1/2 × 15 × 6 = 45 cm².
30. Full mixed challenge
Problem: A school bought some identical exercise books. If each class received 24 books, 36 books would remain. If each class received 27 books, 9 books would be short. How many classes are there, and how many books were bought?
The difference between the two distribution plans is 3 books per class. Moving from the first plan to the second would use the 36 remaining books and require 9 more, a total difference of 45 books. Therefore 3 × number of classes = 45, so there are 15 classes.
Total books = 15 × 24 + 36 = 360 + 36 = 396 books.
Check using the second condition: 15 × 27 = 405, which is 9 more than 396.
The heuristic is to compare the two conditions directly. The unknown total cancels, leaving a difference-per-class relationship.
31. Teaching non-routine problems without teaching dependence on tricks
Present two problems with similar surface stories but different structures. Ask students to explain why the same heuristic works in one but not the other. Then present two different-looking problems with the same underlying structure and ask for the connection.
This contrast teaches strategy choice rather than name recognition.
When a student discovers a shortcut, ask for the invariant or relationship that makes it valid. A shortcut that cannot be explained is fragile under changed conditions.
32. Final checkpoint
A strong Primary 5 non-routine problem solver can identify the obstacle, choose a heuristic that reduces uncertainty, organise cases without duplication, use invariants, reverse reversible processes, test smaller cases and abandon the heuristic once the structure becomes clear.
Continue to Primary 5 Mathematics Learning Guide | Estimation, Calculator Control & Answer Verification.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Detect the obstacle, rotate the representation, preserve invariants, fit-test the strategy against a changed case, and release only the method that remains valid.
