Singapore School Mathematics Operating Manual · Chapter 29
Many hard Mathematics questions are not hard because any one step is impossible. They are hard because several different jobs are packed together.
A geometry question may require one angle, then a side, then an area. A percentage problem may require reconstructing an original value before applying a second change. A statistics question may require cleaning data, calculating a summary, then interpreting it. A modelling problem may require translating words into variables, solving equations, checking feasibility and returning to context.
The operating principle is decomposition: separate the whole problem into smaller subproblems whose outputs can be passed into later stages. Then use recomposition: reconnect the sub-results into one complete answer.
The difficult part is often the interface between subproblems. A correct intermediate answer can still be misused if its unit, domain, sign, approximation status or meaning is forgotten during handoff.
1. Decomposition changes one large question into several smaller jobs
Suppose a composite shape has a rectangle with a semicircle attached.
Total area can be decomposed into rectangle area plus semicircle area.
Each component uses a familiar formula.
The full answer is obtained by recomposition after confirming the shared dimensions are interpreted consistently.
2. A subproblem should have a clear input and output
If a later step needs the radius, one subproblem can be “find the radius”.
If a later step needs the number of whole containers, another subproblem can be “convert total volume into minimum whole containers”.
Good subproblems are not arbitrary fragments. Each one produces something another stage actually needs.
3. Interfaces are where errors travel
Suppose part A finds time = 30 minutes.
Part B uses speed in km/h.
The interface must convert 30 minutes to 0.5 hours before recomposition.
If the handoff ignores the unit, both subproblems can be individually correct while the final answer is wrong.
4. Decomposition can follow mathematical type
A long problem may contain:
an algebra subproblem, a geometry subproblem, a discrete decision subproblem and a final interpretation subproblem.
Separating by mathematical type helps the learner choose the correct tool for each stage.
5. Decomposition can also follow dependency order
If quantity C requires B and B requires A, then solve A first, then B, then C.
The decomposition therefore creates a dependency graph, not merely a list.
The companion chapter Dependency Order, Solving Sequence and Mathematical Workflows develops this in detail.
6. Geometry decomposition uses auxiliary quantities
Suppose a triangle area is required but the perpendicular height is unknown.
One subproblem may find the height using trigonometry.
The second subproblem uses A = 1/2bh.
The height is an interface quantity between two topic tools.
7. Algebra decomposition uses substitutions
Suppose x⁴−5x²+4=0.
Let u=x².
The first subproblem solves u²−5u+4=0, giving u=1 or 4.
The second subproblem reconstructs x from x²=u, producing x=±1,±2.
The substitution compresses a higher-order expression into a simpler subproblem, then recomposition restores the original variable.
8. Word problems often need a translation subproblem first
Before solving, define variables and convert the words into equations.
For example: “two numbers differ by 4 and sum to 20” becomes x−y=4 and x+y=20.
The translation is one subproblem. Equation solving is another.
Interpretation of x and y is the final recomposition step.
9. Percentage chains should be decomposed by base
A price increases 20%, then decreases 20%.
Do not collapse the words into “net zero”.
Subproblem 1: multiply by 1.2.
Subproblem 2: multiply the new value by 0.8.
Recomposition gives multiplier 0.96, a net 4% decrease.
10. Probability trees are decomposition diagrams
Each branch represents one local conditional subproblem.
Path multiplication recomposes events along one branch.
Branch addition recomposes mutually exclusive routes to the same target event.
A tree is therefore a visible decomposition-and-recomposition architecture.
11. Statistics often decomposes into data, calculation and interpretation
Subproblem 1: identify the correct data set and exclusions.
Subproblem 2: calculate mean, median, spread or another statistic.
Subproblem 3: interpret the statistic in context.
Skipping the first or third stage can produce mathematically polished but contextually wrong work.
12. Optimisation decomposes into objective and feasible set
First define what is being maximised or minimised.
Second define which values are admissible.
Third evaluate or compare candidates.
Fourth verify the best candidate is actually feasible.
These subproblems should remain distinct until recomposition.
13. Decomposition reduces cognitive load
A learner cannot always hold every condition, formula and target in working memory at once.
Breaking the task into named subgoals turns one large load into several smaller loads.
This is one reason clean working improves performance: it externalises the architecture.
14. Decomposition also improves error diagnosis
If the final answer is wrong, ask which subproblem failed.
Was the translation wrong? Was the algebra wrong? Was the unit conversion wrong? Was the recomposition wrong?
Modular working localises errors.
15. Independent checks can be attached to each module
After finding a side length, check positivity and scale.
After solving an equation, substitute back.
After calculating probability branches, check total probability structure.
After recomposition, check units and context.
Local checks reduce the chance of carrying a bad intermediate result forward.
16. Decomposition should not destroy global constraints
Suppose x and y must both be positive integers.
If one subproblem solves an algebraic equation over the reals, the integer and positivity constraints must still be carried to the recomposition stage.
Global constraints belong to the entire problem, even when subproblems temporarily ignore them for convenience.
17. A subproblem can create a temporary variable
Using u=x+1/x can simplify a symmetric expression.
But the temporary variable must eventually reconnect to the original x-domain.
Temporary notation is an interface tool, not a replacement of the original problem.
18. Recomposition can reveal missing cases
Suppose a subproblem yields u=4.
If u=x², recomposition gives x=±2.
Failing to restore both branches makes the final solution incomplete.
19. Recomposition can also remove impossible candidates
Suppose a subproblem yields x=−3 and x=5, but x represents a positive length.
The recomposition stage removes −3.
Subproblem outputs are candidates until the full contract is restored.
20. Interfaces should carry metadata
An intermediate result is stronger when written as:
“radius = 4 cm” rather than “4”.
“θ=30° in the acute branch” rather than “30”.
“x≈2.718 to 3 d.p.” rather than “2.718”.
Units, branch conditions and approximation status are metadata needed for safe recomposition.
21. Different decompositions can solve the same problem
A composite area can be found by adding pieces or subtracting a cut-out from a larger shape.
Two valid decompositions should agree.
This creates a powerful independent check through recomposition.
22. The best decomposition minimises difficult interfaces
If one route requires repeated unit conversion and another keeps all quantities in the same system, the second may be safer.
If one algebraic route creates many branches and another preserves equivalence more directly, the second may be cleaner.
Efficiency is not only fewer steps; it is fewer fragile handoffs.
23. Decomposition is not fragmentation for its own sake
Too many tiny steps can obscure the main structure.
A useful module should correspond to a genuine mathematical job.
Strong solvers choose a level of decomposition that makes the route visible without drowning it in bookkeeping.
24. Recomposition should answer the original question, not merely assemble numbers
If subproblems produce area, cost per tile and number of tiles, the final answer should state total cost in the requested unit and context.
Mathematics ends at the original target, not at the last intermediate arithmetic line.
25. A practical decomposition audit
Ask:
What smaller jobs make up this problem? What does each one need as input? What does each one produce? Which results are passed forward? What units, domains or branch conditions must travel with them? Where can local checks be inserted? After recombination, have all original conditions been restored?
26. Independent practice
1. A rectangle and semicircle share the same width 10 cm. Describe a decomposition for total area.
2. Solve x⁴−13x²+36=0 by substitution.
3. A 90-minute journey covers 120 km. Decompose the work needed to find average speed in km/h.
4. A price rises 10% and then falls 10%. Use decomposition to find the net multiplier.
5. Explain why “radius = 4 cm” is a safer interface result than “4”.
6. Give one example where recomposition adds back a missing branch.
7. Give one example where recomposition removes an impossible candidate.
8. Why can two different decompositions provide a useful check?
9. What is the danger of ignoring a global constraint during recomposition?
10. State one advantage of modular error diagnosis.
27. Worked answers
1. Find rectangle area, find semicircle radius 5 cm, find semicircle area 1/2π(5²), then add.
2. Let u=x². Then u²−13u+36=0=(u−4)(u−9), so u=4 or 9. Hence x=±2,±3.
3. Convert 90 min to 1.5 h, then compute 120/1.5=80 km/h.
4. Multiplier =1.1×0.9=0.99, so net 1% decrease.
5. It carries both meaning and unit, reducing interface errors.
6. From x²=9, recomposition after taking u=x² restores x=±3.
7. If an equation gives x=−2 or 5 but x is a positive length, remove −2.
8. Independent routes should recompose to the same final result; disagreement exposes an error.
9. A candidate that was locally valid can become globally inadmissible.
10. It identifies which stage failed instead of forcing a complete restart.
28. Continue through Batch 08
Use Dependency Order and Solving Sequence to decide which subproblem must come first, Normalisation and Rescaling when modules use different scales, and Sampling, Resolution and Aliasing when observed data are only a partial view of the underlying object.
Return to the BTT Mathematics Hub for Batch 08.