PRIMARY 5 MATHEMATICS LEARNING GUIDE · BATCH 13 · GUIDE 50
Some problems become solvable only after the learner stops trying to answer the final question immediately. Instead, the learner identifies one smaller quantity that can already be found, solves that part, and uses the partial result to unlock the next dependency.
This is the Solve Part of the Problem heuristic. It is not the same as simplifying the whole problem. Simplification changes the case temporarily. Solving part of the problem keeps the original problem intact but attacks an available subgoal first.
The method is central to multi-step Primary 5 word problems, geometry, rates, percentages, composite figures, changing states and structured examination questions. The learner’s job is to build a dependency chain: what must I know before I can know the final answer?
Return to the Primary 5 Mathematics Learning Hub. Related owners: Word Problems, Bar Models & Multi-Step Reasoning · Restate the Problem · Before–After Models.
1. The final answer often depends on hidden intermediate answers
A shop has 800 items. Twenty-five percent are sold, then one third of the remainder are packed. How many remain unpacked?
The final answer cannot be found safely until we know the remainder after the first sale. That remainder is the first subgoal.
25% of 800=200 sold. Remainder=600. One third of 600=200 packed. Unpacked remainder=400.
The subgoal is not extra work. It is the bridge to the final state.
2. Ask the dependency question
Before solving, ask:
What quantity must be known immediately before I can answer the question?
Then ask the same question again about that quantity. This produces a chain of dependencies instead of a vague multi-step problem.
3. Work backward to plan, forward to calculate
Suppose the final question asks for the number of cartons. To know cartons, we need number of items to pack. To know items to pack, we may need accepted items. To know accepted items, we may need the rejected amount.
Planning can move backward from target:
cartons ← items to pack ← accepted items ← original total.
Calculation then proceeds forward.
4. A subgoal should reduce uncertainty
Not every intermediate number is useful. Choose a partial result that unlocks a relationship.
If the problem asks for sale price after discount and tax, finding the original price again is not useful because it is already known. Finding the discount amount or discounted price is useful because the next stage depends on it.
5. Name partial results
Write:
- discounted price = $240;
- tax amount = $19.20;
- final price = $259.20.
Labels prevent an intermediate value from being reused as the wrong quantity later.
6. Partial results create checkpoints
If the final answer is wrong, labelled subgoals help locate the first wrong state. This improves correction because the learner can repair the earliest unstable dependency instead of redoing the entire solution.
7. Multi-step percentage problems need state subgoals
Original=500. Remove 20% → subgoal S1=400. Remove 25% of remainder → subgoal S2=300. If the final question asks what fraction of original remains, then 300/500=3/5.
Each percentage attaches to one state. The subgoals keep those bases separate.
8. Rate problems often require finding one missing component first
A machine makes 18 items per minute for some time and produces 270 items. It then operates 5 more minutes at the same rate. Find total output.
First subgoal: initial time=270÷18=15 minutes. Then total time=20 minutes. Total output=18×20=360.
Alternatively, once rate is fixed, additional output=18×5=90 and final=360. The subgoal chosen depends on the easiest route.
9. Geometry needs dimensional subgoals
A composite figure may require a missing length before area can be found. If total width is 15 cm and known horizontal segments are 6 cm and 4 cm, missing length=5 cm. Only then can the relevant rectangle area be calculated.
The missing dimension is a structural subgoal.
10. Triangle problems may require height first
If triangle area is 60 cm² and base is 15 cm, the perpendicular height is found first:
60=1/2×15×h → 120=15h → h=8 cm.
If the final question then asks for a related rectangle or composite area, the recovered height becomes the next input.
11. Volume problems may require base area first
A rectangular tank base is 30 cm by 20 cm. Water volume increases by 3600 cm³. To find water-level rise, first find base area=600 cm². Then height rise=3600÷600=6 cm.
The base area is the essential subgoal.
12. Grouping problems may require usable quantity first
1000 items are produced; 8% are rejected. Accepted items are packed 36 per carton. Find cartons needed.
Subgoal 1: rejected=80. Subgoal 2: accepted=920. Subgoal 3: 920÷36=25 remainder 20. Final decision: 26 cartons if all accepted items must be packed.
13. Excess-and-shortage problems have a gap subgoal
24 per group leaves 36; 27 per group is short 9.
First subgoal: total gap between plans=36+9=45. Second subgoal: difference per group=3. Groups=45÷3=15. Then recover the total if needed.
14. Transfer problems may require final state first
A=90, B=50. Some move A→B until A has 10 more.
Instead of guessing transfer, first solve final pair from total 140 and final difference 10: final values 75 and 65. Then transfer=90−75=15.
The final pair is the decisive subgoal.
15. Solve one branch before another
In a branching problem, a known branch may determine the remaining parent quantity, which then controls another branch. Solve branches in dependency order rather than story order if that is clearer.
16. Solve an easier part to expose a pattern
In a shape pattern, finding the number added from stage 1 to 2 and from stage 2 to 3 may reveal the growth rule before the large-stage total is attempted.
Partial results can reveal structure, not only feed arithmetic.
17. Subgoals can be qualitative
Sometimes the needed partial result is not a number. It may be:
- which quantity is 100%;
- which sides are parallel;
- whether a transfer is internal;
- which group remains unchanged;
- whether the process repeats.
These decisions unlock the numerical work.
18. Find the first certain fact
When a problem feels overwhelming, ask: “What can I determine with certainty right now?”
A known angle sum, a fixed total, an obvious remainder, a unit rate or a missing aligned length may be enough to start the dependency chain.
19. Partial solutions should be reusable
If a subgoal is found, label it so later steps can use it without recalculating. Recomputing the same quantity increases arithmetic risk and wastes time.
20. Avoid premature subgoals
Do not calculate every available number. A subgoal is useful only if it lies on a path to the target.
In a problem containing price, mass and time, mass may be irrelevant if the target depends only on price and quantity.
21. Build a dependency diagram
For a complex problem, draw arrows:
original quantity → remainder → packed amount → cartons.
Or:
total + difference → final A/B → transfer.
The diagram makes the solution architecture visible before arithmetic begins.
22. Dependency chains prevent circular reasoning
If Step A requires Step B and Step B requires Step A, the plan is circular. Look for another given relationship or choose a different representation.
A good dependency chain eventually reaches quantities directly known from the problem.
23. Some problems have parallel subgoals
To compare unit prices, find unit price for Product A and Product B separately. These subgoals are independent and can be solved in either order. The final comparison depends on both.
24. Merge partial results only when units match
Do not add 30 minutes to 4 kilometres or compare $3/item directly with $15 total without converting to compatible quantities.
Subgoals must be expressed in forms that can legitimately combine.
25. Partial results can provide bounds
If a first subgoal shows at most 300 items remain, a later answer of 450 is impossible. Intermediate states create bounds that can detect downstream mistakes.
26. Partial results support estimation
If 40% of 500 is 200 and the final result subtracts something positive from the remainder 300, the final must be below 300. This state-based estimate is stronger than a generic “answer should look reasonable”.
27. Subgoals and examination pacing
On a long-answer question, writing one correct subgoal can preserve useful working even if the full solution is not completed immediately. More importantly, it gives the learner a stable point from which to resume.
28. Know when a subgoal is unnecessary
If 25% of 320 is needed, directly using 1/4×320=80 is enough. Do not create artificial intermediate steps. Decomposition should reduce complexity, not inflate it.
29. Error map
| Visible error | Likely cause | Repair question |
|---|---|---|
| Jumps directly to final operation | Dependency ignored | What must be known immediately before the final answer? |
| Computes many irrelevant values | No target-directed subgoal selection | Which partial result unlocks the next relation? |
| Uses intermediate number without label | State meaning lost | What does this number represent? |
| Combines incompatible units | Subgoals not normalised | Are these quantities comparable? |
| Gets stuck in circular plan | Dependencies point to each other | Which fact can be obtained directly from the givens? |
30. Diagnostic routine
- State the final target.
- Ask what must be known immediately before it.
- Continue backward until reaching given information.
- Calculate forward through the dependency chain.
- Label every partial result.
- Check each state before using it downstream.
31. Practice laboratory A
- 800 items: 25% sold, then 1/3 of remainder packed. Find unpacked remainder.
- 1000 items: 8% rejected, accepted packed 36 per carton. How many cartons needed for all accepted items?
- A=90, B=50, internal transfer until A has 10 more. Find transfer using a final-state subgoal.
- Tank base 25×16 cm; volume rises 2400 cm³. Find water-level rise.
- 24 items per group leaves 36; 27 per group is short 9. Find number of groups.
32. Practice laboratory B
- After a 20% discount, a $300 item receives an additional charge equal to 5% of the discounted price. Find final price.
- A composite rectangle has total width 18 cm; two aligned horizontal sections are 7 cm and 5 cm. Find the missing aligned length before calculating any area.
- A rate is 16 items/min. A first stage makes 240 items; then machine runs 7 more minutes. Find final output.
- Two products: 6 items cost $15, 10 items cost $24. Find the two unit-price subgoals and compare.
- A triangle area is 84 cm² with base 14 cm. Find height, then find area of a rectangle with same base and height.
33. Answers
1. Remainder after sale=600; packed=200; unpacked=400.
2. Rejected=80; accepted=920; 920÷36=25 r20 → 26 cartons.
3. Final pair from total 140 and difference 10 is 75/65; transfer=15.
4. Base area=400 cm²; rise=2400÷400=6 cm.
5. Gap=45; per-group difference=3; groups=15.
6. Discounted=240; charge=12; final=$252.
7. Missing length=18−7−5=6 cm.
8. First-stage time=15 min; extra output=112; final=352 items.
9. $2.50/item versus $2.40/item; second product has lower unit price.
10. 84=1/2×14×h → h=12. Rectangle area=14×12=168 cm².
34. Full dependency-chain problem
A warehouse receives 2400 bottles. Five percent are damaged. Of the usable bottles, 3/8 are sent to Branch A. The remainder is packed into cartons of 45. How many cartons are needed if every remaining bottle must be packed?
Plan backward from target:
cartons ← bottles to pack ← usable bottles after Branch A ← usable bottles after damage ← original 2400.
Calculate forward:
- Damaged=5%×2400=120.
- Usable=2280.
- Sent to A=3/8×2280=855.
- Remaining to pack=2280−855=1425.
- 1425÷45=31 remainder 30.
Thirty bottles would remain after 31 full cartons, so 32 cartons are needed if every bottle must be packed.
Every partial result has a distinct job and supplies the next dependency.
35. Final checkpoint
A strong Primary 5 learner can identify the final target, build a backward dependency plan, choose meaningful subgoals, calculate forward through labelled states, use partial results as bounds and checkpoints, solve parallel branches when needed, and avoid unnecessary intermediate calculations that do not contribute to the target.
Wintour House V1.0 · CivDJ · eduKate Publishing: decompose only along real dependencies, turn each partial result into a stable state, and let the solution advance through a chain of necessary truths rather than a pile of calculations.
