PRIMARY 5 MATHEMATICS LEARNING GUIDE · BATCH 2 · GUIDE 5
A word problem is not an arithmetic problem with extra sentences. It is a translation problem. The learner must turn language into quantities, relationships, representations and operations, then return the numerical result to the situation that created it.
At Primary 5, the arithmetic may involve whole numbers, fractions, decimals, percentages, rate, measurement and geometry. The difficulty often comes from deciding which relationship controls each stage. A student can know every operation individually and still fail a multi-step problem if the wrong quantities are connected.
This guide develops a stable routine for reading, modelling, calculating and checking. Bar models are used when they make a relationship visible; tables, equations, labelled diagrams and unit statements are used when they are better tools. The aim is not to force one representation onto every problem. The aim is to make the hidden structure inspectable.
Series route: return to the Primary 5 Mathematics Learning Hub. Companion guides: Non-Routine Problems, Heuristics & Strategy Choice · Estimation, Calculator Control & Answer Verification · Mixed Practice, Error Analysis & PSLE Runway.
Curriculum boundary: this is an independent learning companion aligned to upper-primary problem solving in the Singapore Primary Mathematics context. For the official syllabus framework, see the MOE Primary Mathematics Syllabus. The examples below are original teaching problems.
1. The first question is not “Which operation?”
Students are often taught to scan for words such as altogether, left, each, of or more. These words can help, but they do not determine the operation reliably.
Compare two statements:
- Ben has 12 more stickers than Ali.
- Ben has 12 times as many stickers as Ali.
The first is additive. The second is multiplicative. The same word “more” can appear in percentage or comparison problems where the reference quantity matters.
The stronger opening question is: What is the relationship between the quantities? Ask whether one quantity is a total, a difference, a part of a whole, a repeated group, a rate, a fraction of another quantity, a percentage of a reference whole, or a geometric measure.
2. Read the question in layers
A useful reading routine has four passes:
- Situation: What is happening?
- Quantities: What numbers and units are given?
- Relationships: How are the quantities connected?
- Target: What exactly must be found?
Do not calculate during the first sentence unless the relationship is already certain. Many wrong answers begin when the student performs an available operation before reading the final condition.
For example: “A shop has 480 notebooks. Three eighths are blue. Twenty-five percent of the blue notebooks are sold. How many blue notebooks remain?” The 25% does not apply to 480. It applies to the blue subset. The learner must preserve the changing reference whole.
3. Units separate quantities that look numerically similar
In “8 boxes contain 36 pencils each”, 8 counts boxes while 36 is pencils per box. Multiplication gives pencils because:
boxes × pencils per box = pencils.
If the question then says “108 pencils are given away”, 108 belongs to the pencil quantity, not the box quantity. Writing units beside numbers prevents the learner from subtracting 108 from 8 or combining quantities that do not share a meaningful relationship.
Units are not decoration added at the end. They are a reasoning tool used before calculation.
4. A bar model is a picture of a relationship
A bar model represents quantities with lengths. It is especially useful for part–whole and comparison relationships.
Suppose Maya has 72 stickers and Ravi has 18 fewer. Draw Maya’s bar as 72. Ravi’s bar is shorter by a segment of 18. Therefore Ravi has 72 − 18 = 54 stickers.
Now reverse the problem: Ravi has 54 stickers, which is 18 fewer than Maya. The same model shows that Maya has 54 + 18 = 72.
The model is not tied to subtraction. It exposes the relationship so the unknown can be found in whichever direction the problem requires.
5. Part–whole models
If a total is divided into known parts, use one full bar with segments.
Example: A class collected 360 cans. Primary 5A collected 145 cans and Primary 5B collected 127 cans. The rest were collected by Primary 5C. How many did Primary 5C collect?
The whole is 360. Two known parts are 145 and 127. The unknown part is:
360 − 145 − 127 = 88 cans.
The bar prevents a common mistake: adding all three numbers simply because they appear in the question. Once 360 is identified as the whole, the known parts must be removed to reveal the missing part.
6. Comparison models
Comparison models use aligned bars. One bar represents the smaller or reference quantity; the other includes the shared amount plus a difference or scale relationship.
Example: Hana has 240 beads. Jia has 35% more beads than Hana. How many beads does Jia have?
Hana represents 100% = 240 beads. Jia represents 135% of Hana. One route is:
35% of 240 = 84, so Jia has 240 + 84 = 324 beads.
Another route is 1.35 × 240 = 324. The bar makes the reference whole explicit: the 35% increase is measured against Hana’s quantity.
7. Equal-part models
Fractions and later ratio reasoning often become clearer when a whole is divided into equal units.
Example: Three fifths of a tank’s water is 180 litres. How much water is in the full tank?
Three equal parts correspond to 180 litres, so one part is 180 ÷ 3 = 60 litres. Five parts give 60 × 5 = 300 litres.
A common error is to multiply 180 by 3/5 again. The model reveals that 180 is already the three-fifths part, not the whole.
8. Before-and-after models
Some problems change a quantity over time. Separate the states.
Example: A shop had some pens. It sold 120 pens in the morning and received 80 new pens in the afternoon. It then had 350 pens. How many did it have at first?
Work backward from the final state. Before receiving 80, it had 350 − 80 = 270 pens. Before selling 120, it had 270 + 120 = 390 pens.
A timeline or before–change–after bar is often more useful than a single bar because the problem is about state transitions.
9. Tables are better when repeated rates are central
A bar model is not always the best representation. If a pump moves 18 litres per minute, a table can expose the constant rate:
| Minutes | Litres |
|---|---|
| 1 | 18 |
| 5 | 90 |
| 10 | 180 |
| 14 | 252 |
The constant multiplier is 18. The table makes scaling direct and allows the student to see whether a proposed answer preserves the same rate.
Choose the cheapest representation that keeps the relationship visible.
10. Labelled diagrams are best for geometry
If a problem asks for an angle, area or volume, the geometry itself is the representation. Do not force a bar model onto a triangle.
Mark known angles, parallel sides, equal sides, right angles and dimensions. Write the required angle or length with a question mark. Then use only the properties guaranteed by the diagram and labels.
Representation choice is part of problem solving. A student who can switch from bar model to table to geometric diagram is stronger than a student who has learned one visual routine mechanically.
11. Multi-step problems should create named intermediate quantities
Consider this original problem:
A warehouse receives 48 cartons. Each carton contains 125 packets. Three tenths of the packets are sent to Store A. Twenty percent of the remaining packets are sent to Store B. How many packets remain?
Step 1: total packets = 48 × 125 = 6,000.
Step 2: Store A receives 3/10 × 6,000 = 1,800. Remaining = 4,200.
Step 3: Store B receives 20% of 4,200 = 840. Final remaining = 3,360 packets.
The key is naming the changing total after each step. The 20% applies to 4,200, not to the original 6,000.
12. The reference whole can change
Percentage and fraction problems often fail because the learner keeps using the first total even after the situation has changed.
In the warehouse example, the original 6,000 packets are the whole for the three-tenths calculation. After Store A receives its share, the remaining 4,200 packets become the whole for the next 20% calculation.
Write “100% = 4,200” before the second percentage if the transition is causing errors. A short label can prevent a long wrong solution.
13. Inverse thinking solves reverse problems
A reverse problem gives the result and asks for an earlier quantity.
Example: After a 20% discount, a bag costs $72. What was the original price?
After a 20% discount, the sale price is 80% of the original. So 80% = $72. One percent is $72 ÷ 80 = $0.90. One hundred percent is $90.
Equivalently, original × 0.8 = 72, so original = 72 ÷ 0.8 = $90.
Subtracting 20% of $72 would answer a different question because $72 is the discounted price, not the reference whole.
14. Work backward when forward steps are reversible
Suppose a number is multiplied by 4, then 30 is added, giving 270. Reverse the steps in reverse order: 270 − 30 = 240, then 240 ÷ 4 = 60.
In story form: “A machine quadruples a quantity, then adds 30 units.” The same logic applies.
Working backward is powerful because many multi-step processes are reversible. But be careful when a step loses information or includes whole-object constraints; then a simple inverse may not uniquely recover the earlier state.
15. Do not round until the problem allows it
Premature rounding can change a later decision. If one unit requires 2.4 m of cable and 17 units are made, exact requirement is 40.8 m. Rounding 2.4 m to 2 m before multiplying gives 34 m, which may lead to ordering too little material.
Keep exact fractions or decimals through intermediate steps where practical. Round only when the question requests it or when the context requires a whole-object decision.
Estimation is still useful, but keep it separate from the exact calculation.
16. Whole-object answers require interpretation
If 169 students travel in buses that hold 24 students each, 169 ÷ 24 is a little more than 7. Seven buses hold only 168 students, so 8 buses are required.
Rounding to the nearest whole number would give 7, which fails the capacity condition. This is not an arithmetic issue. The quotient must be interpreted against the real constraint.
In another problem, 7.04 kg may be a perfectly valid measured amount. The unit and context determine whether a fractional answer is acceptable.
17. A correct calculation can answer the wrong question
Example: A jacket costs $160 and is discounted by 25%. A student calculates 25% of 160 = $40 and stops.
$40 is correct as the discount amount. If the question asks for the sale price, the final answer is 160 − 40 = $120.
After each calculation, name what the number means. “This is the discount.” “This is the remaining water.” “This is the rate per minute.” “This is the number of full buses.” Naming prevents an intermediate answer from being mistaken for the target.
18. One problem can support several representations
Problem: Three fifths of a group are girls. There are 18 more girls than boys. How many children are in the group?
A bar model divides the total into five equal parts: girls occupy three parts, boys two parts. The difference is one part, and that one part equals 18. Therefore the total is five parts = 5 × 18 = 90 children.
An equation route is possible too. Let one part be x. Then girls = 3x, boys = 2x, and 3x − 2x = 18, so x = 18 and total = 5x = 90.
The model and equation encode the same structure. Learning improves when the student can see the equivalence.
19. Test the solution by returning it to the story
For the previous problem, 90 children means 3/5 × 90 = 54 girls and 2/5 × 90 = 36 boys. The difference is 18. The answer recreates all stated conditions.
This return check is stronger than repeating the same arithmetic. It asks whether the proposed answer generates the original situation.
For a rate problem, multiply rate by number of units to recover the total. For a percentage problem, apply the percentage relationship. For geometry, substitute the found dimension back into the formula or angle sum.
20. Error map for word problems
| Visible failure | Likely first cause | Repair question |
|---|---|---|
| All numbers combined immediately | Target and relationships not identified | What does each number represent? |
| Correct percentage applied to wrong quantity | Reference whole lost | What is 100% at this stage? |
| Correct intermediate number reported as final answer | Target not revisited | What did the question ask for? |
| Bar model drawn but not used | Representation became ritual | Which segment is unknown and how is it related? |
| Fractional number of buses accepted | Context not interpreted | Can the object be divided in this situation? |
| Units mixed | Quantity types not preserved | Are these numbers expressed in compatible units? |
21. Practice laboratory
- A school has 960 books. Three eighths are fiction. How many are non-fiction?
- Mei has 84 stickers. Ravi has 27 fewer. How many stickers do they have altogether?
- Five identical boxes contain 360 marbles. Two boxes are given away. How many marbles remain?
- A tank contains 480 ℓ. One quarter is used. Then 20% of the remaining water is used. How much remains?
- After a 25% discount, a toy costs $54. Find the original price.
- Eight tickets cost $120. At the same rate, find the cost of 14 tickets.
- A printer produces 240 pages per hour for 6 hours. Fifteen percent are rejected. How many usable pages remain?
- Three fifths of a group are boys. There are 24 more boys than girls. Find the total group size.
- A hall has 32 rows of 48 seats. 185 seats are blocked. All remaining seats are sold at $15 each. Find the ticket revenue.
- A bus holds 42 students. How many buses are needed for 257 students?
- A 20 cm by 12 cm rectangle has a triangle of base 8 cm and height 5 cm removed. Find the remaining area.
- A tank base is 25 cm by 16 cm. Its water level rises 6 cm. Find the added volume in litres.
22. Explained answers
1. Fiction = 3/8 × 960 = 360. Non-fiction = 600 books.
2. Ravi = 84 − 27 = 57. Together = 141 stickers.
3. One box = 360 ÷ 5 = 72. Three boxes remain, so 216 marbles.
4. First use = 120; remaining = 360. Second use = 72; final = 288 ℓ.
5. $54 is 75% of original. Original = 54 ÷ 0.75 = $72.
6. Unit price = 120 ÷ 8 = $15. Fourteen tickets cost $210.
7. Total = 240 × 6 = 1440. Rejected = 216. Usable = 1224 pages.
8. Boys:girls corresponds to 3 parts:2 parts. Difference = 1 part = 24. Total = 5 parts = 120.
9. Total seats = 1536. Sellable = 1351. Revenue = 1351 × 15 = $20,265.
10. Six buses hold 252, which is not enough. 7 buses are required.
11. Rectangle = 240 cm². Triangle = 20 cm². Remaining = 220 cm².
12. Base area = 400 cm². Added volume = 2400 cm³ = 2.4 ℓ.
23. A full PSLE-runway style mixed problem
Problem: A shop receives 600 bottles. Forty percent are orange drink. The rest are apple drink. During the morning, one quarter of the orange bottles and one fifth of the apple bottles are sold. In the afternoon, 96 more bottles are sold. At the end of the day, how many bottles remain?
Orange = 40% × 600 = 240. Apple = 360.
Morning orange sold = 1/4 × 240 = 60. Morning apple sold = 1/5 × 360 = 72. Total morning sold = 132.
Remaining after morning = 600 − 132 = 468.
After afternoon sales: 468 − 96 = 372 bottles.
Check by totals: sold altogether = 60 + 72 + 96 = 228. 600 − 228 = 372. Two different aggregation routes agree.
24. Teaching sequence for stronger transfer
Start with one relationship at a time and require the student to name it. Then change the surface details while preserving the same structure. Next, place the problem beside a different structure so the student must choose. Finally, use a multi-step question where one intermediate quantity becomes the input to the next relationship.
After an error, do not simply show the full correct solution. Identify the first unstable decision. If the learner chose the wrong whole for a percentage, repair that choice with a smaller contrast problem. If the learner mixed units, repair unit labelling. If the model was drawn incorrectly, rebuild the relationship before returning to the original numbers.
Transfer grows when the learner can explain why the method applies to a changed case.
25. Final checkpoint
A strong Primary 5 word-problem solver can identify the target, label quantities and units, choose a useful representation, preserve changing reference wholes, create named intermediate quantities, interpret whole-object constraints and return the answer to the original conditions.
Continue to Primary 5 Mathematics Learning Guide | Non-Routine Problems, Heuristics & Strategy Choice.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Translate language into mathematical objects, preserve the relationship through each state change, test the answer by returning it to the story, and keep the representation only when it reduces uncertainty.
