Primary 5 Maths: Word Problems
P5 word problems test reading and model drawing as much as arithmetic. The structures to master, why children freeze on them, and worked examples with bar models.
Most schools typically cover this topic in Term 4, around weeks 4–6 — though every school sets its own sequence.
Look at any P5 Maths paper and count where the marks sit: the long word problems at the back of Paper 2 carry 3–5 marks each. A child can be fluent in every calculation and still lose a grade's worth of marks here — because word problems test something extra: turning a paragraph of English into a mathematical structure.
In Singapore Maths, that structure is usually a bar model. P5 is the year the models get genuinely hard: two-variable comparisons, before-and-after situations, and problems that chain fractions, percentages and whole numbers together. "Word Problems" is a cross-topic skill page in our own practice taxonomy — it draws on whichever topic a question is built from, rather than being a syllabus strand of its own.
The structures your child must recognise
- Part-whole and comparison models. "A and B have $120 altogether. A has 3 times as much as B." Two bars, four equal units. This is the bread and butter — it must be automatic before the harder structures make sense.
- Before-and-after. "Siti had twice as much as Raj. After Siti spent $18, they had the same amount." The model changes between two points in time; children who try to hold it in their head instead of drawing usually fail.
- Remainder chains. Fraction-of-remainder problems (covered in depth in P5 Fractions).
- Excess-and-shortage. "If she buys 5 pens she has $2 left; if she buys 7 she is short of $4." Rare in drills, regular in exams.
The mistakes to watch for
- Freezing instead of drawing. The most common failure isn't a wrong model — it's no model. Children who feel a problem is hard skip the diagram, exactly when they need it most.
- Units that don't match the story. Assigning "1 unit" to Adam when the problem says Adam has 3 times as much as Ben makes the arithmetic ugly. Unit choice is a skill worth practising on its own.
- Answering the intermediate number. Solving for one unit and writing that down, when the question asked for the total.
Bar models for the two core structures
Easy: single-step models
Easy: sharing in a given multiple
45 sweets are shared between two children so that one has 4 times as many as the other. How many does the child with fewer sweets have?
Show the worked solution
1 unit + 4 units = 5 units = 45 sweets, so 1 unit = 9 sweets.
Easy: a difference and a total
Jane has $50 more than Kim. Together they have $250. How much does Kim have?
Show the worked solution
If Kim has $x, Jane has $(x + 50). Together: x + (x + 50) = 250.
2x + 50 = 250, so 2x = 200, and x = $100.
Check: Kim $100, Jane $150, total $250. ✓
Easy: times as many, given the total
Peter has 3 times as many stickers as Wei Jie. Together they have 96 stickers. How many does Peter have?
Show the worked solution
1 unit + 3 units = 4 units = 96 stickers, so 1 unit = 24.
Peter = 3 units = 72 stickers.
Easy: straightforward subtraction in context
A shop had 320 books. It sold 145 books in the morning and 98 books in the afternoon. How many books are left?
Show the worked solution
320 − 145 − 98 = 77 books.
Exam: comparison and before-after models
Exam: altogether and times as much
Adam and Ben have $120 altogether. Adam has 3 times as much money as Ben. How much money does Ben have?
Show the worked solution
Draw Ben as 1 unit and Adam as 3 units. Together: 4 units.
4 units = $120, so 1 unit = $120 ÷ 4 = $30.
Check: Ben $30, Adam $90 — Adam is 3 times Ben, total $120. ✓
Exam: before and after
Siti had twice as much money as Raj. After Siti spent $18, both of them had the same amount of money. How much money did Raj have?
Show the worked solution
Before: Siti = 2 units, Raj = 1 unit. Raj's money never changes.
After spending $18, Siti equals Raj: 2 units − $18 = 1 unit.
So 1 unit = $18. Raj had $18.
Check: Siti had $36; after spending $18 she has $18 — same as Raj. ✓
Exam: excess and shortage
If Mrs Wong buys 5 identical storybooks, she has $8 left. If she buys 7 of the same storybooks, she is short of $10. How much money does she have?
Show the worked solution
The 2 extra books cost the $8 she had left plus the $10 she was short: $8 + $10 = $18 for 2 books, so each book costs $9.
Her money = 5 × $9 + $8 = $53.
Check: 7 books cost $63; she has $53, which is $10 short. ✓
Exam: comparison with a change
Container A has 3 times as much water as Container B. After 15 litres is poured from A to B, both containers have the same amount. How much water did Container A have at first?
Show the worked solution
Let B = 1 unit, so A = 3 units. After pouring: A has (3 units − 15), B has (1 unit + 15).
These are equal: 3 units − 15 = 1 unit + 15, so 2 units = 30, and 1 unit = 15.
Container A at first = 3 × 15 = 45 litres.
Check: A 45, B 15; after pouring, A 30, B 30 — equal. ✓
Exam: part-whole with a difference
The total mass of a box of apples and a box of oranges is 18 kg. The apples are 4 kg heavier than the oranges. Find the mass of the oranges.
Show the worked solution
Let the oranges weigh o kg, so the apples weigh (o + 4) kg.
o + (o + 4) = 18, so 2o = 14, and o = 7 kg.
PSLE challenge: multi-step structures
PSLE challenge: three quantities compared
Ali, Beng and Chandra have some money. Ali has twice as much as Beng. Beng has 3 times as much as Chandra. Altogether they have $150. How much does Chandra have?
Show the worked solution
Let Chandra = 1 unit, so Beng = 3 units, and Ali = 2 × 3 = 6 units.
Total: 1 + 3 + 6 = 10 units = $150, so 1 unit = $15.
PSLE challenge: before-after with a transfer
Rui and Mei together have 84 marbles. After Rui gives 12 marbles to Mei, Mei has twice as many marbles as Rui. How many marbles did Rui have at first?
Show the worked solution
Let Rui have R marbles at first, so Mei has (84 − R).
After the transfer: Rui has (R − 12); Mei has (84 − R + 12) = (96 − R).
Mei is now twice Rui: 96 − R = 2(R − 12), so 96 − R = 2R − 24, giving 3R = 120, R = 40.
Check: Rui 40, Mei 44 at first. After: Rui 28, Mei 56 — Mei is twice Rui. ✓
PSLE challenge: remainder chain with two sales
A shopkeeper had some oranges. He sold 2/5 of them in the morning, and 30 more in the afternoon, leaving him with 1/5 of his original number of oranges. How many oranges did he have at first?
Show the worked solution
Let the original number be T. After the morning sale, T − 2/5 T = 3/5 T remain.
After selling 30 more: 3/5 T − 30 = 1/5 T (what's left).
3/5 T − 1/5 T = 30, so 2/5 T = 30, and T = 75 oranges.
Check: 75 → morning sale 30, remainder 45 → afternoon sale 30, left 15 = 1/5 of 75. ✓
How to practise this topic
Word-problem skill is diagnostic gold: a wrong answer tells you which step broke — reading, model, or arithmetic. Practice should do the same. Rather than grinding 50 mixed problems, work on one structure at a time until the model is automatic, then interleave structures so your child learns to choose the model, not just execute it.
What this topic covers
In the MOE Primary Mathematics syllabus, Primary 5 Word Problems includes the concepts below. (P5-M-12 is Test Paper's own topic code, not an MOE reference.)
- Multi-step word problems
- Before-and-after concept
- Model drawing (bar models)
- Heuristic problem-solving strategies
Before this topic
These topics feed into Word Problems — if this one wobbles, check them first: Primary 4 Word Problems · Primary 5 Fractions · Primary 5 Decimals · Primary 5 Percentage.
Where to go next
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