Primary 5 Maths: Word Problems

Updated

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

The mistakes to watch for

Bar models for the two core structures

Adam and Ben, 3 times as much Two stacked bars of equal unit width. Ben's bar is 1 unit long; Adam's bar is 3 units long, drawn directly below Ben's for comparison, together making 4 units worth $120. Ben Adam 1 unit 3 units 4 units altogether = $120, so 1 unit = $30
Ben's bar is 1 unit; Adam's is 3 units drawn to the same scale — 4 units in total represent $120.
Siti and Raj, before and after Two rows show Siti's bar shrinking from 2 units before to 1 unit after spending $18, while Raj's bar of 1 unit stays the same in both rows. Before Siti: 2 units Raj After Siti Raj Siti's bar shrinks by $18 (1 unit) to match Raj's unchanged bar: 1 unit = $18.
Raj's amount never changes — that's what makes "before and after" solvable: only Siti's bar moves.

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.)

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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