Primary 6 Maths: PSLE Problem-Solving

Updated

The hardest PSLE Maths questions test heuristics: assumption method, working backwards, spotting patterns. What they look like and how to practise them without tears.

Most schools typically cover this topic in Term 3, around week 5 — though every school sets its own sequence.

Every PSLE Maths paper ends with a handful of questions that don't belong to any single topic. They're the problem-solving questions — the ones tuition centres build entire courses around — and they test heuristics: the assumption (supposition) method, working backwards, drawing systematic lists, spotting number patterns, and restructuring unfamiliar problems into familiar ones. "PSLE Problem-Solving" is a cross-topic skill in our own practice taxonomy, not a separate MOE syllabus strand — the heuristics it covers are meant to be applied across whichever topic a question draws on.

Two honest facts for parents. First: these questions are a minority of the paper — the bulk of PSLE marks still come from solid topic mastery, which is why chasing heuristics before the foundations are secure is backwards. Most P6 average questions are a good example of the foundation coming first: they are solved by working back from the average to the total, and that step has to be automatic before any heuristic helps. Second: heuristics can be learned. They look like flashes of insight; they're actually a small library of reusable moves.

The heuristics that matter most

The mistakes to watch for

Heuristics as diagrams

Assumption method: legs if every animal were a chicken A bar for the assumed 60 legs (all chickens) next to a taller bar for the actual 84 legs, with the 24-leg gap marked and labelled as the shortfall to be closed by swapping chickens for goats. Assumed: 60 legs Actual: 84 legs gap = 24 legs ÷ 2 per goat = 12 goats
Each chicken "swapped" for a goat adds 2 legs to the assumed total — the 24-leg gap divided by 2 gives the number of goats.
Working backwards through Devi's stickers A left-to-right chain of three labelled boxes: 23 stickers now, reversing the plus-8 step to 15, then reversing the half-kept step by doubling to 30 stickers at first. Now: 23 ←−8 15 ←×2 First: 30 Reverse each step in order, right to left, to undo the story.
Working backwards reverses every operation, in reverse order, starting from the known end state.

Easy: single-heuristic warm-ups

Easy: working backwards

Devi had some stickers. She gave half of them to her brother, then bought 8 more. She now has 23 stickers. How many stickers did she have at first?

Show the worked solution

Reverse the steps from the end.

Before buying 8: 23 − 8 = 15 stickers.

15 is the half she kept, so at first she had 15 × 2 = 30 stickers.

Check: 30 → gives 15 away → 15 → buys 8 → 23. ✓

Easy: systematic listing

Using the digits 3, 5 and 7 exactly once each, list all the possible 2-digit numbers you can form.

Show the worked solution

List systematically by first digit: starting with 3 → 35, 37; starting with 5 → 53, 57; starting with 7 → 73, 75.

35, 37, 53, 57, 73, 75 — six numbers.

Easy: spotting a number pattern

Look at the pattern 4, 7, 10, 13, … Find the 6th term.

Show the worked solution

Each term increases by 3. The terms are 4, 7, 10, 13, 16, 19.

6th term = 19.

Easy: a simple assumption

A farm has 10 animals — all either goats or ducks — with 28 legs altogether. How many goats are there?

Show the worked solution

Assume all 10 are ducks: 10 × 2 = 20 legs. Shortfall: 28 − 20 = 8.

Each duck swapped for a goat adds 4 − 2 = 2 legs, so goats = 8 ÷ 2 = 4 goats.

Check: 4 goats (16 legs) + 6 ducks (12 legs) = 10 animals, 28 legs. ✓

Exam: the assumption method and working backwards

Exam: chickens and goats

A farmer keeps chickens and goats. The animals have 30 heads and 84 legs altogether. How many goats are there?

Show the worked solution

Assume all 30 animals are chickens: 30 × 2 = 60 legs.

Shortfall: 84 − 60 = 24 legs.

Each goat swapped in adds 4 − 2 = 2 legs.

Goats = 24 ÷ 2 = 12 goats.

Check: 12 goats (48 legs) + 18 chickens (36 legs) = 30 heads, 84 legs. ✓

Exam: working backwards through two steps

Raj thought of a number. He multiplied it by 3, then added 7, and got 40. What was his original number?

Show the worked solution

Reverse from the end: before adding 7, the value was 40 − 7 = 33.

Before multiplying by 3, the value was 33 ÷ 3 = 11.

Check: 11 × 3 = 33, +7 = 40. ✓

Exam: systematic listing with a constraint

Using only the digits 2, 4 and 6, how many 3-digit even numbers can be formed if each digit can be used more than once?

Show the worked solution

Every digit available (2, 4, 6) is already even, so any arrangement is an even number.

Each of the 3 positions can be filled in 3 ways: 3 × 3 × 3 = 27 numbers.

Exam: before-after with an unchanged total

Kumar and Devi had $90 altogether. After Kumar gave $10 to Devi, Kumar had half as much money as Devi. How much money did Kumar have at first?

Show the worked solution

The total stays $90 throughout. Let Kumar's amount after giving be k, so Devi's is 2k.

k + 2k = 90, so 3k = 90, and k = 30.

Kumar had $10 more before giving it away: 30 + 10 = $40.

Check: Kumar $40, Devi $50 at first (total $90). After: Kumar $30, Devi $60 — Kumar is half of Devi. ✓

Exam: assumption with money

Amirah buys 15 items, some pens at $2 each and some pencils at $1 each, spending $23 in total. How many pens did she buy?

Show the worked solution

Assume all 15 items are pencils: 15 × $1 = $15. Shortfall: $23 − $15 = $8.

Each pencil swapped for a pen adds $2 − $1 = $1, so pens = 8 ÷ 1 = 8 pens.

Check: 8 pens ($16) + 7 pencils ($7) = 15 items, $23. ✓

PSLE challenge: multi-step heuristics

PSLE challenge: three-way comparison

Aiman has 3 times as many marbles as Bala. Chen has 12 more marbles than Bala. Altogether they have 132 marbles. How many marbles does Chen have?

Show the worked solution

Let Bala = b. Aiman = 3b, Chen = b + 12.

b + 3b + (b + 12) = 132, so 5b + 12 = 132, giving 5b = 120, b = 24.

Chen = 24 + 12 = 36 marbles.

Check: Aiman 72, Bala 24, Chen 36; total 132. ✓

PSLE challenge: a doubling pattern worked backwards

In a sequence, each term after the first is double the term before it. The 4th term is 96. What is the 1st term?

Show the worked solution

Working backwards, halve three times: 96 → 48 → 24 → 12.

1st term = 12.

Check: 12, 24, 48, 96 — each term doubles. ✓

PSLE challenge: the assumption method with two vehicle types

A car park has motorcycles and cars only, 50 vehicles in total, with 168 wheels altogether. Motorcycles have 2 wheels and cars have 4 wheels. How many cars are there?

Show the worked solution

Assume all 50 are motorcycles: 50 × 2 = 100 wheels. Shortfall: 168 − 100 = 68.

Each motorcycle swapped for a car adds 4 − 2 = 2 wheels, so cars = 68 ÷ 2 = 34 cars.

Check: 34 cars (136 wheels) + 16 motorcycles (32 wheels) = 50 vehicles, 168 wheels. ✓

How to practise this topic

Heuristics stick when they're practised as templates across many surface stories — chickens and goats today, cars and motorcycles tomorrow — so the child learns to see the structure underneath. Practise one heuristic at a time, then mix them so the real skill (choosing the right move) gets exercised. And keep perspective: if topic-level gaps exist in fractions, ratio or circles, close those first — they're worth more marks.

What this topic covers

In the MOE Primary Mathematics syllabus, Primary 6 PSLE Problem-Solving includes the concepts below. (P6-M-12 is Test Paper's own topic code, not an MOE reference.)

Before this topic

These topics feed into PSLE Problem-Solving — if this one wobbles, check them first: Primary 5 Word Problems · Primary 6 Fractions · Primary 6 Decimals · Primary 6 Percentage · Primary 6 Ratio · Primary 6 Algebra.

Where to go next

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