Rows of Primary 6 pupils in white school uniforms seen from behind in a Singapore primary-school examination hall during the PSLE Mathematics paper, one open paper in the foreground with a word problem ringed in red.

PSLE Math · 11 min read

The 2026 PSLE Math Paper: the Questions That Trapped Kids, and the Method That Solves Them

IllumiTutor Team·25 September 2026

I spent part of today speaking to a few parents and students who had just come out of the 2026 PSLE Math paper. Almost every conversation reached the same question: the last one. A matchstick pattern, three marks, and a stretch of blank space where the working should have been. "Is this even in the syllabus?" one mother asked me.

It is. And that is the thing worth saying plainly about this year's paper: almost nothing in it was outside what a P6 child has been taught. The difficulty sat somewhere else — in reading the exact wording, and in reaching for the right method instead of the nearest number.

Between those conversations and the questions the children described to me, a clear picture of the paper has come together. I am not going to score the whole thing for you; plenty of centres have done that. What I want to do instead is take four questions that separated the careful child from the rushed one, and show, step by step, how a teacher would solve each. Because the method is the part you can actually practise at home.

What this paper was really testing

The arithmetic load was fair. There was no question that needed a trick you could not have learned in class. What the paper leaned on, again and again, was the set of moves we drill all year: a part-whole bar model, a before-after model, grouping with a limiting factor, and finding a repeating pattern. Percentages, fractions of a remainder, and matchsticks — all familiar, none exotic.

That is quietly good news for parents. A paper that rewards method is a paper you can prepare for. Rote practice alone would not have carried a child through it, but method, taught properly, would.

Paper 1, Question 17: the "more than" that cost a mark

This is a supposition question, and the arithmetic is short. Start by imagining the worst case: all 25 days over two hours. That is 25 × 3 = 75 points deducted, so a score of −75.

Each day you flip from "over" to "under" changes the score by 7 points — the 3-point deduction disappears and a 4-point gain appears. To climb from −75 past 65 is a climb of more than 140 points, and 140 ÷ 7 = 20.

Twenty days lands him on exactly 65. But the question says more than 65, and exactly 65 earns nothing. So the answer is 21.

Paper 2, Question 11: a percentage hidden in a remainder

The word doing the damage here is remaining. This is a before-after model, and it is worth drawing rather than doing in your head.

Take A's original stickers as one whole — 100 units, one for each percent. A uses 30, so 70 remains. The gift is 20% of that 70:

  • 20% of 70 = 14

So A gave B 14% of his original stickers — that is part (a). A is left with 70 − 14 = 56%.

Now the equal ending does the rest. B finishes with his original 63 stickers plus the 14% gift. For the two to be equal, those 63 stickers must be the difference between 56% and 14%, which is 42%. So 42% of A's stickers is 63:

  • 1% = 63 ÷ 42 = 1.5
  • 100% = 150 stickers
A before-after bar model: A at first is 30 percent used, 14 percent given away and 56 percent kept; A now is 56 percent; B now is 63 stickers plus the 14 percent gift, with the 63 labelled as 42 percent.
A at first, split into used, given and kept. Because they end equal, B's 63 stickers are the 42% between the 56% and the 14%.

The common wrong answer is 20% — a child takes 20% of the original instead of 20% of what was left. Same reading error as Question 17, wearing a different costume.

Paper 2, Question 14: fractions, then a limit

Two models stack here, which is why this one shakes students who can only hold one idea at a time.

First, a part-whole model. Eight equal parts, 720 in total, so 1 part is 90. White is 5 parts, red is 3:

  • White = 5 × 90 = 450
  • Red = 3 × 90 = 270
A single bar of 720 beads split into five equal white parts and three equal red parts, with each part marked.
One bar, eight equal parts: five white and three red.

Then a before-after subtraction. Of the white, 130 are used; of the red, 40:

  • White left = 450 − 130 = 320
  • Red left = 270 − 40 = 230

That is part (a). For part (b), find what one necklace costs in each colour: 130 ÷ 10 = 13 white and 40 ÷ 10 = 4 red.

A bar for one necklace split into 13 white beads and 4 red beads, together labelled 17 beads.
One necklace takes 13 white and 4 red — seventeen beads, but the colours cannot be swapped.

Now the phrase as many as possible is doing the real work. You cannot run out of one colour and keep going, so check each colour separately:

  • White: 320 ÷ 13 = 24 necklaces, 8 beads to spare
  • Red: 230 ÷ 4 = 57 necklaces, 2 beads to spare

White runs out first, so she can make only 24 more necklaces. That uses 24 × 13 = 312 white and 24 × 4 = 96 red. Add the leftovers: 8 white + 134 red = 142 beads.

Paper 2, Question 15: the pattern that came back

Four matchstick figures growing in size, labelled Figure 1 to Figure 4, each built from a row of horizontal blue sticks joined by vertical orange sticks.
Figures 1 to 4. The shape grows by a row each time — but it is the stick count that gives the rule.
Fig. No123456
Sticks8101315

Pattern questions reward one habit and punish its opposite. The habit is writing the numbers down and looking for the rule. The punishment is trying to picture Figure 45 and count.

Lay the counts in a row: 8, 10, 13, 15. The jumps are +2, +3, +2, +3. It is a two-step repeating block: every two figures, the number of sticks grows by 5. Figures 5 and 6 are therefore 18 and 20 — part (a).

For part (b), do not count. From Figure 1 to Figure 45 is 44 steps, which is 22 blocks of "+5":

  • 22 × 5 = 110
  • 8 + 110 = 118 sticks

The thread through all four

Put the four side by side and the paper's logic is clear. Question 17 turned on the exact wording of a comparison. Question 11 turned on of the remainder. Question 14 turned on a limiting factor. Question 15 turned on finding a repeating rule rather than counting.

Four different topics, one shared demand: read the question precisely, decide which method it is asking for, and draw it. Not one of them needed algebra. Not one of them needed a formula your child has not already met. This is the same story we tell in our guide to solving problem sums with the model method: the strong P6 child reaches for a pencil and draws, and the marks follow the drawing.

Where children dropped marks this year

  • Reading "more than" as "at least". The screen-time question and every "fewest/most" question punish this instantly.
  • A percentage of the wrong number. 20% of the original instead of 20% of the remainder.
  • Ignoring the limiting factor. Taking the larger of the two counts in a "make as many as possible" question.
  • Counting a pattern instead of finding its rule. Fine for Figure 3, fatal at Figure 45.
  • Skipping the unit line. Writing the answer without "42% = 63, so 1% = 1.5". In Paper 2 the method marks are real marks, and a bare answer can lose them even when it is right. Our notes on careless mistakes that cost PSLE Math marks go deeper on this.

What to do this week

You do not need the whole paper. You need these four moves.

  • Pull one question of each type — a comparison trap, a percentage of a remainder, an "as many as possible" grouping, and an alternating pattern. Four questions, one sitting.
  • Before your child writes anything, ask them to name the method: "It's a before-after." Naming it is half the battle, and it is the step that fails first under pressure.
  • Make them draw the model, even when they can see the answer. The drawing is what earns the method mark.
  • Finish with the check: does the answer satisfy the exact condition the question set?

If you want a structured set of these, our P6 heuristics walkthrough maps them topic by topic.

Keep learning

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