The most demanding questions of the decade, solved and checked twice · Jump to the twelve →

Shown working, not just shown off

The hardest PSLE Math questions, shown working.

The short version

Across the 664 questions actually sat from 2012 to 2025, just 31 carry the top difficulty rating, and almost all of them live in the back third of Paper 2. They are demanding, but they are a small, knowable set, and they cluster in three familiar places. Below are twelve of them, across nine years, each solved cleanly and checked twice. The single idea that runs through every one: a hard question is hard because of its structure, not its arithmetic.

Based on the 664 questions from the 14 papers sat 2012 to 2025, with the MOE Specimen reported separately · every solution worked independently then checked against the verified GPA key · Mrs Eileen Toh signs off the mathematics · last reviewed 22 Jun 2026

Where the hardest cluster

Thirty-one difficulty-five questions in fourteen years. Three structures hold almost all of them.

The notoriously hard tail of the paper is not random. When we tagged every question and counted, the hardest items kept landing in the same three structures. Learn to recognise these three, and most of the scary tail stops being a surprise.

Area & Perimeter

Rings, folds and joined figures, where area is conserved and the boundary has to be rebuilt.

Volume & Rate

Tanks that fill and drain, where the trick is to work in volume, not water level.

Fractions & Part-Whole

A fraction of a fraction, tamed by naming the whole in units that divide everything.

Difficulty-five is a structural label we assign when tagging, not a marketing claim. These questions are demanding; none is a trick. Source: GPA tagged index, 664 sat-paper questions.

How to read these honestly

Name the lock, and the key follows.

These are questions that have already been sat, 2012 to 2025. We chose them because they are among the most demanding in the paper, by marks and by the number of steps, not because we predict anything like them will appear again. Each card names the lock, which of the structures it is and how a child tends to mis-read it, then the key, the one move that opens it, then the clean steps and the answer.

Not one of these is won by faster arithmetic. Each turns on seeing the structure first: the equal spend behind a discount, the matching edges inside a box, the shared numerator across three children, the mirror inside a fold. That is the whole method, and it is why the same handful of locks keep opening with the same handful of keys.

2019 · Paper 2 · Q14 Percentage · reverse percentage Difficulty 5

The two shoppers who spent the same

The question

Kevin and Julie each spent exactly $61.20 on egg tarts. Julie got 6 more tarts than Kevin, because she had a coupon giving her 15% off. (a) How many tarts did Julie get? (b) What was the full price of one tart?

We reproduce this one because it made national news — one of the 2019 PSLE questions widely shared online, covered by Mothership. For other questions our pages point you to your Ten-Year Series instead.

Video: a Genius Plus Academy teacher solving PSLE 2019 Paper 2 Question 14 Teacher video · 2019 P2 Q14

The lock

The trap is to hunt for 15% of a price you do not yet know. The real structure is hidden: both children spent the same total, so the discount did not save Julie money, it bought her more tarts with the same money.

The key

Anchor the percentage to the equal spend. The 15% Julie did not lose on price became 15% of $61.20 worth of extra tarts. That is the whole question.

Worked steps

  1. Julie's extra tarts are worth 15% of her $61.20 spend: \(0.15 \times 61.20 = 9.18\).
  2. Those extra tarts number 6, so each tart at the discounted price costs \(9.18 \div 6 = 1.53\).
  3. (a) Julie's total tarts \(= 61.20 \div 1.53 = 40\) tarts.
  4. The discounted $1.53 is 85% of the full price, so full price \(= 1.53 \div 0.85 = 1.80\).

Answer: (a) 40 tarts. (b) $1.80 full price.

What makes it click. The moment you see "same spend", the discount stops being about money saved and becomes about tarts gained. Find what 15% of the spend bought, and the rest is division.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

2019 · Paper 2 · Q15 Volume & Rate · packing by edges Difficulty 5

The box that two sizes of cube fill exactly

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2019 Paper 2, Q15, parts (a) and (b) (the volume question about packing two sizes of cube) — then follow our worked solution below.

The lock

A child counts the first layer and never notices the hinge: the two cube sizes share the box's height exactly, which only works if 2 large edges equal 3 small edges. Miss that and part (a) is a guess.

The key

Read the geometry off the first layer, then use the volume split. The figure is doing the teaching.

Worked steps

  1. The first layer shows 2 large cubes spanning the same height as 3 small cubes, so 2 large edges \(=\) 3 small edges.
  2. The 8 large cubes make \(8 \div 2 = 4\) large layers; that same height fits 6 small layers, each of 6 small cubes, so small cubes \(= 6 \times 6 = 36\).
  3. (b) Large cubes' volume \(= \frac{3}{7} \times 4032 = 1728\) cm³; small cubes' volume \(= 4032 - 1728 = 2304\) cm³.
  4. One small cube \(= 2304 \div 36 = 64\) cm³, so its edge \(= \sqrt[3]{64} = 4\) cm.

Answer: (a) 36 small cubes. (b) 4 cm.

What makes it click. Once "two large equals three small" is on the page, the count of small cubes is forced, and the cube root at the end is the gentle part. Check: each large cube is \(1728 \div 8 = 216\) cm³, edge 6 cm, and \(2 \times 6 = 3 \times 4\).

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

2019 · Paper 2 · Q17 Number Patterns · figure to formula Difficulty 5

The triangle that grows by odd numbers

The question

A pattern of triangles: Figure 1 is one small triangle, and each new figure adds a row, so Figure \(n\) is a big triangle made of \(n^2\) small triangles, alternately white and grey by row. (a) Complete the table for Figure 5. (b) How many small triangles are in Figure 250 altogether? (c) In Figure 250, what percentage are grey?

We reproduce this one because it made national news — one of the 2019 PSLE questions widely shared online, covered by Mothership. For other questions our pages point you to your Ten-Year Series instead.

Video: a Genius Plus Academy teacher solving PSLE 2019 Paper 2 Question 17 Teacher video · 2019 P2 Q17

The lock

A child can fill the next column by drawing, but Figure 250 cannot be drawn. The lock is spotting that the total is a square number and that the grey-minus-white gap grows in a steady, countable way.

The key

Find the rule, then handle grey and white as a small gap on a large total: grey is half the total plus half the gap.

Worked steps

  1. (a) Figure 5 adds a fifth row of 9 white triangles to Figure 4: white \(= 6 + 9 = 15\), grey stays 10 (total 25).
  2. (b) Figure \(n\) holds \(n^2\) small triangles, so Figure 250 holds \(250^2 = 62\,500\).
  3. (c) For even-numbered figures, grey exceeds white by \(n\) (Figure 2: \(3-1=2\); Figure 4: \(10-6=4\)). For Figure 250 the gap is 250.
  4. Grey \(= (62\,500 + 250) \div 2 = 31\,375\), so grey % \(= 31\,375 \div 62\,500 \times 100\% = 50.2\%\).

Answer: (a) white 15, grey 10. (b) 62 500. (c) 50.2%.

What makes it click. Two ideas, neatly separated: the total is a square, and grey is just half the total plus half the gap. Hard problems often split into one idea about the whole and one about the difference.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

2020 · Paper 2 · Q17 Part-Whole · fraction of a fraction Difficulty 5

Mrs Wu's dress, blouses and watch

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2020 Paper 2, Q17, parts (a) and (b) (the spend-in-fractions money question) — then follow our worked solution below.

Video: a Genius Plus Academy teacher solving PSLE 2020 Paper 2 Question 17 Teacher video · 2020 P2 Q17

The lock

Three fractions sit on different bases: 1/6 of everything, then 3/4 of the remainder. A child who keeps switching the base drowns. Choose one unit for the whole so every fraction becomes a tidy count of units.

The key

Let the whole be a number of units that all the fractions divide cleanly, here 30.

Her money as 30 units:

dress 3u 2 blouses 2u watch 18.75u (3/4 of the 25u remaining) left 6.25u

Worked steps

  1. (a) The dress plus 2 blouses make 1/6 of her money, and the dress is 3 blouses, so that 1/6 is shared as 5 blouse-portions. Each blouse \(= \frac{1}{6} \div 5 = \frac{1}{30}\) of her money.
  2. (b) Let her money be 30 units. Then 2 blouses \(=\) 2u, dress \(=\) 3u, and the first 1/6 is 5u, leaving 25u.
  3. The watch is 3/4 of the remaining 25u \(= 18.75\)u.
  4. Watch minus dress \(= 18.75\text{u} - 3\text{u} = 15.75\text{u} = 220.50\), so 1u \(= 14\).
  5. Her money \(= 30 \times 14 = 420\).

Answer: (a) 1/30. (b) $420 at first.

What makes it click. Choosing 30 units up front turns three competing fractions into whole-number parts. The awkward 3/4 becomes a clean 18.75u, and one subtraction unlocks the value of a unit.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

2021 · Paper 2 · Q16 Area & Perimeter · rearrangement Difficulty 5

The ring cut into quarters and rearranged

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2021 Paper 2, Q16, parts (a) and (b) (the area question where a ring is cut up and rearranged) — then follow our worked solution below.

Video: a Genius Plus Academy teacher solving PSLE 2021 Paper 2 Question 16 Teacher video · 2021 P2 Q16

The lock

Rearranging the pieces alarms children into recomputing everything. Two quiet truths survive the cut: the area does not change, and the new boundary is still made only of the original arcs and the two fresh straight cuts.

The key

Extract the ring width from the 42 cm span, then conserve area and rebuild the boundary.

Worked steps

  1. The 42 cm width is 4 small-circle radii plus 2 ring-widths: \(42 = 4 \times 8 + 2w\), so \(2w = 10\) and \(w = 5\) cm.
  2. Large radius \(= 8 + 5 = 13\) cm.
  3. (a) Area is unchanged by rearranging: \(\pi(13^2) - \pi(8^2) = 3.14 \times (169 - 64) = 3.14 \times 105 = 329.7\) cm².
  4. (b) The boundary is one whole outer circle, one whole inner circle, and two exposed 5 cm cuts: \(2 \times 3.14 \times 13 + 2 \times 3.14 \times 8 + 5 + 5 = 141.88\) cm.

Answer: (a) 329.7 cm². (b) 141.88 cm.

What makes it click. Rearranging is a magician's misdirection. Hold onto the two facts that survive the cut, area is conserved and the arcs still add to two full circles, and the shape's new look stops mattering. This is the same structure as the MOE Specimen P2 Q14.

Independently solved, matches the GPA handwritten key (2021 had a handwritten key, not a published marking scheme). Open the full worked solution →

2021 · Paper 2 · Q17 Volume & Rate · net rate, timing offset Difficulty 5

The tank with one tap in and one tap out

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2021 Paper 2, Q17, parts (a) and (b) (the tank filling-and-draining rate question) — then follow our worked solution below.

Video: a Genius Plus Academy teacher solving PSLE 2021 Paper 2 Question 17 Teacher video · 2021 P2 Q17

The lock

Two taps with different start times tempt a child to track the water level, which lurches up and down. Work in volume in minus volume out over carefully counted minutes, and notice the taps run for different lengths of time.

The key

Count each tap's minutes separately and work in litres. The five-minute head start is the whole trick.

Worked steps

  1. Tank volume \(= 55 \times 30 \times 20 = 33\,000\) cm³ \(= 33\) litres, so half \(= 16.5\) litres.
  2. (a) From 2.00 to 2.15 p.m., tap A ran 15 min: in \(= 15 \times 4.2 = 63\) litres. Tap B ran only 10 min (from 2.05). It must have drained \(63 - 16.5 = 46.5\) litres, so its rate \(= 46.5 \div 10 = 4.65\) litres/min.
  3. (b) From 2.15 to 2.30 p.m. (15 min): in \(= 15 \times 4.2 = 63\) litres; out \(= 15 \times 4.65 = 69.75\) litres.
  4. Water at 2.30 p.m. \(= 16.5 + 63 - 69.75 = 9.75\) litres, so the fraction filled \(= 9.75 \div 33 = \frac{13}{44}\).

Answer: (a) 4.65 litres/min. (b) 13/44.

What makes it click. Tap A ran 15 minutes while tap B ran 10. Get the two clocks right and the rest is "in minus out". This exact question is also drilled in the GPA Intensive corpus.

Independently solved, matches the GPA handwritten key. Open the full worked solution →

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2024 · Paper 2 · Q16 Part-Whole · shared numerator Difficulty 5

Three children, one shared numerator

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2024 Paper 2, Q16, parts (a) and (b) (the fractions-of-money donation question) — then follow our worked solution below.

Video: a Genius Plus Academy teacher solving PSLE 2024 Paper 2 Question 16 Teacher video · 2024 P2 Q16

The lock

Three different fractions of three different totals look impossible to compare. The line "Devi and Eric donated the same amount" lets you rewrite the fractions so they share a numerator, and the comparison becomes obvious.

The key

Match numerators, then read off the units.

Worked steps

  1. Devi gave \(\frac{1}{4} = \frac{2}{8}\) of her money; Eric gave \(\frac{2}{7}\) of his. Same numerator (2), so Devi's money \(=\) 8 units and Eric's \(=\) 7 units, each donating 2 units.
  2. Haziq donated \(3 \times\) Eric's donation \(= 3 \times 2 = 6\) units; that is 2/5 of his money, so Haziq's money \(=\) 15 units.
  3. (a) Most at first: Haziq (15u); least: Eric (7u).
  4. (b) Total \(= 8 + 7 + 15 = 30\) units \(= 1560\), so 1 unit \(= 52\). Haziq donated 6 units \(= 312\).

Answer: (a) Haziq most, Eric least. (b) $312.

What makes it click. Forcing 1/4 into 2/8 so it shares Eric's numerator is the whole trick: equal donations plus equal numerators turn three tangled fractions into one clean string of units.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

2024 · Paper 2 · Q17 Area & Perimeter · fold then cut Difficulty 5

The paper that is folded, then cut

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2024 Paper 2, Q17, parts (a) and (b) (the folding-and-cutting area and perimeter question) — then follow our worked solution below.

Video: a Genius Plus Academy teacher solving PSLE 2024 Paper 2 Question 17 Teacher video · 2024 P2 Q17

The lock

Folding makes two edges equal, and cutting reveals new edges. A child who treats the fold and the cut as unrelated cannot connect the 60 cm to anything. Let BC be one unit, write each perimeter in units, and use the difference.

The key

One unit on the unknown side, build both perimeters, solve.

Worked steps

  1. Let BC \(=\) 1u, so BD \(=\) 4u. The flap X is a right triangle with legs 50 cm and 1u; Y is the remaining piece, whose outline runs along 3u, then 50, then 4u.
  2. P(Y) minus P(X) \(= (3\text{u} + 50 + 4\text{u}) - (50 + 1\text{u}) = 6\text{u} = 60\) cm, so u \(= 10\) cm.
  3. (a) BD \(= 4\text{u} = 40\) cm.
  4. (b) Area of X \(= \frac{1}{2} \times 50 \times 10 = 250\) cm². The whole rectangle is \(50 \times 40 = 2000\) cm², and the fold removes two copies of X, so area of Y \(= 2000 - 2 \times 250 = 1500\) cm².

Answer: (a) BD = 40 cm. (b) 1500 cm².

What makes it click. The 60 cm is not a length you measure; it is a difference of perimeters, and once both perimeters are written in units of BC, that single sentence hands you the unit.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

2017 · Paper 2 · Q17 Geometry · a fold that doubles an angle Difficulty 5

The folded triangle and the angle that doubles

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2017 Paper 2, Q17, parts (a) and (b) (the folded-triangle angles question) — then follow our worked solution below.

The lock

Children know angles in a triangle sum to 180°, but a fold reflects an angle to a new place, and the reflected angle is easy to put in the wrong spot. Treat the fold as a mirror.

The key

Find the base angles first, then track each reflected angle to its true position.

Worked steps

  1. Base angles of the isosceles triangle: angle BAC = angle BCA \(= (180^\circ - 84^\circ) \div 2 = 48^\circ\).
  2. (a) In triangle DEC, angle \(x = 180^\circ - 48^\circ - 67^\circ = 65^\circ\) (the fold reflects this angle to where \(x\) is marked).
  3. (b) The fold places a second 67° beside the first at D, so the straight line at D leaves \(180^\circ - 67^\circ - 67^\circ = 46^\circ\) for the folded edge.
  4. In the small triangle at A, angle \(y = 180^\circ - 48^\circ - 46^\circ = 86^\circ\).

Answer: (a) x = 65°. (b) y = 86°.

What makes it click. A fold is a mirror, nothing more. The 67° appears twice, the base angle 48° travels along, and angle-chasing does the rest. Drawing the reflected angle in its true position is the whole battle.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

2023 · Paper 1 · Q15 Area & Perimeter · the rare hard one in Paper 1 Difficulty 5

The rare hard one in Paper 1

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2023 Paper 1, Q15 (the shaded-region fraction question) — then follow our worked solution below.

The lock

This sits in Paper 1, where the average question is gentle, so it is easy to rush. The 1 : 8 ratio fixes how high up the crossing point G sits, and from there the unshaded triangle is a single area calculation.

The key

Let the side be 3 units, use the ratio to find G's height, then subtract the one unshaded triangle.

Worked steps

  1. Let AD \(=\) 3 units (so AE = EF = FD = 1) and let the rectangle's height be \(h\).
  2. The ratio area EGF : area ABF \(=\) 1 : 8 places G at height \(h/4\) above AD (check: \(\frac{1}{2} \times 1 \times \frac{h}{4} : \frac{1}{2} \times 2 \times h = \frac{h}{8} : h = 1 : 8\)).
  3. The only unshaded part is triangle BGC, with base 3 and height \(\frac{3h}{4}\): area \(= \frac{1}{2} \times 3 \times \frac{3h}{4} = \frac{9h}{8}\).
  4. Rectangle area \(= 3h\), so shaded fraction \(= \frac{3h - \frac{9h}{8}}{3h} = \frac{\frac{15h}{8}}{3h} = \frac{5}{8}\).

Answer: shaded fraction = 5/8.

What makes it click. Two marks, but a genuine difficulty-five: the small ratio is not decoration, it is the lever that fixes G's height, and once G is placed, a single triangle's area finishes the job. A reminder that Paper 1 can bite.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

PSLE Intensive · Type 1 Part-Whole · two ratios, one whole From 2021 PSLE

Two ratios, one shared gift

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2021 PSLE Paper 2 (the shared-cost ratio question) — then follow our worked solution below.

The lock

A child treats the two ratios as separate puzzles. "Alan to the rest = 1 : 3" simply means Alan is 1/4 of the whole, and "Ben to the rest = 1 : 5" means Ben is 1/6 of the same whole. Both ratios speak about one common total.

The key

Convert each ratio to a fraction of the whole, then put everyone on a common number of units (12).

The gift as 12 units:

Alan 3u Ben 2u Chandra 7u

Worked steps

  1. Alan : (Ben + Chandra) \(=\) 1 : 3, so Alan \(= \frac{1}{4}\) of the total \(=\) 3 units out of 12.
  2. Ben : (Alan + Chandra) \(=\) 1 : 5, so Ben \(= \frac{1}{6}\) of the total \(=\) 2 units out of 12.
  3. Chandra \(= 12 - 3 - 2 = 7\) units.
  4. Chandra minus Ben \(= 7 - 2 = 5\) units \(= 80\), so 1 unit \(= 16\).
  5. Gift \(= 12\) units \(= 12 \times 16 = 192\).

Answer: the gift cost $192.

What makes it click. Each "one to the rest" ratio is secretly a fraction of the whole. Translate both onto 12 units and the three shares line up, so a single $80 difference scales the lot. This is the Lock & Key reflex in miniature.

From the GPA PSLE Intensive corpus (based on 2021 PSLE), independently solved, the typed and handwritten keys agree.

2018 · Paper 2 · Q17 Gaps & Differences · buy in fixed boxes Difficulty 5

The candles that only come in boxes

Where to find it

This is a worded PSLE question, so we don’t reproduce it here. Find it in your Ten-Year Series (TYS) or the official paper — 2018 Paper 2, Q17, parts (a) and (b) (the buying-in-fixed-boxes question) — then follow our worked solution below.

Video: a Genius Plus Academy teacher solving PSLE 2018 Paper 2 Question 17 Teacher video · 2018 P2 Q17

The lock

A child divides 19 by 7, gets a decimal, and is unsure whether to round. You can only buy whole boxes, so the counts must be multiples of 7 and 5, and part (b) has exactly one combination that fits both conditions.

The key

Round up to whole boxes, then list box-combinations against the constraints. A short, honest list beats a clever formula.

Worked steps

  1. (a) 3 long candles need 1 box ($3.20). 19 short candles need \(19 \div 7 = 2\) remainder 5, so 3 boxes (21 candles, $2.50 each). Least cost \(= 3.20 + 3 \times 2.50 = 10.70\).
  2. (b) Long come in 5s, short in 7s. We need \(5L - 7S = 21\) with total \(5L + 7S < 50\).
  3. Listing: \(L = 7\) boxes gives 35 long; \(S = 2\) boxes gives 14 short. Then \(35 - 14 = 21\) and \(35 + 14 = 49 < 50\). This is the only combination under 50.
  4. Cost \(= 7 \times 3.20 + 2 \times 2.50 = 22.40 + 5.00 = 27.40\).

Answer: (a) $10.70. (b) $27.40.

What makes it click. "Sold only in boxes" is the whole instruction: you round up to whole boxes, then test box-counts against both conditions until one fits.

Independently solved, matches the GPA marking-scheme key. Open the full worked solution →

What the showcase proves

Twelve questions, nine years, seven structures. Not one won by faster arithmetic.

Each turned on seeing the structure first: the equal spend behind a discount, the matching edges inside a box, the shared numerator across three children, the mirror inside a fold. That is the entire method, and it is why the same handful of locks keep opening with the same handful of keys.

The same lesson appears from the other side too: take questions from different years that look nothing alike, and they share one lock and one key. A child trained to recognise the structure meets far fewer "new" problems than the variety of stories suggests.

From the hardest questions to the reflex

These are hard because of their structure, not their arithmetic.

They cluster in area and perimeter, volume and rate, and fractions. The PSLE Math Intensive trains the reflex of recognising the structure before choosing the method, across all ten question types, with 158 worked examples.

Explore the PSLE Intensive →

Keep reading

The full picture these questions sit inside, and the structures they keep landing on.

Pillar guide

Most-tested PSLE Math topics, counted

What 709 tagged questions show the paper rewards.

 Type guide

Area & Perimeter questions

Conserve area, rebuild the boundary, watch for hidden edges.

 Type guide

Volume & Rate questions

Work in volume, not water level, and count each tap's minutes.

Questions parents ask

How many genuinely hard questions are in a PSLE paper?

Across the 664 questions sat from 2012 to 2025, only 31 carry the top difficulty rating in our tagging, and almost all sit in the back third of Paper 2. The hardest tail is a small, knowable set, not a flood. This is analysis of past papers, not a forecast of the next one.

Where do the hardest PSLE questions tend to come from?

When we tagged and counted, the hardest items clustered in three structures: Area and Perimeter, Volume and Rate, and Fractions or Part-Whole. A child who learns to recognise these three meets far fewer surprises in the demanding tail.

Are these solutions reliable?

Every solution here was worked independently and then checked against the verified GPA key: the official marking scheme where one exists, a handwritten key for 2021, and the Intensive-corpus key for the Type 1 example. All twelve agreed on every answer. Mrs Eileen Toh signs off the mathematics.

Can I get the videos and similar questions to practise?

Yes. We publish a worked solution and a video for PSLE Paper 2 questions, and you can request this set with three similar questions per topic to practise. Browse them all in our worked-solutions library.

See how your child reads the hard ones.

Book a free diagnostic. We will sit with a real paper and show you which structures your child already sees, and which need work.

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