The hardest PSLE 2025 Maths questions

Question 1 of 5

The three overlapping triangles

Why children lost marks. The diagram gives you three lengths, and not one of them is the side of a triangle. Until your child finds one side, there is nothing to calculate with.

Start with the word identical. The three triangles are the same, and each one is equilateral. So all nine sides are equal. Give that length a name: one side.

Part (a). Work along the bottom line. Every number you need sits on it. The middle triangle's base runs from D to F, so D to F is one side, and reading along the line it is made of three pieces: D to C, then C to P, then P to F.

• P to F is 3 cm. That one is given.
• A to C is one side, and A to D is 18 cm. So D to C is one side − 18.
• P to R is one side, and C to R is 41 cm. So C to P is 41 − one side.

Add the three pieces together: (one side − 18) + (41 − one side) + 3. The two lots of one side cancel each other out, and plain numbers are left: 41 + 3 − 18 = 26.

But those three pieces make up D to F, which is one side. So one side = 26 cm, and AB = 26 cm.

Part (b). This is where most of the marks went. It looks like you need the shape of each shaded piece. You don't.

The corner that gets cut off is itself an equilateral triangle. So the shaded shape loses a piece of the bottom, loses the same length up the slanted side, and gains the cut edge back. Two lost, one gained, so the perimeter drops by exactly one cut length.

• Left shape: the cut is 26 − 18 = 8 cm. Perimeter = (3 × 26) − 8 = 70 cm.
• Right shape: the cut is the 3 cm overlap. Perimeter = (3 × 26) − 3 = 75 cm.
• Total = 70 + 75 = 145 cm.

Question 2 of 5

The painted block

Why children lost marks. Before any calculating, your child has to picture which cubes end up with exactly two painted faces. Most of the marks go there, not in the arithmetic.

A cube gets one painted face for each outside surface it touches. So a cube in a corner of the block touches three surfaces and gets three. A cube sitting in the middle of a face touches one. A cube buried inside gets none. The ones we want, with exactly two, are the cubes that run along an edge but are not at either end.

So counting them means going edge by edge. A block has twelve edges, and on every one of them the two end cubes are corners and don't count. A longer edge simply has more in the middle.

Start with the four upright edges, because the 9 cm already fixes them.

• Each upright edge is 9 cubes tall. The top and bottom ones are corners, so 7 count. Four upright edges give 4 × 7 = 28.
• That leaves 68 − 28 = 40 cubes, and they can only be on the top and bottom squares.
• Those two squares have 8 edges between them, so each edge holds 40 ÷ 8 = 5 counting cubes.
• Those 5 sit between the two corner cubes, so the full edge is 5 + 2 = 7 cubes, which is 7 cm.

The base is 7 cm by 7 cm and the block is 9 cm tall, so the volume is 7 × 7 × 9 = 441 cm³, option (4).

A shortcut worth knowing. All four options are the same block with a different base: 4 × 4 × 9 = 144, 5 × 5 × 9 = 225, 6 × 6 × 9 = 324, 7 × 7 × 9 = 441. So the question is really only asking whether the base is 4, 5, 6 or 7, and your child can test them instead of solving backwards.

Question 3 of 5

The folded square

Why children lost marks. There are three statements and no obvious place to start, so it is easy to spend the time guessing. But all three come out of one fact about folding, and once your child has that fact the rest is quick.

The one fact: folding doesn't change any length. When corner Q is folded over to E, the edge RQ lies down exactly on RE. So RE is the same length as RQ, which is one side of the square. The same fold on the other half gives ES the same length as PS, another side. And RS was a side all along.

So take statement C first. RE, ES and RS are all one side of the square, so all three are equal, so RSE is an equilateral triangle. C is true. Notice this needed no numbers at all.

Now the angles fall out of it. Every angle in an equilateral triangle is 60°, so ∠ERS = 60°. The corner of a square at R is 90°, so the part left over is ∠QRE = 90° − 60° = 30°.

Statement A. The crease RM folds Q onto E, so it splits that 30° into two equal halves. Each half is 15°, which makes ∠MRE 15° and A false. This is the one most children get wrong: 30° is the whole angle, not the half. The corner at R is worth drawing, because it shows the square's 90° is really 15 + 15 + 60.

Statement B. Now look at triangle QMR. The corner at Q is 90° because it's a corner of the square, and ∠QRM is the 15° we just found, so ∠QMR = 180° − 90° − 15° = 75°. Q, M and G sit on one straight edge of the paper, so the rest of the 180° is ∠RMG = 105°, and B is true.

So B and C are true and A isn't. That's option (3).

Question 4 of 5

The quarter-circle figure

Why children lost marks. The figure looks complicated and only two lengths are given, so it is tempting to start measuring off the picture. Everything needed is in those two numbers, but your child has to read the radii out of them first.

Get the two radii before anything else. The catch is that each quarter circle is centred on a CORNER of the dashed box, which isn't where a child looks for a centre.

• The two large quarter circles sit side by side and fill the whole width, so together they are two radii across. The large radius is 38 ÷ 2 = 19 cm.
• Each small quarter circle turns through the depth of the lower part, so the small radius is the depth itself, 16 cm.
• That also gives the flat foot at the bottom: the width less the two small radii, 38 − 16 − 16 = 6 cm.

Part (a), the perimeter. Two quarter circles of the same size make a half circle, and that is the shortcut worth having: a half circle's arc is π × r, no halving of anything else needed.

• The two large arcs make a half circle of radius 19: 3.14 × 19 = 59.66 cm.
• The two small arcs make a half circle of radius 16: 3.14 × 16 = 50.24 cm.
• The two upright edges are one large radius each: 19 + 19 = 38 cm.
• The flat foot is 6 cm.

Perimeter = 59.66 + 50.24 + 38 + 6 = 153.9 cm.

Part (b), the area. Cut the figure along the dashed line into the part above and the part below.

• Above the line, the two large quarter circles make a half circle of radius 19: 3.14 × 19 × 19 ÷ 2 = 566.77 cm².
• Below the line, start with the whole 38 by 16 rectangle, which is 608 cm².
• The two small quarter circles are the pieces cut away from it, and together they are a half circle of radius 16: 3.14 × 16 × 16 ÷ 2 = 401.92 cm².
• So the lower part is 608 − 401.92 = 206.08 cm².

Area = 566.77 + 206.08 = 772.85 cm².

Question 5 of 5

The rectangle with two squares

Why children lost marks. Most children go hunting for the side of the square, because it's the thing the diagram keeps hinting at. It can't be found on its own, and the question never needs it.

Give the ratio units. AD and DC are 5 : 2, so call AD 5 units and DC 2 units. Now write each of them a second way, using the square.

• Down the left side: the square reaches the bottom, and there is 10 cm above it. So DC, the height, is 10 cm plus one square.
• Along the top: 29 cm, then the square, then 29 cm. So AD, the length, is 58 cm plus one square.

Now compare the two. Both contain exactly one square, so the difference between them has no square in it at all: 58 − 10 = 48 cm. That is the trick, and it is why the square's side never has to be worked out.

In units, the difference between AD and DC is 5 units − 2 units = 3 units.

• 3 units = 48 cm, so 1 unit = 48 ÷ 3 = 16 cm.
• AD is 5 units, so AD = 5 × 16 = 80 cm.

Worth checking, because it is quick: DC is 2 × 16 = 32 cm, so the square is 32 − 10 = 22 cm, and the top edge is 29 + 22 + 29 = 80 cm, which matches AD.