Geometry

Let's start with one of the most common geometry setups in OC MR: a shape where the perimeter is given and one dimension is a fixed multiple of the other. The goal is to find the area — which means first recovering the two dimensions from the perimeter using a single variable.

Ratio-based perimeter-to-area questions appear in almost every OC MR paper, usually in the first half. They are reliably easy marks for students who know to introduce one variable for both dimensions — and a common trap for those who try to guess or trial-and-error.

The examiner is checking whether you can set up a simple algebraic equation from P = 2(l + b), substitute the ratio relationship, solve for the unknown, and then correctly multiply the two dimensions to get the area — rather than confusing perimeter with area.

A rectangle is described with its perimeter and a length-to-breadth ratio (e.g. length is twice the breadth, or length is 3 more than breadth). You must find the area. The numbers are chosen so that trial-and-error with the options is slow — one-variable algebra is the fastest path.

A wire 80 cm long is bent into a rectangle. Two sides are 10 cm each. What's the length of one longer side? The key: wire length = perimeter. Subtract the two known sides, then halve what's left. Option E (60 cm) is the most common wrong answer — students forget to divide by 2 and give the combined length of both longer sides.

Rectangle perimeter questions where one pair of sides is given and the other must be found appear regularly in the easy band of OC MR. The 'wire = perimeter' framing is a classic presentation. Students who know perimeter = 2L + 2W and rearrange correctly always score here.

The examiner is testing whether students can set up the perimeter equation with one unknown, subtract the known sides, then halve the remaining length to find one side. Option E (60) directly catches students who subtract the short sides but forget to divide by 2. Option A (15) catches students who halve 80 first (to get 40) and then subtract 10 + 10 + 5.

A wire or string of given total length is bent into a rectangle. Two sides (the shorter ones) are given. All wire is used, so wire length = perimeter. Students must subtract the two known sides from the total, then divide by 2 to find each unknown (longer) side.

A rectangle and a rhombus share the same perimeter. The rhombus side is 9 cm → perimeter 36 cm. The rectangle length is 14 cm → width = ? The classic trap is option D (22 cm): students who compute 36 − 14 = 22 forget that a rectangle has TWO lengths, so you must halve the perimeter first: 36 ÷ 2 = 18, then subtract 14 to get 4.

Equal-perimeter problems linking two different shapes appear regularly at the easy level of OC MR. They test two shape properties in sequence: the student must know that a rhombus has 4 equal sides, and that a rectangle's perimeter formula is 2(l + w). Option D (forgetting to halve) is a standard examiner trap.

The examiner tests knowledge of the rhombus (4 equal sides) and the rectangle perimeter formula P = 2(l + w). Option D (22 cm) traps students who subtract the length from the full perimeter rather than the half-perimeter. Option E (36 cm) traps students who report the perimeter instead of the width.

Two shapes are stated to have equal perimeters. One shape's side length is given to find the perimeter. The other shape has one dimension given; students must use the perimeter formula to find the missing dimension.

Now for a classic 3D reasoning question: a large cube built from unit cubes with all outside faces painted. When it is pulled apart, how many small cubes escaped the paint entirely? This is one of those questions that looks daunting until you know the one-line trick — after which it takes under 20 seconds.

Painted-cube questions appear consistently in the medium-to-difficult range of OC MR, almost always involving a cube of side 3, 4, or 5 units. The zero-painted-faces variant is the easiest in the family; the exactly-two-faces and exactly-three-faces variants are harder and appear later in the paper.

The examiner is testing whether you understand that the only unpainted unit cubes are those completely surrounded — the interior cubes that were never touched by the outer layer. The key insight is to remove one layer from each side of every dimension to find the inner core.

An n × n × n cube (where n = 3, 4, or 5) is assembled from unit cubes, all outer faces are painted, and the cube is broken apart. You are asked how many unit cubes have zero (or exactly one, two, or three) faces painted.

Three shape-classification statements. Only #1 is true: every rhombus has 2 pairs of parallel sides, so it IS a parallelogram. #2 fails because a kite needs adjacent equal sides but a rectangle has opposite equal sides. #3 fails because not all 4-sided shapes are squares.

Shape-hierarchy and classification statements appear at the medium level of OC MR. Students must know the defining property of each shape and whether one shape class is a subset of another (e.g. rhombus ⊂ parallelogram, square ⊂ rectangle ⊂ parallelogram).

The examiner tests precise knowledge of shape definitions. Statement 2 ('all rectangles are kites') is the key trap — students who vaguely remember that kites and rectangles both have equal sides may accept it. The distinction is adjacent vs opposite equal sides.

Three statements make claims about quadrilateral hierarchy or properties (e.g. 'all X are Y'). Students evaluate each and identify the true one(s) using a statement-combination option list.

Maxine turns a dial through an acute angle each time. She needs at least one full revolution (360°). What is the fewest turns? The trap is option C (4): four times 90° = 360°, but 90° is a right angle, NOT acute. So 4 turns can never quite make 360°. Five turns of 72° each reaches exactly 360° — and 72° is acute.

Minimum-turns problems involving angle type constraints appear at the difficult tier of OC MR. They require students to know the strict definition of an acute angle (0° < θ < 90°) and reason that the boundary value (90°) is excluded, forcing them up to the next integer.

The examiner is testing precise knowledge of angle classification and logical reasoning about maximisation under a strict constraint. Option C (4) is the main trap: students who think 'almost 90°' × 4 reaches 360° forget that with a strict inequality, 4 acute angles can only get arbitrarily close to 360° but never reach it.

An object is rotated through a specific angle type (acute, obtuse, etc.) repeatedly. A total rotation requirement is given. Students must find the minimum number of turns by identifying the largest possible angle in the given category and determining how many such turns are needed.