Cameroon GCE O-Level Mathematics study materials
Cameroon GCE O-Level Mathematics study resources

Cameroon GCE O-Level Mathematics: Mathematics - Cameroon GCE O-Level

Cameroon GCE O-Level Mathematics · Work through your exam, one topic at a time.

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Mathematics - Cameroon GCE O-Level

Overview

The Mathematics domain of the Cameroon GCE O-Level is the whole assessed syllabus, not a single subtopic. It tests number, algebra, geometry and mensuration, trigonometry, statistics and probability through Paper 1 multiple-choice items and Paper 2 structured questions. Success depends on accurate recall of standard results, correct substitution into formulae, and clear written reasoning in Paper 2.

Core rules and concepts

  • Number: convert between bases, use standard form, apply ratio and proportion, calculate percentage change, simple and compound interest, and manipulate indices.
  • Algebra: expand and factorise; solve linear, simultaneous and quadratic equations; solve linear inequalities, remembering to reverse the sign when multiplying or dividing by a negative; handle direct and inverse variation; recognise AP and GP terms.
  • Geometry and mensuration: use angle properties of parallel lines, triangles and polygons; apply circle theorems with stated reasons; calculate lengths, areas and volumes of prisms, cylinders, cones and spheres.
  • Trigonometry: use SOHCAHTOA in right-angled triangles, the sine and cosine rules, and area = ½ab sin C; interpret bearings as three-figure angles measured clockwise from north.
  • Statistics and probability: calculate mean, median and mode from raw data or grouped frequency tables; use cumulative frequency curves; apply P(A') = 1 − P(A) and the addition rule for mutually exclusive events.

Worked example

A solid cone has base radius 7 cm and slant height 25 cm. Taking π = 22/7, find its total surface area and volume.

The perpendicular height h is not 25 cm; the slant height l is the hypotenuse. By Pythagoras: h² = 25² − 7² = 625 − 49 = 576, so h = 24 cm.

Total surface area = πr(l + r) = (22/7) × 7 × (25 + 7) = 22 × 32 = 704 cm².

Volume = (1/3)πr²h = (1/3) × (22/7) × 49 × 24 = 1,232 cm³.

Check: the volume is about one-third of the enclosing cylinder (πr²h ≈ 3,696 cm³), so 1,232 cm³ is reasonable.

Traps that cost marks

  • Using the slant height as the perpendicular height in cone volume.
  • Omitting the circular base when finding total surface area.
  • Rounding π too early or ignoring an instruction to take π = 22/7.
  • Giving a numerical answer without units or without the required reason in geometry.
  • Dropping the negative sign when solving inequalities or quadratic equations.