Cameroon GCE O-Level Mathematics common mistakes
Cameroon GCE O-Level Mathematics · Understand the traps and how to avoid them.
When solving a linear equation such as 3(x − 2) = 5x + 4, the candidate expands only the first term and writes 3x − 2 = 5x + 4, dropping the multiplication onto the second term inside the bracket.
The bracket is read as a label around the expression rather than as a multiplier applied to every term inside it; the habit of 'removing brackets' is learned as deleting them.
What to do instead
Before expanding, draw an arrow from the number outside to each term inside and write the intermediate line with both products shown, e.g. 3x − 6 = 5x + 4. Practise ten expansions per session and always check by substituting a simple value such as x = 1 into the original and the expanded form.
When adding fractions such as 2/3 + 3/5, the candidate adds numerators and denominators separately and gives 5/8.
Whole-number addition is over-generalised to fractions; the candidate has not internalised that the denominator names the size of the part and cannot be combined by addition.
What to do instead
Use a diagram or fraction strips to show that thirds and fifths are different-sized parts, then require the common-denominator line to be written explicitly: 10/15 + 9/15 = 19/15. Insist on the estimate check (each fraction is under 1 but the answer 5/8 is smaller than either addend), which flags the error immediately.
When simplifying an algebraic fraction such as (x² − 9)/(x − 3), the candidate cancels the x² against the x and the 9 against the 3 to get x − 3.
Cancellation is treated as deleting matching symbols rather than as dividing the whole numerator and denominator by a common factor; the candidate has not connected cancelling to factorisation.
What to do instead
Require factorisation of the numerator first: (x − 3)(x + 3)/(x − 3), then cancel the common bracket, not individual letters. Reinforce with a numerical counterexample showing that (10² − 9)/(10 − 3) is not 10 − 3, so the illegal cancellation is visibly false.
When finding the gradient of the line through (2, 5) and (6, 13), the candidate computes (6 − 2)/(13 − 5) = 4/8 = 1/2, inverting the formula.
The formula is memorised as 'difference of x over difference of y' or as a vague 'change over change' without attaching the change in y to the numerator.
What to do instead
Anchor the formula in the phrase 'rise over run' and always write Δy above Δx before substituting. Check with a sketch: the line rises 8 while running 4, so the gradient must be greater than 1, which exposes the inverted answer.
When solving 2x + 3 > 11, the candidate divides by 2 correctly to get x > 4 but, in a variant such as −2x > 8, writes x > −4 instead of reversing the inequality.
The rule that multiplying or dividing by a negative reverses the sign is memorised as a fact without a model, so it is forgotten under exam pressure.
What to do instead
Test every inequality answer by substituting a value from the claimed solution set back into the original. For −2x > 8, substituting x = 0 gives 0 > 8, which is false, so the direction must be wrong; this self-check works even when the rule is forgotten.
When asked for the probability of drawing two red counters without replacement from a bag, the candidate multiplies 3/7 by 3/7, reusing the first denominator instead of reducing it to 2/6.
The 'and means multiply' rule is applied mechanically without updating the sample space after the first draw, and the phrase 'without replacement' is not translated into an action.
What to do instead
Draw the bag twice, crossing out the counter removed, and write the second fraction only after the crossed-out diagram. Label each branch of a tree diagram with the counts remaining, so the changing denominator is produced by the picture rather than by memory.
When calculating the volume of a cylinder with radius 7 cm and height 10 cm, the candidate uses 2πr instead of πr² for the base area, giving 440 cm³.
Circumference and area formulas are stored together and retrieved by topic ('circle') rather than by purpose ('base area'), so the wrong one surfaces first.
What to do instead
Write a formula card that pairs each formula with what it measures: πr² gives square units (area), 2πr gives linear units (perimeter). Before substituting, state the unit of the quantity required; if the answer must be in cm³, the base factor must be an area.
When converting 0.00045 m to standard form, the candidate writes 4.5 × 10⁴ instead of 4.5 × 10⁻⁴.
The direction of the exponent is decided by counting decimal places moved without recording whether the movement was to the left or the right, so the sign is guessed.
What to do instead
Always write the conversion as a two-step chain: 0.00045 = 4.5 ÷ 10 000 = 4.5 × 10⁻⁴. Check by estimating: the original number is less than 1, so its standard form must have a negative index; a positive index would make it 45 000.
When finding the mean of a grouped frequency table, the candidate adds the class boundaries (or the frequencies) and divides by the number of classes rather than using midpoints and total frequency.
The ungrouped mean procedure (add the data, divide by how many) is applied to the table entries, because the candidate does not see that each class stands for many values represented by its midpoint.
What to do instead
Build a working table with columns for midpoint, frequency and midpoint × frequency, and write the formula as Σfx ÷ Σf above it. Check plausibility: the mean must lie inside the range of the data and near the class with the highest frequency.
When asked to construct a triangle given two sides and the included angle, the candidate draws the angle first and then marks both given lengths along the same ray, producing a degenerate figure, or omits the construction arcs entirely.
Construction is treated as freehand drawing of a lookalike figure; the candidate does not follow the fixed order of steps and does not know that arcs are the evidence of accuracy.
What to do instead
Memorise a fixed recipe: draw the base line, construct the angle at one end with compass arcs, mark the second side along the new ray, then join. Leave all arcs visible and label given lengths and angles on the figure; marks are awarded for the arcs, not only the final shape.