CSEC Mathematics common mistakes
CSEC Mathematics · Understand the traps and how to avoid them.
When solving a linear equation such as 3(2x − 5) = 4x + 7, the candidate distributes only to the first term and writes 6x − 5 = 4x + 7, or drops a minus sign when moving a term across the equals sign, producing a plausible but wrong value of x.
Under time pressure the candidate treats brackets as decoration and mentally 'sees' the answer, skipping the written distribution step. Sign errors arise because subtracting a negative or moving a term is done in the head rather than recorded, so the inverse operation is applied to only one side.
What to do instead
Adopt a written two-line routine: first rewrite the equation with every bracket expanded and every term shown (6x − 15 = 4x + 7), then perform the same operation on both sides in a visible column. Finish by substituting the answer back into the original equation and checking both sides give the same number; if they do not, the error is in the expansion, not the arithmetic.
The candidate factorises a quadratic such as x² − 5x − 24 by writing (x − 8)(x + 3) and then states the solutions as x = 8 and x = 3, or as x = −8 and x = −3, instead of x = 8 and x = −3.
The candidate has learned 'change the sign' as a rule without connecting it to the zero-product property. Having found the factors, the final step is treated as a formality, so the sign flip is applied inconsistently or not at all.
What to do instead
Never write the answer directly from the factors. Write the equation as (x − 8)(x + 3) = 0, then state in words that a product is zero only when one bracket is zero, so x − 8 = 0 or x + 3 = 0. Solve each mini-equation explicitly and check both roots by substitution into the original quadratic.
Asked to change the subject of a formula such as v² = u² + 2as to make s the subject, the candidate divides only the 2as term by 2a, or takes the square root of individual terms and writes v = u + √(2as).
The candidate applies an operation to part of an expression rather than to the whole side, because the formula is read left to right as a sequence of symbols rather than as a balance. Square roots of sums are confused with products because the distributive law is over-generalised.
What to do instead
Treat the formula as a set of nested operations and undo them in reverse order, writing each step underneath the previous one. Before taking a root, isolate the squared quantity completely (v² − u² = 2as, then s = (v² − u²)/(2a)). Test any root result with numbers: pick u = 1, a = 2, s = 3 and confirm the rearranged formula reproduces v.
The candidate solves a pair of simultaneous equations by elimination but forgets to multiply the whole equation, or finds x correctly and then substitutes into the equation that was altered, obtaining a mismatched y.
Elimination requires holding two operations in mind at once (scaling, then adding), and candidates often scale only the coefficient they are targeting. The back-substitution step is rushed because the candidate believes the hard part is finished.
What to do instead
Label the equations (1) and (2), write the multiplier beside each before applying it, and cross out the eliminated variable in both lines. Always substitute the found value into the simpler original equation, then verify in the other equation as well; a pair that fails the second check signals an elimination slip.
In a word problem involving ratio or proportion, the candidate sets up the ratio in the wrong order, for example dividing a sum of money in the ratio 3:5 by treating the 3 as the larger share, or uses direct proportion where inverse proportion applies.
The candidate matches numbers to names by their order of appearance in the sentence rather than by meaning, and does not distinguish 'more workers take less time' from 'more workers do more work'. Ratio notation is treated as two separate quantities rather than as parts of a total.
What to do instead
Before calculating, write a one-line statement of what one part represents and what the total number of parts is (3 + 5 = 8 parts). For proportion, ask whether an increase in one quantity increases or decreases the other, and write 'direct' or 'inverse' explicitly before forming the equation. Check that the shares sum to the original total.
When calculating the area of a compound shape or the surface area of a solid, the candidate adds all visible face areas but omits hidden faces, or uses the diameter as the radius in circle formulas.
Diagrams are read as pictures rather than as three-dimensional objects, so faces at the back or base are not counted. The diameter is the labelled number on most diagrams, and the halving step is skipped because the formula is recalled as 'πr²' without the definition of r.
What to do instead
Redraw the solid as a net or list every face by name (top, bottom, four sides) and tick each one as its area is computed. Write 'r = d ÷ 2' as an explicit line before substituting, and state units squared at the end so an area answer in cm is immediately suspect.
The candidate uses Pythagoras' theorem on a triangle that is not right-angled, or identifies the hypotenuse incorrectly when the right angle is not at the bottom of the diagram, leading to a² + b² = c² applied with the hypotenuse as a leg.
The formula is memorised as a string of letters tied to a standard picture. When the triangle is rotated or the right-angle marker is small, the candidate relies on the longest-looking side rather than on the position of the right angle.
What to do instead
Mark the right angle with a square symbol and label the side opposite it as c before substituting anything. If no right angle is marked, check whether one can be established from the geometry; if not, Pythagoras does not apply. Verify by confirming the computed hypotenuse is the longest side.
In trigonometry, the candidate selects the wrong ratio (using sine where cosine is required) or leaves the calculator in the wrong mode, so an angle of 30° is evaluated as 30 radians.
Ratio selection is done by pattern-matching SOH-CAH-TOA to the letters in the diagram rather than to which sides are actually known. Calculator mode is a setting the candidate never checks because the display looks normal.
What to do instead
Before choosing a ratio, write down which two sides are involved relative to the marked angle (opposite, adjacent, hypotenuse) and only then pick the ratio. Check the mode indicator at the start of every trigonometry question, and sanity-check with a known value such as sin 30° = 0.5.
The candidate computes a mean, median or modal class from grouped data by averaging the class boundaries rather than the midpoints, or reports the modal class as though it were an exact mode.
Grouped frequency tables look like ordinary number lists, and the candidate applies the ungrouped procedure. The distinction between an estimate from grouped data and an exact value is not stated in the answer because it is not understood.
What to do instead
Add a midpoint column to the table before any calculation, multiply each midpoint by its frequency, and divide the total by the total frequency. When reporting, state that the mean is an estimate because individual values are unknown, and describe the mode as a class interval rather than a single value.
On a probability question, the candidate adds probabilities for independent successive events instead of multiplying, or fails to adjust the denominator when selection is without replacement.
The word 'or' in ordinary speech suggests addition and 'and' suggests combination, and this is carried into probability without checking whether events are mutually exclusive. Without replacement is a phrase the candidate reads past rather than a condition to model.
What to do instead
Draw a tree diagram for any two-stage event, writing the fraction on each branch and multiplying along branches, adding only across separate branches. For without replacement, write the reduced denominator on the second branch explicitly and check that probabilities on each set of branches sum to 1.