ASSEB HSLC General Mathematics common mistakes
ASSEB HSLC General Mathematics · Understand the traps and how to avoid them.
When simplifying a surd such as √72, the candidate writes √72 = 6√2 by guessing a 'nice' factor pair, or leaves it as √72 after only partly extracting the largest square factor, instead of checking whether the radicand still contains a perfect square.
Surd simplification is often memorised as a list of examples rather than as the procedure of factoring out the largest perfect-square factor; candidates stop as soon as one factor comes out and do not re-test the remaining radicand.
What to do instead
Always factor the radicand into primes and pair them up: √72 = √(2³·3²) = 3·2·√2 = 6√2. After any extraction, re-check that the leftover radicand has no square factor greater than 1. Practise a mixed set including √50, √98, √180 and non-simplifiable ones like √30 so the stopping condition becomes automatic.
When rationalising a denominator like 1/(√5 − √2), the candidate multiplies by (√5 − √2) instead of the conjugate (√5 + √2), or multiplies numerator only, producing a denominator that is still irrational or a value that is no longer equal to the original expression.
The conjugate rule is remembered as 'multiply top and bottom by the same thing' without the reason: the denominator must be treated as a difference of two squares so the surds vanish. Candidates also lose track that both numerator and denominator must be multiplied.
What to do instead
Write the identity explicitly each time: (a − b)(a + b) = a² − b², so the multiplier is the denominator with the sign of the surd term flipped. Show the full working 1/(√5 − √2) × (√5 + √2)/(√5 + √2) = (√5 + √2)/3, and verify numerically with a calculator that the original and final values agree.
When solving a quadratic by factorising, the candidate finds the factors correctly but then reports the roots as the numbers inside the brackets (e.g. from (x − 3)(x + 5) = 0 writes x = 3 and x = −5 only after reversing signs inconsistently, or writes x = −3, x = 5), or divides by the leading coefficient without adjusting the roots.
The zero-product property is applied mechanically: candidates equate each bracket to zero but then solve in their head and drop a sign, especially when the constant term is negative. The step 'set each factor equal to zero' is not written down.
What to do instead
Never read roots off the brackets. Write the line (x − 3) = 0 or (x + 5) = 0, then solve each linear equation in full. Substitute both roots back into the original quadratic to confirm the result is zero; this catches every sign error in seconds.
The candidate applies the quadratic formula with a, b, c taken directly from the printed equation without first rearranging to the form ax² + bx + c = 0, so for 2x² = 5x − 3 they use a = 2, b = 5, c = −3 and obtain wrong roots.
The formula is associated with 'the numbers in the equation' rather than with standard form. Moving terms across the equals sign, and the sign changes that follow, is the step most often skipped under time pressure.
What to do instead
Make 'rearrange to ax² + bx + c = 0' a compulsory first line of working, then write a = …, b = …, c = … on a separate line with signs included. Check the discriminant b² − 4ac before substituting, and confirm the sum of the roots equals −b/a as a quick audit.
In an arithmetic progression the candidate confuses the nth term with the sum of n terms, using a + (n − 1)d when the question asks for the total, or using n/2[2a + (n − 1)d] when it asks for a single term; also mis-identifying n when the progression starts at a term other than the first.
Both formulas share the block a + (n − 1)d, so they look interchangeable in memory. Candidates also treat 'n' as the value of the last term rather than its position in the sequence.
What to do instead
Label every problem before computing: write 'need tₙ' or 'need Sₙ', then select the formula from that label. For position questions, set up a + (n − 1)d = given term and solve for n, checking that n is a positive whole number. Cross-check Sₙ for small n by adding the terms by hand.
When finding the equation of a straight line, the candidate computes the gradient correctly but then substitutes the wrong point, or uses the y-intercept as if it were a point on the line without verifying it, giving an equation that passes through only one of the given points.
The point–gradient form is memorised as y = mx + c and candidates solve for c using whichever coordinates are nearest, rather than anchoring to a stated point. The final equation is rarely tested against both points.
What to do instead
Use y − y₁ = m(x − x₁) with (x₁, y₁) written explicitly as one of the given points, then rearrange. As a final check, substitute the other point into the finished equation; if it does not satisfy it, the line is wrong. Also confirm the sign of m against a rough sketch.
The candidate solves trigonometric problems in a right triangle by choosing the ratio by habit (often sine) rather than by matching the known and required sides to the angle, or forgets to switch the calculator to degree mode, producing answers that are wildly out of scale.
SOHCAHTOA is recited but not applied as a decision rule: candidates do not first label the sides relative to the given angle as opposite, adjacent or hypotenuse. Calculator mode is an invisible setting that is never checked.
What to do instead
Before touching the calculator, mark the reference angle and label the three sides O, A, H. Write which two are involved (e.g. O and H known/required → sine), then form the equation. State the mode in the working ('degrees') and sanity-check that the answer is plausible: the hypotenuse must be the longest side and sin/cos values must lie between 0 and 1.
In circle theorems, the candidate states the result without naming the theorem, or applies the wrong one - for example treating the angle in a semicircle as 90° when the angle is not subtended by a diameter, or assuming a quadrilateral is cyclic without justification.
Circle theorems are learned as isolated pictures rather than as conditions with premises. Candidates pattern-match to a familiar diagram and skip verifying that the stated configuration actually satisfies the theorem's hypothesis.
What to do instead
For every circle problem, write the theorem name and its condition before using it: 'angle in a semicircle = 90° because AB is a diameter'. If the condition cannot be stated, the theorem cannot be used. Build a one-page table of theorem, condition, and conclusion, and practise identifying which condition is missing in flawed proofs.
When computing the volume or surface area of a combined solid (e.g. a cone on a cylinder), the candidate adds the two total surface areas, double-counting the shared circular face, or uses the slant height of the cone in place of the perpendicular height in the volume formula.
Formulas are applied to each component separately and then summed, without visualising which faces are internal and therefore excluded. The distinction between slant height l and height h is blurred because both appear in cone formulas.
What to do instead
Sketch the solid and shade only the exposed surfaces before calculating. For volume, add the component volumes; for surface area, add only the outer faces, explicitly subtracting the shared circle. Write the Pythagorean relation l² = r² + h² at the top of the working so the correct height is always available.
In probability, the candidate computes P(A or B) by adding P(A) and P(B) for events that can occur together, or calculates 'at least one' by adding single-event probabilities instead of using the complement 1 − P(none).
The addition rule is remembered without its mutually-exclusive condition, and 'at least one' is read literally as 'one', leading to a single-case calculation. Candidates do not check whether outcomes overlap.
What to do instead
Before adding, ask whether the events can happen at the same time; if they can, use P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For 'at least one', always compute 1 − P(none) and verify the result lies between 0 and 1. Test the logic on a small sample space listed out in full.