UPMSP High School Mathematics common mistakes
UPMSP High School Mathematics · Understand the traps and how to avoid them.
When finding HCF and LCM of two numbers by prime factorisation, the candidate multiplies only the common prime factors for the LCM, or takes the highest power of every prime for the HCF, so LCM(12, 18) is written as 6 instead of 36.
The two rules are learned as a single blurry sentence ('take common ones'), and the candidate does not distinguish the 'product of lowest powers of common primes' rule from the 'product of highest powers of all primes' rule. Under time pressure the wrong rule is retrieved.
What to do instead
Write the two definitions side by side in the rough column before starting: HCF = product of the smallest power of each common prime; LCM = product of the greatest power of each prime appearing in either number. Then verify with the identity HCF × LCM = product of the two numbers (6 × 36 = 216 = 12 × 18). If the check fails, the answer is wrong; redo it immediately rather than moving on.
While solving a pair of linear equations by elimination, the candidate multiplies only one equation to match coefficients, or subtracts the equations but keeps the sign of the eliminated term, producing a wrong value of the remaining variable and then a wrong pair.
The candidate is juggling three operations (scaling, adding/subtracting, back-substituting) and drops one. Sign errors concentrate in the subtraction step because the minus sign is applied mentally to a whole expression rather than to each term.
What to do instead
Adopt a fixed written routine: (i) write both equations aligned with x-terms, y-terms and constants in columns; (ii) choose the variable with the smaller LCM of coefficients; (iii) write the multiplier for each equation explicitly above it; (iv) when subtracting, rewrite the second equation with every sign flipped and then add. Finish by substituting the pair back into both original equations. If either check fails, the error is in the arithmetic, not the method.
When a quadratic equation has a negative constant term or a leading coefficient other than 1, the candidate factorises by guessing two numbers that multiply to c but adds them to b without accounting for a, e.g. factorising 6x² − x − 2 as (6x − 2)(x + 1) and losing the correct split.
The splitting-the-middle-term method is practised almost exclusively on monic equations with positive terms, so the candidate memorises 'two numbers multiplying to c' and never internalises the ac-rule. Negative signs and a ≠ 1 expose the gap.
What to do instead
Always apply the ac-method explicitly: compute a × c including its sign, list factor pairs of that product, and pick the pair whose sum is b. Rewrite the middle term as two terms, then group in pairs and factor each pair. Verify by expanding back to the original expression. For 6x² − x − 2, ac = −12, the pair is −4 and +3, giving (3x − 2)(2x + 1).
In word problems on quadratic equations the candidate forms the equation correctly but rejects the positive root, or accepts a negative root as a length, time or speed, and reports an impossible answer.
The candidate treats root-finding as the end of the problem and does not return to the physical meaning of the variable. The rejection step is not written down, so it is simply forgotten.
What to do instead
Before solving, write one line: 'x represents ___ and must satisfy ___' (for example, x > 0 for a side, or x < the total for a remaining quantity). After obtaining both roots, test each against that condition in writing and explicitly state which root is rejected and why. Marks in this exam are usually awarded for the rejection statement, so it must appear in the answer.
Given the first term and a later term, or two terms, the candidate uses the wrong index in aₙ = a + (n − 1)d, for example treating the 10th term as a + 10d, and consequently mis-derives n, d or the required sum.
The formula is recalled as a shape rather than as 'n − 1 steps from the first term'. Off-by-one errors are invisible because the arithmetic still looks clean.
What to do instead
Anchor the formula to a concrete example every time: the 1st term has 0 steps, the 2nd has 1 step, so the nth has n − 1 steps. Write the two given terms as simultaneous equations in a and d with the (n − 1) factor shown, e.g. a + 9d = 25 and a + 4d = 10. Check the recovered d against the difference of the two given terms divided by the difference of their positions.
When asked for the sum of the first n terms of an AP, the candidate substitutes the number of terms into Sₙ = (n/2)[2a + (n − 1)d] but uses the value of the last term as n, or uses n instead of n − 1 inside the bracket.
Two similar formulas (Sₙ with a and d, and Sₙ with a and l) are mixed, and the symbol n is used for two different quantities in the same problem.
What to do instead
Before substituting, write down separately: a = ___, d = ___, n = ___ (the count of terms, not the last term's value), l = ___. Choose the form that matches the data you actually have: use Sₙ = (n/2)(a + l) when the last term is known, and Sₙ = (n/2)[2a + (n − 1)d] when d is known. Cross-check small cases by listing the first three terms and adding them by hand.
In coordinate geometry the candidate computes distance, midpoint or section-ratio by subtracting coordinates in inconsistent order, e.g. writing the distance as √[(x₂ + x₁)² + (y₂ + y₁)²] or mixing x of one point with y of the other, and also forgets to take the square root.
The formula is remembered as a rhythm of symbols rather than as 'difference of x's and difference of y's'. The squared terms look plausible even when the order is scrambled, and the outer square root is easily lost.
What to do instead
Write the formula with named points first: AB = √[(x₂ − x₁)² + (y₂ − y₁)²], and state 'difference in x, difference in y'. Substitute inside brackets, square, add, then take the root as a separate written step. Sanity-check the result against a rough sketch of the two points, since a distance larger than the coordinate spread signals an error.
In similarity problems the candidate writes the correspondence of vertices in the wrong order (e.g. ΔABC ~ ΔDEF when the correct correspondence is ΔABC ~ ΔEDF) and then applies the wrong ratio, or applies the ratio of sides directly to areas instead of its square.
The candidate matches triangles by appearance rather than by equal angles, and does not distinguish the linear ratio from the area ratio. Diagrams are read left-to-right instead of by marked angles.
What to do instead
Before writing any ratio, mark the equal angles on the figure and list the vertex correspondence angle by angle. Then write the proportion with corresponding sides in the same position. Record the two rules explicitly: corresponding sides are in ratio k, areas are in ratio k², and volumes in k³. If areas are given, take the square root first to recover k.
In circle questions the candidate applies the tangent–radius theorem to a chord, or uses the alternate segment theorem with the angle in the wrong segment, and in numerical problems assumes the tangent length equals the radius.
All circle theorems are stored together without their preconditions. The candidate recognises 'circle plus tangent' and grabs the nearest theorem without checking which lines the angle actually sits between.
What to do instead
For every circle problem, first draw the radius to the point of contact and mark the right angle; state the theorem you are using in words before using it. For the alternate segment theorem, identify the chord and the tangent at one endpoint, then name the angle in the alternate segment explicitly. Never assume a tangent length without computing it from a right triangle using r² + t² = (distance from centre)².
In heights-and-distances problems the candidate sets tan θ = adjacent/opposite, or uses the angle of elevation from the wrong end of the baseline, so the computed height is the reciprocal of the true value.
The three ratios are memorised by the mnemonic alone without being tied to a labelled right triangle, and the candidate does not redraw the situation with the observer's eye, the horizontal and the line of sight marked.
What to do instead
Redraw the figure for each problem and label the right angle, the horizontal baseline, the vertical height and the line of sight. Mark the angle of elevation at the observer's eye, between the horizontal and the line of sight. Then choose the ratio from the labelled sides only, and check plausibility: for a small elevation angle the height must be much less than the baseline.