Haryana BSEH Secondary Mathematics - Standard common mistakes
Haryana BSEH Secondary Mathematics - Standard · Understand the traps and how to avoid them.
When finding HCF and LCM by prime factorisation, the candidate takes the product of the highest powers for the HCF and the product of the lowest powers (or of all factors) for the LCM, then records HCF × LCM = product of the two numbers even when the numbers are not coprime or when three numbers are involved.
The textbook identity HCF(a,b) × LCM(a,b) = a × b is memorised as a universal rule, so it is applied blindly to three numbers or to fractions; the candidate never checks the underlying prime-factor structure, and the words 'highest common factor' and 'lowest common multiple' are not re-derived from the factor lists.
What to do instead
For every HCF/LCM item, write the full prime factorisation first, then physically circle the shared factor columns (HCF) and the union of columns (LCM). Verify the answer by dividing: HCF must divide both numbers and LCM must be a multiple of both. Restrict the identity HCF × LCM = product to exactly two positive integers and state that restriction in the working; for three numbers compute pairwise or use prime factors. For fractions, apply HCF of numerators / LCM of denominators and LCM of numerators / HCF of denominators explicitly.
The candidate proves √2 (or √3, √5) is irrational by writing '√2 = p/q, so 2q² = p², therefore p is even and q is even, so √2 is irrational' without ever stating the assumption that p and q are coprime and without deriving the contradiction from that assumption.
The proof is memorised as a chain of sentences rather than as a proof by contradiction; the coprimality condition is treated as decoration, so the final step ('q is also even, which contradicts...') is left dangling or replaced by an assertion. This is the standard Euclid's lemma / irrationality item and markers award method marks for the assumption and the contradiction.
What to do instead
Write the proof with four labelled stages: (1) suppose √2 = p/q with p, q integers, q ≠ 0, and HCF(p, q) = 1; (2) square and rearrange to 2q² = p²; (3) since 2 divides p², by Euclid's lemma 2 divides p, so p = 2m and substitution gives q² = 2m², hence 2 divides q; (4) p and q share the factor 2, contradicting HCF(p, q) = 1, so the supposition is false. Practise the identical skeleton for √3 and √5 and for 'prove 3 + 2√5 is irrational' by assuming the sum is rational and isolating √5.
Given a cubic or quadratic and two of its zeros, the candidate divides by only one factor and then stops, or divides by (x − α)(x − β) without expanding it, and reports the remaining zero incorrectly; alternatively, the candidate uses α + β + γ = −b/a and αβγ = −d/a but forgets the middle relation αβ + βγ + γα = c/a.
The division algorithm is applied mechanically without tracking degree: after one division the quotient is still quadratic and needs a second factorisation. The coefficient–root relations are memorised for quadratics only, so the cubic sum-of-products relation is dropped or its sign is guessed.
What to do instead
For a cubic with known zeros α and β, divide f(x) by (x − α) to get a quadratic quotient, then divide again by (x − β); alternatively divide once by the expanded product (x² − (α+β)x + αβ). For coefficient relations, always write the general forms side by side: quadratic ax² + bx + c gives sum −b/a and product c/a; cubic ax³ + bx² + cx + d gives sum −b/a, sum of products taken two at a time c/a, product −d/a. Check signs by testing a simple factored example such as (x − 1)(x − 2)(x − 3).
When a linear-equation word problem is solved algebraically, the candidate finds a value for x and y and stops, without substituting back into the original equations or checking that the answer satisfies the physical constraints (non-negative ages, whole numbers of articles, speeds positive), and without converting the pair back to the quantities the question asked for.
The algebra is treated as the whole task, so the final 'answer the question' step is skipped; elimination errors in one equation are never caught because no check is performed, and constraint violations (a negative age, a fractional number of tickets) go unnoticed.
What to do instead
Adopt a fixed three-line closing routine for every pair: substitute the pair into both original equations and confirm equality; test each constraint stated in the problem (age positive, count integral, speed positive); and write a sentence naming both quantities. For the graphical method, plot at least two points per line, read the intersection, and confirm it satisfies both equations algebraically. For the substitution method, isolate the variable with coefficient 1 to avoid fractions.
In a quadratic, the candidate computes the discriminant as b² + 4ac, or uses b² − 4ac but then misreads the conclusion - declaring 'real and equal roots' when D > 0, or claiming no real roots when D = 0 - and in word problems omits the negative root that the context rejects or keeps it when the context forbids it.
The formula is recalled with the wrong sign because the quadratic formula's denominator and the minus sign blur together; the four cases (D > 0, D = 0, D < 0, D a perfect square) are memorised as a list without being tied to the shape of the graph. Word problems are solved as pure equations, so the physical meaning of each root is not examined.
What to do instead
Write D = b² − 4ac with the sign of c carried explicitly, then state the case in words before solving: D > 0 two distinct real roots, D = 0 two equal real roots, D < 0 no real roots. Sketch the parabola to confirm the count of x-intercepts. In word problems, after finding both roots, write one line per root explaining why it is accepted or rejected (e.g. 't = −3 rejected, time cannot be negative'), and for 'find k such that roots are equal' set D = 0 as an equation in k and solve, checking both k values.
For an AP the candidate uses aₙ = a + nd instead of a + (n − 1)d, or uses Sₙ = (n/2)[2a + (n − 1)d] but substitutes n as the number of the last term when the question asks for the sum of the first n terms; in 'how many terms' questions the candidate solves for n and discards a fractional or negative root without comment.
The off-by-one in the nth term arises because counting is confused with the number of gaps; the sum formula is applied without identifying what n denotes. When the resulting quadratic in n has a non-integer root, the candidate does not connect it to the impossibility of a partial term.
What to do instead
Anchor the formulas by writing the first few terms explicitly: a, a + d, a + 2d, so the nth term is a + (n − 1)d. For sums, label n as 'number of terms' before substituting and check with S₁ = a and S₂ = 2a + d. In 'how many terms' problems, solve the quadratic in n, keep only the positive integer root, and state that the other root is rejected because n must be a positive integer. For 'which term is 0' or 'which term is negative', set a + (n − 1)d equal to the target and solve for n, again requiring a positive integer.
In similarity problems the candidate writes the ratio of areas as the ratio of corresponding sides (AB/PQ) rather than its square, or applies the Basic Proportionality Theorem with the segments in the wrong order - writing AD/DB = AE/EC but substituting DB/AD - and in the converse direction assumes parallelism from a ratio that has not been shown to match.
The area theorem is remembered as 'ratio of areas equals ratio of sides' with the squaring dropped; BPT is memorised as a string of letters without a diagram convention, so the correspondence between the two transversals is lost, and the difference between a theorem and its converse is not registered.
What to do instead
For similar triangles, write the correspondence explicitly (△ABC ~ △PQR means A↔P, B↔Q, C↔R), then state: ratio of perimeters = ratio of sides k, ratio of areas = k², ratio of volumes of similar solids = k³. For BPT, redraw the triangle with DE ∥ BC, label the two segments on each side, and write AD/DB = AE/EC, then substitute only after matching each segment to its side. When asked to prove DE ∥ BC, use the converse: show AD/DB = AE/EC from the given lengths, then conclude parallelism.
In coordinate geometry the candidate applies the section formula as ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)) with the ratio m₁ : m₂ assigned to the wrong point, or uses the distance formula but forgets to square the second difference, and in area-of-triangle problems drops the ½ or the absolute value and reports a negative area.
The section formula is memorised as a symbol pattern without the rule that m₁ multiplies the coordinates of the point nearer to the second endpoint; the distance formula is written quickly and the y-difference is left unsquared. The area formula is applied as a signed determinant without the modulus, so orientation affects the sign.
What to do instead
Fix the section formula by the memory rule 'the ratio part next to a point multiplies the other point's coordinates', and verify by checking the midpoint case (ratio 1 : 1 must give the average). For distance, write d² = (x₂ − x₁)² + (y₂ − y₁)² and compute each square separately before adding. For area, compute the determinant first, then apply the modulus and the factor ½ as a separate step, and sanity-check with a right triangle of known area. For collinearity, set the area to zero and solve.
In heights-and-distances problems the candidate sets up tan θ = adjacent/opposite, or uses the angle of elevation and the angle of depression as if they were measured from the same reference, and in identity questions replaces sin²θ + cos²θ by 1 in the wrong place or cancels sin θ from a sum rather than a product.
The three ratios are memorised as 'SOH-CAH-TOA' labels but the opposite and adjacent sides are identified from the wrong vertex when the diagram is redrawn; the angle of depression is measured from the horizontal at the observer, not from the vertical, and this is forgotten. Algebraic cancellation is applied to sums because the identity looks like a factor.
What to do instead
For every heights-and-distances item, draw the diagram, mark the right angle, and label the side opposite the given angle and the side adjacent to it before choosing a ratio; write the ratio definition beside the triangle. Convert an angle of depression to the equal angle of elevation at the base by alternate angles and state this. For identities, work on one side only, convert everything to sin and cos, and cancel only common factors of a product; never divide a sum by a term. Verify by substituting θ = 30° numerically.
In tangent and chord problems the candidate assumes the radius drawn to a point of contact is perpendicular to the chord rather than to the tangent, or treats the two tangents from an external point as unequal, and in constructions the candidate draws freehand arcs without preserving the compass radius, so the claimed construction does not satisfy the stated condition.
The theorem 'radius is perpendicular to the tangent at the point of contact' is memorised without its exact contact condition, so it is misapplied to chords; equal-tangent and equal-chord properties are recalled as facts but not tied to the congruent triangles that prove them. Constructions are treated as sketches rather than as verified compass-and-straightedge steps.
What to do instead
State each circle theorem with its exact hypothesis before using it: the perpendicularity is between the radius and the tangent at the point of contact; tangents from an external point are equal because the two right triangles share the hypotenuse and a radius (RHS congruence). For constructions, list the steps (draw circle, mark external point, bisect the segment joining centre and point, draw the circle on that diameter, mark intersections), keep the compass opening fixed and show the arcs, and finish by justifying why the construction works. Leave construction lines visible.
For a combined solid the candidate adds the surface areas of the two constituent solids without subtracting the hidden circular faces where they join, or uses the diameter as the radius in cylinder, cone and sphere formulas, and in 'melting/recasting' problems equates surface areas instead of volumes.
Surface area is treated as additive like volume, and the joining face is not visualised; the radius/diameter confusion comes from reading the diagram's labelled diameter and substituting it directly. Recasting problems are misclassified because the candidate focuses on the shape rather than on the conserved quantity.
What to do instead
Before computing, sketch the combined solid and shade the faces that are exposed, explicitly subtracting the joined circle(s) from the total; for a hemisphere on a cylinder, the exposed area is curved cylinder + curved hemisphere + base circle. Write r = d/2 as a first line whenever a diameter is given. For melting, recasting, and pipe-flow problems, state 'volume is conserved' and equate volumes, never areas. Keep π symbolic until the final step to reduce rounding error.
In grouped-data statistics the candidate computes the mean using class marks but divides by the number of classes instead of the total frequency, or finds the median class correctly but substitutes the cumulative frequency of the median class instead of the cumulative frequency of the class before it into the median formula; in ogive questions the candidate reads the median from the frequency axis rather than from the cumulative-frequency axis.
The formulas are memorised as symbol strings: in median = l + [(n/2 − cf)/f] × h the symbols cf and f are not distinguished, and the mean is confused with a simple average of class marks. The ogive's axes are not identified before reading, so the intersection is projected onto the wrong axis.
What to do instead
Build a full frequency table with columns: class, frequency f, class mark x, f × x, cumulative frequency. Compute mean = Σfx / Σf using the total of the f column, not the count of rows. For the median, write n = Σf, compute n/2, identify the median class as the first class whose cumulative frequency reaches n/2, then label cf as the cumulative frequency of the preceding class and f as the frequency of the median class before substituting. For an ogive, label the horizontal axis as upper class limits and the vertical axis as cumulative frequency, and read the median by projecting from n/2 on the vertical axis to the curve and then down to the horizontal axis. For the mode, confirm the modal class is the one with the highest frequency and apply mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h with f₁ the modal frequency.
In probability the candidate writes the probability of an event as the number of favourable outcomes divided by the number of favourable outcomes, or counts outcomes without listing the sample space, so that for two dice the sum 7 is given probability 1/12 (one outcome) instead of 6/36; the candidate also reports a probability greater than 1 or a negative probability without noticing.
The sample space is assumed rather than written, so equally likely outcomes are not established and the denominator is guessed; the complement and the range 0 ≤ P(E) ≤ 1 are memorised as facts but not used as a check. Cards and dice problems are answered from memory of similar questions, which is unreliable when the wording changes (with or without replacement).
What to do instead
For every probability item, write the sample space explicitly (a 6 × 6 grid for two dice, a list of 52 cards by suit and rank for a deck) and count the total outcomes before counting favourable ones. State the formula P(E) = favourable / total with the total fixed by the sample space. Check 0 ≤ P(E) ≤ 1 and use P(not E) = 1 − P(E) as a cross-check. For 'at least one' events, compute the complement. For drawing without replacement, reduce both numerator and denominator for the second draw.