Bihar BSEB Matric Mathematics common mistakes
Bihar BSEB Matric Mathematics · Understand the traps and how to avoid them.
When finding HCF and LCM by prime factorisation, the candidate takes the highest power of every prime for the HCF and the lowest power for the LCM, then reports HCF × LCM equal to the product of the two numbers even when they are not coprime.
The two rules are memorised as a single slogan ('HCF = smallest, LCM = biggest') without attaching each rule to the right question, and the HCF × LCM = product identity is remembered without its coprimality condition.
What to do instead
Re-derive both on 12 and 18: HCF takes the common primes with the lowest powers (2 × 3 = 6), LCM takes all primes with the highest powers (2² × 3² = 36). Then test the identity only on coprime pairs such as 8 and 15, and state explicitly in writing that HCF × LCM = product holds only for coprime numbers.
While using Euclid's division lemma, the candidate writes a = bq − r or a = bq + r with r allowed to equal b, and stops the chain of divisions before the remainder reaches zero.
The condition 0 ≤ r < b is glossed over because attention goes to the quotient, and the process is treated as finished once a 'small' remainder appears.
What to do instead
Before each division, write the inequality 0 ≤ r < b above the line and check the remainder against it. Practise the full chain for 4052 and 12576 until the last line reads remainder 0, and state that the divisor at that line is the HCF.
After finding the zeros of a quadratic polynomial, the candidate writes the sum as α − β or forgets the minus sign, giving sum = −b/a instead of b/a, and product = −c/a instead of c/a.
The formulas are recalled from the coefficients of the equation ax² + bx + c = 0 rather than the polynomial, and the sign of the constant term is lost when the polynomial is written with a negative middle term.
What to do instead
Always write the polynomial in standard form first, then copy the coefficients onto a three-line table (a, b, c) before substituting. Check on x² − 5x + 6: the zeros 2 and 3 must give sum 5 = −(−5)/1 and product 6 = 6/1; if the signs fail the check, the formula has been misquoted.
When solving a pair of linear equations, the candidate adds the two equations without first making a coefficient equal, or substitutes a value back into the equation it has just rearranged rather than into the other one, and reports a solution that does not satisfy both equations.
The elimination step is done mechanically without checking that one variable will actually cancel, and verification is skipped because the arithmetic feels finished.
What to do instead
Before adding or subtracting, write the multiplier needed for each equation so that one coefficient becomes equal in magnitude. After obtaining x and y, substitute both into the untouched original equations and show the two true statements in the answer. If either fails, the error is in the elimination, not the arithmetic.
The candidate answers 'no solution' or 'infinitely many solutions' for a pair of equations by comparing a₁/a₂ and b₁/b₂ only, without checking c₁/c₂, so a pair with equal ratios of coefficients but different constants is declared consistent.
The three ratio conditions are remembered as two, and the distinction between parallel lines and coincident lines is not linked to the constant terms.
What to do instead
Write all three ratios a₁/a₂, b₁/b₂, c₁/c₂ side by side for every problem. Only a₁/a₂ ≠ b₁/b₂ gives a unique solution; a₁/a₂ = b₁/b₂ = c₁/c₂ gives infinitely many; a₁/a₂ = b₁/b₂ ≠ c₁/c₂ gives none. Confirm with a rough sketch of the two lines.
When solving a quadratic equation, the candidate divides both sides by x (or by the variable factor) and loses one root, and in word problems accepts a negative or fractional root that the physical situation cannot allow.
Cancelling looks like simplification, and the habit of treating every algebraic solution as a valid answer is carried over from pure equation-solving.
What to do instead
Factorise to x(ax + b) = 0 and write both roots from the zero-product rule before cancelling anything. In word problems, state the quantity the variable represents and its permitted range (a length, a number of articles, an age) before solving, then reject any root outside that range with a one-line reason.
In an arithmetic progression, the candidate uses n instead of n − 1 in aₙ = a + (n − 1)d, or finds the number of terms and then uses that count as the last term rather than as n in the sum formula.
The two formulas aₙ = a + (n − 1)d and Sₙ = n/2[2a + (n − 1)d] share symbols, and the role of n as a position rather than a quantity is not fixed in mind.
What to do instead
Label the first term position 1 and compute a few terms by hand to see that the first term has zero differences added. For every sum problem, first find n from the last term using aₙ = a + (n − 1)d, write n in a box, then substitute that boxed value into the sum formula.
When stating a similarity criterion, the candidate writes 'AAA' as a valid criterion for congruence, or names corresponding vertices in the wrong order so that the ratio of sides is set up incorrectly.
Similarity and congruence criteria are learned together and their names overlap, and the order of letters in a triangle name is treated as a label rather than as a statement of correspondence.
What to do instead
Keep two separate lists: for similarity AA, SSS and SAS; for congruence SSS, SAS, ASA, AAS and RHS. Write each triangle name with vertices matched to equal angles, then read the equal ratios straight off the letter order, e.g. AB/PQ = BC/QR = CA/RP.
In coordinate geometry, the candidate substitutes coordinates into the distance or section formula with the x and y values interchanged, or drops the minus signs inside the squared brackets and then takes a square root of a negative-looking expression.
The formulas are written from memory in a single line, and the squared terms are expanded mentally without writing the subtraction of coordinates explicitly.
What to do instead
Write the formula with empty brackets, then fill x₁, y₁, x₂, y₂ from a labelled sketch of the points. Keep the subtraction inside brackets until it is squared, since the square removes the sign. Check the result against the sketch: a horizontal segment must give a distance equal to the difference of x-coordinates.
In trigonometry, the candidate uses the standard values of sin, cos and tan for 30°, 45° and 60° from memory but swaps sin 30° with cos 60°, and in height-and-distance problems measures the angle of elevation from the horizontal at the wrong end of the diagram.
The table of standard values is memorised as a sequence of numbers without the underlying 1:√3:2 triangle, and the angle of elevation is not tied to the observer's eye line.
What to do instead
Reconstruct the values from the 30-60-90 and 45-45-90 triangles each time rather than reciting them. In every height-and-distance problem, draw the horizontal through the observer's eye, mark the angle between that line and the line of sight, and label the right angle before choosing tan (for opposite and adjacent) or sin/cos.
In circle questions, the candidate applies the tangent-radius theorem to a chord, or assumes a line is a tangent because it touches the drawn circle, and in constructions omits the arcs that justify the construction.
The diagram is trusted over the stated conditions, and the difference between a tangent (one point of contact, perpendicular to the radius at that point) and a chord is not checked against the given data.
What to do instead
Before using any circle theorem, write which line is a tangent and which is a chord from the given conditions, and mark the right angle only at the point of contact with the radius. In constructions, leave all arcs visible and add the one-line justification for each step, since marks are awarded for the construction marks, not only the final figure.
In areas related to circles, the candidate adds the areas of a sector and a triangle to find the area of a segment, and uses the diameter in place of the radius in πr² and 2πr.
The segment is treated as a sum of its visible parts rather than as a difference, and the diameter is often the labelled quantity in the figure.
What to do instead
Write the segment as sector minus triangle and check that the answer is smaller than the sector. Immediately after reading any circle problem, write r = d/2 in the margin and use only r thereafter. For a semicircle or quadrant, state the fraction of the circle (1/2 or 1/4) before substituting.
In surface areas and volumes, the candidate adds the curved surface areas of a cone and a hemisphere without removing the common circular base, and mixes units such as cm and m within one calculation.
Each solid's formula is applied correctly in isolation, but the joined solid is not examined for surfaces that disappear at the join, and units are copied from the question without conversion.
What to do instead
Sketch the combined solid and shade the surfaces that remain exposed, crossing out the base that is hidden at the join. Convert every length to one unit before substituting, and state the final answer with its unit and, for volume, in the unit cubed.
In statistics, the candidate uses the wrong class mark or the wrong cumulative frequency when finding the median of a grouped distribution, and in probability writes the probability of an event as a count rather than as a ratio out of the total number of outcomes.
The median formula's f, cf and h are substituted from whichever column looks convenient, and probability is answered as 'number of favourable outcomes' without the denominator.
What to do instead
Build the frequency table with columns for class, frequency, and cumulative frequency, and mark the median class where the cumulative frequency first reaches n/2. In the median formula, label f as the frequency of the median class (not the cumulative frequency) and cf as the cumulative frequency of the class before it. For probability, always write P(E) = favourable outcomes / total outcomes and check that the value lies between 0 and 1.