RBSE Secondary Mathematics common mistakes
RBSE Secondary Mathematics · Understand the traps and how to avoid them.
While finding HCF and LCM by prime factorisation, the candidate takes the LCM as the product of only the highest powers present in one number, or takes HCF as the product of all prime factors of both numbers including the lower powers, so HCF(12, 18) is written as 36 or LCM as 6.
The candidate memorises 'HCF = common factors, LCM = all factors' without the rule that HCF uses the smallest power of each common prime and LCM uses the highest power of every prime appearing in either number. Under exam pressure the two rules get swapped.
What to do instead
Write both numbers in prime factorisation form side by side, then draw two columns: for HCF take only primes present in both with the lower exponent, for LCM take every prime with the higher exponent. Always verify with the identity HCF x LCM = product of the two numbers, and check that the answer divides/ is divisible as expected.
While proving that sqrt(2) or sqrt(3) is irrational, the candidate assumes the fraction p/q is already in lowest terms but never uses this, or concludes 'so p and q are both even' without stating that this contradicts the assumption that HCF(p, q) = 1.
The proof is learned as a sequence of algebraic steps without understanding that the contradiction comes from the coprimality assumption, not from the evenness alone. The candidate also forgets to state p, q integers with q not equal to 0 at the start.
What to do instead
Write the proof as a fixed four-line template: assume rational in lowest terms with HCF 1; square and rearrange; show the prime divides both numerator and denominator; explicitly say 'this contradicts HCF(p, q) = 1, hence our assumption is false'. Practise the template on sqrt(3) and sqrt(5) so the structure, not the numbers, is memorised.
When finding the zeroes of a quadratic and verifying the relationship with coefficients, the candidate writes sum of zeroes as -b/a but substitutes the constant term for b, or forgets the minus sign, giving sum = c/a.
The formulas alpha + beta = -b/a and alpha beta = c/a are recalled as a pair without anchoring which coefficient is which. Signs are especially lost when the equation is written as ax^2 + bx + c = 0 with a negative b.
What to do instead
Before substituting, rewrite the equation and label a, b, c explicitly underneath each coefficient, including signs. Compute sum and product separately, then check against the actual zeroes found by factorisation. If the zeroes are 2 and 3, the sum must be 5, so -b/a must equal 5; use this as a self-check every time.
In word problems on pairs of linear equations (age, boat-stream, or two-digit number problems), the candidate forms only one equation and then solves it as if it were a single-variable problem, or sets up both equations with the same variable on both sides.
The candidate focuses on translating the first sentence and stops, or defines variables vaguely ('let the number be x') without distinguishing the two unknowns such as present age and age after five years.
What to do instead
Always begin by writing two clear 'Let ...' statements, one per unknown, with units. Then translate each sentence into a separate equation and number them (1) and (2). Before solving, confirm there are two unknowns and two independent equations; if only one equation emerges, a second relation has been missed in the question.
When solving a quadratic by the quadratic formula, the candidate computes the discriminant as b^2 - 4ac but substitutes -b as b, or drops the plus-minus and reports only one root, so a two-root equation is answered with a single value.
The formula is recalled in fragments; the negative sign in front of b and the plus-minus symbol are the parts most often lost. The candidate also stops after finding the discriminant instead of completing the root calculation.
What to do instead
Write the formula in full at the top of the working, then substitute inside brackets: x = [-(b) +/- sqrt((b)^2 - 4(a)(c))] / 2(a). Compute the discriminant on a separate line and state whether it is positive, zero, or negative before proceeding. Finish with both roots written explicitly, and check each by substitution.
In arithmetic progression problems, the candidate uses a_n = a + nd instead of a + (n-1)d, or finds the number of terms by dividing the difference of first and last term by d without adding 1, so an AP with 10 terms is reported as having 9.
The off-by-one in the index is a counting error: the first term already corresponds to n = 1, not n = 0. The formula is memorised loosely and the derivation from a, a+d, a+2d is not reconstructed.
What to do instead
Derive the formula once by writing the first four terms and observing that the coefficient of d is one less than the term number. For the count of terms, use n = (last term - first term)/d + 1 and always test on a short AP such as 3, 5, 7, 9 where the answer is visibly 4.
When applying the Basic Proportionality Theorem or similarity criteria, the candidate writes ratios of sides that are not corresponding, for example AB/DE = BC/DF when the triangles are named ABC and DEF but the correspondence is actually ABC ~ EDF, or uses AA without stating which two angles are equal.
The candidate matches sides by position in the diagram rather than by the vertex correspondence declared in the similarity statement. The naming order of similar triangles is treated as cosmetic rather than as the definition of which vertices correspond.
What to do instead
First write the similarity statement with matched vertices, then list corresponding sides underneath in the same order before forming any ratio. For every similarity proof, name the two equal angles explicitly with reasons (given, vertically opposite, alternate angles). Cross-multiply only after the correspondence list is written.
In coordinate geometry, the candidate substitutes coordinates into the distance or section formula with x and y interchanged, or uses the midpoint formula when the point divides the segment in a given ratio, producing the midpoint regardless of the ratio.
The formulas look similar and are recalled as a block of symbols without the meaning of each. The candidate does not pause to identify whether the question asks for a midpoint, a trisection point, or a general ratio division.
What to do instead
Label the two points as (x1, y1) and (x2, y2) in writing before substituting, and underline the words 'midpoint' or 'in the ratio m:n' in the question. Use the section formula with m = n = 1 as a check that it reduces to the midpoint formula, so the general formula is seen to contain the special case.
While evaluating trigonometric ratios of standard angles, the candidate writes sin 30 = 1/2 but cos 60 = 1/2 as well while also writing sin 60 = 1/2, mixing the values of sine and cosine at complementary angles, or treats sin(A+B) as sin A + sin B.
The table of standard values is memorised row by row without the pattern that sine increases from 0 to 1 as the angle goes from 0 to 90 while cosine decreases. The candidate also applies algebraic distributivity to a function that is not linear.
What to do instead
Rebuild the table from the pattern sqrt(0)/2, sqrt(1)/2, sqrt(2)/2, sqrt(3)/2, sqrt(4)/2 for sine at 0, 30, 45, 60, 90 and reverse it for cosine. For identities, always test with a numerical angle, e.g. check whether sin(30+60) equals sin 30 + sin 60, to see that the distributive step is invalid.
In problems on areas related to circles or surface areas and volumes, the candidate uses the diameter as the radius, or leaves the answer in terms of pi when a numerical value is required, or forgets to convert units such as cm to m before combining volumes.
Diagrams give the diameter more often than the radius, and the candidate substitutes the given number directly. Unit conversion and the choice of pi = 22/7 versus 3.14 are treated as afterthoughts rather than as part of the setup.
What to do instead
At the start of every mensuration problem, write 'r = d/2 = ...' explicitly and circle the radius value used. Convert all lengths to one unit before any formula. Decide on the value of pi from the numbers given (if a radius is a multiple of 7, use 22/7) and state it. End by checking that the units of the answer match the quantity asked (square units for area, cubic for volume).