RBSE Praveshika Mathematics common mistakes
RBSE Praveshika Mathematics · Understand the traps and how to avoid them.
While simplifying surds, the candidate multiplies the radicands directly, e.g., writes √2 × √3 = √6 but then extends the rule incorrectly to addition, writing √2 + √3 = √5.
The candidate overgeneralises the product rule for radicals to addition, because both operations look like 'combining' the numbers under the root; the distributive property of radicals over addition does not exist.
What to do instead
Rehearse the two separate rules aloud: √a × √b = √(ab) only for multiplication and division; for addition, first simplify each surd to like terms (e.g., √8 + √18 = 2√2 + 3√2 = 5√2) and only add coefficients of identical radicands. Test numerically with a calculator: √2 + √3 ≈ 3.146, not √5 ≈ 2.236.
When rationalising a denominator such as 1/(√5 − √2), the candidate multiplies numerator and denominator by (√5 − √2) again instead of the conjugate (√5 + √2), leaving a radical in the denominator.
The candidate memorises 'multiply by the denominator' without understanding that the goal is to create a difference of squares; they do not recognise the conjugate as the required factor.
What to do instead
Write the identity (a − b)(a + b) = a² − b² beside every rationalisation problem and explicitly label a = √5, b = √2. Practice five conjugates in both orders: 1/(√a − √b) and 1/(√a + √b), checking that the denominator becomes a − b or a + b with no square root.
When solving a pair of linear equations by elimination, the candidate subtracts the equations term by term but forgets to subtract the constant on the right-hand side, or subtracts only the x-terms.
The candidate treats the equation as a list of separate operations rather than a balance; under time pressure they focus on eliminating the visible variable and neglect the constant column.
What to do instead
Adopt a column format: write both equations aligned as ax + by = c, then draw a vertical line and perform the same operation on every column, including the constant. After eliminating, substitute the found value into the original equation and verify both equations are satisfied before moving on.
While factorising a quadratic such as x² − 5x + 6, the candidate looks for two numbers whose sum is −5 and product is +6, but writes the factors as (x − 5)(x + 6) or (x + 2)(x + 3), mixing up the sum and product conditions.
The candidate memorises 'sum and product' without anchoring them to the expansion (x + p)(x + q) = x² + (p + q)x + pq; they confuse the coefficient of x with the constant term.
What to do instead
Always expand the proposed factors back: (x + 2)(x + 3) = x² + 5x + 6, so for x² − 5x + 6 the numbers must be −2 and −3. Drill ten quadratics by writing the target sum and product at the top of the page before guessing, and check by expansion every time.
When applying the distance formula between two points, the candidate computes √((x₂ − x₁)² + (y₂ − y₁)²) but drops the square root at the end, or subtracts coordinates in inconsistent order (x₂ − x₁ but y₁ − y₂).
The candidate conflates the distance formula with the squared-distance expression used in coordinate proofs; inconsistent order arises because squaring hides the sign, so the candidate never notices the error.
What to do instead
Write the formula with absolute-value brackets first: d = √((x₂ − x₁)² + (y₂ − y₁)²), then substitute. Because squares remove sign, any consistent order works, but choose one order (always second minus first) and state it. Finish by checking the answer against a rough sketch on graph paper.
In circle geometry, the candidate states that the angle in a semicircle is 90° but then applies it to any chord, or claims the angle between a tangent and a chord equals the angle in the alternate segment without identifying the correct alternate segment.
The candidate learns theorems as isolated phrases without the accompanying diagram conditions; they do not check which arc or segment the angle subtends.
What to do instead
For every circle theorem, redraw the figure and label the arc or segment being used before writing the reason. Practise three versions of the alternate segment theorem with the tangent on different sides, and always name the chord and the alternate segment explicitly in the written reason.
When finding the area of a triangle using Heron's formula, the candidate computes s = a + b + c instead of s = (a + b + c)/2, or forgets to take the square root of s(s − a)(s − b)(s − c).
The candidate memorises the formula as a string of symbols without the semi-perimeter definition; under exam pressure the division by 2 is the first element to be dropped.
What to do instead
Write 's = semi-perimeter = (a + b + c) ÷ 2' in words above the calculation, compute s first and box it, then substitute. Check plausibility: the area must be less than the area of a rectangle with the same base and height, and the units must be square units.
In trigonometry, the candidate evaluates sin 30° as 1/√2 or cos 60° as √3/2, swapping the standard values for 30° and 60°, especially when the angle appears in a multi-step problem.
The candidate memorises the table by row rather than by pattern; the values for 30° and 60° are mirror images and are easily interchanged when recall is rushed.
What to do instead
Rebuild the table from the 30-60-90 and 45-45-90 triangles each time rather than reciting it: in a 30-60-90 triangle with sides 1, √3, 2, sin 30° = opposite/hypotenuse = 1/2 and sin 60° = √3/2. Use the memory aid 'sin increases from 0 to 1 as the angle goes 0° to 90°' to catch swaps.
When computing the mean of a grouped frequency distribution, the candidate adds the class marks and divides by the number of classes, ignoring the frequencies, or uses class limits instead of class marks.
The candidate treats the grouped data like ungrouped data and does not internalise that each class mark represents f observations; the formula Σfx/Σf is remembered only partially.
What to do instead
Build a three-column table: class, class mark x (midpoint), and fx. Sum the f column and the fx column separately, then divide. Verify with the assumed-mean method on the same data; both methods must give the same answer.
When calculating simple or compound interest, the candidate substitutes the rate as 5 instead of 0.05 in the compound interest formula, or uses the simple interest formula for compound interest problems.
The candidate does not convert the percentage rate to a decimal, and does not distinguish between interest calculated on the principal only (SI) and interest calculated on the amount (CI).
What to do instead
Write the rate as r% = r/100 in the formula before substituting, and label the formula: SI = PRT/100 versus CI = P(1 + R/100)^T − P. For a two-year check, compute CI year by year and confirm it exceeds SI by P(R/100)².