Maharashtra SSC Mathematics (MSBSHSE) common mistakes
Maharashtra SSC Mathematics (MSBSHSE) · Understand the traps and how to avoid them.
When solving a quadratic equation by the formula, candidates substitute the coefficient of x as b without keeping its sign, so for 2x² − 5x + 3 = 0 they write b = 5 and compute (−5 ± √(25−24))/4 instead of (5 ± √(25−24))/4, producing roots 1 and 3/2 with the wrong signs or a spurious negative root.
The formula is memorised as x = (−b ± √(b² − 4ac))/2a as a letter pattern, and candidates treat the '−' in front of b as already applied, then plug the bare magnitude. The standard form ax² + bx + c = 0 is written with b as a signed number, but students copy b from the printed equation after mentally stripping the minus.
What to do instead
Before any substitution, rewrite the equation and list a, b, c in a three-row table with their signs, then substitute into the formula with brackets: x = (−(b) ± √((b)² − 4ac))/(2a). Do five equations where b is negative and five where c is negative, and check each root by substituting back into the original equation, not the rearranged one.
While solving simultaneous equations by elimination, candidates multiply only one term of an equation by the chosen factor instead of every term, so 3x + 2y = 12 scaled by 2 becomes 6x + 2y = 12 rather than 6x + 4y = 24, and the resulting solution satisfies neither original equation.
The operation is seen as 'multiply the x-term to match coefficients' rather than 'multiply the whole equation'. Because the equals sign invites a left-to-right reading, the constant on the right and the other variable term are overlooked. The error is silent - the arithmetic afterwards is correct - so it is rarely caught by re-reading the working.
What to do instead
Adopt the habit of writing a bracket round the whole equation before scaling: 2(3x + 2y) = 2(12). After finding the solution, substitute both values into both original equations in full and confirm both hold; make this verification a compulsory step in every simultaneous-equation problem, including word problems.
When a problem asks for the value of a polynomial or an algebraic expression at a given value, candidates substitute the number into one term only - for p(x) = x² − 4x + 7 they compute p(2) = 2² − 4·2 = −4 and forget the constant +7, or they evaluate −4x at x = 2 as −42 by writing the multiplication without a sign.
The expression is read term by term and the constant term is felt to be independent of x, so it is dropped. Where a coefficient sits directly against the variable, the multiplication sign is invisible and the digits merge. Both are notation-reading failures rather than algebra failures.
What to do instead
Rewrite the expression with explicit multiplication signs and brackets before substituting: p(2) = (2)² − 4·(2) + 7. Work down one line per term, then add. Practise with expressions containing a constant, a squared term and a negative coefficient, and check the number of terms in the answer line matches the number in the expression.
In a triangle problem involving the basic proportionality theorem, candidates write the ratio of the segments on one side equal to the ratio of the segments on the other side in the wrong order, giving AD/DB = AE/EC as AD/DB = EC/AE, or they apply the theorem when the line is not stated to be parallel to the third side.
The theorem is recalled as a picture rather than as a statement with a condition. Because the diagram looks symmetric, the correspondence between the two sides is guessed. The parallel condition, which is the entire justification, is treated as decoration and omitted from the proof.
What to do instead
Write the theorem statement in full before using it: 'If a line is parallel to one side of a triangle and intersects the other two sides in distinct points, it divides those sides in the same ratio.' Then name the correspondence explicitly - AD/DB = AE/EC - and state the parallel condition as a given line in the proof. Redraw the figure with the parallel line deliberately off-centre so the symmetry cue is removed.
When proving two triangles congruent or similar, candidates list the vertices in arbitrary order, writing ΔABC ≅ ΔDEF when the actual correspondence is A↔D, B↔F, C↔E, and then draw conclusions about the wrong pairs of sides or angles.
The symbolic notation is treated as a label for the pair of triangles rather than as an ordered correspondence. Candidates match triangles visually by position on the page. The convention that order encodes the correspondence is often never made explicit in revision.
What to do instead
Before writing the congruence or similarity statement, write the three vertex pairings in a column (A↔D, B↔F, C↔E) and only then assemble the statement from that column. Mark corresponding angles with matching arc counts on the figure. When using the result, read sides and angles off the pairing column, never off the diagram.
When computing the volume or surface area of a combined solid, candidates add the volumes of the component solids but also add their full surface areas, counting the joining face twice, or they use the diameter in place of the radius in πr²h.
The formula is applied mechanically to each shape and the results summed, because 'combined solid' is read as 'add the two answers'. The hidden internal face is invisible in the formula. The radius/diameter confusion comes from problems that state the diameter, which is the more natural measurement to give.
What to do instead
Draw the combined solid, shade the internal joining face, and write 'subtract 2 × area of joining face' beside the surface-area working before computing anything. At the start of every mensuration problem, write r = d/2 = ... as an explicit line. Do one cylinder-plus-hemisphere and one cone-plus-cylinder problem end to end with the internal-face subtraction shown.
When a problem gives a ratio such as 3 : 4 and a total, candidates divide the total by the ratio (e.g. 140 ÷ 3 : 4) or split it as 3/4 and 1/4 of the total, instead of dividing by the sum of the parts and multiplying by each part.
The ratio is read as a fraction of the whole. The idea that 3 : 4 means five equal parts is a modelling step that is skipped when candidates move straight to arithmetic. The error is common in partnership profit-sharing and mixture problems where the same ratio language appears.
What to do instead
Always write 'total parts = 3 + 4 = 7' as the first line, then 'one part = total ÷ 7', then each share. Check that the shares add back to the given total. Practise with three-term ratios and with ratios that must first be converted to a common base (as in A : B = 2 : 3, B : C = 4 : 5).
In percentage problems involving successive change or reverse percentage, candidates add the two percentage changes, treating a 20% increase followed by a 20% decrease as no change, or they take a discount of 20% on the marked price and then add tax to the marked price rather than to the discounted price.
Percentages are treated as additive quantities because they are written with the same symbol. The base to which each percentage applies is not identified. Successive-change problems are also usually worded in a single long sentence, so the order of operations is lost.
What to do instead
For every percentage, write 'this % is of ___' before calculating. For successive change use the multiplier method: ×1.20 then ×0.80 gives ×0.96, a 4% net decrease. For discount-then-tax problems, draw a short chain: marked price → less discount → plus tax, with a number at each arrow. Verify that the final figure is plausible relative to the original.
When using a trigonometric identity to simplify or prove an expression, candidates cancel sin θ from a sum - writing (sin θ + cos θ)/sin θ = cos θ - or they 'move' a term across the equals sign in an identity proof without operating on both sides.
Algebraic cancellation rules are applied to trigonometric function names as if they were factors. In a sum, sin θ is a term, not a factor, but the visual similarity to 2x/2 = x is strong. Identity proofs also invite the habit of manipulating the equation as a whole rather than one side.
What to do instead
State the rule explicitly: only factors may be cancelled. Convert everything to sin and cos first, then factorise the numerator before cancelling. For identity proofs, work on one side only (usually the more complicated one) and write 'LHS = ... = RHS' as a chain, never crossing terms between sides. Test each step numerically with θ = 30° to catch invalid cancellations.
When finding the mean of a grouped frequency distribution, candidates use the class limits (e.g. 10 and 20) instead of the class marks (15), or they divide the sum of f·x by the number of classes rather than by the total frequency Σf.
The table presents class intervals prominently and the midpoint is an intermediate quantity that candidates skip when working quickly. Dividing by the number of classes is a leftover from the ungrouped mean, where the divisor is the count of items. The two errors often occur together and produce a plausible-looking answer.
What to do instead
Add a class-mark column to every frequency table before starting, and label the final divisor Σf in words ('total frequency'). Check the answer lies inside the range of class marks - a mean outside that range signals a divisor error. Do one problem fully in a five-column table: class, mark, frequency, product, total.
When computing the probability of an event, candidates write the number of favourable outcomes over the number of favourable outcomes plus something else, or they treat outcomes as equally likely when they are not - for two coins, listing {two heads, two tails, one of each} as three equally likely outcomes and giving P(one head) = 1/3 instead of 1/2.
The sample space is written from the description of the event rather than enumerated outcome by outcome, so indistinguishable-looking results are merged. The equally-likely condition of the classical definition is assumed rather than checked. Dice and coin problems are the usual site of this.
What to do instead
Enumerate the sample space explicitly and count it: for two coins write HH, HT, TH, TT and state n(S) = 4. Label each outcome as equally likely or not before applying the formula. Use a table or tree diagram for two-dice problems and count the cells. Always check 0 ≤ P ≤ 1 and that the probabilities of complementary events sum to 1.