Board Exam Prep · Mathematics (Standard)

Mathematics — solved board-pattern questions

The paper

The CBSE Class 10 Mathematics (Standard) theory paper carries 80 marks and runs for 3 hours (the remaining 20 marks are internal assessment). It has 38 compulsory questions in five sections: Section A — 20 questions of 1 mark each (multiple-choice, including two Assertion-Reason items); Section B — 5 very-short-answer questions of 2 marks each; Section C — 6 short-answer questions of 3 marks each; Section D — 4 long-answer questions of 5 marks each; and Section E — 3 case-study / source-based integrated units of 4 marks each, each with internal sub-parts. There is no overall choice, but internal choice is provided in about two questions each of the 2-, 3- and 5-mark sections and in the case-study sub-parts. Calculators are not allowed.

How to score

Budget time at roughly 1 mark per minute: about 25 min on Section A, 20 on B, 35 on C, 45 on D and 35 on E, leaving ~20 min to revise — never overrun a 5-marker chasing one tricky step.
Write the formula first, then substitute. CBSE marking schemes give a separate method mark for the correct formula even if the final arithmetic slips, so never jump straight to numbers.
Carry units through every mensuration, heights-and-distances and statistics answer (cm, cm², m) and end with a one-line conclusion — missing units and missing final statements are the most common avoidable deductions.
Draw a neat, labelled figure for every geometry, circle and heights-and-distances question; half a mark to one mark is reserved for the diagram and it also prevents set-up errors.
Treat Section A as guaranteed marks — attempt every MCQ and Assertion-Reason; for A-R decide each statement's truth first, then test whether the Reason actually explains the Assertion.
In 'prove' questions work from one side to the other and name each identity/theorem used; in case-study questions pick the internal-choice sub-part you are surest of and still show full working.

14 solved questions · 41 marks · tap to reveal the model answer

Q1 1 mark Assertion–Reason Polynomials Assertion (A): The polynomial p(x) = x² + 4x + 5 has no real zeroes. Reason (R): A quadratic polynomial can have at most two zeroes. Choose: (a) Both A and R are true and R is the correct explanation of A; (b) Both A and R are true but R is NOT the correct explanation of A; (c) A is true, R is false; (d) A is false, R is true. Show model answer ▾

Model answer

Discriminant D = b² − 4ac = (4)² − 4(1)(5) = 16 − 20 = −4 < 0, so p(x) = x² + 4x + 5 has no real zeroes → Assertion is TRUE.

A quadratic polynomial does have at most two zeroes → Reason is TRUE.

But a quadratic has no real zeroes because its discriminant is negative, not because it can have at most two zeroes; so R does not explain A.

Correct option: (b) Both A and R are true, but R is not the correct explanation of A.

Scoring tip: Decide each statement's truth independently, then ask whether R actually causes A. Many pick (a) here; the full mark goes only to (b) because the 'no real zero' conclusion comes from D < 0, not from the at-most-two-zeroes fact.

Q2 1 mark MCQ Probability A die is thrown once. The probability of getting a prime number is: (a) 1/2 (b) 1/3 (c) 2/3 (d) 1/6 Show model answer ▾

Model answer

Sample space = {1, 2, 3, 4, 5, 6}, total outcomes = 6.

Prime numbers on a die = {2, 3, 5}, favourable outcomes = 3.

P(prime) = 3/6 = 1/2.

Correct option: (a) 1/2.

Scoring tip: List the sample space and the favourable primes explicitly, then give the answer as a reduced fraction (1/2). Counting 1 as prime, or leaving 3/6 unreduced, loses the mark.

Q3 1 mark MCQ Introduction to Trigonometry The value of 9 sec²A − 9 tan²A is: (a) 9 (b) 1 (c) 0 (d) 8 Show model answer ▾

Model answer

9 sec²A − 9 tan²A = 9(sec²A − tan²A).

Using the identity sec²A − tan²A = 1:

= 9 × 1 = 9.

Correct option: (a) 9.

Scoring tip: Quote the identity sec²A − tan²A = 1 and factor out the 9 first. Trying to evaluate sec²A and tan²A separately for an unknown angle is the trap that loses the mark.

Q4 3 marks Short Real Numbers Prove that the square root of 2 (√2) is an irrational number. Show model answer ▾

Model answer

Step 1: Assume, to the contrary, that √2 is rational. Then √2 = p/q, where p and q are integers, q is not 0, and p, q have no common factor other than 1 (lowest terms).

Step 2: Squaring both sides: 2 = p²/q², so p² = 2q². Hence 2 divides p², which means 2 divides p. Write p = 2m.

Step 3: Substitute: (2m)² = 2q² → 4m² = 2q² → q² = 2m². So 2 divides q², which means 2 divides q.

Step 4: Thus 2 divides both p and q, so p and q have a common factor 2 - contradicting the assumption that p/q is in lowest terms.

Step 5: This contradiction shows the assumption is wrong. Hence √2 is irrational.

Scoring tip: State at the start that p and q are coprime, reach p² = 2q², deduce 2|p then 2|q, and finish with the contradiction - each is a separate marking-scheme step. Skipping the 'coprime' assumption breaks the proof and costs a mark.

Q5 2 marks Short Quadratic Equations Find the roots of the quadratic equation 6x² − x − 2 = 0 by factorisation. Show model answer ▾

Model answer

Step 1: Product (coefficient of x²) × (constant) = 6 × (−2) = −12. Two numbers with product −12 and sum −1 (the coefficient of x) are −4 and +3.

Step 2: 6x² − 4x + 3x − 2 = 0 → 2x(3x − 2) + 1(3x − 2) = 0 → (3x − 2)(2x + 1) = 0.

Step 3: 3x − 2 = 0 or 2x + 1 = 0 → x = 2/3 or x = −1/2.

Roots: x = 2/3 and x = −1/2.

Scoring tip: Show the middle-term split explicitly - the factorisation itself carries a mark before the roots appear. Always give BOTH roots; stopping at one root forfeits half the marks.

Q6 2 marks Short Arithmetic Progressions How many two-digit numbers are divisible by 3? Show model answer ▾

Model answer

Step 1: The two-digit multiples of 3 are 12, 15, 18, ..., 99 - an AP with first term a = 12, common difference d = 3 and last term l = 99.

Step 2: Using l = a + (n − 1)d: 99 = 12 + (n − 1)(3) → 87 = 3(n − 1) → n − 1 = 29 → n = 30.

There are 30 two-digit numbers divisible by 3.

Scoring tip: Identify a, d and l, then write l = a + (n − 1)d before substituting. Getting the boundary terms wrong (12 and 99) or using d = 1 gives a wrong count; the answer must be the whole number 30.

Q7 3 marks Short Pair of Linear Equations Solve for x and y: 2x + 3y = 11 and 2x − 4y = −24. Show model answer ▾

Model answer

Step 1 (elimination): Subtract the second equation from the first: (2x + 3y) − (2x − 4y) = 11 − (−24) → 7y = 35 → y = 5.

Step 2: Substitute y = 5 in 2x + 3y = 11: 2x + 15 = 11 → 2x = −4 → x = −2.

Step 3 (check): 2(−2) − 4(5) = −4 − 20 = −24, which satisfies the second equation.

Solution: x = −2, y = 5.

Scoring tip: Show the elimination/substitution working clearly - a mark is reserved for the method. Verify the pair in one equation to lock in the accuracy mark and end with the explicit answer x = −2, y = 5.

Q8 3 marks Short Triangles In triangle ABC, points D and E lie on AB and AC respectively with DE parallel to BC. If AD = 4 cm, DB = 6 cm and AE = 5 cm, find EC and the ratio of the areas of triangle ADE and triangle ABC. Draw a neat figure. Show model answer ▾

Model answer

Step 1 (Basic Proportionality / Thales Theorem): Since DE is parallel to BC, AD/DB = AE/EC.

Step 2: 4/6 = 5/EC → EC = (5 × 6)/4 = 30/4 = 7.5 cm.

Step 3: AB = AD + DB = 4 + 6 = 10 cm, so AD/AB = 4/10 = 2/5. Because DE is parallel to BC, triangle ADE ~ triangle ABC (AA similarity).

Step 4: Ratio of areas = (AD/AB)² = (2/5)² = 4/25.

Result: EC = 7.5 cm and ar(ADE) : ar(ABC) = 4 : 25.

Scoring tip: Name the theorem (Basic Proportionality/Thales) before writing AD/DB = AE/EC. The areas step needs the SQUARE of the side ratio, (2/5)² = 4/25 - using 2/5 unsquared is the classic error. A neat labelled figure earns its own mark.

Q9 3 marks Short Circles Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller circle. Draw a neat figure. Show model answer ▾

Model answer

Step 1: Let O be the common centre and AB the chord of the larger circle that touches the smaller circle at P. The radius OP (= 3 cm) is perpendicular to the tangent AB at the point of contact P.

Step 2: The perpendicular from the centre to a chord bisects it, so AP = PB and AB = 2 × AP.

Step 3: In right triangle OPA, OA = 5 cm (radius of larger circle) and OP = 3 cm. AP = √(OA² − OP²) = √(25 − 9) = √16 = 4 cm.

Step 4: AB = 2 × 4 = 8 cm.

Length of the chord = 8 cm.

Scoring tip: State the two key facts - tangent is perpendicular to the radius at the point of contact, and the perpendicular from the centre bisects the chord. Forgetting to double AP (writing 4 cm instead of 8 cm) is the usual full-mark loss; keep the unit cm and draw the figure.

Q10 3 marks Short Areas Related to Circles A chord of a circle of radius 14 cm subtends a right angle (90°) at the centre. Find the area of the corresponding minor segment. (Use π = 22/7.) Show model answer ▾

Model answer

Step 1: Area of sector OAB = (θ/360) × π × r² = (90/360) × (22/7) × 14² = (1/4) × (22/7) × 196 = (1/4) × 616 = 154 cm².

Step 2: Triangle OAB is right-angled at O with the two radii as the equal legs, so its area = (1/2) × 14 × 14 = 98 cm².

Step 3: Area of minor segment = area of sector − area of triangle = 154 − 98 = 56 cm².

Area of the minor segment = 56 cm².

Scoring tip: Write the sector formula with values before computing, then subtract the triangle area. For a 90° angle the triangle is simply (1/2)(14)(14); missing the cm² unit or mishandling the triangle area is the common penalty.

Q11 4 marks Case-based Coordinate Geometry Four students sit in a classroom whose floor is marked as a coordinate grid. Their positions are A(3, 4), B(6, 7), C(9, 4) and D(6, 1). (i) Find the distance between A and B. (1) (ii) Find the coordinates of the mid-point of the diagonal AC. (1) (iii) Show that ABCD is a square. (2) Show model answer ▾

Model answer

(i) AB = √[(6 − 3)² + (7 − 4)²] = √(9 + 9) = √18 = 3√2 (approximately 4.24) units.

(ii) Mid-point of AC = ((3 + 9)/2, (4 + 4)/2) = (6, 4).

(iii) BC = √[(9 − 6)² + (4 − 7)²] = √(9 + 9) = 3√2; CD = √[(6 − 9)² + (1 − 4)²] = 3√2; DA = √[(3 − 6)² + (4 − 1)²] = 3√2. So all four sides are equal (each 3√2).

Diagonals: AC = √[(9 − 3)² + (4 − 4)²] = √36 = 6; BD = √[(6 − 6)² + (1 − 7)²] = √36 = 6. The two diagonals are equal.

Since all four sides are equal AND the diagonals are equal, ABCD is a square.

Scoring tip: Each sub-part is marked separately - quote the distance and mid-point formulas before substituting. For (iii) you must show all four sides equal AND both diagonals equal to conclude 'square'; equal sides alone prove only a rhombus.

Q12 5 marks Long Some Applications of Trigonometry From the top of a building 7 m high, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower and the horizontal distance between the building and the tower. (Take √3 = 1.732.) Draw a neat labelled figure. Show model answer ▾

Model answer

Let AB = 7 m be the building (A = top, B = foot) and CD the tower (C = top, D = foot). Draw AE horizontal to meet the tower at E, so AE = BD = x (the horizontal distance) and ED = AB = 7 m.

Step 1 (depression of foot, 45°): In right triangle ABD, tan 45° = AB/BD = 7/x. Since tan 45° = 1, we get x = 7 m.

Step 2 (elevation of top, 60°): In right triangle AEC, tan 60° = CE/AE = (CD − 7)/x. So √3 = (CD − 7)/7 → CD − 7 = 7√3.

Step 3: CD = 7 + 7√3 = 7(1 + √3) = 7(1 + 1.732) = 7 × 2.732 = 19.124, that is about 19.12 m.

Height of the tower = 19.12 m; horizontal distance between building and tower = 7 m.

Scoring tip: Draw and label the figure first (the depression angle of the foot equals its elevation from the ground). Use formula-before-substitution for both tan 45° and tan 60°; the foot fixes x = 7 m before you can find the height. Keep the unit m, state both answers, and do not swap which angle gives the horizontal distance.

Q13 5 marks Long Surface Areas and Volumes A wooden toy is made by mounting a cone on a hemisphere of the same radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy. (Use π = 22/7.) Show model answer ▾

Model answer

Step 1: Radius r = 7 cm. Height of the cone h = total height − radius of hemisphere = 31 − 7 = 24 cm.

Step 2: Slant height of the cone l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm.

Step 3: Total surface area of the toy = curved surface of cone + curved surface of hemisphere = π r l + 2 π r².

Step 4: = (22/7)(7)(25) + 2(22/7)(7²) = (22 × 25) + (2 × 22 × 7) = 550 + 308 = 858 cm².

Total surface area of the toy = 858 cm².

Scoring tip: Find h = 31 − 7 = 24 and then slant l = 25 by Pythagoras - each is a step mark. Add only the CURVED surfaces (cone + hemisphere); the flat circular base is inside the toy, so adding it (or forgetting cm²) is the common deduction. Answer: 858 cm².

Q14 5 marks Long Statistics Find the median of the following frequency distribution. Class interval: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 5, 8, 20, 15, 7 Show model answer ▾

Model answer

Step 1: Cumulative frequencies are 5, 13, 33, 48, 55. Total n = 55, so n/2 = 27.5.

Step 2: The cumulative frequency just greater than 27.5 is 33, so the median class is 20-30. Here l = 20, cf (of the class before the median class) = 13, f = 20 and h = 10.

Step 3: Median = l + [(n/2 − cf)/f] × h = 20 + [(27.5 − 13)/20] × 10 = 20 + (14.5/20) × 10 = 20 + 7.25.

Step 4: Median = 27.25.

Scoring tip: Build the cumulative-frequency column and compute n/2 = 27.5 to locate the median class - this identification carries marks. Write the median formula with l, cf, f, h before substituting; picking the cf or f of the wrong class is the usual error. Final answer: 27.25.

Board-pattern questions modelled on the CBSE Class 10 exam style and NCERT syllabus — not reproductions of any copyrighted paper. Always cross-check with your textbook and the latest CBSE sample paper.

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