CBSE Class 10 Maths Practice Paper 2026-27 Set 1 with Answer Key (PDF)

August 2, 2026

Maths is the paper where a single silly slip costs you a whole mark, and where the difference between 80 and 65 is almost always practice under a clock rather than more theory. This is Set 1 of our original CBSE Class 10 Maths (Standard) practice papers for the 2026-27 session — 38 questions, 80 marks, three hours.

Every question is written from scratch by the PrincipalSaab desk, so you are testing whether you understand the method and not whether you remember a worked example. The paper follows the official CBSE 2026-27 blueprint exactly: Number Systems 6, Algebra 20, Coordinate Geometry 6, Geometry 15, Trigonometry 12, Mensuration 10 and Statistics & Probability 11, laid out in five sections A to E.

How to use this paper

Sit the full 3 hours in one go, with no calculator.
Show every step. In the board exam, method marks are given even when the final answer slips.
Do not open the answer key until the three hours are up — the reveal button is near the bottom.
For each mistake, write down whether it was a concept error or a careless error. The two need different fixes.

Download the paper and the answer key

Prefer writing on paper? Download both PDFs below. The question paper is set in the board format, and the answer key carries a full worked solution and marking scheme for every question, including both options of every internal choice.

⬇ Download Question Paper (PDF)

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General instructions

Time Allowed: 3 Hours | Maximum Marks: 80

This question paper contains 38 questions. All questions are compulsory.
The paper is divided into 5 Sections — A, B, C, D and E.
Section A: Q1–18 are multiple choice questions and Q19–20 are assertion–reason questions, of 1 mark each.
Section B: Q21–25 are very short answer questions of 2 marks each.
Section C: Q26–31 are short answer questions of 3 marks each.
Section D: Q32–35 are long answer questions of 5 marks each.
Section E: Q36–38 are case-study based questions of 4 marks each, with sub-parts of 1, 1 and 2 marks.
There is no overall choice. An internal choice is provided in 2 questions of Section B, 2 of Section C, 2 of Section D, and in the 2-mark sub-part of every question of Section E.
Draw neat figures wherever required. Take π = 22/7 unless stated otherwise.
Use of a calculator is not allowed.

Section A

Q1 to Q20 carry 1 mark each. Q1–Q18 are multiple choice questions; Q19–Q20 are assertion–reason questions. — 20 marks

Q1. Two positive integers are 2³ × 3² × 5 and 2² × 3³ × 7. Their HCF is: [1]

(A) 2² × 3²

(B) 2³ × 3³

(D) 2³ × 3³ × 5 × 7

Q2. If HCF(a, b) = 12 and a × b = 1728, then LCM(a, b) is: [1]

(A) 144

(B) 288

(D) 1728

Q3. Which one of the following rational numbers has a terminating decimal expansion? [1]

(A) 7/45

(B) 9/64

(D) 13/42

Q4. If 3 is a zero of the polynomial p(x) = x² – 7x + k, then the value of k is: [1]

(A) 10

(B) 12

(D) 21

Q5. The quadratic equation 2x² – 4x + 3 = 0 has: [1]

(A) two distinct real roots

(B) two equal real roots

(D) more than two real roots

Q6. The pair of linear equations 3x + 2y = 5 and 6x + 4y = 11 is: [1]

(A) consistent with a unique solution

(B) consistent with infinitely many solutions

(D) consistent with exactly two solutions

Q7. The 15th term of the AP 7, 11, 15, 19, … is: [1]

(A) 59

(B) 63

(D) 71

Q8. If the nth term of an AP is given by a_n = 5 – 3n, then its common difference is: [1]

(A) 5

(B) 3

(D) 2

Q9. If M(3, –2) is the midpoint of the line segment joining A(1, 4) and B(x, y), then the coordinates of B are: [1]

(A) (5, –8)

(B) (2, 1)

(D) (–1, 10)

Q10. In triangles ABC and DEF, ∠A = ∠D and ∠B = ∠E. If AB = 4 cm, DE = 6 cm and BC = 6 cm, then EF is: [1]

(A) 4 cm

(B) 8 cm

(D) 12 cm

Q11. In △ABC and △DEF it is given that AB/DE = BC/EF = CA/FD. The two triangles are similar by the: [1]

(A) AA criterion

(B) SAS criterion

(D) RHS criterion

Q12. The tangents drawn at the two end points of a diameter of a circle are: [1]

(A) perpendicular to each other

(B) parallel to each other

(D) inclined to each other at 60°

Q13. In △ABC, D lies on AB and E lies on AC with DE ∥ BC. If AD = 3 cm, DB = 5 cm and AE = 4.5 cm, then AC is: [1]

(A) 7.5 cm

(B) 9 cm

(D) 13.5 cm

Q14. If sin θ = 3/5, where θ is acute, then tan θ is: [1]

(A) 3/4

(B) 4/3

(D) 5/3

Q15. The value of (tan 25°)/(cot 65°) is: [1]

(A) 0

(B) 1/2

(D) 2

Q16. A circular sheet of radius 10.5 cm is cut into 6 identical sectors. The area of each sector is (take π = 22/7): [1]

(A) 57.75 cm²

(B) 115.5 cm²

(D) 346.5 cm²

Q17. A solid is formed by mounting a hemisphere of radius 7 cm on one flat end of a right circular cylinder of the same radius and height 10 cm. The curved surface area of the solid, excluding its flat circular base, is (take π = 22/7): [1]

(A) 440 cm²

(B) 594 cm²

(D) 902 cm²

Q18. A bag contains 5 red, 4 green and 6 blue balls. One ball is drawn at random. The probability that it is not blue is: [1]

(A) 2/5

(B) 3/5

(D) 4/15

Q19. Assertion (A): If the probability of an event happening is 0.35, then the probability of the event not happening is 0.65. [1]

Reason (R): For any event E, P(E) + P(not E) = 1.

(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.

(d) A is false but R is true.

Q20. Assertion (A): If two triangles are similar, then their corresponding sides are always equal in length. [1]

Reason (R): In two similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio.

(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.

(d) A is false but R is true.

Section B

Q21 to Q25 are very short answer questions carrying 2 marks each. — 10 marks

Q21. Find the value of k for which the quadratic equation kx² – 6x + 2 = 0 has two equal real roots. [2]

Q22. The point P divides the line segment joining A(–2, 3) and B(6, –5) internally in the ratio 3 : 1. Find the coordinates of P. [2]

Find the value of a for which the point P(a, 3) is equidistant from A(4, –1) and B(–2, 5).

Q23. In △ABC, the point D lies on AB and the point E lies on AC. Given AD = 3 cm, DB = 4.5 cm, AE = 4 cm and EC = 6 cm, show that DE is parallel to BC. [2]

Q24. If 5 tan θ = 12, find the value of (5 sin θ – 3 cos θ)/(5 sin θ + 3 cos θ). [2]

If tan θ + cot θ = 3, find the value of tan²θ + cot²θ.

Q25. In a school quiz, the next question will be answered by exactly one of three students — Aarav, Bela or Chetan. The probability that Aarav answers it is 0.4 and the probability that Bela answers it is 0.35. Find (a) the probability that Chetan answers it, (b) the probability that it is not answered by Aarav. [2]

Section C

Q26 to Q31 are short answer questions carrying 3 marks each. — 18 marks

Q26. A stationery shop has 168 pencils and 120 erasers. The owner wants to make identical gift packs using all of them, so that every pack has the same number of pencils and the same number of erasers, with nothing left over. [3]

(a) Write 168 and 120 as products of their prime factors.

(b) Find the greatest number of such gift packs that can be made.

Q27. The points A(–3, 7) and B(5, –1) are the ends of a line segment. [3]

(a) Find the coordinates of the midpoint of AB.

(b) Find the length of AB.

The points A(1, 2), B(4, 6) and C(x, 10) are such that AB = BC. Find the possible values of x, and state the length of AB.

Q28. A circle with centre O has radius 9 cm. From an external point T, two tangents TA and TB are drawn, touching the circle at A and B. The distance OT is 15 cm. [3]

(a) Explain why ∠OAT and ∠OBT are both right angles.

(b) Find the length of each tangent.

Q29. If sec θ + tan θ = 5/3, find the value of sin θ. [3]

Q30. A circular flower bed of radius 42 m has a path marked along a sector of central angle 30°. Taking π = 22/7, find (a) the area of this sector and (b) its perimeter. [3]

A solid is in the shape of a right circular cone of radius 7 cm and height 24 cm, mounted on a right circular cylinder of the same radius and height 10 cm. Find (a) the slant height of the cone and (b) the total volume of the solid. (Take π = 22/7)

Q31. A bag contains 8 red, 6 white and 10 black marbles, all of the same size. One marble is drawn at random. Find the probability that the marble drawn is [3]

(a) black, (b) not white, (c) either red or white.

Section D

Q32 to Q35 are long answer questions carrying 5 marks each. — 20 marks

Q32. A rectangular sheet of card has a perimeter of 44 cm. A square of side 2 cm is cut away from each of its four corners, and the four flaps that remain are folded up to form an open box of height 2 cm. The base of the box has an area of 45 cm². Find the length and the breadth of the original sheet of card. [5]

Q33. (a) Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. (3) [5]

(b) In △ABC, D lies on AB and E lies on AC with DE ∥ BC. If AD = 4 cm, DB = 6 cm and AC = 15 cm, find AE. (2)

(a) Prove that the lengths of the two tangents drawn from an external point to a circle are equal. (3)

(b) Two tangents are drawn from an external point P to a circle with centre O and radius 8 cm. If each tangent is 15 cm long, find the distance OP. (2)

Q34. A lighthouse stands 60 m above sea level. From its top, the angles of depression of two boats floating on the same side of the lighthouse and in line with its foot are 60° and 30°. [5]

Find (a) the distance of the nearer boat from the foot of the lighthouse, and (b) the distance between the two boats. (Take √3 = 1.73)

Q35. A solid is in the form of a right circular cone mounted on a hemisphere of the same radius. The radius is 21 cm and the height of the cone is 72 cm. Taking π = 22/7, find [5]

(a) the slant height of the cone, (b) the total surface area of the solid, and (c) the volume of the solid.

A grain silo is in the shape of a right circular cylinder of radius 21 m and height 30 m, closed at the top by a hemispherical dome of the same radius. Taking π = 22/7, find

(a) the total volume of the silo, and (b) the total curved surface area to be painted (the curved wall of the cylinder together with the dome).

Section E

Q36 to Q38 are case-study based questions carrying 4 marks each, with sub-parts of 1, 1 and 2 marks. An internal choice is provided in the 2-mark sub-part. — 12 marks

Q36. Read the following and answer the questions that follow. [4]

A newly built stadium has 20 rows of seats. The first row has 24 seats, the second row has 28 seats, the third row has 32 seats, and so on — each row has 4 more seats than the row immediately in front of it.

(a) Write the common difference of the arithmetic progression formed by the number of seats in the rows. (1)

(b) How many seats are there in the 10th row? (1)

Q37. Read the following and answer the questions that follow. [4]

At a school fete, a stall sold identical handmade gift boxes. The number of boxes sold that day turned out to be 5 less than the price of one box in rupees, and the stall collected ₹336 in all.

(a) If the price of one box is ₹x, write an expression for the number of boxes sold. (1)

(b) Form the quadratic equation that represents this situation. (1)

(c) The stall had 4 more boxes left unsold. How much more money would it have collected had those been sold at the same price? (2)

Q38. Read the following and answer the questions that follow. [4]

A class teacher recorded the daily pocket money, in rupees, of the 40 students of her class. The grouped frequency distribution is shown below.

Pocket money (₹)	0–20	20–40	40–60	60–80	80–100
Number of students	6	9	13	8	4

(a) Write the modal class of the distribution. (1)

(b) Find the class mark of the class 60–80. (1)

Answer key with full worked solutions

Open this only after you have attempted the whole paper. Every answer is worked out step by step, and both options of each internal choice are solved.

▸ Show the Full Answer Key

SECTION A

Q1 (1 mark · Number Systems)

(A) 2² × 3²

For the HCF you take each common prime raised to the smaller of the two powers. The prime 2 appears as 2³ and 2², so take 2²; the prime 3 appears as 3² and 3³, so take 3². The primes 5 and 7 are not common to both numbers, so they are left out.

HCF = 2² × 3² = 36.

Q2 (1 mark · Number Systems)

(A) 144

For any two positive integers, HCF × LCM = product of the numbers.

LCM = (a × b) ÷ HCF = 1728 ÷ 12 = 144.

(Such a pair really does exist — for example a = 36 and b = 48 give HCF 12, product 1728 and LCM 144.)

Q3 (1 mark · Number Systems)

(B) 9/64

A rational number in lowest terms has a terminating decimal expansion only when its denominator has no prime factor other than 2 and 5.

64 = 2⁶ — only the prime 2, so 9/64 terminates.

45 = 3² × 5, 30 = 2 × 3 × 5 and 42 = 2 × 3 × 7 all contain a prime other than 2 or 5, so those three do not terminate.

Q4 (1 mark · Algebra)

(B) 12

If 3 is a zero then p(3) = 0.

p(3) = 3² – 7(3) + k = 9 – 21 + k = k – 12.

Setting k – 12 = 0 gives k = 12.

Q5 (1 mark · Algebra)

(C) no real roots

Discriminant D = b² – 4ac = (–4)² – 4(2)(3) = 16 – 24 = –8.

Since D < 0, the equation has no real roots.

Q6 (1 mark · Algebra)

(C) inconsistent

Compare the ratios: a₁/a₂ = 3/6 = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = 5/11.

Here a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so the lines are parallel and never meet — the pair is inconsistent and has no solution.

Q7 (1 mark · Algebra)

(B) 63

Here a = 7 and d = 11 – 7 = 4.

a₁₅ = a + 14d = 7 + 14(4) = 7 + 56 = 63.

Q8 (1 mark · Algebra)

(C) –3

a₁ = 5 – 3(1) = 2 and a₂ = 5 – 3(2) = –1.

d = a₂ – a₁ = –1 – 2 = –3.

(In general, when a_n is linear in n the coefficient of n is the common difference.)

Q9 (1 mark · Coordinate Geometry)

(A) (5, –8)

The midpoint of AB is ((1 + x)/2, (4 + y)/2), and this equals (3, –2).

(1 + x)/2 = 3 → 1 + x = 6 → x = 5

(4 + y)/2 = –2 → 4 + y = –4 → y = –8

So B is (5, –8).

Q10 (1 mark · Geometry)

(C) 9 cm

Two pairs of angles are equal, so △ABC ~ △DEF by the AA criterion, and corresponding sides are proportional:

AB/DE = BC/EF → 4/6 = 6/EF → 4 × EF = 36 → EF = 9 cm.

Q11 (1 mark · Geometry)

(C) SSS criterion

All three pairs of corresponding sides are in the same ratio, and nothing is said about the angles.

That is exactly the SSS similarity criterion: if the corresponding sides of two triangles are proportional, then their corresponding angles are equal and the triangles are similar.

AA needs two pairs of equal angles, and SAS needs two pairs of proportional sides with the included angles equal.

Q12 (1 mark · Geometry)

(B) parallel to each other

A tangent is perpendicular to the radius at its point of contact. Both end points of a diameter lie on the same straight line through the centre, so each tangent makes a right angle with that same line.

Two lines that are both perpendicular to the same line are parallel to each other.

Q13 (1 mark · Geometry)

(C) 12 cm

By the Basic Proportionality Theorem, AD/DB = AE/EC.

3/5 = 4.5/EC → 3 × EC = 22.5 → EC = 7.5 cm.

AC = AE + EC = 4.5 + 7.5 = 12 cm.

Q14 (1 mark · Trigonometry)

(A) 3/4

cos θ = √(1 – sin²θ) = √(1 – 9/25) = √(16/25) = 4/5.

tan θ = sin θ / cos θ = (3/5) ÷ (4/5) = 3/4.

Q15 (1 mark · Trigonometry)

(C) 1

Since 65° = 90° – 25°, we have cot 65° = cot(90° – 25°) = tan 25°.

So the expression becomes (tan 25°)/(tan 25°) = 1.

Q16 (1 mark · Mensuration)

(A) 57.75 cm²

Area of the whole sheet = πr² = (22/7) × 10.5 × 10.5 = 346.5 cm².

The six sectors are identical, so each one is a sixth of the whole:

346.5 ÷ 6 = 57.75 cm². (Option D is the area of the whole sheet.)

Q17 (1 mark · Mensuration)

(C) 748 cm²

Curved surface of the cylinder = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².

Curved surface of the hemisphere = 2πr² = 2 × (22/7) × 49 = 308 cm².

The circular face where the two solids join is hidden, so the exposed curved surface = 440 + 308 = 748 cm².

Q18 (1 mark · Statistics and Probability)

(B) 3/5

Total number of balls = 5 + 4 + 6 = 15. The balls that are not blue are the 5 red and 4 green ones, that is 9 balls.

P(not blue) = 9/15 = 3/5.

Q19 (1 mark · Statistics and Probability)

(a) Both A and R are true and R is the correct explanation of A.

An event either happens or it does not, so the two probabilities must add to 1. Applying this, P(not E) = 1 – 0.35 = 0.65, which is exactly what the assertion states. The reason is therefore the correct explanation.

Q20 (1 mark · Geometry)

(d) A is false but R is true.

The reason states the correct meaning of similarity — equal angles and sides in the same ratio.

The assertion wrongly turns “in the same ratio” into “equal”. Corresponding sides of similar triangles are proportional, not equal; they are equal only in the special case when the two triangles are also congruent (ratio 1 : 1). So A is false while R is true.

SECTION B

Q21 (2 marks · Algebra)

For equal roots the discriminant must be zero.

D = b² – 4ac = (–6)² – 4(k)(2) = 36 – 8k

36 – 8k = 0 → 8k = 36 → k = 9/2

(Note that k ≠ 0, otherwise the equation would not be quadratic; k = 9/2 satisfies this.)

Marking: 1 mark for using D = 0, 1 mark for the correct value of k.

Q22 (2 marks · Coordinate Geometry)

By the section formula with m : n = 3 : 1,

x = (3(6) + 1(–2))/(3 + 1) = (18 – 2)/4 = 4

y = (3(–5) + 1(3))/(3 + 1) = (–15 + 3)/4 = –3

So P is (4, –3).

Marking: 1 mark for the x-coordinate, 1 mark for the y-coordinate.

If you attempted the OR option

P is equidistant from A and B, so PA² = PB²:

(a – 4)² + (3 + 1)² = (a + 2)² + (3 – 5)²

a² – 8a + 16 + 16 = a² + 4a + 4 + 4

–8a + 32 = 4a + 8 → 12a = 24 → a = 2

Check: with P(2, 3), PA = √(4 + 16) = √20 and PB = √(16 + 4) = √20 ✓

Marking: 1 mark for setting PA² = PB², 1 mark for a.

Q23 (2 marks · Geometry)

Compute the two ratios in which D and E divide the sides:

AD/DB = 3/4.5 = 30/45 = 2/3

AE/EC = 4/6 = 2/3

The two ratios are equal, so D and E divide AB and AC in the same ratio.

By the converse of the Basic Proportionality Theorem — if a line divides two sides of a triangle in the same ratio, then it is parallel to the third side — we conclude that DE ∥ BC.

Marking: 1 mark for showing the two ratios are equal, 1 mark for quoting the converse and concluding.

Q24 (2 marks · Trigonometry)

We are given tan θ = 12/5.

Divide both the numerator and the denominator by cos θ:

(5 tan θ – 3)/(5 tan θ + 3) = (5 × 12/5 – 3)/(5 × 12/5 + 3) = (12 – 3)/(12 + 3) = 9/15 = 3/5

Marking: 1 mark for dividing through by cos θ, 1 mark for the value 3/5.

If you attempted the OR option

Square both sides of the given equation:

(tan θ + cot θ)² = 3²

tan²θ + 2(tan θ)(cot θ) + cot²θ = 9

Since tan θ and cot θ are reciprocals, (tan θ)(cot θ) = 1, so the middle term is 2:

tan²θ + 2 + cot²θ = 9 → tan²θ + cot²θ = 7

Marking: 1 mark for squaring and using tan θ · cot θ = 1, 1 mark for the value.

Q25 (2 marks · Statistics and Probability)

Exactly one of the three answers the question, so their three probabilities add up to 1.

(a) P(Chetan) = 1 – P(Aarav) – P(Bela) = 1 – 0.4 – 0.35 = 0.25

(b) “Not Aarav” is the complement of “Aarav”:

P(not Aarav) = 1 – 0.4 = 0.6

(As a check, P(Bela) + P(Chetan) = 0.35 + 0.25 = 0.6 ✓)

Marking: 1 mark for each part.

SECTION C

Q26 (3 marks · Number Systems)

(a) 168 = 2³ × 3 × 7 and 120 = 2³ × 3 × 5.

(b) Every pack must divide both totals exactly, so the greatest possible number of packs is the HCF. Take each common prime to the smaller power:

HCF = 2³ × 3 = 24 packs.

(c) Pencils per pack = 168 ÷ 24 = 7; erasers per pack = 120 ÷ 24 = 5.

Check: 24 × 7 = 168 ✓ and 24 × 5 = 120 ✓

Marking: 1 mark for each part.

Q27 (3 marks · Coordinate Geometry)

(a) Midpoint = ((–3 + 5)/2, (7 + (–1))/2) = (2/2, 6/2) = (1, 3)

(b) AB = √((5 – (–3))² + (–1 – 7)²) = √(8² + (–8)²) = √128 = 8√2 units

(c) By the section formula with m : n = 3 : 1,

x = (3(5) + 1(–3))/(3 + 1) = (15 – 3)/4 = 3

y = (3(–1) + 1(7))/(3 + 1) = (–3 + 7)/4 = 1

So P is (3, 1).

Marking: 1 mark for each part.

If you attempted the OR option

First find AB using the distance formula:

AB = √((4 – 1)² + (6 – 2)²) = √(9 + 16) = √25 = 5 units

Now BC must also equal 5:

BC² = (x – 4)² + (10 – 6)² = (x – 4)² + 16 = 25

(x – 4)² = 9 → x – 4 = 3 or x – 4 = –3

So x = 7 or x = 1, and in both cases AB = BC = 5 units.

Marking: 1 mark for AB, 2 marks for both values of x.

Q28 (3 marks · Geometry)

(a) The tangent at any point of a circle is perpendicular to the radius drawn to the point of contact. OA and OB are the radii to the points of contact A and B, so ∠OAT = ∠OBT = 90°.

(b) By part (a), △OAT is right-angled at A, so

TA² = OT² – OA² = 15² – 9² = 225 – 81 = 144

TA = 12 cm, and since the two tangents from an external point are equal, TB = 12 cm as well.

(c) Perimeter of OATB = OA + AT + TB + BO

= 9 + 12 + 12 + 9 = 42 cm

Marking: 1 mark for each part.

Q29 (3 marks · Trigonometry)

We know the identity sec²θ – tan²θ = 1, that is (sec θ + tan θ)(sec θ – tan θ) = 1.

So sec θ – tan θ = 1 ÷ (5/3) = 3/5.

Adding the two equations: 2 sec θ = 5/3 + 3/5 = (25 + 9)/15 = 34/15, so sec θ = 17/15 and cos θ = 15/17.

Subtracting them: 2 tan θ = 5/3 – 3/5 = (25 – 9)/15 = 16/15, so tan θ = 8/15.

sin θ = tan θ × cos θ = (8/15) × (15/17) = 8/17

Check: sin²θ + cos²θ = 64/289 + 225/289 = 289/289 = 1 ✓

Marking: 1 mark for sec θ – tan θ, 1 mark for sec θ and tan θ, 1 mark for sin θ.

Q30 (3 marks · Mensuration)

(a) Area of a sector = (θ/360°) × πr²

= (30/360) × (22/7) × 42 × 42 = (1/12) × 22 × 6 × 42 = (1/12) × 5544 = 462 m²

(b) The sector is bounded by two radii and its arc.

Arc length = (30/360) × 2 × (22/7) × 42 = (1/12) × 264 = 22 m

Perimeter = 2r + arc = 84 + 22 = 106 m

Marking: 2 marks for the area, 1 mark for the perimeter.

If you attempted the OR option

(a) l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm

(b) Volume of the cone = (1/3)πr²h = (1/3) × (22/7) × 49 × 24 = (1/3) × 22 × 7 × 24 = 1232 cm³

Volume of the cylinder = πr²h = (22/7) × 49 × 10 = 1540 cm³

Total volume = 1232 + 1540 = 2772 cm³

Marking: 1 mark for the slant height, 2 marks for the total volume.

Q31 (3 marks · Statistics and Probability)

Total number of marbles = 8 + 6 + 10 = 24.

(a) There are 10 black marbles.

P(black) = 10/24 = 5/12

(b) The marbles that are not white are the 8 red and 10 black ones, that is 18 marbles.

P(not white) = 18/24 = 3/4

(c) Red or white = 8 + 6 = 14 marbles.

P(red or white) = 14/24 = 7/12

Marking: 1 mark for each part.

SECTION D

Q32 (5 marks · Algebra)

Let the length of the sheet be l cm and the breadth be b cm.

The perimeter is 44 cm, so 2(l + b) = 44, giving l + b = 22 and hence b = 22 – l.

Cutting a 2 cm square from each corner removes 2 cm from both ends of each side, so the base of the box measures (l – 4) cm by (b – 4) cm.

Base area = 45, so (l – 4)(b – 4) = 45.

Substituting b = 22 – l:

(l – 4)(22 – l – 4) = 45 → (l – 4)(18 – l) = 45

18l – l² – 72 + 4l = 45

–l² + 22l – 72 = 45 → l² – 22l + 117 = 0

D = (–22)² – 4(1)(117) = 484 – 468 = 16, and √16 = 4.

l = (22 ± 4)/2 → l = 13 or l = 9.

Taking the length as the larger value, l = 13 cm and b = 22 – 13 = 9 cm.

Check: perimeter = 2(13 + 9) = 44 cm ✓, and the base measures 9 cm × 5 cm = 45 cm² ✓

Marking: 2 marks for forming the quadratic equation, 2 marks for solving it, 1 mark for stating both dimensions with the check.

Q33 (5 marks · Geometry)

(a) Basic Proportionality Theorem.

Given: △ABC with DE ∥ BC, D on AB and E on AC. To prove: AD/DB = AE/EC.

Construction: Join BE and CD. Draw EM ⊥ AB and DN ⊥ AC.

Proof: ar(△ADE) = ½ × AD × EM and ar(△DBE) = ½ × DB × EM, so

ar(△ADE)/ar(△DBE) = AD/DB …(i)

Similarly, ar(△ADE) = ½ × AE × DN and ar(△DEC) = ½ × EC × DN, so

ar(△ADE)/ar(△DEC) = AE/EC …(ii)

Now △DBE and △DEC stand on the same base DE and lie between the same parallels DE and BC, so they are equal in area: ar(△DBE) = ar(△DEC) …(iii)

From (i), (ii) and (iii): AD/DB = AE/EC (proved)

(b) Let AE = x cm, so EC = (15 – x) cm.

By the theorem just proved, AD/DB = AE/EC:

4/6 = x/(15 – x) → 4(15 – x) = 6x → 60 – 4x = 6x → 60 = 10x → x = 6

So AE = 6 cm (and EC = 9 cm).

Check: 4/6 = 6/9 = 2/3 ✓

Marking: 3 marks for the proof (1 for given/to prove/construction, 2 for the argument), 2 marks for part (b).

If you attempted the OR option

(a) Equal tangents from an external point.

Given: A circle with centre O and an external point P, with PA and PB the tangents touching the circle at A and B. To prove: PA = PB.

Construction: Join OA, OB and OP.

Proof: A tangent is perpendicular to the radius at the point of contact, so ∠OAP = ∠OBP = 90°.

In the right triangles OAP and OBP:

OA = OB (radii of the same circle), OP = OP (common), and both are right-angled at A and B.

So △OAP ≅ △OBP by the RHS congruence rule.

Hence PA = PB by CPCT (proved)

(b) Let the tangent touch the circle at A, so PA = 15 cm and OA = 8 cm.

The tangent is perpendicular to the radius at the point of contact, so △OAP is right-angled at A:

OP² = OA² + PA² = 8² + 15² = 64 + 225 = 289

OP = 17 cm

Marking: 3 marks for the proof, 2 marks for part (b).

Q34 (5 marks · Trigonometry)

Let the foot of the lighthouse be F, and let the nearer and farther boats be at distances d₁ and d₂ metres from F.

The angle of depression from the top equals the angle of elevation from the boat, so the larger angle belongs to the nearer boat.

(a) Nearer boat (angle 60°): tan 60° = 60/d₁

√3 = 60/d₁ → d₁ = 60/√3 = 20√3

d₁ = 20 × 1.73 = 34.6 m

(b) Farther boat (angle 30°): tan 30° = 60/d₂

1/√3 = 60/d₂ → d₂ = 60√3

Distance between the boats = d₂ – d₁ = 60√3 – 20√3 = 40√3

= 40 × 1.73 = 69.2 m

Marking: 2 marks for the nearer boat, 3 marks for the distance between the boats.

Q35 (5 marks · Mensuration)

(a) l = √(r² + h²) = √(21² + 72²) = √(441 + 5184) = √5625 = 75 cm

(b) The flat circular face of the hemisphere is joined to the base of the cone, so neither is exposed. The surface consists of the curved surface of the cone plus the curved surface of the hemisphere.

Curved surface of the cone = πrl = (22/7)(21)(75) = 22 × 3 × 75 = 4950 cm²

Curved surface of the hemisphere = 2πr² = 2 × (22/7) × 441 = 2 × 22 × 63 = 2772 cm²

Total surface area = 4950 + 2772 = 7722 cm²

(c) Volume of the cone = (1/3)πr²h = (1/3)(22/7)(441)(72) = (1/3)(22)(63)(72) = 33264 cm³

Volume of the hemisphere = (2/3)πr³ = (2/3)(22/7)(9261) = (2/3)(22)(1323) = 19404 cm³

Total volume = 33264 + 19404 = 52668 cm³

Marking: 1 mark for the slant height, 2 marks for the surface area, 2 marks for the volume.

If you attempted the OR option

(a) Volume of the cylindrical part = πr²h = (22/7)(21)(21)(30) = 22 × 3 × 21 × 30 = 41580 m³

Volume of the hemispherical dome = (2/3)πr³ = (2/3)(22/7)(9261) = (2/3)(22)(1323) = 19404 m³

Total volume = 41580 + 19404 = 60984 m³

(b) Curved surface of the cylinder = 2πrh = 2(22/7)(21)(30) = 2 × 22 × 3 × 30 = 3960 m²

Curved surface of the dome = 2πr² = 2(22/7)(441) = 2 × 22 × 63 = 2772 m²

Total area to be painted = 3960 + 2772 = 6732 m²

Marking: 3 marks for the volume, 2 marks for the surface area.

SECTION E

Q36 (4 marks · Algebra)

The number of seats forms an AP with first term a = 24.

(a) Common difference d = 28 – 24 = 4

(b) a₁₀ = a + 9d = 24 + 9(4) = 24 + 36 = 60 seats

(c) Total seats = S₂₀ = (n/2)[2a + (n – 1)d]

= (20/2)[2(24) + 19(4)] = 10[48 + 76] = 10 × 124 = 1240 seats

Marking: 1 mark for (a), 1 mark for (b), 2 marks for (c).

If you attempted the OR option

(c) Let the nth row have 76 seats. Then a_n = 76:

24 + (n – 1)(4) = 76

(n – 1)(4) = 52 → n – 1 = 13 → n = 14

So the 14th row has 76 seats.

Marking: 1 mark for forming the equation, 1 mark for n.

Q37 (4 marks · Algebra)

(a) Number of boxes sold = (x – 5)

(b) Total collected = price × number of boxes = 336, so

x(x – 5) = 336 → x² – 5x – 336 = 0

(c) D = (–5)² – 4(1)(–336) = 25 + 1344 = 1369, and √1369 = 37.

x = (5 ± 37)/2 → x = 21 or x = –16

A price cannot be negative, so x = 21.

Price of one box = ₹21 and the number sold = 21 – 5 = 16.

Check: 21 × 16 = ₹336 ✓

Marking: 1 mark for (a), 1 mark for (b), 2 marks for (c).

If you attempted the OR option

(c) First solve x² – 5x – 336 = 0. With D = 1369 and √1369 = 37,

x = (5 + 37)/2 = 21 (the negative root –16 is rejected), so each box cost ₹21.

Selling 4 more boxes at ₹21 each would bring in

4 × 21 = ₹84 more.

Marking: 1 mark for the price, 1 mark for the extra amount.

Q38 (4 marks · Statistics and Probability)

(a) The modal class is the class with the highest frequency. The greatest frequency is 13, so the modal class is 40–60.

(b) Class mark = (lower limit + upper limit)/2 = (60 + 80)/2 = 70

(c) Using the class marks x_i = 10, 30, 50, 70, 90 with frequencies f_i = 6, 9, 13, 8, 4:

Σf_ix_i = 6(10) + 9(30) + 13(50) + 8(70) + 4(90)

= 60 + 270 + 650 + 560 + 360 = 1900

Σf_i = 40

Mean = 1900 ÷ 40 = ₹47.50

Marking: 1 mark for (a), 1 mark for (b), 2 marks for (c).

If you attempted the OR option

(c) Build the cumulative frequencies: 6, 15, 28, 36, 40.

Here n = 40, so n/2 = 20.

The first cumulative frequency that is greater than or equal to 20 is 28, which belongs to the class 40–60 — this is the median class.

The cumulative frequency of the class just before it (20–40) is 15.

Marking: 1 mark for the median class, 1 mark for the preceding cumulative frequency.

What your score is telling you

65–80: Strong. Now chase the last few marks — presentation, units, and not skipping steps.
50–64: Your methods are right but accuracy is leaking marks. Redo every wrong question on paper without looking.
35–49: Two or three chapters are carrying most of your errors. Revise those, then take Set 2.
Below 35: Do not take another full paper yet. Work through chapter-wise tests first.

Kaizen — one paper, one honest correction, one small improvement. Do that each week between now and the board exam and the gap by February will not be small.

Written & reviewed by Team Principal Saab — Meet the team →