MathsCBSE Class 10 · Standard · Board exam 2023 (session 2022-23)

CBSE Class 10 Mathematics Standard: 2023 board paper, set 30/2/1

The question paper set in the CBSE Class 10 board exam of 2023, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2023/X/MATHEMATICS_STANDARD.zip (a zip; this paper is the file 30_2_1_Maths Standard.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

Practise a paper like this, timedMy board paper sets a full paper in this exact format from real board questions, with a timer.Get your written answers markedThe Answer Coach marks a photo of your handwritten answer step by step, like the examiner.

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 22 and 24
C	26-31	Short answer; internal choice in 26 and 30
D	32-35	Long answer; internal choice in 33 and 35
E	36-38	Case study based; internal choice in each case study (2-mark subparts)

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 case-study questions of Section E. Use π = 22/7 where required unless stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQQuadratic Equations
Which of the following quadratic equations has sum of its roots as $4$ ?
1. (A) $2 x^{2} - 4 x + 8 = 0$
2. (B) $- x^{2} + 4 x + 4 = 0$
3. (C) $2 x^{2} - \frac{4}{2} x + 1 = 0$
4. (D) $4 x^{2} - 4 x + 4 = 0$
Try it first, then show our solutionSolution
1. For $a x^{2} + b x + c = 0$ , sum of roots $= - \frac{b}{a}$ .
2. (A) $\frac{4}{2} = 2$ ; (B) $- \frac{4}{- 1} = 4$ ; (C) $\frac{4/ 2}{2} = 2$ ; (D) $\frac{4}{4} = 1$ . Only (B) gives 4.
  1 mark
Answer: Option (B) $- x^{2} + 4 x + 4 = 0$
Where the marks are
- Uses sum $= - b / a$ and picks (B) (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQAreas Related to Circles
What is the length of the arc of the sector of a circle with radius $14$ cm and of central angle $9 0^{\circ}$ ?
1. (A) $22$ cm
2. (B) $44$ cm
3. (C) $88$ cm
4. (D) $11$ cm
Try it first, then show our solutionSolution
1. Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{90}{360} \times 2 \times \frac{22}{7} \times 14 = 22$ cm.
  1 mark
Answer: 22 cm, option (A)
Where the marks are
- Applies arc formula to get 22 cm (1 mark)
Q3 · 1 mark · MCQTriangles
If $△ A B C \sim △ P QR$ with $∠ A = 3 2^{\circ}$ and $∠ R = 6 5^{\circ}$ , then the measure of $∠ B$ is :
1. (A) $3 2^{\circ}$
2. (B) $6 5^{\circ}$
3. (C) $8 3^{\circ}$
4. (D) $9 7^{\circ}$
Try it first, then show our solutionSolution
1. Similar triangles: $∠ C = ∠ R = 6 5^{\circ}$ .
2. $∠ B = 18 0^{\circ} - 3 2^{\circ} - 6 5^{\circ} = 8 3^{\circ}$ .
  1 mark
Answer: $8 3^{\circ}$ , option (C)
Where the marks are
- Uses corresponding angles and angle sum to get $8 3^{\circ}$ (1 mark)
Q4 · 1 mark · MCQReal Numbers
If ' $p$ ' and ' $q$ ' are natural numbers and ' $p$ ' is the multiple of ' $q$ ', then what is the HCF of ' $p$ ' and ' $q$ ' ?
1. (A) $pq$
2. (B) $p$
3. (C) $q$
4. (D) $p + q$
Try it first, then show our solutionSolution
1. Since $q$ divides $p$ and $q$ divides itself, and no number larger than $q$ divides $q$ , HCF $= q$ .
  1 mark
Answer: $q$ , option (C)
Where the marks are
- Identifies HCF as $q$ (1 mark)
Q5 · 1 mark · MCQCoordinate Geometry
The coordinates of the vertex A of a rectangle ABCD whose three vertices are given as $B (0, 0)$ , $C (3, 0)$ and $D (0, 4)$ are :
1. (A) $(4, 0)$
2. (B) $(0, 3)$
3. (C) $(3, 4)$
4. (D) $(4, 3)$
Try it first, then show our solutionSolution
1. In a rectangle diagonals bisect each other: midpoint of BD $= (0, 2)$ = midpoint of AC.
2. $\frac{x + 3}{2} = 0$ ?? — instead note A is opposite C... Using diagonals AC and BD: $\frac{x + 3}{2} = 0, \frac{y + 0}{2} = 2$ gives $(- 3, 4)$ , not a rectangle in order; correct order B(0,0), C(3,0), D(0,4) means A is opposite B with diagonal AB, CD: midpoint of CD $= (1.5, 2)$ , so A $= (3, 4)$ .
  1 mark
Answer: $(3, 4)$ , option (C)
Where the marks are
- Uses diagonal midpoint (or plotting) to get A(3, 4) (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQPair of Linear Equations in Two Variables
If the pair of equations $3 x - y + 8 = 0$ and $6 x - r y + 16 = 0$ represent coincident lines, then the value of ' $r$ ' is :
1. (A) $- \frac{1}{2}$
2. (B) $\frac{1}{2}$
3. (C) $- 2$
4. (D) $2$
Try it first, then show our solutionSolution
1. Coincident: $\frac{3}{6} = \frac{- 1}{- r} = \frac{8}{16}$ , so $\frac{1}{r} = \frac{1}{2}$ , $r = 2$ .
  1 mark
Answer: $r = 2$ , option (D)
Where the marks are
- Uses equal ratios to get $r = 2$ (1 mark)
Q7 · 1 mark · MCQProbability
A bag contains $100$ cards numbered $1$ to $100$ . A card is drawn at random from the bag. What is the probability that the number on the card is a perfect cube ?
1. (A) $\frac{1}{20}$
2. (B) $\frac{3}{50}$
3. (C) $\frac{1}{25}$
4. (D) $\frac{7}{100}$
Try it first, then show our solutionSolution
1. Perfect cubes 1–100: 1, 8, 27, 64 → 4 outcomes.
2. $P = \frac{4}{100} = \frac{1}{25}$ .
  1 mark
Answer: $\frac{1}{25}$ , option (C)
Where the marks are
- Counts 4 cubes, probability $1/25$ (1 mark)
Q8 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of equations $x = a$ and $y = b$ graphically represents lines which are :
1. (A)parallel
2. (B)intersecting at $(b, a)$
3. (C)coincident
4. (D)intersecting at $(a, b)$
Try it first, then show our solutionSolution
1. $x = a$ is vertical, $y = b$ horizontal; they meet where $x = a, y = b$ , i.e. at $(a, b)$ .
  1 mark
Answer: Intersecting at $(a, b)$ , option (D)
Where the marks are
- Identifies intersection at $(a, b)$ (1 mark)
Q9 · 1 mark · MCQPolynomials
If one zero of the polynomial $6 x^{2} + 37 x - (k - 2)$ is reciprocal of the other, then what is the value of $k$ ?
1. (A) $- 4$
2. (B) $- 6$
3. (C) $6$
4. (D) $4$
Try it first, then show our solutionSolution
1. Zeroes $α, \frac{1}{α}$ , product $= 1 = \frac{c}{a} = \frac{- ( k - 2 )}{6}$ .
2. $- (k - 2) = 6 \Rightarrow k = - 4$ .
  1 mark
Answer: $k = - 4$ , option (A)
Where the marks are
- Sets product of zeroes $= 1$ and solves $k = - 4$ (1 mark)
Q10 · 1 mark · MCQSurface Areas and Volumes
What is the total surface area of a solid hemisphere of diameter ' $d$ ' ?
1. (A) $3 π d^{2}$
2. (B) $2 π d^{2}$
3. (C) $\frac{1}{2} π d^{2}$
4. (D) $\frac{3}{4} π d^{2}$
Try it first, then show our solutionSolution
1. TSA $= 3 π r^{2}$ with $r = \frac{d}{2}$ : $3 π \frac{d ^{2}}{4} = \frac{3}{4} π d^{2}$ .
  1 mark
Answer: $\frac{3}{4} π d^{2}$ , option (D)
Where the marks are
- Uses $3 π r^{2}$ with $r = d /2$ (1 mark)
Q11 · 1 mark · MCQProbability
If three coins are tossed simultaneously, what is the probability of getting at most one tail ?
1. (A) $\frac{3}{8}$
2. (B) $\frac{4}{8}$
3. (C) $\frac{5}{8}$
4. (D) $\frac{7}{8}$
Try it first, then show our solutionSolution
1. 8 outcomes; at most one tail: HHH, HHT, HTH, THH → 4.
2. $P = \frac{4}{8}$ .
  1 mark
Answer: $\frac{4}{8}$ , option (B)
Where the marks are
- Counts 4 favourable of 8 (1 mark)
Q12 · 1 mark · MCQTriangles
In the given figure, $D E ∥ B C$ . If $A D = 2$ units, $D B = A E = 3$ units and $E C = x$ units, then the value of $x$ is :
1. (A) $2$
2. (B) $3$
3. (C) $5$
4. (D) $\frac{9}{2}$
Figure in the paper: Triangle ABC with A at the top. D lies on side AB and E lies on side AC, with DE parallel to BC (arrows marked). Labels: AD = 2, DB = 3, AE = 3 and EC = x. See it in the official paper.
Try it first, then show our solutionSolution
1. By BPT, $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{2}{3} = \frac{3}{x} \Rightarrow x = \frac{9}{2}$ .
  1 mark
Answer: $x = \frac{9}{2}$ , option (D)
Where the marks are
- Applies BPT to get $9/2$ (1 mark)
Q13 · 1 mark · MCQAreas Related to Circles
The hour-hand of a clock is $6$ cm long. The angle swept by it between $7 : 20$ a.m. and $7 : 55$ a.m. is :
1. (A) $(\frac{35}{4})^{\circ}$
2. (B) $(\frac{35}{2})^{\circ}$
3. (C) $3 5^{\circ}$
4. (D) $7 0^{\circ}$
Try it first, then show our solutionSolution
1. Hour hand turns $3 0^{\circ}$ per 60 min = $\frac{1}{2}^{\circ}$ per min.
2. 35 min: $35 \times \frac{1}{2} = (\frac{35}{2})^{\circ}$ .
  1 mark
Answer: $(\frac{35}{2})^{\circ}$ , option (B)
Where the marks are
- Uses $0. 5^{\circ}$ /min for 35 min (1 mark)
Q14 · 1 mark · MCQPolynomials
The zeroes of the polynomial $p (x) = x^{2} + 4 x + 3$ are given by :
1. (A) $1, 3$
2. (B) $- 1, 3$
3. (C) $1, - 3$
4. (D) $- 1, - 3$
Try it first, then show our solutionSolution
1. $x^{2} + 4 x + 3 = (x + 1) (x + 3)$ , zeroes $- 1, - 3$ .
  1 mark
Answer: $- 1, - 3$ , option (D)
Where the marks are
- Factorises to get $- 1, - 3$ (1 mark)
Q15 · 1 mark · MCQCircles
In the given figure, the quadrilateral PQRS circumscribes a circle. Here $P A + C S$ is equal to :
1. (A) $QR$
2. (B) $P R$
3. (C) $P S$
4. (D) $P Q$
Figure in the paper: Quadrilateral PQRS with a circle inscribed inside it, touching PQ at A, QR at B, RS at C and SP at D. P is at the top, Q on the right, R at the bottom and S on the left. See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents from external point are equal: $P A = P D$ , $S C = S D$ .
2. $P A + C S = P D + D S = P S$ .
  1 mark
Answer: $P S$ , option (C)
Where the marks are
- Uses equal tangents to get PS (1 mark)
Q16 · 1 mark · MCQPolynomials
If $α$ and $β$ are the zeroes of the quadratic polynomial $p (x) = x^{2} - a x - b$ , then the value of $α^{2} + β^{2}$ is :
1. (A) $a^{2} - 2 b$
2. (B) $a^{2} + 2 b$
3. (C) $b^{2} - 2 a$
4. (D) $b^{2} + 2 a$
Try it first, then show our solutionSolution
1. $α + β = a$ , $α β = - b$ .
2. $α^{2} + β^{2} = (α + β)^{2} - 2 α β = a^{2} + 2 b$ .
  1 mark
Answer: $a^{2} + 2 b$ , option (B)
Where the marks are
- Uses identity with sum and product (1 mark)
Q17 · 1 mark · MCQCoordinate Geometry
The area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is :
1. (A) $ab$
2. (B) $\frac{1}{2} ab$
3. (C) $\frac{1}{4} ab$
4. (D) $2 ab$
Try it first, then show our solutionSolution
1. Intercepts: $(a, 0)$ and $(0, b)$ ; right triangle with legs $a, b$ .
2. Area $= \frac{1}{2} ab$ .
  1 mark
Answer: $\frac{1}{2} ab$ , option (B)
Where the marks are
- Finds intercepts and area $\frac{1}{2} ab$ (1 mark)
Q18 · 1 mark · MCQTriangles
In the given figure, $A B ∥ P Q$ . If $A B = 6$ cm, $P Q = 2$ cm and $O B = 3$ cm, then the length of $O P$ is :
1. (A) $9$ cm
2. (B) $3$ cm
3. (C) $4$ cm
4. (D) $1$ cm
Figure in the paper: AB and PQ are two parallel segments, AB = 6 cm (A at bottom left, B at top) and PQ = 2 cm (P lower, Q upper right). Lines AQ and BP intersect at O. BO = 3 cm is marked along BP, and OP is the unknown segment. See it in the official paper.
Try it first, then show our solutionSolution
1. $△ A O B \sim △ QO P$ (vertically opposite angles, alternate angles as $A B ∥ P Q$ ).
2. $\frac{O B}{O P} = \frac{A B}{P Q} \Rightarrow \frac{3}{O P} = \frac{6}{2} \Rightarrow O P = 1$ cm.
  1 mark
Answer: 1 cm, option (D)
Where the marks are
- Similarity ratio gives OP = 1 cm (1 mark)
Open Q18 on its own page
Q19 · 1 mark · Assertion-reasonCompetency-basedCircles
Assertion (A): A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R): The lengths of tangents drawn from an external point to a circle are equal.
Codes: (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, but R is false. (d) A is false, but R is true.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A is a standard theorem (true); R is also a standard theorem (true).
2. R is about tangent lengths and does not explain perpendicularity, so (b).
  1 mark
Answer: Option (b)
Where the marks are
- Both true, R not explanation (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedPolynomials
Assertion (A): The polynomial $p (x) = x^{2} + 3 x + 3$ has two real zeroes.
Reason (R): A quadratic polynomial can have at most two real zeroes.
Codes: (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, but R is false. (d) A is false, but R is true.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $D = 3^{2} - 4 (1) (3) = - 3 < 0$ , so no real zeroes; A is false.
2. A degree-2 polynomial has at most 2 zeroes, R true. Option (d).
  1 mark
Answer: Option (d)
Where the marks are
- Discriminant negative → A false, R true (1 mark)

Section B

Very short answer; internal choice in 22 and 24

Q21 · 2 marks · Very short answerReal Numbers
Prove that $2 + 3$ is an irrational number, given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $2 + 3$ is rational. Then $2 + 3 = \frac{p}{q}$ , where $p, q$ are integers and $q \neq = 0$ .
  ½ mark
2. Rearranging gives $3 = \frac{p}{q} - 2 = \frac{p - 2 q}{q}$ .
  ½ mark
3. Since $p$ and $q$ are integers, $p - 2 q$ is an integer and $q \neq = 0$ . So $\frac{p - 2 q}{q}$ is a rational number.
  ½ mark
4. This makes $3$ rational, which contradicts the given fact that $3$ is irrational. Hence our assumption is wrong, and $2 + 3$ is irrational.
  ½ mark
Answer: $2 + 3$ is irrational (proved by contradiction).
Where the marks are
- Assumes $2 + 3 = \frac{p}{q}$ with $q \neq = 0$ , $p, q$ integers (½ mark)
- Expresses $3 = \frac{p - 2 q}{q}$ (½ mark)
- States that $\frac{p - 2 q}{q}$ is rational (½ mark)
- Identifies the contradiction and concludes (½ mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerIntroduction to Trigonometry
If $4 cot^{2} 4 5^{\circ} - sec^{2} 6 0^{\circ} + sin^{2} 6 0^{\circ} + p = \frac{3}{4}$ , then find the value of $p$ .
Try it first, then show our solutionSolution
1. Use $cot 4 5^{\circ} = 1$ , $sec 6 0^{\circ} = 2$ and $sin 6 0^{\circ} = \frac{3}{2}$ :
  $4 (1)^{2} - (2)^{2} + (\frac{3}{2})^{2} + p = \frac{3}{4}$
  1 mark
2. $4 - 4 + \frac{3}{4} + p = \frac{3}{4}$
  ½ mark
3. $\Rightarrow p = 0$
  ½ mark
Answer: $p = 0$
Where the marks are
- Substitutes the correct values of $cot 4 5^{\circ}$ , $sec 6 0^{\circ}$ and $sin 6 0^{\circ}$ (1 mark)
- Simplifies to $4 - 4 + \frac{3}{4} + p = \frac{3}{4}$ (½ mark)
- Gets $p = 0$ (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerIntroduction to Trigonometry
If $cos A + cos^{2} A = 1$ , then find the value of $sin^{2} A + sin^{4} A$ .
Try it first, then show our solutionSolution
1. From $cos A + cos^{2} A = 1$ :
  $cos A = 1 - cos^{2} A = sin^{2} A$ [since $sin^{2} A + cos^{2} A = 1$ ]
  1 mark
2. Square this to get $sin^{4} A = cos^{2} A$ . Therefore
  $sin^{2} A + sin^{4} A = cos A + cos^{2} A = 1$
  1 mark
Answer: $sin^{2} A + sin^{4} A = 1$
Where the marks are
- Obtains $cos A = sin^{2} A$ (1 mark)
- Substitutes to get $sin^{2} A + sin^{4} A = cos A + cos^{2} A = 1$ (1 mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerCoordinate Geometry
Show that the points $(- 2, 3)$ , $(8, 3)$ and $(6, 7)$ are the vertices of a right-angled triangle.
Try it first, then show our solutionSolution
1. Let the points be $A (- 2, 3)$ , $B (8, 3)$ and $C (6, 7)$ . Use the distance formula:
  $A B = (8 + 2)^{2} + (3 - 3)^{2} = 10$
  $B C = (6 - 8)^{2} + (7 - 3)^{2} = 4 + 16 = 20$
  1 mark
2. $A C = (6 + 2)^{2} + (7 - 3)^{2} = 64 + 16 = 80$
  ½ mark
3. $B C^{2} + A C^{2} = 20 + 80 = 100 = A B^{2}$ .
  By the converse of Pythagoras theorem, the triangle is right-angled at $C$ . Hence the points are the vertices of a right-angled triangle.
  ½ mark
Answer: $A B^{2} = B C^{2} + A C^{2}$ ( $100 = 20 + 80$ ), so the triangle is right-angled at C.
Where the marks are
- Finds $A B = 10$ and $B C = 20$ (1 mark)
- Finds $A C = 80$ (½ mark)
- Verifies $A B^{2} = B C^{2} + A C^{2}$ and concludes (½ mark)
Open Q23 on its own page
Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerCompetency-basedSome Applications of Trigonometry
The length of the shadow of a tower on the plane ground is $3$ times the height of the tower. Find the angle of elevation of the sun.
Try it first, then show our solutionSolution
1. Let $A B = h$ be the tower and $A C$ its shadow, so $A C = 3 h$ . Let $θ$ be the angle of elevation of the sun at C.
  In right $△ B A C$ : $tan θ = \frac{A B}{A C} = \frac{h}{3 h}$
  1 mark
2. $tan θ = \frac{1}{3}$
  ½ mark
3. $\Rightarrow θ = 3 0^{\circ}$
  ½ mark
Answer: Angle of elevation of the sun = $3 0^{\circ}$
Where the marks are
- Writes $tan θ = \frac{h}{3 h}$ (1 mark)
- Simplifies to $tan θ = \frac{1}{3}$ (½ mark)
- Concludes $θ = 3 0^{\circ}$ (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerCompetency-basedSome Applications of Trigonometry
The angle of elevation of the top of a tower from a point on the ground which is $30$ m away from the foot of the tower, is $3 0^{\circ}$ . Find the height of the tower.
Try it first, then show our solutionSolution
1. Let $A B$ be the tower and C the point on the ground, with $A C = 30$ m and $∠ A C B = 3 0^{\circ}$ .
  In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{A C} = \frac{A B}{30}$
  1 mark
2. $A B = \frac{30}{3} = 103$
  So the height of the tower is $103$ m (about 17.32 m).
  1 mark
Answer: Height of tower = $103$ m
Where the marks are
- Forms $tan 3 0^{\circ} = \frac{A B}{30}$ (1 mark)
- Finds $A B = 103$ m (1 mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerCircles
In the given figure, O is the centre of the circle. AB and AC are tangents drawn to the circle from point A. If $∠ B A C = 6 5^{\circ}$ , then find the measure of $∠ B O C$ .
Figure in the paper: A circle with centre O. External point A on the left with two tangents AB and AC touching the circle at B (upper) and C (lower). Radii OB and OC are drawn. Angle BAC = 65° is marked at A. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O B A = ∠ O C A = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  The angles of quadrilateral ABOC add up to $36 0^{\circ}$ , so $∠ B A C + ∠ B O C = 36 0^{\circ} - 18 0^{\circ} = 18 0^{\circ}$ .
  1 mark
2. $∠ B O C = 18 0^{\circ} - 6 5^{\circ} = 11 5^{\circ}$
  1 mark
Answer: $∠ B O C = 11 5^{\circ}$
Where the marks are
- Uses $∠ B A C + ∠ B O C = 18 0^{\circ}$ , with the reason tangent ⟂ radius (1 mark)
- Gets $∠ B O C = 11 5^{\circ}$ (1 mark)
Open Q25 on its own page

Section C

Short answer; internal choice in 26 and 30

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerReal Numbers
Find by prime factorisation the LCM of the numbers $18180$ and $7575$ . Also, find the HCF of the two numbers.
Try it first, then show our solutionSolution
1. $18180 = 2^{2} \times 3^{2} \times 5 \times 101$
  ½ mark
2. $7575 = 3 \times 5^{2} \times 101$
  ½ mark
3. LCM = product of the highest powers of all the prime factors
  $= 2^{2} \times 3^{2} \times 5^{2} \times 101 = 90900$
  1 mark
4. HCF = product of the lowest powers of the common prime factors
  $= 3 \times 5 \times 101 = 1515$
  1 mark
Answer: LCM = 90900, HCF = 1515
Where the marks are
- Prime factorisation of 18180 (½ mark)
- Prime factorisation of 7575 (½ mark)
- LCM = 90900 (1 mark)
- HCF = 1515 (1 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerCompetency-basedReal Numbers
Three bells ring at intervals of $6$ , $12$ and $18$ minutes. If all the three bells rang at $6$ a.m., when will they ring together again ?
Try it first, then show our solutionSolution
1. The bells ring together again after LCM(6, 12, 18) minutes.
  $6 = 2 \times 3$ , $12 = 2^{2} \times 3$ , $18 = 2 \times 3^{2}$
  LCM $= 2^{2} \times 3^{2} = 36$
  2 marks
2. So all three bells ring together again 36 minutes after 6 a.m., i.e. at 6:36 a.m.
  1 mark
Answer: At 6:36 a.m.
Where the marks are
- Finds LCM of 6, 12, 18 = 36 (2 marks)
- Concludes they ring together at 6:36 a.m. (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $(\frac{1}{c o s θ} - cos θ) (\frac{1}{s i n θ} - sin θ) = \frac{1}{t a n θ + c o t θ}$ .
Try it first, then show our solutionSolution
1. LHS $= (\frac{1 - cos ^{2} θ}{cos θ}) (\frac{1 - sin ^{2} θ}{sin θ})$
  ½ mark
2. $= \frac{sin ^{2} θ}{cos θ} \times \frac{cos ^{2} θ}{sin θ} = sin θ cos θ$ [using $sin^{2} θ + cos^{2} θ = 1$ ]
  1 mark
3. RHS $= \frac{1}{\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}}$
  ½ mark
4. $= \frac{sin θ cos θ}{sin ^{2} θ + cos ^{2} θ} = sin θ cos θ$
  Therefore LHS = RHS. Hence proved.
  1 mark
Answer: Both sides equal $sin θ cos θ$ ; hence proved.
Where the marks are
- Takes the LCM in each bracket of the LHS (½ mark)
- Simplifies the LHS to $sin θ cos θ$ (1 mark)
- Writes the RHS in terms of sin and cos (½ mark)
- Simplifies the RHS to $sin θ cos θ$ and concludes (1 mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerCoordinate Geometry
If $Q (0, 1)$ is equidistant from $P (5, - 3)$ and $R (x, 6)$ , find the values of $x$ .
Try it first, then show our solutionSolution
1. Since Q is equidistant from P and R, $P Q = QR$ , so $P Q^{2} = Q R^{2}$ :
  $(5 - 0)^{2} + (- 3 - 1)^{2} = (x - 0)^{2} + (6 - 1)^{2}$
  1 mark
2. $25 + 16 = x^{2} + 25$
  1 mark
3. $x^{2} = 16 \Rightarrow x = 4$ or $x = - 4$
  1 mark
Answer: $x = 4$ or $x = - 4$
Where the marks are
- Sets up $P Q^{2} = Q R^{2}$ using the distance formula (1 mark)
- Simplifies to $41 = x^{2} + 25$ (1 mark)
- Obtains $x = 4$ and $x = - 4$ (½ + ½ mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerCompetency-basedAreas Related to Circles
A car has two wipers which do not overlap. Each wiper has a blade of length $21$ cm sweeping through an angle of $12 0^{\circ}$ . Find the total area cleaned at each sweep of the two blades.
Try it first, then show our solutionSolution
1. Each blade sweeps a sector with $r = 21$ cm and $θ = 12 0^{\circ}$ .
  Area of one sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 21 \times 21$
  1½ marks
2. $= 462 cm^{2}$
  1 mark
3. The wipers do not overlap, so total area cleaned $= 2 \times 462 = 924 cm^{2}$ .
  ½ mark
Answer: 924 cm²
Where the marks are
- Uses the sector area formula with correct values (1½ marks)
- Area of one sector = 462 cm² (1 mark)
- Total area = 924 cm² (½ mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerPair of Linear Equations in Two Variables
If the system of linear equations $2 x + 3 y = 7$ and $2 a x + (a + b) y = 28$ have infinite number of solutions, then find the values of ' $a$ ' and ' $b$ '.
Try it first, then show our solutionSolution
1. For infinitely many solutions, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ :
  $\frac{2}{2 a} = \frac{3}{a + b} = \frac{7}{28}$
  1 mark
2. $\frac{1}{a} = \frac{1}{4} \Rightarrow a = 4$
  1 mark
3. $\frac{3}{a + b} = \frac{1}{4} \Rightarrow a + b = 12 \Rightarrow b = 12 - 4 = 8$
  1 mark
Answer: $a = 4$ , $b = 8$
Where the marks are
- Writes the condition $\frac{2}{2 a} = \frac{3}{a + b} = \frac{7}{28}$ (1 mark)
- Finds $a = 4$ (1 mark)
- Finds $b = 8$ (1 mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerPair of Linear Equations in Two Variables
If $217 x + 131 y = 913$ and $131 x + 217 y = 827$ , then solve the equations for the values of $x$ and $y$ .
Try it first, then show our solutionSolution
1. Adding the two equations: $348 x + 348 y = 1740 \Rightarrow x + y = 5$ …(i)
  1 mark
2. Subtracting the second from the first: $86 x - 86 y = 86 \Rightarrow x - y = 1$ …(ii)
  1 mark
3. Adding (i) and (ii): $2 x = 6 \Rightarrow x = 3$ ; then $y = 5 - 3 = 2$ .
  1 mark
Answer: $x = 3$ , $y = 2$
Where the marks are
- Adds the equations to get $x + y = 5$ (1 mark)
- Subtracts to get $x - y = 1$ (1 mark)
- Finds $x = 3$ , $y = 2$ (½ + ½ mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerCircles
In the given figure, O is the centre of the circle and QPR is a tangent to it at P. Prove that $∠ Q A P + ∠ A P R = 9 0^{\circ}$ .
Figure in the paper: A circle with centre O. Points B and A lie on the circle, with chord BA going up-right to A. Q is on the left and R on the right of a tangent line QPR touching the circle at P at the bottom. OP is a dashed radius. Angle QAP is marked at A and angle APR is marked at P. See it in the official paper.
Try it first, then show our solutionSolution
1. $O A = O P$ [radii of the same circle], so in $△ O A P$ , $∠ O P A = ∠ O A P$ …(i) [angles opposite equal sides]
  1 mark
2. $O P ⊥ QR$ [tangent ⟂ radius at the point of contact], so $∠ O P R = 9 0^{\circ}$ , i.e. $∠ O P A + ∠ A P R = 9 0^{\circ}$ .
  1 mark
3. Using (i): $∠ O A P + ∠ A P R = 9 0^{\circ}$
  ½ mark
4. Since O lies on the line AQ, $∠ O A P = ∠ Q A P$ . Hence $∠ Q A P + ∠ A P R = 9 0^{\circ}$ . Proved.
  ½ mark
Answer: $∠ Q A P + ∠ A P R = 9 0^{\circ}$ (proved).
Where the marks are
- Uses $O A = O P$ to get $∠ O P A = ∠ O A P$ (1 mark)
- Uses tangent ⟂ radius: $∠ O P A + ∠ A P R = 9 0^{\circ}$ (1 mark)
- Substitutes from (i) (½ mark)
- Replaces $∠ O A P$ by $∠ Q A P$ and concludes (½ mark)
Open Q31 on its own page

Section D

Long answer; internal choice in 33 and 35

Q32 · 5 marks · Long answerArithmetic Progressions
How many terms of the arithmetic progression $45, 39, 33, \dots$ must be taken so that their sum is $180$ ? Explain the double answer.
Try it first, then show our solutionSolution
1. Here $a = 45$ , $d = 39 - 45 = - 6$ and $S_{n} = 180$ .
  ½ mark
2. Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ :
  $180 = \frac{n}{2} [90 + (n - 1) (- 6)] = \frac{n}{2} [96 - 6 n]$
  1 mark
3. $360 = 96 n - 6 n^{2} \Rightarrow 6 n^{2} - 96 n + 360 = 0$
  1 mark
4. Dividing by 6: $n^{2} - 16 n + 60 = 0 \Rightarrow (n - 6) (n - 10) = 0$
  1 mark
5. $n = 6$ or $n = 10$
  1 mark
6. Both answers are valid. Explanation: $a_{7} = 45 + 6 (- 6) = 9$ , $a_{8} = 3$ , $a_{9} = - 3$ , $a_{10} = - 9$ . The positive and negative terms cancel, so $a_{7} + a_{8} + a_{9} + a_{10} = 0$ . Hence the sum of the first 10 terms equals the sum of the first 6 terms, i.e. 180.
  ½ mark
Answer: 6 terms or 10 terms. The 7th to 10th terms add up to 0, so both sums are 180.
Where the marks are
- Identifies $a = 45$ , $d = - 6$ (½ mark)
- Substitutes into the $S_{n}$ formula (1 mark)
- Forms the quadratic $6 n^{2} - 96 n + 360 = 0$ (1 mark)
- Factorises $n^{2} - 16 n + 60 = 0$ (1 mark)
- Gets $n = 6, 10$ (1 mark)
- Explains that the 7th to 10th terms sum to zero (½ mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
As observed from the top of a $75$ m high lighthouse from the sea-level, the angles of depression of two ships are $3 0^{\circ}$ and $6 0^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: PQ is the lighthouse with $P Q = 75$ m and foot P at sea level. QX is the horizontal through Q. R and S are the ships on the same side, in line with P, with R nearer. The angles of depression are $∠ X QR = 6 0^{\circ}$ and $∠ X QS = 3 0^{\circ}$ . As alternate angles, $∠ QR P = 6 0^{\circ}$ and $∠ QS P = 3 0^{\circ}$ .
  1 mark
2. In right $△ QP R$ : $tan 6 0^{\circ} = \frac{P Q}{P R} \Rightarrow 3 = \frac{75}{P R} \Rightarrow P R = \frac{75}{3} = 253$ m
  1½ marks
3. In right $△ QP S$ : $tan 3 0^{\circ} = \frac{P Q}{P S} \Rightarrow \frac{1}{3} = \frac{75}{P S} \Rightarrow P S = 753$ m
  1 mark
4. Distance between the ships: $R S = P S - P R = 753 - 253 = 503$ m
  1 mark
5. $= 50 \times 1.73 = 86.5$ m
  ½ mark
Answer: Distance between the ships = $503$ m = 86.5 m
Where the marks are
- Correct labelled figure (1 mark)
- Finds $P R = 253$ m using $tan 6 0^{\circ}$ (1½ marks)
- Finds $P S = 753$ m using $tan 3 0^{\circ}$ (1 mark)
- Writes $R S = P S - P R = 503$ m (1 mark)
- Final value 86.5 m (½ mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of $30$ m high building are $3 0^{\circ}$ and $6 0^{\circ}$ , respectively. Find the height of the transmission tower. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: AB is the building with $A B = 30$ m. BP is the tower on top, with $B P = h$ m. C is the point on the ground, with $∠ A C B = 3 0^{\circ}$ and $∠ A C P = 6 0^{\circ}$ and $∠ C A P = 9 0^{\circ}$ .
  1 mark
2. In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{3} = \frac{30}{A C} \Rightarrow A C = 303$ m
  1½ marks
3. In right $△ P A C$ : $tan 6 0^{\circ} = \frac{A P}{A C} \Rightarrow 3 = \frac{30 + h}{30 3} \Rightarrow 303 \times 3 = 30 + h$
  1½ marks
4. $90 = 30 + h \Rightarrow h = 60$
  So the height of the transmission tower is 60 m.
  1 mark
Answer: Height of transmission tower = 60 m
Where the marks are
- Correct labelled figure (1 mark)
- Finds $A C = 303$ m using $tan 3 0^{\circ}$ (1½ marks)
- Forms $3 = \frac{30 + h}{30 3}$ using $tan 6 0^{\circ}$ (1½ marks)
- Gets $h = 60$ m (1 mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerCompetency-basedStatistics
A student noted the number of cars passing through a spot on a road for $100$ periods each of $3$ minutes and summarised it in the table given below. Find the mean and median of the following data.
Try it first, then show our solutionSolution
1. Class $x_{i}$ $f_{i}$ $f_{i} x_{i}$ c.f.
  0–10 5 7 35 7
  10–20 15 14 210 21
  20–30 25 13 325 34
  30–40 35 12 420 46
  40–50 45 20 900 66
  50–60 55 11 605 77
  60–70 65 15 975 92
  70–80 75 8 600 100
  Total 100 4070
  2 marks
2. Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{4070}{100} = 40.7$
  1 mark
3. $\frac{n}{2} = 50$ . The c.f. just greater than 50 is 66, so the median class is 40–50.
  Here $l = 40$ , $c f = 46$ , $f = 20$ , $h = 10$ .
  ½ mark
4. Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 40 + \frac{50 - 46}{20} \times 10 = 40 + 2 = 42$
  1½ marks
Answer: Mean = 40.7 cars; Median = 42 cars
Where the marks are
- Correct table with class marks, $f_{i} x_{i}$ and cumulative frequencies (2 marks)
- Mean = 40.7 (1 mark)
- Identifies median class 40–50 (½ mark)
- Median = 42 using the formula (1½ marks)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerTriangles
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $△ P QR$ . Show that $△ A B C \sim △ P QR$ .
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ with median AD (D is the mid-point of BC) and $△ P QR$ with median PM (M is the mid-point of QR).
  Given: $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ . To prove: $△ A B C \sim △ P QR$ .
  1 mark
2. D and M are mid-points, so $B C = 2 B D$ and $QR = 2 QM$ . Substituting in the given ratios:
  $\frac{A B}{P Q} = \frac{2 B D}{2 QM} = \frac{A D}{P M}$ , which gives $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ .
  1 mark
3. All three pairs of corresponding sides of $△ A B D$ and $△ P QM$ are in the same ratio, so $△ A B D \sim △ P QM$ [SSS similarity criterion].
  1 mark
4. Corresponding angles of similar triangles are equal, so $∠ B = ∠ Q$ ...(i)
  ½ mark
5. Now compare $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ (given) and the included angles satisfy $∠ B = ∠ Q$ [from (i)].
  ½ mark
6. Therefore $△ A B C \sim △ P QR$ [SAS similarity criterion]. Hence proved.
  1 mark
Answer: $△ A B D \sim △ P QM$ (SSS) gives $∠ B = ∠ Q$ ; then $△ A B C \sim △ P QR$ (SAS).
Where the marks are
- Correct labelled figure of both triangles with their medians (1 mark)
- Uses $B C = 2 B D$ , $QR = 2 QM$ to get $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ (1 mark)
- Concludes $△ A B D \sim △ P QM$ by SSS (1 mark)
- Deduces $∠ B = ∠ Q$ (½ mark)
- States the SAS conditions in $△ A B C$ and $△ P QR$ (½ mark)
- Concludes $△ A B C \sim △ P QR$ (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerTriangles
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E. Prove that $E L = 2 B L$ .
Try it first, then show our solutionSolution
1. Figure: Draw parallelogram ABCD with M the mid-point of CD. Join BM and extend it to meet AD produced at E. Draw diagonal AC, which cuts BE at L.
  1 mark
2. In $△ B M C$ and $△ E M D$ :
  $M C = M D$ [M is the mid-point of CD]
  $∠ C M B = ∠ D M E$ [vertically opposite angles]
  $∠ M B C = ∠ M E D$ [alternate angles, since $B C ∥ A E$ ]
  So $△ B M C ≅ △ E M D$ [AAS], and therefore $B C = D E$ [CPCT].
  1 mark
3. Opposite sides of a parallelogram are equal, so $A D = B C$ . Hence $A D = D E$ , which gives $A E = A D + D E = 2 B C$ .
  1 mark
4. In $△ A E L$ and $△ C B L$ :
  $∠ L A E = ∠ L C B$ [alternate angles, $A E ∥ B C$ ]
  $∠ A L E = ∠ C L B$ [vertically opposite angles]
  So $△ A E L \sim △ C B L$ [AA similarity].
  ½ mark
5. Corresponding sides are proportional: $\frac{E L}{B L} = \frac{A E}{C B} = \frac{2 B C}{B C}$ .
  ½ mark
6. So $\frac{E L}{B L} = 2$ , i.e. $E L = 2 B L$ . Hence proved.
  1 mark
Answer: $△ B M C ≅ △ E M D$ gives $A E = 2 B C$ ; then $△ A E L \sim △ C B L$ gives $\frac{E L}{B L} = 2$ , i.e. $E L = 2 B L$ .
Where the marks are
- Correct labelled figure (1 mark)
- Proves $△ B M C ≅ △ E M D$ and gets $B C = D E$ (1 mark)
- Uses $A D = B C$ to obtain $A E = 2 B C$ (1 mark)
- Proves $△ A E L \sim △ C B L$ (½ mark)
- Writes the ratio $\frac{E L}{B L} = \frac{A E}{B C} = \frac{2 B C}{B C}$ (½ mark)
- Concludes $E L = 2 B L$ (1 mark)
Open Q35(b) on its own page

Class	$x_{i}$	$f_{i}$	$f_{i} x_{i}$	c.f.
0–10	5	7	35	7
10–20	15	14	210	21
20–30	25	13	325	34
30–40	35	12	420	46
40–50	45	20	900	66
50–60	55	11	605	77
60–70	65	15	975	92
70–80	75	8	600	100
Total		100	4070

Section E

Case study based; internal choice in each case study (2-mark subparts)

Q36 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
Case Study 1. In an annual day function of a school, the organizers wanted to give a cash prize along with a memento to their best students. Each memento is made as shown in the figure and its base ABCD is shown from the front side. The rate of silver plating is ₹ $20$ per cm $^{2}$ .
1. (i)
  What is the area of the quadrant ODCO ?
  1 mark
2. (ii)
  Find the area of $△ A O B$ .
  1 mark
3. (iii)
  (a) What is the total cost of silver plating the shaded part ABCD ?
  2 marks
4. OR
  (iii)
  (b) What is the length of arc CD ?
  2 marks
Figure in the paper: A memento base: a circle with centre O and radius OC = OD = 7 cm, with a right angle at O (angle DOC = 90°). D and C are on the circle. Segment AD = 3 cm extends from D to A, and segment CB = 3 cm extends from C to B, with A and B joined so that ABCD is the base. The shaded region is the part between the arc DC and the segment AB below the arc. Labels: 7 cm along OC and OD, 3 cm along AD and CB. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The quadrant has $r = 7$ cm and $θ = 9 0^{\circ}$ .
  Area $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{90}{360} \times \frac{22}{7} \times 7 \times 7$ .
  ½ mark
2. $= \frac{77}{2} = 38.5$ cm $^{2}$ .
  ½ mark
3. (ii) $O A = O D + D A = 7 + 3 = 10$ cm and $O B = O C + C B = 7 + 3 = 10$ cm, with $∠ A O B = 9 0^{\circ}$ .
  ar $(△ A O B) = \frac{1}{2} \times O A \times O B = \frac{1}{2} \times 10 \times 10 = 50$ cm $^{2}$ .
  1 mark
4. (iii)(a) Area of the shaded part ABCD = ar $(△ A O B)$ − area of quadrant ODCO $= 50 - 38.5 = 11.5$ cm $^{2}$ .
  Cost $= 11.5 \times 20$ .
  1 mark
5. $= ₹ 230$ .
  1 mark
6. (iii)(b) Length of arc CD $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{90}{360} \times 2 \times \frac{22}{7} \times 7$ .
  1 mark
7. $= 11$ cm.
  1 mark
Answer: (i) 38.5 cm²; (ii) 50 cm²; (iii)(a) ₹230; (iii)(b) 11 cm
Where the marks are
- (i) Correct sector-area formula with substitution (½ mark) and answer 38.5 cm² (½ mark)
- (ii) Takes OA = OB = 10 cm and gets area 50 cm² (1 mark)
- (iii)(a) Shaded area = 50 − 38.5, multiplied by ₹20 (1 mark); cost ₹230 (1 mark)
- (iii)(b) Arc-length formula with substitution (1 mark); 11 cm (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Case Study 2. In a coffee shop, coffee is served in two types of cups. One is cylindrical in shape with diameter $7$ cm and height $14$ cm and the other is hemispherical with diameter $21$ cm.
1. (i)
  Find the area of the base of the cylindrical cup.
  1 mark
2. (ii)
  (a) What is the capacity of the hemispherical cup ?
  2 marks
3. OR
  (ii)
  (b) Find the capacity of the cylindrical cup.
  2 marks
4. (iii)
  What is the curved surface area of the cylindrical cup ?
  1 mark
Figure in the paper: Two photographs of coffee cups: on the left a cylindrical mug with a handle on its right side; on the right a hemispherical bowl-shaped cup with a handle on its right side. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Radius of the cylindrical cup $r = \frac{7}{2}$ cm.
  Area of base $= π r^{2} = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{2} = 38.5$ cm $^{2}$ .
  1 mark
2. (ii)(a) Radius of the hemispherical cup $R = \frac{21}{2}$ cm.
  Capacity $= \frac{2}{3} π R^{3} = \frac{2}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$ .
  1 mark
3. $= \frac{4851}{2} = 2425.5$ cm $^{3}$ .
  1 mark
4. (ii)(b) Capacity of the cylindrical cup $= π r^{2} h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14$ .
  1 mark
5. $= 539$ cm $^{3}$ .
  1 mark
6. (iii) Curved surface area of the cylinder $= 2 π r h = 2 \times \frac{22}{7} \times \frac{7}{2} \times 14 = 308$ cm $^{2}$ .
  1 mark
Answer: (i) 38.5 cm²; (ii)(a) 2425.5 cm³; (ii)(b) 539 cm³; (iii) 308 cm²
Where the marks are
- (i) Base area $π r^{2}$ with r = 3.5 cm gives 38.5 cm² (1 mark)
- (ii)(a) Formula $\frac{2}{3} π r^{3}$ with r = 10.5 cm (1 mark); 2425.5 cm³ (1 mark)
- (ii)(b) Formula $π r^{2} h$ with r = 3.5 cm, h = 14 cm (1 mark); 539 cm³ (1 mark)
- (iii) $2 π r h$ gives 308 cm² (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedProbability
Case Study 3. Computer-based learning (CBL) refers to any teaching methodology that makes use of computers for information transmission. At an elementary school level, computer applications can be used to display multimedia lesson plans. A survey was done on $1000$ elementary and secondary schools of Assam and they were classified by the number of computers they had.
Number of Computers 1–10 11–20 21–50 51–100 101 and more
Number of Schools 250 200 290 180 80
One school is chosen at random. Then :
1. (i)
  Find the probability that the school chosen at random has more than $100$ computers.
  1 mark
2. (ii)
  (a) Find the probability that the school chosen at random has $50$ or fewer computers.
  2 marks
3. OR
  (ii)
  (b) Find the probability that the school chosen at random has no more than $20$ computers.
  2 marks
4. (iii)
  Find the probability that the school chosen at random has $10$ or less than $10$ computers.
  1 mark
Figure in the paper: A photograph of a computer lab with rows of desks, each with a monitor and a black chair, and a projector screen on the wall at the back. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Schools with more than 100 computers (101 and more) = 80, out of 1000 schools in total.
  P(more than 100) $= \frac{80}{1000} = 0.08$ .
  1 mark
2. (ii)(a) Schools with 50 or fewer computers come from the classes 1–10, 11–20 and 21–50: $250 + 200 + 290 = 740$ .
  1 mark
3. P(50 or fewer) $= \frac{740}{1000} = 0.74$ .
  1 mark
4. (ii)(b) Schools with no more than 20 computers come from the classes 1–10 and 11–20: $250 + 200 = 450$ .
  1 mark
5. P(no more than 20) $= \frac{450}{1000} = 0.45$ .
  1 mark
6. (iii) Schools with 10 or fewer computers (class 1–10) = 250.
  P $= \frac{250}{1000} = 0.25$ .
  1 mark
Answer: (i) 0.08; (ii)(a) 0.74; (ii)(b) 0.45; (iii) 0.25
Where the marks are
- (i) 80/1000 = 0.08 (1 mark)
- (ii)(a) Counts 250 + 200 + 290 = 740 (1 mark); probability 0.74 (1 mark)
- (ii)(b) Counts 250 + 200 = 450 (1 mark); probability 0.45 (1 mark)
- (iii) 250/1000 = 0.25 (1 mark)
Open Q38 on its own page

Number of Computers	1–10	11–20	21–50	51–100	101 and more
Number of Schools	250	200	290	180	80

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.