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

CBSE Class 10 Mathematics Basic: 2023 board paper, set 430/5/1

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_BASIC.zip (a zip; this paper is the file 430_5_1_Maths Basic.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 Q22
C	26-31	Short answer; internal choice in Q26 and Q29
D	32-35	Long answer; internal choice in Q33 and Q35
E	36-38	Case study based; internal choice in 3 case-study 2-mark subparts

Instructions in brief: All 38 questions are compulsory; there is no overall choice, but internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 case-study subparts in Section E.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQArithmetic Progressions
The $8^{t h}$ term of an A.P. is 17 and its $1 4^{t h}$ term is 29. The common difference of this A.P. is:
1. (A)3
2. (B)2
3. (C)5
4. (D)-2
Try it first, then show our solutionSolution
1. $a_{14} - a_{8} = (a + 13 d) - (a + 7 d) = 6 d = 29 - 17 = 12$
2. $d = 2$ , option (B)
  1 mark
Answer: d = 2, option (B)
Where the marks are
- Correct option (B) 2 (1 mark)
Q2 · 1 mark · MCQCircles
In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of $5 0^{\circ}$ with PQ. The measure of $∠ P O Q$ is:
1. (A) $13 0^{\circ}$
2. (B) $10 0^{\circ}$
3. (C) $9 0^{\circ}$
4. (D) $7 5^{\circ}$
Figure in the paper: A circle with centre O. A chord PQ is drawn. A tangent PT is drawn at point P, extending to the right. The angle between the tangent PT and chord PQ is marked $5 0^{\circ}$ . OP and OQ are radii joining the centre O to P and Q. The angle POQ is at the centre. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ O P Q = 9 0^{\circ} - 5 0^{\circ} = 4 0^{\circ}$ . Since OP = OQ, $∠ O QP = 4 0^{\circ}$ .
2. $∠ P O Q = 18 0^{\circ} - 4 0^{\circ} - 4 0^{\circ} = 10 0^{\circ}$ , option (B)
  1 mark
Answer: 100°, option (B)
Where the marks are
- Correct option (B) 100° (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQProbability
One card is drawn at random from a well shuffled deck of 52 playing cards. What is the probability of getting '4 of hearts'?
1. (A) $\frac{1}{52}$
2. (B) $\frac{1}{13}$
3. (C) $\frac{1}{26}$
4. (D) $\frac{1}{6}$
Try it first, then show our solutionSolution
1. There is exactly one '4 of hearts' among 52 cards, so $P = \frac{1}{52}$ , option (A).
  1 mark
Answer: 1/52, option (A)
Where the marks are
- Correct option (A) (1 mark)
Q4 · 1 mark · MCQCoordinate Geometry
The distance between the points A(0, 6) and B(-6, 2) is:
1. (A)6 units
2. (B) $26$ units
3. (C) $213$ units
4. (D) $132$ units
Try it first, then show our solutionSolution
1. $A B = (- 6 - 0)^{2} + (2 - 6)^{2} = 36 + 16 = 52$
2. $= 213$ units, option (C)
  1 mark
Answer: $213$ units, option (C)
Where the marks are
- Correct option (C) (1 mark)
Q5 · 1 mark · MCQQuadratic Equations
The value(s) of k for which the roots of quadratic equation $x^{2} + 4 x + k = 0$ are real, is:
1. (A) $k \geq 4$
2. (B) $k \leq 4$
3. (C) $k \geq - 4$
4. (D) $k \leq - 4$
Try it first, then show our solutionSolution
1. Real roots need $D = b^{2} - 4 a c \geq 0$ : $16 - 4 k \geq 0$ .
2. $k \leq 4$ , option (B)
  1 mark
Answer: k ≤ 4, option (B)
Where the marks are
- Correct option (B) (1 mark)
Q6 · 1 mark · MCQReal Numbers
LCM of $(2^{3} \times 3 \times 5)$ and $(2^{4} \times 5 \times 7)$ is:
1. (A)40
2. (B)560
3. (C)1680
4. (D)1120
Try it first, then show our solutionSolution
1. LCM = product of highest powers: $2^{4} \times 3 \times 5 \times 7 = 1680$ , option (C)
  1 mark
Answer: 1680, option (C)
Where the marks are
- Correct option (C) (1 mark)
Q7 · 1 mark · MCQPolynomials
If one zero of the quadratic polynomial $k x^{2} + 3 x + k$ is 2, then the value of k is:
1. (A) $- \frac{6}{5}$
2. (B) $\frac{6}{5}$
3. (C) $\frac{5}{6}$
4. (D) $- \frac{5}{6}$
Try it first, then show our solutionSolution
1. $p (2) = 4 k + 6 + k = 0 \Rightarrow 5 k = - 6$
2. $k = - \frac{6}{5}$ , option (A)
  1 mark
Answer: k = −6/5, option (A)
Where the marks are
- Correct option (A) (1 mark)
Q8 · 1 mark · MCQPair of Linear Equations in Two Variables
If the lines represented by equations $3 x + 2 m y = 2$ and $2 x + 5 y + 1 = 0$ are parallel, then the value of m is:
1. (A) $\frac{2}{5}$
2. (B) $- \frac{5}{4}$
3. (C) $\frac{3}{2}$
4. (D) $\frac{15}{4}$
Try it first, then show our solutionSolution
1. Parallel: $\frac{3}{2} = \frac{2 m}{5} \neq = \frac{c _{1}}{c _{2}}$
2. $4 m = 15 \Rightarrow m = \frac{15}{4}$ , option (D)
  1 mark
Answer: m = 15/4, option (D)
Where the marks are
- Correct option (D) (1 mark)
Q9 · 1 mark · MCQTriangles
$△ A B C \sim △ D E F$ and their perimeters are 32 cm and 24 cm respectively. If AB = 10 cm, then DE equals:
1. (A)8 cm
2. (B)7.5 cm
3. (C)15 cm
4. (D) $53$ cm
Try it first, then show our solutionSolution
1. Ratio of perimeters = ratio of corresponding sides: $\frac{10}{D E} = \frac{32}{24}$
2. $D E = \frac{240}{32} = 7.5$ cm, option (B)
  1 mark
Answer: 7.5 cm, option (B)
Where the marks are
- Correct option (B) (1 mark)
Q10 · 1 mark · MCQQuadratic Equations
The two roots of the equation $3 x^{2} - 26 x + 2 = 0$ are:
1. (A)real and distinct
2. (B)not real
3. (C)real and equal
4. (D)rational
Try it first, then show our solutionSolution
1. $D = (26)^{2} - 4 (3) (2) = 24 - 24 = 0$
2. D = 0, so roots are real and equal, option (C)
  1 mark
Answer: real and equal, option (C)
Where the marks are
- Correct option (C) (1 mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
If $sin θ = \frac{a}{b}$ , then $sec θ$ is equal to $(0 \leq θ \leq 9 0^{\circ})$ :
1. (A) $\frac{a}{b ^{2} - a ^{2}}$
2. (B) $\frac{b}{b ^{2} - a ^{2}}$
3. (C) $\frac{b ^{2} - a ^{2}}{b}$
4. (D) $\frac{b ^{2} - a ^{2}}{a}$
Try it first, then show our solutionSolution
1. $cos θ = 1 - \frac{a ^{2}}{b ^{2}} = \frac{b ^{2} - a ^{2}}{b}$
2. $sec θ = \frac{b}{b ^{2} - a ^{2}}$ , option (B)
  1 mark
Answer: $\frac{b}{b ^{2} - a ^{2}}$ , option (B)
Where the marks are
- Correct option (B) (1 mark)
Q12 · 1 mark · MCQArithmetic Progressions
The sum of the first 100 even natural numbers is:
1. (A)10100
2. (B)2550
3. (C)5050
4. (D)10010
Try it first, then show our solutionSolution
1. $2 + 4 + \dots + 200 = \frac{100}{2} (2 + 200)$
2. $= 50 \times 202 = 10100$ , option (A)
  1 mark
Answer: 10100, option (A)
Where the marks are
- Correct option (A) (1 mark)
Q13 · 1 mark · MCQTriangles
In the given figure, AD = 2 cm, DB = 3 cm, DE = 2.5 cm and DE $∥$ BC. The value of x is:
1. (A)6 cm
2. (B)3.75 cm
3. (C)6.25 cm
4. (D)7.5 cm
Figure in the paper: Triangle ABC with vertex A at top. D lies on AB with AD = 2 cm and DB = 3 cm. E lies on AC. DE = 2.5 cm and DE is parallel to base BC. The base BC is labelled x. See it in the official paper.
Try it first, then show our solutionSolution
1. $△ A D E \sim △ A B C$ (AA, DE ∥ BC), so $\frac{D E}{B C} = \frac{A D}{A B} = \frac{2}{5}$
2. $x = \frac{2.5 \times 5}{2} = 6.25$ cm, option (C)
  1 mark
Answer: 6.25 cm, option (C)
Where the marks are
- Correct option (C) (1 mark)
Q14 · 1 mark · MCQCircles
A circle is of radius 3 cm. The distance between two of its parallel tangents is:
1. (A)12 cm
2. (B)6 cm
3. (C)3 cm
4. (D)4.5 cm
Try it first, then show our solutionSolution
1. Parallel tangents touch at ends of a diameter, so distance = diameter = 2 × 3 = 6 cm, option (B)
  1 mark
Answer: 6 cm, option (B)
Where the marks are
- Correct option (B) (1 mark)
Q15 · 1 mark · MCQStatistics
The median class for the data given below is:
Class 20–40 40–60 60–80 80–100 100–120
Frequency 10 12 14 13 17
1. (A)80–100
2. (B)20–40
3. (C)40–60
4. (D)60–80
Try it first, then show our solutionSolution
1. N = 66, N/2 = 33. Cumulative frequencies: 10, 22, 36, 49, 66.
2. First cf ≥ 33 is 36, so median class is 60–80, option (D)
  1 mark
Answer: 60–80, option (D)
Where the marks are
- Correct option (D) (1 mark)
Q16 · 1 mark · MCQIntroduction to Trigonometry
If $sin θ = \frac{3}{4}$ , then $\frac{( s e c ^{2} θ - 1 ) c o s ^{2} θ}{s i n θ}$ equals:
1. (A) $\frac{3}{5}$
2. (B) $\frac{3}{4}$
3. (C) $\frac{4}{3}$
4. (D) $\frac{9}{16}$
Try it first, then show our solutionSolution
1. $(sec^{2} θ - 1) cos^{2} θ = tan^{2} θ cos^{2} θ = sin^{2} θ$
2. Expression $= \frac{s i n ^{2} θ}{s i n θ} = sin θ = \frac{3}{4}$ , option (B)
  1 mark
Answer: 3/4, option (B)
Where the marks are
- Correct option (B) (1 mark)
Q17 · 1 mark · MCQTriangles
In two triangles $△ P QR$ and $△ A B C$ , it is given that $\frac{A B}{B C} = \frac{P Q}{P R}$ . For these two triangles to be similar, which of the following should be true?
1. (A) $∠ A = ∠ P$
2. (B) $∠ B = ∠ Q$
3. (C) $∠ B = ∠ P$
4. (D)CA = QR
Try it first, then show our solutionSolution
1. For SAS similarity the included angles must be equal: angle between AB, BC is ∠B; between PQ, PR is ∠P.
2. So ∠B = ∠P, option (C)
  1 mark
Answer: ∠B = ∠P, option (C)
Where the marks are
- Correct option (C) (1 mark)
Q18 · 1 mark · MCQStatistics
Mean and median of some data are 32 and 30 respectively. Using empirical relation, mode of the data is:
1. (A)36
2. (B)26
3. (C)30
4. (D)20
Try it first, then show our solutionSolution
1. Mode = 3 Median − 2 Mean = 90 − 64 = 26, option (B)
  1 mark
Answer: 26, option (B)
Where the marks are
- Correct option (B) (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A): When two coins are tossed together, the probability of getting no tail is $\frac{1}{4}$ .
Reason (R): The probability P(E) of an event E satisfies $0 \leq P (E) \leq 1$ .
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. Outcomes HH, HT, TH, TT; no tail only HH, so P = 1/4: A true. R is a true general fact.
2. R does not explain why the value is 1/4, so option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Correct option (B) (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedSurface Areas and Volumes
Assertion (A): The surface area of largest sphere that can be inscribed in a hollow cube of side 'a' cm is $π a^{2}$ cm $^{2}$ .
Reason (R): The surface area of a sphere of radius 'r' is $\frac{4}{3} π r^{3}$ .
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. Largest sphere has radius a/2, surface area $4 π (a /2)^{2} = π a^{2}$ : A true.
2. $\frac{4}{3} π r^{3}$ is volume, not surface area: R false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Correct option (C) (1 mark)

Class	20–40	40–60	60–80	80–100	100–120
Frequency	10	12	14	13	17

Section B

Very short answer; internal choice in Q22

Q21 · 2 marks · Very short answerReal Numbers
Find LCM of 576 and 512 by prime factorization.
Try it first, then show our solutionSolution
1. Prime factorisation of 576: $576 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^{6} \times 3^{2}$
  ½ mark
2. Prime factorisation of 512: $512 = 2^{9}$
  ½ mark
3. LCM = product of the highest powers of all prime factors $= 2^{9} \times 3^{2} = 512 \times 9 = 4608$
  1 mark
Answer: LCM = 4608
Where the marks are
- Writes $576 = 2^{6} \times 3^{2}$ (½ mark)
- Writes $512 = 2^{9}$ (½ mark)
- Takes highest powers to get LCM $= 2^{9} \times 3^{2} = 4608$ (1 mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $\frac{s i n 3 0 ^{\circ} + t a n 4 5 ^{\circ}}{s e c 3 0 ^{\circ} + c o t 4 5 ^{\circ}}$
Try it first, then show our solutionSolution
1. Substitute the standard values $sin 3 0^{\circ} = \frac{1}{2}$ , $tan 4 5^{\circ} = 1$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $cot 4 5^{\circ} = 1$ :
  $\frac{\frac{1}{2} + 1}{\frac{2}{3} + 1}$
  1½ marks
2. Simplify: $\frac{\frac{3}{2}}{\frac{2 + 3}{3}} = \frac{3 3}{2 ( 2 + 3 )}$ (on rationalising, $= \frac{3 3 ( 2 - 3 )}{2} = \frac{6 3 - 9}{2}$ )
  ½ mark
Answer: $\frac{3 3}{2 ( 2 + 3 )}$ , i.e. $\frac{6 3 - 9}{2}$
Where the marks are
- Correct substitution of all four trigonometric values (1½ marks)
- Simplifies to $\frac{3 3}{2 ( 2 + 3 )}$ or equivalent (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerIntroduction to Trigonometry
For $A = 3 0^{\circ}$ and $B = 6 0^{\circ}$ , verify that: $sin (A + B) = sin A cos B + cos A sin B$ .
Try it first, then show our solutionSolution
1. LHS $= sin (3 0^{\circ} + 6 0^{\circ}) = sin 9 0^{\circ} = 1$
  ½ mark
2. RHS $= sin 3 0^{\circ} cos 6 0^{\circ} + cos 3 0^{\circ} sin 6 0^{\circ} = \frac{1}{2} \times \frac{1}{2} + \frac{3}{2} \times \frac{3}{2}$
  1 mark
3. $= \frac{1}{4} + \frac{3}{4} = 1$ . Hence LHS = RHS, verified.
  ½ mark
Answer: LHS = RHS = 1, hence verified
Where the marks are
- LHS $= sin 9 0^{\circ} = 1$ (½ mark)
- Correct substitution in RHS (1 mark)
- RHS = 1, so LHS = RHS (½ mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerCircles
In the given figure, tangents AB and AC are drawn to a circle centred at O. If $∠ O A B = 6 0^{\circ}$ and OB = 5 cm, find lengths OA and AC.
Figure in the paper: A circle with centre O. Two tangents AB and AC are drawn from external point A, touching the circle at B and C. OB = 5 cm is marked as a radius. The angle OAB = $6 0^{\circ}$ is marked at A. O, A and the line joining them are shown. See it in the official paper.
Try it first, then show our solutionSolution
1. $O B ⊥ A B$ [tangent ⟂ radius at the point of contact], so $△ O B A$ is right-angled at B. In it, $sin 6 0^{\circ} = \frac{O B}{O A} \Rightarrow \frac{3}{2} = \frac{5}{O A}$
  ½ mark
2. $O A = \frac{10}{3} = \frac{10 3}{3}$ cm
  ½ mark
3. $tan 6 0^{\circ} = \frac{O B}{A B} \Rightarrow 3 = \frac{5}{A B}$
  ½ mark
4. $A B = \frac{5}{3} = \frac{5 3}{3}$ cm, and $A C = A B = \frac{5 3}{3}$ cm [tangents from an external point are equal]
  ½ mark
Answer: $O A = \frac{10 3}{3}$ cm, $A C = \frac{5 3}{3}$ cm
Where the marks are
- Uses $sin 6 0^{\circ} = O B / O A$ in right $△ O B A$ (½ mark)
- $O A = \frac{10 3}{3}$ cm (½ mark)
- Uses $tan 6 0^{\circ} = O B / A B$ (½ mark)
- $A C = A B = \frac{5 3}{3}$ cm (½ mark)
Open Q23 on its own page
Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerCoordinate Geometry
Show that A(1, 2), B(5, 4), C(3, 8) and D(-1, 6) are vertices of a parallelogram ABCD.
Try it first, then show our solutionSolution
1. Mid-point of diagonal AC $= (\frac{1 + 3}{2}, \frac{2 + 8}{2}) = (2, 5)$
  1 mark
2. Mid-point of diagonal BD $= (\frac{5 + ( - 1 )}{2}, \frac{4 + 6}{2}) = (2, 5)$
  ½ mark
3. The diagonals AC and BD have the same mid-point, i.e. they bisect each other. Hence ABCD is a parallelogram.
  ½ mark
Answer: Diagonals AC and BD have common mid-point (2, 5), so ABCD is a parallelogram.
Where the marks are
- Mid-point of AC = (2, 5) (1 mark)
- Mid-point of BD = (2, 5) (½ mark)
- Concludes diagonals bisect each other, so parallelogram (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerCoordinate Geometry
Show that the points A(3, 0), B(6, 4) and C(-1, 3) are vertices of a right-angled triangle.
Try it first, then show our solutionSolution
1. $A B^{2} = (6 - 3)^{2} + (4 - 0)^{2} = 9 + 16 = 25$
  ½ mark
2. $B C^{2} = (- 1 - 6)^{2} + (3 - 4)^{2} = 49 + 1 = 50$
  ½ mark
3. $A C^{2} = (- 1 - 3)^{2} + (3 - 0)^{2} = 16 + 9 = 25$
  ½ mark
4. $A B^{2} + A C^{2} = 25 + 25 = 50 = B C^{2}$ , so by the converse of Pythagoras theorem $△ A B C$ is right-angled at A.
  ½ mark
Answer: $B C^{2} = A B^{2} + A C^{2}$ , so ABC is right-angled at A.
Where the marks are
- $A B^{2} = 25$ (½ mark)
- $B C^{2} = 50$ (½ mark)
- $A C^{2} = 25$ (½ mark)
- Shows $B C^{2} = A B^{2} + A C^{2}$ and concludes right angle (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerArithmetic Progressions
Find the sum of the first 15 terms of the A.P.: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$
Try it first, then show our solutionSolution
1. $a = \frac{1}{15}$ , $d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$
  ½ mark
2. $S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{15} = \frac{15}{2} [\frac{2}{15} + 14 \times \frac{1}{60}]$
  1 mark
3. $= \frac{15}{2} [\frac{8}{60} + \frac{14}{60}] = \frac{15}{2} \times \frac{22}{60} = \frac{11}{4}$
  ½ mark
Answer: $S_{15} = \frac{11}{4}$
Where the marks are
- Finds $d = \frac{1}{60}$ (½ mark)
- Correct substitution in $S_{n}$ formula (1 mark)
- $S_{15} = \frac{11}{4}$ (½ mark)
Open Q25 on its own page

Section C

Short answer; internal choice in Q26 and Q29

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
Sabina went to a bank ATM to withdraw ₹ 2,000. She received ₹ 50 and ₹ 100 notes only. If Sabina got 25 notes in all, how many notes of ₹ 50 and ₹ 100 did she receive?
Try it first, then show our solutionSolution
1. Let the number of ₹50 notes be $x$ and of ₹100 notes be $y$ . Total notes: $x + y = 25$ …(i)
  1 mark
2. Total amount: $50 x + 100 y = 2000 \Rightarrow x + 2 y = 40$ …(ii)
  1 mark
3. Subtracting (i) from (ii): $y = 15$ ; then $x = 25 - 15 = 10$ . So she received 10 notes of ₹50 and 15 notes of ₹100.
  1 mark
Answer: ₹50 notes = 10, ₹100 notes = 15
Where the marks are
- Forms $x + y = 25$ (1 mark)
- Forms $50 x + 100 y = 2000$ (1 mark)
- Solves to get $x = 10$ , $y = 15$ (1 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
Five years ago, Amit was thrice as old as Baljeet. Ten years hence, Amit shall be twice as old as Baljeet. What are their present ages?
Try it first, then show our solutionSolution
1. Let Amit's present age be $x$ years and Baljeet's be $y$ years. Five years ago: $x - 5 = 3 (y - 5) \Rightarrow x - 3 y = - 10$ …(i)
  1 mark
2. Ten years hence: $x + 10 = 2 (y + 10) \Rightarrow x - 2 y = 10$ …(ii)
  1 mark
3. (ii) − (i): $y = 20$ ; then $x = 10 + 2 (20) = 50$ . Amit is 50 years and Baljeet is 20 years old.
  1 mark
Answer: Amit = 50 years, Baljeet = 20 years
Where the marks are
- Forms $x - 3 y = - 10$ (1 mark)
- Forms $x - 2 y = 10$ (1 mark)
- Solves to get $x = 50$ , $y = 20$ (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerReal Numbers
Prove that $(7 - 23)$ is an irrational number, given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $7 - 23$ is rational. Then $7 - 23 = \frac{a}{b}$ , where $a, b$ are integers and $b \neq = 0$ .
  1 mark
2. Rearranging: $23 = 7 - \frac{a}{b} \Rightarrow 3 = \frac{7 b - a}{2 b}$
  1 mark
3. Since $a, b$ are integers, $\frac{7 b - a}{2 b}$ is rational, so $3$ would be rational. This contradicts the fact that $3$ is irrational. So our assumption is wrong and $7 - 23$ is irrational.
  1 mark
Answer: $7 - 23$ is irrational (proved by contradiction).
Where the marks are
- Assumes $7 - 23 = \frac{a}{b}$ , $b \neq = 0$ (1 mark)
- Expresses $3 = \frac{7 b - a}{2 b}$ (1 mark)
- States contradiction and concludes (1 mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerStatistics
Find mean of the following data:
Class 0–15 15–30 30–45 45–60 60–75 75–90
Frequency 12 15 11 20 16 6
Try it first, then show our solutionSolution
1. Take assumed mean $a = 37.5$ , class size $h = 15$ , $u_{i} = \frac{x _{i} - 37.5}{15}$ .
  Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  0–15 7.5 12 −2 −24
  15–30 22.5 15 −1 −15
  30–45 37.5 11 0 0
  45–60 52.5 20 1 20
  60–75 67.5 16 2 32
  75–90 82.5 6 3 18
  Total 80 31
  2 marks
2. Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 37.5 + \frac{31}{80} \times 15 = 37.5 + 5.8125 = 43.31$ (approx. 43.3)
  1 mark
Answer: Mean ≈ 43.31
Where the marks are
- Correct table with class marks, $u_{i}$ , $\sum f_{i} = 80$ and $\sum f_{i} u_{i} = 31$ (2 marks)
- Applies step-deviation formula to get mean ≈ 43.3 (1 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCoordinate Geometry
Determine the ratio in which the point P(a, -2) divides the line segment joining the points A(-4, 3) and B(2, -4). Also, find the value of 'a'.
Try it first, then show our solutionSolution
1. Let P divide AB in ratio $k : 1$ . By section formula, y-coordinate of P: $\frac{k ( - 4 ) + 1 ( 3 )}{k + 1} = - 2 \Rightarrow - 4 k + 3 = - 2 k - 2 \Rightarrow 2 k = 5 \Rightarrow k = \frac{5}{2}$
  1½ marks
2. So the ratio is $A P : P B = \frac{5}{2} : 1 = 5 : 2$ .
  ½ mark
3. x-coordinate: $a = \frac{5 ( 2 ) + 2 ( - 4 )}{5 + 2} = \frac{10 - 8}{7} = \frac{2}{7}$
  1 mark
Answer: Ratio 5 : 2; $a = \frac{2}{7}$
Where the marks are
- Uses section formula on y-coordinate to get $k = \frac{5}{2}$ (1½ marks)
- Ratio 5 : 2 (½ mark)
- Finds $a = \frac{2}{7}$ (1 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCoordinate Geometry
In the given figure, in $△ A B C$ points D and E are mid-points of sides BC and AC respectively. If given vertices are A(4, -2), B(2, -2) and C(-6, -7), then verify the result $D E = \frac{1}{2} A B$ .
Figure in the paper: Triangle ABC with B at upper left, A at upper right (AB horizontal on top), and C at lower left. D is the midpoint of BC and E is the midpoint of AC; the segment DE is drawn horizontally joining D and E. See it in the official paper.
Try it first, then show our solutionSolution
1. D is the mid-point of BC: $D = (\frac{2 + ( - 6 )}{2}, \frac{- 2 + ( - 7 )}{2}) = (- 2, - \frac{9}{2})$
  1 mark
2. E is the mid-point of AC: $E = (\frac{4 + ( - 6 )}{2}, \frac{- 2 + ( - 7 )}{2}) = (- 1, - \frac{9}{2})$
  1 mark
3. $D E = (- 1 + 2)^{2} + (- \frac{9}{2} + \frac{9}{2})^{2} = 1^{2} + 0^{2} = 1$ and $A B = (4 - 2)^{2} + (- 2 + 2)^{2} = 2^{2} + 0^{2} = 2$
  1 mark
4. Hence $D E = 1 = \frac{1}{2} \times 2 = \frac{1}{2} A B$ . Verified.
Answer: DE = 1, AB = 2, so DE = ½ AB (verified)
Where the marks are
- Finds D $= (- 2, - \frac{9}{2})$ (1 mark)
- Finds E $= (- 1, - \frac{9}{2})$ (1 mark)
- DE = 1 and AB = 2 giving DE = ½AB (½ + ½ marks)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $(cosec A - sin A) (sec A - cos A) = \frac{1}{t a n A + c o t A}$
Try it first, then show our solutionSolution
1. LHS $= (\frac{1}{sin A} - sin A) (\frac{1}{cos A} - cos A)$
  ½ mark
2. $= \frac{1 - sin ^{2} A}{sin A} \times \frac{1 - cos ^{2} A}{cos A} = \frac{cos ^{2} A sin ^{2} A}{sin A cos A} = sin A cos A$
  1 mark
3. $= \frac{sin A cos A}{sin ^{2} A + cos ^{2} A}$ [since $sin^{2} A + cos^{2} A = 1$ ]
  ½ mark
4. Dividing numerator and denominator by $sin A cos A$ : $= \frac{1}{\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}} = \frac{1}{tan A + cot A} =$ RHS
  1 mark
Answer: LHS = RHS, proved.
Where the marks are
- Writes cosec and sec in terms of sin and cos (½ mark)
- Simplifies LHS to $sin A cos A$ (1 mark)
- Writes as $\frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A}$ (½ mark)
- Reaches $\frac{1}{t a n A + c o t A}$ (1 mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCircles
ABC is an isosceles triangle with AB = AC, circumscribed about a circle. Prove that BC is bisected at E.
Figure in the paper: Isosceles triangle ABC with apex A at top and base BC horizontal. A circle is inscribed touching the sides at D (on AB), E (on BC) and F (on AC). E is the midpoint of BC where the circle touches the base. See it in the official paper.
Try it first, then show our solutionSolution
1. The circle touches AB, BC, CA at D, E, F. Since tangents from an external point are equal: $A D = A F$ , $B D = B E$ , $C E = C F$ .
  1 mark
2. Given $A B = A C \Rightarrow A D + D B = A F + F C$
  ½ mark
3. Using $A D = A F$ : $A F + D B = A F + F C$
  ½ mark
4. $\Rightarrow D B = F C$
  ½ mark
5. $\Rightarrow B E = C E$ [as $B D = B E$ and $C F = C E$ ]. Hence E is the mid-point of BC, i.e. BC is bisected at E.
  ½ mark
Answer: BE = EC, so BC is bisected at E.
Where the marks are
- States equal tangent lengths AD = AF, BD = BE, CE = CF (1 mark)
- Writes AB = AC as AD + DB = AF + FC (½ mark)
- Substitutes AD = AF (½ mark)
- Gets DB = FC (½ mark)
- Concludes BE = EC (½ mark)
Open Q31 on its own page

Class	0–15	15–30	30–45	45–60	60–75	75–90
Frequency	12	15	11	20	16	6

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
0–15	7.5	12	−2	−24
15–30	22.5	15	−1	−15
30–45	37.5	11	0	0
45–60	52.5	20	1	20
60–75	67.5	16	2	32
75–90	82.5	6	3	18
Total		80		31

Section D

Long answer; internal choice in Q33 and Q35

Q32 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A person walking 48 m towards a tower in a horizontal line through its base observes that angle of elevation of the top of the tower changes from $4 5^{\circ}$ to $6 0^{\circ}$ . Find the height of the tower and distance of the person, now, from the tower. (Use $3 = 1.732$ )
Try it first, then show our solutionSolution
1. Figure: Let AB = $h$ m be the tower (A at the foot, B the top). D is the first position of the person and C the new position, with D, C, A on a horizontal line, DC = 48 m and CA = $x$ m. $∠ B D A = 4 5^{\circ}$ , $∠ B C A = 6 0^{\circ}$ , $∠ B A D = 9 0^{\circ}$ .
  1 mark
2. In right $△ B A D$ : $tan 4 5^{\circ} = \frac{A B}{A D} = \frac{h}{x + 48} \Rightarrow h = x + 48$ …(i)
  1 mark
3. In right $△ B A C$ : $tan 6 0^{\circ} = \frac{A B}{A C} = \frac{h}{x} \Rightarrow h = 3 x$ …(ii)
  1½ marks
4. From (i) and (ii): $3 x = x + 48 \Rightarrow x = \frac{48}{3 - 1} = \frac{48 ( 3 + 1 )}{2} = 24 (3 + 1) = 24 \times 2.732 = 65.57$ m
  1 mark
5. $h = x + 48 = 65.57 + 48 = 113.57$ m
  ½ mark
Answer: Height of tower ≈ 113.57 m; present distance from tower ≈ 65.57 m
Where the marks are
- Correct labelled figure (1 mark)
- $tan 4 5^{\circ}$ gives $h = x + 48$ (1 mark)
- $tan 6 0^{\circ}$ gives $h = 3 x$ (1½ marks)
- Solves $x = 24 (3 + 1) = 65.57$ m (1 mark)
- $h = 113.57$ m (½ mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerAreas Related to Circles
In the given figure, AB is a chord of a circle of radius 7 cm and centred at O. Find the area of the shaded region if $∠ A O B = 9 0^{\circ}$ . Also, find length of minor arc AB.
Figure in the paper: A circle with centre O. A chord AB is drawn, with OA and OB joined to form $∠ A O B = 9 0^{\circ}$ at the centre. The region between chord AB and the minor arc AB below it is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. Area of sector AOB $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{90}{360} \times \frac{22}{7} \times 7 \times 7$
  1 mark
2. $= \frac{77}{2}$ cm² $= 38.5$ cm²
  ½ mark
3. Area of right $△ A O B = \frac{1}{2} \times O A \times O B = \frac{1}{2} \times 7 \times 7 = \frac{49}{2}$ cm²
  1 mark
4. Area of shaded segment = sector − triangle $= \frac{77}{2} - \frac{49}{2} = \frac{28}{2} = 14$ cm²
  1 mark
5. Length of minor arc AB $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 = 11$ cm
  1½ marks
Answer: Shaded area = 14 cm²; minor arc AB = 11 cm
Where the marks are
- Sector area formula with substitution (1 mark)
- Sector area = 77/2 cm² (½ mark)
- Area of triangle AOB = 49/2 cm² (1 mark)
- Shaded area = 14 cm² (1 mark)
- Arc length = 11 cm (1½ marks)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerAreas Related to Circles
AB and CD are arcs of two concentric circles of radii 3.5 cm and 10.5 cm respectively and centred at O. Find the area of the shaded region if $∠ A O B = 6 0^{\circ}$ . Also, find the length of arc CD.
Figure in the paper: Two concentric circles centred at O. Points A and B lie on the inner arc and C and D lie on the outer arc; OA, OB, OC and OD are joined. $∠ A O B = 6 0^{\circ}$ is marked at O. The region between the two arcs bounded by OC, OD is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. OA = 3.5 cm, OC = 10.5 cm, $θ = 6 0^{\circ}$ . Shaded area = sector OCD − sector OAB $= \frac{60}{360} \times π (10. 5^{2} - 3. 5^{2})$
  2 marks
2. $10. 5^{2} - 3. 5^{2} = (10.5 + 3.5) (10.5 - 3.5) = 14 \times 7 = 98$ , so area $= \frac{22}{7} \times \frac{1}{6} \times 98$
  1 mark
3. $= \frac{154}{3}$ cm² $\approx 51.33$ cm²
  ½ mark
4. Length of arc CD $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 10.5$
  1 mark
5. $= 11$ cm
  ½ mark
Answer: Shaded area = 154/3 cm² ≈ 51.33 cm²; arc CD = 11 cm
Where the marks are
- Expresses shaded area as difference of two sectors $\frac{60}{360} π (R^{2} - r^{2})$ (2 marks)
- Correct simplification with 98 (1 mark)
- Area = 154/3 cm² (½ mark)
- Arc length formula with substitution (1 mark)
- Arc CD = 11 cm (½ mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerTriangles
If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then prove that the other two sides are divided in the same ratio.
Try it first, then show our solutionSolution
1. Given: In $△ A B C$ , a line parallel to $B C$ meets $A B$ at $D$ and $A C$ at $E$ , so $D E ∥ B C$ .
  To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
  1 mark
2. Figure and construction: Draw $△ A B C$ with $D E ∥ B C$ , where $D$ is on $A B$ and $E$ is on $A C$ . Join $B E$ and $C D$ . Draw $E N ⊥ A B$ (N on AB) and $D M ⊥ A C$ (M on AC).
  1 mark
3. Proof: Using area $= \frac{1}{2} \times$ base $\times$ height:
  $ar (A D E) = \frac{1}{2} \cdot A D \cdot E N$ and $ar (B D E) = \frac{1}{2} \cdot D B \cdot E N$
  So $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ ...(1)
  Similarly:
  $ar (A D E) = \frac{1}{2} \cdot A E \cdot D M$ and $ar (D E C) = \frac{1}{2} \cdot E C \cdot D M$
  So $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ...(2)
  1 mark
4. $△ B D E$ and $△ D E C$ stand on the same base $D E$ and lie between the same parallels $D E$ and $B C$ .
  So $ar (B D E) = ar (D E C)$ ...(3)
  1 mark
5. From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ .
  Hence the other two sides are divided in the same ratio. Hence proved.
  1 mark
Answer: $\frac{A D}{D B} = \frac{A E}{E C}$ (Basic Proportionality Theorem proved)
Where the marks are
- Correct Given and To Prove (1 mark)
- Correct figure and construction: join BE and CD, draw perpendiculars (1 mark)
- Area ratios $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ and $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ (1 mark)
- $a r (B D E) = a r (D E C)$ : same base, between same parallels (1 mark)
- Concludes $\frac{A D}{D B} = \frac{A E}{E C}$ (1 mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerQuadratic Equations
The difference of two numbers is 5 and the difference of their reciprocals is $\frac{1}{10}$ . Find the numbers.
Try it first, then show our solutionSolution
1. Let the smaller number be $x$ ; then the larger is $x + 5$ .
  ½ mark
2. Since $\frac{1}{x + 5} < \frac{1}{x}$ (for positive numbers): $\frac{1}{x} - \frac{1}{x + 5} = \frac{1}{10}$
  1 mark
3. $\frac{x + 5 - x}{x ( x + 5 )} = \frac{1}{10} \Rightarrow 50 = x^{2} + 5 x \Rightarrow x^{2} + 5 x - 50 = 0$
  1 mark
4. Factorising: $x^{2} + 10 x - 5 x - 50 = 0 \Rightarrow (x + 10) (x - 5) = 0$
  1 mark
5. $x = - 10$ or $x = 5$
  ½ mark
6. If $x = 5$ , the numbers are 5 and 10; if $x = - 10$ , the numbers are −10 and −5. (Check: $\frac{1}{5} - \frac{1}{10} = \frac{1}{10}$ ; $- \frac{1}{10} + \frac{1}{5} = \frac{1}{10}$ .)
  1 mark
Answer: The numbers are 5 and 10, or −10 and −5
Where the marks are
- Takes numbers as x and x + 5 (½ mark)
- Forms the reciprocal equation $\frac{1}{x} - \frac{1}{x + 5} = \frac{1}{10}$ (1 mark)
- Reduces to $x^{2} + 5 x - 50 = 0$ (1 mark)
- Factorises as $(x + 10) (x - 5) = 0$ (1 mark)
- $x = - 10, 5$ (½ mark)
- States both pairs of numbers (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerQuadratic Equations
Find all the values of k for which the quadratic equation $2 x^{2} + k x + 8 = 0$ has equal roots. Also, find the roots.
Try it first, then show our solutionSolution
1. Compare $2 x^{2} + k x + 8 = 0$ with $a x^{2} + b x + c = 0$ : $a = 2, b = k, c = 8$ . For equal roots the discriminant must be zero: $D = b^{2} - 4 a c = k^{2} - 4 (2) (8) = k^{2} - 64 = 0$ .
  1 mark
2. $k^{2} = 64 \Rightarrow k = 8$ or $k = - 8$ , i.e. $k = \pm 8$ .
  1 mark
3. Substituting these values gives the equations $2 x^{2} + 8 x + 8 = 0$ (for $k = 8$ ) and $2 x^{2} - 8 x + 8 = 0$ (for $k = - 8$ ).
  1 mark
4. For $k = 8$ : $2 (x^{2} + 4 x + 4) = 0 \Rightarrow 2 (x + 2)^{2} = 0 \Rightarrow x = - 2, - 2$ .
  1 mark
5. For $k = - 8$ : $2 (x^{2} - 4 x + 4) = 0 \Rightarrow 2 (x - 2)^{2} = 0 \Rightarrow x = 2, 2$ .
  1 mark
Answer: k = ±8; roots are −2, −2 (for k = 8) and 2, 2 (for k = −8)
Where the marks are
- Writes the equal-roots condition $k^{2} - 64 = 0$ (1 mark)
- Obtains $k = \pm 8$ (1 mark)
- Writes the two resulting equations $2 x^{2} \pm 8 x + 8 = 0$ (1 mark)
- Finds root $x = - 2$ for $k = 8$ (1 mark)
- Finds root $x = 2$ for $k = - 8$ (1 mark)
Open Q35(b) on its own page

Section E

Case study based; internal choice in 3 case-study 2-mark subparts

Q36 · 4 marks · Case-basedCompetency-basedPolynomials
Rainbow is an arch of colours that is visible in the sky after rain or when water droplets are present in the atmosphere. The colours of the rainbow are generally, red, orange, yellow, green, blue, indigo and violet. Each colour of the rainbow makes a parabola. We know that any quadratic polynomial $p (x) = a x^{2} + b x + c$ ( $a \neq = 0$ ) represents a parabola on the graph paper.
Based on the above, answer the following questions:
1. (i)
  The graph of a rainbow $y = f (x)$ is shown in the figure. Write the number of zeroes of the curve.
  1 mark
2. (ii)
  If the graph of a rainbow does not intersect the x-axis but intersects y-axis at one point, then how many zeroes will it have?
  1 mark
3. (iii)(a)
  If a rainbow is represented by the quadratic polynomial $p (x) = x^{2} + (a + 1) x + b$ , whose zeroes are 2 and -3, find the value of a and b.
  2 marks
4. OR
  (iii)(b)
  The polynomial $x^{2} - 2 x - (7 p + 3)$ represents a rainbow. If -4 is a zero of it, find the value of p.
  2 marks
Figure in the paper: Graph of a downward-opening parabola y = f(x) with vertex on the positive y-axis. The curve crosses the x-axis at two points, one on each side of the origin O, and the arms extend downward beyond the x-axis. The x-axis and y-axis are labelled. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The parabola cuts the x-axis at two points (one on each side of O), so the curve has 2 zeroes.
  1 mark
2. (ii) Zeroes are the x-coordinates where the graph meets the x-axis. If the graph does not meet the x-axis, it has no zero (0 zeroes).
  1 mark
3. (iii)(a) Since 2 is a zero: $p (2) = 4 + 2 (a + 1) + b = 0 \Rightarrow 2 a + b = - 6$ …(1). Since $- 3$ is a zero: $p (- 3) = 9 - 3 (a + 1) + b = 0 \Rightarrow - 3 a + b = - 6$ …(2).
  1 mark
4. (iii)(a) Subtracting (2) from (1): $5 a = 0 \Rightarrow a = 0$ . Then from (1), $b = - 6$ . (Check: sum of zeroes $= - (a + 1) = - 1 = 2 + (- 3)$ , product $= b = - 6 = 2 \times (- 3)$ .)
  1 mark
5. (iii)(b) Since $- 4$ is a zero: $(- 4)^{2} - 2 (- 4) - (7 p + 3) = 0 \Rightarrow 16 + 8 - 7 p - 3 = 0 \Rightarrow 21 - 7 p = 0$ .
  1 mark
6. (iii)(b) $7 p = 21 \Rightarrow p = 3$ .
  1 mark
Answer: (i) 2 zeroes; (ii) 0 (no zero); (iii)(a) a = 0, b = −6; (iii)(b) p = 3
Where the marks are
- (i) States 2 zeroes (1 mark)
- (ii) States no zero (1 mark)
- (iii)(a) Forms $2 a + b = - 6$ and $- 3 a + b = - 6$ (½ + ½ marks)
- (iii)(a) Solves to get $a = 0$ and $b = - 6$ (½ + ½ marks)
- (iii)(b) Substitutes $x = - 4$ to get $21 - 7 p = 0$ (1 mark)
- (iii)(b) Obtains $p = 3$ (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Singing bowls (hemispherical in shape) are commonly used in sound healing practices. Mallet (cylindrical in shape) is used to strike the bowl in a sequence to produce sound and vibration.
One such bowl is shown here whose dimensions are: Hemispherical bowl has outer radius 6 cm and inner radius 5 cm. Mallet has height of 10 cm and radius 2 cm.
Based on the above, answer the following questions:
1. (i)
  What is the volume of the material used in making the mallet?
  1 mark
2. (ii)
  The bowl is to be polished from inside. Find the inner surface area of the bowl.
  1 mark
3. (iii)(a)
  Find the volume of metal used to make the bowl.
  2 marks
4. OR
  (iii)(b)
  Find total surface area of the mallet. (Use $π = 3.14$ )
  2 marks
Figure in the paper: Photograph of a hemispherical singing bowl resting on a wooden surface, with a cylindrical mallet lying across the bowl, with its handle end and striking end visible. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Mallet is a cylinder with $r = 2$ cm, $h = 10$ cm. Volume $= π r^{2} h = 3.14 \times 2 \times 2 \times 10 = 125.6 cm^{3}$ .
  1 mark
2. (ii) Inner curved surface of a hemisphere with $r = 5$ cm: $2 π r^{2} = 2 \times 3.14 \times 25 = 157 cm^{2}$ .
  1 mark
3. (iii)(a) Volume of metal $= \frac{2}{3} π (R^{3} - r^{3}) = \frac{2}{3} \times 3.14 \times (6^{3} - 5^{3}) = \frac{2}{3} \times 3.14 \times (216 - 125) = \frac{2}{3} \times 3.14 \times 91$ .
  1 mark
4. (iii)(a) $= \frac{571.48}{3} \approx 190.5 cm^{3}$ .
  1 mark
5. (iii)(b) TSA of cylinder $= 2 π r (r + h) = 2 \times 3.14 \times 2 \times (2 + 10)$ .
  1 mark
6. (iii)(b) $= 12.56 \times 12 = 150.72 \approx 150.7 cm^{2}$ .
  1 mark
Answer: (i) 125.6 cm³; (ii) 157 cm²; (iii)(a) ≈ 190.5 cm³; (iii)(b) ≈ 150.7 cm²
Where the marks are
- (i) Uses $π r^{2} h$ to get 125.6 cm³ (1 mark)
- (ii) Uses $2 π r^{2}$ with r = 5 to get 157 cm² (1 mark)
- (iii)(a) Writes $\frac{2}{3} π (6^{3} - 5^{3})$ (1 mark)
- (iii)(a) Gets 190.5 cm³ (1 mark)
- (iii)(b) Writes $2 π r (r + h)$ with values (1 mark)
- (iii)(b) Gets 150.7 cm² (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedProbability
Some students were asked to list their favourite colour. The measure of each colour is shown by the central angle of a pie chart given below.
Study the pie chart and answer the following questions:
1. (i)
  If a student is chosen at random, then find the probability of his/her favourite colour being white?
  1 mark
2. (ii)
  What is the probability of his/her favourite colour being blue or green?
  1 mark
3. (iii)(a)
  If 15 students liked the colour yellow, how many students participated in the survey?
  2 marks
4. OR
  (iii)(b)
  What is the probability of the favourite colour being red or blue?
  2 marks
Figure in the paper: A pie chart divided into five sectors with central angles: Blue $6 0^{\circ}$ , Red $3 0^{\circ}$ , Yellow $9 0^{\circ}$ , White $12 0^{\circ}$ , Green $6 0^{\circ}$ . Blue and Green are at the upper part, Red at the upper right, Yellow at the lower right, and White at the lower left. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The whole circle (360°) represents all students. P(white) $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} = \frac{1}{3}$ .
  1 mark
2. (ii) P(blue or green) $= \frac{6 0 ^{\circ} + 6 0 ^{\circ}}{36 0 ^{\circ}} = \frac{120}{360} = \frac{1}{3}$ .
  1 mark
3. (iii)(a) Let the total number of students be $x$ . Yellow's share $= \frac{90}{360} = \frac{1}{4}$ , so $\frac{15}{x} = \frac{1}{4}$ .
  1½ marks
4. (iii)(a) $x = 60$ , so 60 students took part in the survey.
  ½ mark
5. (iii)(b) P(red or blue) $= \frac{3 0 ^{\circ} + 6 0 ^{\circ}}{36 0 ^{\circ}}$
  1 mark
6. (iii)(b) $= \frac{90}{360} = \frac{1}{4}$ .
  1 mark
Answer: (i) 1/3; (ii) 1/3; (iii)(a) 60 students; (iii)(b) 1/4
Where the marks are
- (i) P(white) = 120/360 = 1/3 (1 mark)
- (ii) P(blue or green) = 120/360 = 1/3 (1 mark)
- (iii)(a) Sets up $\frac{15}{x} = \frac{1}{4}$ (1½ marks)
- (iii)(a) Gets 60 students (½ mark)
- (iii)(b) Writes (30+60)/360 (1 mark)
- (iii)(b) Simplifies to 1/4 (1 mark)
Open Q38 on its own page

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