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

CBSE Class 10 Mathematics Standard: 2023 board paper, set 30/5/1

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_STANDARD.zip (a zip; this paper is the file 30_5_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
C	26-31	Short answer
D	32-35	Long answer
E	36-38	Case study based

Instructions in brief: All questions compulsory; no overall choice; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (in 2-mark case-study subparts); use π = 22/7 if not stated; calculators not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPolynomials
The number of polynomials having zeroes $- 3$ and $5$ is :
1. (A)only one
2. (B)infinite
3. (C)exactly two
4. (D)at most two
Try it first, then show our solutionSolution
1. Any polynomial $k (x + 3) (x - 5)$ , $k \neq = 0$ , has zeroes $- 3$ and $5$ .
2. Since $k$ can be any non-zero real number, there are infinitely many such polynomials. Option (B).
  1 mark
Answer: Infinite, option (B)
Where the marks are
- Recognises that $k (x^{2} - 2 x - 15)$ works for every $k \neq = 0$ (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of equations $a x + 2 y = 9$ and $3 x + b y = 18$ represent parallel lines, where $a, b$ are integers, if :
1. (A) $a = b$
2. (B) $3 a = 2 b$
3. (C) $2 a = 3 b$
4. (D) $ab = 6$
Try it first, then show our solutionSolution
1. For parallel lines: $\frac{a}{3} = \frac{2}{b} \neq = \frac{9}{18}$ .
2. $\frac{a}{3} = \frac{2}{b} \Rightarrow ab = 6$ . Option (D).
  1 mark
Answer: ab = 6, option (D)
Where the marks are
- Uses $a_{1} / a_{2} = b_{1} / b_{2}$ to get $ab = 6$ (1 mark)
Q3 · 1 mark · MCQArithmetic Progressions
The common difference of the A.P. whose $n^{t h}$ term is given by $a_{n} = 3 n + 7$ , is :
1. (A)7
2. (B)3
3. (C) $3 n$
4. (D)1
Try it first, then show our solutionSolution
1. $a_{1} = 10$ , $a_{2} = 13$ .
2. $d = a_{2} - a_{1} = 3$ . Option (B).
  1 mark
Answer: 3, option (B)
Where the marks are
- Finds $d = a_{n + 1} - a_{n} = 3$ (1 mark)
Q4 · 1 mark · MCQTriangles
In the given figure, $D E ∥ B C$ . The value of x is :
1. (A)6
2. (B)12·5
3. (C)8
4. (D)10
Figure in the paper: Triangle ABC with D on side AB and E on side AC, DE parallel to BC. AD = 2 cm, DB = 3 cm, DE = 4 cm, BC = x cm. See it in the official paper.
Try it first, then show our solutionSolution
1. $D E ∥ B C \Rightarrow △ A D E \sim △ A B C$ (AA), so $\frac{A D}{A B} = \frac{D E}{B C}$ .
2. $\frac{2}{5} = \frac{4}{x} \Rightarrow x = 10$ . Option (D).
  1 mark
Answer: x = 10, option (D)
Where the marks are
- Uses similarity with $A B = 5$ cm to get $x = 10$ (1 mark)
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Q5 · 1 mark · MCQQuadratic Equations
A quadratic equation whose roots are $(2 + 3)$ and $(2 - 3)$ is :
1. (A) $x^{2} - 4 x + 1 = 0$
2. (B) $x^{2} + 4 x + 1 = 0$
3. (C) $4 x^{2} - 3 = 0$
4. (D) $x^{2} - 1 = 0$
Try it first, then show our solutionSolution
1. Sum of roots $= 4$ , product $= 4 - 3 = 1$ .
2. Equation: $x^{2} - 4 x + 1 = 0$ . Option (A).
  1 mark
Answer: $x^{2} - 4 x + 1 = 0$ , option (A)
Where the marks are
- Uses $x^{2} - (sum) x + product = 0$ (1 mark)
Q6 · 1 mark · MCQIntroduction to Trigonometry
If $tan θ = \frac{5}{12}$ , then the value of $\frac{s i n θ + c o s θ}{s i n θ - c o s θ}$ is :
1. (A) $- \frac{17}{7}$
2. (B) $\frac{17}{7}$
3. (C) $\frac{17}{13}$
4. (D) $- \frac{7}{13}$
Try it first, then show our solutionSolution
1. Divide numerator and denominator by $cos θ$ : $\frac{t a n θ + 1}{t a n θ - 1}$ .
2. $= \frac{17/12}{- 7/12} = - \frac{17}{7}$ . Option (A).
  1 mark
Answer: $- \frac{17}{7}$ , option (A)
Where the marks are
- Converts to tan form and evaluates $- 17/7$ (1 mark)
Q7 · 1 mark · MCQCoordinate Geometry
The distance between the points $P (- \frac{11}{3}, 5)$ and $Q (- \frac{2}{3}, 5)$ is :
1. (A)6 units
2. (B)4 units
3. (C)2 units
4. (D)3 units
Try it first, then show our solutionSolution
1. Same y-coordinate, so distance $= - \frac{2}{3} + \frac{11}{3} = \frac{9}{3} = 3$ units. Option (D).
  1 mark
Answer: 3 units, option (D)
Where the marks are
- Uses distance formula to get 3 units (1 mark)
Q8 · 1 mark · MCQTriangles
In the given figure, $A B = B C = 10$ cm. If $A C = 7$ cm, then the length of BP is :
1. (A)3·5 cm
2. (B)7 cm
3. (C)6·5 cm
4. (D)5 cm
Figure in the paper: Isosceles triangle ABC with apex B, AB = BC = 10 cm, base AC = 7 cm. A circle touches sides AB, BC and AC at P, Q and R respectively; P lies on AB, Q on BC, R on AC. See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents from an external point are equal: $A P = A R$ , $C R = C Q$ , $B P = B Q$ . By symmetry $A R = C R = 3.5$ cm, so $A P = 3.5$ cm.
2. $B P = A B - A P = 10 - 3.5 = 6.5$ cm. Option (C).
  1 mark
Answer: 6·5 cm, option (C)
Where the marks are
- Uses equal tangents to get $A P = 3.5$ , $B P = 6.5$ cm (1 mark)
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Q9 · 1 mark · MCQCompetency-basedStatistics
Water in a river which is 3 m deep and 40 m wide is flowing at the rate of 2 km/h. How much water will fall into the sea in 2 minutes ?
1. (A) $800 m^{3}$
2. (B) $4000 m^{3}$
3. (C) $8000 m^{3}$
4. (D) $2000 m^{3}$
Try it first, then show our solutionSolution
1. Length flowing in 2 min $= 2000 \times \frac{2}{60} = \frac{200}{3}$ m.
2. Volume $= 3 \times 40 \times \frac{200}{3} = 8000$ m $^{3}$ . Option (C).
  1 mark
Answer: 8000 m³, option (C)
Where the marks are
- Computes flow length and volume 8000 m³ (1 mark)
Q10 · 1 mark · MCQStatistics
If the mean and the median of a data are 12 and 15 respectively, then its mode is :
1. (A)13·5
2. (B)21
3. (C)6
4. (D)14
Try it first, then show our solutionSolution
1. Mode $= 3 Median - 2 Mean = 45 - 24 = 21$ . Option (B).
  1 mark
Answer: 21, option (B)
Where the marks are
- Applies empirical relation to get 21 (1 mark)
Q11 · 1 mark · MCQCircles
In the given figure, AB is a tangent to the circle centred at O. If $O A = 6$ cm and $∠ O A B = 3 0^{\circ}$ , then the radius of the circle is :
1. (A)3 cm
2. (B) $33$ cm
3. (C)2 cm
4. (D) $3$ cm
Figure in the paper: Circle with centre O. A point A outside the circle, with segment OA = 6 cm. Line AB is tangent to the circle at B. Angle OAB = 30° marked at A. See it in the official paper.
Try it first, then show our solutionSolution
1. $O B ⊥ A B$ [tangent ⟂ radius], so in right $△ O B A$ : $sin 3 0^{\circ} = \frac{O B}{O A}$ .
2. $O B = 6 \times \frac{1}{2} = 3$ cm. Option (A).
  1 mark
Answer: 3 cm, option (A)
Where the marks are
- Uses right angle at B and $sin 3 0^{\circ}$ to get 3 cm (1 mark)
Q12 · 1 mark · MCQIntroduction to Trigonometry
$(\frac{2 t a n 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}})$ is equal to :
1. (A) $sin 6 0^{\circ}$
2. (B) $cos 6 0^{\circ}$
3. (C) $tan 6 0^{\circ}$
4. (D) $sin 3 0^{\circ}$
Try it first, then show our solutionSolution
1. $\frac{2/ 3}{1 + 1/3} = \frac{2}{3} \times \frac{3}{4} = \frac{3}{2}$ .
2. $\frac{3}{2} = sin 6 0^{\circ}$ . Option (A).
  1 mark
Answer: sin 60°, option (A)
Where the marks are
- Evaluates to $3 /2 = sin 6 0^{\circ}$ (1 mark)
Q13 · 1 mark · MCQTriangles
In $△ A B C$ and $△ D E F$ , $\frac{A B}{D E} = \frac{B C}{F D}$ . Which of the following makes the two triangles similar ?
1. (A) $∠ A = ∠ D$
2. (B) $∠ B = ∠ D$
3. (C) $∠ B = ∠ E$
4. (D) $∠ A = ∠ F$
Try it first, then show our solutionSolution
1. For SAS similarity the included angles must be equal: in $△ A B C$ , AB and BC include $∠ B$ ; in $△ D E F$ , DE and FD include $∠ D$ .
2. So $∠ B = ∠ D$ . Option (B).
  1 mark
Answer: ∠B = ∠D, option (B)
Where the marks are
- Identifies the included angles for SAS (1 mark)
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Q14 · 1 mark · MCQArithmetic Progressions
The $1 1^{t h}$ term from the end of the A.P. : $10, 7, 4, \dots\dots, - 62$ is :
1. (A)25
2. (B)16
3. (C) $- 32$
4. (D)0
Try it first, then show our solutionSolution
1. From the end the A.P. is $- 62, - 59, \dots$ with $d = 3$ .
2. $11$ th term $= - 62 + 10 \times 3 = - 32$ . Option (C).
  1 mark
Answer: −32, option (C)
Where the marks are
- Reverses A.P. and gets −32 (1 mark)
Q15 · 1 mark · MCQProbability
Two coins are tossed together. The probability of getting at least one tail is :
1. (A) $\frac{1}{4}$
2. (B) $\frac{1}{2}$
3. (C) $\frac{3}{4}$
4. (D)1
Try it first, then show our solutionSolution
1. Outcomes: HH, HT, TH, TT; at least one tail: HT, TH, TT.
2. $P = \frac{3}{4}$ . Option (C).
  1 mark
Answer: 3/4, option (C)
Where the marks are
- Lists outcomes, gets 3/4 (1 mark)
Q16 · 1 mark · MCQCircles
In the given figure, AC and AB are tangents to a circle centred at O. If $∠ C O D = 12 0^{\circ}$ , then $∠ B A O$ is equal to :
1. (A) $3 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $9 0^{\circ}$
Figure in the paper: A circle with centre O and diameter AD passing through the line from external point A. AB and AC are tangents touching the circle at B and C. Angle COD = 120° marked at O. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ A O C = 18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ (linear pair), $∠ O C A = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ C A O = 3 0^{\circ}$ .
2. $∠ B A O = ∠ C A O = 3 0^{\circ}$ (OA bisects angle between tangents). Option (A).
  1 mark
Answer: 30°, option (A)
Where the marks are
- Finds ∠AOC = 60° and ∠BAO = 30° (1 mark)
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Q17 · 1 mark · MCQProbability
Which of the following numbers cannot be the probability of happening of an event ?
1. (A)0
2. (B) $\frac{7}{0.01}$
3. (C)0·07
4. (D) $\frac{0.07}{3}$
Try it first, then show our solutionSolution
1. Probability lies in $[0, 1]$ . $\frac{7}{0.01} = 700 > 1$ , so it cannot be a probability. Option (B).
  1 mark
Answer: 7/0·01, option (B)
Where the marks are
- Uses $0 \leq P \leq 1$ (1 mark)
Q18 · 1 mark · MCQStatistics
If every term of the statistical data consisting of n terms is decreased by 2, then the mean of the data :
1. (A)decreases by 2
2. (B)remains unchanged
3. (C)decreases by $2 n$
4. (D)decreases by 1
Try it first, then show our solutionSolution
1. New mean $= \frac{\sum ( x _{i} - 2 )}{n} = \overset{x}{ˉ} - \frac{2 n}{n} = \overset{x}{ˉ} - 2$ . Option (A).
  1 mark
Answer: decreases by 2, option (A)
Where the marks are
- Shows new mean $= \overset{x}{ˉ} - 2$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedCoordinate Geometry
Assertion (A): If the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3), then the value of x is 2.
Reason (R): Centre of a circle is the mid-point of each chord of the circle.
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. $O A^{2} = O B^{2}$ : $4 = (x - 2)^{2} + 4 \Rightarrow x = 2$ . A is true.
2. The centre is the midpoint only of diameters, not of every chord, so R is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Verifies A using equal radii and rejects R (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A): The number $5^{n}$ cannot end with the digit 0, where n is a natural number.
Reason (R): Prime factorisation of 5 has only two factors, 1 and 5.
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. To end in 0, a number needs prime factors 2 and 5; $5^{n}$ has no factor 2, so A is true.
2. 1 is not a prime factor; the prime factorisation of 5 is just 5, so R is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Justifies A by absence of factor 2 and identifies R as false (1 mark)
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Section B

Very short answer

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerCoordinate Geometry
The line segment joining the points A(4, $- 5$ ) and B(4, 5) is divided by the point P such that AP : AB = 2 : 5. Find the coordinates of P.
Try it first, then show our solutionSolution
1. Since $A P : A B = 2 : 5$ , we get $P B = A B - A P$ , so $A P : P B = 2 : 3$ . Hence P divides AB internally in the ratio $2 : 3$ .
  ½ mark
2. By the section formula with $m_{1} = 2, m_{2} = 3$ , A(4, −5), B(4, 5):
  $x = \frac{2 ( 4 ) + 3 ( 4 )}{2 + 3} = \frac{8 + 12}{5} = 4$ , and $y = \frac{2 ( 5 ) + 3 ( - 5 )}{5} = \frac{10 - 15}{5} = - 1$ .
  1 mark
3. Therefore the coordinates of P are $(4, - 1)$ .
  ½ mark
Answer: P(4, −1)
Where the marks are
- Converts AP : AB = 2 : 5 into AP : PB = 2 : 3 (½ mark)
- Applies the section formula correctly to get x = 4 and y = −1 (1 mark)
- States P = (4, −1) (½ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerCoordinate Geometry
Point P(x, y) is equidistant from points A(5, 1) and B(1, 5). Prove that $x = y$ .
Try it first, then show our solutionSolution
1. P is equidistant from A and B, so $P A = P B$ , i.e. $P A^{2} = P B^{2}$ . By the distance formula:
  $(x - 5)^{2} + (y - 1)^{2} = (x - 1)^{2} + (y - 5)^{2}$ .
  1 mark
2. Expanding: $x^{2} - 10 x + 25 + y^{2} - 2 y + 1 = x^{2} - 2 x + 1 + y^{2} - 10 y + 25$
  $\Rightarrow - 10 x - 2 y = - 2 x - 10 y \Rightarrow - 8 x = - 8 y \Rightarrow x = y$ . Hence proved.
  1 mark
Answer: PA² = PB² simplifies to x = y (proved).
Where the marks are
- Writes PA² = PB² using the distance formula (1 mark)
- Simplifies correctly to x = y (1 mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerCircles
In the given figure, PT is a tangent to the circle centred at O. OC is perpendicular to chord AB. Prove that $P A \cdot P B = P C^{2} - A C^{2}$ .
Figure in the paper: Circle with centre O. P is outside the circle; PT is a tangent touching the circle at T. Line from P cuts the circle at A and B (B further along the chord). OC is drawn from O perpendicular to chord AB, with C on AB between A and B. OP is also drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. From the figure, A lies between P and C, and C lies between A and B. So $P A = P C - A C$ and $P B = P C + C B$ .
  Hence $P A \cdot P B = (P C - A C) (P C + C B)$ .
  1 mark
2. $O C ⊥ A B$ , and the perpendicular from the centre to a chord bisects the chord, so $A C = C B$ .
  $∴ P A \cdot P B = (P C - A C) (P C + A C)$ .
  ½ mark
3. $= P C^{2} - A C^{2}$ [using $(a - b) (a + b) = a^{2} - b^{2}$ ]. Hence proved.
  ½ mark
Answer: PA · PB = (PC − AC)(PC + AC) = PC² − AC² (proved).
Where the marks are
- Writes PA = PC − AC and PB = PC + CB, so PA·PB = (PC − AC)(PC + CB) (1 mark)
- Uses AC = CB since the perpendicular from the centre bisects the chord (½ mark)
- Simplifies to PC² − AC² (½ mark)
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Q23 · 2 marks · Very short answerReal Numbers
Using prime factorisation, find HCF and LCM of 96 and 120.
Try it first, then show our solutionSolution
1. $96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^{5} \times 3$
  ½ mark
2. $120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^{3} \times 3 \times 5$
  ½ mark
3. HCF = product of the smallest powers of the common prime factors $= 2^{3} \times 3 = 24$
  ½ mark
4. LCM = product of the greatest powers of all the prime factors $= 2^{5} \times 3 \times 5 = 480$
  ½ mark
Answer: HCF = 24, LCM = 480
Where the marks are
- Prime factorisation 96 = 2⁵ × 3 (½ mark)
- Prime factorisation 120 = 2³ × 3 × 5 (½ mark)
- HCF = 24 (½ mark)
- LCM = 480 (½ mark)
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Q24 · 2 marks · Very short answerCoordinate Geometry
Find the ratio in which y-axis divides the line segment joining the points $(5, - 6)$ and $(- 1, - 4)$ .
Try it first, then show our solutionSolution
1. Let the y-axis divide the segment joining A(5, −6) and B(−1, −4) in the ratio $k : 1$ at the point P. Every point on the y-axis has x-coordinate 0, so P = $(0, y)$ .
  ½ mark
2. By the section formula, x-coordinate of P: $0 = \frac{k ( - 1 ) + 1 ( 5 )}{k + 1} \Rightarrow - k + 5 = 0 \Rightarrow k = 5$ .
  1 mark
3. Hence the y-axis divides the segment in the ratio $5 : 1$ (internally).
  ½ mark
Answer: 5 : 1
Where the marks are
- Takes ratio k : 1 and point of division (0, y) on the y-axis (½ mark)
- Equates the x-coordinate from the section formula to 0 and gets k = 5 (1 mark)
- States the ratio 5 : 1 (½ mark)
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Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerIntroduction to Trigonometry
If $a cos θ + b sin θ = m$ and $a sin θ - b cos θ = n$ , then prove that $a^{2} + b^{2} = m^{2} + n^{2}$ .
Try it first, then show our solutionSolution
1. $m^{2} + n^{2} = (a cos θ + b sin θ)^{2} + (a sin θ - b cos θ)^{2}$
  ½ mark
2. $= a^{2} cos^{2} θ + b^{2} sin^{2} θ + 2 ab sin θ cos θ + a^{2} sin^{2} θ + b^{2} cos^{2} θ - 2 ab sin θ cos θ$
  $= a^{2} (cos^{2} θ + sin^{2} θ) + b^{2} (sin^{2} θ + cos^{2} θ)$
  1 mark
3. $= a^{2} + b^{2}$ [since $sin^{2} θ + cos^{2} θ = 1$ ]. Hence $a^{2} + b^{2} = m^{2} + n^{2}$ .
  ½ mark
Answer: m² + n² = a² + b² (proved).
Where the marks are
- Writes m² + n² as the sum of the two squared expressions (½ mark)
- Expands, cancels the 2ab sinθ cosθ terms and groups as a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ) (1 mark)
- Uses sin²θ + cos²θ = 1 to get a² + b² (½ mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerIntroduction to Trigonometry
Prove that : $\frac{s e c A - 1}{s e c A + 1} + \frac{s e c A + 1}{s e c A - 1} = 2 cosec A$
Try it first, then show our solutionSolution
1. LHS $= \frac{sec A - 1}{sec A + 1} + \frac{sec A + 1}{sec A - 1}$ . Taking the LCM $(sec A + 1) (sec A - 1)$ :
  $= \frac{( sec A - 1 ) + ( sec A + 1 )}{sec ^{2} A - 1}$
  1 mark
2. $= \frac{2 sec A}{tan ^{2} A} = \frac{2 sec A}{tan A}$ [since $sec^{2} A - 1 = tan^{2} A$ ]
  ½ mark
3. $= \frac{2/ cos A}{sin A / cos A} = \frac{2}{sin A} = 2 cosec A =$ RHS. Hence proved.
  ½ mark
Answer: LHS = 2 sec A / tan A = 2 cosec A = RHS (proved).
Where the marks are
- Adds the two fractions to get (sec A − 1 + sec A + 1)/√(sec²A − 1) (1 mark)
- Uses sec²A − 1 = tan²A to get 2 sec A / tan A (½ mark)
- Simplifies to 2 cosec A (½ mark)
Open Q25(b) on its own page

Section C

Short answer

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3$ is rational.
  Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are coprime.
  Squaring: $3 q^{2} = p^{2}$ .
  ½ mark
2. So $p^{2}$ is divisible by 3. Since 3 is prime, 3 divides $p^{2}$ means 3 divides $p$ . ... (1)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2}$ , so $q^{2} = 3 a^{2}$ .
  So $q^{2}$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)
  1 mark
4. From (1) and (2), 3 divides both $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong. Hence $3$ is irrational.
  ½ mark
Answer: Assuming $3 = p / q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $3$ is irrational
Where the marks are
- Assumes $3 = \frac{p}{q}$ with p, q coprime and gets $3 q^{2} = p^{2}$ (1/2 mark)
- Shows 3 divides p (1 mark)
- Substitutes $p = 3 a$ and shows 3 divides q (1 mark)
- States the contradiction and concludes (1/2 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerCompetency-basedReal Numbers
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change together next ?
Try it first, then show our solutionSolution
1. The lights change together after a time that is the LCM of 48, 72 and 108 seconds.
  $48 = 2^{4} \times 3$ , $72 = 2^{3} \times 3^{2}$ , $108 = 2^{2} \times 3^{3}$
  1 mark
2. LCM $= 2^{4} \times 3^{3} = 16 \times 27 = 432$ seconds.
  1 mark
3. $432$ s $= \frac{432}{60}$ min $= 7$ min $12$ s. So the lights will next change together at 7 : 07 : 12 a.m. (7 minutes 12 seconds past 7 a.m.).
  1 mark
Answer: LCM = 432 s = 7 min 12 s; next simultaneous change at 7:07:12 a.m.
Where the marks are
- Prime factorises 48, 72 and 108 (1 mark)
- Finds LCM = 432 seconds (1 mark)
- Converts to 7 min 12 s and states the time 7:07:12 a.m. (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerArithmetic Progressions
If $p^{t h}$ term of an A.P. is q and $q^{t h}$ term is p, then prove that its $n^{t h}$ term is $(p + q - n)$ .
Try it first, then show our solutionSolution
1. Let the first term be $a$ and common difference $d$ . Given $a_{p} = q$ : $a + (p - 1) d = q$ ……(i)
  ½ mark
2. Given $a_{q} = p$ : $a + (q - 1) d = p$ ……(ii)
  ½ mark
3. Subtracting (ii) from (i): $(p - q) d = q - p = - (p - q)$ . Since $p \neq = q$ , $d = - 1$ .
  ½ mark
4. Substituting in (i): $a + (p - 1) (- 1) = q \Rightarrow a = p + q - 1$ .
  ½ mark
5. $a_{n} = a + (n - 1) d = (p + q - 1) + (n - 1) (- 1) = p + q - 1 - n + 1 = p + q - n$ . Hence proved.
  1 mark
Answer: d = −1, a = p + q − 1, so aₙ = p + q − n (proved).
Where the marks are
- Writes a + (p − 1)d = q (½ mark)
- Writes a + (q − 1)d = p (½ mark)
- Solves to get d = −1 (½ mark)
- Finds a = p + q − 1 (½ mark)
- Obtains aₙ = p + q − n (1 mark)
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Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerTriangles
In the given figure, CD is the perpendicular bisector of AB. EF is perpendicular to CD. AE intersects CD at G. Prove that $\frac{C F}{C D} = \frac{F G}{D G}$ .
Figure in the paper: Points A and B joined by segment AB; D is the foot on AB (right angle marked at D) and CD is perpendicular to AB, with C on the left. Point E lies on segment CB; EF is perpendicular to CD with F on CD. Segment AE meets CD at G. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ E F G$ and $△ A D G$ : $∠ E F G = ∠ A D G = 9 0^{\circ}$ and $∠ E GF = ∠ A G D$ (vertically opposite angles). So $△ E F G \sim △ A D G$ (AA similarity).
  $\Rightarrow \frac{E F}{A D} = \frac{F G}{D G}$ ……(i)
  1 mark
2. In $△ E F C$ and $△ B D C$ : $∠ E F C = ∠ B D C = 9 0^{\circ}$ and $∠ E C F = ∠ B C D$ (common angle). So $△ E F C \sim △ B D C$ (AA similarity).
  $\Rightarrow \frac{E F}{B D} = \frac{C F}{C D}$
  1 mark
3. CD is the perpendicular bisector of AB, so $B D = A D$ . Hence $\frac{E F}{A D} = \frac{C F}{C D}$ ……(ii)
  ½ mark
4. From (i) and (ii): $\frac{C F}{C D} = \frac{F G}{D G}$ . Hence proved.
  ½ mark
Answer: Using ΔEFG ~ ΔADG and ΔEFC ~ ΔBDC with AD = BD, CF/CD = FG/DG (proved).
Where the marks are
- Proves ΔEFG ~ ΔADG and writes EF/AD = FG/DG (1 mark)
- Proves ΔEFC ~ ΔBDC and writes EF/BD = CF/CD (1 mark)
- Uses BD = AD (D is the midpoint of AB) to get EF/AD = CF/CD (½ mark)
- Combines to get CF/CD = FG/DG (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerTriangles
In the given figure, ABCD is a parallelogram. BE bisects CD at M and intersects AC at L. Prove that $E L = 2 B L$ .
Figure in the paper: Parallelogram ABCD with A bottom-left, B bottom-right, C top-right, D top-left. E is a point above D, with line ED extended from A side; segment EB passes through M (midpoint of DC) and L (on diagonal AC). Diagonal AC drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. E lies on AD produced, and $A D ∥ B C$ , so $A E ∥ B C$ . In $△ A L E$ and $△ C L B$ : $∠ L A E = ∠ L C B$ (alternate angles) and $∠ A L E = ∠ C L B$ (vertically opposite). So $△ A L E \sim △ C L B$ (AA).
  $\Rightarrow \frac{A L}{C L} = \frac{E L}{B L}$ ……(i)
  1 mark
2. Also $D C ∥ A B$ . In $△ C L M$ and $△ A L B$ : $∠ L C M = ∠ L A B$ (alternate angles) and $∠ C L M = ∠ A L B$ (vertically opposite). So $△ C L M \sim △ A L B$ (AA).
  $\Rightarrow \frac{A L}{C L} = \frac{A B}{C M}$
  1 mark
3. Opposite sides of a parallelogram are equal, so $A B = C D$ : $\frac{A L}{C L} = \frac{C D}{C M}$ ……(ii)
  ½ mark
4. From (i) and (ii): $\frac{E L}{B L} = \frac{C D}{C M} = \frac{2 C M}{C M} = 2$ [M is the midpoint of CD]. $\Rightarrow E L = 2 B L$ . Hence proved.
  ½ mark
Answer: Using ΔALE ~ ΔCLB and ΔCLM ~ ΔALB with AB = CD = 2CM, EL = 2BL (proved).
Where the marks are
- Proves ΔALE ~ ΔCLB and writes AL/CL = EL/BL (1 mark)
- Proves ΔCLM ~ ΔALB and writes AL/CL = AB/CM (1 mark)
- Uses AB = CD to get AL/CL = CD/CM (½ mark)
- Uses CD = 2CM to conclude EL = 2BL (½ mark)
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Q29 · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
Two people are 16 km apart on a straight road. They start walking at the same time. If they walk towards each other with different speeds, they will meet in 2 hours. Had they walked in the same direction with same speeds as before, they would have met in 8 hours. Find their walking speeds.
Try it first, then show our solutionSolution
1. Let the speeds be $x$ km/h and $y$ km/h with $x > y$ . Walking towards each other, together they cover 16 km in 2 h:
  $2 x + 2 y = 16 \Rightarrow x + y = 8$ ……(i)
  1 mark
2. Walking in the same direction, the faster one gains 16 km in 8 h:
  $8 x - 8 y = 16 \Rightarrow x - y = 2$ ……(ii)
  1 mark
3. Adding (i) and (ii): $2 x = 10 \Rightarrow x = 5$ ; then from (i), $y = 3$ .
  Their walking speeds are 5 km/h and 3 km/h.
  1 mark
Answer: 5 km/h and 3 km/h
Where the marks are
- Forms 2x + 2y = 16 for the case of walking towards each other (1 mark)
- Forms 8x − 8y = 16 for the case of walking in the same direction (1 mark)
- Solves to get x = 5 km/h (½ mark) and y = 3 km/h (½ mark)
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Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + sec θ cosec θ$
Try it first, then show our solutionSolution
1. Write everything in sin and cos:
  LHS $= \frac{\frac{s i n θ}{c o s θ}}{1 - \frac{c o s θ}{s i n θ}} + \frac{\frac{c o s θ}{s i n θ}}{1 - \frac{s i n θ}{c o s θ}}$
  ½ mark
2. Simplify each fraction:
  $= \frac{sin ^{2} θ}{cos θ ( sin θ - cos θ )} - \frac{cos ^{2} θ}{sin θ ( sin θ - cos θ )}$
  1 mark
3. Take a common denominator:
  $= \frac{1}{sin θ - cos θ} [\frac{sin ^{3} θ - cos ^{3} θ}{sin θ cos θ}]$
  ½ mark
4. Factorise the difference of cubes:
  $= \frac{( sin θ - cos θ ) ( sin ^{2} θ + sin θ cos θ + cos ^{2} θ )}{( sin θ - cos θ ) sin θ cos θ}$
  ½ mark
5. Use $sin^{2} θ + cos^{2} θ = 1$ :
  $= \frac{1 + sin θ cos θ}{sin θ cos θ} = \frac{1}{sin θ cos θ} + 1 = 1 + sec θ cosec θ =$ RHS
  ½ mark
Answer: LHS = 1 + sec θ cosec θ = RHS (proved)
Where the marks are
- Converts tan and cot into sin and cos (½ mark)
- Simplifies to the two fractions with the factor (sin θ − cos θ) (1 mark)
- Combines into (sin³θ − cos³θ)/(sin θ cos θ) (½ mark)
- Factorises using a³ − b³ (½ mark)
- Uses sin²θ + cos²θ = 1 to reach the RHS (½ mark)
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Q31 · 3 marks · Short answerStatistics
Find the mean of the following frequency distribution :
Classes 25–30 30–35 35–40 40–45 45–50 50–55 55–60
Frequency 14 22 16 6 5 3 4
Try it first, then show our solutionSolution
1. Take assumed mean $a = 42.5$ , class size $h = 5$ , $u_{i} = \frac{x _{i} - 42.5}{5}$ .
  Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  25–30 27.5 14 −3 −42
  30–35 32.5 22 −2 −44
  35–40 37.5 16 −1 −16
  40–45 42.5 6 0 0
  45–50 47.5 5 1 5
  50–55 52.5 3 2 6
  55–60 57.5 4 3 12
  Total $Σ f_{i} = 70$ $Σ f_{i} u_{i} = - 79$
  2 marks
2. Mean $\overset{x}{ˉ} = a + \frac{Σ f _{i} u _{i}}{Σ f _{i}} \times h = 42.5 + \frac{- 79}{70} \times 5 = 42.5 - 5.64 = 36.86$ (approx.)
  1 mark
Answer: Mean ≈ 36.86
Where the marks are
- Correct table of class marks, uᵢ and fᵢuᵢ with Σfᵢ = 70 and Σfᵢuᵢ = −79 (2 marks)
- Applies step-deviation formula to get mean ≈ 36.86 (1 mark)
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Classes	25–30	30–35	35–40	40–45	45–50	50–55	55–60
Frequency	14	22	16	6	5	3	4

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
25–30	27.5	14	−3	−42
30–35	32.5	22	−2	−44
35–40	37.5	16	−1	−16
40–45	42.5	6	0	0
45–50	47.5	5	1	5
50–55	52.5	3	2	6
55–60	57.5	4	3	12
Total		$Σ f_{i} = 70$		$Σ f_{i} u_{i} = - 79$

Section D

Long answer

Q32 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
One observer estimates the angle of elevation to the basket of a hot air balloon to be $6 0^{\circ}$ , while another observer 100 m away estimates the angle of elevation to be $3 0^{\circ}$ . Find :
1. (a)
  The height of the basket from the ground.
  0 marks
2. (b)
  The distance of the basket from the first observer's eye.
  0 marks
3. (c)
  The horizontal distance of the second observer from the basket.
  0 marks
Try it first, then show our solutionSolution
1. Figure: Let B be the basket and A the point on the ground directly below it, so $A B = h$ m and $∠ B A C = 9 0^{\circ}$ . The first observer is at D with $∠ B D A = 6 0^{\circ}$ , $A D = x$ m; the second observer is at C on the same side, 100 m beyond D, with $∠ B C A = 3 0^{\circ}$ , so $A C = x + 100$ m. Let $B D = y$ m.
  1 mark
2. In right $△ B A D$ : $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow 3 = \frac{h}{x} \Rightarrow h = 3 x$ ……(i)
  1 mark
3. In right $△ B A C$ : $tan 3 0^{\circ} = \frac{h}{x + 100} \Rightarrow \frac{1}{3} = \frac{h}{x + 100} \Rightarrow x = 3 h - 100$ ……(ii)
  1 mark
4. (a) Substituting (ii) in (i): $h = 3 (3 h - 100) = 3 h - 1003 \Rightarrow 2 h = 1003 \Rightarrow h = 503$ m ( $\approx 86.6$ m).
  ½ mark
5. (b) In $△ B A D$ : $sin 6 0^{\circ} = \frac{h}{y} \Rightarrow \frac{3}{2} = \frac{50 3}{y} \Rightarrow y = 100$ m. So the basket is 100 m from the first observer's eye.
  1 mark
6. (c) $x = \frac{h}{3} = \frac{50 3}{3} = 50$ m, so the horizontal distance of the second observer from the basket $= A C = x + 100 = 150$ m.
  ½ mark
7. Note: if the two observers are taken on opposite sides of the balloon (DA = x, AC = 100 − x), the same method gives $h = 253$ m, BD = 50 m and AC = 75 m; the scheme accepts this version as well.
Answer: (a) 50√3 m ≈ 86.6 m; (b) 100 m; (c) 150 m. (Observers on opposite sides: 25√3 m, 50 m, 75 m.)
Where the marks are
- Correct labelled figure (1 mark)
- tan 60° = h/x giving h = √3x (1 mark)
- tan 30° = h/(x + 100) giving x = √3h − 100 (1 mark)
- (a) h = 50√3 m (½ mark)
- (b) Uses sin 60° = h/y to get 100 m (1 mark)
- (c) x = 50 m, so horizontal distance = 150 m (½ mark)
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Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerCircles
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that area $△ A B C = 90$ cm $^{2}$ .
Figure in the paper: Triangle ABC with incircle centre O of radius 4 cm. The incircle touches BC at D; OD is drawn vertically with length 4 cm. BD = 10 cm and DC = 8 cm are marked on base BC. See it in the official paper.
Try it first, then show our solutionSolution
1. Let the incircle touch AB at F and AC at E. Join OA, OB, OC, and draw $O E ⊥ A C$ , $O F ⊥ A B$ (radius ⟂ tangent), so $O D = O E = O F = 4$ cm.
  Tangents from an external point are equal: $B F = B D = 10$ cm, $C E = C D = 8$ cm; let $A F = A E = x$ cm.
  1½ marks
2. ar $△ A B C$ = ar $△ B O C$ + ar $△ C O A$ + ar $△ A O B$ $= \frac{1}{2} (4) (B C) + \frac{1}{2} (4) (C A) + \frac{1}{2} (4) (A B) = \frac{1}{2} \cdot 4 (B C + C A + A B)$
  1½ marks
3. $B C = 18$ , $C A = 8 + x$ , $A B = 10 + x$ . So $90 = 2 (18 + 8 + x + 10 + x) = 2 (36 + 2 x) = 4 (18 + x)$ $\Rightarrow 18 + x = 22.5 \Rightarrow x = 4.5$
  1 mark
4. $A B = 10 + 4.5 = 14.5$ cm and $A C = 8 + 4.5 = 12.5$ cm.
  1 mark
Answer: AB = 14.5 cm, AC = 12.5 cm
Where the marks are
- Draws radii to points of contact, uses equal tangents BF = 10, CE = 8, AF = AE = x (1½ marks)
- Splits area of ΔABC into ar BOC + ar COA + ar AOB with heights 4 cm (1½ marks)
- Forms 90 = 2(36 + 2x) and gets x = 4.5 (1 mark)
- States AB = 14.5 cm and AC = 12.5 cm (1 mark)
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OR
Q33(b) · 5 marks · Long answerCircles
Two circles with centres O and O' of radii 6 cm and 8 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.
Figure in the paper: Two intersecting circles with centres O (left) and O' (right). P is the upper intersection point and Q the lower one. Segment PQ is the common chord, and it meets line OO' at A. Segments OP and O'P are drawn, with tangent lines at P shown. See it in the official paper.
Try it first, then show our solutionSolution
1. OP is tangent to the circle with centre O' at P, so $O P ⊥ O^{'} P$ [tangent ⟂ radius]. In right $△ O P O^{'}$ : $O O^{'} = 6^{2} + 8^{2} = 100 = 10$ cm.
  ½ mark
2. The line of centres OO' is the perpendicular bisector of the common chord PQ, so $P A ⊥ O O^{'}$ and $P Q = 2 A P$ . Let $O A = x$ cm, then $O^{'} A = (10 - x)$ cm.
  In right $△ P A O$ : $A P^{2} = O P^{2} - O A^{2} = 36 - x^{2}$
  ½ mark
3. In right $△ P A O^{'}$ : $A P^{2} = O^{'} P^{2} - O^{'} A^{2} = 64 - (10 - x)^{2}$
  ½ mark
4. Equating: $36 - x^{2} = 64 - (10 - x)^{2} = 64 - 100 + 20 x - x^{2}$ $\Rightarrow 36 = - 36 + 20 x \Rightarrow 20 x = 72 \Rightarrow x = 3.6$ cm
  2 marks
5. $A P^{2} = 36 - (3.6)^{2} = 36 - 12.96 = 23.04 \Rightarrow A P = 4.8$ cm
  1 mark
6. Length of common chord $P Q = 2 \times A P = 2 \times 4.8 = 9.6$ cm.
  ½ mark
Answer: PQ = 9.6 cm
Where the marks are
- Uses OP ⟂ O'P to get OO' = 10 cm (½ mark)
- Writes AP² = 36 − x² in ΔPAO (½ mark)
- Writes AP² = 64 − (10 − x)² in ΔPAO' (½ mark)
- Equates and solves to get x = 3.6 (2 marks)
- Finds AP = 4.8 cm (1 mark)
- PQ = 2AP = 9.6 cm (½ mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the journey, what was its first average speed ?
Try it first, then show our solutionSolution
1. Let the first average speed be $x$ km/h; the second speed is $(x + 6)$ km/h. Time = distance/speed, and total time is 3 h:
  $\frac{54}{x} + \frac{63}{x + 6} = 3$
  2 marks
2. $\Rightarrow 54 (x + 6) + 63 x = 3 x (x + 6) \Rightarrow 54 x + 324 + 63 x = 3 x^{2} + 18 x$
  $\Rightarrow 3 x^{2} - 99 x - 324 = 0 \Rightarrow x^{2} - 33 x - 108 = 0$
  2 marks
3. $\Rightarrow (x - 36) (x + 3) = 0 \Rightarrow x = 36$ or $x = - 3$ . Speed cannot be negative, so $x = - 3$ is rejected.
  The first average speed of the train was 36 km/h.
  1 mark
Answer: 36 km/h
Where the marks are
- Forms the equation 54/x + 63/(x + 6) = 3 (2 marks)
- Simplifies to x² − 33x − 108 = 0 (2 marks)
- Factorises, rejects −3 and states 36 km/h (1 mark)
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OR
Q34(b) · 5 marks · Long answerCompetency-basedQuadratic Equations
Two pipes together can fill a tank in $\frac{15}{8}$ hours. The pipe with larger diameter takes 2 hours less than the pipe with smaller diameter to fill the tank separately. Find the time in which each pipe can fill the tank separately.
Try it first, then show our solutionSolution
1. Let the smaller-diameter pipe fill the tank in $x$ hours; then the larger one takes $(x - 2)$ hours. In 1 hour they fill $\frac{1}{x}$ and $\frac{1}{x - 2}$ of the tank; together they fill $\frac{8}{15}$ of it:
  $\frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15}$
  2 marks
2. $\Rightarrow 15 (x - 2 + x) = 8 x (x - 2) \Rightarrow 30 x - 30 = 8 x^{2} - 16 x$
  $\Rightarrow 8 x^{2} - 46 x + 30 = 0 \Rightarrow 4 x^{2} - 23 x + 15 = 0$
  1 mark
3. $\Rightarrow 4 x^{2} - 20 x - 3 x + 15 = 0 \Rightarrow (4 x - 3) (x - 5) = 0 \Rightarrow x = \frac{3}{4}$ or $x = 5$
  1 mark
4. If $x = \frac{3}{4}$ , then $x - 2 < 0$ , which is impossible for time, so $x = 5$ .
  Smaller-diameter pipe: 5 hours; larger-diameter pipe: $5 - 2 = 3$ hours.
  1 mark
Answer: Smaller pipe: 5 hours; larger pipe: 3 hours
Where the marks are
- Forms the equation 1/x + 1/(x − 2) = 8/15 (2 marks)
- Simplifies to 4x² − 23x + 15 = 0 (1 mark)
- Factorises to get x = 3/4 or 5 (1 mark)
- Rejects 3/4 and states 5 h and 3 h (1 mark)
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Q35 · 5 marks · Long answerCompetency-basedAreas Related to Circles
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find the area of that part of the field in which the horse can graze. Also, find the increase in grazing area if length of rope is increased to 10 m. (Use $π = 3 \cdot 14$ )
Figure in the paper: A square field of side 15 m. The peg is fixed at one corner, and a quarter-circle arc is drawn from that corner inside the square showing the grazing region of radius 5 m. See it in the official paper.
Try it first, then show our solutionSolution
1. The horse is tied at a corner of the square and the rope (5 m) is shorter than the side (15 m). The corner angle is $9 0^{\circ}$ , so the grazing region is a quadrant (sector of angle $9 0^{\circ}$ ) of radius $r = 5$ m.
  Grazing area $= \frac{9 0 ^{\circ}}{36 0 ^{\circ}} π r^{2} = \frac{1}{4} \times 3.14 \times (5)^{2}$
  1 mark
2. $= \frac{1}{4} \times 3.14 \times 25 = 19.625 m^{2}$
  1 mark
3. With a 10 m rope (still less than 15 m), the region is again a quadrant, now of radius 10 m.
  New grazing area $= \frac{1}{4} \times 3.14 \times (10)^{2}$
  1 mark
4. $= \frac{1}{4} \times 314 = 78.5 m^{2}$
  1 mark
5. Increase in grazing area $= 78.5 - 19.625 = 58.875 m^{2}$
  1 mark
Answer: Grazing area = 19.625 m²; increase in grazing area = 58.875 m²
Where the marks are
- Uses quadrant area $\frac{1}{4} π r^{2}$ with r = 5 m (1 mark)
- Gets 19.625 m² (1 mark)
- Uses quadrant area with r = 10 m (1 mark)
- Gets 78.5 m² (1 mark)
- Increase = 58.875 m² (1 mark)
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Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Case Study – 1
A golf ball is spherical with about 300 – 500 dimples that help increase its velocity while in play. Golf balls are traditionally white but available in colours also. In the given figure, a golf ball has diameter 4·2 cm and the surface has 315 dimples (hemi-spherical) of radius 2 mm.
Based on the above, answer the following questions :
1. (i)
  Find the surface area of one such dimple.
  1 mark
2. (ii)
  Find the volume of the material dug out to make one dimple.
  1 mark
3. (iii)
  (a) Find the total surface area exposed to the surroundings.
  2 marks
4. OR
  (iii)
  (b) Find the volume of the golf ball.
  2 marks
Figure in the paper: Left: a spherical golf ball with a dimpled surface. Right: a circle outline with hemispherical dimples (small semicircular notches) around its edge. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Each dimple is a hemispherical hollow of radius $r = 2$ mm. Its surface is the curved surface of a hemisphere:
  $CSA = 2 π r^{2} = 2 \times \frac{22}{7} \times 4 = \frac{176}{7} mm^{2} \approx 25.14 mm^{2}$
  1 mark
2. (ii) Material dug out = volume of a hemisphere of radius 2 mm:
  $V = \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 8 = \frac{352}{21} mm^{3} \approx 16.76 mm^{3}$
  1 mark
3. (iii)(a) Radius of ball $R = \frac{4.2}{2} = 2.1$ cm $= 21$ mm. For each dimple, a circular patch of area $π r^{2}$ is removed from the sphere's surface and a hemispherical surface $2 π r^{2}$ is added.
  Exposed area $= 4 π R^{2} - 315 π r^{2} + 315 \times 2 π r^{2} = 4 π (21)^{2} + 315 π (2)^{2}$
  1 mark
4. (iii)(a) $= \frac{22}{7} (1764 + 1260) = \frac{22}{7} \times 3024 = 9504 mm^{2}$
  1 mark
5. (iii)(b) Volume of ball = volume of sphere − volume of 315 hemispherical dimples:
  $V = \frac{4}{3} π (21)^{3} - 315 \times \frac{2}{3} π (2)^{3} = π (12348 - 1680)$
  1 mark
6. (iii)(b) $= \frac{22}{7} \times 10668 = 22 \times 1524 = 33528 mm^{3}$
  1 mark
Answer: (i) 176/7 mm² ≈ 25.1 mm²; (ii) 352/21 mm³ ≈ 16.76 mm³; (iii)(a) 9504 mm²; (iii)(b) 33528 mm³
Where the marks are
- (i) CSA of hemisphere $2 π r^{2} = 176/7$ mm² (1 mark)
- (ii) Volume $\frac{2}{3} π r^{3} = 352/21$ mm³ (1 mark)
- (iii)(a) Correct expression $4 π R^{2} - 315 π r^{2} + 315 \cdot 2 π r^{2}$ with R = 21 mm (1 mark)
- (iii)(a) Exposed area = 9504 mm² (1 mark)
- (iii)(b) Correct expression $\frac{4}{3} π R^{3} - 315 \cdot \frac{2}{3} π r^{3}$ (1 mark)
- (iii)(b) Volume = 33528 mm³ (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedProbability
Case Study – 2
A middle school decided to run the following spinner game as a fund-raiser on Christmas Carnival.
Making Purple : Spin each spinner once. Blue and red make purple. So, if one spinner shows Red (R) and another Blue (B), then you 'win'. One such outcome is written as 'RB'.
Based on the above, answer the following questions :
1. (i)
  List all possible outcomes of the game.
  1 mark
2. (ii)
  Find the probability of 'Making Purple'.
  1 mark
3. (iii)
  (a) For each win, a participant gets ₹ 10, but if he/she loses, he/she has to pay ₹ 5 to the school. If 99 participants played, calculate how much fund could the school have collected.
  2 marks
4. OR
  (iii)
  (b) If the same amount of ₹ 5 has been decided for winning or losing the game, then how much fund had been collected by school ? (Number of participants = 99)
  2 marks
Figure in the paper: Spinner I: circle divided into three sections: Red (R) upper left, Green (G) upper right, and Yellow (Y) at the bottom; an arrow pointer at the centre points toward the Red section. Spinner II: circle divided into three sections: Red (R) upper left, Blue (B) upper right, and Green (G) at the bottom; an arrow pointer at the centre points to the right toward the Blue section. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Spinner I shows R, G or Y; Spinner II shows R, B or G. All possible outcomes (first letter from Spinner I):
  RR, RB, RG, GR, GB, GG, YR, YB, YG — 9 equally likely outcomes.
  1 mark
2. (ii) Purple needs one Red and one Blue. Only Spinner II has Blue, so the only favourable outcome is RB.
  $P (Making Purple) = \frac{1}{9}$
  1 mark
3. (iii)(a) Expected number of winners $= \frac{1}{9} \times 99 = 11$
  ½ mark
4. (iii)(a) Number of losers $= 99 - 11 = 88$
  ½ mark
5. (iii)(a) Fund collected = amount received from losers − amount paid to winners $= 88 \times 5 - 11 \times 10 = 440 - 110 = ₹ 330$
  1 mark
6. (iii)(b) As before, with $P (win) = \frac{1}{9}$ : winners $= \frac{1}{9} \times 99 = 11$
  ½ mark
7. (iii)(b) Losers $= 99 - 11 = 88$
  ½ mark
8. (iii)(b) Each winner gets ₹5 and each loser pays ₹5:
  Fund $= 88 \times 5 - 11 \times 5 = 440 - 55 = ₹ 385$
  1 mark
Answer: (i) RR, RB, RG, GR, GB, GG, YR, YB, YG; (ii) 1/9; (iii)(a) ₹330; (iii)(b) ₹385
Where the marks are
- (i) Lists all 9 outcomes correctly (1 mark)
- (ii) P(Making Purple) = 1/9 (1 mark)
- (iii)(a) Number of winners = 11 (½ mark)
- (iii)(a) Number of losers = 88 (½ mark)
- (iii)(a) Fund = 88×5 − 11×10 = ₹330 (1 mark)
- (iii)(b) Winners = 11 (½ mark)
- (iii)(b) Losers = 88 (½ mark)
- (iii)(b) Fund = 88×5 − 11×5 = ₹385 (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedPolynomials
Case Study – 3
In a pool at an aquarium, a dolphin jumps out of the water travelling at 20 cm per second. Its height above water level after t seconds is given by $h = 20 t - 16 t^{2}$ .
Based on the above, answer the following questions :
1. (i)
  Find zeroes of polynomial $p (t) = 20 t - 16 t^{2}$ .
  1 mark
2. (ii)
  Which of the following types of graph represents $p (t)$ ?
  (a) Parabola opening downward from the origin, crossing the time axis at $\frac{5}{4}$ , with the curve above the time axis between $0$ and $\frac{5}{4}$ .
  (b) Parabola with vertex above the time axis at $t = \frac{5}{4}$ , crossing the time axis at $0$ and at a point beyond $\frac{5}{4}$ .
  (c) Parabola opening upward, with vertex below the time axis, crossing the axis at $0$ and $\frac{5}{4}$ .
  (d) Downward-opening parabola with vertex at $t = \frac{5}{4}$ on the time axis, starting above the origin on the height axis.
  1 mark
3. (iii)
  (a) What would be the value of h at $t = \frac{3}{2}$ ? Interpret the result.
  2 marks
4. OR
  (iii)
  (b) How much distance has the dolphin covered before hitting the water level again ?
  2 marks
Figure in the paper: Four graph options labelled (a) to (d), each with axes x′, x (time) and y′, y (height). Option (a): upward-opening arc from origin to a point at 5/4 on the time axis. Option (b): arc peaking at t = 5/4 with dashed vertical line, returning to the time axis. Option (c): upward parabola dipping below the axis between 0 and 5/4. Option (d): downward-opening curve with vertex at 5/4 on the time axis. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Put $p (t) = 0$ : $20 t - 16 t^{2} = 0 \Rightarrow 4 t (5 - 4 t) = 0$
  $\Rightarrow t = 0$ or $t = \frac{5}{4}$ . Zeroes are $0$ and $\frac{5}{4}$ .
  1 mark
2. (ii) The coefficient of $t^{2}$ is $- 16 < 0$ , so the graph is a downward-opening parabola, and it cuts the time axis at $t = 0$ and $t = \frac{5}{4}$ , staying above the axis in between. This matches graph (a).
  1 mark
3. (iii)(a) At $t = \frac{3}{2}$ : $h = 20 \times \frac{3}{2} - 16 \times \frac{9}{4} = 30 - 36 = - 6$ cm
  1 mark
4. (iii)(a) Interpretation: a negative height means the dolphin is below the water surface; after $\frac{3}{2}$ s it is 6 cm below water level (it re-entered the water at $t = \frac{5}{4}$ s).
  1 mark
5. (iii)(b) The dolphin is out of the water from $t = 0$ to $t = \frac{5}{4}$ s (the zeroes of $p (t)$ ). Travelling at 20 cm per second, distance covered $= 20 \times \frac{5}{4} = 25$ cm.
  2 marks
Answer: (i) 0 and 5/4; (ii) graph (a); (iii)(a) h = −6 cm, i.e. 6 cm below water level; (iii)(b) 25 cm
Where the marks are
- (i) Factorises and finds zeroes 0 and 5/4 (1 mark)
- (ii) Chooses graph (a) (1 mark)
- (iii)(a) Computes h = −6 (1 mark)
- (iii)(a) Interprets as 6 cm below water level (1 mark)
- (iii)(b) Uses time 5/4 s with speed 20 cm/s to get 25 cm (2 marks)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.