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

CBSE Class 10 Mathematics Basic: 2023 board paper, set 430/2/1

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_BASIC.zip (a zip; this paper is the file 430_2_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	VSA; internal choice in 22 and 23
C	26-31	SA; internal choice in 28
D	32-35	LA; internal choice in 32 and 35
E	36-38	Case study; internal choice in each case (2-mark sub-parts)

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 studies of Section E. Use π = 22/7 where required if not stated; calculators not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQArithmetic Progressions
How many terms are there in the A.P. given below?
$14, 19, 24, 29, \dots, 119$
1. (A)18
2. (B)14
3. (C)22
4. (D)21
Try it first, then show our solutionSolution
1. Here $a = 14$ , $d = 5$ , $a_{n} = 119$ . Using $a_{n} = a + (n - 1) d$ : $119 = 14 + 5 (n - 1)$ .
2. $5 (n - 1) = 105 \Rightarrow n - 1 = 21 \Rightarrow n = 22$ . Option (C).
  1 mark
Answer: 22, option (C)
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ to get $n = 22$ (1 mark)
Q2 · 1 mark · MCQCoordinate Geometry
In what ratio does x-axis divide the line segment joining the points $A (2, - 3)$ and $B (5, 6)$ ?
1. (A) $2 : 3$
2. (B) $2 : 1$
3. (C) $3 : 4$
4. (D) $1 : 2$
Try it first, then show our solutionSolution
1. Let the x-axis divide AB in the ratio $k : 1$ . On the x-axis the y-coordinate is 0, so $\frac{6 k - 3}{k + 1} = 0$ .
2. $6 k = 3 \Rightarrow k = \frac{1}{2}$ . The ratio is $1 : 2$ , option (D).
  1 mark
Answer: 1 : 2, option (D)
Where the marks are
- Sets the y-coordinate of the section point to 0 and gets $1 : 2$ (1 mark)
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Q3 · 1 mark · MCQIntroduction to Trigonometry
$9 sec^{2} A - 9 tan^{2} A$ is equal to:
1. (A)9
2. (B)0
3. (C)8
4. (D) $\frac{1}{9}$
Try it first, then show our solutionSolution
1. $9 sec^{2} A - 9 tan^{2} A = 9 (sec^{2} A - tan^{2} A) = 9 \times 1 = 9$ [since $sec^{2} A - tan^{2} A = 1$ ]. Option (A).
  1 mark
Answer: 9, option (A)
Where the marks are
- Takes 9 common and uses $sec^{2} A - tan^{2} A = 1$ (1 mark)
Q4 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
The string of a kite in air is 50 m long and it makes an angle of $6 0^{\circ}$ with the horizontal. Assuming the string to be straight, the height of the kite from the ground is:
1. (A) $503$ m
2. (B) $\frac{100}{3}$ m
3. (C) $\frac{50}{3}$ m
4. (D) $253$ m
Try it first, then show our solutionSolution
1. Height $h$ is opposite the $6 0^{\circ}$ angle and the string (50 m) is the hypotenuse, so $sin 6 0^{\circ} = \frac{h}{50}$ .
2. $h = 50 \times \frac{3}{2} = 253$ m. Option (D).
  1 mark
Answer: $253$ m, option (D)
Where the marks are
- Uses $sin 6 0^{\circ} = h /50$ to get $253$ m (1 mark)
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Q5 · 1 mark · MCQCircles
From a point P, two tangents PQ and PR are drawn to a circle with centre at O. T is a point on the major arc QR of the circle. If $∠ QP R = 5 0^{\circ}$ , then $∠ QT R$ equals:
1. (A) $5 0^{\circ}$
2. (B) $13 0^{\circ}$
3. (C) $6 5^{\circ}$
4. (D) $9 0^{\circ}$
Figure in the paper: Circle with centre O. External point P with tangents PQ and PR touching the circle at Q and R. Radii OQ and OR are drawn. T is a point on the major arc QR on the left side; chords TQ and TR are drawn. The angle at P, $∠$ QPR, is marked $5 0^{\circ}$ . See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O QP = ∠ O R P = 9 0^{\circ}$ [tangent ⟂ radius], so in quadrilateral OQPR, $∠ QO R = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 5 0^{\circ} = 13 0^{\circ}$ .
2. Angle subtended at the circumference is half the angle at the centre: $∠ QT R = \frac{13 0 ^{\circ}}{2} = 6 5^{\circ}$ . Option (C).
  1 mark
Answer: $6 5^{\circ}$ , option (C)
Where the marks are
- Finds $∠ QO R = 13 0^{\circ}$ and halves it to get $6 5^{\circ}$ (1 mark)
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Q6 · 1 mark · MCQAreas Related to Circles
The area of a sector of angle $α$ (in degrees) of a circle with radius R is:
1. (A) $\frac{α}{180} \times 2 π R$
2. (B) $\frac{α}{360} \times 2 π R$
3. (C) $\frac{α}{180} \times π R^{2}$
4. (D) $\frac{α}{360} \times π R^{2}$
Try it first, then show our solutionSolution
1. A sector of angle $α$ is the fraction $\frac{α}{360}$ of the whole circle of area $π R^{2}$ , so its area is $\frac{α}{360} \times π R^{2}$ . Option (D).
  1 mark
Answer: $\frac{α}{360} \times π R^{2}$ , option (D)
Where the marks are
- Identifies the sector area formula (1 mark)
Q7 · 1 mark · MCQReal Numbers
If the HCF of 360 and 64 is 8, then their LCM is:
1. (A)2480
2. (B)2780
3. (C)512
4. (D)2880
Try it first, then show our solutionSolution
1. HCF × LCM = product of the numbers, so LCM $= \frac{360 \times 64}{8} = 360 \times 8 = 2880$ . Option (D).
  1 mark
Answer: 2880, option (D)
Where the marks are
- Uses HCF × LCM = product to get 2880 (1 mark)
Q8 · 1 mark · MCQSurface Areas and Volumes
The curved surface area of a right circular cylinder of height 14 cm is $88 c m^{2}$ . The diameter of its circular base is:
1. (A)2 cm
2. (B)1 cm
3. (C)4 cm
4. (D)7 cm
Try it first, then show our solutionSolution
1. CSA $= 2 π r h$ : $2 \times \frac{22}{7} \times r \times 14 = 88 \Rightarrow 88 r = 88 \Rightarrow r = 1$ cm.
2. Diameter $= 2 r = 2$ cm. Option (A).
  1 mark
Answer: 2 cm, option (A)
Where the marks are
- Uses $2 π r h = 88$ to get r = 1 cm, diameter 2 cm (1 mark)
Q9 · 1 mark · MCQProbability
A die is rolled once. The probability that a composite number comes up, is:
1. (A) $\frac{1}{2}$
2. (B) $\frac{2}{3}$
3. (C) $\frac{1}{3}$
4. (D)0
Try it first, then show our solutionSolution
1. Outcomes: 1 to 6 (6 outcomes). Composite numbers among them: 4, 6 (2 outcomes).
2. P(composite) $= \frac{2}{6} = \frac{1}{3}$ . Option (C).
  1 mark
Answer: $\frac{1}{3}$ , option (C)
Where the marks are
- Identifies 4 and 6 as composite and gets $\frac{1}{3}$ (1 mark)
Q10 · 1 mark · MCQQuadratic Equations
If the quadratic equation $9 x^{2} + b x + \frac{1}{4} = 0$ has equal roots, then the value of b is:
1. (A)0
2. (B) $- 3$ only
3. (C) $3$ only
4. (D) $\pm 3$
Try it first, then show our solutionSolution
1. Equal roots ⇒ $D = b^{2} - 4 a c = 0$ : $b^{2} - 4 \times 9 \times \frac{1}{4} = 0 \Rightarrow b^{2} = 9$ .
2. $b = \pm 3$ . Option (D).
  1 mark
Answer: $b = \pm 3$ , option (D)
Where the marks are
- Puts discriminant = 0 and gets $b = \pm 3$ (1 mark)
Q11 · 1 mark · MCQSurface Areas and Volumes
A solid is of the form of a cone of radius 'r' surmounted on a hemisphere of the same radius. If the height of the cone is the same as the diameter of its base, then the volume of the solid is:
1. (A) $π r^{3}$
2. (B) $\frac{4}{3} π r^{3}$
3. (C) $3 π r^{3}$
4. (D) $\frac{2}{3} π r^{3}$
Figure in the paper: A cone with its base circle of radius r resting on the flat face of a hemisphere of the same radius r; the radius r is marked from the centre of the base to its edge. The cone's apex is directly above the centre. See it in the official paper.
Try it first, then show our solutionSolution
1. Cone height $h = 2 r$ , so cone volume $= \frac{1}{3} π r^{2} (2 r) = \frac{2}{3} π r^{3}$ ; hemisphere volume $= \frac{2}{3} π r^{3}$ .
2. Total $= \frac{2}{3} π r^{3} + \frac{2}{3} π r^{3} = \frac{4}{3} π r^{3}$ . Option (B).
  1 mark
Answer: $\frac{4}{3} π r^{3}$ , option (B)
Where the marks are
- Adds cone ( $h = 2 r$ ) and hemisphere volumes to get $\frac{4}{3} π r^{3}$ (1 mark)
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Q12 · 1 mark · MCQPolynomials
Graph of a polynomial p(x) is given in the figure. The number of zeroes of p(x) is:
1. (A)2
2. (B)3
3. (C)4
4. (D)5
Figure in the paper: Coordinate axes x'Ox and y'Oy. A curve rises from the left below the x-axis, crosses the x-axis on the negative side, peaks above the axis near the y-axis, dips to a trough above the x-axis, rises to a second peak, and then falls, crossing the x-axis on the positive side. See it in the official paper.
Try it first, then show our solutionSolution
1. Zeroes are the x-coordinates where the graph meets the x-axis. The curve crosses the x-axis once on the negative side and once on the positive side; the trough stays above the axis.
2. Number of zeroes = 2. Option (A).
  1 mark
Answer: 2, option (A)
Where the marks are
- Counts the points where the graph meets the x-axis: 2 (1 mark)
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Q13 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of linear equations $x + 2 y - 5 = 0$ and $2 x - 4 y + 6 = 0$ :
1. (A)is inconsistent
2. (B)is consistent with many solutions
3. (C)is consistent with a unique solution
4. (D)is consistent with two solutions
Try it first, then show our solutionSolution
1. $\frac{a _{1}}{a _{2}} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{- 4} = - \frac{1}{2}$ .
2. Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect: consistent with a unique solution. Option (C).
  1 mark
Answer: Consistent with a unique solution, option (C)
Where the marks are
- Compares ratios, $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ ⇒ unique solution (1 mark)
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Q14 · 1 mark · MCQProbability
Which of the following numbers cannot be the probability of an event?
1. (A)0·5
2. (B)5%
3. (C) $\frac{1}{0 \cdot 5}$
4. (D) $\frac{0 \cdot 5}{14}$
Try it first, then show our solutionSolution
1. Probability must lie between 0 and 1. $\frac{1}{0.5} = 2 > 1$ , while 0.5, 5% = 0.05 and $\frac{0.5}{14}$ all lie in [0, 1].
2. So $\frac{1}{0.5}$ cannot be a probability. Option (C).
  1 mark
Answer: $\frac{1}{0.5}$ , option (C)
Where the marks are
- Uses $0 \leq P \leq 1$ to reject $\frac{1}{0.5} = 2$ (1 mark)
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Q15 · 1 mark · MCQIntroduction to Trigonometry
The value of $2 sin^{2} 3 0^{\circ} + 3 tan^{2} 6 0^{\circ} - cos^{2} 4 5^{\circ}$ is:
1. (A) $33$
2. (B) $\frac{19}{2}$
3. (C) $\frac{9}{4}$
4. (D)9
Try it first, then show our solutionSolution
1. $2 (\frac{1}{2})^{2} + 3 (3)^{2} - (\frac{1}{2})^{2} = \frac{1}{2} + 9 - \frac{1}{2}$ .
2. $= 9$ . Option (D).
  1 mark
Answer: 9, option (D)
Where the marks are
- Substitutes standard values and simplifies to 9 (1 mark)
Q16 · 1 mark · MCQPair of Linear Equations in Two Variables
The lines represented by the linear equations $y = x$ and $x = 4$ intersect at P. The coordinates of the point P are:
1. (A)(4, 0)
2. (B)(4, 4)
3. (C)(0, 4)
4. (D)(−4, 4)
Figure in the paper: Coordinate axes with the line y = x passing through the origin and the vertical line x = 4 meeting it at point P in the first quadrant. See it in the official paper.
Try it first, then show our solutionSolution
1. At P, $x = 4$ , and from $y = x$ , $y = 4$ . So P = (4, 4). Option (B).
  1 mark
Answer: (4, 4), option (B)
Where the marks are
- Solves the two equations to get (4, 4) (1 mark)
Q17 · 1 mark · MCQStatistics
Median and Mode of a distribution are 25 and 21 respectively. Mean of the data using empirical relationship is:
1. (A)27
2. (B)29
3. (C)18
4. (D) $\frac{29}{3}$
Try it first, then show our solutionSolution
1. Empirical relation: $3 Median = Mode + 2 Mean \Rightarrow 75 = 21 + 2 Mean$ .
2. Mean $= \frac{54}{2} = 27$ . Option (A).
  1 mark
Answer: 27, option (A)
Where the marks are
- Uses 3 Median = Mode + 2 Mean to get 27 (1 mark)
Q18 · 1 mark · MCQIntroduction to Trigonometry
If $tan A = \frac{2}{5}$ , then the value of $\frac{1 - c o s ^{2} A}{1 - s i n ^{2} A}$ is:
1. (A) $\frac{25}{4}$
2. (B) $\frac{4}{25}$
3. (C) $\frac{4}{5}$
4. (D) $\frac{5}{4}$
Try it first, then show our solutionSolution
1. $\frac{1 - c o s ^{2} A}{1 - s i n ^{2} A} = \frac{s i n ^{2} A}{c o s ^{2} A} = tan^{2} A$ .
2. $= (\frac{2}{5})^{2} = \frac{4}{25}$ . Option (B).
  1 mark
Answer: $\frac{4}{25}$ , option (B)
Where the marks are
- Simplifies to $tan^{2} A$ and gets $\frac{4}{25}$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedPolynomials
Assertion (A): Polynomial $x^{2} + 4 x$ has two real zeroes.
Reason (R): Zeroes of the polynomial $x^{2} + a x$ ( $a \neq = 0$ ) are 0 and a.
Select the correct answer from the codes (a), (b), (c) and (d) given below: (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. $x^{2} + 4 x = x (x + 4)$ , zeroes 0 and −4: two real zeroes, so A is true.
2. $x^{2} + a x = x (x + a)$ has zeroes 0 and $- a$ , not 0 and $a$ , so R is false. Option (C).
  1 mark
Answer: Option (C): A true, R false
Where the marks are
- Shows zeroes of $x^{2} + 4 x$ are 0, −4 and that R's zeroes should be 0, −a (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A): The probability of getting a prime number, when a die is thrown once, is $\frac{2}{3}$ .
Reason (R): On the faces of a die, prime numbers are 2, 3, 5.
Select the correct answer from the codes (a), (b), (c) and (d) given below.
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. Primes on a die: 2, 3, 5, so R is true.
2. P(prime) $= \frac{3}{6} = \frac{1}{2} \neq = \frac{2}{3}$ , so A is false. Option (D).
  1 mark
Answer: Option (D): A false, R true
Where the marks are
- Finds P(prime) = 1/2, so A false and R true (1 mark)

Section B

VSA; internal choice in 22 and 23

Q21 · 2 marks · Very short answerTriangles
In the given figure, $D E ∥ A C$ and $\frac{B E}{E C} = \frac{B C}{C P}$ . Prove that $D C ∥ A P$ .
Figure in the paper: Triangle ABP with A at the top. Point D lies on side AB. Points B, E, C, P lie in order on the base line BP. Segment DE is drawn parallel to AC (both arrows marked), and segment AC joins A to C on the base. Segment DC is drawn from D to C. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ , $D E ∥ A C$ . By the Basic Proportionality Theorem, $\frac{B E}{E C} = \frac{B D}{D A}$ ...(1)
  1 mark
2. It is given that $\frac{B E}{E C} = \frac{B C}{C P}$ ...(2). From (1) and (2), $\frac{B D}{D A} = \frac{B C}{C P}$ .
  ½ mark
3. In $△ A B P$ , D lies on AB and C lies on BP, and the line DC divides these two sides in the same ratio. Therefore $D C ∥ A P$ [converse of BPT]. Hence proved.
  ½ mark
Answer: $\frac{B D}{D A} = \frac{B C}{C P}$ , so $D C ∥ A P$ by the converse of BPT.
Where the marks are
- Applies BPT in $△ A B C$ to get $\frac{B E}{E C} = \frac{B D}{D A}$ (1 mark)
- Combines with the given ratio to get $\frac{B D}{D A} = \frac{B C}{C P}$ (1/2 mark)
- Concludes $D C ∥ A P$ by the converse of BPT in $△ A B P$ (1/2 mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerReal Numbers
Find the HCF of the numbers 540 and 630, using prime factorization method.
Try it first, then show our solutionSolution
1. $540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^{2} \times 3^{3} \times 5$
  ½ mark
2. $630 = 2 \times 3 \times 3 \times 5 \times 7 = 2 \times 3^{2} \times 5 \times 7$
  ½ mark
3. HCF is the product of the smallest powers of the common prime factors: $HCF = 2 \times 3^{2} \times 5 = 90$
  1 mark
Answer: HCF(540, 630) = 90
Where the marks are
- Correct prime factorisation of 540 (1/2 mark)
- Correct prime factorisation of 630 (1/2 mark)
- HCF $= 2 \times 3^{2} \times 5 = 90$ (1 mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerReal Numbers
Show that $(15)^{n}$ cannot end with the digit 0 for any natural number 'n'.
Try it first, then show our solutionSolution
1. $1 5^{n} = (3 \times 5)^{n} = 3^{n} \times 5^{n}$ , so the only prime factors of $1 5^{n}$ are 3 and 5.
  1 mark
2. A number ending in 0 is divisible by 10 $= 2 \times 5$ , so 2 must appear in its prime factorisation. Since 2 is not a prime factor of $1 5^{n}$ (by uniqueness of prime factorisation, Fundamental Theorem of Arithmetic), $1 5^{n}$ cannot end with the digit 0 for any natural number $n$ .
  1 mark
Answer: $1 5^{n} = 3^{n} \times 5^{n}$ has no factor 2, so it cannot end with 0.
Where the marks are
- Writes $1 5^{n} = 3^{n} \times 5^{n}$ (1 mark)
- Argues that ending in 0 needs prime factors 2 and 5, and 2 is absent, so $1 5^{n}$ cannot end in 0 (1 mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerCoordinate Geometry
Find the value(s) of 'x' so that $P Q = QR$ , where the coordinates of P, Q and R are $(6, - 1)$ , $(1, 3)$ and $(x, 8)$ respectively.
Try it first, then show our solutionSolution
1. Using the distance formula, $P Q = QR$ gives $(6 - 1)^{2} + (- 1 - 3)^{2} = (x - 1)^{2} + (8 - 3)^{2}$
  1 mark
2. Squaring: $25 + 16 = (x - 1)^{2} + 25 \Rightarrow (x - 1)^{2} = 16 \Rightarrow x - 1 = \pm 4$
  ½ mark
3. $x = 5$ or $x = - 3$
  ½ mark
Answer: x = 5 or x = −3
Where the marks are
- Sets up PQ = QR with the distance formula (1 mark)
- Simplifies to $(x - 1)^{2} = 16$ (1/2 mark)
- Gets $x = 5$ or $- 3$ (1/2 mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerCoordinate Geometry
The vertices of a triangle are $(- 2, 0)$ , $(2, 3)$ and $(1, - 3)$ . Is the triangle equilateral, isosceles or scalene?
Try it first, then show our solutionSolution
1. Let A(−2, 0), B(2, 3), C(1, −3). $A B = (2 + 2)^{2} + (3 - 0)^{2} = 16 + 9 = 5$
  ½ mark
2. $B C = (1 - 2)^{2} + (- 3 - 3)^{2} = 1 + 36 = 37$
  ½ mark
3. $C A = (- 2 - 1)^{2} + (0 + 3)^{2} = 9 + 9 = 32$
  ½ mark
4. All three sides are unequal, so $△ A B C$ is a scalene triangle.
  ½ mark
Answer: Sides 5, $37$ , $32$ are all different, so the triangle is scalene.
Where the marks are
- AB = 5 (1/2 mark)
- BC = $37$ (1/2 mark)
- CA = $32$ (1/2 mark)
- Concludes scalene (1/2 mark)
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Q24 · 2 marks · Very short answerPolynomials
Find the value of 'k' such that the polynomial $p (x) = 3 x^{2} + 2 k x + x - k - 5$ has the sum of zeroes equal to half of their product.
Try it first, then show our solutionSolution
1. $p (x) = 3 x^{2} + (2 k + 1) x - (k + 5)$ , so $a = 3$ , $b = 2 k + 1$ , $c = - (k + 5)$ . Sum of zeroes $= - \frac{b}{a} = - \frac{2 k + 1}{3}$
  ½ mark
2. Product of zeroes $= \frac{c}{a} = - \frac{k + 5}{3}$
  ½ mark
3. Given sum $= \frac{1}{2} \times$ product: $- \frac{2 k + 1}{3} = - \frac{1}{2} \cdot \frac{k + 5}{3}$
  ½ mark
4. $2 (2 k + 1) = k + 5 \Rightarrow 4 k + 2 = k + 5 \Rightarrow 3 k = 3 \Rightarrow k = 1$
  ½ mark
Answer: k = 1
Where the marks are
- Sum of zeroes $- \frac{2 k + 1}{3}$ (1/2 mark)
- Product of zeroes $- \frac{k + 5}{3}$ (1/2 mark)
- Forms the equation sum = half of product (1/2 mark)
- Solves to k = 1 (1/2 mark)
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Q25 · 2 marks · Very short answerProbability
From a well-shuffled deck of 52 playing cards, all diamond cards are removed. Now, a card is drawn from the remaining pack at random. Find the probability that the selected card is a king.
Try it first, then show our solutionSolution
1. After removing the 13 diamond cards, the number of cards left $= 52 - 13 = 39$ , so there are 39 equally likely outcomes.
  ½ mark
2. One king (king of diamonds) has been removed, so favourable outcomes = number of kings left = 3.
  ½ mark
3. $P (king) = \frac{3}{39} = \frac{1}{13}$
  1 mark
Answer: $\frac{1}{13}$
Where the marks are
- Total cards remaining = 39 (1/2 mark)
- Kings remaining = 3 (1/2 mark)
- Probability $= \frac{3}{39} = \frac{1}{13}$ (1 mark)
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Section C

SA; internal choice in 28

Q26 · 3 marks · Short answerAreas Related to Circles
In the given figure, two concentric circles with centre O are shown. Radii of the circles are 2 cm and 5 cm respectively. Find the area of the shaded region.
Figure in the paper: Two concentric circles with centre O; inner radius 2 cm, outer radius 5 cm. Two radii OA and OC go to the outer circle and OE and OD to the inner circle, with $∠$ EOD = $6 0^{\circ}$ marked at O. The shaded region is bounded by segment EA, arc AC of the outer circle (through B), segment CD and arc ED of the inner circle. See it in the official paper.
Try it first, then show our solutionSolution
1. Area of sector OAC of the outer circle (r = 5 cm, θ = 60°) $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times π \times 5^{2} = \frac{25 π}{6} cm^{2}$
  1 mark
2. Area of sector OED of the inner circle (r = 2 cm, θ = 60°) $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times π \times 2^{2} = \frac{4 π}{6} cm^{2}$
  1 mark
3. Area of shaded region $= \frac{25 π}{6} - \frac{4 π}{6} = \frac{21 π}{6} = \frac{21}{6} \times \frac{22}{7} = 11 cm^{2}$
  1 mark
Answer: 11 cm²
Where the marks are
- Area of outer sector $\frac{25 π}{6}$ cm² (1 mark)
- Area of inner sector $\frac{4 π}{6}$ cm² (1 mark)
- Shaded area = difference = 11 cm² (1 mark)
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Q27 · 3 marks · Short answerReal Numbers
Prove that $4 + 23$ is an irrational number, given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $4 + 23$ is rational. Then $4 + 23 = \frac{p}{q}$ , where p, q are integers and $q \neq = 0$ .
  1 mark
2. Rearranging: $23 = \frac{p}{q} - 4 = \frac{p - 4 q}{q} \Rightarrow 3 = \frac{p - 4 q}{2 q}$
  1 mark
3. Since p, q are integers, $\frac{p - 4 q}{2 q}$ is rational, so $3$ would be rational. This contradicts the given fact that $3$ is irrational. So our assumption is wrong, and $4 + 23$ is irrational.
  1 mark
Answer: $4 + 23$ is irrational (proved by contradiction).
Where the marks are
- Assumes $4 + 23 = \frac{p}{q}$ , p, q integers, $q \neq = 0$ (1 mark)
- Expresses $3 = \frac{p - 4 q}{2 q}$ (1 mark)
- States the contradiction and concludes (1 mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerCircles
A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that $A B + C D = A D + B C$ .
Figure in the paper: Quadrilateral ABCD with a circle inscribed inside it, touching all four sides. Vertices A (bottom left), B (bottom right), C (top right), D (top left). See it in the official paper.
Try it first, then show our solutionSolution
1. Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively. Tangents drawn from an external point to a circle are equal, so $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ .
  1 mark
2. $A B + C D = (A P + P B) + (C R + R D)$
  ½ mark
3. $= (A S + B Q) + (C Q + D S)$ [using the equal tangents]
  ½ mark
4. $= (A S + D S) + (B Q + C Q)$
  ½ mark
5. $= A D + B C$ . Hence $A B + C D = A D + B C$ .
  ½ mark
Answer: AB + CD = AD + BC (proved using equal tangents from external points).
Where the marks are
- Names points of contact and writes the four pairs of equal tangents (1 mark)
- Splits AB + CD into tangent segments (1/2 mark)
- Replaces with equal tangents (1/2 mark)
- Regroups the terms (1/2 mark)
- Concludes AD + BC (1/2 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerCircles
Prove that the parallelogram circumscribing a circle is a rhombus.
Try it first, then show our solutionSolution
1. Figure: Draw parallelogram ABCD with all four sides touching a circle. The points of contact are P on AB, Q on BC, R on CD and S on DA.
  Given: ABCD is a parallelogram circumscribing the circle.
  To prove: ABCD is a rhombus.
  ½ mark
2. Tangents drawn from an external point to a circle are equal in length. So:
  $A P = A S, B P = B Q, C R = C Q, D R = D S$
  1 mark
3. Adding these four equations:
  $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$
  $\Rightarrow A B + C D = A D + B C$
  ½ mark
4. In a parallelogram opposite sides are equal: $A B = C D$ and $B C = A D$ .
  So $2 A B = 2 B C$ , which gives $A B = B C$ .
  ½ mark
5. Therefore $A B = B C = C D = D A$ . A parallelogram with all sides equal is a rhombus, so ABCD is a rhombus. Hence proved.
  ½ mark
Answer: AB + CD = AD + BC together with AB = CD and BC = AD gives AB = BC = CD = DA, so ABCD is a rhombus
Where the marks are
- Correct labelled figure (1/2 mark)
- Writes the four equal-tangent relations (1 mark)
- Adds them to get $A B + C D = A D + B C$ (1/2 mark)
- Uses the opposite sides of the parallelogram to get $A B = B C$ (1/2 mark)
- Concludes all sides are equal, so ABCD is a rhombus (1/2 mark)
Open Q28(b) on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $\frac{1 - c o s θ}{1 + c o s θ} = (cosec θ - cot θ)^{2}$
Try it first, then show our solutionSolution
1. LHS $= \frac{1 - cos θ}{1 + cos θ} = \frac{( 1 - cos θ ) ( 1 - cos θ )}{( 1 + cos θ ) ( 1 - cos θ )} = \frac{( 1 - cos θ ) ^{2}}{1 - cos ^{2} θ}$
  1 mark
2. $= \frac{( 1 - cos θ ) ^{2}}{sin ^{2} θ}$ [since $1 - cos^{2} θ = sin^{2} θ$ ] $= (\frac{1 - cos θ}{sin θ})^{2}$
  1 mark
3. $= (\frac{1}{sin θ} - \frac{cos θ}{sin θ})^{2}$
  ½ mark
4. $= (cosec θ - cot θ)^{2} =$ RHS. Hence proved.
  ½ mark
Answer: LHS = $(cosec θ - cot θ)^{2}$ = RHS
Where the marks are
- Multiplies numerator and denominator by $1 - cos θ$ (1 mark)
- Uses $1 - cos^{2} θ = sin^{2} θ$ and writes as a square (1 mark)
- Splits into $\frac{1}{s i n θ} - \frac{c o s θ}{s i n θ}$ (1/2 mark)
- Concludes $(cosec θ - cot θ)^{2}$ (1/2 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $(1 + \frac{1}{t a n ^{2} A}) (1 + \frac{1}{c o t ^{2} A}) = \frac{1}{s i n ^{2} A - s i n ^{4} A}$
Try it first, then show our solutionSolution
1. LHS $= (1 + \frac{cos ^{2} A}{sin ^{2} A}) (1 + \frac{sin ^{2} A}{cos ^{2} A})$ [since $\frac{1}{t a n A} = \frac{c o s A}{s i n A}$ , $\frac{1}{c o t A} = \frac{s i n A}{c o s A}$ ]
  1 mark
2. $= (\frac{sin ^{2} A + cos ^{2} A}{sin ^{2} A}) (\frac{cos ^{2} A + sin ^{2} A}{cos ^{2} A}) = \frac{1}{sin ^{2} A} \times \frac{1}{cos ^{2} A}$ [since $sin^{2} A + cos^{2} A = 1$ ]
  1 mark
3. $= \frac{1}{sin ^{2} A ( 1 - sin ^{2} A )}$ [since $cos^{2} A = 1 - sin^{2} A$ ]
  ½ mark
4. $= \frac{1}{sin ^{2} A - sin ^{4} A} =$ RHS. Hence proved.
  ½ mark
Answer: LHS = $\frac{1}{s i n ^{2} A - s i n ^{4} A}$ = RHS
Where the marks are
- Rewrites in terms of sin and cos (1 mark)
- Uses $sin^{2} A + cos^{2} A = 1$ to get $\frac{1}{s i n ^{2} A c o s ^{2} A}$ (1 mark)
- Replaces $cos^{2} A$ by $1 - sin^{2} A$ (1/2 mark)
- Expands to the RHS (1/2 mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerPolynomials
Find the zeroes of the polynomial $p (x) = 2 x^{2} - 7 x - 15$ and verify the relationship between its coefficients and zeroes.
Try it first, then show our solutionSolution
1. $2 x^{2} - 7 x - 15 = 2 x^{2} - 10 x + 3 x - 15 = 2 x (x - 5) + 3 (x - 5) = (2 x + 3) (x - 5)$
  1 mark
2. Setting $p (x) = 0$ : zeroes are $α = - \frac{3}{2}$ and $β = 5$ .
  1 mark
3. Sum: $α + β = - \frac{3}{2} + 5 = \frac{7}{2}$ and $- \frac{coefficient of x}{coefficient of x ^{2}} = - \frac{( - 7 )}{2} = \frac{7}{2}$ . Verified.
  ½ mark
4. Product: $α β = - \frac{3}{2} \times 5 = - \frac{15}{2}$ and $\frac{constant term}{coefficient of x ^{2}} = \frac{- 15}{2}$ . Verified.
  ½ mark
Answer: Zeroes: $- \frac{3}{2}$ and 5; sum $= \frac{7}{2}$ , product $= - \frac{15}{2}$ (relationships verified).
Where the marks are
- Factorises as $(2 x + 3) (x - 5)$ (1 mark)
- Zeroes $- \frac{3}{2}$ and 5 (1 mark)
- Verifies sum of zeroes $= - b / a$ (1/2 mark)
- Verifies product of zeroes $= c / a$ (1/2 mark)
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Q31 · 3 marks · Short answerCoordinate Geometry
Prove that the points $A (- 1, 0)$ , $B (3, 1)$ , $C (2, 2)$ and $D (- 2, 1)$ are the vertices of a parallelogram ABCD. Is it also a rectangle?
Try it first, then show our solutionSolution
1. Mid-point of AC $= (\frac{- 1 + 2}{2}, \frac{0 + 2}{2}) = (\frac{1}{2}, 1)$
  ½ mark
2. Mid-point of BD $= (\frac{3 - 2}{2}, \frac{1 + 1}{2}) = (\frac{1}{2}, 1)$
  ½ mark
3. The diagonals AC and BD have the same mid-point, i.e. they bisect each other, so ABCD is a parallelogram.
  1 mark
4. $A C = (2 + 1)^{2} + (2 - 0)^{2} = 13$ , $B D = (- 2 - 3)^{2} + (1 - 1)^{2} = 25 = 5$ . Since $A C \neq = B D$ , the diagonals are not equal, so ABCD is not a rectangle.
  1 mark
Answer: ABCD is a parallelogram (diagonals bisect each other) but not a rectangle ( $A C = 13 \neq = B D = 5$ ).
Where the marks are
- Mid-point of AC = $(\frac{1}{2}, 1)$ (1/2 mark)
- Mid-point of BD = $(\frac{1}{2}, 1)$ (1/2 mark)
- Concludes parallelogram since diagonals bisect each other (1 mark)
- Finds AC = $13$ , BD = 5, so not a rectangle (1 mark)
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Section D

LA; internal choice in 32 and 35

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower. Also, find the distance between the building and the tower. (Use $3 = 1 \cdot 732$ )
Try it first, then show our solutionSolution
1. Figure: Let AB = 60 m be the building (A at the foot, B at the top) and CD = h m be the tower (C at the foot, D at the top), with AC = x m. Draw DE ⟂ AB, so DE = AC = x and AE = h, BE = 60 − h. Angle of depression of C from B is 60°, so ∠BCA = 60°; angle of depression of D from B is 30°, so ∠BDE = 30° (alternate angles).
  1 mark
2. In right $△ B A C$ : $tan 6 0^{\circ} = \frac{A B}{A C} = \frac{60}{x} \Rightarrow 3 = \frac{60}{x} \Rightarrow x = \frac{60}{3} = 203$ m ...(i)
  1½ marks
3. In right $△ B E D$ : $tan 3 0^{\circ} = \frac{B E}{D E} = \frac{60 - h}{x} \Rightarrow 60 - h = \frac{x}{3} = \frac{20 3}{3} = 20$ ...(ii)
  1½ marks
4. Distance between building and tower $= x = 203 = 20 \times 1.732 = 34.64$ m
  ½ mark
5. From (ii), $h = 60 - 20 = 40$ m. Height of the tower = 40 m.
  ½ mark
Answer: Height of tower = 40 m; distance between building and tower = $203$ = 34.64 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ} = \frac{60}{x}$ to get $x = 203$ (1½ marks)
- Uses $tan 3 0^{\circ} = \frac{60 - h}{x}$ to get $60 - h = 20$ (1½ marks)
- Distance = 34.64 m (1/2 mark)
- Height of tower = 40 m (1/2 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
The angle of elevation of the top of a building from a point A on the ground is $3 0^{\circ}$ . On moving a distance of 30 m towards its base to the point B, the angle of elevation changes to $4 5^{\circ}$ . Find the height of the building and the distance of its base from point A. (Use $3 = 1 \cdot 732$ )
Try it first, then show our solutionSolution
1. Figure: Let CD = h m be the building with foot C and top D. A and B are points on the ground in line with C, AB = 30 m, BC = x m, so AC = x + 30 m. ∠DAC = 30°, ∠DBC = 45°, ∠DCA = 90°.
  1 mark
2. In right $△ D C A$ : $tan 3 0^{\circ} = \frac{h}{x + 30} \Rightarrow \frac{1}{3} = \frac{h}{x + 30}$ ...(i)
  1½ marks
3. In right $△ D C B$ : $tan 4 5^{\circ} = \frac{h}{x} \Rightarrow h = x$ ...(ii)
  1 mark
4. Putting (ii) in (i): $3 h = h + 30 \Rightarrow h (3 - 1) = 30 \Rightarrow h = \frac{30}{3 - 1} \times \frac{3 + 1}{3 + 1} = 15 (3 + 1)$
  ½ mark
5. $h = 15 \times 2.732 = 40.98$ m. Height of the building = 40.98 m.
  ½ mark
6. Distance of the base from A $= x + 30 = 40.98 + 30 = 70.98$ m.
  ½ mark
Answer: Height of building = $15 (3 + 1)$ = 40.98 m; distance of base from A = 70.98 m
Where the marks are
- Correct labelled figure (1 mark)
- Forms $tan 3 0^{\circ} = \frac{h}{x + 30}$ (1½ marks)
- Forms $tan 4 5^{\circ} = \frac{h}{x}$ , so h = x (1 mark)
- Solves to $h = 15 (3 + 1)$ (1/2 mark)
- Height = 40.98 m (1/2 mark)
- Distance from A = 70.98 m (1/2 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerStatistics
Find the mean and the median of the following data:
Marks Number of Students
0 – 10 3
10 – 20 5
20 – 30 16
30 – 40 12
40 – 50 13
50 – 60 20
60 – 70 6
70 – 80 5
Try it first, then show our solutionSolution
1. Take assumed mean $a = 35$ , class size $h = 10$ , $u_{i} = \frac{x _{i} - 35}{10}$ .
  Marks $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$ cf
  0–10 5 3 −3 −9 3
  10–20 15 5 −2 −10 8
  20–30 25 16 −1 −16 24
  30–40 35 12 0 0 36
  40–50 45 13 1 13 49
  50–60 55 20 2 40 69
  60–70 65 6 3 18 75
  70–80 75 5 4 20 80
  Total 80 56
  2 marks
2. Mean $= a + h \cdot \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 35 + 10 \times \frac{56}{80} = 35 + 7 = 42$
  1½ marks
3. $\frac{n}{2} = \frac{80}{2} = 40$ . The cumulative frequency just greater than 40 is 49, so the median class is 40–50, with $l = 40$ , $c f = 36$ , $f = 13$ , $h = 10$ .
  ½ mark
4. Median $= l + (\frac{\frac{n}{2} - c f}{f}) \times h = 40 + \frac{40 - 36}{13} \times 10 = 40 + \frac{40}{13} \approx 40 + 3.08 = 43.08$ (≈ 43.1)
  1 mark
Answer: Mean = 42; Median ≈ 43.08 (about 43.1)
Where the marks are
- Correct table with class marks, $f_{i} u_{i}$ and cumulative frequencies (2 marks)
- Mean $= 35 + 10 \times \frac{56}{80} = 42$ (1½ marks)
- Identifies median class 40–50 (1/2 mark)
- Median $= 40 + \frac{4}{13} \times 10 \approx 43.1$ (1 mark)
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Q34 · 5 marks · Long answerTriangles
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ . Take a line parallel to $B C$ that meets $A B$ at $D$ and $A C$ at $E$ . Join $B E$ and $C D$ . Draw $E N ⊥ A B$ (N on AB) and $D M ⊥ A C$ (M on AC).
  ½ mark
2. Given: In $△ A B C$ , $D E ∥ B C$ , with $D E$ meeting $A B$ at $D$ and $A C$ at $E$ .
  ½ mark
3. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  ½ mark
4. Construction: Join $B E$ and $C D$ . Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  ½ mark
5. Proof: Use area $= \frac{1}{2} \times$ base $\times$ height.
  In $△ A D E$ and $△ B D E$ , the height from $E$ is $E N$ for both:
  $\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} \cdot A D \cdot E N}{\frac{1}{2} \cdot D B \cdot E N} = \frac{A D}{D B}$ ... (1)
  1 mark
6. In $△ A D E$ and $△ D E C$ , the height from $D$ is $D M$ for both:
  $\frac{a r ( A D E )}{a r ( D E C )} = \frac{\frac{1}{2} \cdot A E \cdot D M}{\frac{1}{2} \cdot E C \cdot D M} = \frac{A E}{E C}$ ... (2)
  1 mark
7. $△ B D E$ and $△ D E C$ have the same base $D E$ and lie between the same parallels $D E$ and $B C$ . So $a r (B D E) = a r (D E C)$ ... (3)
  From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: Proved: if DE ∥ BC in △ABC, then AD/DB = AE/EC (Basic Proportionality Theorem).
Where the marks are
- Draws a correct labelled figure (½ mark)
- States what is given (½ mark)
- States what is to be proved (½ mark)
- States the construction: join BE and CD, draw EN ⊥ AB and DM ⊥ AC (½ mark)
- Expresses $a r (A D E) / a r (B D E)$ as $A D / D B$ (1 mark)
- Expresses $a r (A D E) / a r (D E C)$ as $A E / E C$ (1 mark)
- Uses equal areas of triangles on the same base between the same parallels to conclude (1 mark)
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Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerArithmetic Progressions
If the sum of the first 7 terms of an A.P. is $- 14$ and that of 11 terms is $- 55$ , then find the sum of its first 'n' terms.
Try it first, then show our solutionSolution
1. Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ : $S_{7} = \frac{7}{2} (2 a + 6 d) = - 14 \Rightarrow 2 a + 6 d = - 4 \Rightarrow a + 3 d = - 2$ ...(i)
  1 mark
2. $S_{11} = \frac{11}{2} (2 a + 10 d) = - 55 \Rightarrow 2 a + 10 d = - 10 \Rightarrow a + 5 d = - 5$ ...(ii)
  1 mark
3. Subtracting (i) from (ii): $2 d = - 3 \Rightarrow d = - \frac{3}{2}$
  1 mark
4. From (i): $a = - 2 - 3 (- \frac{3}{2}) = - 2 + \frac{9}{2} = \frac{5}{2}$
  1 mark
5. $S_{n} = \frac{n}{2} [2 \cdot \frac{5}{2} + (n - 1) (- \frac{3}{2})] = \frac{n}{2} [\frac{10 - 3 n + 3}{2}] = \frac{n}{4} (13 - 3 n)$
  1 mark
Answer: $S_{n} = \frac{n}{4} (13 - 3 n)$ (with $a = \frac{5}{2}$ , $d = - \frac{3}{2}$ )
Where the marks are
- Forms equation for $S_{7}$ (1 mark)
- Forms equation for $S_{11}$ (1 mark)
- Finds $d = - \frac{3}{2}$ (1 mark)
- Finds $a = \frac{5}{2}$ (1 mark)
- Writes $S_{n} = \frac{n}{4} (13 - 3 n)$ (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerArithmetic Progressions
In an A.P., the sum of the first 'n' terms is $3 n^{2} + n$ . Find the first term and the common difference of the A.P. Hence, find its 15th term.
Try it first, then show our solutionSolution
1. $S_{n} = 3 n^{2} + n$ . First term $a_{1} = S_{1} = 3 (1)^{2} + 1 = 4$
  1 mark
2. $S_{2} = a_{1} + a_{2} = 3 (2)^{2} + 2 = 14$
  1 mark
3. $a_{2} = S_{2} - S_{1} = 14 - 4 = 10$
  1 mark
4. Common difference $d = a_{2} - a_{1} = 10 - 4 = 6$
  1 mark
5. $a_{15} = a + 14 d = 4 + 14 \times 6 = 4 + 84 = 88$
  1 mark
Answer: First term = 4, common difference = 6, 15th term = 88
Where the marks are
- $a_{1} = S_{1} = 4$ (1 mark)
- $S_{2} = 14$ (1 mark)
- $a_{2} = 10$ (1 mark)
- $d = 6$ (1 mark)
- $a_{15} = 4 + 14 (6) = 88$ (1 mark)
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Marks	Number of Students
0 – 10	3
10 – 20	5
20 – 30	16
30 – 40	12
40 – 50	13
50 – 60	20
60 – 70	6
70 – 80	5

Marks	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$	cf
0–10	5	3	−3	−9	3
10–20	15	5	−2	−10	8
20–30	25	16	−1	−16	24
30–40	35	12	0	0	36
40–50	45	13	1	13	49
50–60	55	20	2	40	69
60–70	65	6	3	18	75
70–80	75	5	4	20	80
Total		80		56

Section E

Case study; internal choice in each case (2-mark sub-parts)

Q36 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
Case Study 1. Flower beds look beautiful growing in gardens. One such circular park of radius 'r' m, has two segments with flowers. One segment which subtends an angle of $9 0^{\circ}$ at the centre is full of red roses, while the other segment with central angle $6 0^{\circ}$ is full of yellow coloured flowers. [See figure]
It is given that the combined area of the two segments (of flowers) is $256 \frac{2}{3}$ sq m.
Based on the above, answer the following questions:
1. (i)
  Write an equation representing the total area of the two segments in terms of 'r'.
  1 mark
2. (ii)
  Find the value of 'r'.
  1 mark
3. (iii)
  (a) Find the area of the segment with red roses.
  2 marks
4. OR
  (iii)
  (b) Find the area of the segment with yellow flowers.
  2 marks
Figure in the paper: A circle with centre O. A horizontal chord near the top bounds a small segment shaded and labelled 'Yellow flowers'. A horizontal chord near the bottom bounds a segment shaded and labelled 'Red roses'. The red segment subtends $9 0^{\circ}$ at the centre and the yellow segment subtends $6 0^{\circ}$ at the centre. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Area of a segment of central angle $θ$ = area of sector − area of triangle $= \frac{θ}{36 0 ^{\circ}} π r^{2} - \frac{1}{2} r^{2} sin θ$ .
  Red segment ( $9 0^{\circ}$ ): $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2}$ . Yellow segment ( $6 0^{\circ}$ , triangle is equilateral): $\frac{1}{6} π r^{2} - \frac{3}{4} r^{2}$ .
  Equation: $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2} + \frac{1}{6} π r^{2} - \frac{3}{4} r^{2} = 256 \frac{2}{3} = \frac{770}{3}$
  1 mark
2. (ii) $(\frac{5 π}{12} - \frac{1}{2} - \frac{3}{4}) r^{2} = \frac{770}{3}$ . With $π = \frac{22}{7}$ , $3 = 1.732$ : $(1.3095 - 0.5 - 0.4330) r^{2} = 256.67 \Rightarrow 0.3765 r^{2} = 256.67 \Rightarrow r^{2} \approx 681.7$ , so $r \approx 26.1$ m.
  1 mark
3. (iii)(a) Area of red segment $= (\frac{π}{4} - \frac{1}{2}) r^{2} = (\frac{11}{14} - \frac{1}{2}) r^{2} = \frac{2}{7} r^{2}$
  1 mark
4. $= \frac{2}{7} \times 681.7 \approx 194.8$ sq m (≈ 194.63 sq m if $r = 26.1$ is used).
  1 mark
5. (iii)(b) Area of yellow segment $= (\frac{π}{6} - \frac{3}{4}) r^{2} = (0.5238 - 0.4330) r^{2} = 0.0908 r^{2}$
  1 mark
6. $= 0.0908 \times 681.7 \approx 61.9$ sq m (check: $194.8 + 61.9 \approx 256.7$ sq m).
  1 mark
Answer: (i) $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2} + \frac{1}{6} π r^{2} - \frac{3}{4} r^{2} = 256 \frac{2}{3}$ ; (ii) r ≈ 26.1 m; (iii)(a) ≈ 194.8 sq m; (iii)(b) ≈ 61.9 sq m
Where the marks are
- Writes the sum of the two segment areas (sector − triangle) equated to $256 \frac{2}{3}$ (1 mark)
- Simplifies and finds r ≈ 26.1 m (1 mark)
- (iii)(a) Correct expression $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2}$ and value ≈ 194.6–194.8 sq m (2 marks)
- (iii)(b) Correct expression $\frac{1}{6} π r^{2} - \frac{3}{4} r^{2}$ and value ≈ 62 sq m (2 marks)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedCircles
Case Study 2. Circles play an important part in our life. When a circular object is hung on the wall with a cord at nail N, the cords NA and NB work like tangents. Observe the figure, given that $∠$ ANO = $3 0^{\circ}$ and OA = 5 cm.
Based on the above, answer the following questions:
1. (i)
  Find the distance AN.
  1 mark
2. (ii)
  Find the measure of $∠$ AOB.
  1 mark
3. (iii)
  (a) Find the total length of cords NA, NB and the chord AB.
  2 marks
4. OR
  (iii)
  (b) If $∠$ ANO is $4 5^{\circ}$ , then name the type of quadrilateral OANB. Justify your answer.
  2 marks
Figure in the paper: A circle with centre O. External point N is above the circle; tangents NA and NB touch the circle at A and B. Radii OA and OB are drawn. Dashed segment NO goes from N to O and meets chord AB at its midpoint M. Chord AB is drawn horizontally above O. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $O A ⊥ N A$ [tangent ⟂ radius], so in right $△ O A N$ : $tan 3 0^{\circ} = \frac{O A}{A N} \Rightarrow \frac{1}{3} = \frac{5}{A N}$
  ½ mark
2. $\Rightarrow A N = 53$ cm
  ½ mark
3. (ii) NO bisects $∠ A N B$ [tangents from an external point], so $∠ B N O = 3 0^{\circ}$ and $∠ A N B = 6 0^{\circ}$ . In quadrilateral OANB, $∠ O A N = ∠ O B N = 9 0^{\circ}$ , so $∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
  1 mark
4. (iii)(a) $N A = N B$ [tangents from N] and $∠ A N B = 6 0^{\circ}$ , so $∠ N A B = ∠ N B A = 6 0^{\circ}$ ; $△ N A B$ is equilateral.
  ½ mark
5. Hence $A B = N A = N B = 53$ cm.
  1 mark
6. Total length $= N A + N B + A B = 3 \times 53 = 153$ cm $\approx 25.98$ cm.
  ½ mark
7. (iii)(b) If $∠ A N O = 4 5^{\circ}$ , then $∠ A N B = 9 0^{\circ}$ , and $∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ .
  ½ mark
8. So every angle of OANB is $9 0^{\circ}$ ( $∠ O A N = ∠ O B N = 9 0^{\circ}$ as tangent ⟂ radius).
  1 mark
9. Also $O A = O B$ (radii) and $N A = N B$ ; in right $△ O A N$ with $4 5^{\circ}$ , $N A = O A = 5$ cm. All sides equal and all angles right, so OANB is a square.
  ½ mark
Answer: (i) AN = 5√3 cm; (ii) ∠AOB = 120°; (iii)(a) 15√3 cm (≈ 25.98 cm); (iii)(b) OANB is a square
Where the marks are
- Uses $tan 3 0^{\circ} = O A / A N$ (½ mark) and gets AN = 5√3 cm (½ mark)
- Finds ∠ANB = 60° and ∠AOB = 120° (1 mark)
- (iii)(a) Shows △NAB equilateral (½ mark), AB = 5√3 cm (1 mark), total 15√3 cm (½ mark)
- (iii)(b) ∠AOB = 90° (½ mark), all angles 90° (1 mark), adjacent sides equal ⇒ square (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Case Study 3. A wooden toy is shown in the picture. This is a cuboidal wooden block of dimensions $14 cm \times 17 cm \times 4 cm$ . On its top there are seven cylindrical hollows for bees to fit in. Each cylindrical hollow is of height 3 cm and radius 2 cm.
Based on the above, answer the following questions:
1. (i)
  Find the volume of wood carved out to make one cylindrical hollow.
  1 mark
2. (ii)
  Find the lateral surface area of the cuboid to paint it with green colour.
  1 mark
3. (iii)
  (a) Find the volume of wood in the remaining cuboid after carving out seven cylindrical hollows.
  2 marks
4. OR
  (iii)
  (b) Find the surface area of the top surface of the cuboid to be painted yellow.
  2 marks
Figure in the paper: Photograph of a yellow cuboidal wooden toy with a honeycomb pattern on its front face. Seven circular hollows are arranged in a honeycomb pattern on the top face, each hollow is coloured-outlined, and colourful bee-shaped rods/pegs stand below the block. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Volume of one hollow $= π r^{2} h = \frac{22}{7} \times 2 \times 2 \times 3 = \frac{264}{7}$ cm³ $\approx 37.71$ cm³.
  1 mark
2. (ii) Lateral surface area of cuboid $= 2 (l + b) h = 2 (14 \times 4 + 17 \times 4) = 2 (56 + 68) = 248$ cm².
  1 mark
3. (iii)(a) Volume of 7 hollows $= 7 \times \frac{264}{7} = 264$ cm³.
  ½ mark
4. Volume of the full cuboid $= l \times b \times h = 14 \times 17 \times 4 = 952$ cm³.
  1 mark
5. Remaining wood $= 952 - 264 = 688$ cm³.
  ½ mark
6. (iii)(b) Top area to be painted = area of top face − area of 7 circular openings $= (14 \times 17) - 7 \times \frac{22}{7} \times 2^{2} = 238 - 88$
  1 mark
7. $= 150$ cm².
  1 mark
Answer: (i) 264/7 cm³ ≈ 37.71 cm³; (ii) 248 cm²; (iii)(a) 688 cm³; (iii)(b) 150 cm²
Where the marks are
- Volume of one cylindrical hollow = 264/7 cm³ (1 mark)
- LSA = 2(14×4 + 17×4) = 248 cm² (1 mark)
- (iii)(a) Volume of 7 hollows 264 cm³ (½ mark), cuboid volume 952 cm³ (1 mark), remaining 688 cm³ (½ mark)
- (iii)(b) Expression 14×17 − 7πr² with values (1 mark), answer 150 cm² (1 mark)
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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.