MathsCBSE Class 10 · Standard · Board exam 2024 (session 2023-24)

CBSE Class 10 Mathematics Standard: 2024 board paper, set 30/2/1

The question paper set in the CBSE Class 10 board exam of 2024, 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/2024/X/MATHEMATICS_STANDARD.zip (a zip; this paper is the file 30-2-1(Mathematics Standard).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 23 and 25
C	26-31	Short answer; internal choice in 26 and 28
D	32-35	Long answer; internal choice in 32 and 33
E	36-38	Case study; internal choice in 2-mark subparts of each case

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choices are given in 2 questions in Section B, 2 in Section C, 2 in Section D and 3 in Section E (2-mark subparts of each case study).

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of $k$ for which the system of equations $3 x - y + 8 = 0$ and $6 x - k y + 16 = 0$ has infinitely many solutions, is
1. (A) $- 2$
2. (B) $2$
3. (C) $\frac{1}{2}$
4. (D) $- \frac{1}{2}$
Try it first, then show our solutionSolution
1. For infinitely many solutions: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , i.e. $\frac{3}{6} = \frac{- 1}{- k} = \frac{8}{16}$ .
2. $\frac{1}{k} = \frac{1}{2} \Rightarrow k = 2$ . Option (B).
  1 mark
Answer: k = 2, option (B)
Where the marks are
- Uses equal ratios of coefficients to get $k = 2$ (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQCoordinate Geometry
Point P divides the line segment joining the points $A (4, - 5)$ and $B (1, 2)$ in the ratio $5 : 2$ . Co-ordinates of point P are
1. (A) $(\frac{5}{2}, \frac{- 3}{2})$
2. (B) $(\frac{11}{7}, 0)$
3. (C) $(\frac{13}{7}, 0)$
4. (D) $(0, \frac{13}{7})$
Try it first, then show our solutionSolution
1. Section formula: $x = \frac{5 ( 1 ) + 2 ( 4 )}{7} = \frac{13}{7}$ , $y = \frac{5 ( 2 ) + 2 ( - 5 )}{7} = 0$ .
  1 mark
2. $P (\frac{13}{7}, 0)$ , option (C).
Answer: (13/7, 0), option (C)
Where the marks are
- Correct use of section formula giving $(13/7, 0)$ (1 mark)
Q3 · 1 mark · MCQArithmetic Progressions
The common difference of an A.P. in which $a_{15} - a_{11} = 48$ , is
1. (A) $12$
2. (B) $16$
3. (C) $- 12$
4. (D) $- 16$
Try it first, then show our solutionSolution
1. $a_{15} - a_{11} = (a + 14 d) - (a + 10 d) = 4 d = 48$
2. $d = 12$ , option (A).
  1 mark
Answer: 12, option (A)
Where the marks are
- Gets $4 d = 48$ , $d = 12$ (1 mark)
Q4 · 1 mark · MCQQuadratic Equations
The quadratic equation $x^{2} + x + 1 = 0$ has ______ roots.
1. (A)real and equal
2. (B)irrational
3. (C)real and distinct
4. (D)not-real
Try it first, then show our solutionSolution
1. $D = b^{2} - 4 a c = 1 - 4 = - 3 < 0$
2. Negative discriminant ⇒ no real roots. Option (D).
  1 mark
Answer: not-real, option (D)
Where the marks are
- Finds $D = - 3 < 0$ , so roots not real (1 mark)
Q5 · 1 mark · MCQReal Numbers
If the HCF $(2520, 6600) = 40$ and LCM $(2520, 6600) = 252 \times k$ , then the value of $k$ is
1. (A) $1650$
2. (B) $1600$
3. (C) $165$
4. (D) $1625$
Try it first, then show our solutionSolution
1. HCF × LCM = product: $40 \times 252 k = 2520 \times 6600$
2. $k = \frac{2520 \times 6600}{40 \times 252} = 10 \times 165 = 1650$ . Option (A).
  1 mark
Answer: 1650, option (A)
Where the marks are
- Uses HCF×LCM = product to get k = 1650 (1 mark)
Q6 · 1 mark · MCQTriangles
In the given figure $△ A B C$ is shown. $D E$ is parallel to $B C$ . If $A D = 5$ cm, $D B = 2.5$ cm and $B C = 12$ cm, then $D E$ is equal to
1. (A) $10$ cm
2. (B) $6$ cm
3. (C) $8$ cm
4. (D) $7.5$ cm
Figure in the paper: Triangle ABC with vertex A at the top, B bottom-left, C bottom-right. D lies on side AB and E lies on side AC, with segment DE drawn parallel to base BC. Labels: AD = 5 cm (along AD), DB = 2.5 cm (along DB), BC = 12 cm along the base. See it in the official paper.
Try it first, then show our solutionSolution
1. $D E ∥ B C$ ⇒ $△ A D E \sim △ A B C$ (AA), so $\frac{D E}{B C} = \frac{A D}{A B} = \frac{5}{7.5} = \frac{2}{3}$ .
2. $D E = \frac{2}{3} \times 12 = 8$ cm. Option (C).
  1 mark
Answer: 8 cm, option (C)
Where the marks are
- Uses similarity ratio AD/AB to get DE = 8 cm (1 mark)
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Q7 · 1 mark · MCQIntroduction to Trigonometry
If $sin θ = cos θ$ , $(0^{\circ} < θ < 9 0^{\circ})$ , then value of $(sec θ \cdot sin θ)$ is :
1. (A) $\frac{1}{2}$
2. (B) $2$
3. (C) $1$
4. (D) $0$
Try it first, then show our solutionSolution
1. $sec θ sin θ = \frac{s i n θ}{c o s θ} = tan θ$
2. $sin θ = cos θ \Rightarrow tan θ = 1$ . Option (C).
  1 mark
Answer: 1, option (C)
Where the marks are
- Recognises expression = tan θ = 1 (1 mark)
Q8 · 1 mark · MCQProbability
Two dice are rolled together. The probability of getting the sum of the two numbers to be more than 10, is
1. (A) $\frac{1}{9}$
2. (B) $\frac{1}{6}$
3. (C) $\frac{7}{12}$
4. (D) $\frac{1}{12}$
Try it first, then show our solutionSolution
1. Sum > 10: (5,6), (6,5), (6,6) → 3 outcomes of 36.
2. $P = \frac{3}{36} = \frac{1}{12}$ . Option (D).
  1 mark
Answer: 1/12, option (D)
Where the marks are
- Counts 3 favourable of 36 outcomes (1 mark)
Q9 · 1 mark · MCQPolynomials
If $α$ and $β$ are zeroes of the polynomial $5 x^{2} + 3 x - 7$ , the value of $\frac{1}{α} + \frac{1}{β}$ is
1. (A) $- \frac{3}{7}$
2. (B) $\frac{3}{5}$
3. (C) $\frac{3}{7}$
4. (D) $- \frac{5}{7}$
Try it first, then show our solutionSolution
1. $α + β = - \frac{3}{5}$ , $α β = - \frac{7}{5}$
2. $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{3}{7}$ . Option (C).
  1 mark
Answer: 3/7, option (C)
Where the marks are
- Uses sum and product of zeroes to get 3/7 (1 mark)
Q10 · 1 mark · MCQTriangles
The perimeters of two similar triangles $A B C$ and $P QR$ are $56$ cm and $48$ cm respectively. $\frac{P Q}{A B}$ is equal to
1. (A) $\frac{7}{8}$
2. (B) $\frac{6}{7}$
3. (C) $\frac{7}{6}$
4. (D) $\frac{8}{7}$
Try it first, then show our solutionSolution
1. For similar triangles, ratio of perimeters = ratio of corresponding sides.
2. $\frac{P Q}{A B} = \frac{48}{56} = \frac{6}{7}$ . Option (B).
  1 mark
Answer: 6/7, option (B)
Where the marks are
- Uses perimeter ratio = side ratio: 6/7 (1 mark)
Q11 · 1 mark · MCQTriangles
$A B$ and $C D$ are two chords of a circle intersecting at $P$ . Choose the correct statement from the following:
1. (A) $△ A D P \sim △ C B A$
2. (B) $△ A D P \sim △ B P C$
3. (C) $△ A D P \sim △ B C P$
4. (D) $△ A D P \sim △ C B P$
Figure in the paper: Circle with chords AB and CD intersecting at point P inside the circle. Points on the circle are labelled D (upper left), B (right), C (lower, near P) and A (bottom). Segments AD, AC, BC and the chord lines meet at P. The centre of the circle is marked with a dot but is not labelled. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A D P$ and $△ C B P$ : $∠ A P D = ∠ C P B$ (vertically opposite), $∠ D A P = ∠ D A B = ∠ D C B = ∠ B C P$ (angles in same segment on arc DB).
2. By AA, $△ A D P \sim △ C B P$ (A↔C, D↔B, P↔P). Option (D).
  1 mark
Answer: △ADP ~ △CBP, option (D)
Where the marks are
- Matches vertices correctly via AA similarity (1 mark)
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Q12 · 1 mark · MCQStatistics
If value of each observation in a data is increased by 2, then median of the new data
1. (A)increases by 2
2. (B)increases by $2 n$
3. (C)remains same
4. (D)decreases by 2
Try it first, then show our solutionSolution
1. Adding 2 to every value keeps their order, so the middle value also increases by 2. Option (A).
  1 mark
Answer: increases by 2, option (A)
Where the marks are
- Median shifts by the same constant (1 mark)
Q13 · 1 mark · MCQProbability
A box contains cards numbered 6 to 55. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is
1. (A) $\frac{7}{50}$
2. (B) $\frac{7}{55}$
3. (C) $\frac{1}{10}$
4. (D) $\frac{5}{49}$
Try it first, then show our solutionSolution
1. Total cards = 55 − 6 + 1 = 50. Perfect squares: 9, 16, 25, 36, 49 → 5.
2. $P = \frac{5}{50} = \frac{1}{10}$ . Option (C).
  1 mark
Answer: 1/10, option (C)
Where the marks are
- 50 cards, 5 perfect squares, probability 1/10 (1 mark)
Q14 · 1 mark · MCQCircles
In the given figure, tangents $P A$ and $P B$ to the circle centred at $O$ , from point $P$ are perpendicular to each other. If $P A = 5$ cm, then length of $A B$ is equal to
1. (A) $5$ cm
2. (B) $52$ cm
3. (C) $25$ cm
4. (D) $10$ cm
Figure in the paper: Circle centred at O. From external point P, two tangents touch the circle at A (upper) and B (lower). The angle at P is marked as a right angle. Radii OA and OB are drawn. The chord AB is drawn dashed. PA = 5 cm is the tangent length from P to A. See it in the official paper.
Try it first, then show our solutionSolution
1. $P A = P B = 5$ cm [tangents from external point equal], $∠ A P B = 9 0^{\circ}$ .
2. $A B = 5^{2} + 5^{2} = 52$ cm. Option (B).
  1 mark
Answer: 5√2 cm, option (B)
Where the marks are
- Uses PA = PB and Pythagoras to get 5√2 cm (1 mark)
Q15 · 1 mark · MCQCoordinate Geometry
$X O Y Z$ is a rectangle with vertices $X (- 3, 0)$ , $O (0, 0)$ , $Y (0, 4)$ and $Z (x, y)$ . The length of its each diagonal is
1. (A) $5$ units
2. (B) $5$ units
3. (C) $x^{2} + y^{2}$ units
4. (D) $4$ units
Try it first, then show our solutionSolution
1. Diagonal XY $= (0 + 3)^{2} + (4 - 0)^{2} = 25 = 5$ units; diagonals of a rectangle are equal. Option (A).
  1 mark
Answer: 5 units, option (A)
Where the marks are
- Computes diagonal XY = 5 units (1 mark)
Q16 · 1 mark · MCQArithmetic Progressions
Which term of the A.P. $- 29, - 26, - 23, \dots, 61$ is $16$ ?
1. (A) $1 1^{th}$
2. (B) $1 6^{th}$
3. (C) $1 0^{th}$
4. (D) $3 1^{st}$
Try it first, then show our solutionSolution
1. $a = - 29$ , $d = 3$ : $- 29 + (n - 1) 3 = 16 \Rightarrow n - 1 = 15$
2. $n = 16$ , option (B).
  1 mark
Answer: 16th term, option (B)
Where the marks are
- Solves $a_{n} = 16$ to get n = 16 (1 mark)
Q17 · 1 mark · MCQCircles
In the given figure, $A T$ is tangent to a circle centred at $O$ . If $∠ C A T = 4 0^{\circ}$ , then $∠ C B A$ is equal to
1. (A) $7 0^{\circ}$
2. (B) $5 0^{\circ}$
3. (C) $6 5^{\circ}$
4. (D) $4 0^{\circ}$
Figure in the paper: Circle with centre O. AB is a diameter drawn vertically, with B at the top and A at the bottom of the circle. The line AT is a tangent at A, extending horizontally to the right through T. C is a point on the circle; chord AC and chord BC are drawn. The angle between AC and tangent AT, at A, is marked as 40 degrees. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ B A T = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ B A C = 5 0^{\circ}$ ; $∠ A C B = 9 0^{\circ}$ [angle in semicircle].
2. $∠ C B A = 18 0^{\circ} - 9 0^{\circ} - 5 0^{\circ} = 4 0^{\circ}$ . Option (D).
  1 mark
Answer: 40°, option (D)
Where the marks are
- Uses tangent ⟂ radius and semicircle angle to get 40° (1 mark)
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Q18 · 1 mark · MCQStatistics
After an examination, a teacher wants to know the marks obtained by maximum number of the students in her class. She requires to calculate ______ of marks.
1. (A)median
2. (B)mode
3. (C)mean
4. (D)range
Try it first, then show our solutionSolution
1. The value occurring most often (highest frequency) is the mode. Option (B).
  1 mark
Answer: mode, option (B)
Where the marks are
- Identifies mode (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Directions: In Question 19 and 20, Assertion (A) and Reason (R) are given. Select the correct option from the following: (A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true. Reason (R) does not give correct explanation of (A). (C) Assertion (A) is true but Reason (R) is not true. (D) Assertion (A) is not true but Reason (R) is true.
Assertion (A): If $sin A = \frac{1}{3}$ $(0^{\circ} < A < 9 0^{\circ})$ , then the value of $cos A$ is $\frac{2 2}{3}$ .
Reason (R): For every angle $θ$ , $sin^{2} θ + cos^{2} θ = 1$ .
1. (A)Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true. Reason (R) does not give correct explanation of (A).
3. (C)Assertion (A) is true but Reason (R) is not true.
4. (D)Assertion (A) is not true but Reason (R) is true.
Try it first, then show our solutionSolution
1. $cos A = 1 - \frac{1}{9} = \frac{2 2}{3}$ , so A is true.
2. R is the identity used to get this, so R correctly explains A. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both true and R explains A (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedSurface Areas and Volumes
Directions: In Question 19 and 20, Assertion (A) and Reason (R) are given. Select the correct option from the following: (A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true. Reason (R) does not give correct explanation of (A). (C) Assertion (A) is true but Reason (R) is not true. (D) Assertion (A) is not true but Reason (R) is true.
Assertion (A): Two cubes each of edge length 10 cm are joined together. The total surface area of newly formed cuboid is $1200$ cm $^{2}$ .
Reason (R): Area of each surface of a cube of side 10 cm is $100$ cm $^{2}$ .
1. (A)Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true. Reason (R) does not give correct explanation of (A).
3. (C)Assertion (A) is true but Reason (R) is not true.
4. (D)Assertion (A) is not true but Reason (R) is true.
Try it first, then show our solutionSolution
1. Cuboid 20 × 10 × 10: TSA $= 2 (200 + 100 + 200) = 1000$ cm², not 1200, so A is false.
2. Each face of the cube = $1 0^{2} = 100$ cm², R true. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- TSA = 1000 cm², A false, R true (1 mark)

Section B

Very short answer; internal choice in 23 and 25

Q21 · 2 marks · Very short answerReal Numbers
Can the number $(15)^{n}$ , $n$ being a natural number, end with the digit $0$ ? Give reasons.
Try it first, then show our solutionSolution
1. Prime factorisation: $1 5^{n} = (3 \times 5)^{n} = 3^{n} \times 5^{n}$ .
  1 mark
2. A number ends with the digit $0$ only if it is divisible by $10 = 2 \times 5$ , so its prime factorisation must contain both $2$ and $5$ . By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), $1 5^{n}$ has only the primes $3$ and $5$ , with no $2$ . Hence $1 5^{n}$ can never end with the digit $0$ for any natural number $n$ .
  1 mark
Answer: No; $1 5^{n} = 3^{n} \cdot 5^{n}$ has no prime factor 2, so it cannot end with 0.
Where the marks are
- Writes $1 5^{n} = 3^{n} \times 5^{n}$ (1 mark)
- Reasons that ending in 0 needs both 2 and 5 as prime factors, and 2 is absent, so it cannot end in 0 (1 mark)
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Q22 · 2 marks · Very short answerCoordinate Geometry
Find the type of triangle $A B C$ formed whose vertices are $A (1, 0)$ , $B (- 5, 0)$ and $C (- 2, 5)$ .
Try it first, then show our solutionSolution
1. $A B = (- 5 - 1)^{2} + (0 - 0)^{2} = 36 = 6$ units
  ½ mark
2. $B C = (- 2 + 5)^{2} + (5 - 0)^{2} = 9 + 25 = 34$ units
  ½ mark
3. $C A = (1 + 2)^{2} + (0 - 5)^{2} = 9 + 25 = 34$ units
  ½ mark
4. Since $B C = C A = 34$ , two sides are equal. Also $34 + 34 \neq = 36$ , so it is not right-angled. Hence $△ A B C$ is an isosceles triangle.
  ½ mark
Answer: Isosceles triangle ( $B C = C A = 34$ , $A B = 6$ )
Where the marks are
- $A B = 6$ (½ mark)
- $B C = 34$ (½ mark)
- $C A = 34$ (½ mark)
- Concludes isosceles since $B C = C A$ (½ mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $2 sin^{2} 3 0^{\circ} sec 6 0^{\circ} + tan^{2} 6 0^{\circ}$ .
Try it first, then show our solutionSolution
1. Substitute $sin 3 0^{\circ} = \frac{1}{2}$ , $sec 6 0^{\circ} = 2$ , $tan 6 0^{\circ} = 3$ : $2 (\frac{1}{2})^{2} \times 2 + (3)^{2}$
  1½ marks
2. $= 2 \times \frac{1}{4} \times 2 + 3 = 1 + 3 = 4$
  ½ mark
Answer: 4
Where the marks are
- Correct substitution of standard values (1½ marks)
- Final value 4 (½ mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerIntroduction to Trigonometry
If $2 sin (A + B) = 3$ and $cos (A - B) = 1$ , then find the measures of angles $A$ and $B$ . $0^{\circ} \leq A, B, (A + B) \leq 9 0^{\circ}$ .
Try it first, then show our solutionSolution
1. $sin (A + B) = \frac{3}{2} = sin 6 0^{\circ} \Rightarrow A + B = 6 0^{\circ}$ …(1)
  ½ mark
2. $cos (A - B) = 1 = cos 0^{\circ} \Rightarrow A - B = 0^{\circ}$ …(2)
  ½ mark
3. Adding (1) and (2): $2 A = 6 0^{\circ} \Rightarrow A = 3 0^{\circ}$ ; then $B = 6 0^{\circ} - 3 0^{\circ} = 3 0^{\circ}$ .
  1 mark
Answer: $A = 3 0^{\circ}$ , $B = 3 0^{\circ}$
Where the marks are
- $A + B = 6 0^{\circ}$ (½ mark)
- $A - B = 0^{\circ}$ (½ mark)
- Solves to get $A = B = 3 0^{\circ}$ (1 mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerCircles
In the given figure, $A B$ and $C D$ are tangents to a circle centred at $O$ . Is $∠ B A C = ∠ D C A$ ? Justify your answer.
Figure in the paper: Circle with centre O. Two tangent lines touch the circle at A (lower-left) and C (lower-right). Line AB extends from A up-left to point B, and line CD extends from C up-right to point D. Chord AC is drawn horizontally between the two points of contact. See it in the official paper.
Try it first, then show our solutionSolution
1. Yes. Join $O A$ and $O C$ . $O A = O C$ [radii of the same circle].
  ½ mark
2. In $△ O A C$ , $∠ O A C = ∠ O C A$ [angles opposite equal sides].
  ½ mark
3. $∠ O A B = ∠ O C D = 9 0^{\circ}$ [tangent ⟂ radius at point of contact]. Adding: $∠ O A B + ∠ O A C = ∠ O C D + ∠ O C A$ .
  ½ mark
4. $\Rightarrow ∠ B A C = ∠ D C A$ . Hence the two angles are equal.
  ½ mark
Answer: Yes, $∠ B A C = ∠ D C A$ .
Where the marks are
- Joins OA, OC and uses OA = OC (½ mark)
- $∠ O A C = ∠ O C A$ (½ mark)
- Uses $∠ O A B = ∠ O C D = 9 0^{\circ}$ and adds angles (½ mark)
- Concludes $∠ B A C = ∠ D C A$ (½ mark)
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Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerCoordinate Geometry
In what ratio is the line segment joining the points $(3, - 5)$ and $(- 1, 6)$ divided by the line $y = x$ ?
Try it first, then show our solutionSolution
1. Let the line $y = x$ divide the join of $A (3, - 5)$ and $B (- 1, 6)$ at $P$ in the ratio $k : 1$ . By the section formula, $P = (\frac{- k + 3}{k + 1}, \frac{6 k - 5}{k + 1})$ .
  1 mark
2. $P$ lies on $y = x$ : $\frac{- k + 3}{k + 1} = \frac{6 k - 5}{k + 1} \Rightarrow - k + 3 = 6 k - 5 \Rightarrow 7 k = 8 \Rightarrow k = \frac{8}{7}$ .
  ½ mark
3. Required ratio $= \frac{8}{7} : 1 = 8 : 7$ (internally).
  ½ mark
Answer: 8 : 7
Where the marks are
- Coordinates of dividing point by section formula (1 mark)
- Uses x-coordinate = y-coordinate to get $k = 8/7$ (½ mark)
- Ratio 8 : 7 (½ mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerCoordinate Geometry
$A (3, 0)$ , $B (6, 4)$ and $C (- 1, 3)$ are vertices of a triangle $A B C$ . Find length of its median $B E$ .
Try it first, then show our solutionSolution
1. Median $B E$ joins $B$ to the mid-point $E$ of $A C$ : $E = (\frac{3 + ( - 1 )}{2}, \frac{0 + 3}{2}) = (1, \frac{3}{2})$ .
  1 mark
2. $B E = (6 - 1)^{2} + (4 - \frac{3}{2})^{2} = 25 + \frac{25}{4} = \frac{125}{4} = \frac{5 5}{2}$ units.
  1 mark
Answer: $B E = \frac{5 5}{2}$ units $(= 125/4)$
Where the marks are
- Mid-point E of AC is $(1, 3/2)$ (1 mark)
- Distance BE $= \frac{5 5}{2}$ (1 mark)
Open Q25(b) on its own page

Section C

Short answer; internal choice in 26 and 28

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerArithmetic Progressions
If the sum of first $m$ terms of an A.P. is same as sum of its first $n$ terms $(m \neq = n)$ , then show that the sum of its first $(m + n)$ terms is zero.
Try it first, then show our solutionSolution
1. Let first term $= a$ , common difference $= d$ . Given $S_{m} = S_{n}$ : $\frac{m}{2} [2 a + (m - 1) d] = \frac{n}{2} [2 a + (n - 1) d]$ .
  1 mark
2. $\Rightarrow 2 a (m - n) + d [(m^{2} - n^{2}) - (m - n)] = 0 \Rightarrow 2 a (m - n) + d (m - n) (m + n - 1) = 0$ .
  1 mark
3. Since $m \neq = n$ , divide by $(m - n)$ : $2 a + (m + n - 1) d = 0$ .
  ½ mark
4. $S_{m + n} = \frac{m + n}{2} [2 a + (m + n - 1) d] = \frac{m + n}{2} \times 0 = 0$ . Hence proved.
  ½ mark
Answer: $S_{m + n} = 0$ (shown)
Where the marks are
- Writes $S_{m} = S_{n}$ using sum formula (1 mark)
- Simplifies to $2 a (m - n) = - d (m - n) (m + n - 1)$ (1 mark)
- Gets $2 a + (m + n - 1) d = 0$ (½ mark)
- Concludes $S_{m + n} = 0$ (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerArithmetic Progressions
In an A.P., the sum of three consecutive terms is $24$ and the sum of their squares is $194$ . Find the numbers.
Try it first, then show our solutionSolution
1. Let the three consecutive terms be $a - d, a, a + d$ .
  ½ mark
2. Sum: $(a - d) + a + (a + d) = 24 \Rightarrow 3 a = 24 \Rightarrow a = 8$ .
  ½ mark
3. Sum of squares: $(8 - d)^{2} + 8^{2} + (8 + d)^{2} = 194 \Rightarrow 192 + 2 d^{2} = 194$ .
  1 mark
4. $\Rightarrow d^{2} = 1 \Rightarrow d = \pm 1$ .
  ½ mark
5. Numbers are $7, 8, 9$ (for $d = 1$ ) or $9, 8, 7$ (for $d = - 1$ ).
  ½ mark
Answer: 7, 8, 9 (or 9, 8, 7)
Where the marks are
- Takes terms as $a - d, a, a + d$ (½ mark)
- Finds $a = 8$ (½ mark)
- Forms the sum-of-squares equation (1 mark)
- $d = \pm 1$ (½ mark)
- Numbers 7, 8, 9 (½ mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerReal Numbers
Prove that $5$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $5$ is rational.
  Then $5 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are co-prime.
  ½ mark
2. Squaring gives $5 q^{2} = p^{2}$ . So $p^{2}$ is divisible by 5.
  Since 5 is prime, $p$ is divisible by 5 [if a prime divides $p^{2}$ , it divides $p$ ].
  So $p = 5 a$ for some integer $a$ .
  1 mark
3. Substitute: $25 a^{2} = 5 q^{2} \Rightarrow q^{2} = 5 a^{2}$ . So $q^{2}$ is divisible by 5, and hence $q$ is divisible by 5.
  ½ mark
4. Now 5 is a common factor of $p$ and $q$ . This contradicts the assumption that $p$ and $q$ are co-prime.
  Hence our assumption is wrong, and $5$ is irrational.
  1 mark
Answer: $5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 = p / q$ with $p, q$ co-prime (½ mark)
- Shows 5 divides $p$ , writes $p = 5 a$ (1 mark)
- Shows 5 divides $q$ (½ 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
In the given figure, $P Q$ is tangent to a circle centred at $O$ and $∠ B A Q = 3 0^{\circ}$ ; show that $B P = B Q$ .
Figure in the paper: Circle with centre O. AB is a diameter drawn horizontally, with A on the left and B on the right, and O at the midpoint. Point P lies on the extension of AB beyond B. PQ is a tangent touching the circle at Q (upper part). Chord AQ and chord BQ are drawn. The angle at A between AB and AQ is marked 30 degrees. See it in the official paper.
Try it first, then show our solutionSolution
1. Join $O Q$ . $O Q = O A$ [radii].
  ½ mark
2. $∴ ∠ O Q A = ∠ O A Q = 3 0^{\circ}$ [angles opposite equal sides].
  ½ mark
3. $∠ A QB = 9 0^{\circ}$ [angle in a semicircle], so $∠ O QB = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .
  ½ mark
4. $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ B QP = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ .
  ½ mark
5. $∠ QO B = ∠ O A Q + ∠ O Q A = 3 0^{\circ} + 3 0^{\circ} = 6 0^{\circ}$ [exterior angle of $△ A O Q$ ].
  ½ mark
6. In $△ O QP$ : $∠ QP O = 18 0^{\circ} - 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ . So in $△ B QP$ , $∠ B P Q = ∠ B QP = 3 0^{\circ}$ , hence $B P = B Q$ [sides opposite equal angles].
  ½ mark
Answer: $∠ B QP = ∠ B P Q = 3 0^{\circ}$ , so $B P = B Q$ .
Where the marks are
- Joins OQ, OQ = OA (½ mark)
- $∠ O Q A = 3 0^{\circ}$ (½ mark)
- $∠ O QB = 6 0^{\circ}$ (½ mark)
- $∠ B QP = 3 0^{\circ}$ (½ mark)
- $∠ QO B = 6 0^{\circ}$ (½ mark)
- $∠ B P Q = 3 0^{\circ} = ∠ B QP$ , hence BP = BQ (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerCircles
In the given figure, $A B$ , $B C$ , $C D$ and $D A$ are tangents to the circle with centre $O$ forming a quadrilateral $A B C D$ . Show that $∠ A O B + ∠ C O D = 18 0^{\circ}$ .
Figure in the paper: Quadrilateral ABCD circumscribing a circle with centre O, with each side AB, BC, CD and DA tangent to the circle. The diagonals AC and BD are drawn as segments passing through the centre O. See it in the official paper.
Try it first, then show our solutionSolution
1. Let the circle touch $A B, B C, C D, D A$ at $P, Q, R, S$ . Join $O P, O Q, O R, O S$ . Name $∠ P O B = ∠1$ , $∠ B O Q = ∠2$ , $∠ QO C = ∠3$ , $∠ C O R = ∠4$ , $∠ R O D = ∠5$ , $∠ D O S = ∠6$ , $∠ S O A = ∠7$ , $∠ A O P = ∠8$ .
  ½ mark
2. In $△ O P B$ and $△ O QB$ : $O P = O Q$ [radii], $O B$ common, $∠ O P B = ∠ O QB = 9 0^{\circ}$ [tangent ⟂ radius]. So $△ O P B ≅ △ O QB$ (RHS) $\Rightarrow ∠1 = ∠2$ (CPCT).
  1 mark
3. Similarly $∠3 = ∠4$ , $∠5 = ∠6$ , $∠7 = ∠8$ .
  ½ mark
4. Angles round point $O$ : $∠1 + ∠2 + \dots + ∠8 = 36 0^{\circ} \Rightarrow 2 (∠1 + ∠8 + ∠4 + ∠5) = 36 0^{\circ}$ .
  ½ mark
5. $\Rightarrow (∠1 + ∠8) + (∠4 + ∠5) = 18 0^{\circ}$ , i.e. $∠ A O B + ∠ C O D = 18 0^{\circ}$ . Hence proved.
  ½ mark
Answer: $∠ A O B + ∠ C O D = 18 0^{\circ}$ (proved)
Where the marks are
- Joins O to points of contact (½ mark)
- Proves $△ O P B ≅ △ O QB$ giving $∠1 = ∠2$ (1 mark)
- States the other three pairs of equal angles (½ mark)
- Sum of angles at O is $36 0^{\circ}$ (½ mark)
- Concludes $∠ A O B + ∠ C O D = 18 0^{\circ}$ (½ mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{1 + s e c θ - t a n θ}{1 + s e c θ + t a n θ} = \frac{1 - s i n θ}{c o s θ}$ .
Try it first, then show our solutionSolution
1. Replace $1$ in the numerator by $sec^{2} θ - tan^{2} θ$ : LHS $= \frac{( s e c ^{2} θ - t a n ^{2} θ ) + ( s e c θ - t a n θ )}{1 + s e c θ + t a n θ}$
  1 mark
2. $= \frac{( s e c θ - t a n θ ) ( s e c θ + t a n θ ) + ( s e c θ - t a n θ )}{1 + s e c θ + t a n θ} = \frac{( s e c θ - t a n θ ) ( s e c θ + t a n θ + 1 )}{1 + s e c θ + t a n θ} = sec θ - tan θ$
  1 mark
3. $= \frac{1}{c o s θ} - \frac{s i n θ}{c o s θ} = \frac{1 - s i n θ}{c o s θ} =$ RHS. Hence proved.
  1 mark
Answer: LHS = $sec θ - tan θ = \frac{1 - s i n θ}{c o s θ}$ = RHS
Where the marks are
- Uses $1 = sec^{2} θ - tan^{2} θ$ in the numerator (1 mark)
- Factorises and cancels to get $sec θ - tan θ$ (1 mark)
- Converts to $\frac{1 - s i n θ}{c o s θ}$ (1 mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerStatistics
In a test, the marks obtained by 100 students (out of 50) are given below:
Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Number of students 12 23 34 25 6
Find the mean marks of the students.
Try it first, then show our solutionSolution
1. Table with class marks $x_{i}$ and $f_{i} x_{i}$ :
  Marks $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  0–10 12 5 60
  10–20 23 15 345
  20–30 34 25 850
  30–40 25 35 875
  40–50 6 45 270
  Total 100 2400
  1½ marks
2. Mean $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2400}{100}$
  1 mark
3. $= 24$ marks
  ½ mark
Answer: Mean marks = 24
Where the marks are
- Correct table with $x_{i}$ , $f_{i} x_{i}$ , $\sum f_{i} = 100$ , $\sum f_{i} x_{i} = 2400$ (1½ marks)
- Uses mean formula $\sum f_{i} x_{i} / \sum f_{i}$ (1 mark)
- Mean = 24 (½ mark)
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Q31 · 3 marks · Short answerPair of Linear Equations in Two Variables
In a 2-digit number, the digit at the unit's place is $5$ less than the digit at the ten's place. The product of the digits is $36$ . Find the number.
Try it first, then show our solutionSolution
1. Let the ten's digit be $x$ ; then the unit's digit is $x - 5$ .
  ½ mark
2. Product of digits: $x (x - 5) = 36$
  ½ mark
3. $\Rightarrow x^{2} - 5 x - 36 = 0$
  ½ mark
4. $\Rightarrow (x - 9) (x + 4) = 0$
  ½ mark
5. $x = 9$ or $x = - 4$ ; a digit cannot be negative, so $x = 9$ and unit's digit $= 4$ .
  ½ mark
6. Required number $= 10 \times 9 + 4 = 94$ .
  ½ mark
Answer: 94
Where the marks are
- Unit's digit $= x - 5$ (½ mark)
- Equation $x (x - 5) = 36$ (½ mark)
- Quadratic $x^{2} - 5 x - 36 = 0$ (½ mark)
- Factorisation $(x - 9) (x + 4) = 0$ (½ mark)
- Rejects $- 4$ , $x = 9$ (½ mark)
- Number 94 (½ mark)
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Marks obtained	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Number of students	12	23	34	25	6

Marks	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0–10	12	5	60
10–20	23	15	345
20–30	34	25	850
30–40	25	35	875
40–50	6	45	270
Total	100		2400

Section D

Long answer; internal choice in 32 and 33

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerPair of Linear Equations in Two Variables
Using graphical method, solve the following system of equations: $3 x + y + 4 = 0$ and $3 x - y + 2 = 0$ .
Try it first, then show our solutionSolution
1. Line 1: $3 x + y + 4 = 0 \Rightarrow y = - 3 x - 4$ . Table: $x = 0, - 1, - 2$ give $y = - 4, - 1, 2$ . Plot $(0, - 4), (- 1, - 1), (- 2, 2)$ on graph paper and join them to get a straight line.
  2 marks
2. Line 2: $3 x - y + 2 = 0 \Rightarrow y = 3 x + 2$ . Table: $x = 0, - 1, 1$ give $y = 2, - 1, 5$ . Plot $(0, 2), (- 1, - 1), (1, 5)$ and join them to get the second straight line.
  2 marks
3. The two lines intersect at $(- 1, - 1)$ . Hence the solution is $x = - 1, y = - 1$ . (Check: $3 (- 1) + (- 1) + 4 = 0$ and $3 (- 1) - (- 1) + 2 = 0$ .)
  1 mark
Answer: $x = - 1$ , $y = - 1$
Where the marks are
- Correct table and correctly drawn line for $3 x + y + 4 = 0$ (2 marks)
- Correct table and correctly drawn line for $3 x - y + 2 = 0$ (2 marks)
- Reads intersection point: $x = - 1, y = - 1$ (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerPair of Linear Equations in Two Variables
Tara scored $40$ marks in a test, getting $3$ marks for each right answer and losing $1$ mark for each wrong answer. Had $4$ marks been awarded for each correct answer and $2$ marks been deducted for each wrong answer, then Tara would have scored $50$ marks. Assuming that Tara attempted all questions, find the total number of questions in the test.
Try it first, then show our solutionSolution
1. Let the number of right answers be $x$ and wrong answers be $y$ . First scheme: $3 x - y = 40$ …(1)
  1½ marks
2. Second scheme: $4 x - 2 y = 50$ , i.e. $2 x - y = 25$ …(2)
  1½ marks
3. Subtracting (2) from (1): $x = 15$ . From (1): $y = 3 (15) - 40 = 5$ .
  1 mark
4. Since all questions were attempted, total number of questions $= x + y = 15 + 5 = 20$ .
  1 mark
Answer: 20 questions
Where the marks are
- Equation $3 x - y = 40$ (1½ marks)
- Equation $4 x - 2 y = 50$ (1½ marks)
- Solves $x = 15$ , $y = 5$ (1 mark)
- Total questions = 20 (1 mark)
Open Q32(b) on its own page
Answer one of these 2 alternatives.
Q33(a) · 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)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerTriangles
Sides $A B$ and $A C$ and median $A D$ to $△ A B C$ are respectively proportional to sides $P Q$ and $P R$ and median $P M$ of another triangle $P QR$ . Show that $△ A B C \sim △ P QR$ .
Try it first, then show our solutionSolution
1. Figure: $△ A B C$ with median $A D$ ( $D$ mid-point of $B C$ ) and $△ P QR$ with median $P M$ ( $M$ mid-point of $QR$ ). Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ . Mark $∠ B A D = ∠3$ , $∠ D A C = ∠1$ , $∠ QP M = ∠4$ , $∠ M P R = ∠2$ .
  1 mark
2. Construction: Produce $A D$ to $E$ so that $D E = A D$ and join $E C$ ; produce $P M$ to $N$ so that $M N = P M$ and join $N R$ .
  ½ mark
3. In $△ A D B$ and $△ E D C$ : $A D = E D$ , $B D = C D$ , $∠ A D B = ∠ E D C$ (vertically opposite). So $△ A D B ≅ △ E D C$ (SAS) $\Rightarrow A B = E C$ and $∠ B A D = ∠ C E D$ . Similarly $P Q = N R$ and $∠ QP M = ∠ R N M$ .
  1 mark
4. Now $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M} \Rightarrow \frac{E C}{N R} = \frac{A C}{P R} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ . Hence $△ A E C \sim △ P N R$ (SSS similarity).
  1 mark
5. $\Rightarrow ∠ E A C = ∠ N P R$ , i.e. $∠1 = ∠2$ (corresponding angles of similar triangles).
  ½ mark
6. Also $∠ A E C = ∠ P N R$ ; since $∠3 = ∠ A E C$ and $∠4 = ∠ P N R$ , we get $∠3 = ∠4$ . Adding: $∠1 + ∠3 = ∠2 + ∠4$ , i.e. $∠ B A C = ∠ QP R$ .
  ½ mark
7. In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{A C}{P R}$ and $∠ A = ∠ P$ . Hence $△ A B C \sim △ P QR$ (SAS similarity).
  ½ mark
Answer: $△ A B C \sim △ P QR$ (SAS similarity) – proved
Where the marks are
- Correct figure with given data (1 mark)
- Construction: produce medians to double length and join EC, NR (½ mark)
- $△ A D B ≅ △ E D C$ giving AB = EC (and PQ = NR) (1 mark)
- $△ A E C \sim △ P N R$ by SSS (1 mark)
- $∠1 = ∠2$ (½ mark)
- $∠3 = ∠4$ and hence $∠ A = ∠ P$ (½ mark)
- Concludes $△ A B C \sim △ P QR$ by SAS (½ mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
From the top of a $45$ m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ . If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: $A B = 45$ m is the lighthouse with foot $A$ and top $B$ . Ships at $C$ and $D$ on opposite sides of $A$ , on a horizontal line through $A$ . Angles of depression from $B$ are $3 0^{\circ}$ (to $C$ ) and $6 0^{\circ}$ (to $D$ ), so $∠ B C A = 3 0^{\circ}$ , $∠ B D A = 6 0^{\circ}$ [alternate angles]. Let $C A = x$ , $A D = y$ .
  1 mark
2. In right $△ B A D$ : $tan 6 0^{\circ} = \frac{A B}{A D} \Rightarrow 3 = \frac{45}{y}$
  1 mark
3. $\Rightarrow y = \frac{45}{3} = 153$ m
  ½ mark
4. In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{C A} \Rightarrow \frac{1}{3} = \frac{45}{x}$
  1 mark
5. $\Rightarrow x = 453$ m
  ½ mark
6. Distance between ships $= x + y = 453 + 153 = 603 = 60 \times 1.73 = 103.8$ m
  1 mark
Answer: 103.8 m
Where the marks are
- Correct figure (1 mark)
- $tan 6 0^{\circ} = 45/ y$ (1 mark)
- $y = 153$ (½ mark)
- $tan 3 0^{\circ} = 45/ x$ (1 mark)
- $x = 453$ (½ mark)
- Distance $= 603 = 103.8$ m (1 mark)
Open Q34 on its own page
Q35 · 5 marks · Long answerAreas Related to Circles
The perimeter of a certain sector of a circle of radius $5.6$ m is $20.0$ m. Find the area of the sector.
Try it first, then show our solutionSolution
1. Perimeter of sector $= 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r = 20$ $\Rightarrow 11.2 + 2 \times \frac{22}{7} \times 5.6 \times \frac{θ}{36 0 ^{\circ}} = 20$
  2 marks
2. $\Rightarrow 35.2 \times \frac{θ}{36 0 ^{\circ}} = 8.8 \Rightarrow θ = \frac{8.8 \times 36 0 ^{\circ}}{35.2} = 9 0^{\circ}$
  1 mark
3. Area of sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{90}{360} \times \frac{22}{7} \times 5.6 \times 5.6$
  1 mark
4. $= \frac{1}{4} \times \frac{22}{7} \times 31.36 = 24.64 m^{2}$
  1 mark
Answer: 24.64 m²
Where the marks are
- Perimeter equation $2 r + \frac{2 π r θ}{360} = 20$ with values (2 marks)
- Finds $θ = 9 0^{\circ}$ (1 mark)
- Area formula with substitution (1 mark)
- Area $= 24.64$ m² (1 mark)
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Section E

Case study; internal choice in 2-mark subparts of each case

Q36 · 4 marks · Case-basedCompetency-basedPolynomials
A ball is thrown in the air so that $t$ seconds after it is thrown, its height $h$ metre above its starting point is given by the polynomial $h = 25 t - 5 t^{2}$ . Observe the graph of the polynomial and answer the following questions:
1. (i)
  Write zeroes of the given polynomial.
  1 mark
2. (ii)
  Find the maximum height achieved by ball.
  1 mark
3. (iii)
  (a) After throwing upward, how much time did the ball take to reach to the height of $30$ m?
  2 marks
4. OR
  (iii)
  (b) Find the two different values of $t$ when the height of the ball was $20$ m.
  2 marks
Figure in the paper: Graph of h(t) on vertical axis against t on horizontal axis. The curve is a downward-opening parabola starting at the origin 0, rising to a maximum point, and returning to the t-axis. A dashed vertical line marks the peak, labelled (5/2, 0) on the t-axis below it. The curve meets the t-axis again at (5, 0). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The graph of $h = 25 t - 5 t^{2} = 5 t (5 - t)$ meets the $t$ -axis at $t = 0$ and $t = 5$ . So the zeroes of the polynomial are $0$ and $5$ .
  1 mark
2. (ii) The maximum occurs at the vertex of the parabola, at $t = \frac{5}{2}$ (the dashed line in the graph). Substituting: $h = 25 \times \frac{5}{2} - 5 \times (\frac{5}{2})^{2}$
  ½ mark
3. $h = \frac{125}{2} - \frac{125}{4} = \frac{125}{4} = 31.25$ m. So the maximum height is $31.25$ m.
  ½ mark
4. (iii)(a) Put $h = 30$ : $25 t - 5 t^{2} = 30$
  ½ mark
5. Dividing by $- 5$ and rearranging: $t^{2} - 5 t + 6 = 0$
  ½ mark
6. Factorising: $(t - 2) (t - 3) = 0$ , so $t = 2$ or $t = 3$ .
  ½ mark
7. The ball goes up first and reaches 30 m at $t = 2$ s. It is at 30 m again at $t = 3$ s while coming down. So the ball takes 2 seconds to reach 30 m on the way up.
  ½ mark
8. (iii)(b) Put $h = 20$ : $25 t - 5 t^{2} = 20$
  ½ mark
9. Dividing by $- 5$ and rearranging: $t^{2} - 5 t + 4 = 0$
  ½ mark
10. Factorising: $(t - 1) (t - 4) = 0$
  ½ mark
11. $t = 1$ s (going up) and $t = 4$ s (coming down).
  ½ mark
Answer: (i) 0 and 5; (ii) 125/4 m = 31.25 m; (iii)(a) 2 seconds; (iii)(b) t = 1 s and t = 4 s
Where the marks are
- (i) Writes the zeroes as 0 and 5 (1 mark)
- (ii) Substitutes $t = \frac{5}{2}$ into $h$ (½ mark)
- (ii) Gets the maximum height $\frac{125}{4}$ m or 31.25 m (½ mark)
- (iii)(a) Forms $25 t - 5 t^{2} = 30$ (½ mark)
- (iii)(a) Reduces it to $t^{2} - 5 t + 6 = 0$ (½ mark)
- (iii)(a) Factorises as $(t - 2) (t - 3) = 0$ (½ mark)
- (iii)(a) Rejects $t = 3$ for the upward journey and states $t = 2$ s (½ mark)
- (iii)(b) Forms $25 t - 5 t^{2} = 20$ (½ mark)
- (iii)(b) Reduces it to $t^{2} - 5 t + 4 = 0$ (½ mark)
- (iii)(b) Factorises as $(t - 1) (t - 4) = 0$ (½ mark)
- (iii)(b) States $t = 1$ s and $t = 4$ s (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
The word 'circus' has the same root as 'circle'. In a closed circular area, various entertainment acts including human skill and animal training are presented before the crowd.
A circus tent is cylindrical upto a height of $8$ m and conical above it. The diameter of the base is $28$ m and total height of tent is $18.5$ m.
Based on the above, answer the following questions:
1. (i)
  Find slant height of the conical part.
  1 mark
2. (ii)
  Determine the floor area of the tent.
  1 mark
3. (iii)
  (a) Find area of the cloth used for making tent.
  2 marks
4. OR
  (iii)
  (b) Find total volume of air inside an empty tent.
  2 marks
Figure in the paper: Photograph of a circus tent. The lower part is a cylinder with vertical striped canvas walls, and the upper part is a conical roof with a pointed top, with the canvas stripes radiating from the apex. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Radius $r = \frac{28}{2} = 14$ m. Height of the conical part $h = 18.5 - 8 = 10.5$ m.
  ½ mark
2. Slant height $l = h^{2} + r^{2} = (10.5)^{2} + 1 4^{2} = 110.25 + 196 = 306.25 = 17.5$ m.
  ½ mark
3. (ii) Floor area $= π r^{2} = \frac{22}{7} \times 14 \times 14 = 616 m^{2}$ .
  1 mark
4. (iii)(a) Cloth used = CSA of cylinder + CSA of cone $= 2 π r H + π r l = 2 \times \frac{22}{7} \times 14 \times 8 + \frac{22}{7} \times 14 \times 17.5$
  1 mark
5. $= 704 + 770 = 1474 m^{2}$
  1 mark
6. (iii)(b) Volume of air = volume of cylinder + volume of cone $= π r^{2} H + \frac{1}{3} π r^{2} h = \frac{22}{7} \times 14 \times 14 \times 8 + \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times 10.5$
  1 mark
7. $= 4928 + 2156 = 7084 m^{3}$
  1 mark
Answer: (i) 17.5 m; (ii) 616 m²; (iii)(a) 1474 m²; (iii)(b) 7084 m³
Where the marks are
- (i) Finds the height of the cone as 10.5 m and the radius as 14 m (½ mark)
- (i) Gets slant height 17.5 m using $l = h^{2} + r^{2}$ (½ mark)
- (ii) Gets floor area $π r^{2} = 616$ m² (1 mark)
- (iii)(a) Writes cloth area = CSA of cylinder + CSA of cone, with correct substitution (1 mark)
- (iii)(a) Gets 1474 m² (1 mark)
- (iii)(b) Writes volume = cylinder + cone, with correct substitution (1 mark)
- (iii)(b) Gets 7084 m³ (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedProbability
In a survey on holidays, $120$ people were asked to state which type of transport they used on their last holiday. The following pie chart shows the results of the survey.
Observe the pie chart and answer the following questions:
1. (i)
  If one person is selected at random, find the probability that he/she travelled by bus or ship.
  1 mark
2. (ii)
  Which is most favourite mode of transport and how many people used it?
  1 mark
3. (iii)
  (a) A person is selected at random. If the probability that he did not use train is $\frac{4}{5}$ , find the number of people who used train.
  2 marks
4. OR
  (iii)
  (b) The probability that randomly selected person used aeroplane is $\frac{7}{60}$ . Find the revenue collected by air company at the rate of ₹ $5, 000$ per person.
  2 marks
Figure in the paper: Pie chart of 120 people by mode of transport. Sectors in order: Ship 33°, Aeroplane (no angle printed), Train (no angle printed), Bus 36°, and Car 177°. The Car sector is at the bottom of the circle, and the Ship, Aeroplane and Train sectors lie in the upper-left and upper portion. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The angles of the bus and ship sectors together are $3 6^{\circ} + 3 3^{\circ} = 6 9^{\circ}$ out of $36 0^{\circ}$ . So P(bus or ship) $= \frac{69}{360} = \frac{23}{120}$ .
  1 mark
2. (ii) Car has the largest sector ( $17 7^{\circ}$ ), so car is the most favourite mode of transport.
  ½ mark
3. Number of people who used car $= \frac{177}{360} \times 120 = 59$ .
  ½ mark
4. (iii)(a) P(used train) $= 1 - P (did not use train) = 1 - \frac{4}{5} = \frac{1}{5}$ .
  1 mark
5. Number of people who used train $= \frac{1}{5} \times 120 = 24$ .
  1 mark
6. (iii)(b) Number of people who used aeroplane $= \frac{7}{60} \times 120 = 14$ .
  1 mark
7. Revenue collected $= 14 \times ₹5000 = ₹70, 000$ .
  1 mark
Answer: (i) 23/120; (ii) Car, 59 people; (iii)(a) 24 people; (iii)(b) ₹70,000
Where the marks are
- (i) Gets P(bus or ship) $= \frac{69}{360} = \frac{23}{120}$ (1 mark)
- (ii) Identifies car as the most favourite mode (½ mark)
- (ii) Finds that 59 people used car (½ mark)
- (iii)(a) Finds P(train) $= \frac{1}{5}$ using the complement (1 mark)
- (iii)(a) Finds that 24 people used train (1 mark)
- (iii)(b) Finds that 14 people used aeroplane (1 mark)
- (iii)(b) Finds the revenue ₹70,000 (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.