MathsCBSE Class 10 · Standard · Board exam 2025 (session 2024-25)

CBSE Class 10 Mathematics Standard: 2025 board paper, set 30/3/1

The question paper set in the CBSE Class 10 board exam of 2025, 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/2025/X/041_Mathematics_Standard.zip (a zip; this paper is the file 30-3-1_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 24
C	26-31	Short answer; internal choice in 26 and 28
D	32-35	Long answer; internal choice in 33 and 34
E	36-38	Case study based; internal choice in 36, 37 and 38 (2-mark subparts)

Instructions in brief: All 38 questions are compulsory; there is no overall choice, but internal choices are provided in 2 questions in Section B, 2 in Section C, 2 in Section D and 3 in Section E (Case Studies).

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQIntroduction to Trigonometry
If $tan 3 θ = 3$ , then $\frac{θ}{2}$ equals :
1. (A) $6 0^{\circ}$
2. (B) $3 0^{\circ}$
3. (C) $2 0^{\circ}$
4. (D) $1 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan 3 θ = 3 = tan 6 0^{\circ} \Rightarrow 3 θ = 6 0^{\circ} \Rightarrow θ = 2 0^{\circ}$ , so $\frac{θ}{2} = 1 0^{\circ}$ . Option (D).
  1 mark
Answer: $1 0^{\circ}$ , option (D)
Where the marks are
- Uses $tan 6 0^{\circ} = 3$ to get $θ = 2 0^{\circ}$ , so $θ /2 = 1 0^{\circ}$ (1 mark)
Q2 · 1 mark · MCQReal Numbers
If x is the LCM of 4, 6, 8 and y is the LCM of 3, 5, 7 and p is the LCM of x and y, then which of the following is true ?
1. (A) $p = 35 x$
2. (B) $p = 4 y$
3. (C) $p = 8 x$
4. (D) $p = 16 y$
Try it first, then show our solutionSolution
1. $x = LCM (4, 6, 8) = 24$ , $y = LCM (3, 5, 7) = 105$ .
  ½ mark
2. $p = LCM (24, 105) = 2^{3} \times 3 \times 5 \times 7 = 840 = 35 \times 24 = 35 x$ . Option (A).
  ½ mark
Answer: $p = 35 x$ , option (A)
Where the marks are
- Finds x = 24, y = 105 (½ mark)
- p = 840 = 35x (½ mark)
Q3 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of ' $k$ ' for which the system of linear equations $6 x + y = 3 k$ and $36 x + 6 y = 3$ have infinitely many solutions is :
1. (A) $6$
2. (B) $\frac{1}{6}$
3. (C) $\frac{1}{2}$
4. (D) $\frac{1}{3}$
Try it first, then show our solutionSolution
1. For infinitely many solutions $\frac{6}{36} = \frac{1}{6} = \frac{3 k}{3}$ .
  ½ mark
2. So $k = \frac{1}{6}$ . Option (B).
  ½ mark
Answer: $k = \frac{1}{6}$ , option (B)
Where the marks are
- Uses $a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}$ (½ mark)
- Gets $k = 1/6$ (½ mark)
Q4 · 1 mark · MCQPolynomials
If $α$ and $β$ are the zeroes of the polynomial $p (x) = x^{2} - a x - b$ , then the value of $(α + β + α β)$ is equal to :
1. (A) $a + b$
2. (B) $- a - b$
3. (C) $a - b$
4. (D) $- a + b$
Try it first, then show our solutionSolution
1. $α + β = a$ , $α β = - b$ .
  ½ mark
2. $α + β + α β = a - b$ . Option (C).
  ½ mark
Answer: $a - b$ , option (C)
Where the marks are
- Sum $= a$ , product $= - b$ (½ mark)
- Result $a - b$ (½ mark)
Q5 · 1 mark · MCQQuadratic Equations
If $\frac{x}{12} - \frac{3}{x} = 0$ , then the values of x are :
1. (A) $\pm 6$
2. (B) $\pm 4$
3. (C) $\pm 12$
4. (D) $\pm 3$
Try it first, then show our solutionSolution
1. Multiply both sides by $12 x$ ( $x \neq = 0$ ): $x^{2} - 36 = 0$ .
2. So $x^{2} = 36$ , giving $x = \pm 6$ . The answer is option (A).
  1 mark
Answer: $x = \pm 6$ , option (A)
Where the marks are
- Clears the fractions to get $x^{2} = 36$ and chooses $\pm 6$ (1 mark)
Q6 · 1 mark · MCQCoordinate Geometry
The line represented by $\frac{x}{4} + \frac{y}{6} = 1$ , intersects x-axis and y-axis respectively at P and Q. The coordinates of the mid-point of line segment PQ are :
1. (A) $(2, 3)$
2. (B) $(3, 2)$
3. (C) $(2, 0)$
4. (D) $(0, 3)$
Try it first, then show our solutionSolution
1. Put $y = 0$ : P $(4, 0)$ ; put $x = 0$ : Q $(0, 6)$ .
  ½ mark
2. Mid-point $= (\frac{4 + 0}{2}, \frac{0 + 6}{2}) = (2, 3)$ . Option (A).
  ½ mark
Answer: $(2, 3)$ , option (A)
Where the marks are
- Finds P(4,0), Q(0,6) (½ mark)
- Mid-point (2,3) (½ mark)
Q7 · 1 mark · MCQCoordinate Geometry
Two of the vertices of $Δ$ PQR are P $(- 1, 5)$ and Q $(5, 2)$ . The coordinates of a point which divides PQ in the ratio $2 : 1$ are :
1. (A) $(3, - 3)$
2. (B) $(5, 5)$
3. (C) $(3, 3)$
4. (D) $(5, 1)$
Try it first, then show our solutionSolution
1. Section formula with $m : n = 2 : 1$ : $x = \frac{2 ( 5 ) + 1 ( - 1 )}{3} = \frac{9}{3} = 3$ .
2. $y = \frac{2 ( 2 ) + 1 ( 5 )}{3} = \frac{9}{3} = 3$ . The point is $(3, 3)$ , option (C).
  1 mark
Answer: $(3, 3)$ , option (C)
Where the marks are
- Applies the section formula correctly to get $(3, 3)$ (1 mark)
Q8 · 1 mark · MCQCircles
If tangents PA and PB drawn from an external point P to the circle with centre O are inclined to each other at an angle of $8 0^{\circ}$ as shown in the given figure, then the measure of $∠$ POA is :
1. (A) $4 0^{\circ}$
2. (B) $5 0^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $8 0^{\circ}$
Figure in the paper: A circle with centre O. External point P on the left. Two tangents from P touch the circle at A (upper) and B (lower). The angle APB is marked $8 0^{\circ}$ at P. The line PO is drawn horizontally from P through to O. OA and OB are radii. See it in the official paper.
Try it first, then show our solutionSolution
1. PO bisects $∠ A P B$ , so $∠ A P O = 4 0^{\circ}$ ; $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius].
  ½ mark
2. In $△ O A P$ : $∠ P O A = 18 0^{\circ} - 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ}$ . Option (B).
  ½ mark
Answer: $5 0^{\circ}$ , option (B)
Where the marks are
- Uses $∠ A P O = 4 0^{\circ}$ and $∠ O A P = 9 0^{\circ}$ (½ mark)
- Angle sum gives $5 0^{\circ}$ (½ mark)
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Q9 · 1 mark · MCQIntroduction to Trigonometry
$(cot θ + tan θ)$ equals :
1. (A) $cosec θ sec θ$
2. (B) $sin θ sec θ$
3. (C) $cos θ tan θ$
4. (D) $sin θ cos θ$
Try it first, then show our solutionSolution
1. Write in sines and cosines: $\frac{c o s θ}{s i n θ} + \frac{s i n θ}{c o s θ} = \frac{c o s ^{2} θ + s i n ^{2} θ}{s i n θ c o s θ}$ .
2. Since $sin^{2} θ + cos^{2} θ = 1$ , this equals $\frac{1}{s i n θ c o s θ} = cosec θ sec θ$ . The answer is option (A).
  1 mark
Answer: $cosec θ sec θ$ , option (A)
Where the marks are
- Simplifies the sum to $\frac{1}{s i n θ c o s θ}$ and identifies option (A) (1 mark)
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Q10 · 1 mark · MCQTriangles
If in two triangles $Δ$ DEF and $Δ$ PQR, $∠$ D $= ∠$ Q and $∠$ R $= ∠$ E, then which of the following is not true ?
1. (A) $\frac{D E}{QR} = \frac{D F}{P Q}$
2. (B) $\frac{E F}{P R} = \frac{D F}{P Q}$
3. (C) $\frac{E F}{R P} = \frac{D E}{QR}$
4. (D) $\frac{D E}{P Q} = \frac{E F}{R P}$
Try it first, then show our solutionSolution
1. $∠ D = ∠ Q$ , $∠ E = ∠ R$ , so $△ D E F \sim △ QR P$ (D↔Q, E↔R, F↔P), giving $\frac{D E}{QR} = \frac{E F}{R P} = \frac{D F}{QP}$ .
  ½ mark
2. (A), (B), (C) follow from this; (D) pairs DE with PQ, which is not a corresponding side, so (D) is not true.
  ½ mark
Answer: Option (D)
Where the marks are
- Correct correspondence DEF ↔ QRP (½ mark)
- Identifies (D) as false (½ mark)
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Q11 · 1 mark · MCQTriangles
The measurements of $Δ$ LMN and $Δ$ ABC are shown in the figure given below. The length of side AC is :
1. (A)16 cm
2. (B)7 cm
3. (C)8 cm
4. (D)4 cm
Figure in the paper: Triangle LMN: LM = 45 cm, LN = 72 cm, MN = 63 cm, angle at M = $13 0^{\circ}$ , angle at N = $2 8^{\circ}$ . Triangle ABC: AB = 5 cm, angle at A = $2 2^{\circ}$ , angle at B = $13 0^{\circ}$ , with side AC to be found and BC the base. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ L M N$ , $∠ L = 18 0^{\circ} - 13 0^{\circ} - 2 8^{\circ} = 2 2^{\circ}$ . So $∠ A = ∠ L$ , $∠ B = ∠ M$ : $△ A B C \sim △ L M N$ (AA).
  ½ mark
2. $\frac{A C}{L N} = \frac{A B}{L M} \Rightarrow A C = \frac{5}{45} \times 72 = 8$ cm. Option (C).
  ½ mark
Answer: 8 cm, option (C)
Where the marks are
- Establishes similarity by AA (½ mark)
- AC = 8 cm (½ mark)
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Q12 · 1 mark · MCQSurface Areas and Volumes
If the volumes of two cubes are in the ratio $8 : 125$ , then the ratio of their surface areas is :
1. (A) $8 : 125$
2. (B) $4 : 25$
3. (C) $2 : 5$
4. (D) $16 : 25$
Try it first, then show our solutionSolution
1. $a_{1}^{3} : a_{2}^{3} = 8 : 125 \Rightarrow a_{1} : a_{2} = 2 : 5$ .
  ½ mark
2. Surface areas $6 a_{1}^{2} : 6 a_{2}^{2} = 4 : 25$ . Option (B).
  ½ mark
Answer: $4 : 25$ , option (B)
Where the marks are
- Sides ratio 2:5 (½ mark)
- Surface area ratio 4:25 (½ mark)
Q13 · 1 mark · MCQAreas Related to Circles
If the area of a sector of circle of radius 36 cm is $54 π$ cm $^{2}$ , then the length of the corresponding arc of the sector is :
1. (A) $8 π$ cm
2. (B) $6 π$ cm
3. (C) $4 π$ cm
4. (D) $3 π$ cm
Try it first, then show our solutionSolution
1. Area of sector $= \frac{1}{2} l r$ , so $54 π = \frac{1}{2} \times l \times 36$ .
  ½ mark
2. $l = 3 π$ cm. Option (D).
  ½ mark
Answer: $3 π$ cm, option (D)
Where the marks are
- Uses area $= \frac{1}{2} l r$ (½ mark)
- $l = 3 π$ cm (½ mark)
Q14 · 1 mark · MCQProbability
A die is thrown once. The probability of getting a number which is not a factor of 36, is :
1. (A) $\frac{1}{2}$
2. (B) $\frac{2}{3}$
3. (C) $\frac{1}{6}$
4. (D) $\frac{5}{6}$
Try it first, then show our solutionSolution
1. A die has 6 outcomes: 1, 2, 3, 4, 5, 6. Of these, 1, 2, 3, 4 and 6 are factors of 36; only 5 is not.
2. So P(not a factor of 36) $= \frac{1}{6}$ . The answer is option (C).
  1 mark
Answer: $\frac{1}{6}$ , option (C)
Where the marks are
- Identifies 5 as the only outcome that is not a factor of 36, giving $\frac{1}{6}$ (1 mark)
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Q15 · 1 mark · MCQStatistics
If the mean of $2, 9, x + 6, 2 x + 3, 5, 10, 5$ is $7$ , then the value of x is :
1. (A) $9$
2. (B) $6$
3. (C) $5$
4. (D) $3$
Try it first, then show our solutionSolution
1. Sum $= 2 + 9 + x + 6 + 2 x + 3 + 5 + 10 + 5 = 3 x + 40$ ; mean $= \frac{3 x + 40}{7} = 7$ .
  ½ mark
2. $3 x + 40 = 49 \Rightarrow x = 3$ . Option (D).
  ½ mark
Answer: $x = 3$ , option (D)
Where the marks are
- Forms $(3 x + 40) /7 = 7$ (½ mark)
- x = 3 (½ mark)
Q16 · 1 mark · MCQCoordinate Geometry
AOBC is a rectangle whose three vertices are A $(0, 2)$ , O $(0, 0)$ and B $(4, 0)$ . The square of the length of its diagonal is equal to :
1. (A) $36$
2. (B) $20$
3. (C) $16$
4. (D) $4$
Try it first, then show our solutionSolution
1. The fourth vertex is C $(4, 2)$ . Diagonal AB joins A $(0, 2)$ and B $(4, 0)$ .
2. $A B^{2} = (4 - 0)^{2} + (0 - 2)^{2} = 16 + 4 = 20$ . The answer is option (B).
  1 mark
Answer: $20$ , option (B)
Where the marks are
- Uses the distance formula on a diagonal to get $A B^{2} = 20$ (1 mark)
Q17 · 1 mark · MCQPolynomials
Zeroes of the polynomial $p (x) = x^{2} - 32 x + 4$ are :
1. (A) $2, 2$
2. (B) $22, 2$
3. (C) $42, - 2$
4. (D) $2, 2$
Try it first, then show our solutionSolution
1. Split the middle term: $x^{2} - 22 x - 2 x + 4 = x (x - 22) - 2 (x - 22) = (x - 22) (x - 2)$ .
2. So the zeroes are $22$ and $2$ . Check: sum $= 32$ and product $= 4$ . The answer is option (B).
  1 mark
Answer: $22, 2$ , option (B)
Where the marks are
- Factorises the polynomial (or uses sum and product of zeroes) to obtain $22, 2$ (1 mark)
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Q18 · 1 mark · MCQTriangles
In the given figure, in $Δ$ ABC, AD $⊥$ BC and $∠$ BAC $= 9 0^{\circ}$ . If BC $= 16$ cm and DC $= 4$ cm, then the value of x is :
1. (A)4 cm
2. (B)5 cm
3. (C)8 cm
4. (D)3 cm
Figure in the paper: Triangle ABC with right angle at A (marked). AD is the altitude from A to base BC, meeting BC at D (right angle marked at D). BC = 16 cm is the base, with BD + DC; DC = 4 cm is marked. The side AC is labelled x. See it in the official paper.
Try it first, then show our solutionSolution
1. $△ A D C \sim △ B A C$ (AA), so $\frac{A C}{B C} = \frac{D C}{A C} \Rightarrow A C^{2} = B C \cdot D C$ .
  ½ mark
2. $x^{2} = 16 \times 4 = 64 \Rightarrow x = 8$ cm. Option (C).
  ½ mark
Answer: 8 cm, option (C)
Where the marks are
- Uses $A C^{2} = B C \cdot D C$ from similarity (½ mark)
- x = 8 cm (½ mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedSome Applications of Trigonometry
Assertion (A): A ladder leaning against a wall, stands at a horizontal distance of 6 m from the wall. If the height of the wall up to which the ladder reaches is 8 m, then the length of the ladder is 10 m.
Reason (R): The ladder makes an angle of $6 0^{\circ}$ with the ground.
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. Length $= 6^{2} + 8^{2} = 10$ m, so A is true.
  ½ mark
2. $cos θ = 6/10 = 0.6 \neq = cos 6 0^{\circ} = 0.5$ , so R is false. Option (C).
  ½ mark
Answer: Option (C)
Where the marks are
- Pythagoras gives 10 m (½ mark)
- Shows angle is not 60° (½ mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedCircles
Assertion (A): If two tangents are drawn to a circle from an external point, then they subtend equal angles at the centre of the circle.
Reason (R): A parallelogram circumscribing a circle is a rhombus.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A is true: $△ O A P ≅ △ O B P$ (RHS), so $∠ A O P = ∠ B O P$ .
  ½ mark
2. R is true (equal tangents make adjacent sides equal), but it is a separate result and does not explain A. Option (B).
  ½ mark
Answer: Option (B)
Where the marks are
- A true via congruent triangles (½ mark)
- R true but not explanation (½ mark)
Open Q20 on its own page

Section B

Very short answer; internal choice in 23 and 24

Q21 · 2 marks · Very short answerIntroduction to Trigonometry
If $4 k = tan^{2} 6 0^{\circ} - 2 cosec^{2} 3 0^{\circ} - 2 tan^{2} 3 0^{\circ}$ , then find the value of k.
Try it first, then show our solutionSolution
1. Use the standard values: $tan 6 0^{\circ} = 3$ , $cosec 3 0^{\circ} = 2$ , $tan 3 0^{\circ} = \frac{1}{3}$ .
  $4 k = (3)^{2} - 2 (2)^{2} - 2 (\frac{1}{3})^{2} = 3 - 8 - \frac{2}{3} = - \frac{17}{3}$
  1½ marks
2. $k = - \frac{17}{3} \div 4 = - \frac{17}{12}$
  ½ mark
Answer: $k = - \frac{17}{12}$
Where the marks are
- Substitutes the correct trigonometric values and simplifies to $4 k = - \frac{17}{3}$ (1½ marks)
- Obtains $k = - \frac{17}{12}$ (½ mark)
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Q22 · 2 marks · Very short answerProbability
The probability of guessing the correct answer of a certain test question is $\frac{x}{12}$ . If the probability of not guessing the correct answer is $\frac{5}{6}$ , then find the value of x.
Try it first, then show our solutionSolution
1. Guessing correctly and not guessing correctly are complementary events, so $P (E) + P (\overset{ˉ}{E}) = 1$ :
  $\frac{x}{12} + \frac{5}{6} = 1$
  1 mark
2. $\frac{x}{12} = 1 - \frac{5}{6} = \frac{1}{6}$ , so $x = \frac{12}{6} = 2$
  1 mark
Answer: $x = 2$
Where the marks are
- Writes $\frac{x}{12} + \frac{5}{6} = 1$ using complementary events (1 mark)
- Solves to get $x = 2$ (1 mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerReal Numbers
Find the smallest number which is divisible by both 644 and 462.
Try it first, then show our solutionSolution
1. Prime factorisation: $462 = 2 \times 3 \times 7 \times 11$
  ½ mark
2. $644 = 2^{2} \times 7 \times 23$
  ½ mark
3. The smallest number divisible by both is their LCM. Take each prime factor with its highest power:
  $LCM = 2^{2} \times 3 \times 7 \times 11 \times 23 = 21252$
  So the required smallest number is 21252.
  1 mark
Answer: 21252
Where the marks are
- Prime factorisation of 462 (½ mark)
- Prime factorisation of 644 (½ mark)
- LCM = 21252 identified as the required number (1 mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerReal Numbers
Two numbers are in the ratio $4 : 5$ and their HCF is 11. Find the LCM of these numbers.
Try it first, then show our solutionSolution
1. Let the numbers be $4 x$ and $5 x$ . Since 4 and 5 are co-prime, HCF $= x$ . So $x = 11$ and the numbers are 44 and 55.
  ½ mark
2. Use HCF × LCM = product of the numbers:
  $LCM = \frac{44 \times 55}{11}$
  1 mark
3. $LCM = 4 \times 55 = 220$
  ½ mark
Answer: LCM = 220
Where the marks are
- Finds the numbers as 44 and 55 (½ mark)
- Uses HCF × LCM = product of numbers (1 mark)
- LCM = 220 (½ mark)
Open Q23(b) on its own page
Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerQuadratic Equations
Find the value(s) of ' $k$ ' so that the quadratic equation $4 x^{2} + k x + 1 = 0$ has real and equal roots.
Try it first, then show our solutionSolution
1. For real and equal roots, the discriminant must be zero: $D = b^{2} - 4 a c = 0$ . Here $a = 4$ , $b = k$ , $c = 1$ .
  ½ mark
2. $k^{2} - 4 (4) (1) = 0 \Rightarrow k^{2} - 16 = 0$
  1 mark
3. $k^{2} = 16 \Rightarrow k = \pm 4$
  ½ mark
Answer: $k = 4$ or $k = - 4$
Where the marks are
- States the condition $D = 0$ (½ mark)
- Forms $k^{2} - 16 = 0$ (1 mark)
- Gets $k = \pm 4$ (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerPolynomials
If ' $α$ ' and ' $β$ ' are the zeroes of the polynomial $p (y) = y^{2} - 5 y + 3$ , then find the value of $α^{4} β^{3} + α^{3} β^{4}$ .
Try it first, then show our solutionSolution
1. Sum of zeroes: $α + β = - \frac{- 5}{1} = 5$
  ½ mark
2. Product of zeroes: $α β = \frac{3}{1} = 3$
  ½ mark
3. Take out the common factor: $α^{4} β^{3} + α^{3} β^{4} = α^{3} β^{3} (α + β) = (α β)^{3} (α + β)$
  ½ mark
4. $= 3^{3} \times 5 = 27 \times 5 = 135$
  ½ mark
Answer: 135
Where the marks are
- $α + β = 5$ (½ mark)
- $α β = 3$ (½ mark)
- Writes the expression as $(α β)^{3} (α + β)$ (½ mark)
- Value = 135 (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerCircles
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O and diameter AB. Line PQ is the tangent at A (P and Q on either side of A), and line RS is the tangent at B (R above P, S above Q).
  ½ mark
2. $∠ O A P = 9 0^{\circ}$ and $∠ O B R = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]
  ½ mark
3. $∠ O A P + ∠ O B R = 9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ . These are co-interior angles made by the transversal AB with the lines PQ and RS, and they are supplementary.
  ½ mark
4. Therefore $P Q ∥ R S$ , i.e. the tangents at the ends of a diameter are parallel.
  ½ mark
Answer: The co-interior angles at A and B add to $18 0^{\circ}$ , so the tangents PQ and RS are parallel.
Where the marks are
- Correct figure (½ mark)
- Both angles $= 9 0^{\circ}$ with reason (½ mark)
- Sum $= 18 0^{\circ}$ , co-interior angles supplementary (½ mark)
- Concludes $P Q ∥ R S$ (½ mark)
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Section C

Short answer; internal choice in 26 and 28

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerCoordinate Geometry
If the mid-point of the line segment joining the points A $(3, 4)$ and B $(k, 6)$ is P $(x, y)$ and $x + y - 10 = 0$ , find the value of k.
Try it first, then show our solutionSolution
1. By the mid-point formula, $x = \frac{3 + k}{2}$ .
  ½ mark
2. $y = \frac{4 + 6}{2} = 5$ .
  ½ mark
3. P lies on $x + y - 10 = 0$ , so:
  $\frac{3 + k}{2} + 5 - 10 = 0$
  1 mark
4. $\frac{3 + k}{2} = 5 \Rightarrow 3 + k = 10 \Rightarrow k = 7$ .
  1 mark
Answer: k = 7
Where the marks are
- Finds $x = \frac{3 + k}{2}$ (1/2 mark)
- Finds $y = 5$ (1/2 mark)
- Substitutes into $x + y - 10 = 0$ (1 mark)
- Solves to get $k = 7$ (1 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerCoordinate Geometry
Find the coordinates of the points which divide the line segment joining A $(- 2, 2)$ and B $(2, 8)$ into four equal parts.
Try it first, then show our solutionSolution
1. Let P, Q, R divide AB into four equal parts, so AP = PQ = QR = RB. Then Q is the mid-point of AB:
  $Q = (\frac{- 2 + 2}{2}, \frac{2 + 8}{2}) = (0, 5)$
  1 mark
2. P is the mid-point of AQ:
  $P = (\frac{- 2 + 0}{2}, \frac{2 + 5}{2}) = (- 1, \frac{7}{2})$
  1 mark
3. R is the mid-point of QB:
  $R = (\frac{0 + 2}{2}, \frac{5 + 8}{2}) = (1, \frac{13}{2})$
  1 mark
Answer: $(- 1, \frac{7}{2})$ , $(0, 5)$ , $(1, \frac{13}{2})$
Where the marks are
- Q = (0, 5), the mid-point of AB (1 mark)
- P = $(- 1, \frac{7}{2})$ , the mid-point of AQ (1 mark)
- R = $(1, \frac{13}{2})$ , the mid-point of QB (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerReal Numbers
Prove that $(53 + \frac{2}{3})$ is an irrational number given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $53 + \frac{2}{3}$ is rational. Then $53 + \frac{2}{3} = \frac{a}{b}$ , where $a, b$ are integers and $b \neq = 0$ .
  1 mark
2. Rearrange: $53 = \frac{a}{b} - \frac{2}{3} = \frac{3 a - 2 b}{3 b}$ , so $3 = \frac{3 a - 2 b}{15 b}$
  1 mark
3. $3 a - 2 b$ and $15 b$ are integers and $15 b \neq = 0$ . So $\frac{3 a - 2 b}{15 b}$ is rational, which makes $3$ rational.
  ½ mark
4. This contradicts the given fact that $3$ is irrational. So our assumption is wrong, and $53 + \frac{2}{3}$ is irrational.
  ½ mark
Answer: $53 + \frac{2}{3}$ is irrational (proved by contradiction).
Where the marks are
- Assumes the number is rational and writes it as $\frac{a}{b}$ , $b \neq = 0$ (1 mark)
- Isolates $3 = \frac{3 a - 2 b}{15 b}$ (1 mark)
- States that the RHS is rational (½ mark)
- Concludes by contradiction (½ mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{s e c A - 1}{s e c A + 1} + \frac{s e c A + 1}{s e c A - 1} = 2 cosec A$
Try it first, then show our solutionSolution
1. Write both terms over the common denominator $(sec A + 1) (sec A - 1)$ :
  $LHS = \frac{( s e c A - 1 ) + ( s e c A + 1 )}{( s e c A + 1 ) ( s e c A - 1 )} = \frac{2 s e c A}{s e c ^{2} A - 1}$
  1 mark
2. Since $sec^{2} A - 1 = tan^{2} A$ :
  $LHS = \frac{2 s e c A}{t a n A}$
  1 mark
3. $= \frac{2}{c o s A} \times \frac{c o s A}{s i n A} = \frac{2}{s i n A} = 2 cosec A = RHS$
  1 mark
Answer: LHS = $2 cosec A$ = RHS. Hence proved.
Where the marks are
- Combines the two terms into $\frac{2 s e c A}{s e c ^{2} A - 1}$ (1 mark)
- Uses $sec^{2} A - 1 = tan^{2} A$ to get $\frac{2 s e c A}{t a n A}$ (1 mark)
- Simplifies to $2 cosec A$ (1 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $(\frac{1}{c o s A} - cos A) (\frac{1}{s i n A} - sin A) = \frac{1}{t a n A + c o t A}$
Try it first, then show our solutionSolution
1. $LHS = (\frac{1 - c o s ^{2} A}{c o s A}) (\frac{1 - s i n ^{2} A}{s i n A})$
  1 mark
2. Use $sin^{2} A + cos^{2} A = 1$ :
  $LHS = \frac{s i n ^{2} A}{c o s A} \cdot \frac{c o s ^{2} A}{s i n A}$
  ½ mark
3. $= sin A cos A$
  ½ mark
4. $RHS = \frac{1}{\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}} = \frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A}$
  ½ mark
5. $= \frac{s i n A c o s A}{1} = sin A cos A$ . So LHS = RHS.
  ½ mark
Answer: Both sides equal $sin A cos A$ . Hence proved.
Where the marks are
- Takes LCM in each bracket (1 mark)
- Uses the identity to get $\frac{s i n ^{2} A}{c o s A} \cdot \frac{c o s ^{2} A}{s i n A}$ (½ mark)
- LHS = $sin A cos A$ (½ mark)
- RHS written as $\frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A}$ (½ mark)
- RHS = $sin A cos A$ , so LHS = RHS (½ mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerAreas Related to Circles
A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use $π = 3 \cdot 14$ ]
Try it first, then show our solutionSolution
1. Let chord AB subtend $∠ A O B = 9 0^{\circ}$ at centre O, with $r = 10$ cm.
  Area of minor segment = Area of sector OAB − Area of $△ O A B$ .
  Area of sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{90}{360} \times 3.14 \times 10 \times 10$
  ½ mark
2. $= 78.5 cm^{2}$
  ½ mark
3. $△ O A B$ is right-angled at O, so its area $= \frac{1}{2} \times O A \times O B = \frac{1}{2} \times 10 \times 10$
  ½ mark
4. $= 50 cm^{2}$
  ½ mark
5. Area of minor segment $= 78.5 - 50$
  ½ mark
6. $= 28.5 cm^{2}$
  ½ mark
Answer: 28.5 cm²
Where the marks are
- Correct sector formula with substitution (½ mark)
- Sector area = 78.5 cm² (½ mark)
- Triangle area formula $\frac{1}{2} \times 10 \times 10$ (½ mark)
- Triangle area = 50 cm² (½ mark)
- Segment = sector − triangle (½ mark)
- Answer 28.5 cm² (½ mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerProbability
Three unbiased coins are tossed simultaneously. Find the probability of getting :
1. (a)
  exactly two tails
  1 mark
2. (b)
  at least one head
  1 mark
3. (c)
  at most two heads
  1 mark
Try it first, then show our solutionSolution
1. Sample space: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, so there are 8 equally likely outcomes.
2. (a) Exactly two tails: HTT, THT, TTH, which is 3 outcomes.
  $P = \frac{3}{8}$
  1 mark
3. (b) At least one head: every outcome except TTT, which is 7 outcomes.
  $P = \frac{7}{8}$
  1 mark
4. (c) At most two heads: every outcome except HHH, which is 7 outcomes.
  $P = \frac{7}{8}$
  1 mark
Answer: (a) $\frac{3}{8}$ ; (b) $\frac{7}{8}$ ; (c) $\frac{7}{8}$
Where the marks are
- (a) P(exactly two tails) = $\frac{3}{8}$ (1 mark)
- (b) P(at least one head) = $\frac{7}{8}$ (1 mark)
- (c) P(at most two heads) = $\frac{7}{8}$ (1 mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCircles
In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find $∠$ CBA and $∠$ BCO, if $∠$ PCA $= 11 0^{\circ}$ .
Figure in the paper: A circle with centre O. Diameter AOB lies horizontally, with B on the left and A on the right; O between them. The line AB extended beyond B meets the tangent at P on the far left. PC is the tangent touching the circle at C (top). Chords/segments CA, CB and radius CO are drawn. $∠$ PCA $= 11 0^{\circ}$ is marked at C. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ A C B = 9 0^{\circ}$ [angle in a semicircle, since AB is a diameter]
  ½ mark
2. $∠ P C A = ∠ P C B + ∠ B C A$ , so $11 0^{\circ} = ∠ P C B + 9 0^{\circ}$ .
  Hence $∠ P C B = 2 0^{\circ}$ .
  1 mark
3. $∠ O C P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  So $∠ B C O = 9 0^{\circ} - ∠ P C B = 9 0^{\circ} - 2 0^{\circ} = 7 0^{\circ}$ .
  1 mark
4. OB = OC [radii], so $∠ O B C = ∠ O C B$ [angles opposite equal sides].
  Hence $∠ C B A = ∠ O B C = 7 0^{\circ}$ .
  ½ mark
Answer: $∠ C B A = 7 0^{\circ}$ , $∠ B C O = 7 0^{\circ}$
Where the marks are
- $∠ A C B = 9 0^{\circ}$ (angle in semicircle) (½ mark)
- $∠ P C B = 2 0^{\circ}$ (1 mark)
- $∠ B C O = 7 0^{\circ}$ using tangent ⟂ radius (1 mark)
- $∠ C B A = 7 0^{\circ}$ using OB = OC (½ mark)
Open Q31 on its own page

Section D

Long answer; internal choice in 33 and 34

Q32 · 5 marks · Long answerTriangles
The perimeter of an isosceles triangle is 32 cm. If each equal side is $\frac{5}{6}$ th of the base, find the area of the triangle.
Try it first, then show our solutionSolution
1. Let each equal side be $x$ cm and the base be $y$ cm.
  Perimeter: $x + x + y = 32 \Rightarrow 2 x + y = 32$ …(1)
  1 mark
2. Given condition: $x = \frac{5}{6} y$ …(2)
  1 mark
3. Substitute (2) in (1): $\frac{5}{3} y + y = 32 \Rightarrow \frac{8}{3} y = 32 \Rightarrow y = 12$ , and $x = \frac{5}{6} \times 12 = 10$ .
  So the sides are 10 cm, 10 cm and 12 cm.
  1 mark
4. Semi-perimeter: $s = \frac{32}{2} = 16$ cm
  ½ mark
5. By Heron's formula:
  Area $= s (s - a) (s - b) (s - c) = 16 \times 6 \times 6 \times 4$
  1 mark
6. $= 2304 = 48 cm^{2}$
  ½ mark
Answer: Area = 48 cm²
Where the marks are
- Forms $2 x + y = 32$ (1 mark)
- Forms $x = \frac{5}{6} y$ (1 mark)
- Solves: $x = 10$ , $y = 12$ (½ + ½ marks)
- Semi-perimeter = 16 cm (½ mark)
- Heron's formula with substitution $16 \times 6 \times 6 \times 4$ (1 mark)
- Area = 48 cm² (½ mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerArithmetic Progressions
The sum of the third term and the seventh term of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.
Try it first, then show our solutionSolution
1. Let the first term be $a$ and the common difference be $d$ .
  $a_{3} + a_{7} = (a + 2 d) + (a + 6 d) = 6 \Rightarrow a + 4 d = 3 \Rightarrow a = 3 - 4 d$
  ½ mark
2. $a_{3} \cdot a_{7} = (a + 2 d) (a + 6 d) = 8$
  ½ mark
3. Substitute $a = 3 - 4 d$ : $(3 - 2 d) (3 + 2 d) = 8 \Rightarrow 9 - 4 d^{2} = 8 \Rightarrow d^{2} = \frac{1}{4} \Rightarrow d = \pm \frac{1}{2}$
  1 mark
4. Case 1: $d = \frac{1}{2}$ gives $a = 3 - 2 = 1$ .
  ½ mark
5. $S_{16} = \frac{16}{2} [2 \times 1 + 15 \times \frac{1}{2}]$
  ½ mark
6. $= 8 \times 9.5 = 76$
  ½ mark
7. Case 2: $d = - \frac{1}{2}$ gives $a = 3 + 2 = 5$ .
  ½ mark
8. $S_{16} = \frac{16}{2} [2 \times 5 + 15 \times (- \frac{1}{2})]$
  ½ mark
9. $= 8 \times 2.5 = 20$
  ½ mark
Answer: $S_{16} = 76$ (when $d = \frac{1}{2}$ ) or $S_{16} = 20$ (when $d = - \frac{1}{2}$ )
Where the marks are
- Forms $a + 4 d = 3$ from the sum condition (½ mark)
- Forms the product equation $(a + 2 d) (a + 6 d) = 8$ (½ mark)
- Solves to get $d = \pm \frac{1}{2}$ (1 mark)
- For $d = \frac{1}{2}$ : $a = 1$ , correct substitution, $S_{16} = 76$ (½ + ½ + ½ marks)
- For $d = - \frac{1}{2}$ : $a = 5$ , correct substitution, $S_{16} = 20$ (½ + ½ + ½ marks)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerCompetency-basedArithmetic Progressions
The minimum age of children eligible to participate in a painting competition is 8 years. It is observed that the age of the youngest boy was 8 years and the ages of the participants, when seated in order of age, have a common difference of 4 months. If the sum of the ages of all the participants is 168 years, find the age of the eldest participant in the painting competition.
Try it first, then show our solutionSolution
1. 4 months $= \frac{4}{12} = \frac{1}{3}$ year. The ages (in years) form an AP: $8, 8 \frac{1}{3}, 8 \frac{2}{3}, 9, \dots$ with $a = 8$ and $d = \frac{1}{3}$ .
  1 mark
2. Let there be $n$ participants. Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ :
  $\frac{n}{2} [16 + \frac{n - 1}{3}] = 168$
  1 mark
3. $\frac{n ( 48 + n - 1 )}{6} = 168 \Rightarrow n^{2} + 47 n - 1008 = 0$
  1 mark
4. $(n + 63) (n - 16) = 0 \Rightarrow n = 16$ [n cannot be negative, so reject $n = - 63$ ]
  1 mark
5. Age of the eldest $= a_{16} = 8 + 15 \times \frac{1}{3} = 8 + 5 = 13$ years
  1 mark
Answer: The eldest participant is 13 years old.
Where the marks are
- Identifies the AP with $a = 8$ , $d = \frac{1}{3}$ year (1 mark)
- Sets up $\frac{n}{2} [16 + (n - 1) \frac{1}{3}] = 168$ (1 mark)
- Forms $n^{2} + 47 n - 1008 = 0$ (1 mark)
- Solves to get $n = 16$ (1 mark)
- Eldest age $= a_{16} = 13$ years (1 mark)
Open Q33(b) on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerTriangles
In the given figure, PA, QB and RC are perpendicular to AC. If PA $= x$ units, QB $= y$ units and RC $= z$ units, prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ .
Figure in the paper: Points A and C on a horizontal line with B between them on AC. PA is a vertical segment at A of length x, going up to P. QB is a vertical segment at B of length y, with Q on it. RC is a vertical segment at C of length z, going up to R. P, Q, R are collinear with lines PC and AR crossing at Q; line AR passes through Q, and line PC passes through Q. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B Q$ and $△ A C R$ : $∠ A$ is common, and $∠ A B Q = ∠ A C R = 9 0^{\circ}$ .
  So $△ A B Q \sim △ A C R$ [AA similarity].
  1 mark
2. Corresponding sides are proportional:
  $\frac{A B}{A C} = \frac{QB}{R C} = \frac{y}{z}$ …(i)
  ½ mark
3. Similarly, in $△ C B Q$ and $△ C A P$ : $∠ C$ is common, and $∠ C B Q = ∠ C A P = 9 0^{\circ}$ .
  So $△ C B Q \sim △ C A P$ [AA similarity].
  ½ mark
4. $\frac{B C}{A C} = \frac{QB}{P A} = \frac{y}{x}$ …(ii)
  ½ mark
5. Adding (i) and (ii):
  $\frac{A B}{A C} + \frac{B C}{A C} = \frac{y}{z} + \frac{y}{x}$
  1 mark
6. $\frac{A B + B C}{A C} = y (\frac{1}{x} + \frac{1}{z})$ . Since $A B + B C = A C$ , this gives $1 = y (\frac{1}{x} + \frac{1}{z})$ .
  1 mark
7. Dividing by $y$ : $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ . Hence proved.
  ½ mark
Answer: $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ (proved)
Where the marks are
- Proves $△ A B Q \sim △ A C R$ (1 mark)
- Writes $\frac{A B}{A C} = \frac{y}{z}$ (½ mark)
- Proves $△ C B Q \sim △ C A P$ (½ mark)
- Writes $\frac{B C}{A C} = \frac{y}{x}$ (½ mark)
- Adds the two ratios (1 mark)
- Uses $A B + B C = A C$ (1 mark)
- Concludes the result (½ mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerTriangles
Sides AB and BC and median AD of triangle ABC are respectively proportional to sides PQ and QR and median PM of $Δ$ PQR. Show that $Δ$ ABC $\sim$ $Δ$ PQR.
Try it first, then show our solutionSolution
1. Figure: draw $△ A B C$ with median AD (D is the mid-point of BC), and $△ P QR$ with median PM (M is the mid-point of QR).
  Given: $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ . To prove: $△ A B C \sim △ P QR$ .
  1 mark
2. Since D and M are mid-points, $B C = 2 B D$ and $QR = 2 QM$ . So
  $\frac{A B}{P Q} = \frac{2 B D}{2 QM} = \frac{A D}{P M} \Rightarrow \frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$
  1 mark
3. Therefore $△ A B D \sim △ P QM$ [SSS similarity].
  1 mark
4. So $∠ B = ∠ Q$ [corresponding angles of similar triangles].
  ½ mark
5. In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ (given) and the included angles are equal, $∠ B = ∠ Q$ .
  1 mark
6. Hence $△ A B C \sim △ P QR$ [SAS similarity].
  ½ mark
Answer: $△ A B C \sim △ P QR$ (proved using SSS, then SAS similarity)
Where the marks are
- Correct labelled figure with both medians (1 mark)
- Uses $B C = 2 B D$ , $QR = 2 QM$ to get $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ (1 mark)
- $△ A B D \sim △ P QM$ by SSS (1 mark)
- Deduces $∠ B = ∠ Q$ (½ mark)
- States $\frac{A B}{P Q} = \frac{B C}{QR}$ with included $∠ B = ∠ Q$ (1 mark)
- Concludes $△ A B C \sim △ P QR$ by SAS (½ mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerSurface Areas and Volumes
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0 \cdot 5$ cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Try it first, then show our solutionSolution
1. Cone: $r = 5$ cm, $h = 8$ cm. Volume of water in the full cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} π \times 5^{2} \times 8$
  1 mark
2. $= \frac{200 π}{3}$ cm $^{3}$
  ½ mark
3. The water that flows out equals the total volume of the lead shots dropped in. Volume of water that flows out $= \frac{1}{4} \times \frac{200 π}{3} = \frac{50 π}{3}$ cm $^{3}$
  1 mark
4. Radius of one lead shot $= 0.5 = \frac{1}{2}$ cm. Volume of one shot $= \frac{4}{3} π r^{3} = \frac{4}{3} π (\frac{1}{2})^{3}$
  1 mark
5. $= \frac{π}{6}$ cm $^{3}$
  ½ mark
6. Number of lead shots $= \frac{50 π /3}{π /6} = \frac{50 π}{3} \times \frac{6}{π} = 100$
  1 mark
Answer: 100 lead shots
Where the marks are
- Writes the volume of the cone as $\frac{1}{3} π (5)^{2} (8)$ (1 mark)
- Evaluates it as $\frac{200 π}{3}$ cm $^{3}$ (½ mark)
- Finds the water displaced as $\frac{50 π}{3}$ cm $^{3}$ (1 mark)
- Writes the volume of one shot as $\frac{4}{3} π (0.5)^{3}$ (1 mark)
- Evaluates it as $\frac{π}{6}$ cm $^{3}$ (½ mark)
- Divides the two volumes to get 100 shots (1 mark)
Open Q35 on its own page

Section E

Case study based; internal choice in 36, 37 and 38 (2-mark subparts)

Q36 · 4 marks · Case-basedCompetency-basedQuadratic Equations
Case Study 1. A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway. The total area of the lawn and the walkway is 360 square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are 12 metres by 10 metres.
Based on the information given above, answer the following questions :
1. (i)
  Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway $= x$ m.
  1 mark
2. (ii)
  (a) Solve the quadratic equation to find the width of the walkway 'x'.
  2 marks
3. OR
  (ii)
  (b) If the cost of paving the walkway at the rate of $₹ 50$ per square metre is $₹ 12, 000$ , calculate the area of the walkway.
  2 marks
4. (iii)
  Find the perimeter of the lawn.
  1 mark
Figure in the paper: A photograph of a rectangular green lawn surrounded by a stone walkway with plants around it, and a line diagram: a rectangle representing the lawn, labelled 10 m (vertical side) and 12 m (horizontal side), surrounded by a uniform walkway of width x on all sides, with x marked on the left and bottom edges. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) With the walkway of width $x$ m on every side, the outer rectangle measures $(12 + 2 x)$ m by $(10 + 2 x)$ m. So $(12 + 2 x) (10 + 2 x) = 360$ .
  ½ mark
2. (i) Expanding: $120 + 44 x + 4 x^{2} = 360$ , i.e. $4 x^{2} + 44 x - 240 = 0$ , or $x^{2} + 11 x - 60 = 0$ .
  ½ mark
3. (ii)(a) Splitting the middle term: $x^{2} + 15 x - 4 x - 60 = 0 \Rightarrow (x + 15) (x - 4) = 0$ .
  1 mark
4. (ii)(a) This gives $x = 4$ or $x = - 15$ . A width cannot be negative, so the width of the walkway is $x = 4$ m.
  1 mark
5. (ii)(b) Area of walkway $= \frac{Total cost}{Rate} = \frac{12000}{50}$
  1 mark
6. (ii)(b) $= 240$ m $^{2}$ . (Check: $360 - 12 \times 10 = 240$ m $^{2}$ .)
  1 mark
7. (iii) Perimeter of the lawn $= 2 (l + b) = 2 (12 + 10) = 44$ m.
  1 mark
Answer: (i) $x^{2} + 11 x - 60 = 0$ ; (ii)(a) 4 m; (ii)(b) 240 m²; (iii) 44 m
Where the marks are
- Writes $(12 + 2 x) (10 + 2 x) = 360$ (½ mark)
- Simplifies to $x^{2} + 11 x - 60 = 0$ (½ mark)
- (ii)(a) Factorises as $(x + 15) (x - 4) = 0$ (1 mark)
- (ii)(a) Rejects the negative root and states the width as 4 m (1 mark)
- (ii)(b) Divides the total cost by the rate, $12000 \div 50$ (1 mark)
- (ii)(b) Gets the area as 240 m² (1 mark)
- (iii) Perimeter $= 2 (12 + 10) = 44$ m (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Case Study 2. A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be $3 0^{\circ}$ and $4 5^{\circ}$ as it moves from point P to point Q. The height of the lighthouse is 50 metres.
Based on the information given above, answer the following questions :
1. (i)
  Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is $4 5^{\circ}$ .
  1 mark
2. (ii)
  Find the measures of $∠$ PBA and $∠$ QBA.
  1 mark
3. (iii)
  (a) Find the distance travelled by the ship between points P and Q.
  2 marks
4. OR
  (iii)
  (b) If the ship continues moving towards the shore and takes 10 minutes to travel from Q to A, calculate the speed of the ship in km/h, from Q to A.
  2 marks
Figure in the paper: A photograph of a lighthouse on a rocky cliff by the sea, and a line diagram: B is the top of the lighthouse and A is its base, with BA vertical on the right. P and Q lie on the horizontal ground to the left of A, with Q nearer to A. A horizontal line BX goes from B to the left. The angle between BX and BP is $3 0^{\circ}$ and the angle between BX and BQ is $4 5^{\circ}$ (both angles of depression at B). The angle at P between the ground and PB is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $B X ∥ P A$ , so $∠ B Q A = ∠ X B Q = 4 5^{\circ}$ (alternate angles). In right $△ B A Q$ : $tan 4 5^{\circ} = \frac{A B}{A Q} = \frac{50}{A Q}$
  ½ mark
2. (i) $1 = \frac{50}{A Q} \Rightarrow A Q = 50$ m
  ½ mark
3. (ii) $∠ X B A = 9 0^{\circ}$ . So $∠ P B A = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .
  ½ mark
4. (ii) $∠ QB A = 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$ .
  ½ mark
5. (iii)(a) $∠ B P A = 3 0^{\circ}$ (alternate angles). In right $△ B A P$ : $tan 3 0^{\circ} = \frac{50}{A P} \Rightarrow \frac{1}{3} = \frac{50}{A P}$
  ½ mark
6. (iii)(a) $A P = 503$ m
  ½ mark
7. (iii)(a) $P Q = A P - A Q = 503 - 50 = 50 (3 - 1)$ m $\approx 50 \times 0.732 = 36.6$ m
  1 mark
8. (iii)(b) Distance $Q A = 50$ m $= 0.05$ km. Time $= 10$ min $= \frac{1}{6}$ h. Speed $= \frac{distance}{time} = \frac{0.05}{1/6}$
  1 mark
9. (iii)(b) $= 0.3$ km/h
  1 mark
Answer: (i) 50 m; (ii) ∠PBA = 60°, ∠QBA = 45°; (iii)(a) $50 (3 - 1)$ m ≈ 36.6 m; (iii)(b) 0.3 km/h
Where the marks are
- (i) Uses $tan 4 5^{\circ} = \frac{50}{A Q}$ (½ mark)
- (i) Gets AQ = 50 m (½ mark)
- (ii) ∠PBA = 60° (½ mark)
- (ii) ∠QBA = 45° (½ mark)
- (iii)(a) Uses $tan 3 0^{\circ} = \frac{50}{A P}$ (½ mark)
- (iii)(a) Gets AP = $503$ m (½ mark)
- (iii)(a) Gets PQ = $50 (3 - 1)$ m (1 mark)
- (iii)(b) Writes speed as 50 m ÷ 10 min with the units converted (1 mark)
- (iii)(b) Gets 0.3 km/h (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedStatistics
Case Study 3. The India Meteorological Department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results.
The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023.
Rainfall (mm) No. of Sub-divisions
200 – 400 3
400 – 600 4
600 – 800 7
800 – 1000 4
1000 – 1200 3
1200 – 1400 3
Based on the information given above, answer the following questions :
1. (i)
  Write the modal class.
  1 mark
2. (ii)
  (a) Find the median of the given data.
  2 marks
3. OR
  (ii)
  (b) Find the mean rainfall in the season.
  2 marks
4. (iii)
  If a sub-division having at least 800 mm rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall ?
  1 mark
Try it first, then show our solutionSolution
1. (i) The highest frequency is 7, which belongs to the class 600–800. So the modal class is 600–800.
  1 mark
2. (ii)(a) Cumulative frequencies: 200–400: 3; 400–600: 7; 600–800: 14; 800–1000: 18; 1000–1200: 21; 1200–1400: 24. So $N = 24$ .
  ½ mark
3. (ii)(a) $\frac{N}{2} = 12$ . The first cumulative frequency of at least 12 is 14, so the median class is 600–800. Here $l = 600$ , $c f = 7$ , $f = 7$ , $h = 200$ .
  ½ mark
4. (ii)(a) Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 600 + \frac{12 - 7}{7} \times 200$
  ½ mark
5. (ii)(a) $= 600 + \frac{1000}{7} = \frac{5200}{7} \approx 742.86$ mm
  ½ mark
6. (ii)(b) Class marks $x_{i}$ : 300, 500, 700, 900, 1100, 1300. Products $f_{i} x_{i}$ : 900, 2000, 4900, 3600, 3300, 3900. So $\sum f_{i} = 24$ and $\sum f_{i} x_{i} = 18600$ .
  1 mark
7. (ii)(b) Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{18600}{24} = 775$ mm
  1 mark
8. (iii) The classes with at least 800 mm are 800–1000, 1000–1200 and 1200–1400. Number of sub-divisions $= 4 + 3 + 3 = 10$ .
  1 mark
Answer: (i) 600–800; (ii)(a) Median = 5200/7 ≈ 742.86 mm; (ii)(b) Mean = 775 mm; (iii) 10
Where the marks are
- (i) States the modal class as 600–800 (1 mark)
- (ii)(a) Draws up the correct cumulative frequency table (½ mark)
- (ii)(a) Identifies the median class 600–800 (½ mark)
- (ii)(a) Substitutes correctly into the median formula (½ mark)
- (ii)(a) Gets the median as 5200/7 ≈ 742.86 mm (½ mark)
- (ii)(b) Draws up the correct $x_{i}$ , $f_{i} x_{i}$ table with $\sum f_{i} x_{i} = 18600$ (1 mark)
- (ii)(b) Gets the mean as 775 mm (1 mark)
- (iii) Counts 10 sub-divisions (1 mark)
Open Q38 on its own page

Rainfall (mm)	No. of Sub-divisions
200 – 400	3
400 – 600	4
600 – 800	7
800 – 1000	4
1000 – 1200	3
1200 – 1400	3

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