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

CBSE Class 10 Mathematics Basic: 2024 board paper, set 430/5/1

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_BASIC.zip (a zip; this paper is the file 430-5-1MATHEMATICS (BASIC).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 22 and 25
C	26-31	Short answer; internal choice in 29 and 30
D	32-35	Long answer; internal choice in 32 and 34
E	36-38	Case study based; internal choice in 2-mark subparts of each case study

Instructions in brief: There is no overall choice; internal choice is given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 case-study subparts in Section E.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
HCF $\times$ LCM for the numbers 40 and 30 is :
1. (A)12
2. (B)120
3. (C)1200
4. (D)40
Try it first, then show our solutionSolution
1. For any two positive integers, HCF $\times$ LCM = product of the numbers.
2. HCF $\times$ LCM $= 40 \times 30 = 1200$ .
  1 mark
Answer: 1200, option (C)
Where the marks are
- Uses HCF × LCM = product of the numbers to get 1200 (1 mark)
Q2 · 1 mark · MCQQuadratic Equations
The roots of the quadratic equation $x^{2} + 3 x - 10 = 0$ are :
1. (A)5, 2
2. (B) $- 5, 2$
3. (C)5, $- 2$
4. (D) $- 5, - 2$
Try it first, then show our solutionSolution
1. Split the middle term: $x^{2} + 5 x - 2 x - 10 = (x + 5) (x - 2) = 0$ .
2. So $x = - 5$ or $x = 2$ .
  1 mark
Answer: $- 5, 2$ , option (B)
Where the marks are
- Factorises and gets roots $- 5$ and $2$ (1 mark)
Q3 · 1 mark · MCQPair of Linear Equations in Two Variables
The pair of linear equations $2 k x + 5 y = 7$ , $6 x + 5 y = 11$ have a unique solution, if
1. (A) $k \neq = 3$
2. (B) $k \neq = - 3$
3. (C) $k \neq = \frac{1}{3}$
4. (D) $k \neq = - \frac{1}{3}$
Try it first, then show our solutionSolution
1. Unique solution requires $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ : $\frac{2 k}{6} \neq = \frac{5}{5} = 1$ .
2. So $2 k \neq = 6$ , i.e. $k \neq = 3$ .
  1 mark
Answer: $k \neq = 3$ , option (A)
Where the marks are
- Applies $a_{1} / a_{2} \neq = b_{1} / b_{2}$ to get $k \neq = 3$ (1 mark)
Q4 · 1 mark · MCQStatistics
If the mean and mode of a frequency distribution are 28 and 16 respectively, then its median is :
1. (A)22
2. (B)23·5
3. (C)24
4. (D)24·5
Try it first, then show our solutionSolution
1. Empirical relation: $3 Median = Mode + 2 Mean = 16 + 56 = 72$ .
2. Median $= 24$ .
  1 mark
Answer: 24, option (C)
Where the marks are
- Uses 3 Median = Mode + 2 Mean to get 24 (1 mark)
Q5 · 1 mark · MCQProbability
A die is rolled once. What is the probability of getting an odd prime number ?
1. (A) $\frac{1}{6}$
2. (B) $\frac{1}{3}$
3. (C) $\frac{2}{3}$
4. (D) $\frac{3}{4}$
Try it first, then show our solutionSolution
1. Odd primes on a die: 3 and 5, i.e. 2 favourable outcomes out of 6.
2. P $= \frac{2}{6} = \frac{1}{3}$ .
  1 mark
Answer: $\frac{1}{3}$ , option (B)
Where the marks are
- Identifies 3, 5 and gets $\frac{1}{3}$ (1 mark)
Q6 · 1 mark · MCQAreas Related to Circles
If the area of a sector of a circle is $\frac{1}{8}$ of the area of the circle, then the central angle of the sector is :
1. (A) $3 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $\frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{1}{8} π r^{2} \Rightarrow θ = \frac{36 0 ^{\circ}}{8} = 4 5^{\circ}$ .
  1 mark
Answer: $4 5^{\circ}$ , option (B)
Where the marks are
- Equates sector fraction $θ /36 0^{\circ}$ to $1/8$ , gets $4 5^{\circ}$ (1 mark)
Q7 · 1 mark · MCQReal Numbers
Prime factorisation of 424 is :
1. (A) $2 \times 5^{3} \times 4$
2. (B) $2 \times 5^{3} \times 2$
3. (C) $2^{3} \times 5^{3}$
4. (D) $2^{4} \times 5^{3}$
Try it first, then show our solutionSolution
1. $424 = 2 \times 212 = 2 \times 2 \times 106 = 2 \times 2 \times 2 \times 53$ .
2. So $424 = 2^{3} \times 53$ (53 is prime), which is option (C).
  1 mark
Answer: $2^{3} \times 53$ , option (C)
Where the marks are
- Correct prime factorisation $2^{3} \times 53$ (1 mark)
Q8 · 1 mark · MCQReal Numbers
If x is a whole number, then $8^{x}$ ends with an even digit, except for which value of x ?
1. (A)6
2. (B)4
3. (C)2
4. (D)0
Try it first, then show our solutionSolution
1. For $x \geq 1$ , $8^{x}$ is even, so it ends in an even digit.
2. For $x = 0$ , $8^{0} = 1$ , which ends in the odd digit 1.
  1 mark
Answer: 0, option (D)
Where the marks are
- Notes $8^{0} = 1$ ends in an odd digit (1 mark)
Q9 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A tree casts a shadow 7 m long on the ground, when the angle of elevation of the Sun is $4 5^{\circ}$ . The height of the tree is :
1. (A) $73$ m
2. (B) $\frac{7}{3} 3$ m
3. (C)7 m
4. (D)3·5 m
Try it first, then show our solutionSolution
1. $tan 4 5^{\circ} = \frac{h}{7} \Rightarrow h = 7 \times 1 = 7$ m.
  1 mark
Answer: 7 m, option (C)
Where the marks are
- Uses $tan 4 5^{\circ} = h /7$ to get 7 m (1 mark)
Q10 · 1 mark · MCQIntroduction to Trigonometry
$(\frac{5}{c o t ^{2} θ} - \frac{5}{c o s ^{2} θ})$ is equal to :
1. (A)1
2. (B)5
3. (C) $- 5$
4. (D)0
Try it first, then show our solutionSolution
1. $\frac{5}{c o t ^{2} θ} - \frac{5}{c o s ^{2} θ} = 5 tan^{2} θ - 5 sec^{2} θ = 5 (tan^{2} θ - sec^{2} θ)$ .
2. Since $sec^{2} θ - tan^{2} θ = 1$ , value $= 5 (- 1) = - 5$ .
  1 mark
Answer: $- 5$ , option (C)
Where the marks are
- Converts to $5 (tan^{2} θ - sec^{2} θ)$ and uses identity to get $- 5$ (1 mark)
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Q11 · 1 mark · MCQIntroduction to Trigonometry
If $tan^{2} θ = 3$ , where $θ$ is an acute angle, then the value of $θ$ is :
1. (A) $3 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $0^{\circ}$
4. (D) $4 5^{\circ}$
Try it first, then show our solutionSolution
1. $tan θ = 3$ (acute, so positive) $\Rightarrow θ = 6 0^{\circ}$ .
  1 mark
Answer: $6 0^{\circ}$ , option (B)
Where the marks are
- Takes $tan θ = 3$ , gets $6 0^{\circ}$ (1 mark)
Q12 · 1 mark · MCQAreas Related to Circles
The length of an arc of a circle with radius 12 cm is $10 π$ cm. The central angle subtended by this arc at the centre, is :
1. (A) $12 0^{\circ}$
2. (B) $6^{\circ}$
3. (C) $7 5^{\circ}$
4. (D) $15 0^{\circ}$
Try it first, then show our solutionSolution
1. $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r \Rightarrow 10 π = \frac{θ}{36 0 ^{\circ}} \times 24 π$ .
2. $θ = \frac{10 \times 36 0 ^{\circ}}{24} = 15 0^{\circ}$ .
  1 mark
Answer: $15 0^{\circ}$ , option (D)
Where the marks are
- Uses arc length formula to get $15 0^{\circ}$ (1 mark)
Q13 · 1 mark · MCQCircles
If the angle between the two radii of a circle is $13 0^{\circ}$ , then the angle between the tangents at the ends of these radii, is :
1. (A) $5 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $9 0^{\circ}$
4. (D) $13 0^{\circ}$
Try it first, then show our solutionSolution
1. The two radii and two tangents form a quadrilateral with two $9 0^{\circ}$ angles [tangent ⟂ radius].
2. Angle between tangents $= 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 13 0^{\circ} = 5 0^{\circ}$ .
  1 mark
Answer: $5 0^{\circ}$ , option (A)
Where the marks are
- Uses supplementary angles in the quadrilateral to get $5 0^{\circ}$ (1 mark)
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Q14 · 1 mark · MCQQuadratic Equations
The discriminant of the quadratic equation $x^{2} - 4 x + 3 = 0$ is :
1. (A)28
2. (B) $- 8$
3. (C)4
4. (D)2
Try it first, then show our solutionSolution
1. $D = b^{2} - 4 a c = (- 4)^{2} - 4 (1) (3) = 16 - 12 = 4$ .
  1 mark
Answer: 4, option (C)
Where the marks are
- Computes $b^{2} - 4 a c = 4$ (1 mark)
Q15 · 1 mark · MCQCoordinate Geometry
The mid-point of the line segment AB joining A( $- 2$ , 8) and B( $- 6$ , 4) is :
1. (A)(2, 6)
2. (B)( $- 4$ , 12)
3. (C)( $- 4$ , 6)
4. (D)(4, $- 6$ )
Try it first, then show our solutionSolution
1. Mid-point $= (\frac{- 2 - 6}{2}, \frac{8 + 4}{2}) = (- 4, 6)$ .
  1 mark
Answer: $(- 4, 6)$ , option (C)
Where the marks are
- Applies mid-point formula to get $(- 4, 6)$ (1 mark)
Q16 · 1 mark · MCQCircles
AP and AQ are tangents drawn from an external point A to a circle with centre O and inclined to each other at an angle of $9 0^{\circ}$ . If the length of each tangent is 2 cm, then the radius of the circle is :
1. (A)4 cm
2. (B)2 cm
3. (C) $22$ cm
4. (D)1 cm
Figure in the paper: Circle with centre O. External point A on the left with two tangent segments AP and AQ, each labelled 2 cm. P and Q are the points of contact on the circle; the angle at A is marked $9 0^{\circ}$ . Radii OP and OQ are drawn to the points of contact, and O is joined to the circle's centre. See it in the official paper.
Try it first, then show our solutionSolution
1. In APOQ, $∠ A = 9 0^{\circ}$ , $∠ P = ∠ Q = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ O = 9 0^{\circ}$ and APOQ is a rectangle with $A P = A Q$ , i.e. a square.
2. Hence radius $O P = A Q = 2$ cm.
  1 mark
Answer: 2 cm, option (B)
Where the marks are
- Recognises APOQ as a square, radius = 2 cm (1 mark)
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Q17 · 1 mark · MCQTriangles
It is given that $△$ ABC $\sim$ $△$ DEF. If $∠$ A = $5 5^{\circ}$ , $∠$ E = $4 5^{\circ}$ , then $∠$ C is :
1. (A) $8 0^{\circ}$
2. (B) $9 0^{\circ}$
3. (C) $5 5^{\circ}$
4. (D) $4 5^{\circ}$
Try it first, then show our solutionSolution
1. Corresponding angles are equal: $∠ B = ∠ E = 4 5^{\circ}$ .
2. $∠ C = 18 0^{\circ} - 5 5^{\circ} - 4 5^{\circ} = 8 0^{\circ}$ .
  1 mark
Answer: $8 0^{\circ}$ , option (A)
Where the marks are
- Uses $∠ B = ∠ E$ and angle sum to get $8 0^{\circ}$ (1 mark)
Q18 · 1 mark · MCQTriangles
In the given figure, in $△$ ABC, DE $∥$ BC. If AD = $2 x$ cm, AE = $(x + 2)$ cm, DB = 4 cm, EC = 3 cm, then the value of x is :
1. (A)3
2. (B)2
3. (C)6
4. (D)4
Figure in the paper: Triangle ABC with A at the top. D lies on AB and E lies on AC, with DE parallel to BC (arrows marked). Labels: AD = $2 x$ cm, AE = $(x + 2)$ cm, DB = 4 cm, EC = 3 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ : $\frac{2 x}{4} = \frac{x + 2}{3}$ .
2. $6 x = 4 x + 8 \Rightarrow x = 4$ .
  1 mark
Answer: $x = 4$ , option (D)
Where the marks are
- Applies BPT and solves to get $x = 4$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A) : The probability of getting number 8 on rolling a die is zero (0).
Reason (R) : The probability of an impossible event is zero (0).
Codes: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of 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 die shows only 1 to 6, so getting 8 is an impossible event; its probability is 0 — Assertion true.
2. Reason is a true fact and it directly explains the Assertion, so option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both true and R explains A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedArithmetic Progressions
Assertion (A) : Common difference of the A.P. 5, 1, $- 3$ , $- 7$ .... is 4.
Reason (R) : Common difference of the A.P. $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ is obtained by $d = a_{n} - a_{n - 1}$ .
Codes: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of 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. $d = 1 - 5 = - 4$ , not 4, so Assertion is false.
2. Reason: $d = a_{n} - a_{n - 1}$ is true. So option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Finds $d = - 4$ so A false, R true (1 mark)

Section B

Very short answer; internal choice in 22 and 25

Q21 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $sin^{2} 3 0^{\circ} + cos^{2} 4 5^{\circ} - cos 0^{\circ} \cdot tan 4 5^{\circ}$
Try it first, then show our solutionSolution
1. Substitute the standard values $sin 3 0^{\circ} = \frac{1}{2}$ , $cos 4 5^{\circ} = \frac{1}{2}$ , $cos 0^{\circ} = 1$ , $tan 4 5^{\circ} = 1$ :
  $(\frac{1}{2})^{2} + (\frac{1}{2})^{2} - 1 \times 1$
  1½ marks
2. $= \frac{1}{4} + \frac{1}{2} - 1 = \frac{1 + 2 - 4}{4} = - \frac{1}{4}$
  ½ mark
Answer: $- \frac{1}{4}$
Where the marks are
- Substitutes correct values of $sin 3 0^{\circ}$ , $cos 4 5^{\circ}$ , $cos 0^{\circ}$ and $tan 4 5^{\circ}$ (1½ marks)
- Simplifies to $- \frac{1}{4}$ (½ mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerCoordinate Geometry
Find the ratio in which the point (3, y) divides the line segment joining the points (–2, –5) and (6, 3). Also, find the value of y.
Try it first, then show our solutionSolution
1. Let $P (3, y)$ divide the join of $(- 2, - 5)$ and $(6, 3)$ in the ratio $k : 1$ . By the section formula:
  $3 = \frac{6 k + 1 ( - 2 )}{k + 1}$ , $y = \frac{3 k + 1 ( - 5 )}{k + 1}$
  1 mark
2. $3 k + 3 = 6 k - 2 \Rightarrow 3 k = 5 \Rightarrow k = \frac{5}{3}$ , so the ratio is $\frac{5}{3} : 1 = 5 : 3$ .
  ½ mark
3. $y = \frac{3 \cdot \frac{5}{3} - 5}{\frac{5}{3} + 1} = \frac{0}{\frac{8}{3}} = 0$
  ½ mark
Answer: Ratio 5 : 3; y = 0
Where the marks are
- Applies section formula with ratio $k : 1$ for both coordinates (1 mark)
- Finds $k = \frac{5}{3}$ , i.e. ratio 5 : 3 (½ mark)
- Finds $y = 0$ (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerCoordinate Geometry
Find a point on y-axis which is equidistant from the points A(6, 5) and B( $- 4$ , 3).
Try it first, then show our solutionSolution
1. Any point on the y-axis has x-coordinate 0; let it be $P (0, y)$ .
  ½ mark
2. $P A = P B \Rightarrow P A^{2} = P B^{2}$
  $(0 - 6)^{2} + (y - 5)^{2} = (0 + 4)^{2} + (y - 3)^{2}$
  1 mark
3. $36 + y^{2} - 10 y + 25 = 16 + y^{2} - 6 y + 9 \Rightarrow 61 - 10 y = 25 - 6 y \Rightarrow 4 y = 36 \Rightarrow y = 9$
  Required point is $(0, 9)$ .
  ½ mark
Answer: (0, 9)
Where the marks are
- Takes the point as $(0, y)$ (½ mark)
- Uses distance formula with $P A^{2} = P B^{2}$ (1 mark)
- Solves to get $y = 9$ , point $(0, 9)$ (½ mark)
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Q23 · 2 marks · Very short answerQuadratic Equations
If $\frac{2}{3}$ is a root of the quadratic equation $k x^{2} - x - 2 = 0$ , then find the value of k.
Try it first, then show our solutionSolution
1. Since $\frac{2}{3}$ is a root, it satisfies the equation:
  $k (\frac{2}{3})^{2} - \frac{2}{3} - 2 = 0$
  ½ mark
2. $\frac{4}{9} k = \frac{2}{3} + 2 = \frac{8}{3}$
  1 mark
3. $k = \frac{8}{3} \times \frac{9}{4} = 6$
  ½ mark
Answer: k = 6
Where the marks are
- Substitutes $x = \frac{2}{3}$ in the equation (½ mark)
- Obtains $\frac{4}{9} k = \frac{8}{3}$ (1 mark)
- Gets $k = 6$ (½ mark)
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Q24 · 2 marks · Very short answerCircles
The length of a tangent drawn to a circle from a point A, at a distance of 10 cm from the centre of the circle, is 6 cm. Find the radius of the circle.
Try it first, then show our solutionSolution
1. Let O be the centre and AT the tangent at T. Then $O T ⊥ A T$ [tangent ⟂ radius at point of contact], so in right $△ O T A$ , by Pythagoras theorem:
  $O T = O A^{2} - A T^{2} = 1 0^{2} - 6^{2}$
  1 mark
2. $= 100 - 36 = 64 = 8$ cm
  1 mark
Answer: Radius = 8 cm
Where the marks are
- Uses radius ⟂ tangent and Pythagoras: $r = 1 0^{2} - 6^{2}$ (1 mark)
- Gets $r = 8$ cm (1 mark)
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Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerPolynomials
If $α$ , $β$ are zeroes of the quadratic polynomial $2 x^{2} + 7 x + 5$ , then find the value of $α^{2} + β^{2} + α β$ .
Try it first, then show our solutionSolution
1. For $2 x^{2} + 7 x + 5$ : $α + β = - \frac{b}{a} = - \frac{7}{2}$ , $α β = \frac{c}{a} = \frac{5}{2}$
  1 mark
2. $α^{2} + β^{2} + α β = (α + β)^{2} - α β = (- \frac{7}{2})^{2} - \frac{5}{2} = \frac{49}{4} - \frac{10}{4} = \frac{39}{4}$
  1 mark
Answer: $\frac{39}{4}$
Where the marks are
- Writes sum $- \frac{7}{2}$ and product $\frac{5}{2}$ of zeroes (1 mark)
- Uses $(α + β)^{2} - α β$ to get $\frac{39}{4}$ (1 mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerPolynomials
If one zero of the quadratic polynomial $6 x^{2} + 37 x - (p - 2)$ is reciprocal of the other, then find the value of p.
Try it first, then show our solutionSolution
1. Here $a = 6$ , $b = 37$ , $c = - (p - 2)$ . Let the zeroes be $α$ and $\frac{1}{α}$ .
  ½ mark
2. Product of zeroes $= α \times \frac{1}{α} = \frac{c}{a} \Rightarrow 1 = \frac{- ( p - 2 )}{6}$
  1 mark
3. $6 = - p + 2 \Rightarrow p = - 4$
  ½ mark
Answer: p = −4
Where the marks are
- Takes zeroes as $α, \frac{1}{α}$ (½ mark)
- Equates product $1 = \frac{- ( p - 2 )}{6}$ (1 mark)
- Gets $p = - 4$ (½ mark)
Open Q25(b) on its own page

Section C

Short answer; internal choice in 29 and 30

Q26 · 3 marks · Short answerReal Numbers
Prove that $6 + 32$ is an irrational number, given that $2$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $6 + 32$ is rational. Then $6 + 32 = x$ , where $x$ is a rational number.
  1 mark
2. $32 = x - 6 \Rightarrow 2 = \frac{x - 6}{3}$
  1 mark
3. Since $x$ is rational, $\frac{x - 6}{3}$ is rational, so $2$ would be rational. This contradicts the given fact that $2$ is irrational. Hence our assumption is wrong, and $6 + 32$ is irrational.
  1 mark
Answer: $6 + 32$ is irrational (proved by contradiction).
Where the marks are
- Assumes $6 + 32$ is rational, equal to $x$ (1 mark)
- Expresses $2 = \frac{x - 6}{3}$ (1 mark)
- States the contradiction and concludes (1 mark)
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Q27 · 3 marks · Short answerPolynomials
Find the zeroes of the quadratic polynomial $6 x^{2} - 7 x - 3$ and verify the relationship between the zeroes and the coefficients of the polynomial.
Try it first, then show our solutionSolution
1. $6 x^{2} - 7 x - 3 = 6 x^{2} - 9 x + 2 x - 3 = 3 x (2 x - 3) + 1 (2 x - 3) = (2 x - 3) (3 x + 1)$
  1 mark
2. Zeroes: $2 x - 3 = 0 \Rightarrow x = \frac{3}{2}$ ; $3 x + 1 = 0 \Rightarrow x = - \frac{1}{3}$
  1 mark
3. Sum of zeroes $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6}$ and $- \frac{coeff. of x}{coeff. of x ^{2}} = - \frac{- 7}{6} = \frac{7}{6}$ . Verified.
  ½ mark
4. Product of zeroes $= \frac{3}{2} \times (- \frac{1}{3}) = - \frac{1}{2}$ and $\frac{constant term}{coeff. of x ^{2}} = \frac{- 3}{6} = - \frac{1}{2}$ . Verified.
  ½ mark
Answer: Zeroes $\frac{3}{2}$ and $- \frac{1}{3}$ ; sum $\frac{7}{6}$ and product $- \frac{1}{2}$ verified.
Where the marks are
- Factorises as $(2 x - 3) (3 x + 1)$ (1 mark)
- Finds zeroes $\frac{3}{2}, - \frac{1}{3}$ (1 mark)
- Verifies sum of zeroes (½ mark)
- Verifies product of zeroes (½ mark)
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Q28 · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $(sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2} = 7 + tan^{2} θ + cot^{2} θ$ .
Try it first, then show our solutionSolution
1. LHS $= sin^{2} θ + cosec^{2} θ + 2 sin θ cosec θ + cos^{2} θ + sec^{2} θ + 2 cos θ sec θ$
  1 mark
2. $= (sin^{2} θ + cos^{2} θ) + cosec^{2} θ + sec^{2} θ + 2 sin θ \cdot \frac{1}{s i n θ} + 2 cos θ \cdot \frac{1}{c o s θ}$
  ½ mark
3. $= 1 + (1 + cot^{2} θ) + (1 + tan^{2} θ) + 2 + 2$ [using $sin^{2} θ + cos^{2} θ = 1$ , $cosec^{2} θ = 1 + cot^{2} θ$ , $sec^{2} θ = 1 + tan^{2} θ$ ]
  1 mark
4. $= 7 + tan^{2} θ + cot^{2} θ =$ RHS. Hence proved.
  ½ mark
Answer: LHS = $7 + tan^{2} θ + cot^{2} θ$ = RHS, proved.
Where the marks are
- Expands both squares correctly (1 mark)
- Groups terms and uses reciprocal relations (½ mark)
- Applies the three identities (1 mark)
- Simplifies to RHS (½ mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCircles
Prove that the parallelogram circumscribing a circle is a rhombus.
Try it first, then show our solutionSolution
1. Figure: Draw parallelogram ABCD with all four sides touching a circle. The points of contact are P on AB, Q on BC, R on CD and S on DA.
  Given: ABCD is a parallelogram circumscribing the circle.
  To prove: ABCD is a rhombus.
  ½ mark
2. Tangents drawn from an external point to a circle are equal in length. So:
  $A P = A S, B P = B Q, C R = C Q, D R = D S$
  1 mark
3. Adding these four equations:
  $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$
  $\Rightarrow A B + C D = A D + B C$
  ½ mark
4. In a parallelogram opposite sides are equal: $A B = C D$ and $B C = A D$ .
  So $2 A B = 2 B C$ , which gives $A B = B C$ .
  ½ mark
5. Therefore $A B = B C = C D = D A$ . A parallelogram with all sides equal is a rhombus, so ABCD is a rhombus. Hence proved.
  ½ mark
Answer: AB + CD = AD + BC together with AB = CD and BC = AD gives AB = BC = CD = DA, so ABCD is a rhombus
Where the marks are
- Correct labelled figure (1/2 mark)
- Writes the four equal-tangent relations (1 mark)
- Adds them to get $A B + C D = A D + B C$ (1/2 mark)
- Uses the opposite sides of the parallelogram to get $A B = B C$ (1/2 mark)
- Concludes all sides are equal, so ABCD is a rhombus (1/2 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCircles
Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O, an external point P, and tangents PA and PB touching the circle at A and B.
  Given: PA and PB are tangents from P. To prove: $P A = P B$ .
  ½ mark
2. Construction: Join OA, OB and OP.
  ½ mark
3. In $△ O A P$ and $△ O B P$ : $O A = O B$ [radii], $O P = O P$ [common], $∠ O A P = ∠ O B P = 9 0^{\circ}$ [tangent ⟂ radius at point of contact]
  1 mark
4. $∴ △ O A P ≅ △ O B P$ [RHS congruence]
  ½ mark
5. $\Rightarrow P A = P B$ [CPCT]. Hence proved.
  ½ mark
Answer: PA = PB, proved by RHS congruence.
Where the marks are
- Correct figure with given/to prove (½ mark)
- Construction: join OA, OB, OP (½ mark)
- Lists equal parts with right angles from tangent ⟂ radius (1 mark)
- States RHS congruence (½ mark)
- Concludes PA = PB by CPCT (½ mark)
Open Q29(b) on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerCompetency-basedAreas Related to Circles
A horse is tied with a 14 m long rope at one corner of an equilateral triangular field having side 20 m. Find the area of the field where the horse cannot graze.
Try it first, then show our solutionSolution
1. Each angle of an equilateral triangle is $6 0^{\circ}$ , so the horse grazes a sector with $r = 14$ m and $θ = 6 0^{\circ}$ .
  ½ mark
2. Area grazed $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{60}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{308}{3} \approx 102.67$ m²
  1 mark
3. Area of field $= \frac{3}{4} a^{2} = \frac{3}{4} \times 400 = 1003 \approx 173.2$ m²
  1 mark
4. Area not grazed $= 1003 - \frac{308}{3} \approx 173.2 - 102.67 = 70.53$ m²
  ½ mark
Answer: $(1003 - \frac{308}{3})$ m² ≈ 70.53 m²
Where the marks are
- Identifies sector angle as 60° (½ mark)
- Area of sector $\frac{308}{3}$ m² (1 mark)
- Area of triangle $1003$ m² (1 mark)
- Subtracts to get the ungrazed area (½ mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerCompetency-basedAreas Related to Circles
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand between 8:00 am and 8:05 am.
Try it first, then show our solutionSolution
1. The minute hand turns $36 0^{\circ}$ in 60 minutes, so in 5 minutes it turns $\frac{36 0 ^{\circ}}{60} \times 5 = 3 0^{\circ}$ .
  1 mark
2. Area swept $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$
  1½ marks
3. $= \frac{154}{3}$ cm² $\approx 51.33$ cm²
  ½ mark
Answer: $\frac{154}{3}$ cm² ≈ 51.33 cm²
Where the marks are
- Angle swept in 5 minutes = 30° (1 mark)
- Uses sector area formula with correct substitution (1½ marks)
- Gets $\frac{154}{3}$ cm² (½ mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerCoordinate Geometry
Find the coordinates of the points of trisection of the line segment joining the points A(5, $- 3$ ) and B( $- 4$ , 3).
Try it first, then show our solutionSolution
1. Let P and Q be the points of trisection, with $A P : P B = 1 : 2$ . By section formula:
  $P (\frac{1 ( - 4 ) + 2 ( 5 )}{1 + 2}, \frac{1 ( 3 ) + 2 ( - 3 )}{1 + 2})$
  1 mark
2. $P = (\frac{6}{3}, \frac{- 3}{3}) = (2, - 1)$
  ½ mark
3. Q divides AB with $A Q : QB = 2 : 1$ :
  $Q (\frac{2 ( - 4 ) + 1 ( 5 )}{2 + 1}, \frac{2 ( 3 ) + 1 ( - 3 )}{2 + 1})$
  1 mark
4. $Q = (\frac{- 3}{3}, \frac{3}{3}) = (- 1, 1)$
  ½ mark
Answer: (2, −1) and (−1, 1)
Where the marks are
- Section formula for P in ratio 1 : 2 (1 mark)
- P = (2, −1) (½ mark)
- Section formula for Q in ratio 2 : 1 (1 mark)
- Q = (−1, 1) (½ mark)
Open Q31 on its own page

Section D

Long answer; internal choice in 32 and 34

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerSurface Areas and Volumes
A toy is in the form of a cone of radius 7 cm mounted on a hemisphere of same radius. The total height of the toy is 31 cm. Find the surface area of the toy.
Try it first, then show our solutionSolution
1. Radius of cone = radius of hemisphere $= r = 7$ cm. Height of cone $h = 31 - 7 = 24$ cm.
  1 mark
2. Slant height $l = r^{2} + h^{2} = 7^{2} + 2 4^{2} = 625 = 25$ cm
  1 mark
3. Surface area of toy = CSA of hemisphere + CSA of cone $= 2 π r^{2} + π r l = π r (2 r + l)$
  ½ mark
4. $= \frac{22}{7} \times 7 \times (14 + 25) = 22 \times 39$
  2 marks
5. $= 858$ cm²
  ½ mark
Answer: 858 cm²
Where the marks are
- Height of cone 24 cm (1 mark)
- Slant height 25 cm (1 mark)
- Surface area = CSA of hemisphere + CSA of cone (½ mark)
- Correct substitution $\frac{22}{7} \times 7 \times 39$ (2 marks)
- Answer 858 cm² (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerSurface Areas and Volumes
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 15 cm and its base is of radius 4·2 cm, then find the total surface area of the article.
Figure in the paper: A solid cylinder standing upright. A hemispherical cavity is scooped out from the top base and another from the bottom base, shown shaded, so the article has hollow hemispherical ends. See it in the official paper.
Try it first, then show our solutionSolution
1. Height of cylinder $h = 15$ cm; radius of cylinder = radius of each hemisphere $= r = 4.2$ cm.
  1 mark
2. TSA of article = CSA of cylinder + CSA of 2 hemispheres $= 2 π r h + 2 (2 π r^{2}) = 2 π r h + 4 π r^{2}$
  1 mark
3. $= 2 π r (h + 2 r) = 2 \times \frac{22}{7} \times 4.2 \times (15 + 2 \times 4.2)$
  2 marks
4. $= 2 \times \frac{22}{7} \times 4.2 \times 23.4 = 26.4 \times 23.4 = 617.76$ cm²
  1 mark
Answer: 617.76 cm²
Where the marks are
- Identifies h = 15 cm and r = 4.2 cm (1 mark)
- TSA = CSA of cylinder + CSA of two hemispheres (1 mark)
- Correct substitution (2 marks)
- Answer 617.76 cm² (1 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables
The ratio of monthly incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹ 2,000 per month, then find their monthly incomes.
Try it first, then show our solutionSolution
1. Let the monthly incomes be ₹ $9 x$ and ₹ $7 x$ , and the expenditures be ₹ $4 y$ and ₹ $3 y$ .
  1 mark
2. Saving = income − expenditure, so for the first person: $9 x - 4 y = 2000$ ...(i)
  1 mark
3. For the second person: $7 x - 3 y = 2000$ ...(ii)
  1 mark
4. (i)×3: $27 x - 12 y = 6000$ ; (ii)×4: $28 x - 12 y = 8000$ . Subtracting: $x = 2000$ (and then $y = 4000$ ).
  1 mark
5. Monthly incomes: $9 \times 2000 = ₹18000$ and $7 \times 2000 = ₹14000$ .
  1 mark
Answer: ₹18,000 and ₹14,000
Where the marks are
- Takes incomes 9x, 7x and expenditures 4y, 3y (1 mark)
- Forms $9 x - 4 y = 2000$ (1 mark)
- Forms $7 x - 3 y = 2000$ (1 mark)
- Solves to get $x = 2000$ (1 mark)
- States incomes ₹18000 and ₹14000 (1 mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerTriangles
Sides AB, BC and median AD of $△$ ABC are respectively proportional to sides PQ, QR and median PM of $△$ PQR. Show that $△$ ABC $\sim$ $△$ PQR.
Try it first, then show our solutionSolution
1. Figure: $△ A B C$ with median AD (D mid-point of BC) and $△ P QR$ with median PM (M mid-point of QR).
  ½ mark
2. Given: $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ , AD and PM medians. To prove: $△ A B C \sim △ P QR$ .
  1 mark
3. Since D and M are mid-points, $B D = \frac{1}{2} B C$ , $QM = \frac{1}{2} QR$ . So $\frac{A B}{P Q} = \frac{\frac{1}{2} B C}{\frac{1}{2} QR} = \frac{A D}{P M} \Rightarrow \frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$
  1 mark
4. $∴ △ A B D \sim △ P QM$ [SSS similarity criterion]
  ½ mark
5. $\Rightarrow ∠ B = ∠ Q$ [corresponding angles of similar triangles]
  ½ mark
6. In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ [given] and $∠ B = ∠ Q$ [proved]. $∴ △ A B C \sim △ P QR$ [SAS similarity criterion].
  1½ marks
Answer: $△ A B C \sim △ P QR$ (via $△ A B D \sim △ P QM$ by SSS, then SAS).
Where the marks are
- Correct figure (½ mark)
- Given and to prove written (1 mark)
- Shows $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ (1 mark)
- $△ A B D \sim △ P QM$ by SSS (½ mark)
- Deduces $∠ B = ∠ Q$ (½ mark)
- Concludes $△ A B C \sim △ P QR$ by SAS (1½ marks)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerTriangles
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.
Try it first, then show our solutionSolution
1. Figure: $△ A B C$ with a line DE ∥ BC meeting AB at D and AC at E.
  Given: In $△ A B C$ , $D E ∥ B C$ . To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
  ½ mark
2. Construction: Join BE and CD. Draw $D M ⊥ A C$ and $E N ⊥ A B$ .
  1 mark
3. $\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{\frac{1}{2} \times A D \times E N}{\frac{1}{2} \times D B \times E N} = \frac{A D}{D B}$ ...(i)
  1 mark
4. $\frac{a r ( △ A D E )}{a r ( △ C D E )} = \frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times E C \times D M} = \frac{A E}{E C}$ ...(ii)
  1 mark
5. $△ B D E$ and $△ C D E$ lie on the same base DE and between the same parallels DE and BC, so $a r (△ B D E) = a r (△ C D E)$ ...(iii)
  1 mark
6. From (i), (ii) and (iii): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  ½ mark
Answer: $\frac{A D}{D B} = \frac{A E}{E C}$ (Basic Proportionality Theorem proved).
Where the marks are
- Figure with given and to prove (½ mark)
- Construction: join BE, CD; draw DM ⟂ AC, EN ⟂ AB (1 mark)
- Ratio $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ (1 mark)
- Ratio $\frac{a r ( A D E )}{a r ( C D E )} = \frac{A E}{E C}$ (1 mark)
- $a r (B D E) = a r (C D E)$ with reason (1 mark)
- Conclusion (½ mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerStatistics
The following frequency distribution table gives the monthly consumption of electricity of 70 consumers of a locality. Find the median of the data.
Monthly consumption (in units) Number of consumers
65 – 85 7
85 – 105 8
105 – 125 7
125 – 145 20
145 – 165 14
165 – 185 9
185 – 205 5
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Monthly consumption (units) No. of consumers $f$ $c f$
  65 – 85 7 7
  85 – 105 8 15
  105 – 125 7 22
  125 – 145 20 42
  145 – 165 14 56
  165 – 185 9 65
  185 – 205 5 70
  2 marks
2. $N = 70$ , so $\frac{N}{2} = 35$ . The first $c f$ greater than 35 is 42, so the median class is 125 – 145. Hence $l = 125$ , $c f = 22$ , $f = 20$ and $h = 20$ .
  1 mark
3. Median $= l + (\frac{\frac{N}{2} - c f}{f}) \times h = 125 + \frac{35 - 22}{20} \times 20$
  1½ marks
4. $= 125 + 13 = 138$ . The median monthly consumption is 138 units.
  ½ mark
Answer: Median = 138 units
Where the marks are
- Correct cumulative frequency column (2 marks)
- Identifies $N /2 = 35$ , median class 125–145 and $l, c f, f, h$ (1 mark)
- Correct substitution in the median formula (1½ marks)
- Median = 138 (½ mark)
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Monthly consumption (in units)	Number of consumers
65 – 85	7
85 – 105	8
105 – 125	7
125 – 145	20
145 – 165	14
165 – 185	9
185 – 205	5

Monthly consumption (units)	No. of consumers $f$	$c f$
65 – 85	7	7
85 – 105	8	15
105 – 125	7	22
125 – 145	20	42
145 – 165	14	56
165 – 185	9	65
185 – 205	5	70

Section E

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

Q36 · 4 marks · Case-basedCompetency-basedProbability
Case Study – 1
Family structure : In a recent survey of this year, 51% of the families in the United States of America had no children, 20% had one child, 19% had two children, 7% had three children and 3% had four or more children. A family is selected at random.
(The page shows four illustrations of families with different numbers of children.)
Based on the above information, answer the following questions :
1. (i)
  Find the probability that the selected family has two or three children.
  1 mark
2. (ii)
  Find the probability that the selected family has more than one child.
  1 mark
3. (iii)
  (a) Find the probability that the selected family has less than three children.
  2 marks
4. OR
  (iii)
  (b) Find the probability that the selected family has more than two children.
  2 marks
Try it first, then show our solutionSolution
1. (i) Families with two or three children make up $19% + 7% = 26%$ . So P(2 or 3 children) $= \frac{19 + 7}{100}$
  ½ mark
2. (i) $= \frac{26}{100} = \frac{13}{50}$
  ½ mark
3. (ii) More than one child means 2, 3, or 4 or more children. So P $= \frac{19 + 7 + 3}{100}$
  ½ mark
4. (ii) $= \frac{29}{100}$
  ½ mark
5. (iii)(a) Less than three children means 0, 1 or 2 children. So P $= \frac{51 + 20 + 19}{100}$
  1½ marks
6. (iii)(a) $= \frac{90}{100} = \frac{9}{10}$
  ½ mark
7. (iii)(b) More than two children means 3, or 4 or more children. So P $= \frac{7 + 3}{100}$
  1½ marks
8. (iii)(b) $= \frac{10}{100} = \frac{1}{10}$
  ½ mark
Answer: (i) 13/50; (ii) 29/100; (iii)(a) 9/10; (iii)(b) 1/10
Where the marks are
- (i) Adds 19% and 7% as favourable cases (½ mark)
- (i) Gets 13/50 (½ mark)
- (ii) Adds 19 + 7 + 3 (½ mark)
- (ii) Gets 29/100 (½ mark)
- (iii)(a) Adds 51 + 20 + 19 out of 100 (1½ marks)
- (iii)(a) Gets 9/10 (½ mark)
- (iii)(b) Adds 7 + 3 out of 100 (1½ marks)
- (iii)(b) Gets 1/10 (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Case Study – 2
Sumant's mother started a new shoe shop. To display the shoes, she put 3 pairs of shoes in the 1st row, 5 pairs in the 2nd row, 7 pairs in the 3rd row and so on.
(The page shows a 'New Collection Big Sale' social media advertisement for shoes.)
Based on the above information, answer the following questions :
1. (i)
  How many pairs of shoes are displayed in the 6th row ?
  1 mark
2. (ii)
  What is the difference of pairs of shoes in the 1st row and the 6th row ?
  1 mark
3. (iii)
  (a) Find the total number of pairs of shoes displayed in the first 15 rows.
  2 marks
4. OR
  (iii)
  (b) If the pairs of shoes displayed in the 4th row are 'on sale' at price of ₹ 500 for each pair, then find the total amount (money) earned by Sumant's mother if all shoes displayed in the 4th row are sold out.
  2 marks
Try it first, then show our solutionSolution
1. (i) The number of pairs in successive rows is 3, 5, 7, ... This is an AP with $a = 3$ and $d = 2$ .
  ½ mark
2. (i) $a_{6} = a + 5 d = 3 + 5 (2) = 13$ pairs.
  ½ mark
3. (ii) Difference $= a_{6} - a_{1} = 13 - 3 = 10$ pairs.
  1 mark
4. (iii)(a) Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with $n = 15$ : $S_{15} = \frac{15}{2} [2 (3) + 14 (2)]$
  1 mark
5. (iii)(a) $= \frac{15}{2} \times 34 = 255$ pairs.
  1 mark
6. (iii)(b) $a_{4} = a + 3 d = 3 + 3 (2) = 9$ pairs.
  1 mark
7. (iii)(b) Money earned $= 9 \times ₹500 = ₹4500$
  1 mark
Answer: (i) 13 pairs; (ii) 10; (iii)(a) 255 pairs; (iii)(b) ₹4500
Where the marks are
- (i) Identifies $a = 3$ , $d = 2$ (½ mark)
- (i) Finds $a_{6} = 13$ (½ mark)
- (ii) Finds the difference 10 (1 mark)
- (iii)(a) Correct substitution in $S_{15}$ (1 mark)
- (iii)(a) Gets 255 (1 mark)
- (iii)(b) Finds $a_{4} = 9$ (1 mark)
- (iii)(b) Money earned ₹4500 (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Case Study – 3
Two poles of different heights stand on level ground and at a distance of 40 m. Both poles are supported by wires attached from the top of each pole to the bottom of the other. A coupling is placed at point C, where the two wires cross (as shown in the figure).
Based on the above information, answer the following questions :
1. (i)
  Find the height of pole AB.
  1 mark
2. (ii)
  Find the height of pole PQ.
  1 mark
3. (iii)
  (a) If the angle of elevation of the top of the pole PQ from the top of the pole AB is $3 0^{\circ}$ , find the distance BQ.
  2 marks
4. OR
  (iii)
  (b) If the coupling is at a height of 20 m from the ground, how far down the wire from the smaller pole AB is the coupling ?
  2 marks
Figure in the paper: Two poles AB and PQ stand on level ground, with A and P at the base, 40 m apart (marked on the ground line). AB is the shorter pole (B at top) and PQ is the taller pole (Q at top). A wire runs from A to Q and another from P to B; the two wires cross at point C. At A, the angle between ground AP and wire AQ is $6 0^{\circ}$ . At P, the angle between ground PA and wire PB is $3 0^{\circ}$ . Right angles are at A and P. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In right $△ B A P$ (right angle at A): $tan 3 0^{\circ} = \frac{A B}{A P}$
  ½ mark
2. (i) $\frac{1}{3} = \frac{A B}{40} \Rightarrow A B = \frac{40}{3} = \frac{40 3}{3}$ m
  ½ mark
3. (ii) In right $△ QP A$ (right angle at P): $tan 6 0^{\circ} = \frac{P Q}{A P}$
  ½ mark
4. (ii) $3 = \frac{P Q}{40} \Rightarrow P Q = 403$ m
  ½ mark
5. (iii)(a) Figure: draw $B M ⊥ P Q$ , so ABMP is a rectangle and $B M = A P = 40$ m. Then $∠ QB M = 3 0^{\circ}$ in right $△ B M Q$ .
  ½ mark
6. (iii)(a) $cos 3 0^{\circ} = \frac{B M}{B Q}$
  ½ mark
7. (iii)(a) $\frac{3}{2} = \frac{40}{B Q} \Rightarrow B Q = \frac{80}{3} = \frac{80 3}{3}$ m
  1 mark
8. (iii)(b) Figure: draw $C D ⊥ A P$ , so $C D = 20$ m. In right $△ A D C$ , $∠ C A D = 6 0^{\circ}$ .
  ½ mark
9. (iii)(b) $sin 6 0^{\circ} = \frac{C D}{A C}$
  ½ mark
10. (iii)(b) $\frac{3}{2} = \frac{20}{A C} \Rightarrow A C = \frac{40}{3} = \frac{40 3}{3}$ m $(\approx 23.09$ m $)$ . So the coupling is $\frac{40 3}{3}$ m along the wire from the foot A of pole AB.
  1 mark
Answer: (i) AB = 40√3/3 m; (ii) PQ = 40√3 m; (iii)(a) BQ = 80√3/3 m; (iii)(b) AC = 40√3/3 m
Where the marks are
- (i) Uses $tan 3 0^{\circ} = A B / A P$ (½ mark)
- (i) Gets $A B = 403 /3$ m (½ mark)
- (ii) Uses $tan 6 0^{\circ} = P Q / A P$ (½ mark)
- (ii) Gets $P Q = 403$ m (½ mark)
- (iii)(a) Figure with $B M ⊥ P Q$ , $B M = 40$ m (½ mark)
- (iii)(a) Uses $cos 3 0^{\circ} = B M / B Q$ (½ mark)
- (iii)(a) Gets $B Q = 803 /3$ m (1 mark)
- (iii)(b) Figure with $C D = 20$ m (½ mark)
- (iii)(b) Uses $sin 6 0^{\circ} = C D / A C$ (½ mark)
- (iii)(b) Gets $A C = 403 /3$ m (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.