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

CBSE Class 10 Mathematics Basic: 2025 board paper, set 430/1/1

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.

43 of 44 questions (counting each OR alternative) have our step-by-step solution so far. We show a solution only after an independent re-solve agrees with it; the rest say "Solution being checked".

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2025/X/241_Mathematics_Basic.zip (a zip; this paper is the file 430-1-1_Mathematics 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 24
C	26-31	Short answer; internal choice in 26 and 28
D	32-35	Long answer; internal choice in 32 and 34
E	36-38	Case study based; internal choice in the 2-mark subpart of each case

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 case studies in Section E (the 2-mark subpart of each case). Use of calculator is not allowed. Take π = 22/7 where required unless stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If the HCF of two positive integers a and b is 1, then their LCM is :
1. (A)a + b
2. (B)a
3. (C)b
4. (D)ab
Try it first, then show our solutionSolution
1. For any two positive integers, $HCF \times LCM = a \times b$ .
2. Here HCF $= 1$ , so LCM $= \frac{ab}{1} = ab$ . The answer is option (D).
  1 mark
Answer: ab, option (D)
Where the marks are
- Uses HCF × LCM = ab to get LCM = ab (1 mark)
Q2 · 1 mark · MCQReal Numbers
The number $3 + 2$ is :
1. (A)a rational number
2. (B)an irrational number
3. (C)an integer
4. (D)a natural number
Try it first, then show our solutionSolution
1. $2$ is irrational. If $3 + 2$ were rational, then $2 = (3 + 2) - 3$ would also be rational, which is false.
2. So $3 + 2$ is an irrational number. The answer is option (B).
  1 mark
Answer: An irrational number, option (B)
Where the marks are
- Identifies $3 + 2$ as irrational (1 mark)
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Q3 · 1 mark · MCQQuadratic Equations
The discriminant of the quadratic equation $x^{2} - 3 x - 2 = 0$ is :
1. (A)1
2. (B)17
3. (C) $17$
4. (D) $- 17$
Try it first, then show our solutionSolution
1. Here $a = 1$ , $b = - 3$ , $c = - 2$ .
2. $D = b^{2} - 4 a c = 9 - 4 (1) (- 2) = 9 + 8 = 17$ . The answer is option (B).
  1 mark
Answer: 17, option (B)
Where the marks are
- Computes $D = b^{2} - 4 a c = 17$ (1 mark)
Q4 · 1 mark · MCQQuadratic Equations
The equation $x + \frac{1}{x} = 3$ $(x \neq = 0)$ is expressed as a quadratic equation in the form of $a x^{2} + b x + c = 0$ . The value of $a - b + c$ is :
1. (A)5
2. (B)2
3. (C)1
4. (D) $- 1$
Try it first, then show our solutionSolution
1. Multiply both sides by $x$ : $x^{2} + 1 = 3 x$ , so $x^{2} - 3 x + 1 = 0$ . Hence $a = 1$ , $b = - 3$ , $c = 1$ .
2. $a - b + c = 1 + 3 + 1 = 5$ . The answer is option (A).
  1 mark
Answer: 5, option (A)
Where the marks are
- Forms $x^{2} - 3 x + 1 = 0$ and gets $a - b + c = 5$ (1 mark)
Q5 · 1 mark · MCQCoordinate Geometry
For a point $(3, - 5)$ , the value of (abscissa $-$ ordinate) is :
1. (A) $- 8$
2. (B) $- 2$
3. (C)2
4. (D)8
Try it first, then show our solutionSolution
1. For the point $(3, - 5)$ , abscissa $= 3$ and ordinate $= - 5$ .
2. Abscissa − ordinate $= 3 - (- 5) = 8$ . The answer is option (D).
  1 mark
Answer: 8, option (D)
Where the marks are
- Computes $3 - (- 5) = 8$ (1 mark)
Q6 · 1 mark · MCQCoordinate Geometry
The mid-point of a line segment divides the line segment in the ratio :
1. (A)1 : 2
2. (B)2 : 1
3. (C)1 : 1
4. (D) $\frac{1}{2} : 2$
Try it first, then show our solutionSolution
1. The mid-point M of segment AB satisfies $A M = M B$ , so $A M : M B = 1 : 1$ . Hence option (C).
  1 mark
Answer: 1 : 1, option (C)
Where the marks are
- Chooses 1 : 1 since the mid-point is equidistant from both ends (1 mark)
Q7 · 1 mark · MCQTriangles
Which of the following is not the criterion for similarity of triangles ?
1. (A)AAA
2. (B)SSS
3. (C)SAS
4. (D)RHS
Solution being checked. Our step-by-step solution is written and is waiting for an independent check before we show it.
Q8 · 1 mark · MCQTriangles
From the figures given below, which of the following is true about the measure of $∠ P$ ?
1. (A) $∠ P = 6 0^{\circ}$
2. (B) $∠ P = 8 0^{\circ}$
3. (C) $∠ P = 4 0^{\circ}$
4. (D)The measure of $∠ P$ cannot be determined
Figure in the paper: Two triangles side by side. Triangle ABC: AB = 3.8 cm, BC = 6 cm, AC = $33$ cm, angle at A = 80°, angle at B = 60°. Triangle PQR: PQ = 12 cm, PR = $63$ cm, QR = 7.6 cm, with P, Q, R as vertices (R at the top). See it in the official paper.
Try it first, then show our solutionSolution
1. Compare the sides: $\frac{P Q}{B C} = \frac{12}{6} = 2$ , $\frac{P R}{A C} = \frac{6 3}{3 3} = 2$ , $\frac{QR}{A B} = \frac{7.6}{3.8} = 2$ . So $△ A B C \sim △ R QP$ by SSS, with $A \leftrightarrow R$ , $B \leftrightarrow Q$ , $C \leftrightarrow P$ .
2. $∠ C = 18 0^{\circ} - 8 0^{\circ} - 6 0^{\circ} = 4 0^{\circ}$ [angle sum property].
3. So $∠ P = ∠ C = 4 0^{\circ}$ [corresponding angles of similar triangles]. The answer is option (C).
  1 mark
Answer: ∠P = 40°, option (C)
Where the marks are
- Establishes similarity and gets ∠P = ∠C = 40° (1 mark)
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Q9 · 1 mark · MCQCircles
In the given figure, PA is a tangent to a circle with centre O. If OP = 10 cm, then the length of AP is :
1. (A) $103$ cm
2. (B)20 cm
3. (C)5 cm
4. (D) $53$ cm
Figure in the paper: Circle with centre O; P outside the circle; PA is a tangent touching the circle at A. OA is a radius. Angle APO = 30°. OP = 10 cm is marked along the line from O to P. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
2. In right $△ O A P$ : $cos 3 0^{\circ} = \frac{A P}{O P}$ , so $A P = 10 \times \frac{3}{2} = 53$ cm. The answer is option (D).
  1 mark
Answer: 5√3 cm, option (D)
Where the marks are
- Uses ∠OAP = 90° and cos 30° to get AP = 5√3 cm (1 mark)
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Q10 · 1 mark · MCQIntroduction to Trigonometry
Which of the following statements is false ?
1. (A) $tan 4 5^{\circ} = cot 4 5^{\circ}$
2. (B) $sin 9 0^{\circ} = tan 4 5^{\circ}$
3. (C) $sin 3 0^{\circ} = cos 3 0^{\circ}$
4. (D) $sin 4 5^{\circ} = cos 4 5^{\circ}$
Try it first, then show our solutionSolution
1. $tan 4 5^{\circ} = cot 4 5^{\circ} = 1$ , $sin 9 0^{\circ} = tan 4 5^{\circ} = 1$ , and $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}$ , so (A), (B) and (D) are true.
2. $sin 3 0^{\circ} = \frac{1}{2}$ but $cos 3 0^{\circ} = \frac{3}{2}$ , so statement (C) is false. The answer is option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Identifies sin 30° ≠ cos 30° (1 mark)
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Q11 · 1 mark · MCQIntroduction to Trigonometry
The value of $(tan^{2} A - \frac{1}{c o s ^{2} A})$ is :
1. (A)more than 1
2. (B)1
3. (C)0
4. (D) $- 1$
Try it first, then show our solutionSolution
1. $\frac{1}{cos ^{2} A} = sec^{2} A$ and $sec^{2} A = 1 + tan^{2} A$ .
2. So $tan^{2} A - sec^{2} A = - 1$ . The answer is option (D).
  1 mark
Answer: −1, option (D)
Where the marks are
- Uses sec²A − tan²A = 1 to get −1 (1 mark)
Q12 · 1 mark · MCQSome Applications of Trigonometry
In the given figure, which of the following angles represents the angle of depression ?
1. (A)x
2. (B)y
3. (C)z
4. (D)a
Figure in the paper: A right-angled figure with observer at the top-left and object at the bottom-right. A dashed horizontal line from the observer. A line of sight joins observer to object. Angles marked: z between the horizontal line and the top of the vertical line at the observer, y between the vertical line and the line of sight at the observer, x at the object between the line of sight and the horizontal ground, and a at the bottom-left corner (right angle mark). See it in the official paper.
Try it first, then show our solutionSolution
1. The angle of depression is the angle between the horizontal line through the observer's eye and the line of sight to an object below the observer.
2. In the figure this is angle $z$ . The answer is option (C).
  1 mark
Answer: z, option (C)
Where the marks are
- Identifies z as the angle between the horizontal and the line of sight at the observer (1 mark)
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Q13 · 1 mark · MCQAreas Related to Circles
The perimeter of the shaded region in the given figure is :
1. (A) $l$
2. (B) $l + a$
3. (C) $l + 2 r$
4. (D) $l + 2 r + a$
Figure in the paper: A circle with centre O. A chord of length a joins two points on the circle. The shaded region is the minor segment-plus-triangle region bounded by two radii OA and OB (each labelled r), the chord AB (labelled a), and the arc AB (labelled l) on the lower side of the chord. See it in the official paper.
Try it first, then show our solutionSolution
1. The shaded region is the sector bounded by the two radii $r$ and the arc $l$ . The chord $a$ lies inside the sector and is not part of its boundary.
2. Perimeter $= l + r + r = l + 2 r$ . The answer is option (C).
  1 mark
Answer: l + 2r, option (C)
Where the marks are
- Perimeter of the sector = l + 2r (1 mark)
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Q14 · 1 mark · MCQAreas Related to Circles
The ratio of the area of a quadrant of a circle to the area of the same circle is :
1. (A)1 : 2
2. (B)2 : 1
3. (C)1 : 4
4. (D)4 : 1
Try it first, then show our solutionSolution
1. Area of a quadrant $= \frac{1}{4} π r^{2}$ and area of the circle $= π r^{2}$ .
2. Ratio $= \frac{1}{4} π r^{2} : π r^{2} = 1 : 4$ . Hence option (C).
  1 mark
Answer: 1 : 4, option (C)
Where the marks are
- Uses quadrant area $\frac{1}{4} π r^{2}$ to get ratio 1 : 4 (1 mark)
Q15 · 1 mark · MCQSurface Areas and Volumes
For which of the following solids is the lateral / curved surface area and total surface area the same ?
1. (A)Cube
2. (B)Cuboid
3. (C)Hemisphere
4. (D)Sphere
Try it first, then show our solutionSolution
1. A sphere has no flat faces, so its whole surface is curved: CSA = TSA $= 4 π r^{2}$ .
2. Cube, cuboid and hemisphere all have extra faces (top/bottom or flat base), so their TSA exceeds their LSA/CSA. Hence option (D).
  1 mark
Answer: Sphere, option (D)
Where the marks are
- Identifies sphere as CSA = TSA = $4 π r^{2}$ (1 mark)
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Q16 · 1 mark · MCQStatistics
The class mark of the median class of the following data is :
Class Interval 10 – 25 25 – 40 40 – 55 55 – 70 70 – 85 85 – 100
Frequency 2 3 7 6 6 6
1. (A)40
2. (B)55
3. (C)47·5
4. (D)62·5
Try it first, then show our solutionSolution
1. $n = 2 + 3 + 7 + 6 + 6 + 6 = 30$ , so $\frac{n}{2} = 15$ . Cumulative frequencies are 2, 5, 12, 18, … .
2. The first cumulative frequency greater than 15 is 18, so the median class is 55 – 70.
3. Class mark $= \frac{55 + 70}{2} = 62.5$ . The answer is option (D).
  1 mark
Answer: 62.5, option (D)
Where the marks are
- Finds median class 55–70 and its class mark 62.5 (1 mark)
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Q17 · 1 mark · MCQStatistics
The following distribution shows the number of runs scored by some batsmen in test matches :
Runs Scored 3000 – 4000 4000 – 5000 5000 – 6000 6000 – 7000
Number of Batsmen 5 10 9 8
The lower limit of the modal class is :
1. (A)3000
2. (B)4000
3. (C)5000
4. (D)6000
Try it first, then show our solutionSolution
1. Highest frequency is 10, for class 4000 – 5000, so this is the modal class.
2. Its lower limit is 4000. Hence option (B).
  1 mark
Answer: 4000, option (B)
Where the marks are
- Identifies modal class 4000–5000 and lower limit 4000 (1 mark)
Q18 · 1 mark · MCQProbability
In a random experiment of throwing a die, which of the following is a sure event ?
1. (A)Getting a number between 1 and 6
2. (B)Getting an odd number $< 7$
3. (C)Getting an even number $< 7$
4. (D)Getting a natural number $< 7$
Try it first, then show our solutionSolution
1. Every outcome of a die (1 to 6) is a natural number less than 7, so this event always happens.
2. 'Between 1 and 6' leaves out 1 and 6, and the odd or even events leave out half the outcomes. So the sure event is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Identifies the event that contains all six outcomes (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A): For any two natural numbers a and b, the HCF of a and b is a factor of the LCM of a and b.
Reason (R): HCF of any two natural numbers divides both the numbers.
Codes: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of 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. R is true: the HCF divides both $a$ and $b$ .
2. Since the HCF divides $a$ , and $a$ divides the LCM, the HCF divides the LCM. So A is true, and R is the correct explanation of A. The answer is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both statements true and R explains A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedPair of Linear Equations in Two Variables
Assertion (A): The value of p for which the system of equations $4 x + p y + 8 = 0$ and $2 x + 2 y + 2 = 0$ is consistent is 4.
Reason (R): The system of equations $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ is consistent with infinitely many solutions, if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .
Codes: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of 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. If $p = 4$ : $\frac{a _{1}}{a _{2}} = \frac{4}{2} = 2$ , $\frac{b _{1}}{b _{2}} = \frac{4}{2} = 2$ , $\frac{c _{1}}{c _{2}} = \frac{8}{2} = 4$ . Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the system is inconsistent (parallel lines). So A is false.
2. R is the standard condition for infinitely many solutions, so R is true. The answer is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Shows p = 4 gives an inconsistent system and R is true (1 mark)
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Class Interval	10 – 25	25 – 40	40 – 55	55 – 70	70 – 85	85 – 100
Frequency	2	3	7	6	6	6

Runs Scored	3000 – 4000	4000 – 5000	5000 – 6000	6000 – 7000
Number of Batsmen	5	10	9	8

Section B

Very short answer; internal choice in 22 and 24

Q21 · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve the following system of equations for x and y :
$\frac{x}{2} + \frac{2 y}{3} = - 1$ and $x - \frac{y}{3} = 3$
Try it first, then show our solutionSolution
1. Clear the fractions. Multiply the first equation by 6: $3 x + 4 y = - 6$ ...(1). Multiply the second equation by 3: $3 x - y = 9$ ...(2).
2. Subtract (2) from (1): $5 y = - 15 \Rightarrow y = - 3$ .
  1 mark
3. Substitute in (2): $3 x - (- 3) = 9 \Rightarrow 3 x = 6 \Rightarrow x = 2$ . Hence $x = 2, y = - 3$ .
  1 mark
Answer: x = 2, y = −3
Where the marks are
- Obtains one correct value, e.g. $y = - 3$ (1 mark)
- Obtains the other value, $x = 2$ (1 mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerTriangles
In the given figure, if $P Q ∥ R S$ , then prove that $△ P O Q \sim △ S O R$ .
Figure in the paper: Lines PQ and RS are parallel (arrows marked). Segments PS and QR (or RQ) cross at point O, which lies between P and S and between Q and R. P is at top-left, Q below P, R at top-right, S at bottom-right. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ P O Q$ and $△ S O R$ : $P Q ∥ R S$ and PS is a transversal, so $∠ QP O = ∠ R S O$ [alternate interior angles].
  1 mark
2. QR is also a transversal, so $∠ P QO = ∠ S R O$ [alternate interior angles]. Therefore $△ P O Q \sim △ S O R$ [AA similarity criterion].
  1 mark
Answer: △POQ ~ △SOR by AA similarity
Where the marks are
- States one pair of equal alternate angles with the reason (1 mark)
- States the second pair of equal angles and concludes similarity by AA (1 mark)
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OR
Q22(b) · 2 marks · Very short answerTriangles
In the given figure, $△ O S R \sim △ O QP$ , $∠ R O Q = 12 5^{\circ}$ and $∠ O R S = 7 0^{\circ}$ . Find the measures of $∠ O S R$ and $∠ O QP$ .
Figure in the paper: Two parallel horizontal lines: S and R on the upper line (SR), P and Q on the lower line (PQ). Segments SQ and RP cross at O. Angle at R between RS and RO marked 70°. Angle ROQ = 125° marked at O on the right side. See it in the official paper.
Try it first, then show our solutionSolution
1. SOQ is a straight line, so $∠ R O Q$ is an exterior angle of $△ O R S$ . Then $∠ R O Q = ∠ O R S + ∠ O S R$ [exterior angle property], so $12 5^{\circ} = 7 0^{\circ} + ∠ O S R$ , giving $∠ O S R = 5 5^{\circ}$ .
  1 mark
2. Since $△ O S R \sim △ O QP$ with $S \leftrightarrow Q$ , we get $∠ O QP = ∠ O S R$ [corresponding angles of similar triangles].
  ½ mark
3. $∴ ∠ O QP = 5 5^{\circ}$ .
  ½ mark
Answer: ∠OSR = 55°, ∠OQP = 55°
Where the marks are
- Uses the exterior angle property to get $∠ O S R = 5 5^{\circ}$ (1 mark)
- States that corresponding angles of similar triangles are equal (½ mark)
- Concludes $∠ O QP = 5 5^{\circ}$ (½ mark)
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Q23 · 2 marks · Very short answerCircles
Two concentric circles are of radii 6 cm and 10 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Try it first, then show our solutionSolution
1. Figure: two concentric circles with centre O. Chord AB of the larger circle touches the smaller circle at C. Join OC and OB. Here $O C = 6$ cm and $O B = 10$ cm.
  ½ mark
2. $O C ⊥ A B$ [tangent ⟂ radius at the point of contact]. In right $△ O C B$ : $B C^{2} = O B^{2} - O C^{2} = 1 0^{2} - 6^{2} = 64$ , so $B C = 8$ cm.
  1 mark
3. The perpendicular from the centre bisects the chord, so $A B = 2 \times B C = 16$ cm.
  ½ mark
Answer: 16 cm
Where the marks are
- Draws a correct figure (½ mark)
- Uses $O C ⊥ A B$ and Pythagoras to get $B C = 8$ cm (1 mark)
- Gets $A B = 16$ cm (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerIntroduction to Trigonometry
Find the values of A and B ( $0^{\circ} \leq A < 9 0^{\circ}$ , $0^{\circ} \leq B < 9 0^{\circ}$ ), if $tan (A + B) = 1$ and $tan (A - B) = \frac{1}{3}$ .
Try it first, then show our solutionSolution
1. $tan (A + B) = 1 = tan 4 5^{\circ} \Rightarrow A + B = 4 5^{\circ}$ ...(1)
  ½ mark
2. $tan (A - B) = \frac{1}{3} = tan 3 0^{\circ} \Rightarrow A - B = 3 0^{\circ}$ ...(2)
  ½ mark
3. Adding (1) and (2): $2 A = 7 5^{\circ} \Rightarrow A = 37. 5^{\circ}$ . Then $B = 4 5^{\circ} - 37. 5^{\circ} = 7. 5^{\circ}$ .
  1 mark
Answer: A = 37.5°, B = 7.5°
Where the marks are
- Writes $A + B = 4 5^{\circ}$ (½ mark)
- Writes $A - B = 3 0^{\circ}$ (½ mark)
- Solves to get $A = 37. 5^{\circ}$ and $B = 7. 5^{\circ}$ (1 mark)
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OR
Q24(b) · 2 marks · Very short answerIntroduction to Trigonometry
Prove that $tan 4 5^{\circ} = 1$ geometrically.
Try it first, then show our solutionSolution
1. Take $△ A B C$ right-angled at B with $A B = B C = x$ , so it is an isosceles right triangle. Since $A B = B C$ , $∠ A = ∠ C$ [angles opposite equal sides].
  ½ mark
2. By the angle sum property, $∠ A + ∠ C = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ , so $∠ A = ∠ C = 4 5^{\circ}$ .
  ½ mark
3. In right $△ A B C$ , $tan C = \frac{side opposite to C}{side adjacent to C} = \frac{A B}{B C} = \frac{x}{x} = 1$ . Hence $tan 4 5^{\circ} = 1$ .
  1 mark
Answer: tan 45° = AB/BC = x/x = 1
Where the marks are
- Takes an isosceles right triangle with equal legs (½ mark)
- Shows that both acute angles are $4 5^{\circ}$ using the angle sum property (½ mark)
- Computes $tan 4 5^{\circ} = \frac{x}{x} = 1$ (1 mark)
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Q25 · 2 marks · Very short answerAreas Related to Circles
A chord of a circle of diameter 20 cm subtends an angle of $6 0^{\circ}$ at the centre of the circle. Find the area of the corresponding minor segment of the circle. (Use $π = 3.14$ and $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Radius $r = \frac{20}{2} = 10$ cm. Let chord PQ subtend $∠ P O Q = 6 0^{\circ}$ at O. Since $O P = O Q$ and $∠ P O Q = 6 0^{\circ}$ , $△ O P Q$ is equilateral.
  ½ mark
2. Area of minor segment = area of sector − area of $△ O P Q$ $= \frac{60}{360} \times 3.14 \times 1 0^{2} - \frac{3}{4} \times 1 0^{2} = 52.33 - 1.73 \times 25 = 52.33 - 43.25$
  1 mark
3. $= \frac{109}{12}$ cm² $\approx 9.08$ cm²
  ½ mark
Answer: 109/12 cm² ≈ 9.08 cm²
Where the marks are
- Takes $r = 10$ cm and recognises $△ O P Q$ as equilateral (½ mark)
- Writes area of segment = sector − triangle with correct substitution (1 mark)
- Gets $\approx 9.08$ cm² (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 answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3$ is rational.
  Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are coprime.
  Squaring: $3 q^{2} = p^{2}$ .
  ½ mark
2. So $p^{2}$ is divisible by 3. Since 3 is prime, 3 divides $p^{2}$ means 3 divides $p$ . ... (1)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2}$ , so $q^{2} = 3 a^{2}$ .
  So $q^{2}$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)
  1 mark
4. From (1) and (2), 3 divides both $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong. Hence $3$ is irrational.
  ½ mark
Answer: Assuming $3 = p / q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $3$ is irrational
Where the marks are
- Assumes $3 = \frac{p}{q}$ with p, q coprime and gets $3 q^{2} = p^{2}$ (1/2 mark)
- Shows 3 divides p (1 mark)
- Substitutes $p = 3 a$ and shows 3 divides q (1 mark)
- States the contradiction and concludes (1/2 mark)
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OR
Q26(b) · 3 marks · Short answerReal Numbers
The factor tree of a number x is shown below. Find the values of x, y, a and b. Hence, write the product of the prime factors of the number x so obtained.
Figure in the paper: Factor tree: top box x splits into 2 and y. y splits into 2 and 210. 210 splits into a and 70. 70 splits into 2 and 35. 35 splits into 5 and b. See it in the official paper.
Try it first, then show our solutionSolution
1. $35 = 5 \times b \Rightarrow b = 7$
  ½ mark
2. $210 = a \times 70 \Rightarrow a = 3$
  ½ mark
3. $y = 2 \times 210 = 420$
  ½ mark
4. $x = 2 \times y = 2 \times 420 = 840$
  ½ mark
5. Product of prime factors: $840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7 = 2^{3} \times 3 \times 5 \times 7$
  1 mark
Answer: b = 7, a = 3, y = 420, x = 840; 840 = 2³ × 3 × 5 × 7
Where the marks are
- Finds $b = 7$ (½ mark)
- Finds $a = 3$ (½ mark)
- Finds $y = 420$ (½ mark)
- Finds $x = 840$ (½ mark)
- Writes $840 = 2^{3} \times 3 \times 5 \times 7$ (1 mark)
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Q27 · 3 marks · Short answerPolynomials
Find a quadratic polynomial whose sum and product of zeroes are 0 and $- 9$ , respectively. Also, find the zeroes of the polynomial so obtained.
Try it first, then show our solutionSolution
1. A quadratic polynomial with sum of zeroes S and product P is $x^{2} - S x + P$ . So the polynomial is $x^{2} - 0 \cdot x + (- 9) = x^{2} - 9$ .
  1 mark
2. For the zeroes: $x^{2} - 9 = (x + 3) (x - 3) = 0$ .
  1 mark
3. Zeroes are $x = - 3$ and $x = 3$ . Check: sum $= 0$ and product $= - 9$ .
  1 mark
Answer: x² − 9; zeroes −3 and 3
Where the marks are
- Forms the polynomial $x^{2} - 9$ (1 mark)
- Factorises it as $(x + 3) (x - 3)$ (1 mark)
- States the zeroes $- 3, 3$ (1 mark)
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Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve the following system of equations graphically : $x + 3 y = 6$ ; $2 x - 3 y = 12$
Try it first, then show our solutionSolution
1. For $x + 3 y = 6$ , i.e. $y = \frac{6 - x}{3}$ : table $x$ : 0, 3, 6 and $y$ : 2, 1, 0. Plot (0, 2), (3, 1), (6, 0) and join them to get the first line.
  1 mark
2. For $2 x - 3 y = 12$ , i.e. $y = \frac{2 x - 12}{3}$ : table $x$ : 0, 3, 6 and $y$ : −4, −2, 0. Plot (0, −4), (3, −2), (6, 0) and join them to get the second line.
  1 mark
3. The two lines intersect at (6, 0), which lies on the x-axis. Hence the solution is $x = 6, y = 0$ .
  1 mark
Answer: x = 6, y = 0, i.e. the point (6, 0)
Where the marks are
- Correct table and graph of $x + 3 y = 6$ (1 mark)
- Correct table and graph of $2 x - 3 y = 12$ (1 mark)
- Reads the intersection point (6, 0) as the solution (1 mark)
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OR
Q28(b) · 3 marks · Short answerPair of Linear Equations in Two Variables
$x$ and $y$ are complementary angles such that $x : y = 1 : 2$ . Express the given information as a system of linear equations in two variables and hence solve it.
Try it first, then show our solutionSolution
1. Complementary angles add up to $9 0^{\circ}$ : $x + y = 9 0^{\circ}$ ...(1)
  1 mark
2. $x : y = 1 : 2 \Rightarrow y = 2 x$ , i.e. $2 x - y = 0$ ...(2)
  1 mark
3. Substitute (2) in (1): $x + 2 x = 9 0^{\circ} \Rightarrow x = 3 0^{\circ}$ and $y = 2 \times 3 0^{\circ} = 6 0^{\circ}$ .
  1 mark
Answer: x + y = 90°, y = 2x; x = 30°, y = 60°
Where the marks are
- Writes $x + y = 9 0^{\circ}$ (1 mark)
- Writes $y = 2 x$ (1 mark)
- Gets $x = 3 0^{\circ}$ (½ mark) and $y = 6 0^{\circ}$ (½ mark)
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Q29 · 3 marks · Short answerCircles
Prove that a rectangle circumscribing a circle is a square.
Try it first, then show our solutionSolution
1. Figure: rectangle ABCD circumscribing a circle with centre O. The sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.
  ½ mark
2. Tangents drawn from an external point to a circle are equal, so $A P = A S$ , $B P = B Q$ , $C R = C Q$ and $D R = D S$ . Adding: $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$ , i.e. $A B + C D = A D + B C$ .
  1 mark
3. In a rectangle, opposite sides are equal: $C D = A B$ and $B C = A D$ . So $2 A B = 2 A D \Rightarrow A B = A D$ .
  1 mark
4. Adjacent sides of rectangle ABCD are equal, so ABCD is a square.
  ½ mark
Answer: AB + CD = AD + BC with AB = CD and AD = BC gives AB = AD, so ABCD is a square
Where the marks are
- Draws a correct figure (½ mark)
- Uses equal tangents to get $A B + C D = A D + B C$ (1 mark)
- Uses opposite sides of the rectangle to show $A B = A D$ (1 mark)
- Concludes ABCD is a square (½ mark)
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Q30 · 3 marks · Short answerIntroduction to Trigonometry
Prove that :
$\frac{1 + cot ^{2} A}{1 + tan ^{2} A} = (\frac{1 - cot A}{1 - tan A})^{2}$
Try it first, then show our solutionSolution
1. LHS $= \frac{1 + \frac{c o s ^{2} A}{s i n ^{2} A}}{1 + \frac{s i n ^{2} A}{c o s ^{2} A}} = \frac{\frac{s i n ^{2} A + c o s ^{2} A}{s i n ^{2} A}}{\frac{c o s ^{2} A + s i n ^{2} A}{c o s ^{2} A}} = \frac{cos ^{2} A}{sin ^{2} A} = cot^{2} A$ [using $sin^{2} A + cos^{2} A = 1$ ]
  1 mark
2. RHS: $\frac{1 - cot A}{1 - tan A} = \frac{\frac{s i n A - c o s A}{s i n A}}{\frac{c o s A - s i n A}{c o s A}} = \frac{sin A - cos A}{sin A} \times \frac{cos A}{- ( sin A - cos A )} = - \frac{cos A}{sin A} = - cot A$
  1½ marks
3. So RHS $= (- cot A)^{2} = cot^{2} A =$ LHS. Hence proved.
  ½ mark
Answer: Both sides equal cot²A; hence proved
Where the marks are
- Simplifies LHS to $cot^{2} A$ (1 mark)
- Simplifies $\frac{1 - c o t A}{1 - t a n A}$ to $- cot A$ (1½ marks)
- Concludes LHS = RHS (½ mark)
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Q31 · 3 marks · Short answerCompetency-basedProbability
A lot consists of 200 pens of which 180 are good and the rest are defective. A customer will buy a pen if it is not defective. The shopkeeper draws a pen at random and gives it to the customer. What is the probability that the customer will not buy it ? Another lot of 100 pens containing 80 good pens is mixed with the previous lot of 200 pens. The shopkeeper now draws one pen at random from the entire lot and gives it to the customer. What is the probability that the customer will buy the pen ?
Try it first, then show our solutionSolution
1. Defective pens in the first lot $= 200 - 180 = 20$ . The customer will not buy a defective pen, so P(customer will not buy) $= \frac{20}{200} = \frac{1}{10}$ .
  1 mark
2. After mixing the two lots: total pens $= 200 + 100 = 300$ and good pens $= 180 + 80 = 260$ .
  1 mark
3. P(customer will buy) $= \frac{260}{300} = \frac{13}{15}$ .
  1 mark
Answer: 1/10; 13/15
Where the marks are
- P(will not buy) $= \frac{1}{10}$ (1 mark)
- Total pens 300 and good pens 260 (1 mark)
- P(will buy) $= \frac{13}{15}$ (1 mark)
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Section D

Long answer; internal choice in 32 and 34

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerQuadratic Equations
The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.
Try it first, then show our solutionSolution
1. Let the greater number be $x$ and the smaller number be $y$ . Then $x^{2} - y^{2} = 180$ ...(1)
  1 mark
2. Also $y^{2} = 8 x$ ...(2)
  1 mark
3. Substitute (2) in (1): $x^{2} - 8 x = 180 \Rightarrow x^{2} - 8 x - 180 = 0$ .
  1 mark
4. $x^{2} - 18 x + 10 x - 180 = 0 \Rightarrow (x - 18) (x + 10) = 0$ , so $x = 18$ or $x = - 10$ . Reject $x = - 10$ because the numbers are positive, so $x = 18$ .
  1 mark
5. $y^{2} = 8 \times 18 = 144 \Rightarrow y = 12$ (positive). The numbers are 18 and 12. Check: $324 - 144 = 180$ .
  1 mark
Answer: 18 and 12
Where the marks are
- Forms $x^{2} - y^{2} = 180$ (1 mark)
- Forms $y^{2} = 8 x$ (1 mark)
- Gets the quadratic $x^{2} - 8 x - 180 = 0$ (1 mark)
- Solves to get $x = 18$ and rejects $- 10$ (1 mark)
- Finds the smaller number 12 and states both numbers (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerQuadratic Equations
Find the value(s) of k for which the equation $2 x^{2} + k x + 3 = 0$ has real and equal roots. Hence, find the roots of the equations so obtained.
Try it first, then show our solutionSolution
1. For $2 x^{2} + k x + 3 = 0$ : $a = 2$ , $b = k$ , $c = 3$ . For real and equal roots, $D = b^{2} - 4 a c = 0$ , so $k^{2} - 4 (2) (3) = 0 \Rightarrow k^{2} - 24 = 0$ .
  1 mark
2. $k = \pm 24 = \pm 26$
  1 mark
3. The equations obtained are $2 x^{2} + 26 x + 3 = 0$ and $2 x^{2} - 26 x + 3 = 0$ .
  1 mark
4. For $k = 26$ : $(2 x + 3)^{2} = 0$ , so the equal roots are $x = - \frac{b}{2 a} = - \frac{2 6}{4} = - \frac{6}{2} = - \frac{3}{2}$ , i.e. $- \frac{3}{2}, - \frac{3}{2}$ .
  1 mark
5. For $k = - 26$ : $(2 x - 3)^{2} = 0$ , so the equal roots are $x = \frac{2 6}{4} = \frac{3}{2}$ , i.e. $\frac{3}{2}, \frac{3}{2}$ .
  1 mark
Answer: k = ±2√6; roots −√(3/2), −√(3/2) (for k = 2√6) and √(3/2), √(3/2) (for k = −2√6)
Where the marks are
- Uses $b^{2} - 4 a c = 0$ to get $k^{2} - 24 = 0$ (1 mark)
- Gets $k = \pm 26$ (1 mark)
- Writes both equations (½ + ½ marks)
- Finds the roots of each equation (1 + 1 marks)
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Q33 · 5 marks · Long answerTriangles
State "Basic Proportionality Theorem" and use it to prove the following : In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{A O}{B O} = \frac{C O}{D O}$ as shown in the given figure. Prove that ABCD is a trapezium.
Figure in the paper: Quadrilateral ABCD with D and C at the top, A and B at the bottom (AB is the base). Diagonals AC and BD intersect at O inside the quadrilateral. See it in the official paper.
Try it first, then show our solutionSolution
1. BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
  1 mark
2. Given: quadrilateral ABCD whose diagonals AC and BD meet at O with $\frac{A O}{B O} = \frac{C O}{D O}$ . To prove: $A B ∥ D C$ , i.e. ABCD is a trapezium. Construction: through O draw $O E ∥ A B$ , meeting AD at E. Figure: ABCD with diagonals meeting at O and OE parallel to AB.
  1 mark
3. In $△ D A B$ , $O E ∥ A B$ , so $\frac{D E}{E A} = \frac{D O}{O B}$ ...(1) [by BPT]
  1 mark
4. From the given condition, $A O \cdot D O = B O \cdot C O \Rightarrow \frac{D O}{B O} = \frac{C O}{A O}$ . With (1), $\frac{D E}{E A} = \frac{C O}{O A}$ ...(2)
  ½ mark
5. In $△ A D C$ , E lies on AD and O lies on AC with $\frac{D E}{E A} = \frac{C O}{O A}$ , so $O E ∥ D C$ [by converse of BPT].
  1 mark
6. Now $O E ∥ A B$ and $O E ∥ D C$ , so $A B ∥ D C$ . One pair of opposite sides is parallel, so ABCD is a trapezium.
  ½ mark
Answer: AB ∥ DC, so ABCD is a trapezium
Where the marks are
- Correct statement of BPT (1 mark)
- Figure, given, to prove and construction $O E ∥ A B$ (1 mark)
- Applies BPT in $△ D A B$ (1 mark)
- Combines with the given ratio to get $\frac{D E}{E A} = \frac{C O}{O A}$ (½ mark)
- Applies the converse of BPT in $△ A D C$ to get $O E ∥ D C$ (1 mark)
- Concludes $A B ∥ D C$ , so ABCD is a trapezium (½ mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerSurface Areas and Volumes
A toy is in the form of a cone surmounted on a hemisphere. The cone and hemisphere have the same radii. The height of the conical part of the toy is equal to the diameter of its base. If the radius of the conical part is 5 cm, find the volume of the toy.
Try it first, then show our solutionSolution
1. Radius $r = 5$ cm. Height of cone $h$ = diameter $= 10$ cm. Volume of toy = volume of hemisphere + volume of cone $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h$ .
2. Volume of hemisphere $= \frac{2}{3} \times \frac{22}{7} \times 5 \times 5 \times 5 = \frac{5500}{21}$ cm³
  2 marks
3. Volume of cone $= \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 10 = \frac{5500}{21}$ cm³
  2 marks
4. Volume of toy $= \frac{5500}{21} + \frac{5500}{21} = \frac{11000}{21}$ cm³ $\approx 523.81$ cm³
  1 mark
Answer: 11000/21 cm³ ≈ 523.81 cm³
Where the marks are
- Correct volume of the hemisphere, $\frac{5500}{21}$ cm³ (2 marks)
- Correct volume of the cone with $h = 10$ cm, $\frac{5500}{21}$ cm³ (2 marks)
- Total volume $\frac{11000}{21} \approx 523.81$ cm³ (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerSurface Areas and Volumes
A cubical block is surmounted by a hemisphere of radius 3·5 cm. What is the smallest possible length of the edge of the cube so that the hemisphere can totally lie on the cube ? Find the total surface area of the solid so formed.
Try it first, then show our solutionSolution
1. For the hemisphere to lie totally on the cube, the edge must be at least the diameter of the hemisphere. Smallest edge $a = 2 \times 3.5 = 7$ cm.
  1 mark
2. TSA of solid = TSA of cube − base area of hemisphere + CSA of hemisphere $= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$ .
  1 mark
3. $6 a^{2} = 6 \times 7 \times 7 = 294$ cm²
  1½ marks
4. $π r^{2} = \frac{22}{7} \times 3.5 \times 3.5 = 38.5$ cm²
  1½ marks
5. TSA $= 294 + 38.5 = 332.5$ cm² $= \frac{665}{2}$ cm²
  1 mark
Answer: Edge = 7 cm; TSA = 332.5 cm²
Where the marks are
- Smallest edge $= 7$ cm (1 mark)
- Correct expression $6 a^{2} + 2 π r^{2} - π r^{2}$ (1 mark)
- Computes $6 a^{2} = 294$ cm² (1½ marks)
- Computes $π r^{2} = 38.5$ cm² (1½ marks)
- TSA $= 332.5$ cm² (1 mark)
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Q35 · 5 marks · Long answerStatistics
The following data gives the information on the observed lifetime (in hours) of 200 electrical components :
Lifetime (in hours) Number of electrical components
0 – 20 10
20 – 40 35
40 – 60 50
60 – 80 60
80 – 100 30
100 – 120 15
Find the mean lifetime (in hours) of the electrical components.
Try it first, then show our solutionSolution
1. Use the step-deviation method. Take the assumed mean $a = 50$ and class size $h = 20$ , so $u_{i} = \frac{x _{i} - 50}{20}$ .
  Lifetime $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  0 – 20 10 10 –2 –20
  20 – 40 30 35 –1 –35
  40 – 60 50 50 0 0
  60 – 80 70 60 1 60
  80 – 100 90 30 2 60
  100 – 120 110 15 3 45
  Total $\sum f_{i} = 200$ $\sum f_{i} u_{i} = 110$
  2 marks
2. Apply the formula:
  $\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 50 + \frac{110}{200} \times 20$
  2 marks
3. $\overset{x}{ˉ} = 50 + 11 = 61$
  So the mean lifetime of the electrical components is 61 hours.
  1 mark
Answer: Mean lifetime = 61 hours
Where the marks are
- Correct table with class marks $x_{i}$ , $u_{i}$ and $f_{i} u_{i}$ , giving $\sum f_{i} = 200$ and $\sum f_{i} u_{i} = 110$ (2 marks)
- Substitutes correctly into $\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h$ (2 marks)
- Obtains mean = 61 hours (1 mark)
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Lifetime (in hours)	Number of electrical components
0 – 20	10
20 – 40	35
40 – 60	50
60 – 80	60
80 – 100	30
100 – 120	15

Lifetime	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
0 – 20	10	10	–2	–20
20 – 40	30	35	–1	–35
40 – 60	50	50	0	0
60 – 80	70	60	1	60
80 – 100	90	30	2	60
100 – 120	110	15	3	45
Total		$\sum f_{i} = 200$		$\sum f_{i} u_{i} = 110$

Section E

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

Q36 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Case Study – 1. An injured bird was found on the roof of a building. The building is 15 m high. A fireman was called to rescue the bird. The fireman used an adjustable ladder to reach the roof. He placed the ladder in such a way that the ladder makes an angle of $6 0^{\circ}$ with the ground in order to reach the roof. Based on the above information, answer the following questions :
1. (i)
  Find the length of the ladder used by the fireman to reach the roof.
  1 mark
2. (ii)
  Find the distance of the point on the ground at which the ladder was fixed from the bottom of the building.
  1 mark
3. (iii)(a)
  In order to avoid skidding, the fireman placed the ladder in such a way that the bottom of the ladder touches the base of the wall which is opposite to the building, making an angle of $3 0^{\circ}$ with the ground. Draw a neat diagram to represent the above situation and hence find the width of the road between the building and the wall.
  2 marks
4. OR
  (iii)(b)
  In order to avoid skidding, the fireman placed the ladder in such a way that the bottom of the ladder touches the base of the wall which is opposite to the building, making an angle of $3 0^{\circ}$ with the ground. Find the length of the ladder used by the fireman in this case.
  2 marks
Figure in the paper: Photograph of a bird on a roof and a fireman on a ladder; a sketch shows a ladder leaning against a vertical wall, with its foot on the ground and making an angle of 60° with the ground. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let AB be the building of height 15 m and AC the ladder of length $a$ , with $∠ A C B = 6 0^{\circ}$ and $∠ B = 9 0^{\circ}$ . In right $△ A B C$ :
  $sin 6 0^{\circ} = \frac{A B}{A C} \Rightarrow \frac{3}{2} = \frac{15}{a}$
  ½ mark
2. (i) $a = \frac{30}{3} = 103$
  So the length of the ladder is $103$ m (about 17.32 m).
  ½ mark
3. (ii) Let BC $= x$ be the distance of the foot of the ladder from the building. Then
  $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow 3 = \frac{15}{x}$
  ½ mark
4. (ii) $x = \frac{15}{3} = 53$
  So the foot of the ladder is $53$ m (about 8.66 m) from the bottom of the building.
  ½ mark
5. (iii)(a) Diagram: Draw a vertical segment AB = 15 m for the building, with B on the ground. Draw the horizontal road BC, where C is the base of the opposite wall, and draw a short vertical segment at C for the wall. Join the ladder AC and mark $∠ A C B = 3 0^{\circ}$ and $∠ A B C = 9 0^{\circ}$ . Let BC $= y$ be the width of the road.
  1 mark
6. (iii)(a) In right $△ A B C$ :
  $tan 3 0^{\circ} = \frac{A B}{B C} \Rightarrow \frac{1}{3} = \frac{15}{y}$
  ½ mark
7. (iii)(a) $y = 153$
  So the width of the road is $153$ m (about 25.98 m).
  ½ mark
8. (iii)(b) Use the same right $△ A B C$ with AB = 15 m, $∠ A C B = 3 0^{\circ}$ and ladder AC $= l$ :
  $sin 3 0^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{2} = \frac{15}{l}$
  1 mark
9. (iii)(b) $l = 30$
  So the length of the ladder in this case is 30 m.
  1 mark
Answer: (i) $103$ m; (ii) $53$ m; (iii)(a) $153$ m; (iii)(b) 30 m
Where the marks are
- (i) Uses $sin 6 0^{\circ} = 15/ a$ (½ mark)
- (i) Gets ladder length $103$ m (½ mark)
- (ii) Uses $tan 6 0^{\circ} = 15/ x$ (½ mark)
- (ii) Gets distance $53$ m (½ mark)
- (iii)(a) Draws a correct labelled diagram (1 mark)
- (iii)(a) Uses $tan 3 0^{\circ} = 15/ y$ (½ mark)
- (iii)(a) Gets road width $153$ m (½ mark)
- (iii)(b) Uses $sin 3 0^{\circ} = 15/ l$ (1 mark)
- (iii)(b) Gets ladder length 30 m (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Case Study – 2. In a garden, saplings of rose flowers were planted at equal intervals to form a spiral pattern. The spiral is made up of successive semicircles, with centres alternatively at A and B, starting with centre at A, of radii 50 cm, 100 cm, 150 cm, ....... as shown in the figure given below. Spiral 1 has 10 flowers, Spiral 2 has 20 flowers, Spiral 3 has 30 flowers and so on. Based on the above information, answer the following questions :
1. (i)
  What is the radius of the 13th spiral ?
  1 mark
2. (ii)
  If the radius of the nth spiral is 500 cm, find the value of n.
  1 mark
3. (iii)(a)
  Find the total number of saplings till the 11th spiral.
  2 marks
4. OR
  (iii)(b)
  Till which spiral, will there be a total of 450 saplings ?
  2 marks
Figure in the paper: A horizontal line with points A and B close together near the centre. Semicircular arcs labelled l1 (above, small, centred at A), l2 (below, centred at B), l3 (above, larger), l4 (below, larger) forming an expanding spiral; a dotted continuation arc is shown at the upper right. Photograph of rose saplings in a spiral pattern is also shown. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The radii 50, 100, 150, … form an AP with $a = 50$ and $d = 50$ .
  $a_{13} = a + 12 d = 50 + 12 \times 50 = 650$
  So the radius of the 13th spiral is 650 cm.
  1 mark
2. (ii) Set $a_{n} = 500$ :
  $50 + (n - 1) \times 50 = 500 \Rightarrow 50 n = 500 \Rightarrow n = 10$
  1 mark
3. (iii)(a) The numbers of saplings 10, 20, 30, … form an AP with $a = 10$ and $d = 10$ . Then
  $S_{11} = \frac{11}{2} [2 \times 10 + (11 - 1) \times 10] = \frac{11}{2} [20 + 100]$
  1½ marks
4. (iii)(a) $S_{11} = \frac{11}{2} \times 120 = 660$
  So there are 660 saplings till the 11th spiral.
  ½ mark
5. (iii)(b) With $a = 10$ and $d = 10$ , set $S_{n} = 450$ :
  $\frac{n}{2} [20 + (n - 1) 10] = 450$
  1 mark
6. (iii)(b) Simplify: $5 n (n + 1) = 450$ , so $n^{2} + n - 90 = 0$ .
  ½ mark
7. (iii)(b) Factorise: $(n + 10) (n - 9) = 0$ , so $n = 9$ ( $n = - 10$ is rejected because $n$ must be a natural number).
  So there are 450 saplings in total till the 9th spiral.
  ½ mark
Answer: (i) 650 cm; (ii) n = 10; (iii)(a) 660 saplings; (iii)(b) 9th spiral
Where the marks are
- (i) Finds $a_{13} = 650$ cm using $a = 50$ , $d = 50$ (1 mark)
- (ii) Solves $50 + (n - 1) 50 = 500$ to get $n = 10$ (1 mark)
- (iii)(a) Writes $S_{11} = \frac{11}{2} [20 + 10 \times 10]$ (1½ marks)
- (iii)(a) Gets 660 saplings (½ mark)
- (iii)(b) Forms $\frac{n}{2} [20 + (n - 1) 10] = 450$ (1 mark)
- (iii)(b) Reduces it to $n^{2} + n - 90 = 0$ (½ mark)
- (iii)(b) Gets $n = 9$ (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Case Study – 3. In a society, there is a circular park having two gates. The gates are placed at points A(10, 20) and B(50, 50), as shown in the figure below. Two fountains are installed at points P and Q on AB such that AP = PQ = QB. Based on the above information, answer the following questions :
1. (i)
  Find the coordinates of the centre C.
  1 mark
2. (ii)
  Find the radius of the circular park.
  1 mark
3. (iii)(a)
  Find the coordinates of the point P.
  2 marks
4. OR
  (iii)(b)
  Find the distance of the fountain at Q from gate A.
  2 marks
Figure in the paper: A circle with a diameter AB drawn as a straight line through the centre C. A is at the lower-left on the circle, B at the upper-right on the circle. Points P and Q lie on AB between A and B, with P closer to A, C between P and Q (near the middle), and Q closer to B. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) AB is a diameter, so the centre C is the mid-point of AB:
  $C = (\frac{10 + 50}{2}, \frac{20 + 50}{2}) = (30, 35)$
  1 mark
2. (ii) Radius $= A C = (30 - 10)^{2} + (35 - 20)^{2} = 400 + 225 = 625 = 25$ units.
  1 mark
3. (iii)(a) Since AP = PQ = QB, P divides AB in the ratio $1 : 2$ .
  ½ mark
4. (iii)(a) By the section formula:
  $P = (\frac{1 \times 50 + 2 \times 10}{1 + 2}, \frac{1 \times 50 + 2 \times 20}{1 + 2})$
  1 mark
5. (iii)(a) $P = (\frac{70}{3}, \frac{90}{3}) = (\frac{70}{3}, 30)$
  ½ mark
6. (iii)(b) The diameter is $A B = 2 \times 25 = 50$ units.
  ½ mark
7. (iii)(b) Since AP = PQ = QB, we get $A Q = \frac{2}{3} A B = \frac{2}{3} \times 50$ .
  1 mark
8. (iii)(b) $A Q = \frac{100}{3}$ units (about 33.33 units).
  Check: Q divides AB in the ratio 2 : 1, so $Q = (\frac{110}{3}, 40)$ . Then $A Q = (\frac{80}{3})^{2} + 2 0^{2} = \frac{100}{3}$ , which agrees.
  ½ mark
Answer: (i) C(30, 35); (ii) 25 units; (iii)(a) P(70/3, 30); (iii)(b) AQ = 100/3 units
Where the marks are
- (i) Uses the mid-point formula to get C(30, 35) (1 mark)
- (ii) Uses the distance formula to get radius 25 (1 mark)
- (iii)(a) Recognises that P divides AB in the ratio 1 : 2 (½ mark)
- (iii)(a) Applies the section formula correctly to both coordinates (½ + ½ marks)
- (iii)(a) Gets P(70/3, 30) (½ mark)
- (iii)(b) Finds AB = 50 (½ mark)
- (iii)(b) Writes AQ = (2/3)AB (1 mark)
- (iii)(b) Gets AQ = 100/3 (½ 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.