MathsCBSE Class 10 · Basic · Board exam 2026 (session 2025-26)

CBSE Class 10 Mathematics Basic: 2026 board paper, set 430/1/1

The question paper set in the CBSE Class 10 board exam of 2026, 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/2026/X/Mathematics_Standard.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 type
C	26-31	Short Answer type
D	32-35	Long Answer type
E	36-38	Case study based, 2-mark subparts have internal choice

Instructions in brief: All questions compulsory, no overall choice; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (in the 2-mark subparts of case studies).

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)
Open Q2 on its own page
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 is equidistant from both ends, so the two parts are equal and the ratio is $1 : 1$ .
  1 mark
Answer: 1 : 1, option (C)
Where the marks are
- Identifies equal parts, ratio $1 : 1$ (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. Triangle ABC: AB = 3.8 cm, BC = 6 cm, AC = $33$ cm, angle at A = 80°, angle at B = 60°. Triangle PQR: PR = $63$ cm, RQ = 7.6 cm, PQ = 12 cm; the angle at P is marked as the angle to be determined. See it in the official paper.
Try it first, then show our solutionSolution
1. Compare sides: $\frac{R Q}{A B} = \frac{7.6}{3.8} = 2$ , $\frac{P Q}{B C} = \frac{12}{6} = 2$ , $\frac{P R}{A C} = \frac{6 3}{3 3} = 2$ . So $△ A B C \sim △ R QP$ [SSS similarity], which gives $∠ P = ∠ C$ .
  ½ mark
2. $∠ C = 18 0^{\circ} - 8 0^{\circ} - 6 0^{\circ} = 4 0^{\circ}$ , so $∠ P = 4 0^{\circ}$ .
  ½ mark
Answer: ∠P = 40°, option (C)
Where the marks are
- Uses SSS similarity with the correct vertex correspondence (½ mark)
- Finds $∠ P = 4 0^{\circ}$ (½ mark)
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Q9 · 1 mark · MCQCircles
In the given figure, PA is a tangent to a circle with centre O. If $O P = 10$ 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; A is a point of contact on the circle; P is outside the circle; PA is the tangent at A; OA is joined; OP = 10 cm; angle APO = 30° marked at P. See it in the official paper.
Try it first, then show our solutionSolution
1. $O A ⊥ P A$ [tangent ⟂ radius], so $△ O A P$ is right-angled at A.
  ½ mark
2. $A P = O P cos 3 0^{\circ} = 10 \times \frac{3}{2} = 53$ cm.
  ½ mark
Answer: 5√3 cm, option (D)
Where the marks are
- Uses the right angle at A (½ mark)
- Gets $A P = 53$ cm (½ mark)
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: Observer at top left; a horizontal dashed line from the observer; a vertical line from observer down to the ground; a line of sight from observer to the object on the ground at the right. Angle z is between the horizontal line and the line of sight (at observer); angle y is between the vertical line and the line of sight (at observer); angle a is the right angle at the base of the vertical line; angle x is at the object between the line of sight and the ground. 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 it. That angle is $z$ .
  1 mark
Answer: z, option (C)
Where the marks are
- Identifies the angle between the horizontal and the line of sight at the observer (1 mark)
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: Circle with centre O and radius r. A chord of length a joins two points on the circle; the shaded region is the segment-like region bounded by the two radii OA and OB (each labelled r) and the chord a; the major arc of length l is marked on the outside of the shaded part. See it in the official paper.
Try it first, then show our solutionSolution
1. The boundary of the shaded sector-type region is made up of the arc of length $l$ and the two radii, each of length $r$ .
  ½ mark
2. Perimeter $= l + r + r = l + 2 r$ .
  ½ mark
Answer: l + 2r, option (C)
Where the marks are
- Identifies the arc and the two radii as the boundary (½ mark)
- Perimeter $= l + 2 r$ (½ mark)
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 $= \frac{9 0 ^{\circ}}{36 0 ^{\circ}} π r^{2} = \frac{1}{4} π r^{2}$ , area of the circle $= π r^{2}$ .
  ½ mark
2. Ratio $= \frac{1}{4} π r^{2} : π r^{2} = 1 : 4$ , so option (C).
  ½ mark
Answer: $1 : 4$ , option (C)
Where the marks are
- Identifies the correct ratio $1 : 4$ , option (C) (1 mark)
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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) $47 \cdot 5$
4. (D) $62 \cdot 5$
Try it first, then show our solutionSolution
1. $N = 30$ , so $\frac{N}{2} = 15$ . The cumulative frequencies are 2, 5, 12, 18, …
  ½ mark
2. The first cumulative frequency above 15 is 18, so the median class is 55–70. Its class mark is $\frac{55 + 70}{2} = 62.5$ .
  ½ mark
Answer: 62.5, option (D)
Where the marks are
- Finds $N /2 = 15$ and the cumulative frequencies (½ mark)
- Median class 55–70, class mark 62.5 (½ 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.
Select the correct answer from the codes (A), (B), (C) and (D): (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. R is true: the HCF divides both $a$ and $b$ .
  ½ mark
2. Since HCF divides $a$ , and $a$ divides the LCM, the HCF divides the LCM. So A is true, and R explains it.
  ½ mark
Answer: Option (A)
Where the marks are
- Shows both A and R are true (½ mark)
- R explains A (½ mark)
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 $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .
Select the correct answer from the codes (A), (B), (C) and (D) as given in Questions 19 and 20.
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. For $p = 4$ : $\frac{a _{1}}{a _{2}} = 2$ , $\frac{b _{1}}{b _{2}} = \frac{4}{2} = 2$ , $\frac{c _{1}}{c _{2}} = \frac{8}{2} = 4$ . Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the system is inconsistent, so A is false.
  ½ mark
2. R is the standard condition for infinitely many solutions, so R is true.
  ½ mark
Answer: Option (D)
Where the marks are
- Shows the system is inconsistent at $p = 4$ , so A is false (½ mark)
- States R is true (½ mark)
Open Q20 on its own page

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 type

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 (arrow marks). Segments PS and QR cross at O; P and Q are on one side, R and S on the other, with P-O-S and Q-O-R. 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 $P S$ is a transversal, so $∠ O P Q = ∠ O S R$ [alternate interior angles].
  1 mark
2. $∠ P O Q = ∠ S O R$ [vertically opposite angles]. (Also $∠ O QP = ∠ O R S$ , alternate angles with transversal $QR$ .)
  ½ mark
3. Hence $△ P O Q \sim △ S O R$ [AA similarity criterion].
  ½ mark
Answer: $△ P O Q \sim △ S O R$ by AA similarity
Where the marks are
- Uses alternate interior angles $∠ O P Q = ∠ O S R$ (1 mark)
- Uses vertically opposite angles $∠ P O Q = ∠ S O R$ (½ mark)
- Concludes similarity by AA criterion (½ mark)
Open Q22(a) on its own page
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, P and Q on the lower line. Segments SQ and RP cross at O. Angle ORS = 70° marked at R; angle ROQ = 125° marked at O. See it in the official paper.
Try it first, then show our solutionSolution
1. $S$ , $O$ , $Q$ are collinear, so $∠ S O R = 18 0^{\circ} - ∠ R O Q = 18 0^{\circ} - 12 5^{\circ} = 5 5^{\circ}$ [linear pair].
  ½ mark
2. In $△ O S R$ : $∠ O S R = 18 0^{\circ} - (∠ S O R + ∠ O R S) = 18 0^{\circ} - (5 5^{\circ} + 7 0^{\circ}) = 5 5^{\circ}$ [angle sum property].
  1 mark
3. Since $△ O S R \sim △ O QP$ , corresponding angles are equal: $∠ O QP = ∠ O S R = 5 5^{\circ}$ .
  ½ mark
Answer: $∠ O S R = 5 5^{\circ}$ , $∠ O QP = 5 5^{\circ}$
Where the marks are
- Finds $∠ S O R = 5 5^{\circ}$ using linear pair (½ mark)
- Finds $∠ O S R = 5 5^{\circ}$ by angle sum property (1 mark)
- Uses similarity to get $∠ O QP = 5 5^{\circ}$ (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): $2 A = 7 5^{\circ} \Rightarrow A = 37. 5^{\circ}$ .
  ½ mark
4. Subtracting (2) from (1): $2 B = 1 5^{\circ} \Rightarrow B = 7. 5^{\circ}$ .
  ½ mark
Answer: $A = 37. 5^{\circ}$ , $B = 7. 5^{\circ}$
Where the marks are
- Gets $A + B = 4 5^{\circ}$ (½ mark)
- Gets $A - B = 3 0^{\circ}$ (½ mark)
- Finds $A = 37. 5^{\circ}$ (½ mark)
- Finds $B = 7. 5^{\circ}$ (½ mark)
Open Q24(a) on its own page
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 $π = 3 \cdot 14$ and $3 = 1 \cdot 73$ )
Try it first, then show our solutionSolution
1. Radius $r = \frac{20}{2} = 10$ cm, $θ = 6 0^{\circ}$ . Area of sector $= \frac{60}{360} \times 3.14 \times 1 0^{2} = \frac{314}{6} \approx 52.33$ cm².
  ½ mark
2. The two radii are equal and the included angle is $6 0^{\circ}$ , so the triangle is equilateral with side 10 cm. Area of triangle $= \frac{3}{4} \times 1 0^{2} = \frac{1.73 \times 100}{4} = 43.25$ cm².
  1 mark
3. Area of minor segment $=$ area of sector $-$ area of triangle $= 52.33 - 43.25 = 9.08$ cm² (approx.).
  ½ mark
Answer: Area of minor segment ≈ 9.08 cm²
Where the marks are
- Area of sector ≈ 52.33 cm² (½ mark)
- Area of equilateral triangle = 43.25 cm² (1 mark)
- Area of segment ≈ 9.08 cm² (½ mark)
Open Q25 on its own page

Section C

Short Answer type

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)
Open Q26(a) on its own page
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: root 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. Prime factorisation: $x = 840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7 = 2^{3} \times 3 \times 5 \times 7$ .
  1 mark
Answer: $x = 840$ , $y = 420$ , $a = 3$ , $b = 7$ ; $840 = 2^{3} \times 3 \times 5 \times 7$
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)
Open Q26(b) on its own page
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)
Open Q28(a) on its own page
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)
Open Q28(b) on its own page
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 : $\frac{1 + c o t ^{2} A}{1 + t a n ^{2} A} = (\frac{1 - c o t A}{1 - t a n A})^{2}$
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 type

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)
Open Q32(b) on its own page
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 diagonals AC and BD meeting at O. Vertices in order: A bottom left, B bottom right, C top right, D top left. See it in the official paper.
Try it first, then show our solutionSolution
1. Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle so that it meets the other two sides at distinct points, then it divides those two sides in the same ratio.
  1 mark
2. Given: ABCD is a quadrilateral whose diagonals AC and BD meet at O, with $\frac{A O}{B O} = \frac{C O}{D O}$ . To prove: ABCD is a trapezium. Construction: Through O, draw $O E ∥ A B$ meeting AD at E. (Figure: quadrilateral ABCD with diagonals meeting at O and the line OE drawn parallel to AB.)
  1 mark
3. In $△ A B D$ , $E O ∥ A B$ , so by BPT: $\frac{D E}{E A} = \frac{D O}{O B}$ …(1)
  1 mark
4. Given $\frac{A O}{B O} = \frac{C O}{D O} \Rightarrow \frac{A O}{C O} = \frac{B O}{D O} \Rightarrow \frac{C O}{A O} = \frac{D O}{B O}$ …(2). From (1) and (2): $\frac{D E}{E A} = \frac{C O}{O A}$ .
  1 mark
5. In $△ A D C$ , E and O divide sides AD and AC in the same ratio, so $E O ∥ D C$ [converse of BPT]. Since $E O ∥ A B$ and $E O ∥ D C$ , we get $A B ∥ D C$ . Hence ABCD has a pair of parallel opposite sides, so ABCD is a trapezium.
  1 mark
Answer: BPT stated. $A B ∥ D C$ , so ABCD is a trapezium.
Where the marks are
- Correct statement of BPT (1 mark)
- Given, to prove, construction $O E ∥ A B$ , with figure (1 mark)
- Applies BPT in $△ A B D$ : $\frac{D E}{E A} = \frac{D O}{O B}$ (1 mark)
- Uses the given ratio to get $\frac{D E}{E A} = \frac{C O}{O A}$ (1 mark)
- Converse of BPT gives $E O ∥ D C$ , hence $A B ∥ D C$ and ABCD is a trapezium (1 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 $3 \cdot 5$ 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)
Open Q34(b) on its own page
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, 2-mark subparts have internal choice

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) Find the length of the ladder used by the fireman in this case.
  2 marks
Figure in the paper: Photo of a fireman on a ladder reaching a bird on a roof; schematic: a vertical building wall of height 15 m, a ladder leaning against it making 60° with the ground, with its foot on the ground away from the wall. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let the ladder have length $l$ . Then $sin 6 0^{\circ} = \frac{15}{l} \Rightarrow \frac{3}{2} = \frac{15}{l} \Rightarrow l = \frac{30}{3} = 103 \approx 17.32$ m.
  1 mark
2. (ii) Let the distance be $d$ . Then $tan 6 0^{\circ} = \frac{15}{d} \Rightarrow 3 = \frac{15}{d} \Rightarrow d = \frac{15}{3} = 53 \approx 8.66$ m.
  1 mark
3. (iii)(a) Diagram: draw a vertical line AB = 15 m for the building (B on the ground) and mark C, the base of the opposite wall, on the ground. Join AC as the ladder, with $∠ A C B = 3 0^{\circ}$ and $∠ A B C = 9 0^{\circ}$ . BC is the width of the road.
  1 mark
4. (iii)(a) $tan 3 0^{\circ} = \frac{A B}{B C} \Rightarrow \frac{1}{3} = \frac{15}{B C} \Rightarrow B C = 153 \approx 25.98$ m.
  1 mark
5. (iii)(b) In the same right triangle, $sin 3 0^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{2} = \frac{15}{A C}$ .
  1 mark
6. (iii)(b) Therefore $A C = 30$ m, so the ladder is 30 m long.
  1 mark
Answer: (i) $103 \approx 17.32$ m; (ii) $53 \approx 8.66$ m; (iii)(a) width $= 153 \approx 25.98$ m; (iii)(b) 30 m
Where the marks are
- (i) Uses $sin 6 0^{\circ}$ to get ladder length $103$ m (1 mark)
- (ii) Uses $tan 6 0^{\circ}$ to get distance $53$ m (1 mark)
- (iii)(a) Correct labelled diagram (1 mark)
- (iii)(a) Uses $tan 3 0^{\circ}$ to get width $153$ m (1 mark)
- (iii)(b) Uses $sin 3 0^{\circ} = \frac{15}{A C}$ (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 $1 3^{t h}$ spiral ?
  1 mark
2. (ii)
  If the radius of the $n^{t h}$ spiral is 500 cm, find the value of n.
  1 mark
3. (iii)
  (a) Find the total number of saplings till the $1 1^{t h}$ spiral.
  2 marks
4. OR
  (iii)
  (b) Till which spiral, will there be a total of 450 saplings ?
  2 marks
Figure in the paper: Spiral made of successive semicircles on a horizontal line with points A and B on it, centres alternating. Arcs labelled l1 (small, above line, around A-B), l2 (below line), l3 (above, larger), l4 (below, larger), and further arcs continuing outward as a dotted curve on the right. 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 $d = 50$ , so $r_{n} = 50 n$ . Hence $r_{13} = 50 + 12 \times 50 = 650$ cm.
  1 mark
2. (ii) $50 + (n - 1) 50 = 500 \Rightarrow 50 n = 500 \Rightarrow n = 10$ .
  1 mark
3. (iii)(a) The numbers of flowers 10, 20, 30, … form an AP with $a = 10$ and $d = 10$ . Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ : $S_{11} = \frac{11}{2} [2 (10) + 10 (10)]$ .
  1 mark
4. (iii)(a) $S_{11} = \frac{11}{2} \times 120 = 660$ saplings.
  1 mark
5. (iii)(b) $\frac{n}{2} [20 + (n - 1) 10] = 450 \Rightarrow 5 n (n + 1) = 450 \Rightarrow n^{2} + n - 90 = 0$ .
  1 mark
6. (iii)(b) $(n + 10) (n - 9) = 0 \Rightarrow n = 9$ , since $n = - 10$ is rejected because $n$ must be positive. So there are 450 saplings till the 9th spiral.
  1 mark
Answer: (i) 650 cm; (ii) n = 10; (iii)(a) 660 saplings; (iii)(b) 9th spiral
Where the marks are
- (i) Uses the AP of radii to get $r_{13} = 650$ cm (1 mark)
- (ii) Solves $50 n = 500$ to get $n = 10$ (1 mark)
- (iii)(a) Correct substitution in $S_{11}$ (1 mark)
- (iii)(a) Total = 660 saplings (1 mark)
- (iii)(b) Forms $n^{2} + n - 90 = 0$ (1 mark)
- (iii)(b) Solves to get $n = 9$ , rejecting the negative root (1 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 $A P = P Q = QB$ .
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: Circle representing the park with a diameter AB; A is at lower left on the circle and B at upper right on the circle. C is the centre on AB. P and Q lie on AB between A and C and between C and B respectively, with AP = PQ = QB. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) AB is a diameter, so C is its midpoint: $C = (\frac{10 + 50}{2}, \frac{20 + 50}{2}) = (30, 35)$ .
  1 mark
2. (ii) $A B = (50 - 10)^{2} + (50 - 20)^{2} = 1600 + 900 = 50$ units, so radius $= \frac{50}{2} = 25$ units.
  1 mark
3. (iii)(a) Since $A P = P Q = QB$ , P divides AB in the ratio $1 : 2$ . By the section formula, $x = \frac{1 ( 50 ) + 2 ( 10 )}{3}$ and $y = \frac{1 ( 50 ) + 2 ( 20 )}{3}$ .
  1 mark
4. (iii)(a) $P = (\frac{70}{3}, 30) \approx (23.33, 30)$ .
  1 mark
5. (iii)(b) Q divides AB in the ratio $2 : 1$ , so $Q = (\frac{2 ( 50 ) + 10}{3}, \frac{2 ( 50 ) + 20}{3}) = (\frac{110}{3}, 40)$ .
  1 mark
6. (iii)(b) $A Q = (\frac{110}{3} - 10)^{2} + (40 - 20)^{2} = \frac{6400}{9} + \frac{3600}{9} = \frac{100}{3} \approx 33.33$ units. (Check: $A Q = \frac{2}{3} A B = \frac{2}{3} \times 50$ .)
  1 mark
Answer: (i) C(30, 35); (ii) 25 units; (iii)(a) P(70/3, 30); (iii)(b) AQ = 100/3 ≈ 33.33 units
Where the marks are
- (i) Midpoint formula gives C(30, 35) (1 mark)
- (ii) AB = 50 by distance formula, so radius = 25 units (1 mark)
- (iii)(a) Section formula with ratio 1:2 (1 mark)
- (iii)(a) $P = (70/3, 30)$ (1 mark)
- (iii)(b) Finds $Q = (110/3, 40)$ (1 mark)
- (iii)(b) $A Q = 100/3$ units (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.