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

CBSE Class 10 Mathematics Standard: 2026 board paper, set 30/1/1

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.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2026/X/Mathematics_Standard.zip (a zip; this paper is the file 30-1-1P-Maths (Std)_X.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 (VSA)
C	26-31	Short answer (SA)
D	32-35	Long answer (LA)
E	36-38	Case study based

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choice (OR) is given in Q22, Q24, Q28, Q29, Q33, Q34, and in one sub-part of each case study (Q36, Q37, Q38).

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The HCF of 960 and 432 is :
1. (A)48
2. (B)54
3. (C)72
4. (D)36
Try it first, then show our solutionSolution
1. Prime factorisation: $960 = 2^{6} \times 3 \times 5$ and $432 = 2^{4} \times 3^{3}$ .
2. HCF = product of the smallest powers of the common primes $= 2^{4} \times 3 = 16 \times 3 = 48$ . So option (a) is correct.
  1 mark
Answer: 48, option (a)
Where the marks are
- Selects 48, option (a) (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQReal Numbers
The natural number 2 is :
1. (A)a prime number
2. (B)a composite number
3. (C)prime as well as composite
4. (D)neither prime nor composite
Try it first, then show our solutionSolution
1. The only factors of 2 are 1 and 2 itself.
2. A number with exactly two factors is prime, so 2 is a prime number (the only even prime). Option (a).
  1 mark
Answer: a prime number, option (a)
Where the marks are
- Selects option (a) a prime number (1 mark)
Q3 · 1 mark · MCQReal Numbers
For any natural number $n$ , $6^{n}$ ends with the digit :
1. (A)0
2. (B)6
3. (C)3
4. (D)2
Try it first, then show our solutionSolution
1. $6^{1} = 6$ , $6^{2} = 36$ , $6^{3} = 216$ , and so on.
2. A number ending in 6 multiplied by 6 again ends in 6, because $6 \times 6 = 36$ . So $6^{n}$ always ends in 6. Option (b).
  1 mark
Answer: 6, option (b)
Where the marks are
- Selects 6, option (b) (1 mark)
Open Q3 on its own page
Q4 · 1 mark · MCQPolynomials
The graph of $y = f (x)$ is given. The number of zeroes of $f (x)$ is :
1. (A)0
2. (B)1
3. (C)2
4. (D)4
Figure in the paper: Cartesian axes with X'OX horizontal and Y'OY vertical. A wavy curve is drawn that rises on the left, has a hump (local maximum) near the centre, dips, and rises again on the right, lying entirely above the x-axis. See it in the official paper.
Try it first, then show our solutionSolution
1. The zeroes of $f (x)$ are the $x$ -coordinates of the points where the graph meets the $x$ -axis.
2. The curve lies entirely above the $x$ -axis and never cuts or touches it. So the number of zeroes is 0. Option (a).
  1 mark
Answer: 0, option (a)
Where the marks are
- Selects 0, option (a) (1 mark)
Open Q4 on its own page
Q5 · 1 mark · MCQPair of Linear Equations in Two Variables
If a pair of linear equations in two variables is represented by two coincident lines, then the pair of equations has :
1. (A)a unique solution
2. (B)two solutions
3. (C)no solution
4. (D)an infinite number of solutions
Try it first, then show our solutionSolution
1. Coincident lines lie on top of each other, so every point of one line is also on the other.
2. Every common point is a solution, so the pair has infinitely many solutions. Option (d).
  1 mark
Answer: an infinite number of solutions, option (d)
Where the marks are
- Selects option (d) (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQArithmetic Progressions
The common difference of the AP : $2, 22, 32, 42, \dots$ is :
1. (A) $2$
2. (B)1
3. (C) $22$
4. (D) $- 2$
Try it first, then show our solutionSolution
1. Common difference $d = a_{2} - a_{1} = 22 - 2 = 2$ , and the same holds for every pair of consecutive terms ( $32 - 22 = 2$ ). Hence option (A).
  1 mark
Answer: $d = 2$ , option (A)
Where the marks are
- Finds $d = 22 - 2 = 2$ and chooses (A) (1 mark)
Q7 · 1 mark · MCQTriangles
If $Δ$ ABC and $Δ$ DEF are similar such that $2$ AB = DE and BC = 8 cm, then EF is equal to :
1. (A)4 cm
2. (B)8 cm
3. (C)12 cm
4. (D)16 cm
Try it first, then show our solutionSolution
1. Since $Δ A B C \sim Δ D E F$ , $\frac{A B}{D E} = \frac{B C}{E F}$ .
2. From $2 A B = D E$ we get $\frac{A B}{D E} = \frac{1}{2}$ . So $\frac{8}{E F} = \frac{1}{2}$ , which gives $E F = 16$ cm. Option (d).
  1 mark
Answer: 16 cm, option (d)
Where the marks are
- Uses the ratio of corresponding sides to get EF = 16 cm, option (d) (1 mark)
Open Q7 on its own page
Q8 · 1 mark · MCQCoordinate Geometry
The mid-point of the line segment joining the points $(5, - 4)$ and $(6, 4)$ lies on :
1. (A) $x$ -axis
2. (B) $y$ -axis
3. (C)origin
4. (D)neither $x$ -axis nor $y$ -axis
Try it first, then show our solutionSolution
1. Mid-point $= (\frac{5 + 6}{2}, \frac{- 4 + 4}{2}) = (\frac{11}{2}, 0)$ .
2. Its $y$ -coordinate is 0 (and its $x$ -coordinate is not 0), so the point lies on the $x$ -axis. Option (a).
  1 mark
Answer: $x$ -axis, option (a)
Where the marks are
- Gets mid-point (11/2, 0) and concludes it lies on the x-axis (1 mark)
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Q9 · 1 mark · MCQIntroduction to Trigonometry
Given that $sin θ = \frac{a}{b}$ , then $cos θ$ is equal to :
1. (A) $\frac{b}{b ^{2} - a ^{2}}$
2. (B) $\frac{b}{a}$
3. (C) $\frac{b ^{2} - a ^{2}}{b}$
4. (D) $\frac{a}{b ^{2} - a ^{2}}$
Try it first, then show our solutionSolution
1. $cos^{2} θ = 1 - sin^{2} θ = 1 - \frac{a ^{2}}{b ^{2}} = \frac{b ^{2} - a ^{2}}{b ^{2}}$ .
2. So $cos θ = \frac{b ^{2} - a ^{2}}{b}$ . Option (c).
  1 mark
Answer: $\frac{b ^{2} - a ^{2}}{b}$ , option (c)
Where the marks are
- Uses $sin^{2} θ + cos^{2} θ = 1$ to get option (c) (1 mark)
Q10 · 1 mark · MCQIntroduction to Trigonometry
If $cos A = \frac{1}{2}$ , then the value of $sin^{2} A + 2 cos^{2} A$ is :
1. (A) $\frac{3}{2}$
2. (B) $\frac{5}{4}$
3. (C) $- 1$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. $cos^{2} A = \frac{1}{4}$ , so $sin^{2} A = 1 - \frac{1}{4} = \frac{3}{4}$ .
2. $sin^{2} A + 2 cos^{2} A = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$ . Option (b).
  1 mark
Answer: $\frac{5}{4}$ , option (b)
Where the marks are
- Correct substitution giving 5/4, option (b) (1 mark)
Q11 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $103$ m away from the base of the tower, is :
1. (A) $3 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $9 0^{\circ}$
4. (D) $6 0^{\circ}$
Try it first, then show our solutionSolution
1. Let the angle of elevation be $θ$ . Then $tan θ = \frac{height}{distance} = \frac{30}{10 3} = 3$ .
2. $tan θ = 3$ gives $θ = 6 0^{\circ}$ . Option (d).
  1 mark
Answer: $6 0^{\circ}$ , option (d)
Where the marks are
- Finds $tan θ = 3$ , so $θ = 6 0^{\circ}$ , option (d) (1 mark)
Open Q11 on its own page
Q12 · 1 mark · MCQCircles
If TP and TQ are two tangents to a circle with centre O from an external point T so that $∠ P O Q = 12 0^{\circ}$ , then $∠ P T Q$ is equal to :
1. (A) $6 0^{\circ}$
2. (B) $7 0^{\circ}$
3. (C) $8 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $∠ O P T = ∠ O QT = 9 0^{\circ}$ [tangent ⟂ radius].
2. The angles of quadrilateral OPTQ add up to $36 0^{\circ}$ , so $∠ P T Q = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ . Option (a).
  1 mark
Answer: $6 0^{\circ}$ , option (a)
Where the marks are
- Uses the angle sum of quadrilateral OPTQ to get 60°, option (a) (1 mark)
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Q13 · 1 mark · MCQCircles
In the given figure, PA is a tangent from an external point P to a circle with centre O. If $∠ P O B = 12 5^{\circ}$ , then $∠ A P O$ is equal to :
1. (A) $2 5^{\circ}$
2. (B) $6 5^{\circ}$
3. (C) $9 0^{\circ}$
4. (D) $3 5^{\circ}$
Figure in the paper: A circle with centre O. P is an external point on the left. PA is a tangent touching the circle at A (top). A straight line passes through A, O and continues to B on the lower right of the circle, so AB is a diameter. The line PO is drawn from P to O. The angle POB = 125° is marked at O between OP and OB. See it in the official paper.
Try it first, then show our solutionSolution
1. AOB is a straight line, so $∠ A O P = 18 0^{\circ} - 12 5^{\circ} = 5 5^{\circ}$ .
2. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius]. In $Δ O A P$ : $∠ A P O = 18 0^{\circ} - 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ . Option (d).
  1 mark
Answer: $3 5^{\circ}$ , option (d)
Where the marks are
- Finds ∠AOP = 55°, uses ∠OAP = 90° and gets 35°, option (d) (1 mark)
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Q14 · 1 mark · MCQAreas Related to Circles
The length of the arc of the sector of a circle with radius 21 cm and central angle $6 0^{\circ}$ , is :
1. (A)22 cm
2. (B)44 cm
3. (C)88 cm
4. (D)11 cm
Try it first, then show our solutionSolution
1. Arc length $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$ .
  ½ mark
2. $= \frac{1}{6} \times 132 = 22$ cm. Hence option (A).
  ½ mark
Answer: 22 cm, option (A)
Where the marks are
- Uses $l = \frac{θ}{360} \times 2 π r$ and gets 22 cm, option (A) (1 mark)
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Q15 · 1 mark · MCQAreas Related to Circles
The hour hand of a clock is 7 cm long. The angle swept by it between 7:00 a.m. and 8:10 a.m. is :
1. (A) $(\frac{35}{4})^{\circ}$
2. (B) $(\frac{35}{2})^{\circ}$
3. (C) $3 5^{\circ}$
4. (D) $7 0^{\circ}$
Try it first, then show our solutionSolution
1. The hour hand turns $3 0^{\circ}$ in 60 minutes, which is $\frac{1}{2}^{\circ}$ per minute.
2. From 7:00 to 8:10 is 70 minutes, so the angle swept $= 70 \times \frac{1}{2} = 3 5^{\circ}$ . Option (c).
  1 mark
Answer: $3 5^{\circ}$ , option (c)
Where the marks are
- Gets 70 min × 0.5° = 35°, option (c) (1 mark)
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Q16 · 1 mark · MCQSurface Areas and Volumes
The total surface area of a solid hemisphere of diameter ' $2 d$ ' is :
1. (A) $3 π d^{2}$
2. (B) $2 π d^{2}$
3. (C) $\frac{1}{2} π d^{2}$
4. (D) $\frac{3}{4} π d^{2}$
Try it first, then show our solutionSolution
1. The diameter is $2 d$ , so the radius is $r = d$ .
2. TSA of a solid hemisphere $= 3 π r^{2} = 3 π d^{2}$ . Option (a).
  1 mark
Answer: $3 π d^{2}$ , option (a)
Where the marks are
- Uses TSA = 3πr² with r = d, option (a) (1 mark)
Q17 · 1 mark · MCQStatistics
If the mean and mode of a data are 12 and 21 respectively, then its median is :
1. (A)6
2. (B)13.5
3. (C)15
4. (D)14
Try it first, then show our solutionSolution
1. Empirical relation: $3 Median = Mode + 2 Mean$ .
  ½ mark
2. $3 Median = 21 + 2 (12) = 45 \Rightarrow Median = 15$ . Hence option (C).
  ½ mark
Answer: Median = 15, option (C)
Where the marks are
- Applies $3 Median = Mode + 2 Mean$ to get 15, option (C) (1 mark)
Q18 · 1 mark · MCQProbability
A die is thrown once. Probability of getting a number other than 3 is :
1. (A) $\frac{1}{6}$
2. (B) $\frac{3}{6}$
3. (C) $\frac{5}{6}$
4. (D)1
Try it first, then show our solutionSolution
1. Total outcomes = 6. Outcomes other than 3 are 1, 2, 4, 5, 6, which is 5 outcomes.
  ½ mark
2. $P (not 3) = \frac{5}{6}$ (equivalently $1 - \frac{1}{6}$ ). Hence option (C).
  ½ mark
Answer: $\frac{5}{6}$ , option (C)
Where the marks are
- Counts 5 favourable out of 6 outcomes, giving $\frac{5}{6}$ , option (C) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Directions: Question numbers 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
Assertion (A): The probability that a leap year has 53 Mondays is $\frac{2}{7}$ .
Reason (R): The probability that a non-leap year has 53 Mondays is $\frac{5}{7}$ .
1. (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (c)Assertion (A) is true, but Reason (R) is false.
4. (d)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A leap year has 366 days, which is 52 weeks + 2 extra days. The 2 extra days can be one of 7 consecutive pairs (Sun–Mon, Mon–Tue, …, Sat–Sun). Two of these pairs contain a Monday, so P = $\frac{2}{7}$ . Assertion is true.
2. A non-leap year has 365 days, which is 52 weeks + 1 extra day. That day is Monday in 1 case out of 7, so P = $\frac{1}{7}$ , not $\frac{5}{7}$ . Reason is false. Option (c).
  1 mark
Answer: Option (c): A is true, R is false
Where the marks are
- Identifies A true (2/7) and R false (1/7), option (c) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedPolynomials
Assertion (A): The polynomial $p (y) = y^{2} + 4 y + 3$ has two zeroes.
Reason (R): A quadratic polynomial can have at most two zeroes.
1. (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (c)Assertion (A) is true, but Reason (R) is false.
4. (d)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $y^{2} + 4 y + 3 = (y + 1) (y + 3)$ , so the zeroes are $- 1$ and $- 3$ . Assertion is true.
2. A polynomial of degree 2 has at most 2 zeroes, so Reason is true. However, R only gives an upper limit; it does not explain why this polynomial actually has two zeroes. So R is not the correct explanation of A. Option (b).
  1 mark
Answer: Option (b)
Where the marks are
- Finds both statements true but R not explaining A, option (b) (1 mark)
Open Q20 on its own page

Section B

Very short answer (VSA)

Q21 · 2 marks · Very short answerPolynomials
If $α, β$ are the zeroes of the polynomial $p (x) = x^{2} - 3 x - 1$ , then find the value of $\frac{1}{α} + \frac{1}{β}$ .
Try it first, then show our solutionSolution
1. For $p (x) = x^{2} - 3 x - 1$ : $α + β = - \frac{b}{a} = 3$ and $α β = \frac{c}{a} = - 1$ .
  1 mark
2. $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{3}{- 1}$ .
  ½ mark
3. $= - 3$ .
  ½ mark
Answer: $- 3$
Where the marks are
- Writes sum of zeroes = 3 and product = −1 (1 mark)
- Rewrites the expression as (α+β)/(αβ) and substitutes (½ mark)
- Gets the answer −3 (½ mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(A) · 2 marks · Very short answerTriangles
In $Δ$ ABC, DE $∥$ BC. If AD = $x$ , DB = $x - 2$ , AE = $x + 2$ and EC = $x - 1$ , then find the value of $x$ .
Try it first, then show our solutionSolution
1. Since DE ∥ BC, by the Basic Proportionality Theorem $\frac{A D}{D B} = \frac{A E}{E C}$ .
  1 mark
2. $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$ , so $x (x - 1) = (x - 2) (x + 2)$ , which gives $x^{2} - x = x^{2} - 4$ .
  ½ mark
3. $- x = - 4$ , so $x = 4$ .
  ½ mark
Answer: $x = 4$
Where the marks are
- Applies BPT: AD/DB = AE/EC (1 mark)
- Substitutes and cross-multiplies (½ mark)
- Gets x = 4 (½ mark)
Open Q22(A) on its own page
OR
Q22(B) · 2 marks · Very short answerTriangles
In the figure given above, $Δ$ ABC $\sim$ $Δ$ XYZ, then find the values of $x$ and $y$ .
Figure in the paper: Two similar triangles. Triangle ABC: A at top, B bottom-left, C bottom-right; AB = 4 cm, BC = 6 cm, side AC labelled y. Triangle XYZ: X at top, Y bottom-left, Z bottom-right; XZ = 6 cm, YZ = 7.2 cm, side XY labelled x. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $Δ A B C \sim Δ X Y Z$ : $\frac{A B}{X Y} = \frac{B C}{Y Z} = \frac{A C}{X Z}$ , so $\frac{4}{x} = \frac{6}{7.2} = \frac{y}{6}$ .
  1 mark
2. $x = \frac{4 \times 7.2}{6} = 4.8$ cm.
  ½ mark
3. $y = \frac{6 \times 6}{7.2} = 5$ cm.
  ½ mark
Answer: $x = 4.8$ cm, $y = 5$ cm
Where the marks are
- Writes the ratios of corresponding sides with values (1 mark)
- Gets x = 4.8 cm (½ mark)
- Gets y = 5 cm (½ mark)
Open Q22(B) on its own page
Q23 · 2 marks · Very short answerCoordinate Geometry
The coordinates of the centre of a circle are $(x - 7, 2 x)$ . Find the value(s) of ' $x$ ', if the circle passes through the point $(- 9, 11)$ and has radius $52$ units.
Try it first, then show our solutionSolution
1. Distance from the centre $(x - 7, 2 x)$ to $(- 9, 11)$ equals the radius: $(x - 7 + 9)^{2} + (2 x - 11)^{2} = 52$ .
  ½ mark
2. Squaring: $(x + 2)^{2} + (2 x - 11)^{2} = 50$ , so $x^{2} + 4 x + 4 + 4 x^{2} - 44 x + 121 = 50$ , which gives $5 x^{2} - 40 x + 75 = 0$ , i.e. $x^{2} - 8 x + 15 = 0$ .
  ½ mark
3. $(x - 3) (x - 5) = 0$ , so $x = 3$ or $x = 5$ .
  1 mark
Answer: $x = 3$ or $x = 5$
Where the marks are
- Sets up the distance formula equal to $52$ (½ mark)
- Forms the quadratic $x^{2} - 8 x + 15 = 0$ (½ mark)
- Gets x = 3 and x = 5 (½ + ½ mark)
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Answer one of these 2 alternatives.
Q24(A) · 2 marks · Very short answerIntroduction to Trigonometry
If $tan θ = \frac{24}{7}$ , then find the value of $sin θ + cos θ$ .
Try it first, then show our solutionSolution
1. Since $tan θ = \frac{perpendicular}{base} = \frac{24}{7}$ , take a right triangle with perpendicular $= 24 k$ and base $= 7 k$ .
  ½ mark
2. By Pythagoras theorem, hypotenuse $= (24 k)^{2} + (7 k)^{2} = 576 k^{2} + 49 k^{2} = 625 k^{2} = 25 k$ .
  So $sin θ = \frac{24 k}{25 k} = \frac{24}{25}$ and $cos θ = \frac{7 k}{25 k} = \frac{7}{25}$ .
  1 mark
3. $sin θ + cos θ = \frac{24}{25} + \frac{7}{25} = \frac{31}{25}$
  ½ mark
Answer: $sin θ + cos θ = \frac{31}{25}$
Where the marks are
- Finds the hypotenuse 25 and gets $sin θ = \frac{24}{25}$ and $cos θ = \frac{7}{25}$ (1½ marks)
- Adds them correctly to get $\frac{31}{25}$ (½ mark)
Open Q24(A) on its own page
OR
Q24(B) · 2 marks · Very short answerIntroduction to Trigonometry
If $cot θ = \frac{7}{8}$ , then find the value of $\frac{( 1 + s i n θ ) ( 1 - s i n θ )}{( 1 + c o s θ ) ( 1 - c o s θ )}$ .
Try it first, then show our solutionSolution
1. Multiply out the brackets using $(1 + a) (1 - a) = 1 - a^{2}$ :
  $\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )} = \frac{1 - sin ^{2} θ}{1 - cos ^{2} θ}$
  1 mark
2. Since $sin^{2} θ + cos^{2} θ = 1$ , we have $1 - sin^{2} θ = cos^{2} θ$ and $1 - cos^{2} θ = sin^{2} θ$ . So the expression is
  $\frac{cos ^{2} θ}{sin ^{2} θ} = cot^{2} θ$
  ½ mark
3. Put $cot θ = \frac{7}{8}$ :
  $cot^{2} θ = (\frac{7}{8})^{2} = \frac{49}{64}$
  ½ mark
Answer: $\frac{49}{64}$
Where the marks are
- Writes numerator and denominator as $1 - sin^{2} θ$ and $1 - cos^{2} θ$ (1 mark)
- Uses the identity to reduce the expression to $cot^{2} θ$ (½ mark)
- Substitutes to get $\frac{49}{64}$ (½ mark)
Open Q24(B) on its own page
Q25 · 2 marks · Very short answerCircles
Two concentric circles are of radii 5 cm and 4 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 circles with common centre O. Inner radius 4 cm, outer radius 5 cm. Chord AB of the larger circle touches the smaller circle at M. Join OA and OM.
  ½ mark
2. OM is the radius of the smaller circle and AB is a tangent to it at M, so $O M ⊥ A B$ [tangent ⟂ radius]. Also OM = 4 cm and OA = 5 cm.
  In right $△ O M A$ : $A M = O A^{2} - O M^{2} = 5^{2} - 4^{2} = 9 = 3$ cm.
  1 mark
3. The perpendicular from the centre to a chord bisects the chord, so M is the mid-point of AB.
  $A B = 2 \times A M = 2 \times 3 = 6$ cm.
  ½ mark
Answer: Length of chord = 6 cm
Where the marks are
- Draws a correct labelled figure (½ mark)
- Uses OM ⟂ AB and Pythagoras to get AM = 3 cm (1 mark)
- Uses the perpendicular bisecting the chord to get AB = 6 cm (½ mark)
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Section C

Short answer (SA)

Q26 · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Suppose, to the contrary, that $3$ is rational. Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are co-prime (no common factor other than 1).
  ½ mark
2. Squaring, $3 = \frac{p ^{2}}{q ^{2}}$ , so $p^{2} = 3 q^{2}$ .
  Thus 3 divides $p^{2}$ . Since 3 is prime, 3 divides $p$ . ...(i)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2} \Rightarrow q^{2} = 3 a^{2}$ .
  So 3 divides $q^{2}$ , and since 3 is prime, 3 divides $q$ . ...(ii)
  1 mark
4. From (i) and (ii), 3 is a common factor of $p$ and $q$ . This contradicts the fact that $p$ and $q$ are co-prime.
  The contradiction comes from our assumption, so $3$ is an irrational number.
  ½ mark
Answer: $3$ is irrational (proved by contradiction).
Where the marks are
- Assumes $3 = \frac{p}{q}$ with $p, q$ co-prime and $q \neq = 0$ (½ mark)
- Gets $p^{2} = 3 q^{2}$ and concludes 3 divides $p$ (1 mark)
- Substitutes $p = 3 a$ and concludes 3 divides $q$ (1 mark)
- States the contradiction and concludes $3$ is irrational (½ mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerCoordinate Geometry
Find the ratio in which the $x$ -axis divides the line segment joining the points $(- 6, 5)$ and $(- 4, - 1)$ . Also, find the point of intersection.
Try it first, then show our solutionSolution
1. Let A(−6, 5) and B(−4, −1). Let the x-axis meet AB at P(x, 0), and let P divide AB in the ratio $k : 1$ .
  ½ mark
2. By the section formula,
  $P = (\frac{k ( - 4 ) + 1 ( - 6 )}{k + 1}, \frac{k ( - 1 ) + 1 ( 5 )}{k + 1}) = (\frac{- 4 k - 6}{k + 1}, \frac{- k + 5}{k + 1})$
  1 mark
3. P lies on the x-axis, so its y-coordinate is 0:
  $\frac{- k + 5}{k + 1} = 0 \Rightarrow k = 5$
  ½ mark
4. Hence the x-axis divides the segment in the ratio $5 : 1$ .
  ½ mark
5. x-coordinate of P $= \frac{- 4 ( 5 ) - 6}{5 + 1} = \frac{- 26}{6} = - \frac{13}{3}$ .
  So the point of intersection is $(- \frac{13}{3}, 0)$ .
  ½ mark
Answer: Ratio 5 : 1; point of intersection $(- \frac{13}{3}, 0)$
Where the marks are
- Takes P(x, 0) dividing AB in ratio k : 1 (½ mark)
- Writes the coordinates of P by the section formula correctly (1 mark)
- Sets the y-coordinate to 0 and gets k = 5 (½ mark)
- States the ratio 5 : 1 (½ mark)
- Finds the point $(- \frac{13}{3}, 0)$ (½ mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(A) · 3 marks · Short answerIntroduction to Trigonometry
If $x = h + a cos θ$ , $y = k + b sin θ$ , then prove that : $(\frac{x - h}{a})^{2} + (\frac{y - k}{b})^{2} = 1$
Try it first, then show our solutionSolution
1. From $x = h + a cos θ$ : $x - h = a cos θ \Rightarrow \frac{x - h}{a} = cos θ$
  1 mark
2. From $y = k + b sin θ$ : $y - k = b sin θ \Rightarrow \frac{y - k}{b} = sin θ$
  1 mark
3. LHS $= (\frac{x - h}{a})^{2} + (\frac{y - k}{b})^{2} = cos^{2} θ + sin^{2} θ = 1 =$ RHS [since $sin^{2} θ + cos^{2} θ = 1$ ]
  Hence proved.
  1 mark
Answer: $(\frac{x - h}{a})^{2} + (\frac{y - k}{b})^{2} = cos^{2} θ + sin^{2} θ = 1$ (proved)
Where the marks are
- Gets $\frac{x - h}{a} = cos θ$ (1 mark)
- Gets $\frac{y - k}{b} = sin θ$ (1 mark)
- Squares, adds and uses $sin^{2} θ + cos^{2} θ = 1$ to complete the proof (1 mark)
Open Q28(A) on its own page
OR
Q28(B) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{t a n A}{1 + s e c A} - \frac{t a n A}{1 - s e c A} = 2 cosec A$
Try it first, then show our solutionSolution
1. Write everything in terms of sin A and cos A:
  $LHS = \frac{tan A}{1 + sec A} - \frac{tan A}{1 - sec A} = \frac{\frac{s i n A}{c o s A}}{1 + \frac{1}{c o s A}} - \frac{\frac{s i n A}{c o s A}}{1 - \frac{1}{c o s A}}$
  1 mark
2. Simplify each fraction by multiplying its numerator and denominator by $cos A$ :
  $= \frac{sin A}{cos A + 1} - \frac{sin A}{cos A - 1}$
  ½ mark
3. Take the LCM:
  $= sin A [\frac{( cos A - 1 ) - ( cos A + 1 )}{cos ^{2} A - 1}] = sin A (\frac{- 2}{- sin ^{2} A})$
  [since $cos^{2} A - 1 = - sin^{2} A$ ]
  1 mark
4. $= \frac{2}{sin A} = 2 cosec A = RHS$
  Hence proved.
  ½ mark
Answer: LHS simplifies to $\frac{2}{s i n A} = 2 cosec A$ (proved)
Where the marks are
- Converts tan A and sec A into sin A and cos A (1 mark)
- Simplifies to $\frac{s i n A}{c o s A + 1} - \frac{s i n A}{c o s A - 1}$ (½ mark)
- Combines the fractions and uses $cos^{2} A - 1 = - sin^{2} A$ (1 mark)
- Reaches $\frac{2}{s i n A} = 2 cosec A$ (½ mark)
Open Q28(B) on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerCircles
In the given figure, $Δ$ ABC is a right triangle in which $∠$ B = $9 0^{\circ}$ , AB = 4 cm and BC = 3 cm. Find the radius of the circle inscribed in the triangle ABC.
Figure in the paper: Right triangle ABC with right angle at B (bottom right). A is at the top, C at bottom left, B at bottom right. A circle is inscribed inside the triangle touching all three sides. See it in the official paper.
Try it first, then show our solutionSolution
1. In right $△ A B C$ , $∠ B = 9 0^{\circ}$ .
  By Pythagoras theorem, $A C = A B^{2} + B C^{2} = 4^{2} + 3^{2} = 5$ cm.
  ½ mark
2. Let the incircle (centre O) touch AB at D, BC at E and CA at F. Let $B D = B E = x$ cm [tangents from an external point are equal].
  ½ mark
3. Then $A D = A F = 4 - x$ and $C E = C F = 3 - x$ [tangents from A and from C are equal].
  ½ mark
4. Since $A F + C F = A C$ :
  $(4 - x) + (3 - x) = 5 \Rightarrow 7 - 2 x = 5 \Rightarrow x = 1$
  1 mark
5. In quadrilateral ODBE, $∠ B = 9 0^{\circ}$ , $∠ O D B = ∠ O E B = 9 0^{\circ}$ [tangent ⟂ radius] and $B D = B E$ . So ODBE is a square, and radius $O D = B D = x = 1$ cm.
  Hence the radius of the inscribed circle is 1 cm.
  ½ mark
Answer: Radius = 1 cm
Where the marks are
- Finds AC = 5 cm (½ mark)
- Takes BD = BE = x using equal tangents (½ mark)
- Writes AD = AF = 4 − x and CE = CF = 3 − x (½ mark)
- Forms (4 − x) + (3 − x) = 5 and gets x = 1 (1 mark)
- Concludes radius = 1 cm, since ODBE is a square (½ mark)
- Alternative: equating area 6 cm² with ½·r·(4 + 3 + 5) to get r = 1 cm earns full marks
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerCircles
In the given figure, if a circle touches the side QR of $Δ$ PQR at S and extended sides PQ and PR at M and N respectively, then prove that : $P M = \frac{1}{2} (P Q + QR + P R)$
Figure in the paper: Triangle PQR with apex P at top, Q bottom-left, R bottom-right. A circle lies below side QR, touching QR at S, and touching the extended sides PQ (at M, on the lower left extension) and PR (at N, on the lower right extension). See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents drawn from an external point to a circle are equal in length. So:
  from P: $P M = P N$
  from Q: $QS = QM$
  from R: $R S = R N$
  1½ marks
2. Since M lies on PQ extended and N lies on PR extended, $P M = P Q + QM$ and $P N = P R + R N$ . Adding:
  $P M + P N = P Q + QM + P R + R N$
  ½ mark
3. Using $P N = P M$ , $QM = QS$ and $R N = R S$ :
  $2 P M = P Q + QS + P R + R S = P Q + (QS + S R) + P R$
  ½ mark
4. Since S lies on QR, $QS + S R = QR$ . So
  $2 P M = P Q + QR + P R \Rightarrow P M = \frac{1}{2} (P Q + QR + P R)$
  Hence proved.
  ½ mark
Answer: $P M = \frac{1}{2} (P Q + QR + P R)$ (proved)
Where the marks are
- States the three pairs of equal tangents PM = PN, QS = QM, RS = RN (1½ marks)
- Writes PM + PN = PQ + QM + PR + RN (½ mark)
- Substitutes to get 2PM = PQ + QS + RS + PR (½ mark)
- Uses QS + SR = QR to conclude the result (½ mark)
Open Q29(B) on its own page
Q30 · 3 marks · Short answerSurface Areas and Volumes
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use $π = \frac{22}{7}$ )
Try it first, then show our solutionSolution
1. Radius of cylinder = radius of each hemisphere $= r = \frac{7}{2}$ cm.
  ½ mark
2. Total height = 20 cm. Each hemispherical end adds height $r$ .
  Height of cylinder $h = 20 - \frac{7}{2} - \frac{7}{2} = 13$ cm.
  ½ mark
3. Volume of solid = volume of cylinder + volume of 2 hemispheres:
  $V = π r^{2} h + 2 \times \frac{2}{3} π r^{3} = \frac{22}{7} \times (\frac{7}{2})^{2} \times 13 + 2 \times \frac{2}{3} \times \frac{22}{7} \times (\frac{7}{2})^{3}$
  1 mark
4. Take $π r^{2}$ common:
  $V = \frac{22}{7} \times \frac{49}{4} \times (13 + \frac{4}{3} \times \frac{7}{2}) = \frac{77}{2} \times (13 + \frac{14}{3}) = \frac{77}{2} \times \frac{53}{3}$
  $V = \frac{4081}{6} \approx 680.17 cm^{3}$
  1 mark
Answer: Volume = $\frac{4081}{6}$ cm³ ≈ 680.17 cm³
Where the marks are
- Takes radius = 7/2 cm (½ mark)
- Finds height of cylinder = 13 cm (½ mark)
- Writes volume = cylinder + two hemispheres with correct substitution (1 mark)
- Simplifies to $\frac{4081}{6} \approx 680.17$ cm³ (1 mark)
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Q31 · 3 marks · Short answerProbability
Two dice of different colours are thrown at the same time. Write down all the possible outcomes. What is the probability that :
1. (i)
  same number appears on both the dice?
  0 marks
2. (ii)
  different number appears on both the dice?
  0 marks
Try it first, then show our solutionSolution
1. Write (first die, second die). All possible outcomes:
  (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
  (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
  (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
  (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
  (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
  (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
  Total number of outcomes = 36.
  1 mark
2. (i) Outcomes with the same number on both dice: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), so 6 outcomes.
  P(same number) $= \frac{6}{36} = \frac{1}{6}$
  1 mark
3. (ii) "Different numbers" is the complement of "same number".
  P(different numbers) $= 1 - \frac{1}{6} = \frac{5}{6}$ (that is, $\frac{30}{36}$ )
  1 mark
Answer: 36 outcomes; (i) $\frac{1}{6}$ ; (ii) $\frac{5}{6}$
Where the marks are
- Lists all 36 outcomes correctly (1 mark)
- (i) Finds P(same number) = 6/36 = 1/6 (1 mark)
- (ii) Finds P(different numbers) = 5/6 (1 mark)
Open Q31 on its own page

Section D

Long answer (LA)

Q32 · 5 marks · Long answerPair of Linear Equations in Two Variables
Determine graphically, the coordinates of vertices of a triangle whose equations are $2 x - 3 y + 6 = 0$ ; $2 x + 3 y - 18 = 0$ and $x = 0$ . Also, find the area of this triangle.
Try it first, then show our solutionSolution
1. Line 1: $2 x - 3 y + 6 = 0$ , i.e. $y = \frac{2 x + 6}{3}$ .
  x 0 3 −3
  y 2 4 0
  Plot (0, 2), (3, 4), (−3, 0) and join them to draw the line.
  1 mark
2. Line 2: $2 x + 3 y - 18 = 0$ , i.e. $y = \frac{18 - 2 x}{3}$ .
  x 0 3 9
  y 6 4 0
  Plot (0, 6), (3, 4), (9, 0) and join them to draw the line.
  1 mark
3. Line 3: $x = 0$ is the y-axis itself. Mark it as one side of the triangle.
  ½ mark
4. From the graph:
  Line 1 meets the y-axis at A(0, 2).
  Line 1 and line 2 intersect at B(3, 4).
  Line 2 meets the y-axis at C(0, 6).
  So the vertices of the triangle are A(0, 2), B(3, 4) and C(0, 6).
  1½ marks
5. Take base AC on the y-axis: $A C = 6 - 2 = 4$ units.
  Height = perpendicular distance of B from the y-axis = 3 units.
  Area of $△ A B C = \frac{1}{2} \times 4 \times 3 = 6$ sq. units.
  1 mark
Answer: Vertices A(0, 2), B(3, 4), C(0, 6); area = 6 sq. units
Where the marks are
- Correct table and graph of $2 x - 3 y + 6 = 0$ (1 mark)
- Correct table and graph of $2 x + 3 y - 18 = 0$ (1 mark)
- Shows $x = 0$ as the y-axis (½ mark)
- Reads the vertices (0, 2), (3, 4), (0, 6) from the graph (1½ marks)
- Finds area = ½ × 4 × 3 = 6 sq. units (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerQuadratic Equations
A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains.
Try it first, then show our solutionSolution
1. Let the speed of the faster train be $x$ km/h. Then the speed of the slower train is $(x - 10)$ km/h.
  Time = distance/speed. The slower train takes 1 hour more for 200 km:
  $\frac{200}{x - 10} - \frac{200}{x} = 1$
  2 marks
2. Multiply both sides by $x (x - 10)$ :
  $200 x - 200 (x - 10) = x (x - 10)$
  $2000 = x^{2} - 10 x \Rightarrow x^{2} - 10 x - 2000 = 0$
  1 mark
3. Factorise by splitting the middle term:
  $x^{2} - 50 x + 40 x - 2000 = 0 \Rightarrow (x - 50) (x + 40) = 0$
  1 mark
4. So $x = 50$ or $x = - 40$ . A speed cannot be negative, so we reject $x = - 40$ . Hence $x = 50$ .
  ½ mark
5. Speed of faster train = 50 km/h.
  Speed of slower train = 50 − 10 = 40 km/h.
  ½ mark
Answer: Faster train: 50 km/h; slower train: 40 km/h
Where the marks are
- Forms the equation $\frac{200}{x - 10} - \frac{200}{x} = 1$ (2 marks)
- Reduces it to $x^{2} - 10 x - 2000 = 0$ (1 mark)
- Factorises as $(x - 50) (x + 40) = 0$ (1 mark)
- Rejects the negative root and takes x = 50 (½ mark)
- States both speeds, 50 km/h and 40 km/h (½ mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerQuadratic Equations
The sum of the areas of two squares is $640 m^{2}$ . If the difference in their perimeters is 64 m, find the sides of the two squares.
Try it first, then show our solutionSolution
1. Let the sides of the two squares be $x$ m and $y$ m, with $x > y$ .
  The sum of the areas is 640 m², so
  $x^{2} + y^{2} = 640 ...(1)$
  1 mark
2. The perimeters are $4 x$ and $4 y$ , and their difference is 64 m:
  $4 x - 4 y = 64 \Rightarrow x - y = 16 \Rightarrow y = x - 16 ...(2)$
  1 mark
3. Substitute (2) in (1):
  $x^{2} + (x - 16)^{2} = 640$
  $2 x^{2} - 32 x + 256 - 640 = 0$
  $x^{2} - 16 x - 192 = 0$
  1 mark
4. Split the middle term ( $- 24 \times 8 = - 192$ and $- 24 + 8 = - 16$ ):
  $x^{2} - 24 x + 8 x - 192 = 0 \Rightarrow (x - 24) (x + 8) = 0$
  1 mark
5. So $x = 24$ or $x = - 8$ .
  A side cannot be negative, so $x = - 8$ is rejected. Hence $x = 24$ .
  ½ mark
6. Then $y = 24 - 16 = 8$ .
  Check: $2 4^{2} + 8^{2} = 576 + 64 = 640$ ✓, and $4 (24) - 4 (8) = 64$ ✓.
  The sides of the two squares are 24 m and 8 m.
  ½ mark
Answer: Sides are 24 m and 8 m
Where the marks are
- Writes the area equation $x^{2} + y^{2} = 640$ (1 mark)
- Uses the perimeter condition to get $y = x - 16$ (1 mark)
- Forms the quadratic $x^{2} - 16 x - 192 = 0$ (1 mark)
- Factorises as $(x - 24) (x + 8) = 0$ (1 mark)
- Takes $x = 24$ and rejects $- 8$ (½ mark)
- Finds $y = 8$ and states both sides (½ mark)
Open Q33(B) on its own page
Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerTriangles
State and prove Basic Proportionality Theorem.
Try it first, then show our solutionSolution
1. Statement (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. Figure: Draw $△ A B C$ . Draw a line parallel to $B C$ that meets $A B$ at $D$ and $A C$ at $E$ .
  Given: In $△ A B C$ , $D E ∥ B C$ , with $D$ on $A B$ and $E$ on $A C$ .
  To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  Construction: Join $B E$ and $C D$ . Draw $D M ⊥ A C$ (meeting $A C$ at $M$ ) and $E N ⊥ A B$ (meeting $A B$ at $N$ ).
  2 marks
3. Proof: Use area $= \frac{1}{2} \times base \times height$ .
  Take $E N$ as the height:
  $ar (A D E) = \frac{1}{2} \times A D \times E N, ar (B D E) = \frac{1}{2} \times D B \times E N$
  $∴ \frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B} ...(1)$
  Take $D M$ as the height:
  $ar (A D E) = \frac{1}{2} \times A E \times D M, ar (D E C) = \frac{1}{2} \times E C \times D M$
  $∴ \frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C} ...(2)$
  1 mark
4. $△ B D E$ and $△ D E C$ stand on the same base $D E$ and lie between the same parallels $D E$ and $B C$ . Hence
  $ar (B D E) = ar (D E C) ...(3)$
  From (1), (2) and (3), the left-hand sides of (1) and (2) are equal. Therefore
  $\frac{A D}{D B} = \frac{A E}{E C}$
  Hence proved.
  1 mark
Answer: Statement of BPT with proof that $\frac{A D}{D B} = \frac{A E}{E C}$
Where the marks are
- Correct statement of BPT (1 mark)
- Correct figure, given, to prove and construction (2 marks)
- Area ratios $\frac{A D}{D B}$ and $\frac{A E}{E C}$ using the perpendiculars (1 mark)
- Uses ar(BDE) = ar(DEC) [same base, same parallels] and concludes (1 mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerTriangles
In the given figure, CM and RN are respectively the medians of $Δ$ ABC and $Δ$ PQR. If $Δ$ ABC $\sim$ $Δ$ PQR, then prove that :
1. (i)
  $Δ$ AMC $\sim$ $Δ$ PNR
  0 marks
2. (ii)
  $Δ$ CMB $\sim$ $Δ$ RNQ
  0 marks
Figure in the paper: Two triangles side by side. Triangle ABC with apex C at top, base AB at bottom, M the midpoint of AB on the base with median CM drawn. Triangle PQR with P top-left, Q top-right, R at bottom, N the midpoint of PQ on the top side with median RN drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Since $△ A B C \sim △ P QR$ , corresponding angles are equal:
  $∠ A = ∠ P ...(1)$
  ½ mark
2. (i) Corresponding sides are proportional:
  $\frac{A B}{P Q} = \frac{A C}{P R}$
  $C M$ and $R N$ are medians, so $M$ is the midpoint of $A B$ and $N$ is the midpoint of $P Q$ . Hence $A B = 2 A M$ and $P Q = 2 P N$ :
  $\frac{2 A M}{2 P N} = \frac{A C}{P R} \Rightarrow \frac{A M}{P N} = \frac{A C}{P R} ...(2)$
  1½ marks
3. (i) In $△ A M C$ and $△ P N R$ , $∠ A = ∠ P$ from (1), and the sides containing these angles are proportional from (2).
  By the SAS similarity criterion, $△ A M C \sim △ P N R$ .
  ½ mark
4. (ii) Since $△ A B C \sim △ P QR$ :
  $∠ B = ∠ Q ...(3)$
  ½ mark
5. (ii) Also
  $\frac{A B}{P Q} = \frac{B C}{QR}$
  As $A B = 2 M B$ and $P Q = 2 N Q$ (because $M$ and $N$ are midpoints):
  $\frac{2 M B}{2 N Q} = \frac{B C}{QR} \Rightarrow \frac{M B}{N Q} = \frac{B C}{QR} ...(4)$
  1½ marks
6. (ii) In $△ C M B$ and $△ R N Q$ , $∠ B = ∠ Q$ from (3), and the sides containing these angles are proportional from (4).
  By SAS similarity, $△ C M B \sim △ R N Q$ .
  ½ mark
Answer: (i) $△ A M C \sim △ P N R$ (SAS); (ii) $△ C M B \sim △ R N Q$ (SAS)
Where the marks are
- (i) $∠ A = ∠ P$ from similarity (½ mark)
- (i) $\frac{A M}{P N} = \frac{A C}{P R}$ using the medians (1½ marks)
- (i) Concludes $△ A M C \sim △ P N R$ by SAS (½ mark)
- (ii) $∠ B = ∠ Q$ (½ mark)
- (ii) $\frac{M B}{N Q} = \frac{B C}{QR}$ using the medians (1½ marks)
- (ii) Concludes $△ C M B \sim △ R N Q$ by SAS (½ mark)
Open Q34(B) on its own page
Q35 · 5 marks · Long answerStatistics
The mean of the following frequency distribution is 35. Find the values of $x$ and $y$ , if the sum of frequencies is 25 :
Class Frequency
0-10 1
10-20 $x$
20-30 5
30-40 7
40-50 $y$
50-60 3
60-70 1
Try it first, then show our solutionSolution
1. Take the class marks $x_{i}$ and find $f_{i} x_{i}$ for each class:
  Class $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  0-10 1 5 5
  10-20 $x$ 15 $15 x$
  20-30 5 25 125
  30-40 7 35 245
  40-50 $y$ 45 $45 y$
  50-60 3 55 165
  60-70 1 65 65
  Total $17 + x + y$ $605 + 15 x + 45 y$
  2 marks
2. The sum of frequencies is 25:
  $17 + x + y = 25 \Rightarrow x + y = 8 ...(i)$
  ½ mark
3. Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ :
  $35 = \frac{605 + 15 x + 45 y}{25}$
  ½ mark
4. $605 + 15 x + 45 y = 875 \Rightarrow 15 x + 45 y = 270 \Rightarrow x + 3 y = 18 ...(ii)$
  1 mark
5. Subtract (i) from (ii): $2 y = 10$ , so $y = 5$ .
  Then $x = 8 - 5 = 3$ .
  ** $x = 3$ , $y = 5$ **
  1 mark
Answer: $x = 3$ , $y = 5$
Where the marks are
- Correct table of class marks and $f_{i} x_{i}$ (2 marks)
- $x + y = 8$ from the total frequency (½ mark)
- Sets up the mean equation (½ mark)
- Simplifies to $x + 3 y = 18$ (1 mark)
- Solves to get $x = 3$ and $y = 5$ (½ + ½ marks)
Open Q35 on its own page

x	0	3	−3
y	2	4	0

x	0	3	9
y	6	4	0

Class	Frequency
0-10	1
10-20	$x$
20-30	5
30-40	7
40-50	$y$
50-60	3
60-70	1

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0-10	1	5	5
10-20	$x$	15	$15 x$
20-30	5	25	125
30-40	7	35	245
40-50	$y$	45	$45 y$
50-60	3	55	165
60-70	1	65	65
Total	$17 + x + y$		$605 + 15 x + 45 y$

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato. The other potatoes are arranged 3 m apart in a straight line, with a total of 10 potatoes, as shown in the figure. A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.
Based on the above information, answer the following questions :
1. (i)
  What is the distance covered to pick up the first potato and drop it in bucket ?
  1 mark
2. (ii)
  What is the distance covered to pick up the second potato and drop it in bucket ?
  1 mark
3. (iii)(a)
  What is the total distance the competitor has to run ?
  2 marks
4. OR
  (iii)(b)
  If average speed of competitor is 5 m/s, then find the average time taken by competitor to put all the potatoes in the bucket.
  2 marks
Figure in the paper: A bucket with potatoes at the left end of a horizontal arrow line. The distance from the bucket to the first potato is 5 m; the next nine potatoes are placed in a straight line, each 3 m apart (labelled 3m between consecutive potatoes). Total 10 potatoes. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The first potato is 5 m from the bucket. The competitor runs 5 m to it and 5 m back:
  $5 + 5 = 10 m$
  1 mark
2. (ii) The second potato is $5 + 3 = 8$ m from the bucket. Distance there and back:
  $8 + 8 = 16 m$
  1 mark
3. (iii)(a) Each later potato is 3 m farther, so each round trip is $2 \times 3 = 6$ m longer. The round-trip distances are $10, 16, 22, \dots$
  This is an AP with $a = 10$ , $d = 6$ , $n = 10$ .
  1 mark
4. (iii)(a) $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
  $S_{10} = \frac{10}{2} [2 (10) + 9 (6)] = 5 \times 74 = 370 m$
  1 mark
5. (iii)(b) The round-trip distances form an AP with $a = 10$ , $d = 6$ , $n = 10$ .
  ½ mark
6. (iii)(b) $S_{10} = \frac{10}{2} [2 (10) + 9 (6)] = 5 \times 74 = 370 m$
  1 mark
7. (iii)(b) $Time = \frac{Distance}{Speed} = \frac{370}{5} = 74 seconds$
  ½ mark
Answer: (i) 10 m; (ii) 16 m; (iii)(a) 370 m, (iii)(b) 74 seconds
Where the marks are
- (i) 10 m (1 mark)
- (ii) 16 m (1 mark)
- (iii)(a) Recognises an AP with $a = 10$ , $d = 6$ (1 mark)
- (iii)(a) $S_{10} = 370$ m (1 mark)
- (iii)(b) Identifies the AP (½ mark)
- (iii)(b) Total distance 370 m (1 mark)
- (iii)(b) Time = 74 s (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O' (as shown in figure).
Distance between the base of the tower and point 'O' is 6 m. From point 'O', the angle of elevation of the top of the section 'B' is $3 0^{\circ}$ and the angle of elevation of the top of section 'A' is $6 0^{\circ}$ .
Based on the above information, answer the following questions :
1. (i)
  Find the length of the wire from the point 'O' to the top of section 'B'.
  1 mark
2. (ii)
  Find the length of the wire from the point 'O' to the top of section 'A'.
  1 mark
3. (iii)(a)
  Find the distance AB.
  2 marks
4. OR
  (iii)(b)
  Find the area of $Δ$ OPB.
  2 marks
Figure in the paper: A vertical tower AP with P at the base on the ground and A at the top. Point B lies on the tower between P and A (section boundary). A point O on the ground is 6 m horizontally from P. Wires join O to A (angle of elevation 60° at O) and O to B (angle of elevation 30° at O). PO = 6 m is marked along the ground. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In right $△ O P B$ (right angle at $P$ ), $O P = 6$ m and $∠ B O P = 3 0^{\circ}$ :
  $cos 3 0^{\circ} = \frac{O P}{O B} \Rightarrow \frac{3}{2} = \frac{6}{O B}$
  ½ mark
2. (i) $O B = \frac{12}{3} = 43 m$
  ½ mark
3. (ii) In right $△ O P A$ , $∠ A O P = 6 0^{\circ}$ :
  $cos 6 0^{\circ} = \frac{O P}{O A} \Rightarrow \frac{1}{2} = \frac{6}{O A}$
  ½ mark
4. (ii) $O A = 12 m$
  ½ mark
5. (iii)(a) $tan 3 0^{\circ} = \frac{B P}{O P} \Rightarrow \frac{1}{3} = \frac{B P}{6} \Rightarrow B P = \frac{6}{3} = 23 m$
  1 mark
6. (iii)(a) $tan 6 0^{\circ} = \frac{A P}{O P} \Rightarrow 3 = \frac{A P}{6} \Rightarrow A P = 63 m$
  ½ mark
7. (iii)(a) $A B = A P - B P = 63 - 23 = 43 m$
  ½ mark
8. (iii)(b) $tan 3 0^{\circ} = \frac{B P}{6} \Rightarrow B P = 23 m$
  1 mark
9. (iii)(b) $△ O P B$ is right-angled at $P$ , so
  $ar (△ O P B) = \frac{1}{2} \times O P \times B P = \frac{1}{2} \times 6 \times 23 = 63 m^{2}$
  1 mark
Answer: (i) $43$ m; (ii) 12 m; (iii)(a) $43$ m, (iii)(b) $63$ m²
Where the marks are
- (i) Uses $cos 3 0^{\circ} = 6/ O B$ (½ mark)
- (i) $O B = 43$ m (½ mark)
- (ii) Uses $cos 6 0^{\circ} = 6/ O A$ (½ mark)
- (ii) $O A = 12$ m (½ mark)
- (iii)(a) $B P = 23$ m (1 mark)
- (iii)(a) $A P = 63$ m (½ mark)
- (iii)(a) $A B = 43$ m (½ mark)
- (iii)(b) $B P = 23$ m (1 mark)
- (iii)(b) Area $= 63$ m² (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 cm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors as shown in figure.
Based on the above information, answer the following questions :
1. (i)
  What is the radius of circle ?
  1 mark
2. (ii)
  What is the circumference of the brooch ?
  1 mark
3. (iii)(a)
  What is the total length of silver wire required ?
  2 marks
4. OR
  (iii)(b)
  What is the area of each sector of the brooch ?
  2 marks
Figure in the paper: A circle divided into 10 equal sectors by 5 diameters, all passing through the centre, giving 10 equal angles at the centre. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $r = \frac{diameter}{2} = \frac{35}{2} = 17.5 cm$
  1 mark
2. (ii) $Circumference = 2 π r = 2 \times \frac{22}{7} \times \frac{35}{2} = 110 cm$
  1 mark
3. (iii)(a) The wire forms the circle plus 5 diameters, each 35 cm long:
  $Total length = 5 \times 35 + 110$
  1 mark
4. (iii)(a) $= 175 + 110 = 285 cm$
  1 mark
5. (iii)(b) The 10 equal sectors share the full angle at the centre:
  $Central angle = \frac{36 0 ^{\circ}}{10} = 3 6^{\circ}$
  1 mark
6. (iii)(b) $Area of a sector = \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{36}{360} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = \frac{385}{4} = 96.25 cm^{2}$
  1 mark
Answer: (i) 17.5 cm; (ii) 110 cm; (iii)(a) 285 cm, (iii)(b) 96.25 cm²
Where the marks are
- (i) $r = 17.5$ cm (1 mark)
- (ii) Circumference $= 110$ cm (1 mark)
- (iii)(a) Writes total length $= 5 \times 35 + 110$ (1 mark)
- (iii)(a) Gets 285 cm (1 mark)
- (iii)(b) Central angle $= 3 6^{\circ}$ (1 mark)
- (iii)(b) Sector area $= 96.25$ cm² (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.