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

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

The question paper set in the CBSE Class 10 board exam of 2025, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

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

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2025/X/041_Mathematics_Standard.zip (a zip; this paper is the file 30-1-1_Mathematics Standard.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer
C	26-31	Short answer
D	32-35	Long answer
E	36-38	Case study based

Instructions in brief: All questions compulsory; no overall choice, internal choice in 2 questions of Section B, 2 of C, 2 of D and 3 of E; calculator not allowed; take pi = 22/7 where required.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPolynomials
If $α$ and $β$ are the zeroes of polynomial $3 x^{2} + 6 x + k$ such that $α + β + α β = - \frac{2}{3}$ , then the value of k is :
1. (A) $- 8$
2. (B)8
3. (C) $- 4$
4. (D)4
Try it first, then show our solutionSolution
1. For $3 x^{2} + 6 x + k$ : $a = 3, b = 6, c = k$ . So $α + β = - \frac{b}{a} = - \frac{6}{3} = - 2$ and $α β = \frac{c}{a} = \frac{k}{3}$ .
2. Substitute in the given condition: $- 2 + \frac{k}{3} = - \frac{2}{3}$ , so $\frac{k}{3} = 2 - \frac{2}{3} = \frac{4}{3}$ .
3. Hence $k = 4$ , which is option (D).
  1 mark
Answer: k = 4, option (D)
Where the marks are
- Uses sum $= - 2$ and product $= k /3$ to get $k = 4$ , option (D) (1 mark)
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Q2 · 1 mark · MCQPair of Linear Equations in Two Variables
If $x = 1$ and $y = 2$ is a solution of the pair of linear equations $2 x - 3 y + a = 0$ and $2 x + 3 y - b = 0$ , then :
1. (A) $a = 2 b$
2. (B) $2 a = b$
3. (C) $a + 2 b = 0$
4. (D) $2 a + b = 0$
Try it first, then show our solutionSolution
1. Put $x = 1, y = 2$ in the first equation: $2 - 6 + a = 0 \Rightarrow a = 4$ .
2. Put $x = 1, y = 2$ in the second equation: $2 + 6 - b = 0 \Rightarrow b = 8$ .
3. Since $b = 8 = 2 \times 4$ , we get $2 a = b$ , which is option (B).
  1 mark
Answer: 2a = b, option (B)
Where the marks are
- Finds $a = 4$ and $b = 8$ and concludes $2 a = b$ , option (B) (1 mark)
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Q3 · 1 mark · MCQCoordinate Geometry
The mid-point of the line segment joining the points P( $- 4$ , 5) and Q(4, 6) 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 $= (\frac{- 4 + 4}{2}, \frac{5 + 6}{2}) = (0, \frac{11}{2})$ .
2. The x-coordinate is 0 and the y-coordinate is not 0, so the point lies on the y-axis. This is option (B).
  1 mark
Answer: $(0, 11/2)$ lies on the y-axis, option (B)
Where the marks are
- Finds the mid-point $(0, 11/2)$ and identifies the y-axis, option (B) (1 mark)
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Q4 · 1 mark · MCQIntroduction to Trigonometry
If $θ$ is an acute angle and $7 + 4 sin θ = 9$ , then the value of $θ$ is :
1. (A) $9 0^{\circ}$
2. (B) $3 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $6 0^{\circ}$
Try it first, then show our solutionSolution
1. From $7 + 4 sin θ = 9$ we get $4 sin θ = 2$ , so $sin θ = \frac{1}{2}$ .
2. $sin 3 0^{\circ} = \frac{1}{2}$ and $θ$ is acute, so $θ = 3 0^{\circ}$ . This is option (B).
  1 mark
Answer: θ = 30°, option (B)
Where the marks are
- Gets $sin θ = 1/2$ and hence $θ = 3 0^{\circ}$ , option (B) (1 mark)
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Q5 · 1 mark · MCQIntroduction to Trigonometry
The value of $tan^{2} θ - (\frac{1}{c o s θ} \times sec θ)$ is :
1. (A)1
2. (B)0
3. (C) $- 1$
4. (D)2
Try it first, then show our solutionSolution
1. Since $\frac{1}{cos θ} = sec θ$ , the bracket equals $sec θ \times sec θ = sec^{2} θ$ .
2. So the expression is $tan^{2} θ - sec^{2} θ = - (sec^{2} θ - tan^{2} θ) = - 1$ , using $sec^{2} θ - tan^{2} θ = 1$ . This is option (C).
  1 mark
Answer: −1, option (C)
Where the marks are
- Writes the bracket as $sec^{2} θ$ and uses $sec^{2} θ - tan^{2} θ = 1$ to get $- 1$ , option (C) (1 mark)
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Q6 · 1 mark · MCQReal Numbers
If HCF(98, 28) = m and LCM(98, 28) = n, then the value of $n - 7 m$ is :
1. (A)0
2. (B)28
3. (C)98
4. (D)198
Try it first, then show our solutionSolution
1. Prime factors: $98 = 2 \times 7^{2}$ and $28 = 2^{2} \times 7$ .
2. HCF $m = 2 \times 7 = 14$ . LCM $n = 2^{2} \times 7^{2} = 196$ .
3. $n - 7 m = 196 - 98 = 98$ , which is option (C).
  1 mark
Answer: 98, option (C)
Where the marks are
- Finds HCF $= 14$ and LCM $= 196$ , giving $n - 7 m = 98$ , option (C) (1 mark)
Q7 · 1 mark · MCQCircles
The tangents drawn at the extremities of the diameter of a circle are always :
1. (A)parallel
2. (B)perpendicular
3. (C)equal
4. (D)intersecting
Try it first, then show our solutionSolution
1. A tangent is perpendicular to the radius at the point of contact. So both tangents at the ends of a diameter are perpendicular to that same diameter.
2. Two lines perpendicular to the same line are parallel. So the tangents are parallel, which is option (A).
  1 mark
Answer: Parallel, option (A)
Where the marks are
- Uses tangent ⟂ radius to conclude the tangents are parallel, option (A) (1 mark)
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Q8 · 1 mark · MCQTriangles
In triangles ABC and DEF, $∠ B = ∠ E$ , $∠ F = ∠ C$ and $A B = 3 D E$ . Then, the two triangles are :
1. (A)congruent but not similar
2. (B)congruent as well as similar
3. (C)neither congruent nor similar
4. (D)similar but not congruent
Try it first, then show our solutionSolution
1. $∠ B = ∠ E$ and $∠ C = ∠ F$ , so $△ A B C \sim △ D E F$ [AA similarity].
2. $A B = 3 D E$ , so corresponding sides are not equal (scale factor 3). The triangles are therefore not congruent: they are similar but not congruent, which is option (D).
  1 mark
Answer: Similar but not congruent, option (D)
Where the marks are
- Uses AA similarity and $A B = 3 D E$ (sides unequal) to choose option (D) (1 mark)
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Q9 · 1 mark · MCQReal Numbers
If $(- 1)^{n} + (- 1)^{8} = 0$ , then n is :
1. (A)any positive integer
2. (B)any negative integer
3. (C)any odd number
4. (D)any even number
Try it first, then show our solutionSolution
1. $(- 1)^{8} = 1$ , so the condition becomes $(- 1)^{n} = - 1$ .
2. $(- 1)^{n} = - 1$ only when $n$ is odd. So n is any odd number, which is option (C).
  1 mark
Answer: Any odd number, option (C)
Where the marks are
- Reduces to $(- 1)^{n} = - 1$ and concludes n is odd, option (C) (1 mark)
Q10 · 1 mark · MCQCompetency-basedPolynomials
Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is :
1. (A)3
2. (B)5
3. (C)2
4. (D)4
Figure in the paper: Graph with x'x and y'y axes showing two parabolas: one opening downward (peak on the positive y-axis) and one opening upward (minimum on the negative y-axis). Both cross the x-axis at the same two points, one on each side of the y-axis. See it in the official paper.
Try it first, then show our solutionSolution
1. The zeroes of a polynomial are the x-coordinates of the points where its graph meets the x-axis.
2. Both parabolas cross the x-axis at the same two points, one on each side of the y-axis. So the number of distinct zeroes is 2, which is option (C).
  1 mark
Answer: 2, option (C)
Where the marks are
- Notes both graphs cut the x-axis at the same two points, giving 2 distinct zeroes, option (C) (1 mark)
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Q11 · 1 mark · MCQArithmetic Progressions
If the sum of first m terms of an AP is $2 m^{2} + 3 m$ , then its second term is :
1. (A)10
2. (B)9
3. (C)12
4. (D)4
Try it first, then show our solutionSolution
1. $S_{1} = 2 (1)^{2} + 3 (1) = 5$ and $S_{2} = 2 (2)^{2} + 3 (2) = 14$ .
2. $a_{2} = S_{2} - S_{1} = 14 - 5 = 9$ , which is option (B).
  1 mark
Answer: 9, option (B)
Where the marks are
- Uses $a_{2} = S_{2} - S_{1} = 14 - 5 = 9$ , option (B) (1 mark)
Q12 · 1 mark · MCQStatistics
Mode and Mean of a data are $15 x$ and $18 x$ , respectively. Then the median of the data is :
1. (A) $x$
2. (B) $11 x$
3. (C) $17 x$
4. (D) $34 x$
Try it first, then show our solutionSolution
1. Use the empirical relation: Mode $= 3$ Median $- 2$ Mean.
2. $15 x = 3$ Median $- 2 (18 x) \Rightarrow 3$ Median $= 15 x + 36 x = 51 x$ .
3. Median $= 17 x$ , which is option (C).
  1 mark
Answer: 17x, option (C)
Where the marks are
- Applies Mode = 3 Median − 2 Mean to get Median $= 17 x$ , option (C) (1 mark)
Q13 · 1 mark · MCQProbability
A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is :
1. (A) $\frac{3}{13}$
2. (B) $\frac{2}{13}$
3. (C) $\frac{1}{2}$
4. (D) $\frac{3}{26}$
Try it first, then show our solutionSolution
1. Total number of outcomes = 52. Each suit has 3 face cards (J, Q, K), and there are 2 red suits (hearts and diamonds).
2. Number of red face cards $= 2 \times 3 = 6$ .
3. $P (red face card) = \frac{6}{52} = \frac{3}{26}$ , so the answer is option (D).
  1 mark
Answer: $\frac{3}{26}$ , option (D)
Where the marks are
- Chooses option (D) $\frac{3}{26}$ (1 mark)
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Q14 · 1 mark · MCQReal Numbers
Which of the following is a rational number between $3$ and $5$ ?
1. (A) $1 \cdot 4142387954012 \dots$
2. (B) $2 \cdot 3 \overline{26}$
3. (C) $π$
4. (D) $1 \cdot 857142$
Try it first, then show our solutionSolution
1. $3 \approx 1.732$ and $5 \approx 2.236$ . The required number must be rational and lie between these two values.
2. (A) $1.4142 \dots$ is less than 1.732 and is non-terminating with no repeating pattern. (B) $2.3 \overline{26} = 2.32626 \dots$ is greater than 2.236. (C) $π$ is irrational and is about 3.14.
3. (D) $1.857142$ is a terminating decimal, so it is rational. Also $1.732 < 1.857142 < 2.236$ . The answer is option (D).
  1 mark
Answer: 1.857142, option (D)
Where the marks are
- Chooses option (D) 1.857142 (1 mark)
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Q15 · 1 mark · MCQAreas Related to Circles
If a sector of a circle has an area of $40 π$ sq. units and a central angle of $7 2^{\circ}$ , the radius of the circle is :
1. (A)200 units
2. (B)100 units
3. (C)20 units
4. (D) $102$ units
Try it first, then show our solutionSolution
1. Area of a sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2}$ .
2. $\frac{72}{360} \times π r^{2} = 40 π \Rightarrow \frac{r ^{2}}{5} = 40 \Rightarrow r^{2} = 200$ .
3. $r = 200 = 102$ units, so the answer is option (D).
  1 mark
Answer: $102$ units, option (D)
Where the marks are
- Chooses option (D) $102$ units (1 mark)
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Q16 · 1 mark · MCQCircles
In the given figure, PA is a tangent from an external point P to a circle with centre O. If $∠ P O B = 11 5^{\circ}$ , 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: Circle with centre O. AB is a diameter through O (A at top, B at bottom right). PA is a tangent at A from external point P on the left. PO is drawn to O. Angle POB is marked 115°. See it in the official paper.
Try it first, then show our solutionSolution
1. AOB is a straight line (a diameter), so $∠ A O P = 18 0^{\circ} - 11 5^{\circ} = 6 5^{\circ}$ [linear pair].
2. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
3. In $△ O A P$ : $∠ A P O = 18 0^{\circ} - 9 0^{\circ} - 6 5^{\circ} = 2 5^{\circ}$ , so the answer is option (A).
  1 mark
Answer: $2 5^{\circ}$ , option (A)
Where the marks are
- Chooses option (A) $2 5^{\circ}$ (1 mark)
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Q17 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A kite is flying at a height of 150 m from the ground. It is attached to a string inclined at an angle of $3 0^{\circ}$ to the horizontal. The length of the string is :
1. (A) $1003$ m
2. (B)300 m
3. (C) $1502$ m
4. (D) $1503$ m
Try it first, then show our solutionSolution
1. Let the string length be $L$ . The height (150 m) is the side opposite the $3 0^{\circ}$ angle, and the string is the hypotenuse.
2. $sin 3 0^{\circ} = \frac{150}{L} \Rightarrow \frac{1}{2} = \frac{150}{L}$ .
3. $L = 300$ m, so the answer is option (B).
  1 mark
Answer: 300 m, option (B)
Where the marks are
- Chooses option (B) 300 m (1 mark)
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Q18 · 1 mark · MCQCompetency-basedAreas Related to Circles
A piece of wire 20 cm long is bent into the form of an arc of a circle of radius $\frac{60}{π}$ cm. The angle subtended by the arc at the centre of the circle is :
1. (A) $3 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $9 0^{\circ}$
4. (D) $5 0^{\circ}$
Try it first, then show our solutionSolution
1. Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r$ . So $20 = \frac{θ}{360} \times 2 π \times \frac{60}{π} = \frac{θ}{360} \times 120$ .
2. $\frac{θ}{3} = 20 \Rightarrow θ = 6 0^{\circ}$ , so the answer is option (B).
  1 mark
Answer: $6 0^{\circ}$ , option (B)
Where the marks are
- Uses the arc-length formula to get $θ = 6 0^{\circ}$ , option (B) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A) : The probability of selecting a number at random from the numbers 1 to 20 is 1.
Reason (R) : For any event E, if P(E) = 1, then E is called a sure event.
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. Any number picked from 1 to 20 is certainly a number from 1 to 20. So the event is certain and its probability $= \frac{20}{20} = 1$ . Assertion (A) is true.
2. An event with probability 1 is called a sure (certain) event, so Reason (R) is true. It explains why the probability in (A) is 1. So the answer is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises the sure event: A and R are both true and R explains A, option (A) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedSurface Areas and Volumes
Assertion (A) : If we join two hemispheres of same radius along their bases, then we get a sphere.
Reason (R) : Total Surface Area of a sphere of radius r is $3 π r^{2}$ .
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. Two equal hemispheres joined along their circular bases form a complete sphere. Assertion (A) is true.
2. The surface area of a sphere is $4 π r^{2}$ , not $3 π r^{2}$ ( $3 π r^{2}$ is the TSA of a solid hemisphere). So Reason (R) is false, and the answer is option (C).
  1 mark
Answer: Option (C)
Where the marks are
- States that A is true and R is false (sphere's area is $4 π r^{2}$ ), option (C) (1 mark)
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Section B

Very short answer

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerIntroduction to Trigonometry
If $x cos 6 0^{\circ} + y cos 0^{\circ} + sin 3 0^{\circ} - cot 4 5^{\circ} = 5$ , then find the value of $x + 2 y$ .
Try it first, then show our solutionSolution
1. Use $cos 6 0^{\circ} = \frac{1}{2}$ , $cos 0^{\circ} = 1$ , $sin 3 0^{\circ} = \frac{1}{2}$ and $cot 4 5^{\circ} = 1$ :
  $x (\frac{1}{2}) + y (1) + \frac{1}{2} - 1 = 5$
  1½ marks
2. $\frac{x}{2} + y = \frac{11}{2}$ . Multiplying by 2: $x + 2 y = 11$ .
  ½ mark
Answer: $x + 2 y = 11$
Where the marks are
- Substitutes the correct values of all four trigonometric ratios ( $1 \frac{1}{2}$ marks)
- Simplifies to $x + 2 y = 11$ ( $\frac{1}{2}$ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $\frac{t a n ^{2} 6 0 ^{\circ}}{s i n ^{2} 6 0 ^{\circ} + c o s ^{2} 3 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. Substitute the standard values $tan 6 0^{\circ} = 3$ , $sin 6 0^{\circ} = \frac{3}{2}$ and $cos 3 0^{\circ} = \frac{3}{2}$ :
  $\frac{( 3 ) ^{2}}{( \frac{3}{2} ) ^{2} + ( \frac{3}{2} ) ^{2}} = \frac{3}{\frac{3}{4} + \frac{3}{4}} = \frac{3}{\frac{3}{2}}$
  1½ marks
2. $= 2$
  ½ mark
Answer: 2
Where the marks are
- Substitutes correct values of $tan 6 0^{\circ}$ , $sin 6 0^{\circ}$ , $cos 3 0^{\circ}$ (1½ marks)
- Simplifies to 2 (½ mark)
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Q22 · 2 marks · Very short answerPolynomials
Find the zeroes of the polynomial $p (x) = x^{2} + \frac{4}{3} x - \frac{4}{3}$ .
Try it first, then show our solutionSolution
1. $p (x) = x^{2} + \frac{4}{3} x - \frac{4}{3} = \frac{1}{3} (3 x^{2} + 4 x - 4)$
  ½ mark
2. Split the middle term: $= \frac{1}{3} (3 x^{2} + 6 x - 2 x - 4)$
  ½ mark
3. $= \frac{1}{3} [3 x (x + 2) - 2 (x + 2)] = \frac{1}{3} (3 x - 2) (x + 2)$
  ½ mark
4. $p (x) = 0 \Rightarrow x = \frac{2}{3}$ or $x = - 2$ . The zeroes are $\frac{2}{3}$ and $- 2$ .
  ½ mark
Answer: $\frac{2}{3}$ and $- 2$
Where the marks are
- Takes $\frac{1}{3}$ common to get $3 x^{2} + 4 x - 4$ (½ mark)
- Splits the middle term correctly (½ mark)
- Factorises as $(3 x - 2) (x + 2)$ (½ mark)
- States zeroes $\frac{2}{3}, - 2$ (½ mark)
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Q23 · 2 marks · Very short answerCoordinate Geometry
The coordinates of the centre of a circle are $(2 a, a - 7)$ . Find the value(s) of 'a' if the circle passes through the point $(11, - 9)$ and has diameter $102$ units.
Try it first, then show our solutionSolution
1. Diameter $= 102$ , so radius $= 52$ units and radius $^{2} = 50$ .
  ½ mark
2. The distance from the centre $(2 a, a - 7)$ to the point $(11, - 9)$ equals the radius. By the distance formula:
  $(2 a - 11)^{2} + (a - 7 + 9)^{2} = 50$ , i.e. $(2 a - 11)^{2} + (a + 2)^{2} = 50$
  ½ mark
3. $4 a^{2} - 44 a + 121 + a^{2} + 4 a + 4 = 50 \Rightarrow 5 a^{2} - 40 a + 75 = 0 \Rightarrow a^{2} - 8 a + 15 = 0$
  ½ mark
4. $(a - 5) (a - 3) = 0 \Rightarrow a = 5$ or $a = 3$
  ½ mark
Answer: a = 5 or a = 3
Where the marks are
- Finds radius $52$ units (½ mark)
- Applies the distance formula equated to the radius (½ mark)
- Obtains $a^{2} - 8 a + 15 = 0$ (½ mark)
- Gets $a = 5, 3$ (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerTriangles
If $△ A B C \sim △ P QR$ in which $A B = 6$ cm, $B C = 4$ cm, $A C = 8$ cm and $P R = 6$ cm, then find the length of $(P Q + QR)$ .
Try it first, then show our solutionSolution
1. Since $△ A B C \sim △ P QR$ , corresponding sides are proportional:
  $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R} \Rightarrow \frac{6}{P Q} = \frac{4}{QR} = \frac{8}{6}$
  ½ mark
2. $P Q = \frac{6 \times 6}{8} = \frac{9}{2} = 4.5$ cm
  ½ mark
3. $QR = \frac{4 \times 6}{8} = 3$ cm
  ½ mark
4. $P Q + QR = 4.5 + 3 = 7.5$ cm
  ½ mark
Answer: PQ + QR = 7.5 cm
Where the marks are
- Writes the ratio of corresponding sides (½ mark)
- PQ = 4.5 cm (½ mark)
- QR = 3 cm (½ mark)
- PQ + QR = 7.5 cm (½ mark)
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OR
Q24(b) · 2 marks · Very short answerTriangles
In the given figure, $\frac{QR}{QS} = \frac{QT}{P R}$ and $∠1 = ∠2$ , show that $△ P QS \sim △ T QR$ .
Figure in the paper: Points Q, S, R lie on a horizontal line in that order. P is above, joined to Q, S and R. T is above P, with T joined to Q (line QT passes through P) and to R. Angle 1 is marked at Q (between QR and QP) and angle 2 is marked at R (angle PRS/between RQ and RP). See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ P QR$ , $∠1 = ∠2$ , i.e. $∠ P QR = ∠ P R Q$ . Therefore $P R = P Q$ [sides opposite equal angles are equal].
  ½ mark
2. Given $\frac{QR}{QS} = \frac{QT}{P R}$ . Replacing PR by PQ gives $\frac{QR}{QS} = \frac{QT}{P Q}$ , i.e. $\frac{P Q}{T Q} = \frac{QS}{QR}$ .
  ½ mark
3. In $△ P QS$ and $△ T QR$ : $∠ P QS = ∠ T QR = ∠1$ (common angle at Q).
  ½ mark
4. The included angle is equal and the sides containing it are proportional, so $△ P QS \sim △ T QR$ [SAS similarity criterion].
  ½ mark
Answer: $△ P QS \sim △ T QR$ (SAS similarity)
Where the marks are
- Deduces PR = PQ from $∠1 = ∠2$ (½ mark)
- Rewrites the ratio as $\frac{QR}{QS} = \frac{QT}{P Q}$ (½ mark)
- Identifies the common angle at Q (½ mark)
- Concludes similarity by SAS (½ mark)
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Q25 · 2 marks · Very short answerCircles
A person is standing at P outside a circular ground at a distance of 26 m from the centre of the ground. He found that his distances from the points A and B on the ground are 10 m (PA and PB are tangents to the circle). Find the radius of the circular ground.
Figure in the paper: Circle with centre O. External point P on the left, PO = 26 m drawn. Tangents PA and PB (each 10 m) touch the circle at A (top) and B (bottom). OA and OB are joined. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  ½ mark
2. In right $△ O A P$ , by Pythagoras theorem: $O P^{2} = O A^{2} + P A^{2} \Rightarrow 2 6^{2} = O A^{2} + 1 0^{2}$
  1 mark
3. $O A^{2} = 676 - 100 = 576 \Rightarrow O A = 24$ . Radius of the ground $= 24$ m.
  ½ mark
Answer: Radius = 24 m
Where the marks are
- States $∠ O A P = 9 0^{\circ}$ (½ mark)
- Applies Pythagoras: $2 6^{2} = O A^{2} + 1 0^{2}$ (1 mark)
- Radius = 24 m (½ mark)
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Section C

Short answer

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerCircles
In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $∠ B A C + ∠ A C D = 9 0^{\circ}$ .
Figure in the paper: Circle with centre O. Line BCD is tangent at C (B left, D right). Line from B passes through P (on circle) and O to A on the circle. OC (dashed) and AC are joined. Angle BAC is marked at A. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ O A C$ , $O A = O C$ (radii of the same circle). Therefore $∠ O C A = ∠ O A C$ [angles opposite equal sides].
  1 mark
2. $O C ⊥ B D$ [tangent ⟂ radius at the point of contact], so $∠ O C D = 9 0^{\circ}$ .
  1 mark
3. $∠ O C D = ∠ O C A + ∠ A C D$ , so $∠ O C A + ∠ A C D = 9 0^{\circ}$ .
  ½ mark
4. Using $∠ O C A = ∠ O A C$ : $∠ O A C + ∠ A C D = 9 0^{\circ}$ . Since O lies on AB, $∠ O A C = ∠ B A C$ . Hence $∠ B A C + ∠ A C D = 9 0^{\circ}$ . Proved.
  ½ mark
Answer: $∠ B A C + ∠ A C D = 9 0^{\circ}$ (proved)
Where the marks are
- Uses OA = OC to get $∠ O C A = ∠ O A C$ (1 mark)
- States $∠ O C D = 9 0^{\circ}$ (tangent ⟂ radius) (1 mark)
- Writes $∠ O C A + ∠ A C D = 9 0^{\circ}$ (½ mark)
- Substitutes to conclude $∠ B A C + ∠ A C D = 9 0^{\circ}$ (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerCircles
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Try it first, then show our solutionSolution
1. Figure: Draw quadrilateral ABCD circumscribing a circle with centre O. The circle touches AB at P, BC at Q, CD at R and DA at S. Join OA, OB, OC, OD and OP, OQ, OR, OS. Name the eight angles at O: $∠1 = ∠ A O P$ , $∠2 = ∠ A O S$ , $∠3 = ∠ D O S$ , $∠4 = ∠ D O R$ , $∠5 = ∠ C O R$ , $∠6 = ∠ C O Q$ , $∠7 = ∠ B O Q$ , $∠8 = ∠ B O P$ .
  ½ mark
2. In $△ O A P$ and $△ O A S$ :
  $O P = O S$ [radii of the same circle]
  $A P = A S$ [tangents from an external point are equal]
  $O A = O A$ [common side]
  So $△ O A P ≅ △ O A S$ [SSS], and hence $∠1 = ∠2$ [CPCT].
  1 mark
3. In the same way, using the vertices D, C and B, we get $∠3 = ∠4$ , $∠5 = ∠6$ and $∠7 = ∠8$ .
  ½ mark
4. The angles round the point O add up to $36 0^{\circ}$ :
  $∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 36 0^{\circ}$
  Using the equal pairs, this becomes $2 (∠1 + ∠8 + ∠4 + ∠5) = 36 0^{\circ}$ , so $∠1 + ∠8 + ∠4 + ∠5 = 18 0^{\circ}$ .
  ½ mark
5. Now $∠1 + ∠8 = ∠ A O B$ and $∠4 + ∠5 = ∠ C O D$ .
  So $∠ A O B + ∠ C O D = 18 0^{\circ}$ . Similarly, $∠ B O C + ∠ A O D = 18 0^{\circ}$ .
  Hence opposite sides subtend supplementary angles at the centre. Proved.
  ½ mark
Answer: $∠ A O B + ∠ C O D = 18 0^{\circ}$ and $∠ B O C + ∠ A O D = 18 0^{\circ}$ (proved)
Where the marks are
- Draws a correct labelled figure with the points of contact and the lines from O (½ mark)
- Proves $△ O A P ≅ △ O A S$ and gets $∠1 = ∠2$ (1 mark)
- States the other three pairs of equal angles (½ mark)
- Uses the angle sum of $36 0^{\circ}$ at O to get $2 (∠1 + ∠4 + ∠5 + ∠8) = 36 0^{\circ}$ (½ mark)
- Concludes $∠ A O B + ∠ C O D = 18 0^{\circ}$ , and similarly for the other pair (½ mark)
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Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + sec θ cosec θ$
Try it first, then show our solutionSolution
1. Write everything in sin and cos:
  LHS $= \frac{\frac{s i n θ}{c o s θ}}{1 - \frac{c o s θ}{s i n θ}} + \frac{\frac{c o s θ}{s i n θ}}{1 - \frac{s i n θ}{c o s θ}}$
  ½ mark
2. Simplify each fraction:
  $= \frac{sin ^{2} θ}{cos θ ( sin θ - cos θ )} - \frac{cos ^{2} θ}{sin θ ( sin θ - cos θ )}$
  1 mark
3. Take a common denominator:
  $= \frac{1}{sin θ - cos θ} [\frac{sin ^{3} θ - cos ^{3} θ}{sin θ cos θ}]$
  ½ mark
4. Factorise the difference of cubes:
  $= \frac{( sin θ - cos θ ) ( sin ^{2} θ + sin θ cos θ + cos ^{2} θ )}{( sin θ - cos θ ) sin θ cos θ}$
  ½ mark
5. Use $sin^{2} θ + cos^{2} θ = 1$ :
  $= \frac{1 + sin θ cos θ}{sin θ cos θ} = \frac{1}{sin θ cos θ} + 1 = 1 + sec θ cosec θ =$ RHS
  ½ mark
Answer: LHS = 1 + sec θ cosec θ = RHS (proved)
Where the marks are
- Converts tan and cot into sin and cos (½ mark)
- Simplifies to the two fractions with the factor (sin θ − cos θ) (1 mark)
- Combines into (sin³θ − cos³θ)/(sin θ cos θ) (½ mark)
- Factorises using a³ − b³ (½ mark)
- Uses sin²θ + cos²θ = 1 to reach the RHS (½ mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{2 s i n ^{2} A - 1}$
Try it first, then show our solutionSolution
1. Combine the two fractions over a common denominator:
  LHS $= \frac{( sin A + cos A ) ^{2} + ( sin A - cos A ) ^{2}}{( sin A - cos A ) ( sin A + cos A )}$
  1 mark
2. Expand the numerator and the denominator:
  $= \frac{sin ^{2} A + cos ^{2} A + 2 sin A cos A + sin ^{2} A + cos ^{2} A - 2 sin A cos A}{sin ^{2} A - cos ^{2} A}$
  1 mark
3. Use $sin^{2} A + cos^{2} A = 1$ and $cos^{2} A = 1 - sin^{2} A$ :
  $= \frac{1 + 1}{sin ^{2} A - ( 1 - sin ^{2} A )}$
  ½ mark
4. $= \frac{2}{2 sin ^{2} A - 1} =$ RHS
  ½ mark
Answer: LHS = 2/(2sin²A − 1) = RHS (proved)
Where the marks are
- Takes the LCM correctly (1 mark)
- Expands the squares and uses (a − b)(a + b) for the denominator (1 mark)
- Applies the identity sin²A + cos²A = 1 (½ mark)
- Reaches 2/(2sin²A − 1) (½ mark)
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Q28 · 3 marks · Short answerCoordinate Geometry
Find the ratio in which the y-axis divides the line segment joining the points $(5, - 6)$ and $(- 1, - 4)$ . Also find the point of intersection.
Try it first, then show our solutionSolution
1. Let $A (5, - 6)$ and $B (- 1, - 4)$ . Any point on the y-axis has x-coordinate 0, so let the y-axis cut $A B$ at $P (0, y)$ . Let $P$ divide $A B$ in the ratio $k : 1$ .
  ½ mark
2. Section formula for the x-coordinate: $x = \frac{k ( - 1 ) + 1 ( 5 )}{k + 1}$ . Since $P$ is on the y-axis, $\frac{- k + 5}{k + 1} = 0$ . So $- k + 5 = 0$ , which gives $k = 5$ .
  1 mark
3. Hence the y-axis divides the segment in the ratio $5 : 1$ (internally).
  ½ mark
4. Section formula for the y-coordinate: $y = \frac{5 ( - 4 ) + 1 ( - 6 )}{5 + 1} = \frac{- 20 - 6}{6} = \frac{- 26}{6} = - \frac{13}{3}$ .
  ½ mark
5. Point of intersection is $P (0, - \frac{13}{3})$ .
  ½ mark
Answer: Ratio 5 : 1; point of intersection $(0, - \frac{13}{3})$
Where the marks are
- Takes the point as $(0, y)$ and the ratio as $k : 1$ (½ mark)
- Sets the x-coordinate from the section formula equal to 0 and gets $k = 5$ (1 mark)
- States the ratio as 5 : 1 (½ mark)
- Finds $y = - \frac{13}{3}$ (½ mark)
- States the point $(0, - \frac{13}{3})$ (½ mark)
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Q29 · 3 marks · Short answerReal Numbers
Prove that $\frac{1}{5}$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $\frac{1}{5}$ is rational. Then $\frac{1}{5} = \frac{p}{q}$ , where $p, q$ are positive integers, $q \neq = 0$ , and $p, q$ are co-prime.
  ½ mark
2. Then $q = 5 p$ . Squaring gives $q^{2} = 5 p^{2}$ . So 5 divides $q^{2}$ . Since 5 is prime, 5 divides $q$ (if a prime divides $a^{2}$ , it divides $a$ ). ...(i)
  1 mark
3. Let $q = 5 a$ for some integer $a$ . Then $25 a^{2} = 5 p^{2}$ , so $p^{2} = 5 a^{2}$ . So 5 divides $p^{2}$ , and therefore 5 divides $p$ . ...(ii)
  1 mark
4. From (i) and (ii), 5 is a common factor of $p$ and $q$ . This contradicts the fact that $p$ and $q$ are co-prime. So our assumption is wrong, and $\frac{1}{5}$ is an irrational number.
  ½ mark
Answer: $\frac{1}{5}$ is irrational (proved by contradiction).
Where the marks are
- Assumes $\frac{1}{5} = \frac{p}{q}$ with $p, q$ co-prime and $q \neq = 0$ (½ mark)
- Gets $q^{2} = 5 p^{2}$ and concludes 5 divides $q$ (1 mark)
- Substitutes $q = 5 a$ and concludes 5 divides $p$ (1 mark)
- States the contradiction with co-primeness and concludes irrationality (½ mark)
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Q30 · 3 marks · Short answerCompetency-basedSurface Areas and Volumes
A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains $\frac{1408}{21} m^{3}$ of air, find the height of the cylindrical part. (Use $π = \frac{22}{7}$ ).
Try it first, then show our solutionSolution
1. Let $r$ be the radius of the hemisphere (which is also the radius of the cylinder) and $h$ the height of the cylinder. Given $r = \frac{h}{2}$ , so $h = 2 r$ .
  ½ mark
2. Volume of air = volume of cylinder + volume of hemisphere $= π r^{2} h + \frac{2}{3} π r^{3}$ . So $\frac{1408}{21} = π r^{2} (2 r) + \frac{2}{3} π r^{3}$ .
  ½ mark
3. Simplify: $2 π r^{3} + \frac{2}{3} π r^{3} = \frac{8}{3} π r^{3}$ . So $\frac{8}{3} \times \frac{22}{7} \times r^{3} = \frac{1408}{21}$ , which gives $\frac{176}{21} r^{3} = \frac{1408}{21}$ . So $r^{3} = 8$ .
  1 mark
4. Therefore $r = 2$ m.
  ½ mark
5. Height of the cylindrical part $h = 2 r = 4$ m.
  ½ mark
Answer: Height of the cylindrical part = 4 m
Where the marks are
- Writes $h = 2 r$ (½ mark)
- Writes the total volume as cylinder + hemisphere and equates it to $\frac{1408}{21}$ (½ mark)
- Simplifies to get $r^{3} = 8$ (1 mark)
- Gets $r = 2$ m (½ mark)
- Gets $h = 4$ m (½ mark)
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Q31 · 3 marks · Short answerProbability
Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.
Try it first, then show our solutionSolution
1. When two dice are thrown together, the total number of possible outcomes is $6 \times 6 = 36$ .
  1 mark
2. Favourable outcomes (difference of the numbers is 2): $(1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)$ . Number of favourable outcomes = 8.
  1 mark
3. $P (difference is 2) = \frac{8}{36} = \frac{2}{9}$ .
  1 mark
Answer: $\frac{2}{9}$
Where the marks are
- Total outcomes = 36 (1 mark)
- Lists the 8 favourable outcomes (1 mark)
- Probability $= \frac{8}{36} = \frac{2}{9}$ (1 mark)
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Section D

Long answer

Q32 · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables
Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received ₹ 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received ₹ 20 more as annual interest. How much money did he invest in each scheme ?
Try it first, then show our solutionSolution
1. Let Vijay invest ₹ $x$ in scheme A (8% p.a.) and ₹ $y$ in scheme B (9% p.a.). Total interest: $\frac{8 x}{100} + \frac{9 y}{100} = 1860$ , so $8 x + 9 y = 186000$ ...(i)
  1½ marks
2. If the amounts are interchanged, the interest is $1860 + 20 = 1880$ . So $\frac{9 x}{100} + \frac{8 y}{100} = 1880$ , which gives $9 x + 8 y = 188000$ ...(ii)
  1½ marks
3. Adding (i) and (ii): $17 x + 17 y = 374000$ , so $x + y = 22000$ ...(iii). Subtracting (i) from (ii): $x - y = 2000$ ...(iv). Adding (iii) and (iv): $2 x = 24000$ , so $x = 12000$ .
  1 mark
4. From (iii): $y = 22000 - 12000 = 10000$ . Check: $960 + 900 = 1860$ and $1080 + 800 = 1880$ . Hence Vijay invested ₹ 12000 in scheme A and ₹ 10000 in scheme B.
  1 mark
Answer: Scheme A: ₹ 12,000; Scheme B: ₹ 10,000
Where the marks are
- Forms the first equation $8 x + 9 y = 186000$ (1½ marks)
- Forms the second equation $9 x + 8 y = 188000$ (1½ marks)
- Solves to get $x = 12000$ (1 mark)
- Gets $y = 10000$ and states both investments (1 mark)
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Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerTriangles
The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that $D F \times E F = F B \times F A$ .
Try it first, then show our solutionSolution
1. Figure: Draw parallelogram $A B C D$ with $A B$ at the bottom and $D C$ at the top. Take a point $E$ on side $B C$ and join $A E$ . Draw diagonal $B D$ so that it cuts $A E$ at $F$ .
  1 mark
2. Since $A B C D$ is a parallelogram, $A D ∥ B C$ , so $A D ∥ B E$ (because $E$ lies on $B C$ ).
3. In $△ A D F$ and $△ E B F$ :
  $∠ D F A = ∠ B F E$ [vertically opposite angles]
  $∠ A D F = ∠ E B F$ [alternate interior angles, $A D ∥ B E$ , transversal $B D$ ]
  So $△ A D F \sim △ E B F$ [AA similarity criterion].
  2 marks
4. Corresponding sides of similar triangles are proportional:
  $\frac{D F}{B F} = \frac{A F}{E F} = \frac{A D}{E B}$
  1 mark
5. From $\frac{D F}{F B} = \frac{F A}{E F}$ , cross-multiply to get $D F \times E F = F B \times F A$ . Hence proved.
  1 mark
Answer: $△ A D F \sim △ E B F$ (AA), so $D F / F B = F A / E F$ , i.e. $D F \times E F = F B \times F A$ . Proved.
Where the marks are
- Correct labelled figure of the parallelogram with E on BC, AE and BD meeting at F (1 mark)
- Proves $△ A D F \sim △ E B F$ using vertically opposite angles and alternate angles (AA), with reasons (2 marks)
- Writes the ratio of corresponding sides $D F / F B = F A / E F$ (1 mark)
- Cross-multiplies to reach $D F \times E F = F B \times F A$ (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerTriangles
In $△ A B C$ , if $A D ⊥ B C$ and $A D^{2} = B D \times D C$ , then prove that $∠ B A C = 9 0^{\circ}$ .
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ with $B C$ as the base. Draw $A D ⊥ B C$ with foot $D$ on $B C$ , so $∠ A D B = ∠ A D C = 9 0^{\circ}$ .
  1 mark
2. Given $A D^{2} = B D \times D C$ . Dividing both sides by $A D \times D C$ gives
  $\frac{B D}{A D} = \frac{A D}{D C} .$
  1 mark
3. In $△ B D A$ and $△ A D C$ :
  $\frac{B D}{A D} = \frac{A D}{D C}$ (shown above)
  $∠ B D A = ∠ A D C = 9 0^{\circ}$ (included angles)
  So $△ B D A \sim △ A D C$ [SAS similarity criterion], with $B \leftrightarrow A$ , $D \leftrightarrow D$ , $A \leftrightarrow C$ .
  1 mark
4. Corresponding angles of similar triangles are equal:
  $∠ D B A = ∠ D A C and ∠ B A D = ∠ D C A .$
  1 mark
5. Adding these: $∠ D B A + ∠ D C A = ∠ D A C + ∠ B A D = ∠ B A C$ .
  So $∠ B + ∠ C = ∠ B A C$ .
  In $△ A B C$ , $∠ B + ∠ C + ∠ B A C = 18 0^{\circ}$ [angle sum property].
  Substituting, $2∠ B A C = 18 0^{\circ}$ , so $∠ B A C = 9 0^{\circ}$ . Hence proved.
  1 mark
Answer: $△ B D A \sim △ A D C$ (SAS), so $∠ B + ∠ C = ∠ B A C$ , which gives $∠ B A C = 9 0^{\circ}$ . Proved.
Where the marks are
- Correct figure with $A D ⊥ B C$ (1 mark)
- Rewrites $A D^{2} = B D \times D C$ as $B D / A D = A D / D C$ (1 mark)
- Proves $△ B D A \sim △ A D C$ by SAS using the right angles at D (1 mark)
- States the pairs of equal corresponding angles (1 mark)
- Adds the angles and uses the angle sum property to get $∠ B A C = 9 0^{\circ}$ (1 mark)
Open Q33(b) on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerQuadratic Equations
The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle.
Try it first, then show our solutionSolution
1. Let the other two sides be $x$ cm and $y$ cm.
  From the perimeter: $x + y + 25 = 60$ .
  1 mark
2. So $y = 35 - x$ .
  ½ mark
3. By Pythagoras theorem, $x^{2} + y^{2} = 2 5^{2}$ .
  Substituting: $x^{2} + (35 - x)^{2} = 625$ .
  1 mark
4. Expand: $x^{2} + 1225 - 70 x + x^{2} = 625$ , so $2 x^{2} - 70 x + 600 = 0$ .
  Dividing by 2: $x^{2} - 35 x + 300 = 0$ .
  1 mark
5. Factorise: $x^{2} - 20 x - 15 x + 300 = 0$ , so $(x - 20) (x - 15) = 0$ .
  Therefore $x = 20$ or $x = 15$ .
  1 mark
6. If $x = 20$ , then $y = 15$ . If $x = 15$ , then $y = 20$ .
  Hence the other two sides are 15 cm and 20 cm.
  Check: $15 + 20 + 25 = 60$ and $1 5^{2} + 2 0^{2} = 625 = 2 5^{2}$ .
  ½ mark
Answer: The other two sides are 15 cm and 20 cm.
Where the marks are
- Writes the perimeter equation $x + y + 25 = 60$ (1 mark)
- Expresses $y = 35 - x$ (½ mark)
- Applies Pythagoras: $x^{2} + (35 - x)^{2} = 625$ (1 mark)
- Simplifies to $x^{2} - 35 x + 300 = 0$ (1 mark)
- Solves to get $x = 20$ or $15$ (1 mark)
- States the sides as 15 cm and 20 cm (½ mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerCompetency-basedQuadratic Equations
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
Try it first, then show our solutionSolution
1. Let the uniform speed of the train be $x$ km/h.
  The reduced speed is $(x - 8)$ km/h.
  ½ mark
2. Time $= \frac{distance}{speed}$ . The slower journey takes 3 hours more, so
  $\frac{480}{x - 8} - \frac{480}{x} = 3.$
  1½ marks
3. Take the LCM: $480 [\frac{x - ( x - 8 )}{x ( x - 8 )}] = 3$ .
  So $\frac{3840}{x ( x - 8 )} = 3$ , which gives $x (x - 8) = 1280$ .
  Hence $x^{2} - 8 x - 1280 = 0$ .
  1 mark
4. Factorise: $x^{2} - 40 x + 32 x - 1280 = 0$ , so $(x - 40) (x + 32) = 0$ .
  Therefore $x = 40$ or $x = - 32$ .
  1 mark
5. Speed cannot be negative, so we reject $x = - 32$ .
  The speed of the train is 40 km/h.
  Check: $480/40 = 12$ h and $480/32 = 15$ h, a difference of 3 h.
  1 mark
Answer: Speed of the train = 40 km/h.
Where the marks are
- Takes the speed as $x$ and the reduced speed as $x - 8$ (½ mark)
- Forms the equation $\frac{480}{x - 8} - \frac{480}{x} = 3$ (1½ marks)
- Simplifies to $x^{2} - 8 x - 1280 = 0$ (1 mark)
- Factorises to get $x = 40$ or $- 32$ (1 mark)
- Rejects the negative root and states the speed as 40 km/h (1 mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerStatistics
Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode.
Daily Allowance Number of Children
11 – 13 7
13 – 15 6
15 – 17 9
17 – 19 13
19 – 21 f
21 – 23 5
23 – 25 4
Try it first, then show our solutionSolution
1. Make the table using class marks $x_{i}$ :
  Class $x_{i}$ $f_{i}$ $f_{i} x_{i}$
  11–13 12 7 84
  13–15 14 6 84
  15–17 16 9 144
  17–19 18 13 234
  19–21 20 f 20f
  21–23 22 5 110
  23–25 24 4 96
  Total 44 + f 752 + 20f
  1½ marks
2. Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , so
  $18 = \frac{752 + 20 f}{44 + f}$
  1 mark
3. Cross-multiply: $792 + 18 f = 752 + 20 f$
  So $2 f = 40$ , which gives $f = 20$ .
  ½ mark
4. The highest frequency is now 20, so the modal class is 19–21.
  Here $l = 19$ , $f_{1} = 20$ , $f_{0} = 13$ , $f_{2} = 5$ , and the class width is $h = 2$ .
  ½ mark
5. Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
  $= 19 + \frac{20 - 13}{40 - 13 - 5} \times 2 = 19 + \frac{14}{22}$
  1 mark
6. $= 19 + 0.636 \approx 19.64$
  So the mode is $19 \frac{7}{11} \approx 19.64$ .
  ½ mark
Answer: f = 20; Mode ≈ 19.64 (= 19 7/11)
Where the marks are
- Writes a correct table of $x_{i}$ and $f_{i} x_{i}$ with totals $44 + f$ and $752 + 20 f$ (1½ marks)
- Sets up $\frac{752 + 20 f}{44 + f} = 18$ (1 mark)
- Solves to get $f = 20$ (½ mark)
- Identifies the modal class 19–21 (½ mark)
- Substitutes correctly into the mode formula with $h = 2$ (1 mark)
- Gets mode ≈ 19.64 (½ mark)
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Daily Allowance	Number of Children
11 – 13	7
13 – 15	6
15 – 17	9
17 – 19	13
19 – 21	f
21 – 23	5
23 – 25	4

Class	$x_{i}$	$f_{i}$	$f_{i} x_{i}$
11–13	12	7	84
13–15	14	6	84
15–17	16	9	144
17–19	18	13	234
19–21	20	f	20f
21–23	22	5	110
23–25	24	4	96
Total		44 + f	752 + 20f

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Case Study – 1
A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.
Based on the information given above, answer the following questions :
1. (i)
  Write the fourth, fifth and sixth term of the Arithmetic Progression so formed.
  1 mark
2. (ii)
  Determine the distance of the $8^{t h}$ round.
  1 mark
3. (iii)(a)
  Find the total distance run after completing all 10 rounds.
  2 marks
4. OR
  (iii)(b)
  If a runner completes only the first 6 rounds, what is the total distance run by the runner ?
  2 marks
Figure in the paper: Illustration of ten coloured runner silhouettes running around a circular track. See it in the official paper.
Try it first, then show our solutionSolution
1. The round distances form an AP: 300, 350, 400, …, with $a = 300$ and $d = 50$ .
  (i) $a_{4} = 400 + 50 = 450$ m, $a_{5} = 500$ m, $a_{6} = 550$ m.
  1 mark
2. (ii) $a_{n} = a + (n - 1) d$ , so $a_{8} = 300 + 7 \times 50$
  ½ mark
3. (ii) $a_{8} = 300 + 350 = 650$ m
  ½ mark
4. (iii)(a) $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , so $S_{10} = \frac{10}{2} [2 (300) + 9 (50)] = 5 (600 + 450)$
  1 mark
5. (iii)(a) $S_{10} = 5 \times 1050 = 5250$ m
  1 mark
6. (iii)(b) $S_{6} = \frac{6}{2} [2 (300) + 5 (50)] = 3 (600 + 250)$
  1 mark
7. (iii)(b) $S_{6} = 3 \times 850 = 2550$ m
  1 mark
Answer: (i) 450 m, 500 m, 550 m; (ii) 650 m; (iii)(a) 5250 m, (iii)(b) 2550 m
Where the marks are
- (i) Writes the 4th, 5th and 6th terms as 450, 500 and 550 (1 mark)
- (ii) Uses $a_{8} = 300 + 7 \times 50$ (½ mark)
- (ii) Gets 650 m (½ mark)
- (iii)(a) Substitutes correctly into $S_{10}$ (1 mark)
- (iii)(a) Gets 5250 m (1 mark)
- (iii)(b) Substitutes correctly into $S_{6}$ (1 mark)
- (iii)(b) Gets 2550 m (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
Case Study – 2
A brooch is a decorative piece often worn on clothing like jackets, blouses or dresses to add elegance. Made from precious metals and decorated with gemstones, brooches come in many shapes and designs.
One such brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure.
Based on the above given information, answer the following questions :
1. (i)
  Find the central angle of each sector.
  1 mark
2. (ii)
  Find the length of the arc ACB.
  1 mark
3. (iii)(a)
  Find the area of each sector of the brooch.
  2 marks
4. OR
  (iii)(b)
  Find the total length of the silver wire used.
  2 marks
Figure in the paper: Photo of 25 decorative brooches, and a circle divided by 5 diameters into 10 equal sectors with decorative swirls. Points A, C and B are marked on the circumference at the upper right, with C between A and B; arc ACB spans one sector. See it in the official paper.
Try it first, then show our solutionSolution
1. Radius $r = \frac{35}{2}$ mm. The 5 diameters divide the circle into 10 equal sectors.
2. (i) Central angle of each sector $= \frac{36 0 ^{\circ}}{10} = 3 6^{\circ}$ .
  1 mark
3. (ii) Arc ACB is the arc of one sector, so it is $\frac{1}{10}$ of the circumference:
  $\frac{1}{10} \times 2 \times \frac{22}{7} \times \frac{35}{2} = \frac{110}{10} = 11$ mm.
  1 mark
4. (iii)(a) Area of a sector $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}$
  1 mark
5. (iii)(a) $= \frac{1}{10} \times \frac{3850}{4} = \frac{385}{4} = 96.25 mm^{2}$ .
  1 mark
6. (iii)(b) Wire used = circumference + 5 diameters $= 2 \times \frac{22}{7} \times \frac{35}{2} + 5 \times 35$
  1 mark
7. (iii)(b) $= 110 + 175 = 285$ mm.
  1 mark
Answer: (i) 36°; (ii) 11 mm; (iii)(a) 96.25 mm², (iii)(b) 285 mm
Where the marks are
- (i) Central angle $36 0^{\circ} /10 = 3 6^{\circ}$ (1 mark)
- (ii) Arc length $= \frac{1}{10} \times 2 π r = 11$ mm (1 mark)
- (iii)(a) Correct sector-area expression (1 mark)
- (iii)(a) Area $= 385/4 = 96.25$ mm² (1 mark)
- (iii)(b) Expression: circumference + $5 \times 35$ (1 mark)
- (iii)(b) Total wire 285 mm (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Case Study – 3
Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be $6 0^{\circ}$ . Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be $4 5^{\circ}$ .
Based on the above given information, answer the following questions :
1. (i)
  If CD is h metres, find the distance BD in terms of 'h'.
  1 mark
2. (ii)
  Find distance BC in terms of 'h'.
  1 mark
3. (iii)(a)
  Find the height CE of the lighthouse [Use $3 = 1 \cdot 73$ ]
  2 marks
4. OR
  (iii)(b)
  Find distance AE, if AC = 100 m.
  2 marks
Figure in the paper: Photo of a lighthouse on a cliff, and a diagram: vertical lighthouse CE with C at top, E at base; D lies on CE with CD = h. A is on the ground to the left of E, B is directly above A with AB = 40 m, and BD is horizontal to D. Angle of elevation at B to C is 45° and at A to C (line AC) is 60°. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In right $△ B D C$ (right angle at D): $tan 4 5^{\circ} = \frac{C D}{B D} = \frac{h}{B D}$
  ½ mark
2. (i) Since $tan 4 5^{\circ} = 1$ , $B D = h$ m.
  ½ mark
3. (ii) In $△ B D C$ : $sin 4 5^{\circ} = \frac{C D}{B C}$ , so $\frac{1}{2} = \frac{h}{B C}$
  ½ mark
4. (ii) $B C = 2 h$ m
  ½ mark
5. (iii)(a) ABDE is a rectangle, so $A E = B D = h$ and $D E = A B = 40$ m. This gives $C E = h + 40$ .
  In right $△ A E C$ : $tan 6 0^{\circ} = \frac{C E}{A E}$ , so $3 = \frac{h + 40}{h}$
  1 mark
6. (iii)(a) $3 h - h = 40$ , so $h = \frac{40}{3 - 1} = \frac{40 ( 3 + 1 )}{2} = 20 (3 + 1) = 20 \times 2.73 = 54.6$ m
  ½ mark
7. (iii)(a) $C E = 54.6 + 40 = 94.6$ m
  ½ mark
8. (iii)(b) In right $△ A E C$ : $cos 6 0^{\circ} = \frac{A E}{A C}$ , so $\frac{1}{2} = \frac{A E}{100}$
  1 mark
9. (iii)(b) $A E = 50$ m
  1 mark
Answer: (i) BD = h m; (ii) BC = √2 h m; (iii)(a) CE = 94.6 m, (iii)(b) AE = 50 m
Where the marks are
- (i) Uses $tan 4 5^{\circ} = h / B D$ (½ mark)
- (i) Gets $B D = h$ (½ mark)
- (ii) Uses $sin 4 5^{\circ} = h / B C$ (½ mark)
- (ii) Gets $B C = 2 h$ (½ mark)
- (iii)(a) Forms $3 = \frac{h + 40}{h}$ (1 mark)
- (iii)(a) Finds $h = 20 (3 + 1) = 54.6$ m (½ mark)
- (iii)(a) Gets $C E = 94.6$ m (½ mark)
- (iii)(b) Uses $cos 6 0^{\circ} = A E / A C$ (1 mark)
- (iii)(b) Gets $A E = 50$ m (1 mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.