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

CBSE Class 10 Mathematics Standard: 2026 board paper, set 30/2/1

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_2_1_Mathematics Std._cl.10.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; internal choice in 2-mark subparts

Instructions in brief: No overall choice; internal choice given in 2 questions of Section B (22, 24), 2 of Section C (28, 29), 2 of Section D (33, 34) and 3 case studies (36, 37, 38) in the 2-mark subparts. Calculator not allowed; take π = 22/7 unless stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The LCM of 960 and 240 is :
1. (A)960
2. (B)240
3. (C)60
4. (D)15
Try it first, then show our solutionSolution
1. $960 = 4 \times 240$ , so 240 is a factor of 960.
2. Hence LCM $(960, 240) = 960$ . Option (A).
  1 mark
Answer: 960, option (A)
Where the marks are
- Identifies LCM as 960 (1 mark)
Q2 · 1 mark · MCQReal Numbers
The natural number 1 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. A prime has exactly two factors and a composite has more than two; 1 has only one factor.
2. So 1 is neither prime nor composite. Option (D).
  1 mark
Answer: Neither prime nor composite, option (D)
Where the marks are
- Correct option (D) (1 mark)
Q3 · 1 mark · MCQReal Numbers
For any natural number n, $5^{n}$ ends with the digit :
1. (A)0
2. (B)5
3. (C)3
4. (D)2
Try it first, then show our solutionSolution
1. $5^{1} = 5, 5^{2} = 25, 5^{3} = 125, \dots$ Since $5 \times 5$ ends in 5, every power ends in 5.
2. Option (B).
  1 mark
Answer: 5, option (B)
Where the marks are
- Correct option (B) (1 mark)
Q4 · 1 mark · MCQPolynomials
The graph of $y = f (x)$ is given. The number of distinct zeroes of $y = f (x)$ is :
1. (A)0
2. (B)1
3. (C)2
4. (D)3
Figure in the paper: Graph of y = f(x) with axes x'Ox and yOy'. The curve starts bottom-left, crosses the x-axis at a point labelled A on the negative x-side, rises to a local maximum above the origin on the y-axis, falls crossing the positive x-axis, touches the x-axis at a local minimum (tangent) on the positive side, then rises upward to the right. See it in the official paper.
Try it first, then show our solutionSolution
1. Zeroes are the x-values where the graph meets the x-axis. The curve cuts the axis at A and meets it again where it falls and touches it at the minimum.
2. As in the official figure, the curve meets the x-axis at 2 distinct points. Option (C).
  1 mark
Answer: 2, option (C)
Where the marks are
- Correct option (C) (1 mark)
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Q5 · 1 mark · MCQPolynomials
If $α$ and $β$ are two zeroes of a polynomial $f (x) = p x^{2} - 2 x + 3 p$ and $α + β = α β$ , then value of p is :
1. (A) $- \frac{2}{3}$
2. (B) $\frac{2}{3}$
3. (C) $\frac{1}{3}$
4. (D) $- \frac{1}{3}$
Try it first, then show our solutionSolution
1. $α + β = \frac{2}{p}$ , $α β = \frac{3 p}{p} = 3$ .
2. $\frac{2}{p} = 3 \Rightarrow p = \frac{2}{3}$ . Option (B).
  1 mark
Answer: $p = \frac{2}{3}$ , option (B)
Where the marks are
- Equates sum and product to get $p = 2/3$ (1 mark)
Q6 · 1 mark · MCQPair of Linear Equations in Two Variables
If the pair of linear equations : $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ is consistent and dependent, then
1. (A) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$
2. (B) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
3. (C) $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$
4. (D) $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
Try it first, then show our solutionSolution
1. Consistent and dependent means coincident lines (infinitely many solutions), which occurs when all three ratios are equal.
2. Option (D).
  1 mark
Answer: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , option (D)
Where the marks are
- Correct option (D) (1 mark)
Q7 · 1 mark · MCQArithmetic Progressions
Which of the following sequence is not an A.P. ?
1. (A) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$
2. (B) $- 1 \cdot 2, - 3 \cdot 2, - 5 \cdot 2, - 7 \cdot 2, \dots$
3. (C) $2, 8, 18, \dots$
4. (D) $1^{2}, 3^{2}, 5^{2}, 7^{2}, \dots$
Try it first, then show our solutionSolution
1. (A) d = 1/2; (B) d = −2; (C) $2, 22, 32$ , d = $2$ .
2. (D) 1, 9, 25, 49: differences 8, 16, 24 are not equal, so not an A.P. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Identifies (D) as non-A.P. (1 mark)
Q8 · 1 mark · MCQTriangles
In triangles ABC and PQR, $∠ A = ∠ Q$ and $∠ B = ∠ R$ , then AB : AC is equal to :
1. (A)PQ : PR
2. (B)PQ : QR
3. (C)QR : QP
4. (D)PR : QR
Try it first, then show our solutionSolution
1. A↔Q, B↔R, so C↔P; △ABC ~ △QRP (AA).
2. AB ↔ QR and AC ↔ QP, so AB : AC = QR : QP. Option (C).
  1 mark
Answer: QR : QP, option (C)
Where the marks are
- Correct correspondence giving QR : QP (1 mark)
Q9 · 1 mark · MCQCoordinate Geometry
The distance of the point A(4a, 3a) from x-axis is :
1. (A)3a
2. (B) $- 3 a$
3. (C)4a
4. (D) $- 4 a$
Try it first, then show our solutionSolution
1. The distance of a point $(x, y)$ from the x-axis is $∣ y ∣$ , its y-coordinate taken as a length.
2. For $A (4 a, 3 a)$ the y-coordinate is $3 a$ , so the distance from the x-axis is $3 a$ (taking $a > 0$ ). A distance cannot be negative, so $- 3 a$ is ruled out. Option (A).
  1 mark
Answer: 3a, option (A)
Where the marks are
- Uses the fact that the distance from the x-axis equals the y-coordinate and selects $3 a$ (1 mark)
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Q10 · 1 mark · MCQIntroduction to Trigonometry
If $cos A = \frac{4}{5}$ , then the value of $tan A$ is :
1. (A) $\frac{3}{5}$
2. (B) $\frac{3}{4}$
3. (C) $\frac{4}{3}$
4. (D) $\frac{5}{3}$
Try it first, then show our solutionSolution
1. $sin A = 1 - \frac{16}{25} = \frac{3}{5}$ .
2. $tan A = \frac{3/5}{4/5} = \frac{3}{4}$ . Option (B).
  1 mark
Answer: $\frac{3}{4}$ , option (B)
Where the marks are
- Finds $tan A = 3/4$ (1 mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
If $2 sin A = 1$ , then the value of $tan A + cot A$ is :
1. (A) $3$
2. (B) $\frac{4}{3}$
3. (C) $\frac{3}{2}$
4. (D)1
Try it first, then show our solutionSolution
1. $sin A = \frac{1}{2} \Rightarrow A = 3 0^{\circ}$ .
2. $tan 3 0^{\circ} + cot 3 0^{\circ} = \frac{1}{3} + 3 = \frac{4}{3}$ . Option (B).
  1 mark
Answer: $\frac{4}{3}$ , option (B)
Where the marks are
- Gets A = 30° and value $4/ 3$ (1 mark)
Q12 · 1 mark · MCQSome Applications of Trigonometry
From a point on the ground, which is 60 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be $4 5^{\circ}$ . The height (in metres) of the tower is :
1. (A) $103$
2. (B) $303$
3. (C)60
4. (D)30
Try it first, then show our solutionSolution
1. Let $h$ be the height of the tower. In the right triangle formed with the ground, $tan 4 5^{\circ} = \frac{h}{60}$ .
2. Since $tan 4 5^{\circ} = 1$ , we get $h = 60$ m. Option (C).
  1 mark
Answer: 60 m, option (C)
Where the marks are
- Uses $tan 4 5^{\circ} = h /60$ and gets $h = 60$ m (1 mark)
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Q13 · 1 mark · MCQCircles
In the given figure, PA and PB are tangents to a circle centred at O. If $∠ O A B = 1 5^{\circ}$ , then $∠ A P B$ equals :
1. (A) $3 0^{\circ}$
2. (B) $1 5^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $1 0^{\circ}$
Figure in the paper: Circle with centre O. Points A and B on the circle; tangents from external point P touch the circle at A and B. Chord AB is drawn, and radius OA is drawn. The angle between OA and AB at A is marked as $∠ O A B$ . Angle at P is marked $∠ A P B$ . See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ P A B = 9 0^{\circ} - 1 5^{\circ} = 7 5^{\circ}$ .
2. PA = PB, so $∠ P B A = 7 5^{\circ}$ ; $∠ A P B = 18 0^{\circ} - 15 0^{\circ} = 3 0^{\circ}$ . Option (A).
  1 mark
Answer: 30°, option (A)
Where the marks are
- Finds $∠ A P B = 3 0^{\circ}$ (1 mark)
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Q14 · 1 mark · MCQCircles
In the given figure, PA and PB are tangents to a circle centred at O. If $∠ A O B = 13 0^{\circ}$ , then $∠ A P B$ is equal to :
1. (A) $13 0^{\circ}$
2. (B) $5 0^{\circ}$
3. (C) $12 0^{\circ}$
4. (D) $9 0^{\circ}$
Figure in the paper: Circle with centre O. Tangents from external point P touch the circle at A (upper) and B (lower). Radii OA and OB are drawn, and the central angle AOB is marked as $13 0^{\circ}$ . Angle at P is marked $∠ A P B$ . See it in the official paper.
Try it first, then show our solutionSolution
1. In quadrilateral OAPB, $∠ O A P = ∠ O B P = 9 0^{\circ}$ .
2. $∠ A P B = 36 0^{\circ} - 18 0^{\circ} - 13 0^{\circ} = 5 0^{\circ}$ . Option (B).
  1 mark
Answer: 50°, option (B)
Where the marks are
- Uses angle sum of quadrilateral to get 50° (1 mark)
Q15 · 1 mark · MCQAreas Related to Circles
Area of a segment of a circle of radius 'r' and central angle $6 0^{\circ}$ is :
1. (A) $\frac{π r ^{2}}{2} - \frac{1}{2} r^{2}$
2. (B) $\frac{2 π r}{4} - \frac{3}{4} r^{2}$
3. (C) $\frac{π r ^{2}}{6} - \frac{3}{4} r^{2}$
4. (D) $\frac{2 π r}{4} - r^{2} sin 6 0^{\circ}$
Try it first, then show our solutionSolution
1. Sector = $\frac{60}{360} π r^{2} = \frac{π r ^{2}}{6}$ ; triangle is equilateral with area $\frac{3}{4} r^{2}$ .
2. Segment = $\frac{π r ^{2}}{6} - \frac{3}{4} r^{2}$ . Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Sector minus equilateral triangle area (1 mark)
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Q16 · 1 mark · MCQSurface Areas and Volumes
A hemispherical bowl is made of steel of thickness 1 cm. The outer radius of the bowl is 6 cm. The volume of steel used (in $c m^{3}$ ) is :
1. (A) $182 π$
2. (B) $\frac{182}{3} π$
3. (C) $\frac{682}{3} π$
4. (D) $\frac{364}{3} π$
Try it first, then show our solutionSolution
1. Inner radius = 5 cm. Volume = $\frac{2}{3} π (6^{3} - 5^{3}) = \frac{2}{3} π (91)$ .
2. $= \frac{182}{3} π$ cm³. Option (B).
  1 mark
Answer: $\frac{182}{3} π$ cm³, option (B)
Where the marks are
- Difference of hemisphere volumes = $182 π /3$ (1 mark)
Q17 · 1 mark · MCQStatistics
The mean and median of a frequency distribution are 43 and 43·4 respectively. The mode of the distribution is :
1. (A)43·4
2. (B)42·4
3. (C)44·2
4. (D)49·3
Try it first, then show our solutionSolution
1. Use the empirical relation: Mode $= 3 Median - 2 Mean$ .
2. Mode $= 3 (43.4) - 2 (43) = 130.2 - 86 = 44.2$ . Option (C).
  1 mark
Answer: 44.2, option (C)
Where the marks are
- Applies Mode = 3 Median − 2 Mean and gets 44.2 (1 mark)
Q18 · 1 mark · MCQProbability
The probability for a randomly selected number out of 1, 2, 3, 4, ..., 25 to be a composite number is :
1. (A) $\frac{15}{25}$
2. (B) $\frac{10}{25}$
3. (C) $\frac{11}{25}$
4. (D) $\frac{9}{25}$
Try it first, then show our solutionSolution
1. Primes up to 25: 2, 3, 5, 7, 11, 13, 17, 19, 23 (9). Excluding these and 1: composites = 25 − 9 − 1 = 15.
2. P = $\frac{15}{25}$ . Option (A).
  1 mark
Answer: $\frac{15}{25}$ , option (A)
Where the marks are
- Counts 15 composites, probability 15/25 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedSurface Areas and Volumes
Assertion (A): The surface area of the cuboid formed by joining two cubes of sides 4 cm each, end-to-end, is $160 c m^{2}$ .
Reason (R): The surface area of a cuboid of dimensions $l \times b \times h$ is $(l b + bh + h l)$ .
Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the options (A), (B), (C) and (D) as given below.
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. Cuboid 8 × 4 × 4: SA = 2(32+16+32) = 160 cm², so A is true.
2. Correct formula is $2 (l b + bh + h l)$ , so R is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- A true, R false (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedArithmetic Progressions
Assertion (A): The mean of first 'n' natural numbers is $\frac{n - 1}{2}$ .
Reason (R): The sum of first 'n' natural numbers is $\frac{n ( n + 1 )}{2}$ .
Questions number 19 and 20 are Assertion and Reason based questions. Select the correct answer from the options (A), (B), (C) and (D) as given in question 19.
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. R is true: sum = $\frac{n ( n + 1 )}{2}$ .
2. Mean = sum ÷ n = $\frac{n + 1}{2}$ , not $\frac{n - 1}{2}$ , so A is false. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- A false, R true (1 mark)

Section B

Very short answer (VSA)

Q21 · 2 marks · Very short answerPolynomials
If $α, β$ are the zeroes of the quadratic polynomial $p x^{2} + q x + r$ , then find the value of $α^{3} β + β^{3} α$ .
Try it first, then show our solutionSolution
1. For $p x^{2} + q x + r$ , by the relation between zeroes and coefficients: $α + β = - \frac{q}{p}$ and $α β = \frac{r}{p}$
  1 mark
2. $α^{3} β + β^{3} α = α β (α^{2} + β^{2}) = α β [(α + β)^{2} - 2 α β] = \frac{r}{p} [\frac{q ^{2}}{p ^{2}} - \frac{2 r}{p}]$
  ½ mark
3. $= \frac{r}{p} \cdot \frac{q ^{2} - 2 p r}{p ^{2}} = \frac{r ( q ^{2} - 2 p r )}{p ^{3}}$
  ½ mark
Answer: $α^{3} β + β^{3} α = \frac{r ( q ^{2} - 2 p r )}{p ^{3}}$
Where the marks are
- Writes the sum and product of zeroes correctly (1 mark)
- Factorises as $α β [(α + β)^{2} - 2 α β]$ and substitutes (½ mark)
- Simplifies to $\frac{r ( q ^{2} - 2 p r )}{p ^{3}}$ (½ mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerTriangles
In the given figure, $△ A H K \sim △ A B C$ . If AK = 10 cm, BC = 3·5 cm and HK = 7 cm, find the length of AC.
Figure in the paper: Figure: Points H (top right), A (on a line), K (bottom right) form right triangle AHK with HK vertical and AK horizontal. Points C (left), A and K lie on one horizontal line; B lies below C, and segment BA extended meets H. Triangle ABC has right angle at C; BC is vertical and CA is horizontal. Labels: C, A, B, H, K. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $△ A H K \sim △ A B C$ (A↔A, H↔B, K↔C), the corresponding sides are proportional: $\frac{A K}{A C} = \frac{H K}{B C}$
  1 mark
2. $\frac{10}{A C} = \frac{7}{3.5}$
  ½ mark
3. $A C = \frac{10 \times 3.5}{7} = 5$ cm
  ½ mark
Answer: AC = 5 cm
Where the marks are
- Writes the correct ratio of corresponding sides $\frac{A K}{A C} = \frac{H K}{B C}$ (1 mark)
- Substitutes the given values (½ mark)
- Gets AC = 5 cm (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerTriangles
In the given figure, XY || QR, $\frac{P Q}{X Q} = \frac{7}{3}$ and PR = 6·3 cm. Find the length of YR.
Figure in the paper: Triangle PQR with apex P at top. X lies on side PQ and Y lies on side PR; segment XY is drawn parallel to base QR (arrows marked on XY and QR to show parallel). See it in the official paper.
Try it first, then show our solutionSolution
1. Since $X Y ∥ QR$ , by the Basic Proportionality Theorem $\frac{P X}{X Q} = \frac{P Y}{Y R}$ . Adding 1 to both sides gives $\frac{P Q}{X Q} = \frac{P R}{Y R}$
  1 mark
2. $\frac{7}{3} = \frac{6.3}{Y R}$
  ½ mark
3. $Y R = \frac{3 \times 6.3}{7} = 2.7$ cm
  ½ mark
Answer: YR = 2.7 cm
Where the marks are
- Uses BPT to get $\frac{P Q}{X Q} = \frac{P R}{Y R}$ (1 mark)
- Substitutes the values (½ mark)
- Gets YR = 2.7 cm (½ mark)
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Q23 · 2 marks · Very short answerCoordinate Geometry
If the points A(4, 5), B(m, 6), C(4, 3) and D(1, n) taken in this order are the vertices of a parallelogram ABCD, then find the values of m and n.
Try it first, then show our solutionSolution
1. The diagonals of a parallelogram bisect each other, so mid-point of AC = mid-point of BD: $(\frac{4 + 4}{2}, \frac{5 + 3}{2}) = (\frac{m + 1}{2}, \frac{6 + n}{2})$
  1 mark
2. $\frac{m + 1}{2} = 4 \Rightarrow m = 7$
  ½ mark
3. $\frac{6 + n}{2} = 4 \Rightarrow n = 2$
  ½ mark
Answer: m = 7, n = 2
Where the marks are
- Equates the mid-points of diagonals AC and BD (1 mark)
- Finds m = 7 (½ mark)
- Finds n = 2 (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerIntroduction to Trigonometry
If $tan θ + \frac{1}{t a n θ} = 2$ , find the value of $tan^{2} θ + \frac{1}{t a n ^{2} θ}$ .
Try it first, then show our solutionSolution
1. Squaring both sides: $(tan θ + \frac{1}{tan θ})^{2} = 2^{2}$
  ½ mark
2. $tan^{2} θ + \frac{1}{tan ^{2} θ} + 2 \cdot tan θ \cdot \frac{1}{tan θ} = 4 \Rightarrow tan^{2} θ + \frac{1}{tan ^{2} θ} + 2 = 4$
  1 mark
3. $tan^{2} θ + \frac{1}{tan ^{2} θ} = 2$
  ½ mark
Answer: 2
Where the marks are
- Squares both sides (½ mark)
- Expands correctly with middle term 2 (1 mark)
- Gets the value 2 (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerIntroduction to Trigonometry
Prove that : $\frac{1 - s i n θ}{1 + s i n θ} = sec θ - tan θ$
Try it first, then show our solutionSolution
1. LHS $= \frac{1 - sin θ}{1 + sin θ} \times \frac{1 - sin θ}{1 - sin θ}$ (rationalising)
  ½ mark
2. $= \frac{( 1 - sin θ ) ^{2}}{1 - sin ^{2} θ}$
  ½ mark
3. $= \frac{( 1 - sin θ ) ^{2}}{cos ^{2} θ}$ [since $1 - sin^{2} θ = cos^{2} θ$ ]
  ½ mark
4. $= \frac{1 - sin θ}{cos θ} = \frac{1}{cos θ} - \frac{sin θ}{cos θ} = sec θ - tan θ =$ RHS (θ acute)
  ½ mark
Answer: Proved: $\frac{1 - s i n θ}{1 + s i n θ} = sec θ - tan θ$
Where the marks are
- Multiplies numerator and denominator by $(1 - sin θ)$ (½ mark)
- Gets $(1 - sin θ)^{2} / (1 - sin^{2} θ)$ (½ mark)
- Uses $1 - sin^{2} θ = cos^{2} θ$ (½ mark)
- Takes the square root and splits to $sec θ - tan θ$ (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerCircles
In the given figure, O is the centre of the circle. PQ and PR are tangents. Show that the quadrilateral PQOR is cyclic.
Figure in the paper: Circle with centre O. P is outside the circle on the left. PR is a tangent touching the circle at R (upper left); PQ is a tangent touching the circle at Q (bottom). Radii OR and OQ are drawn; OQ is vertical and OR is drawn to the tangent point R. Ray from P through R and horizontal ray from P through Q are shown with arrowheads. See it in the official paper.
Try it first, then show our solutionSolution
1. The tangent at any point of a circle is perpendicular to the radius through the point of contact, so $∠ P QO = 9 0^{\circ}$ and $∠ P R O = 9 0^{\circ}$
  1 mark
2. $∴ ∠ P QO + ∠ P R O = 9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$
  ½ mark
3. In quadrilateral PQOR, one pair of opposite angles is supplementary (then the other pair also sums to $18 0^{\circ}$ ), so PQOR is cyclic.
  ½ mark
Answer: $∠ P QO + ∠ P R O = 18 0^{\circ}$ , hence PQOR is cyclic.
Where the marks are
- States $∠ P QO = ∠ P R O = 9 0^{\circ}$ with reason tangent ⟂ radius (1 mark)
- Adds to get $18 0^{\circ}$ (½ mark)
- Concludes cyclic since opposite angles are supplementary (½ mark)
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Section C

Short answer (SA)

Q26 · 3 marks · Short answerReal Numbers
Prove that $5$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $5$ is rational.
  Then $5 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are co-prime.
  ½ mark
2. Squaring gives $5 q^{2} = p^{2}$ . So $p^{2}$ is divisible by 5.
  Since 5 is prime, $p$ is divisible by 5 [if a prime divides $p^{2}$ , it divides $p$ ].
  So $p = 5 a$ for some integer $a$ .
  1 mark
3. Substitute: $25 a^{2} = 5 q^{2} \Rightarrow q^{2} = 5 a^{2}$ . So $q^{2}$ is divisible by 5, and hence $q$ is divisible by 5.
  ½ mark
4. Now 5 is a common factor of $p$ and $q$ . This contradicts the assumption that $p$ and $q$ are co-prime.
  Hence our assumption is wrong, and $5$ is irrational.
  1 mark
Answer: $5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 = p / q$ with $p, q$ co-prime (½ mark)
- Shows 5 divides $p$ , writes $p = 5 a$ (1 mark)
- Shows 5 divides $q$ (½ mark)
- States the contradiction and concludes (1 mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerCoordinate Geometry
Find the coordinates of the points of trisection of the line segment joining the points A(−1, 4) and B(−3, −2).
Try it first, then show our solutionSolution
1. Let P and Q be the points of trisection, so A, P, Q, B in order with AP : PB = 1 : 2. By the section formula, $P = (\frac{1 ( - 3 ) + 2 ( - 1 )}{1 + 2}, \frac{1 ( - 2 ) + 2 ( 4 )}{1 + 2})$
  1 mark
2. $P = (\frac{- 5}{3}, \frac{6}{3}) = (- \frac{5}{3}, 2)$
  ½ mark
3. Q is the mid-point of PB: $Q = (\frac{- \frac{5}{3} + ( - 3 )}{2}, \frac{2 + ( - 2 )}{2})$
  1 mark
4. $Q = (\frac{- 14/3}{2}, 0) = (- \frac{7}{3}, 0)$
  ½ mark
Answer: $(- \frac{5}{3}, 2)$ and $(- \frac{7}{3}, 0)$
Where the marks are
- Applies section formula for ratio 1 : 2 (1 mark)
- Gets $P (- \frac{5}{3}, 2)$ (½ mark)
- Uses mid-point of PB or ratio 2 : 1 for Q (1 mark)
- Gets $Q (- \frac{7}{3}, 0)$ (½ mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{s e c ^{3} θ}{s e c ^{2} θ - 1} + \frac{cosec ^{3} θ}{cosec ^{2} θ - 1} = sec θ \cdot cosec θ (sec θ + cosec θ)$
Try it first, then show our solutionSolution
1. LHS $= \frac{sec ^{3} θ}{tan ^{2} θ} + \frac{cosec ^{3} θ}{cot ^{2} θ}$ [since $sec^{2} θ - 1 = tan^{2} θ$ , $cosec^{2} θ - 1 = cot^{2} θ$ ]
  1 mark
2. $= \frac{1}{cos ^{3} θ} \times \frac{cos ^{2} θ}{sin ^{2} θ} + \frac{1}{sin ^{3} θ} \times \frac{sin ^{2} θ}{cos ^{2} θ}$
  ½ mark
3. $= \frac{1}{cos θ sin ^{2} θ} + \frac{1}{sin θ cos ^{2} θ}$
  ½ mark
4. $= \frac{1}{sin θ cos θ} [\frac{1}{sin θ} + \frac{1}{cos θ}]$
  ½ mark
5. $= sec θ \cdot cosec θ (cosec θ + sec θ) =$ RHS
  ½ mark
Answer: Proved: LHS = $sec θ cosec θ (sec θ + cosec θ)$
Where the marks are
- Uses $sec^{2} θ - 1 = tan^{2} θ$ and $cosec^{2} θ - 1 = cot^{2} θ$ (1 mark)
- Converts to sin and cos (½ mark)
- Simplifies each term (½ mark)
- Takes $\frac{1}{s i n θ c o s θ}$ common (½ mark)
- Reaches RHS (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerIntroduction to Trigonometry
If $\frac{s e c α}{cosec β} = p$ and $\frac{t a n α}{cosec β} = q$ , then prove that $(p^{2} - q^{2}) sec^{2} α = p^{2}$ .
Try it first, then show our solutionSolution
1. LHS $= (p^{2} - q^{2}) sec^{2} α = (\frac{sec ^{2} α}{cosec ^{2} β} - \frac{tan ^{2} α}{cosec ^{2} β}) sec^{2} α$
  ½ mark
2. $= (\frac{sec ^{2} α - tan ^{2} α}{cosec ^{2} β}) sec^{2} α$
  1 mark
3. $= \frac{1}{cosec ^{2} β} \times sec^{2} α$ [since $sec^{2} α - tan^{2} α = 1$ ]
  1 mark
4. $= \frac{sec ^{2} α}{cosec ^{2} β} = p^{2} =$ RHS
  ½ mark
Answer: Proved: $(p^{2} - q^{2}) sec^{2} α = p^{2}$
Where the marks are
- Substitutes $p^{2}$ and $q^{2}$ (½ mark)
- Combines over common denominator $cosec^{2} β$ (1 mark)
- Uses $sec^{2} α - tan^{2} α = 1$ (1 mark)
- Identifies the result as $p^{2}$ (½ mark)
Open Q28(b) on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerCircles
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Try it first, then show our solutionSolution
1. Figure: circle with centre O; from external point P draw tangents PQ and PR touching the circle at Q and R; join OP, OQ, OR.
  1 mark
2. Given: PQ and PR are tangents from external point P to a circle with centre O. To prove: $P Q = P R$ . Construction: join OP, OQ, OR.
  ½ mark
3. In $△ O QP$ and $△ O R P$ : $O Q = O R$ [radii]; $O P = O P$ [common]; $∠ O QP = ∠ O R P = 9 0^{\circ}$ [tangent ⟂ radius at point of contact]. So $△ O QP ≅ △ O R P$ [RHS congruence].
  1 mark
4. Hence $P Q = P R$ [CPCT].
  ½ mark
Answer: PQ = PR (proved)
Where the marks are
- Correct labelled figure (1 mark)
- Given, To prove, Construction (½ mark)
- Proves △OQP ≅ △ORP by RHS with reasons (1 mark)
- Concludes PQ = PR by CPCT (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCircles
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $∠ P T Q = 2∠ O P Q$ .
Try it first, then show our solutionSolution
1. Given: TP and TQ are tangents from T to a circle with centre O. To prove: $∠ P T Q = 2∠ O P Q$ .
  Let $∠ T P Q = θ$ . Since $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact], we get $∠ O P Q = 9 0^{\circ} - θ$ ...(1)
  ½ mark
2. $T P = T Q$ [tangents from an external point are equal]
  ½ mark
3. So $∠ T QP = ∠ T P Q = θ$ [angles opposite equal sides of a triangle].
  ½ mark
4. In $△ P T Q$ : $∠ T P Q + ∠ T QP + ∠ P T Q = 18 0^{\circ}$ [angle sum property]
  $\Rightarrow θ + θ + ∠ P T Q = 18 0^{\circ}$
  ½ mark
5. $\Rightarrow ∠ P T Q = 18 0^{\circ} - 2 θ = 2 (9 0^{\circ} - θ)$
  ½ mark
6. Using (1), $∠ P T Q = 2∠ O P Q$ . Hence proved.
  ½ mark
Answer: Proved: ∠PTQ = 2∠OPQ
Where the marks are
- Uses tangent ⟂ radius to write ∠OPQ = 90° − θ (1/2 mark)
- States TP = TQ, giving the reason (1/2 mark)
- Gets ∠TQP = ∠TPQ = θ (1/2 mark)
- Applies the angle sum property in triangle PTQ (1/2 mark)
- Gets ∠PTQ = 180° − 2θ = 2(90° − θ) (1/2 mark)
- Concludes ∠PTQ = 2∠OPQ (1/2 mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerAreas Related to Circles
Find the area of the sector of a circle of radius 42 cm and of central angle $3 0^{\circ}$ . Also, find the area of the corresponding major sector. [Use $π = \frac{22}{7}$ ]
Try it first, then show our solutionSolution
1. Area of minor sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{30}{360} \times \frac{22}{7} \times 42 \times 42$
  1 mark
2. $= \frac{1}{12} \times 5544 = 462$ cm²
  ½ mark
3. Angle of the major sector $= 36 0^{\circ} - 3 0^{\circ} = 33 0^{\circ}$
  ½ mark
4. Area of major sector $= \frac{330}{360} \times \frac{22}{7} \times 42 \times 42$
  ½ mark
5. $= \frac{11}{12} \times 5544 = 5082$ cm² (equivalently $5544 - 462$ )
  ½ mark
Answer: Minor sector = 462 cm²; major sector = 5082 cm²
Where the marks are
- Uses sector formula with 30° (1 mark)
- Gets 462 cm² (½ mark)
- Finds major sector angle 330° (½ mark)
- Sets up major sector area (½ mark)
- Gets 5082 cm² (½ mark)
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Q31 · 3 marks · Short answerProbability
Two dice are thrown at the same time. Determine the probability that the (i) sum of the numbers on the two dice is 5, and (ii) difference of the numbers on the two dice is 3.
Try it first, then show our solutionSolution
1. Total number of outcomes when two dice are thrown $= 6 \times 6 = 36$
  ½ mark
2. (i) Favourable outcomes for sum 5: (1,4), (4,1), (2,3), (3,2) → 4. $P (sum 5) = \frac{4}{36} = \frac{1}{9}$
  1 mark
3. (ii) Favourable outcomes for difference 3: (1,4), (4,1), (2,5), (5,2), (3,6), (6,3) → 6. $P (difference 3) = \frac{6}{36} = \frac{1}{6}$
  1½ marks
Answer: (i) $\frac{1}{9}$ ; (ii) $\frac{1}{6}$
Where the marks are
- Total outcomes = 36 (½ mark)
- Lists 4 outcomes and gets $\frac{1}{9}$ (1 mark)
- Lists 6 outcomes and gets $\frac{1}{6}$ (1½ marks)
Open Q31 on its own page

Section D

Long answer (LA)

Q32 · 5 marks · Long answerCompetency-basedPair of Linear Equations in Two Variables
Aarush bought 2 pencils and 3 chocolates for ₹ 11 and Tanish bought 1 pencil and 2 chocolates for ₹ 7 from the same shop. Represent this situation in the form of a pair of linear equations. Find the price of 1 pencil and 1 chocolate, graphically.
Try it first, then show our solutionSolution
1. Let the cost of 1 pencil be ₹ x and of 1 chocolate be ₹ y. Aarush: $2 x + 3 y = 11$
  1 mark
2. Tanish: $x + 2 y = 7$
  1 mark
3. Table for $2 x + 3 y = 11$ , i.e. $y = \frac{11 - 2 x}{3}$ : x = 1, 4, −2 gives y = 3, 1, 5. Plot (1, 3), (4, 1), (−2, 5) on graph paper and join to get a straight line.
  1 mark
4. Table for $x + 2 y = 7$ , i.e. $y = \frac{7 - x}{2}$ : x = 1, 3, 7 gives y = 3, 2, 0. Plot (1, 3), (3, 2), (7, 0) and join to get the second line.
  1 mark
5. The two lines intersect at (1, 3), so $x = 1$ , $y = 3$ : cost of 1 pencil = ₹ 1 and cost of 1 chocolate = ₹ 3.
  1 mark
Answer: Equations: $2 x + 3 y = 11$ , $x + 2 y = 7$ ; pencil = ₹ 1, chocolate = ₹ 3
Where the marks are
- Forms $2 x + 3 y = 11$ (1 mark)
- Forms $x + 2 y = 7$ (1 mark)
- Correct graph of $2 x + 3 y = 11$ (1 mark)
- Correct graph of $x + 2 y = 7$ (1 mark)
- Reads intersection: x = 1 (½ mark) and y = 3 (½ mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
A person on a tour has ₹ 4,200 for expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by ₹ 70. Find the original duration of the tour.
Try it first, then show our solutionSolution
1. Let the original duration be x days. Original daily expense $= ₹ \frac{4200}{x}$ ; new daily expense $= ₹ \frac{4200}{x + 3}$ . Given: $\frac{4200}{x} - \frac{4200}{x + 3} = 70$
  2 marks
2. $\frac{4200 \cdot 3}{x ( x + 3 )} = 70 \Rightarrow x (x + 3) = 180 \Rightarrow x^{2} + 3 x - 180 = 0$
  1½ marks
3. $x^{2} + 15 x - 12 x - 180 = 0 \Rightarrow (x + 15) (x - 12) = 0 \Rightarrow x = - 15$ or $x = 12$
  1 mark
4. Number of days cannot be negative, so reject $x = - 15$ . Original duration of the tour = 12 days.
  ½ mark
Answer: 12 days
Where the marks are
- Forms equation $\frac{4200}{x} - \frac{4200}{x + 3} = 70$ (2 marks)
- Reduces to $x^{2} + 3 x - 180 = 0$ (1½ marks)
- Factorises to $(x + 15) (x - 12) = 0$ (1 mark)
- Rejects −15 and gives 12 days (½ mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerQuadratic Equations
The area of a right-angled triangle is $600 c m^{2}$ . If the base of the triangle exceeds the altitude by 10 cm, find all the three dimensions of the triangle.
Try it first, then show our solutionSolution
1. Let the altitude be x cm, so base = (x + 10) cm. Area: $\frac{1}{2} \times x \times (x + 10) = 600$
  1½ marks
2. $x^{2} + 10 x - 1200 = 0$
  1 mark
3. $x^{2} + 40 x - 30 x - 1200 = 0 \Rightarrow (x + 40) (x - 30) = 0 \Rightarrow x = - 40$ or $x = 30$
  1 mark
4. Length cannot be negative, so $x = 30$ ; altitude = 30 cm
  ½ mark
5. Base $= 30 + 10 = 40$ cm
  ½ mark
6. Hypotenuse $= 3 0^{2} + 4 0^{2} = 2500 = 50$ cm [Pythagoras theorem]
  ½ mark
Answer: Altitude = 30 cm, base = 40 cm, hypotenuse = 50 cm
Where the marks are
- Forms $\frac{1}{2} x (x + 10) = 600$ (1½ marks)
- Gets $x^{2} + 10 x - 1200 = 0$ (1 mark)
- Factorises correctly (1 mark)
- Altitude 30 cm, rejecting −40 (½ mark)
- Base 40 cm (½ mark)
- Hypotenuse 50 cm (½ mark)
Open Q33(b) on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerTriangles
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Try it first, then show our solutionSolution
1. Figure: Triangle ABC with D on AB and E on AC such that DE ∥ BC; BE and CD joined; EM ⟂ AB (M on AB) and DN ⟂ AC (N on AC).
  Given: In △ABC, DE ∥ BC, meeting AB at D and AC at E.
  To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  Construction: Join BE and CD; draw EM ⟂ AB and DN ⟂ AC.
  2 marks
2. $\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{\frac{1}{2} \times A D \times E M}{\frac{1}{2} \times D B \times E M} = \frac{A D}{D B}$ ...(1) [same height EM]
  1 mark
3. $\frac{a r ( △ A D E )}{a r ( △ D E C )} = \frac{\frac{1}{2} \times A E \times D N}{\frac{1}{2} \times E C \times D N} = \frac{A E}{E C}$ ...(2) [same height DN]
  1 mark
4. △BDE and △DEC are on the same base DE and between the same parallels DE and BC, so $a r (△ B D E) = a r (△ D E C)$ ...(3). From (1), (2), (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: Proved: $\frac{A D}{D B} = \frac{A E}{E C}$ (Basic Proportionality Theorem)
Where the marks are
- Correct figure, given, to prove and construction (2 marks)
- Area ratio of △ADE and △BDE equals AD/DB (1 mark)
- Area ratio of △ADE and △DEC equals AE/EC (1 mark)
- Equal areas of △BDE and △DEC with reason, and conclusion (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
As shown in the given figure, a girl of height 90 cm is walking away from the base of a lamp post at a speed of 1·2 m/s. If the lamp is 3·6 m above the ground, find the length of her shadow after 4 seconds.
Figure in the paper: Right triangle ABE: A at top of lamp post, B at base (right angle at B), AB = 3·6 m vertical. E is on the ground to the right of B. Point C lies on AE; CD is vertical with D on BE, right angle at D, and CD = 90 cm representing the girl. BD is the girl's distance from lamp base; DE is her shadow. See it in the official paper.
Try it first, then show our solutionSolution
1. Distance walked in 4 s: $B D = speed \times time = 1.2 \times 4 = 4.8$ m. Girl's height CD = 90 cm = 0.9 m.
  1 mark
2. In △ABE and △CDE: $∠ A B E = ∠ C D E = 9 0^{\circ}$ and $∠ E$ is common, so $△ A B E \sim △ C D E$ (AA similarity).
  2 marks
3. $\frac{B E}{D E} = \frac{A B}{C D}$
  ½ mark
4. $\frac{4.8 + D E}{D E} = \frac{3.6}{0.9} = 4 \Rightarrow 4.8 + D E = 4 D E$
  1 mark
5. $3 D E = 4.8 \Rightarrow D E = 1.6$ m. Length of her shadow after 4 s = 1.6 m.
  ½ mark
Answer: 1.6 m
Where the marks are
- Finds BD = 4.8 m (1 mark)
- Proves △ABE ∼ △CDE by AA (2 marks)
- Writes ratio BE/DE = AB/CD (½ mark)
- Substitutes with BE = 4.8 + DE (1 mark)
- Gets DE = 1.6 m (½ mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerCompetency-basedStatistics
An SBI health insurance agent found the following data for distribution of ages of 100 policy holders. The health insurance policies are given to persons of age 15 years and onwards, but less than 60 years.
Age (in yrs) Number of policy holders
15 – 20 2
20 – 25 4
25 – 30 18
30 – 35 21
35 – 40 33
40 – 45 11
45 – 50 3
50 – 55 6
55 – 60 2
Find the modal age and median age of the policy holders.
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Age f cf
  15–20 2 2
  20–25 4 6
  25–30 18 24
  30–35 21 45
  35–40 33 78
  40–45 11 89
  45–50 3 92
  50–55 6 98
  55–60 2 100
  $n = 100$
  1 mark
2. Highest frequency is 33, so modal class = 35–40; $l = 35, f_{1} = 33, f_{0} = 21, f_{2} = 11, h = 5$
  ½ mark
3. Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 35 + \frac{33 - 21}{66 - 21 - 11} \times 5$
  1 mark
4. $= 35 + \frac{60}{34} = \frac{625}{17} \approx 36.76$ years
  ½ mark
5. $\frac{n}{2} = 50$ ; cf just greater than 50 is 78, so median class = 35–40; $l = 35, c f = 45, f = 33, h = 5$
  ½ mark
6. Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 35 + \frac{50 - 45}{33} \times 5$
  1 mark
7. $= 35 + \frac{25}{33} = \frac{1180}{33} \approx 35.76$ years
  ½ mark
Answer: Modal age ≈ 36.76 years; median age ≈ 35.76 years
Where the marks are
- Correct cumulative frequency table (1 mark)
- Identifies modal class 35–40 (½ mark)
- Substitutes correctly in mode formula (1 mark)
- Mode = 625/17 ≈ 36.76 years (½ mark)
- Identifies median class 35–40 using n/2 = 50 (½ mark)
- Substitutes correctly in median formula (1 mark)
- Median = 1180/33 ≈ 35.76 years (½ mark)
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Age (in yrs)	Number of policy holders
15 – 20	2
20 – 25	4
25 – 30	18
30 – 35	21
35 – 40	33
40 – 45	11
45 – 50	3
50 – 55	6
55 – 60	2

Age	f	cf
15–20	2	2
20–25	4	6
25–30	18	24
30–35	21	45
35–40	33	78
40–45	11	89
45–50	3	92
50–55	6	98
55–60	2	100

Section E

Case study based; internal choice in 2-mark subparts

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Case Study – 1
Your elder brother wants to buy a car and plans to take a loan from a bank for his car. He repays his total loan of ₹ 1,18,000 by paying every month, starting with the first instalment of ₹ 1,000 and he increases the instalment by ₹ 100 every month.
Based on the information given above, answer the following questions :
1. (i)
  Find the amount paid by him in the 30th instalment.
  1 mark
2. (ii)
  If the total number of instalments is 40, what is the amount paid in the last instalment?
  1 mark
3. (iii)(a)
  What amount does he still have to pay after the 30th instalment?
  2 marks
4. OR
  (iii)(b)
  Find the ratio of the tenth instalment to the last instalment.
  2 marks
Try it first, then show our solutionSolution
1. The instalments form an A.P.: $1000, 1100, 1200, \dots$ with first term $a = 1000$ and common difference $d = 100$ .
2. (i) $a_{30} = a + 29 d = 1000 + 29 (100) = 3900$ . So the 30th instalment is ₹ 3900.
  1 mark
3. (ii) $a_{40} = a + 39 d = 1000 + 39 (100) = 4900$ . So the last (40th) instalment is ₹ 4900.
  1 mark
4. (iii)(a) Amount paid in the first 30 instalments: $S_{30} = \frac{30}{2} [2 (1000) + 29 (100)] = 15 \times 4900 = 73500$ , i.e. ₹ 73500.
  1 mark
5. (iii)(a) Amount still to be paid $= 118000 - 73500 = ₹ 44500$ .
  1 mark
6. (iii)(b) $a_{10} = 1000 + 9 (100) = 1900$ and $a_{40} = 4900$ , so $\frac{a _{10}}{a _{40}} = \frac{1900}{4900}$ .
  1 mark
7. (iii)(b) $\frac{1900}{4900} = \frac{19}{49}$ , so the ratio is $19 : 49$ .
  1 mark
Answer: (i) ₹ 3900; (ii) ₹ 4900; (iii)(a) ₹ 44500; (iii)(b) 19 : 49
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ with $a = 1000$ , $d = 100$ to get $a_{30} = 3900$ (1 mark)
- Finds $a_{40} = 4900$ (1 mark)
- (iii)(a) Finds $S_{30} = 73500$ (1 mark)
- (iii)(a) Subtracts from 118000 to get ₹ 44500 (1 mark)
- (iii)(b) Writes $a_{10} / a_{40} = 1900/4900$ (1 mark)
- (iii)(b) Simplifies to 19 : 49 (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Case Study – 2
Tejas is standing at the top of a building and observes a car at an angle of depression of $3 0^{\circ}$ as it approaches the base of the building at a uniform speed. 6 seconds later, the angle of depression increases to $6 0^{\circ}$ , and at that moment, the car is 25 m away from the building.
The figure shows a building on the left with a car at ground level on the far left. On the right, A is the top of the building, B is the base of the building (right angle at B), and the ground line runs through D, C and B. Angle of depression $3 0^{\circ}$ from A to the first car position on the line AD... more precisely: from A, a line goes down to the car's first position at D making $6 0^{\circ}$ with the horizontal dashed line, and a line to the second position C making $3 0^{\circ}$ with the horizontal dashed line. CB = 25 m.
Based on the information given above, answer the following questions :
1. (i)
  What is the height of the building?
  1 mark
2. (ii)
  What is the distance between the two positions of the car?
  1 mark
3. (iii)(a)
  What would be the total time taken by the car to reach the foot of the building from the starting point?
  2 marks
4. OR
  (iii)(b)
  What is the distance of the observer from the car when it makes an angle of $6 0^{\circ}$ ?
  2 marks
Figure in the paper: Left: a brick building (tall rectangle) with a small car on the ground at far left. Right: a right-angled figure with observer A at top right, vertical side AB down to base B (right angle at B). Ground line D—C—B. Dashed horizontal line from A; angle of depression $3 0^{\circ}$ between the horizontal and line A to C, and $6 0^{\circ}$ between the horizontal and line A to D. CB is marked 25 m. See it in the official paper.
Try it first, then show our solutionSolution
1. Let AB be the building, with A the observer at the top and B its foot. D is the first position of the car and C the second, with $C B = 25$ m. Using alternate angles with the horizontal through A, $∠ A D B = 3 0^{\circ}$ and $∠ A C B = 6 0^{\circ}$ .
2. (i) In right $△ A B C$ : $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow 3 = \frac{A B}{25}$ .
  ½ mark
3. (i) So $A B = 253$ m. The height of the building is $253$ m (about 43.3 m).
  ½ mark
4. (ii) In right $△ A B D$ : $tan 3 0^{\circ} = \frac{A B}{B D} \Rightarrow \frac{1}{3} = \frac{25 3}{B D}$ , so $B D = 75$ m.
  ½ mark
5. (ii) Distance between the two positions: $C D = B D - B C = 75 - 25 = 50$ m.
  ½ mark
6. (iii)(a) The car covers 50 m in 6 s at uniform speed. Time to cover the whole distance $D B = 75$ m is $\frac{6}{50} \times 75$ .
  1 mark
7. (iii)(a) $\frac{6}{50} \times 75 = 9$ s. The car takes 9 seconds in total to reach the foot of the building.
  1 mark
8. (iii)(b) In right $△ A B C$ : $cos 6 0^{\circ} = \frac{B C}{A C} \Rightarrow \frac{1}{2} = \frac{25}{A C}$ .
  1 mark
9. (iii)(b) So $A C = 50$ m. When the angle of depression is $6 0^{\circ}$ , the observer is 50 m from the car.
  1 mark
Answer: (i) $253$ m; (ii) 50 m; (iii)(a) 9 s; (iii)(b) 50 m
Where the marks are
- (i) Uses $tan 6 0^{\circ} = A B /25$ (½ mark)
- (i) Gets $A B = 253$ m (½ mark)
- (ii) Uses $tan 3 0^{\circ} = 253 / B D$ to get $B D = 75$ m (½ mark)
- (ii) Gets $C D = 50$ m (½ mark)
- (iii)(a) Sets up time for 75 m as $\frac{6}{50} \times 75$ using uniform speed (1 mark)
- (iii)(a) Gets 9 s (1 mark)
- (iii)(b) Uses $cos 6 0^{\circ} = B C / A C$ (1 mark)
- (iii)(b) Gets $A C = 50$ m (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Case Study – 3
On a Sunday your parents took you to a fair. You could see lot of toys displayed and you wanted them to buy a Rubik's cube and a strawberry ice-cream for you.
The figure shows a cube (Rubik's cube) on the left and a cone-shaped ice-cream with a hemispherical top on the right.
Based on the information given above, answer the following questions :
1. (i)
  Find the length of the diagonal of Rubik's cube if each edge measures 6 cm.
  1 mark
2. (ii)
  Find the volume of Rubik's cube if the length of the edge is 7 cm.
  1 mark
3. (iii)(a)
  What is the curved surface area of hemisphere (ice-cream) if the base radius is 7 cm?
  2 marks
4. OR
  (iii)(b)
  If two cubes of edges 4 cm are joined end-to-end, then find the surface area of the resulting cuboid.
  2 marks
Figure in the paper: Left: a cube (Rubik's cube) drawn in perspective with dashed hidden edges. Right: an ice-cream cone with a hemispherical top shown with hatching; the hemisphere sits on top of a cone pointing downward. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Diagonal of a cube $= 3 \times edge = 3 \times 6 = 63$ cm (about 10.39 cm).
  1 mark
2. (ii) Volume of a cube $= (edge)^{3} = 7^{3} = 343 cm^{3}$ .
  1 mark
3. (iii)(a) CSA of a hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 7 \times 7$ .
  1 mark
4. (iii)(a) $= 308 cm^{2}$ .
  1 mark
5. (iii)(b) Two cubes of edge 4 cm joined end-to-end form a cuboid with $l = 8$ cm, $b = 4$ cm, $h = 4$ cm. Surface area $= 2 (l b + bh + h l) = 2 (8 \times 4 + 4 \times 4 + 4 \times 8)$ .
  1 mark
6. (iii)(b) $= 2 (32 + 16 + 32) = 2 \times 80 = 160 cm^{2}$ .
  1 mark
Answer: (i) $63$ cm; (ii) 343 cm³; (iii)(a) 308 cm²; (iii)(b) 160 cm²
Where the marks are
- (i) Uses diagonal $= 3 a$ to get $63$ cm (1 mark)
- (ii) Gets volume $7^{3} = 343$ cm³ (1 mark)
- (iii)(a) Writes CSA $= 2 π r^{2}$ with $r = 7$ substituted (1 mark)
- (iii)(a) Gets 308 cm² (1 mark)
- (iii)(b) Identifies the cuboid as 8 × 4 × 4 and applies $2 (l b + bh + h l)$ , or counts $10$ faces of $4^{2}$ (1 mark)
- (iii)(b) Gets 160 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.