MathsCBSE Class 10 · Basic · Board exam 2024 (session 2023-24)

CBSE Class 10 Mathematics Basic: 2024 board paper, set 430/3/1

The question paper set in the CBSE Class 10 board exam of 2024, 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/2024/X/MATHEMATICS_BASIC.zip (a zip; this paper is the file 430-3-1MATHEMATICS (BASIC).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very Short Answer; internal choice in 21 and 23
C	26-31	Short Answer; internal choice in 28 and 31
D	32-35	Long Answer; internal choice in 33 and 35
E	36-38	Case study; internal choice in the 2-mark part of each case

Instructions in brief: All questions are compulsory with no overall choice; internal choices are given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (one 2-mark sub-part in each case study). Use of calculators is not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQCircles
In the given figure, tangents $P A$ and $P B$ drawn from $P$ to circle are inclined to each other at an angle of $8 0^{\circ}$ . The measure of $∠ P A B$ is
1. (A) $8 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $5 0^{\circ}$
4. (D) $4 0^{\circ}$
Figure in the paper: A circle with centre O. An external point P has two tangents touching the circle at A and B. The chord AB is drawn. The angle at P between the two tangents is marked $8 0^{\circ}$ . The angle $∠ P A B$ is to be found. See it in the official paper.
Try it first, then show our solutionSolution
1. $P A = P B$ [tangents from an external point are equal], so $△ P A B$ is isosceles and $∠ P A B = ∠ P B A$ .
2. $∠ P A B = \frac{18 0 ^{\circ} - 8 0 ^{\circ}}{2} = 5 0^{\circ}$ , option (C).
  1 mark
Answer: $5 0^{\circ}$ , option (C)
Where the marks are
- Uses $P A = P B$ and the angle sum to get $5 0^{\circ}$ (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQQuadratic Equations
The value(s) of $k$ for which the quadratic equation $5 x^{2} -9 k x + 5 = 0$ has real and equal roots, is/are :
1. (A) $\frac{- 10}{9}$
2. (B) $\pm \frac{9}{10}$
3. (C) $\frac{10}{9}$
4. (D) $\pm \frac{10}{9}$
Try it first, then show our solutionSolution
1. For equal roots $D = b^{2} - 4 a c = 0$ : $81 k^{2} - 4 (5) (5) = 0$ .
2. $k^{2} = \frac{100}{81}$ , so $k = \pm \frac{10}{9}$ , option (D).
  1 mark
Answer: $k = \pm \frac{10}{9}$ , option (D)
Where the marks are
- Sets discriminant zero and gets $k = \pm 10/9$ (1 mark)
Q3 · 1 mark · MCQCoordinate Geometry
The distance between the points $A (-1, 5)$ and $B (6, -2)$ is :
1. (A) $27$
2. (B) $72$
3. (C) $49$
4. (D) $14$
Try it first, then show our solutionSolution
1. $A B = (6 + 1)^{2} + (- 2 - 5)^{2} = 49 + 49 = 98$ .
2. $A B = 72$ , option (B).
  1 mark
Answer: $72$ , option (B)
Where the marks are
- Applies the distance formula correctly to get $72$ (1 mark)
Q4 · 1 mark · MCQPolynomials
A quadratic polynomial whose zeroes are $3$ and $-2$ , is :
1. (A) $x^{2} - x -6$
2. (B) $x^{2} + x -6$
3. (C) $2 x^{2} - x -12$
4. (D) $x^{2} + x + 6$
Try it first, then show our solutionSolution
1. Sum of zeroes $= 3 + (- 2) = 1$ , product $= - 6$ .
2. Polynomial $= x^{2} - (sum) x + product = x^{2} - x - 6$ , option (A).
  1 mark
Answer: $x^{2} - x - 6$ , option (A)
Where the marks are
- Forms $x^{2} - x - 6$ from sum and product (1 mark)
Q5 · 1 mark · MCQPair of Linear Equations in Two Variables
The lines represented by linear equations $x = a$ and $y = b$ ( $a \neq = b$ ) are
1. (A)intersecting at $(a, b)$ .
2. (B)intersecting at $(b, a)$ .
3. (C)parallel.
4. (D)coincident.
Try it first, then show our solutionSolution
1. $x = a$ is a vertical line and $y = b$ is a horizontal line, so they are perpendicular and meet once.
2. The common point has $x = a$ , $y = b$ , i.e. $(a, b)$ , option (A).
  1 mark
Answer: Intersecting at $(a, b)$ , option (A)
Where the marks are
- Identifies intersection at $(a, b)$ (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQTriangles
If in the given figure, $D E ∥ B C$ . If $A D = 2.8$ cm, $D B = 2.1$ cm and $E C = 4.8$ cm, then the value of $x$ is :
1. (A) $3.6$ cm
2. (B) $2.4$ cm
3. (C) $6.4$ cm
4. (D) $4.8$ cm
Figure in the paper: Triangle ABC with D on side AB and E on side AC. DE is parallel to BC (arrow marks). AD = 2.8 cm, DB = 2.1 cm, EC = 4.8 cm. The length AE is labelled $x$ . See it in the official paper.
Try it first, then show our solutionSolution
1. By BPT [ $D E ∥ B C$ ]: $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{2.8}{2.1} = \frac{x}{4.8}$ .
2. $x = \frac{2.8 \times 4.8}{2.1} = 6.4$ cm, option (C).
  1 mark
Answer: $x = 6.4$ cm, option (C)
Where the marks are
- Uses BPT to get $x = 6.4$ cm (1 mark)
Q7 · 1 mark · MCQIntroduction to Trigonometry
In a right-angled triangle ABC, $∠ A = 9 0^{\circ}$ and $A B = A C$ . The value of $sin C$ is :
1. (A) $0$
2. (B) $\frac{3}{2}$
3. (C) $\frac{1}{2}$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. $A B = A C$ with $∠ A = 9 0^{\circ}$ gives $∠ B = ∠ C = 4 5^{\circ}$ .
2. $sin C = sin 4 5^{\circ} = \frac{1}{2}$ , option (D).
  1 mark
Answer: $\frac{1}{2}$ , option (D)
Where the marks are
- Finds $∠ C = 4 5^{\circ}$ and $sin C = 1/ 2$ (1 mark)
Q8 · 1 mark · MCQProbability
A die is thrown once. The probability of getting a number less than 6, is :
1. (A) $0$
2. (B) $\frac{5}{6}$
3. (C) $\frac{1}{6}$
4. (D) $1$
Try it first, then show our solutionSolution
1. Favourable outcomes: 1, 2, 3, 4, 5, i.e. 5 out of 6.
2. $P = \frac{5}{6}$ , option (B).
  1 mark
Answer: $\frac{5}{6}$ , option (B)
Where the marks are
- Counts 5 favourable outcomes, $P = 5/6$ (1 mark)
Q9 · 1 mark · MCQCircles
From an external point P, a tangent PA is drawn to a circle. The number of tangents through P parallel to PA is :
1. (A) $2$
2. (B)more than 2
3. (C) $1$
4. (D) $0$
Try it first, then show our solutionSolution
1. Any line through P parallel to PA would coincide with PA itself, so no other tangent through P can be parallel to PA.
  1 mark
Answer: 0, option (D)
Where the marks are
- Recognises no distinct parallel line passes through P (1 mark)
Q10 · 1 mark · MCQSurface Areas and Volumes
If the volume of a sphere is $\frac{11}{21}$ cm $^{3}$ , then the radius of the sphere is :
1. (A) $2$ cm
2. (B) $4$ cm
3. (C) $\frac{1}{2}$ cm
4. (D) $\frac{1}{4}$ cm
Try it first, then show our solutionSolution
1. $\frac{4}{3} \times \frac{22}{7} r^{3} = \frac{11}{21} \Rightarrow r^{3} = \frac{11}{21} \times \frac{21}{88} = \frac{1}{8}$ .
2. $r = \frac{1}{2}$ cm, option (C).
  1 mark
Answer: $\frac{1}{2}$ cm, option (C)
Where the marks are
- Uses $V = \frac{4}{3} π r^{3}$ to get $r = 1/2$ cm (1 mark)
Q11 · 1 mark · MCQPolynomials
In the given figure, graph of a polynomial $f (x)$ is shown. The number of zeroes of polynomial $f (x)$ is :
1. (A) $3$
2. (B) $1$
3. (C) $0$
4. (D) $2$
Figure in the paper: Cartesian axes X'OX and Y'OY. A curve labelled f(x) rises from the lower left, crosses the X-axis near the origin, reaches a single peak to the right of the origin above the X-axis, and then falls and crosses the X-axis again on the right. See it in the official paper.
Try it first, then show our solutionSolution
1. The number of zeroes equals the number of points where the graph meets the X-axis.
2. The curve cuts the X-axis at the origin and at one point on the right: 2 zeroes, option (D).
  1 mark
Answer: 2, option (D)
Where the marks are
- Counts 2 intersections with the X-axis (1 mark)
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Q12 · 1 mark · MCQAreas Related to Circles
In the given figure, two concentric circles of radii $5$ cm and $3$ cm have their centre $O$ . $O A B$ is a sector of outer circle making an angle of $6 0^{\circ}$ at the centre while $O C D$ is the sector of smaller circle. The area of the shaded region is :
1. (A) $\frac{7 π}{2}$ cm $^{2}$
2. (B) $\frac{8 π}{3}$ cm $^{2}$
3. (C) $\frac{25 π}{6}$ cm $^{2}$
4. (D) $\frac{3 π}{2}$ cm $^{2}$
Figure in the paper: Two concentric circles with centre O. Radius OA of the outer circle is 5 cm and radius OC of the inner circle is 3 cm. The sector OAB of the outer circle has central angle $6 0^{\circ}$ . The sector OCD of the inner circle lies in the same direction. The region between the two arcs AB and CD within the sector is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. Shaded area $= \frac{60}{360} π (5^{2} - 3^{2}) = \frac{1}{6} π \times 16$ .
2. $= \frac{8 π}{3}$ cm $^{2}$ , option (B).
  1 mark
Answer: $\frac{8 π}{3}$ cm $^{2}$ , option (B)
Where the marks are
- Subtracts the two sector areas to get $8 π /3$ cm $^{2}$ (1 mark)
Q13 · 1 mark · MCQStatistics
If the mean and median of a data are $10$ and $11$ respectively, then mode of the data is :
1. (A) $12$
2. (B) $8$
3. (C) $20$
4. (D) $13$
Try it first, then show our solutionSolution
1. Empirical relation: Mode $= 3 Median - 2 Mean$ .
2. Mode $= 33 - 20 = 13$ , option (D).
  1 mark
Answer: 13, option (D)
Where the marks are
- Applies Mode = 3 Median − 2 Mean to get 13 (1 mark)
Q14 · 1 mark · MCQProbability
Two fair coins are tossed together. The probability of getting 2 heads, is :
1. (A) $\frac{1}{2}$
2. (B) $\frac{3}{4}$
3. (C) $\frac{1}{4}$
4. (D) $\frac{3}{8}$
Try it first, then show our solutionSolution
1. Outcomes: HH, HT, TH, TT; only HH is favourable.
2. $P = \frac{1}{4}$ , option (C).
  1 mark
Answer: $\frac{1}{4}$ , option (C)
Where the marks are
- Lists 4 outcomes, 1 favourable, $P = 1/4$ (1 mark)
Q15 · 1 mark · MCQIntroduction to Trigonometry
If $cos A = \frac{1}{2}$ , then $tan A$ is equal to
1. (A) $\frac{1}{3}$
2. (B) $3$
3. (C) $3$
4. (D) $\frac{3}{2}$
Try it first, then show our solutionSolution
1. $cos A = \frac{1}{2} \Rightarrow A = 6 0^{\circ}$ .
2. $tan 6 0^{\circ} = 3$ , option (B).
  1 mark
Answer: $3$ , option (B)
Where the marks are
- Identifies $A = 6 0^{\circ}$ , $tan A = 3$ (1 mark)
Q16 · 1 mark · MCQSurface Areas and Volumes
From a solid cube of side $14$ cm, a sphere of maximum diameter is carved out. The radius of sphere is :
1. (A) $7$ cm
2. (B) $14$ cm
3. (C) $\frac{7}{2}$ cm
4. (D) $14$ cm
Try it first, then show our solutionSolution
1. Largest sphere has diameter equal to the cube's side, 14 cm.
2. Radius $= 7$ cm, option (A).
  1 mark
Answer: 7 cm, option (A)
Where the marks are
- Diameter = side, radius 7 cm (1 mark)
Q17 · 1 mark · MCQArithmetic Progressions
The common difference of an A.P. whose $n^{th}$ term is given by $a_{n} = 5 n -1$ , is :
1. (A) $1$
2. (B) $6$
3. (C) $5$
4. (D) $4$
Try it first, then show our solutionSolution
1. $d = a_{n + 1} - a_{n} = [5 (n + 1) - 1] - [5 n - 1] = 5$ , option (C).
  1 mark
Answer: 5, option (C)
Where the marks are
- Computes $a_{n + 1} - a_{n} = 5$ (1 mark)
Q18 · 1 mark · MCQStatistics
If for a distribution, $\sum_{i = 1}^{n} f_{i} x_{i} = 132 + 5 p$ , $\sum_{i = 1}^{n} f_{i} = 20$ and the mean of the distribution is $8.1$ , then the value of $p$ is :
1. (A) $3$
2. (B) $6$
3. (C) $4$
4. (D) $5$
Try it first, then show our solutionSolution
1. Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ : $\frac{132 + 5 p}{20} = 8.1 \Rightarrow 132 + 5 p = 162$ .
2. $5 p = 30 \Rightarrow p = 6$ , option (B).
  1 mark
Answer: $p = 6$ , option (B)
Where the marks are
- Forms mean equation and gets $p = 6$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedCoordinate Geometry
Directions: In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following: (A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true. Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): The distance of $P (a, b)$ from origin is $a^{2} + b^{2}$ .
Reason (R): The distance between two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ is $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
1. (A)Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true. Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Reason (R) is the distance formula $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ , which is a true statement.
2. Using R with $O (0, 0)$ and $P (a, b)$ : $O P = (a - 0)^{2} + (b - 0)^{2} = a^{2} + b^{2}$ , not $a^{2} + b^{2}$ . So the Assertion is false while the Reason is true. Option (D).
  1 mark
Answer: (D) Assertion (A) is false, but Reason (R) is true.
Where the marks are
- Recognises that OP = √(a² + b²) ≠ a² + b², so A is false while R (distance formula) is true; chooses option (D) (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Directions: In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following: (A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true. Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): $2 (5- 2)$ is an irrational number.
Reason (R): Product of two irrational numbers is always irrational.
1. (A)Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true. Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $2 (5 - 2) = 52 - 2$ , which is irrational, so A is true.
2. R is false: e.g. $2 \times 2 = 2$ is rational. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows A true and gives counter-example to R (1 mark)

Section B

Very Short Answer; internal choice in 21 and 23

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerCoordinate Geometry
If $Q (0, 2)$ is equidistant from $P (5, -3)$ and $R (x, 7)$ , find the value(s) of $x$ .
Try it first, then show our solutionSolution
1. Since $Q$ is equidistant from $P$ and $R$ , we have $Q P^{2} = Q R^{2}$ . By the distance formula:
  $(0 - 5)^{2} + (2 + 3)^{2} = (0 - x)^{2} + (2 - 7)^{2}$
  1 mark
2. $25 + 25 = x^{2} + 25 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$
  1 mark
Answer: $x = 5$ or $x = - 5$
Where the marks are
- Writes $Q P^{2} = Q R^{2}$ correctly using the distance formula (1 mark)
- Simplifies to $x^{2} = 25$ and gets $x = \pm 5$ (1 mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerCoordinate Geometry
If $A (1, 1)$ and $B (7, 9)$ are the end points of a diameter of a circle, then find the co-ordinates of the centre of the circle.
Try it first, then show our solutionSolution
1. The centre of a circle is the mid-point of any diameter. By the mid-point formula, centre $= (\frac{1 + 7}{2}, \frac{1 + 9}{2})$
  1 mark
2. $= (\frac{8}{2}, \frac{10}{2}) = (4, 5)$
  1 mark
Answer: Centre is $(4, 5)$
Where the marks are
- Uses the mid-point formula with the end points of the diameter (1 mark)
- Gets the centre $(4, 5)$ (1 mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $4 sin^{2} 6 0^{\circ} tan^{2} 4 5^{\circ} -2 sec^{2} 3 0^{\circ} tan^{2} 6 0^{\circ}$
Try it first, then show our solutionSolution
1. Substituting $sin 6 0^{\circ} = \frac{3}{2}$ , $tan 4 5^{\circ} = 1$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $tan 6 0^{\circ} = 3$ :
  $4 (\frac{3}{2})^{2} (1)^{2} - 2 (\frac{2}{3})^{2} (3)^{2} = 4 \cdot \frac{3}{4} \cdot 1 - 2 \cdot \frac{4}{3} \cdot 3$
  1½ marks
2. $= 3 - 8 = - 5$
  ½ mark
Answer: $- 5$
Where the marks are
- Substitutes correct values of all four trigonometric ratios (1½ marks)
- Simplifies to $- 5$ (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerReal Numbers
Prove that $- 7-2 3$ is an irrational number, given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $- 7 - 23$ is rational, say $- 7 - 23 = r$ , where $r$ is rational. Then $23 = - 7 - r \Rightarrow 3 = \frac{- 7 - r}{2}$
  1 mark
2. Since $r$ is rational, $\frac{- 7 - r}{2}$ is rational, so $3$ would be rational. This contradicts the given fact that $3$ is irrational. Hence our assumption is wrong, and $- 7 - 23$ is irrational.
  1 mark
Answer: $- 7 - 23$ is irrational (proved by contradiction).
Where the marks are
- Assumes the number is rational and expresses $3 = \frac{- 7 - r}{2}$ (1 mark)
- Points out the contradiction (RHS rational, $3$ irrational) and concludes (1 mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerReal Numbers
Explain why $(7 \times 11 \times 13 + 2 \times 11)$ is not a prime number.
Try it first, then show our solutionSolution
1. Taking 11 common: $7 \times 11 \times 13 + 2 \times 11 = 11 \times (7 \times 13 + 2) = 11 \times 93$
  1 mark
2. So the number has factors other than 1 and itself (for example 11 and 93). Hence it is a composite number, not a prime number.
  1 mark
Answer: It equals $11 \times 93 = 1023$ , so it has factors other than 1 and itself; hence it is composite, not prime.
Where the marks are
- Writes the number as $11 \times (7 \times 13 + 2)$ (1 mark)
- Concludes it has more than two factors, so it is not prime (1 mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerCoordinate Geometry
Find the ratio in which the Y-axis divides the line segment joining the points $A (5, -6)$ and $B (-1, -4)$ . Also, find the point of intersection.
Try it first, then show our solutionSolution
1. Any point on the Y-axis has x-coordinate 0. Let $P (0, y)$ divide $A B$ in the ratio $k : 1$ .
  ½ mark
2. By the section formula, x-coordinate of $P = \frac{k ( - 1 ) + 1 ( 5 )}{k + 1} = 0 \Rightarrow - k + 5 = 0 \Rightarrow k = 5$ . Required ratio is $5 : 1$ .
  1 mark
3. $y = \frac{5 ( - 4 ) + 1 ( - 6 )}{5 + 1} = \frac{- 26}{6} = - \frac{13}{3}$ . So the point of intersection is $(0, - \frac{13}{3})$ .
  ½ mark
Answer: Ratio $5 : 1$ ; point $(0, - \frac{13}{3})$
Where the marks are
- Takes the point as $(0, y)$ dividing AB in ratio $k : 1$ (½ mark)
- Equates x-coordinate to 0 and gets $k = 5$ , ratio $5 : 1$ (1 mark)
- Finds the point $(0, - 13/3)$ (½ mark)
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Q25 · 2 marks · Very short answerProbability
There are $80$ cards numbered from $1$ to $80$ . One card is drawn at random from them. Find the probability that the number on the selected card is not divisible by $8$ .
Try it first, then show our solutionSolution
1. Total outcomes = 80. Numbers divisible by 8 are $8, 16, 24, \dots, 80$ , i.e. 10 numbers. $P (divisible by 8) = \frac{10}{80} = \frac{1}{8}$
  1 mark
2. $P (not divisible by 8) = 1 - \frac{1}{8} = \frac{7}{8}$
  1 mark
Answer: $\frac{7}{8}$
Where the marks are
- Counts 10 multiples of 8 and finds probability $\frac{1}{8}$ (1 mark)
- Uses complement to get $\frac{7}{8}$ (1 mark)
Open Q25 on its own page

Section C

Short Answer; internal choice in 28 and 31

Q26 · 3 marks · Short answerReal Numbers
Find LCM and HCF of two numbers $336$ and $54$ , using prime-factorisation method.
Try it first, then show our solutionSolution
1. $336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^{4} \times 3 \times 7$
  1 mark
2. $54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^{3}$
  1 mark
3. HCF = product of smallest powers of common prime factors $= 2 \times 3 = 6$
  ½ mark
4. LCM = product of greatest powers of all prime factors $= 2^{4} \times 3^{3} \times 7 = 16 \times 27 \times 7 = 3024$
  ½ mark
Answer: HCF = 6, LCM = 3024
Where the marks are
- Prime factorisation $336 = 2^{4} \times 3 \times 7$ (1 mark)
- Prime factorisation $54 = 2 \times 3^{3}$ (1 mark)
- HCF = 6 (½ mark)
- LCM = 3024 (½ mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerQuadratic Equations
The altitude of a right-angled triangle is $7$ cm less than its base. If its hypotenuse is $17$ cm long, then
1. (a)
  represent the above information in the form of a quadratic equation;
  1 mark
2. (b)
  find the length of the sides of the triangle.
  2 marks
Try it first, then show our solutionSolution
1. (a) Let the base be $x$ cm; then the altitude is $(x - 7)$ cm. By Pythagoras theorem: $x^{2} + (x - 7)^{2} = 1 7^{2}$
  ½ mark
2. (a) $x^{2} + x^{2} - 14 x + 49 = 289 \Rightarrow 2 x^{2} - 14 x - 240 = 0 \Rightarrow x^{2} - 7 x - 120 = 0$ , the required quadratic equation.
  ½ mark
3. (b) $x^{2} - 15 x + 8 x - 120 = 0 \Rightarrow (x - 15) (x + 8) = 0 \Rightarrow x = 15$ or $x = - 8$
  ½ mark
4. (b) Length cannot be negative, so $x \neq = - 8$ ; hence $x = 15$ .
  ½ mark
5. (b) Base = 15 cm, altitude = $15 - 7 = 8$ cm, hypotenuse = 17 cm.
  1 mark
Answer: (a) $x^{2} - 7 x - 120 = 0$ ; (b) sides 15 cm, 8 cm and 17 cm
Where the marks are
- Takes base $x$ and altitude $x - 7$ and applies Pythagoras theorem (½ mark)
- Obtains $x^{2} - 7 x - 120 = 0$ (½ mark)
- Factorises as $(x - 15) (x + 8) = 0$ (½ mark)
- Rejects negative root, $x = 15$ (½ mark)
- States the sides 15 cm and 8 cm (with hypotenuse 17 cm) (1 mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerCompetency-basedSurface Areas and Volumes
A vessel is in the form of a hollow hemisphere surmounted by a hollow cylinder. The diameter of the hemisphere is $14$ cm and the total height of the vessel is $13$ cm. Find the inner surface area of the vessel.
Try it first, then show our solutionSolution
1. Radius $r = \frac{14}{2} = 7$ cm. Height of cylindrical part $h = 13 - 7 = 6$ cm.
  ½ mark
2. Inner curved surface area of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 6 = 264$ cm²
  1 mark
3. Inner curved surface area of hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 7 \times 7 = 308$ cm²
  1 mark
4. Inner surface area of vessel $= 264 + 308 = 572$ cm²
  ½ mark
Answer: 572 cm²
Where the marks are
- Finds $r = 7$ cm and cylinder height 6 cm (½ mark)
- CSA of cylinder $2 π r h = 264$ cm² (1 mark)
- CSA of hemisphere $2 π r^{2} = 308$ cm² (1 mark)
- Total 572 cm² (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerCompetency-basedSurface Areas and Volumes
A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2$ cm and the diameter of the base is $4$ cm. Determine the volume of the toy.
Try it first, then show our solutionSolution
1. Radius of base (of both cone and hemisphere) $r = \frac{4}{2} = 2$ cm; height of cone $h = 2$ cm.
  ½ mark
2. Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 4 \times 2 = \frac{176}{21}$ cm³
  1 mark
3. Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 8 = \frac{352}{21}$ cm³
  1 mark
4. Volume of toy $= \frac{176}{21} + \frac{352}{21} = \frac{528}{21} = \frac{176}{7} \approx 25.14$ cm³
  ½ mark
Answer: $\frac{176}{7}$ cm³ ≈ 25.14 cm³
Where the marks are
- Radius 2 cm (½ mark)
- Volume of cone $\frac{1}{3} π r^{2} h$ correctly substituted (1 mark)
- Volume of hemisphere $\frac{2}{3} π r^{3}$ correctly substituted (1 mark)
- Total $\frac{176}{7}$ cm³ (½ mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerPolynomials
Find the zeroes of the quadratic polynomial $5 x^{2} + 3 x -2$ and verify the relationship between the zeroes and the co-efficients.
Try it first, then show our solutionSolution
1. Splitting the middle term: $5 x^{2} + 3 x - 2 = 5 x^{2} + 5 x - 2 x - 2 = 5 x (x + 1) - 2 (x + 1) = (5 x - 2) (x + 1)$
  1 mark
2. Zeroes: $5 x - 2 = 0 \Rightarrow x = \frac{2}{5}$ ; $x + 1 = 0 \Rightarrow x = - 1$
  1 mark
3. Sum of zeroes $= \frac{2}{5} + (- 1) = - \frac{3}{5}$ , and $- \frac{coefficient of x}{coefficient of x ^{2}} = - \frac{3}{5}$ . Verified.
  ½ mark
4. Product of zeroes $= \frac{2}{5} \times (- 1) = - \frac{2}{5}$ , and $\frac{constant term}{coefficient of x ^{2}} = - \frac{2}{5}$ . Verified.
  ½ mark
Answer: Zeroes are $\frac{2}{5}$ and $- 1$ ; sum $= - \frac{3}{5}$ , product $= - \frac{2}{5}$ (relationships verified)
Where the marks are
- Factorises correctly (1 mark)
- Finds zeroes $\frac{2}{5}$ and $- 1$ (1 mark)
- Verifies sum $= - b / a$ (½ mark)
- Verifies product $= c / a$ (½ mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerTriangles
In the given figure, $∠ A B C = ∠ A C B$ and $\frac{B C}{B E} = \frac{B D}{A C}$ . Show that $△ A B E \sim △ D B C$ and $A E ∥ D C$ .
Figure in the paper: Points B and C are at the base. Point D is above and to the left of the figure, with line segments from B to D and from D to C. Point A lies inside the figure, joined to B and to C. Point E lies on segment BC between B and C, and A is joined to E. Segments BD and AC are in the figure as described in the given ratio. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ , $∠ A B C = ∠ A C B \Rightarrow A C = A B$ [sides opposite equal angles]. Given $\frac{B C}{B E} = \frac{B D}{A C} = \frac{B D}{A B}$ , so $\frac{A B}{D B} = \frac{B E}{B C}$ .
  1 mark
2. In $△ A B E$ and $△ D B C$ : $\frac{A B}{D B} = \frac{B E}{B C}$ and $∠ A B E = ∠ D B C$ [common angle B]. Therefore $△ A B E \sim △ D B C$ [SAS similarity criterion].
  1 mark
3. Corresponding angles of similar triangles are equal, so $∠ B A E = ∠ B D C$ .
  ½ mark
4. These are corresponding angles made by lines AE and DC with transversal BD, and they are equal. Hence $A E ∥ D C$ .
  ½ mark
Answer: $△ A B E \sim △ D B C$ (SAS) and hence $A E ∥ D C$ .
Where the marks are
- Uses $A B = A C$ to get $\frac{A B}{D B} = \frac{B E}{B C}$ (1 mark)
- Common $∠ B$ and SAS similarity concluded (1 mark)
- $∠ B A E = ∠ B D C$ (½ mark)
- Corresponding angles equal, so $A E ∥ D C$ (½ mark)
Open Q30 on its own page
Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $(sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2} = 7 + tan^{2} θ + cot^{2} θ$ .
Try it first, then show our solutionSolution
1. LHS $= sin^{2} θ + cosec^{2} θ + 2 sin θ cosec θ + cos^{2} θ + sec^{2} θ + 2 cos θ sec θ$
  1 mark
2. $= (sin^{2} θ + cos^{2} θ) + cosec^{2} θ + sec^{2} θ + 2 sin θ \cdot \frac{1}{sin θ} + 2 cos θ \cdot \frac{1}{cos θ}$
  ½ mark
3. Using $sin^{2} θ + cos^{2} θ = 1$ , $cosec^{2} θ = 1 + cot^{2} θ$ , $sec^{2} θ = 1 + tan^{2} θ$ :
  $= 1 + (1 + cot^{2} θ) + (1 + tan^{2} θ) + 2 + 2$
  1 mark
4. $= 7 + tan^{2} θ + cot^{2} θ =$ RHS. Hence proved.
  ½ mark
Answer: LHS simplifies to $7 + tan^{2} θ + cot^{2} θ$ = RHS; proved.
Where the marks are
- Expands both squares correctly (1 mark)
- Groups terms and uses reciprocal relations (½ mark)
- Applies the three identities (1 mark)
- Reaches RHS (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerIntroduction to Trigonometry
If $cos A = \frac{5}{13}$ , then verify that $\frac{c o s A}{1- t a n A} + \frac{s i n A}{1- c o t A} = cos A + sin A$ .
Try it first, then show our solutionSolution
1. $cos A = \frac{5}{13} \Rightarrow sin A = 1 - \frac{25}{169} = \frac{12}{13}$ ; $tan A = \frac{12}{5}$ ; $cot A = \frac{5}{12}$
  1½ marks
2. LHS $= \frac{\frac{5}{13}}{1 - \frac{12}{5}} + \frac{\frac{12}{13}}{1 - \frac{5}{12}} = \frac{\frac{5}{13}}{- \frac{7}{5}} + \frac{\frac{12}{13}}{\frac{7}{12}} = - \frac{25}{91} + \frac{144}{91} = \frac{119}{91} = \frac{17}{13}$
  1 mark
3. RHS $= \frac{5}{13} + \frac{12}{13} = \frac{17}{13}$ . Hence LHS = RHS, verified.
  ½ mark
Answer: Both sides equal $\frac{17}{13}$ ; verified.
Where the marks are
- Finds $sin A = \frac{12}{13}$ , $tan A = \frac{12}{5}$ , $cot A = \frac{5}{12}$ (½ + ½ + ½ marks)
- Evaluates LHS $= \frac{17}{13}$ (1 mark)
- Evaluates RHS $= \frac{17}{13}$ and concludes (½ mark)
Open Q31(b) on its own page

Section D

Long Answer; internal choice in 33 and 35

Q32 · 5 marks · Long answerStatistics
Consider the following distribution of hourly wages of $50$ workers of a factory:
Hourly wages (in ₹) 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10
Find the mean and the median of the above data.
Try it first, then show our solutionSolution
1. Take assumed mean $a = 150$ , class size $h = 20$ , $u_{i} = \frac{x _{i} - 150}{20}$ .
  C.I. $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$ c.f.
  100-120 110 12 −2 −24 12
  120-140 130 14 −1 −14 26
  140-160 150 8 0 0 34
  160-180 170 6 1 6 40
  180-200 190 10 2 20 50
  Total 50 −12
  2½ marks
2. Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 150 + \frac{- 12}{50} \times 20 = 150 - 4.8 = 145.2$ . Mean wage = ₹145.20
  1½ marks
3. $\frac{N}{2} = 25$ ; median class is 120-140 with $l = 120$ , $c f = 12$ , $f = 14$ , $h = 20$ .
  Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 120 + \frac{25 - 12}{14} \times 20 = 120 + 18.57 = 138.57$ (approx.). Median wage ≈ ₹138.57
  1 mark
Answer: Mean = ₹145.2; Median ≈ ₹138.57
Where the marks are
- Correct table with class marks, $u_{i}$ , $f_{i} u_{i}$ and cumulative frequencies (2½ marks)
- Mean by step-deviation $= 145.2$ (1½ marks)
- Median using formula $= 138.57$ (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerCircles
In the given figure, $A B$ is chord of length $6$ cm of a circle of radius $5$ cm. The tangents at $A$ and $B$ intersect at a point $P$ . Find the length of $P B$ .
Figure in the paper: A circle with centre O and radius 5 cm. Chord AB of length 6 cm is drawn. Tangents at B and A meet at external point P. The radius OB and OA are joined, and the dashed line OP is drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. Let OP meet AB at Q. Since $P A = P B$ [tangents from an external point] and $O A = O B$ [radii], OP is the perpendicular bisector of AB. So $O Q ⊥ A B$ and $A Q = QB = 3$ cm.
  1 mark
2. In right $△ O QB$ : $O Q = O B^{2} - Q B^{2} = 25 - 9 = 4$ cm.
  ½ mark
3. Let $P Q = x$ cm and $P B = y$ cm. In right $△ P QB$ : $x^{2} + 3^{2} = y^{2}$ ...(i)
  1 mark
4. $O B ⊥ P B$ [tangent ⟂ radius], so in right $△ O B P$ : $O P^{2} = O B^{2} + P B^{2} \Rightarrow (x + 4)^{2} = 25 + y^{2}$ ...(ii)
  1 mark
5. Substituting (i) in (ii): $x^{2} + 8 x + 16 = 25 + x^{2} + 9 \Rightarrow 8 x = 18 \Rightarrow x = \frac{9}{4}$ cm
  1 mark
6. From (i): $y^{2} = \frac{81}{16} + 9 = \frac{225}{16} \Rightarrow y = \frac{15}{4}$ . So $P B = \frac{15}{4}$ cm $= 3.75$ cm.
  ½ mark
Answer: $P B = \frac{15}{4}$ cm = 3.75 cm
Where the marks are
- OP bisects AB at right angles, so $A Q = QB = 3$ cm (1 mark)
- $O Q = 4$ cm by Pythagoras (½ mark)
- Equation $x^{2} + 9 = y^{2}$ from $△ P QB$ (1 mark)
- Equation $(x + 4)^{2} = y^{2} + 25$ from $△ O B P$ (1 mark)
- Solves to get $x = \frac{9}{4}$ (1 mark)
- Gets $P B = \frac{15}{4}$ cm (½ mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerCircles
Prove that the parallelogram circumscribing a circle is a rhombus. Also, find area of the rhombus, if radius of circle is $3$ cm and length of one side of the rhombus is $10$ cm.
Try it first, then show our solutionSolution
1. Figure: Parallelogram ABCD circumscribes a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S respectively; OA, OB and OP are joined.
  1 mark
2. Tangents from an external point are equal: $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ .
  1 mark
3. Adding: $A P + B P + C R + D R = A S + B Q + C Q + D S$
  ½ mark
4. $\Rightarrow A B + C D = A D + B C$
  ½ mark
5. In a parallelogram $A B = C D$ and $A D = B C$ , so $2 A B = 2 B C \Rightarrow A B = B C$ . Adjacent sides equal, hence ABCD is a rhombus.
  1 mark
6. $O P ⊥ A B$ [tangent ⟂ radius], $O P = 3$ cm. The four triangles AOB, BOC, COD, DOA each have base 10 cm and height 3 cm. Area of rhombus $= 4 \times \frac{1}{2} \times 10 \times 3 = 60$ cm²
  1 mark
Answer: ABCD is a rhombus; area = 60 cm²
Where the marks are
- Correct labelled figure (1 mark)
- Equal tangent lengths from each vertex (1 mark)
- Adding the equalities (½ mark)
- $A B + C D = A D + B C$ (½ mark)
- Using opposite sides equal to get $A B = B C$ , so rhombus (1 mark)
- Area $= 4 \times \frac{1}{2} \times 10 \times 3 = 60$ cm² (1 mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
Two poles of equal height are standing opposite each other on either side of a road, which is $100$ m wide. From a point somewhere between them on the road, the angles of elevation of the top of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ respectively. Find the height of the poles and the distances of the point from the poles.
Try it first, then show our solutionSolution
1. Figure: Let AB and CD be the poles, each of height $h$ m, standing at A and C on opposite sides of the road, $A C = 100$ m. P is a point on AC with $P A = y$ m and $P C = x$ m, $∠ B P A = 6 0^{\circ}$ , $∠ D P C = 3 0^{\circ}$ , and $∠ A = ∠ C = 9 0^{\circ}$ .
  1 mark
2. In right $△ B A P$ : $tan 6 0^{\circ} = \frac{h}{y} \Rightarrow h = 3 y$ ...(i)
  1 mark
3. In right $△ D C P$ : $tan 3 0^{\circ} = \frac{h}{x} \Rightarrow h = \frac{x}{3}$ ...(ii)
  1 mark
4. From (i) and (ii): $3 y = \frac{x}{3} \Rightarrow x = 3 y$ . Also $x + y = 100$ .
  1 mark
5. $3 y + y = 100 \Rightarrow y = 25$ m, $x = 75$ m
  ½ mark
6. $h = 253$ m ( $\approx 43.3$ m). So each pole is $253$ m high; the point is 25 m from the pole whose top is seen at $6 0^{\circ}$ and 75 m from the other pole.
  ½ mark
Answer: Height of each pole = $253$ m (≈ 43.3 m); distances of the point from the poles: 25 m and 75 m
Where the marks are
- Correct labelled figure (1 mark)
- $tan 6 0^{\circ} = h / y$ giving $h = 3 y$ (1 mark)
- $tan 3 0^{\circ} = h / x$ giving $h = x / 3$ (1 mark)
- Obtains $x = 3 y$ with $x + y = 100$ (1 mark)
- $x = 75$ m, $y = 25$ m (½ mark)
- $h = 253$ m (½ mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerPair of Linear Equations in Two Variables
Using graphical method, solve the following pair of equations: $x + 2 y = 8$ and $3 x -2 y = 12$
Try it first, then show our solutionSolution
1. **Table for $x + 2 y = 8$ **, i.e. $y = \frac{8 - x}{2}$ :
  $x$ 0 8 2
  $y$ 4 0 3
  1 mark
2. **Table for $3 x - 2 y = 12$ **, i.e. $y = \frac{3 x - 12}{2}$ :
  $x$ 0 4 2
  $y$ −6 0 −3
  1 mark
3. On graph paper, plot $(0, 4)$ , $(8, 0)$ and $(2, 3)$ . Join them to get the line $x + 2 y = 8$ .
  1 mark
4. On the same axes, plot $(0, - 6)$ , $(4, 0)$ and $(2, - 3)$ . Join them to get the line $3 x - 2 y = 12$ .
  1 mark
5. The two lines intersect at the point $(5, 1.5)$ .
  Check: $5 + 2 (1.5) = 8$ ✓ and $3 (5) - 2 (1.5) = 12$ ✓.
  Hence the solution is $x = 5$ , $y = 1.5$ .
  1 mark
Answer: x = 5, y = 1.5 (the lines meet at (5, 1.5))
Where the marks are
- Correct table of values for $x + 2 y = 8$ (1 mark)
- Correct table of values for $3 x - 2 y = 12$ (1 mark)
- Correct graph of $x + 2 y = 8$ (1 mark)
- Correct graph of $3 x - 2 y = 12$ (1 mark)
- Reads the intersection point and states $x = 5$ , $y = 1.5$ (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerPair of Linear Equations in Two Variables
The sum of the digits of a 2-digit number is $9$ . Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Try it first, then show our solutionSolution
1. Let the digit in the ones place be $x$ and the digit in the tens place be $y$ .
  Then the number is $10 y + x$ .
  Sum of the digits: $x + y = 9$ ...(1)
  1 mark
2. The reversed number is $10 x + y$ .
  Given condition: $9 (10 y + x) = 2 (10 x + y)$
  1 mark
3. Simplify the condition:
  $90 y + 9 x = 20 x + 2 y$
  $88 y = 11 x$
  $\Rightarrow x = 8 y$ ...(2)
  1 mark
4. Substitute (2) in (1):
  $8 y + y = 9 \Rightarrow y = 1$ , so $x = 8$ .
  1 mark
5. The required number is $10 (1) + 8 = 18$ .
  Check: $9 \times 18 = 162 = 2 \times 81$ ✓
  1 mark
Answer: The number is 18
Where the marks are
- Forms $x + y = 9$ (1 mark)
- Forms $9 (10 y + x) = 2 (10 x + y)$ (1 mark)
- Simplifies to $x = 8 y$ (1 mark)
- Solves to get $x = 8$ , $y = 1$ (1 mark)
- States the number is 18 (1 mark)
Open Q35(b) on its own page

Hourly wages (in ₹)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

C.I.	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$	c.f.
100-120	110	12	−2	−24	12
120-140	130	14	−1	−14	26
140-160	150	8	0	0	34
160-180	170	6	1	6	40
180-200	190	10	2	20	50
Total		50		−12

$x$	0	8	2
$y$	4	0	3

$x$	0	4	2
$y$	−6	0	−3

Section E

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

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
While preparing for a competitive examination, Akbar came across a match-stick pattern based question. The pattern is given below: Fig. (1), Fig. (2), Fig. (3), ...
Based on the above information, answer the following questions:
1. (i)
  Write first term and common difference of the A.P. formed by number of squares in each figure.
  1 mark
2. (ii)
  Write first term and common difference of the A.P. formed by number of sticks used in each figure.
  1 mark
3. (iii)
  (a) How many squares are there in Fig. (10)? Also, write the number of sticks used in Fig. (10).
  2 marks
4. OR
  (iii)
  (b) If $88$ sticks are used to make $m^{th}$ figure (Fig. (m)), find the value of $m$ . How many squares are formed in this figure?
  2 marks
Figure in the paper: Match-stick figures of squares. Fig. (1) is a single square. Fig. (2) is a row of three squares in a cross-like arrangement with one square above and one below the middle square. Fig. (3) is a larger cross-shaped arrangement of squares, with a row of squares horizontally and a column vertically, extended with more squares. A dotted ellipsis follows Fig. (3). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Count the squares in each figure: Fig. (1) has 1, Fig. (2) has 5, Fig. (3) has 9. Each new figure adds 4 squares.
  So the A.P. is $1, 5, 9, \dots$ with first term $a = 1$ and common difference $d = 4$ .
  1 mark
2. (ii) Count the sticks: Fig. (1) uses 4, Fig. (2) uses 16, Fig. (3) uses 28. Each new figure adds 12 sticks.
  So the A.P. is $4, 16, 28, \dots$ with first term $a = 4$ and common difference $d = 12$ .
  1 mark
3. (iii)(a) Use $a_{n} = a + (n - 1) d$ .
  Squares in Fig. (10): $a_{10} = 1 + 9 \times 4 = 37$
  1 mark
4. (iii)(a) Sticks in Fig. (10): $a_{10} = 4 + 9 \times 12 = 112$
  1 mark
5. (iii)(b) For the sticks: $88 = 4 + (m - 1) \times 12$
  1 mark
6. (iii)(b) $(m - 1) \times 12 = 84 \Rightarrow m - 1 = 7 \Rightarrow m = 8$
  ½ mark
7. (iii)(b) Squares in Fig. (8): $1 + 7 \times 4 = 29$
  ½ mark
Answer: (i) a = 1, d = 4; (ii) a = 4, d = 12; (iii)(a) 37 squares, 112 sticks; (iii)(b) m = 8, 29 squares
Where the marks are
- (i) First term 1 and common difference 4 (1 mark)
- (ii) First term 4 and common difference 12 (1 mark)
- (iii)(a) 37 squares in Fig. (10) (1 mark)
- (iii)(a) 112 sticks in Fig. (10) (1 mark)
- (iii)(b) Sets up $88 = 4 + (m - 1) 12$ (1 mark)
- (iii)(b) Gets $m = 8$ (½ mark)
- (iii)(b) Gets 29 squares (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
NSS (National Service Scheme) aims to connect the students to the community and to involve them in problem solving process. NSS symbol is based on the 'Rath' wheel of the Konark Sun Temple situated in Odisha. The wheel signifies the progress cycle of life. The diagrammatic representation of the symbol is given below. Observe the figure given above. The diameters of inner circle are equally placed. Given that $O P = 21$ cm, $O S = 10$ cm.
Based on the above information, answer the following questions:
1. (i)
  Find $m ∠ R O S$ .
  1 mark
2. (ii)
  Find the perimeter of sector $O P Q$ .
  1 mark
3. (iii)
  (a) Find the area of shaded region $P QR S$ .
  2 marks
4. OR
  (iii)
  (b) Find the area of shaded region $A C B$ i.e. the segment $A C B$ .
  2 marks
Figure in the paper: A large outer circle with centre O and a smaller inner circle with centre O. Diameters of the inner circle are drawn at equal angular spacing, forming spokes of the wheel. Points on the outer circle are labelled P (on the right), Q (upper right), and the inner circle has points R, S, A, B and C. OP is a radius of the outer circle of length 21 cm and OS is a radius of the inner circle of length 10 cm. Sector OPQ is shaded at the upper right and the region PQRS is shaded between the two circles. The segment ACB of the inner circle is shaded on the left side, bounded by chord AB and arc ACB. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The 4 equally spaced diameters divide the full angle into 8 equal parts.
  $∠ R O S = \frac{36 0 ^{\circ}}{8} = 4 5^{\circ}$
  1 mark
2. (ii) Perimeter of sector $= 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r$
  $= 21 + 21 + \frac{45}{360} \times 2 \times \frac{22}{7} \times 21$
  $= 42 + 16.5 = 58.5$ cm
  1 mark
3. (iii)(a) Area of $P QR S$ = area of sector $O P Q$ − area of sector $O S R$
  $= \frac{45}{360} \times \frac{22}{7} \times (2 1^{2} - 1 0^{2})$
  1 mark
4. (iii)(a) $= \frac{1}{8} \times \frac{22}{7} \times 341 = \frac{3751}{28}$ cm² $\approx 133.96$ cm²
  1 mark
5. (iii)(b) Chord $A B$ spans two $4 5^{\circ}$ sectors, so $∠ A O B = 9 0^{\circ}$ .
  Area of segment = area of sector $O A C B$ − area of $△ O A B$
  $= \frac{90}{360} \times \frac{22}{7} \times 1 0^{2} - \frac{1}{2} \times 10 \times 10$
  1 mark
6. (iii)(b) $= \frac{550}{7} - 50 = \frac{200}{7}$ cm² $\approx 28.57$ cm²
  1 mark
Answer: (i) 45°; (ii) 58.5 cm; (iii)(a) 3751/28 cm² ≈ 133.96 cm²; (iii)(b) 200/7 cm² ≈ 28.57 cm²
Where the marks are
- (i) $∠ R O S = 36 0^{\circ} /8 = 4 5^{\circ}$ (1 mark)
- (ii) Perimeter $= 42 + 16.5 = 58.5$ cm (1 mark)
- (iii)(a) Writes area as difference of sectors, $\frac{45}{360} \cdot \frac{22}{7} (2 1^{2} - 1 0^{2})$ (1 mark)
- (iii)(a) Gets $\frac{3751}{28}$ cm² ≈ 133.96 cm² (1 mark)
- (iii)(b) Sector of $9 0^{\circ}$ minus right triangle, correct expression (1 mark)
- (iii)(b) Gets $\frac{200}{7}$ cm² ≈ 28.57 cm² (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedTriangles
In the figure given below, a folding table is shown. The legs of the table are represented by line segments $A B$ and $C D$ intersecting at $O$ . Join $A C$ and $B D$ . Considering table top is parallel to the ground, and $O B = x$ , $O D = x + 3$ , $O C = 3 x + 19$ and $O A = 3 x + 4$ , answer the following questions:
1. (i)
  Prove that $△ O A C$ is similar to $△ O B D$ .
  1 mark
2. (ii)
  Prove that $\frac{O A}{A C} = \frac{O B}{B D}$ .
  1 mark
3. (iii)
  (a) Observe the figure and find the value of $x$ . Hence, find the length of $O C$ .
  2 marks
4. OR
  (iii)
  (b) Observe the figure and find $\frac{B D}{A C}$ .
  2 marks
Figure in the paper: A bow-tie shaped figure formed by two line segments AB and CD intersecting at O. D is at the upper left, B at the upper right, A at the lower left and C at the lower right. Segments AC and BD are joined forming the two triangles OAC and OBD. Labels: OD = x + 3, OB = x, OA = 3x + 4, OC = 3x + 19. The top edge DB is horizontal, parallel to the ground line AC. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $D B ∥ A C$ because the table top is parallel to the ground.
  In $△ O A C$ and $△ O B D$ :
  $∠ O A C = ∠ O B D$ [alternate interior angles]
  $∠ O C A = ∠ O D B$ [alternate interior angles]
  $∴ △ O A C \sim △ O B D$ [AA similarity]
  1 mark
2. (ii) Since $△ O A C \sim △ O B D$ , corresponding sides are proportional:
  $\frac{O A}{O B} = \frac{A C}{B D}$
  Rearranging gives $\frac{O A}{A C} = \frac{O B}{B D}$ .
  1 mark
3. (iii)(a) From the similarity, $\frac{O A}{O B} = \frac{O C}{O D}$ , so $\frac{3 x + 4}{x} = \frac{3 x + 19}{x + 3}$ .
  Cross-multiplying: $(3 x + 4) (x + 3) = x (3 x + 19)$
  $3 x^{2} + 13 x + 12 = 3 x^{2} + 19 x$
  $6 x = 12 \Rightarrow x = 2$
  1½ marks
4. (iii)(a) $O C = 3 (2) + 19 = 25$ units
  ½ mark
5. (iii)(b) From the similarity, $\frac{O B}{O A} = \frac{O D}{O C} = \frac{B D}{A C}$ , so $\frac{x}{3 x + 4} = \frac{x + 3}{3 x + 19}$ .
  Cross-multiplying: $x (3 x + 19) = (x + 3) (3 x + 4)$ , which gives $6 x = 12$ , so $x = 2$ .
  1½ marks
6. (iii)(b) $\frac{B D}{A C} = \frac{O B}{O A} = \frac{2}{10} = \frac{1}{5}$
  ½ mark
Answer: (i) △OAC ~ △OBD by AA; (ii) OA/AC = OB/BD proved; (iii)(a) x = 2, OC = 25; (iii)(b) BD/AC = 1/5
Where the marks are
- (i) Two pairs of equal alternate angles (from $D B ∥ A C$ ) and the AA similarity conclusion (1 mark)
- (ii) Uses proportional corresponding sides $\frac{O A}{O B} = \frac{A C}{B D}$ and rearranges (1 mark)
- (iii)(a) Sets up $\frac{3 x + 4}{x} = \frac{3 x + 19}{x + 3}$ and solves $x = 2$ (1½ marks)
- (iii)(a) $O C = 25$ (½ mark)
- (iii)(b) Sets up the proportion and gets $x = 2$ (1½ marks)
- (iii)(b) $\frac{B D}{A C} = \frac{1}{5}$ (½ mark)
Open Q38 on its own page

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.