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

CBSE Class 10 Mathematics Standard: 2026 board paper, set 30/3/1

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_3_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

Instructions in brief: No overall choice; internal choice given in 2 questions of Section B, 2 in Section C, 2 in Section D and 3 in Section E (2-mark subparts of each case study). Calculator not allowed; take π = 22/7 where required if not stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQQuadratic Equations
The roots of the quadratic equation $(x - 1)^{2} = 16$ are :
1. (A)5, 3
2. (B)4, -4
3. (C)5, -3
4. (D)-5, 3
Try it first, then show our solutionSolution
1. Take the square root of both sides: $x - 1 = \pm 4$ .
2. So $x = 1 + 4 = 5$ or $x = 1 - 4 = - 3$ . The roots are $5, - 3$ , which is option (C).
  1 mark
Answer: 5, −3; option (C)
Where the marks are
- Chooses option (C) 5, −3 (1 mark)
Q2 · 1 mark · MCQCircles
In the given figure, PQ and PR are tangents to a circle with centre O and radius 3 cm. If $∠ QP R = 6 0^{\circ}$ , then the length of each tangent is :
1. (A) $33$ cm
2. (B)3 cm
3. (C)6 cm
4. (D) $3$ cm
Figure in the paper: Circle with centre O and radius 3 cm. External point P with two tangents PQ and PR touching the circle at Q and R. Radii OQ and OR are drawn. The angle at P, QPR, is marked 60°. See it in the official paper.
Try it first, then show our solutionSolution
1. $O Q ⊥ P Q$ [tangent ⟂ radius]. $O P$ bisects $∠ QP R$ , so $∠ O P Q = 3 0^{\circ}$ .
2. In right $△ O QP$ : $tan 3 0^{\circ} = \frac{O Q}{P Q}$ , so $\frac{1}{3} = \frac{3}{P Q}$ .
3. Hence $P Q = 33$ cm, which is option (A).
  1 mark
Answer: $33$ cm; option (A)
Where the marks are
- Chooses option (A) $33$ cm (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQTriangles
In $Δ$ DEF, AB $∥$ EF. The value of x is :
1. (A)0, 2
2. (B)2 only
3. (C)-2
4. (D)1
Figure in the paper: Triangle DEF with A on side DE and B on side DF; AB is parallel to EF. Labels: DA = 2x, AE = 3x + 1, DB = x, BF = 2x - 1/2. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $A B ∥ E F$ , BPT gives $\frac{D A}{A E} = \frac{D B}{B F}$ , that is $\frac{2 x}{3 x + 1} = \frac{x}{2 x - \frac{1}{2}}$ .
2. Cross-multiply: $4 x^{2} - x = 3 x^{2} + x$ , so $x^{2} - 2 x = 0$ , giving $x = 0$ or $x = 2$ .
3. $x = 0$ would make $D A = 0$ , which is not possible for a length. So $x = 2$ only, which is option (B).
  1 mark
Answer: x = 2 only; option (B)
Where the marks are
- Chooses option (B) 2 only (1 mark)
Open Q3 on its own page
Q4 · 1 mark · MCQReal Numbers
$(3 \times 11 \times 13 + 3)$ is :
1. (A)a prime number
2. (B)divisible by 13
3. (C)a composite number
4. (D)an odd number
Try it first, then show our solutionSolution
1. Take 3 common: $3 \times 11 \times 13 + 3 = 3 (143 + 1) = 3 \times 144 = 432$ .
2. 432 has factors other than 1 and itself (for example 3), so it is composite. It is even, and $432 \div 13$ is not a whole number. The answer is option (C).
  1 mark
Answer: A composite number; option (C)
Where the marks are
- Chooses option (C) a composite number (1 mark)
Open Q4 on its own page
Q5 · 1 mark · MCQProbability
Two dice are rolled together. The probability that the sum of the numbers obtained is divisible by 6, is :
1. (A) $\frac{1}{6}$
2. (B) $\frac{11}{36}$
3. (C) $\frac{1}{12}$
4. (D) $\frac{1}{4}$
Try it first, then show our solutionSolution
1. There are 36 outcomes in total. The sums divisible by 6 are 6 and 12.
2. Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1). Sum 12: (6,6). That makes 6 favourable outcomes.
3. $P = \frac{6}{36} = \frac{1}{6}$ , which is option (A).
  1 mark
Answer: $\frac{1}{6}$ ; option (A)
Where the marks are
- Chooses option (A) $\frac{1}{6}$ (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQArithmetic Progressions
The number of multiples of 4 lying between 12 and 250 is :
1. (A)59
2. (B)59.5
3. (C)60
4. (D)61
Try it first, then show our solutionSolution
1. The multiples of 4 between 12 and 250 are $16, 20, \dots, 248$ . This is an AP with $a = 16$ and $d = 4$ .
2. $248 = 16 + (n - 1) 4 \Rightarrow n - 1 = 58 \Rightarrow n = 59$ . The answer is option (A).
  1 mark
Answer: 59; option (A)
Where the marks are
- Chooses option (A) 59 (1 mark)
Open Q6 on its own page
Q7 · 1 mark · MCQSurface Areas and Volumes
A cone of maximum size is carved out from a solid cube of edge length $l$ . The volume of the cone is :
1. (A) $\frac{π l ^{3}}{12}$
2. (B) $\frac{π l ^{3}}{3}$
3. (C) $l^{3} (1 - \frac{π}{3})$
4. (D) $\frac{π l ^{3}}{8}$
Figure in the paper: A cube with a cone carved out of it: the cone's circular top base lies inscribed in the top face of the cube and its vertex touches the centre of the bottom face. See it in the official paper.
Try it first, then show our solutionSolution
1. For the largest cone, the base diameter is $l$ (so $r = \frac{l}{2}$ ) and the height is $h = l$ .
2. $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \cdot \frac{l ^{2}}{4} \cdot l = \frac{π l ^{3}}{12}$ , which is option (A).
  1 mark
Answer: $\frac{π l ^{3}}{12}$ ; option (A)
Where the marks are
- Chooses option (A) $\frac{π l ^{3}}{12}$ (1 mark)
Open Q7 on its own page
Q8 · 1 mark · MCQPair of Linear Equations in Two Variables
Equation of another line parallel to the line represented by $2 x - 6 y = 7$ is :
1. (A) $y = 3 x - 7$
2. (B) $2 x = 9 - 6 y$
3. (C) $x - 3 y = 7$
4. (D) $x = \frac{7}{2} - 3 y$
Try it first, then show our solutionSolution
1. Lines are parallel when $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
2. For $2 x - 6 y = 7$ and $x - 3 y = 7$ : $\frac{2}{1} = \frac{- 6}{- 3} = 2$ , but $\frac{7}{7} = 1 \neq = 2$ . So they are parallel.
3. The other options have slopes $3$ or $- \frac{1}{3}$ , not $\frac{1}{3}$ , so they are not parallel. The answer is option (C).
  1 mark
Answer: $x - 3 y = 7$ ; option (C)
Where the marks are
- Chooses option (C) $x - 3 y = 7$ (1 mark)
Open Q8 on its own page
Q9 · 1 mark · MCQSome Applications of Trigonometry
If the length of the shadow of a tower is $3$ times that of its height, then altitude of the Sun is :
1. (A) $4 5^{\circ}$
2. (B) $3 0^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $1 5^{\circ}$
Try it first, then show our solutionSolution
1. Let the height be $h$ , so the shadow is $3 h$ . Then $tan θ = \frac{h}{3 h} = \frac{1}{3}$ .
2. So $θ = 3 0^{\circ}$ , which is option (B).
  1 mark
Answer: 30°; option (B)
Where the marks are
- Chooses option (B) 30° (1 mark)
Q10 · 1 mark · MCQQuadratic Equations
If the roots of the quadratic equation $3 x^{2} - k x + 23 = 0$ are real and equal, then the value(s) of k is/are :
1. (A) $\pm 24$
2. (B)0
3. (C)4
4. (D)-5
Try it first, then show our solutionSolution
1. For equal roots, $D = b^{2} - 4 a c = 0$ : $k^{2} - 4 (3) (23) = 0$ .
2. $k^{2} - 24 = 0 \Rightarrow k = \pm 24$ , which is option (A).
  1 mark
Answer: $k = \pm 24$ ; option (A)
Where the marks are
- Chooses option (A) $\pm 24$ (1 mark)
Q11 · 1 mark · MCQArithmetic Progressions
The $n^{t h}$ term of the A.P. $\frac{- 1}{3}, \frac{2}{3}, \frac{5}{3}, \frac{8}{3}, \dots$ is :
1. (A) $3 n - 4$
2. (B) $n - \frac{4}{3}$
3. (C) $\frac{n - 2}{3}$
4. (D) $\frac{n - 4}{3}$
Try it first, then show our solutionSolution
1. $a = - \frac{1}{3}$ and $d = \frac{2}{3} - (- \frac{1}{3}) = 1$ .
2. $a_{n} = - \frac{1}{3} + (n - 1) \cdot 1 = n - \frac{4}{3}$ , which is option (B).
  1 mark
Answer: $n - \frac{4}{3}$ ; option (B)
Where the marks are
- Chooses option (B) $n - \frac{4}{3}$ (1 mark)
Q12 · 1 mark · MCQCircles
In the given figure, PT is a tangent to the circle with centre O and radius r. If $∠ P O T = 4 5^{\circ}$ , then the length of OP is :
1. (A) $r 2$
2. (B) $2 r$
3. (C) $2 r$
4. (D) $r^{2}$
Figure in the paper: Circle with centre O and radius r. Point T on the circle with tangent PT at T; P is an external point, OP is drawn, and the angle POT is marked 45°. OT is labelled r. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O T P = 9 0^{\circ}$ [tangent ⟂ radius].
2. $cos 4 5^{\circ} = \frac{O T}{O P} \Rightarrow \frac{1}{2} = \frac{r}{O P} \Rightarrow O P = r 2$ , which is option (A).
  1 mark
Answer: $r 2$ ; option (A)
Where the marks are
- Chooses option (A) $r 2$ (1 mark)
Open Q12 on its own page
Q13 · 1 mark · MCQTriangles
It is given that $Δ$ ABC $\sim Δ$ QRP such that AB = 9 cm, BC = 5 cm and PR = 2 cm. Length of side QR is :
1. (A)0.9 cm
2. (B) $\frac{5}{18}$ cm
3. (C) $\frac{10}{9}$ cm
4. (D)3.6 cm
Try it first, then show our solutionSolution
1. From $△ A B C \sim △ QR P$ : $A \leftrightarrow Q$ , $B \leftrightarrow R$ , $C \leftrightarrow P$ . So $\frac{A B}{QR} = \frac{B C}{R P}$ .
2. $\frac{9}{QR} = \frac{5}{2} \Rightarrow QR = \frac{18}{5} = 3.6$ cm, which is option (D).
  1 mark
Answer: 3.6 cm; option (D)
Where the marks are
- Chooses option (D) 3.6 cm (1 mark)
Open Q13 on its own page
Q14 · 1 mark · MCQPolynomials
The sum and product of zeroes of a quadratic polynomial p(x) are $\frac{- 1}{3}$ and 2 respectively. The polynomial p(x) is :
1. (A) $3 x^{2} - x + 6$
2. (B) $x^{2} + \frac{1}{3} x - 2$
3. (C) $3 x^{2} - x + 2$
4. (D) $- 3 x^{2} - x - 6$
Try it first, then show our solutionSolution
1. $p (x) = k [x^{2} - (sum) x + product] = k (x^{2} + \frac{1}{3} x + 2)$ .
2. Taking $k = - 3$ gives $- 3 x^{2} - x - 6$ . Check: sum $= - \frac{- 1}{- 3} = - \frac{1}{3}$ and product $= \frac{- 6}{- 3} = 2$ . The answer is option (D).
  1 mark
Answer: $- 3 x^{2} - x - 6$ ; option (D)
Where the marks are
- Chooses option (D) $- 3 x^{2} - x - 6$ (1 mark)
Open Q14 on its own page
Q15 · 1 mark · MCQIntroduction to Trigonometry
Given that $sin 2 α = \frac{3}{2}$ , the value of $sin 3 α$ is :
1. (A) $\frac{3 3}{4}$
2. (B) $\frac{1}{2}$
3. (C)1
4. (D) $\frac{3}{4}$
Try it first, then show our solutionSolution
1. $sin 2 α = \frac{3}{2} = sin 6 0^{\circ}$ , so $2 α = 6 0^{\circ}$ and $α = 3 0^{\circ}$ .
2. $sin 3 α = sin 9 0^{\circ} = 1$ , which is option (C).
  1 mark
Answer: 1; option (C)
Where the marks are
- Chooses option (C) 1 (1 mark)
Q16 · 1 mark · MCQStatistics
The median and mode of a distribution are 25.2 and 26.1 respectively. The mean of the distribution is :
1. (A)24.75
2. (B)24.25
3. (C)24.3
4. (D)25.5
Try it first, then show our solutionSolution
1. Empirical relation: Mode $= 3$ Median $- 2$ Mean.
2. $26.1 = 3 (25.2) - 2$ Mean $\Rightarrow 2$ Mean $= 75.6 - 26.1 = 49.5 \Rightarrow$ Mean $= 24.75$ , which is option (A).
  1 mark
Answer: 24.75; option (A)
Where the marks are
- Chooses option (A) 24.75 (1 mark)
Open Q16 on its own page
Q17 · 1 mark · MCQCoordinate Geometry
In the given figure, $Δ$ ABC is an equilateral triangle. AD is a median of the triangle joining the points $A (0, \frac{5 3}{2})$ , $D (0, 0)$ . Points B and C are (in same order) :
1. (A)(-5, 0), (5, 0)
2. (B) $(- \frac{5}{2}, 0), (\frac{5}{2}, 0)$
3. (C)(-10, 0), (10, 0)
4. (D) $(- 53, 0), (53, 0)$
Figure in the paper: Equilateral triangle ABC with apex A at $(0, \frac{5 3}{2})$ on the y-axis, base BC on the x-axis, and D(0,0) the foot of the median AD on BC; B is to the left and C to the right of D. See it in the official paper.
Try it first, then show our solutionSolution
1. In an equilateral triangle the median is also the altitude: $A D = \frac{3}{2} a = \frac{5 3}{2}$ , so $a = 5$ .
2. $B D = D C = \frac{5}{2}$ . So $B (- \frac{5}{2}, 0)$ and $C (\frac{5}{2}, 0)$ , which is option (B).
  1 mark
Answer: $(- \frac{5}{2}, 0), (\frac{5}{2}, 0)$ ; option (B)
Where the marks are
- Chooses option (B) (1 mark)
Open Q17 on its own page
Q18 · 1 mark · MCQIntroduction to Trigonometry
The value of $(\frac{1}{2} tan^{2} 4 5^{\circ} - cos^{2} 6 0^{\circ})$ is :
1. (A)0
2. (B) $- \frac{1}{2}$
3. (C) $\frac{1}{4}$
4. (D) $- \frac{1}{4}$
Try it first, then show our solutionSolution
1. $tan 4 5^{\circ} = 1$ and $cos 6 0^{\circ} = \frac{1}{2}$ .
2. $\frac{1}{2} (1)^{2} - (\frac{1}{2})^{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ , which is option (C).
  1 mark
Answer: $\frac{1}{4}$ ; option (C)
Where the marks are
- Chooses option (C) $\frac{1}{4}$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedCircles
Assertion (A): Radius is the smallest distance of a tangent from the centre of the circle. Reason (R): Radius is perpendicular to the tangent.
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: the tangent at any point is perpendicular to the radius through the point of contact.
2. (A) is true: the perpendicular is the shortest segment from a point to a line. Since the radius is perpendicular to the tangent, it is the shortest distance from the centre. So (R) correctly explains (A), which is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Chooses option (A) (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Assertion (A): $tan 2 θ$ is not defined at $θ = 4 5^{\circ}$ . Reason (R): $sin 9 0^{\circ} \neq = cos 9 0^{\circ}$ .
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. At $θ = 4 5^{\circ}$ , $tan 2 θ = tan 9 0^{\circ} = \frac{sin 9 0 ^{\circ}}{cos 9 0 ^{\circ}} = \frac{1}{0}$ , which is not defined. So (A) is true.
2. $sin 9 0^{\circ} = 1$ and $cos 9 0^{\circ} = 0$ , so (R) is true. However, $tan 9 0^{\circ}$ is undefined because $cos 9 0^{\circ} = 0$ , not because $sin 9 0^{\circ} \neq = cos 9 0^{\circ}$ . So (R) does not explain (A), which is option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Chooses option (B) (1 mark)
Open Q20 on its own page

Section B

Very short answer (VSA)

Q21 · 2 marks · Very short answerReal Numbers
Find the length of the plank that can be used to measure the lengths 4 m 20 cm and 5 m 4 cm exactly, in the least time.
Try it first, then show our solutionSolution
1. Convert to the same unit: 4 m 20 cm $= 420$ cm and 5 m 4 cm $= 504$ cm.
  ½ mark
2. To measure in the least time, the plank must be as long as possible while dividing both lengths exactly. So we need HCF(420, 504).
  $420 = 2^{2} \times 3 \times 5 \times 7$ and $504 = 2^{3} \times 3^{2} \times 7$ .
  1 mark
3. HCF $= 2^{2} \times 3 \times 7 = 84$ . The required length of the plank is 84 cm.
  ½ mark
Answer: 84 cm
Where the marks are
- Converts both lengths to cm: 420 cm and 504 cm (½ mark)
- Writes the prime factorisations of 420 and 504 (1 mark)
- Gets HCF = 84 cm (½ mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerArithmetic Progressions
In an A.P., the first term is 32 and the last term is -10. If the common difference is -2, then find the number of terms and their sum.
Try it first, then show our solutionSolution
1. Here $a = 32$ , $l = a_{n} = - 10$ , $d = - 2$ . Using $a_{n} = a + (n - 1) d$ : $32 + (n - 1) (- 2) = - 10$ .
  ½ mark
2. $(n - 1) (- 2) = - 42 \Rightarrow n - 1 = 21 \Rightarrow n = 22$ .
  ½ mark
3. $S_{n} = \frac{n}{2} (a + l)$ , so $S_{22} = \frac{22}{2} [32 + (- 10)]$ .
  ½ mark
4. $S_{22} = 11 \times 22 = 242$ . There are 22 terms and their sum is 242.
  ½ mark
Answer: n = 22, sum = 242
Where the marks are
- Writes $32 + (n - 1) (- 2) = - 10$ (½ mark)
- Gets n = 22 (½ mark)
- Uses $S_{22} = \frac{22}{2} (32 - 10)$ (½ mark)
- Gets sum = 242 (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerArithmetic Progressions
Find the sum of the first 28 terms of an A.P. whose $n^{t h}$ term is given by $a_{n} = 3 n - 2$ .
Try it first, then show our solutionSolution
1. First term: $a_{1} = 3 (1) - 2 = 1$ .
  ½ mark
2. 28th term: $a_{28} = 3 (28) - 2 = 82$ .
  ½ mark
3. $S_{28} = \frac{28}{2} (a_{1} + a_{28}) = 14 (1 + 82)$ .
  ½ mark
4. $S_{28} = 14 \times 83 = 1162$ .
  ½ mark
Answer: $S_{28} = 1162$
Where the marks are
- Finds $a_{1} = 1$ (½ mark)
- Finds $a_{28} = 82$ (½ mark)
- Uses $S_{28} = \frac{28}{2} (1 + 82)$ (½ mark)
- Gets 1162 (½ mark)
Open Q22(b) on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerCoordinate Geometry
Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are (9, -2) and (1, 6) respectively.
1. (i)
  Find the co-ordinates of point P.
  1 mark
2. (ii)
  Find the length of the side of the square.
  1 mark
Figure in the paper: Square ABCD with A at bottom-left, B at bottom-right (9, -2), C at top-right, D at top-left (1, 6). Diagonals AC and BD cross at P. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The diagonals of a square bisect each other, so P is the midpoint of BD: $P = (\frac{9 + 1}{2}, \frac{- 2 + 6}{2}) = (5, 2)$ .
  1 mark
2. (ii) $B D = (9 - 1)^{2} + (- 2 - 6)^{2} = 64 + 64 = 128 = 82$ . In a square, diagonal $= 2 \times$ side, so $2 A B^{2} = B D^{2} = 128$ .
  ½ mark
3. (ii) $A B^{2} = 64 \Rightarrow A B = 8$ . The side of the square is 8 units.
  ½ mark
Answer: (i) P(5, 2); (ii) 8 units
Where the marks are
- (i) Uses the midpoint formula to get P(5, 2) (1 mark)
- (ii) Uses $2 A B^{2} = B D^{2}$ with $B D^{2} = 128$ (½ mark)
- (ii) Gets side = 8 units (½ mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerCoordinate Geometry
Find the coordinates of a point on the line $x + y = 5$ which is equidistant from (6, 4) and (5, 2).
Try it first, then show our solutionSolution
1. Let the required point be $P (x, y)$ . It is equidistant from $(6, 4)$ and $(5, 2)$ , so by the distance formula:
  $(x - 6)^{2} + (y - 4)^{2} = (x - 5)^{2} + (y - 2)^{2}$
  $\Rightarrow x^{2} - 12 x + 36 + y^{2} - 8 y + 16 = x^{2} - 10 x + 25 + y^{2} - 4 y + 4$
  $\Rightarrow 52 - 12 x - 8 y = 29 - 10 x - 4 y$
  $\Rightarrow 2 x + 4 y = 23 ...(i)$
  1 mark
2. P lies on the line, so $x + y = 5$ , which gives $2 x + 2 y = 10 ...(ii)$ .
  Subtracting (ii) from (i): $2 y = 13 \Rightarrow y = \frac{13}{2}$ .
  Then $x = 5 - \frac{13}{2} = - \frac{3}{2}$ .
  ½ mark
3. Hence the required point is $(- \frac{3}{2}, \frac{13}{2})$ .
  (Check: both squared distances equal $62.5$ .)
  ½ mark
Answer: $(- \frac{3}{2}, \frac{13}{2})$
Where the marks are
- Equates the squared distances from (x, y) to (6, 4) and (5, 2) and simplifies to $2 x + 4 y = 23$ (1 mark)
- Solves this with $x + y = 5$ to get $x = - \frac{3}{2}$ , $y = \frac{13}{2}$ (½ mark)
- States the required point $(- \frac{3}{2}, \frac{13}{2})$ (½ mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerTriangles
The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{A O}{O C} = \frac{B O}{O D}$ . Show that quadrilateral ABCD is a trapezium.
Try it first, then show our solutionSolution
1. Figure: Draw quadrilateral ABCD with AB at the bottom and DC at the top. The diagonals AC and BD cross at O.
  ½ mark
2. Given $\frac{A O}{O C} = \frac{B O}{O D}$ , so $\frac{A O}{B O} = \frac{O C}{O D}$ .
  In $△ A O B$ and $△ C O D$ :
  $\frac{A O}{O C} = \frac{B O}{O D}$ [given]
  $∠ A O B = ∠ C O D$ [vertically opposite angles]
  So $△ A O B \sim △ C O D$ [SAS similarity criterion].
  1 mark
3. Corresponding angles of similar triangles are equal, so $∠ O A B = ∠ O C D$ , that is, $∠ C A B = ∠ A C D$ .
  These are alternate angles made by the transversal AC with lines AB and DC. Hence $A B ∥ D C$ .
  So ABCD has one pair of parallel sides, and ABCD is a trapezium.
  ½ mark
Answer: $△ A O B \sim △ C O D$ (SAS), so $∠ O A B = ∠ O C D$ (alternate angles). Hence $A B ∥ D C$ and ABCD is a trapezium.
Where the marks are
- Correct labelled figure (½ mark)
- Proves $△ A O B \sim △ C O D$ by SAS, using vertically opposite angles (1 mark)
- Deduces $∠ O A B = ∠ O C D$ , so $A B ∥ D C$ , and concludes ABCD is a trapezium (½ mark)
- Alternative: draws $E O ∥ A B$ meeting AD at E, uses BPT in $△ A B D$ and its converse in $△ A D C$ to get $E O ∥ D C$ , hence $A B ∥ D C$ (½ + ½ + ½ + ½)
Open Q24 on its own page
Q25 · 2 marks · Very short answerSurface Areas and Volumes
A toy is in the form of a cone mounted on a hemisphere of radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.
Figure in the paper: A toy: a cone with vertical axis sitting on the flat circular face of a hemisphere (hemisphere below, its rim visible as an ellipse at the junction). See it in the official paper.
Try it first, then show our solutionSolution
1. Radius $r = 7$ cm.
  Height of the cone: $h = 31 - 7 = 24$ cm.
  Slant height: $l = r^{2} + h^{2} = 7^{2} + 2 4^{2} = 625 = 25$ cm.
  ½ mark
2. The flat circular faces are joined together, so they are hidden. Therefore:
  TSA of toy = CSA of cone + CSA of hemisphere $= π r l + 2 π r^{2}$
  $= \frac{22}{7} \times 7 \times 25 + 2 \times \frac{22}{7} \times 7 \times 7 = 550 + 308$
  1 mark
3. $= 858$ cm².
  So the total surface area of the toy is $858 cm^{2}$ .
  ½ mark
Answer: 858 cm²
Where the marks are
- Finds the cone height 24 cm and slant height 25 cm (½ mark)
- Writes TSA = $π r l + 2 π r^{2}$ and substitutes correctly (1 mark)
- Gets 858 cm² (½ mark)
Open Q25 on its own page

Section C

Short answer (SA)

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerIntroduction to Trigonometry
If $sin θ + cos θ = 3$ , then prove that $tan θ + cot θ = 1$ .
Try it first, then show our solutionSolution
1. Given $sin θ + cos θ = 3$ . Squaring both sides:
  $(sin θ + cos θ)^{2} = 3$
  ½ mark
2. $sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 3$
  $\Rightarrow 1 + 2 sin θ cos θ = 3$ [since $sin^{2} θ + cos^{2} θ = 1$ ]
  $\Rightarrow sin θ cos θ = 1 ...(i)$
  1 mark
3. LHS $= tan θ + cot θ = \frac{sin θ}{cos θ} + \frac{cos θ}{sin θ} = \frac{sin ^{2} θ + cos ^{2} θ}{sin θ cos θ} = \frac{1}{sin θ cos θ}$
  1 mark
4. Using (i): $\frac{1}{sin θ cos θ} = \frac{1}{1} = 1 =$ RHS.
  Hence proved.
  ½ mark
Answer: Shown: $sin θ cos θ = 1$ , so $tan θ + cot θ = \frac{1}{s i n θ c o s θ} = 1$ .
Where the marks are
- Squares the given relation (½ mark)
- Uses $sin^{2} θ + cos^{2} θ = 1$ to get $sin θ cos θ = 1$ (1 mark)
- Writes $tan θ + cot θ = \frac{1}{s i n θ c o s θ}$ (1 mark)
- Substitutes to get 1 = RHS (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $(sin A + sec A)^{2} + (cos A + cosec A)^{2} = (1 + sec A cosec A)^{2}$
Try it first, then show our solutionSolution
1. LHS $= (sin A + \frac{1}{cos A})^{2} + (cos A + \frac{1}{sin A})^{2}$
  ½ mark
2. Expanding both squares:
  $= sin^{2} A + \frac{1}{cos ^{2} A} + \frac{2 sin A}{cos A} + cos^{2} A + \frac{1}{sin ^{2} A} + \frac{2 cos A}{sin A}$
  $= (sin^{2} A + cos^{2} A) + (\frac{1}{sin ^{2} A} + \frac{1}{cos ^{2} A}) + 2 (\frac{sin A}{cos A} + \frac{cos A}{sin A})$
  1 mark
3. Simplifying each bracket:
  $= 1 + \frac{cos ^{2} A + sin ^{2} A}{sin ^{2} A cos ^{2} A} + 2 \cdot \frac{sin ^{2} A + cos ^{2} A}{sin A cos A}$
  $= 1 + \frac{1}{sin ^{2} A cos ^{2} A} + \frac{2}{sin A cos A}$
  $= 1 + sec^{2} A cosec^{2} A + 2 sec A cosec A$
  1 mark
4. This has the form $1 + 2 t + t^{2}$ with $t = sec A cosec A$ :
  $= (1 + sec A cosec A)^{2} =$ RHS.
  Hence proved.
  ½ mark
Answer: LHS simplifies to $1 + 2 sec A cosec A + sec^{2} A cosec^{2} A = (1 + sec A cosec A)^{2}$ = RHS.
Where the marks are
- Writes sec and cosec in terms of cos and sin (½ mark)
- Expands correctly and groups the terms (1 mark)
- Simplifies to $1 + sec^{2} A cosec^{2} A + 2 sec A cosec A$ (1 mark)
- Recognises the perfect square and concludes (½ mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerAreas Related to Circles
In the given figure, chord AB subtends an angle of $12 0^{\circ}$ at the centre of the circle with radius 7 cm. Find (i) perimeter of major sector OACB, and (ii) area of the shaded segment, if area of $Δ$ OAB = $21.2$ cm $^{2}$ .
Figure in the paper: Circle with centre O and radius 7 cm. Chord AB with central angle AOB = 120°. Arc ACB (major arc) passes through C at the top. The minor segment between chord AB and minor arc is shaded. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The angle of the major sector is $36 0^{\circ} - 12 0^{\circ} = 24 0^{\circ}$ .
  Perimeter of major sector OACB = length of major arc ACB + $2 r$
  $= \frac{240}{360} \times 2 \times \frac{22}{7} \times 7 + 2 \times 7 = \frac{88}{3} + 14$
  1 mark
2. (i) $= \frac{130}{3}$ cm $\approx 43.3$ cm.
  ½ mark
3. (ii) Area of shaded segment = area of minor sector OAB − area of $△ O A B$
  $= \frac{120}{360} \times \frac{22}{7} \times 7 \times 7 - 21.2 = \frac{154}{3} - 21.2 \approx 51.33 - 21.2$
  1 mark
4. (ii) $\approx 30.1$ cm².
  ½ mark
Answer: (i) $\frac{130}{3}$ cm ≈ 43.3 cm; (ii) ≈ 30.1 cm²
Where the marks are
- Major arc length with angle 240° plus two radii, correctly substituted (1 mark)
- Perimeter = 130/3 cm ≈ 43.3 cm (½ mark)
- Segment area = sector area (120°) − 21.2, correctly substituted (1 mark)
- Segment area ≈ 30.1 cm² (½ mark)
Open Q27 on its own page
Q28 · 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 coprime.
  ½ mark
2. Squaring: $5 q^{2} = p^{2}$ . So $p^{2}$ is divisible by 5.
  Since 5 is prime, $p$ is divisible by 5 ...(1) [if a prime divides $p^{2}$ , it divides $p$ ].
  1 mark
3. Let $p = 5 a$ for some integer $a$ . Then $5 q^{2} = 25 a^{2} \Rightarrow q^{2} = 5 a^{2}$ .
  So $q^{2}$ is divisible by 5, and hence $q$ is divisible by 5 ...(2).
  1 mark
4. From (1) and (2), 5 is a common factor of $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong, and $5$ is irrational.
  ½ mark
Answer: $5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 = p / q$ with p, q coprime and q ≠ 0 (½ mark)
- Gets $5 q^{2} = p^{2}$ and deduces 5 divides p (1 mark)
- Substitutes p = 5a and deduces 5 divides q (1 mark)
- States the contradiction and concludes (½ mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerQuadratic Equations
Find two consecutive negative integers, sum of whose squares is 481.
Try it first, then show our solutionSolution
1. Let the two consecutive negative integers be $x$ and $x + 1$ .
  Given: $x^{2} + (x + 1)^{2} = 481$
  1 mark
2. $2 x^{2} + 2 x + 1 = 481 \Rightarrow 2 x^{2} + 2 x - 480 = 0 \Rightarrow x^{2} + x - 240 = 0$
  ½ mark
3. Splitting the middle term: $x^{2} + 16 x - 15 x - 240 = 0$
  $\Rightarrow (x + 16) (x - 15) = 0 \Rightarrow x = - 16$ or $x = 15$
  1 mark
4. The integers must be negative, so $x = 15$ is rejected and $x = - 16$ .
  Required integers: $- 16$ and $- 15$ .
  (Check: $256 + 225 = 481$ .)
  ½ mark
Answer: −16 and −15
Where the marks are
- Forms the equation $x^{2} + (x + 1)^{2} = 481$ (1 mark)
- Reduces it to $x^{2} + x - 240 = 0$ (½ mark)
- Factorises to get x = −16 or 15 (1 mark)
- Rejects 15 and states the integers −16, −15 (½ mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerCoordinate Geometry
A point P(x, 7) divides a line segment joining the points A(-5, 4) and B(7, 9) in a certain ratio. Find the ratio and hence find the value of x.
Try it first, then show our solutionSolution
1. Let P divide AB in the ratio $k : 1$ , that is, $A P : P B = k : 1$ .
  ½ mark
2. Using the section formula for the y-coordinate:
  $7 = \frac{k \cdot 9 + 1 \cdot 4}{k + 1}$
  $\Rightarrow 7 k + 7 = 9 k + 4 \Rightarrow k = \frac{3}{2}$
  1 mark
3. So the ratio is $\frac{3}{2} : 1 = 3 : 2$ .
  ½ mark
4. Using the section formula for the x-coordinate with ratio $3 : 2$ :
  $x = \frac{3 \times 7 + 2 \times ( - 5 )}{3 + 2} = \frac{21 - 10}{5} = \frac{11}{5}$
  1 mark
Answer: Ratio 3 : 2; $x = \frac{11}{5}$
Where the marks are
- Takes the ratio as k : 1 (½ mark)
- Uses the y-coordinate in the section formula to get k = 3/2 (1 mark)
- States the ratio 3 : 2 (½ mark)
- Finds x = 11/5 (1 mark)
Open Q30 on its own page
Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerCircles
Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that $∠ A P B = 2∠ O A B$ .
Try it first, then show our solutionSolution
1. Figure: Draw a circle with centre O and an external point P. Draw tangents PA and PB touching the circle at A and B. Join OA and AB.
  ½ mark
2. $O A ⊥ A P$ [tangent ⟂ radius at the point of contact], so $∠ O A P = 9 0^{\circ}$ .
  Therefore $∠ P A B = 9 0^{\circ} - ∠ O A B ...(i)$
  ½ mark
3. $P A = P B$ [tangents from an external point are equal], so $∠ P A B = ∠ P B A$ [angles opposite equal sides].
  In $△ P A B$ , by the angle sum property:
  $∠ A P B = 18 0^{\circ} - 2∠ P A B$
  1 mark
4. Using (i): $∠ A P B = 18 0^{\circ} - 2 (9 0^{\circ} - ∠ O A B)$
  ½ mark
5. $\Rightarrow ∠ A P B = 2∠ O A B$ . Hence proved.
  ½ mark
Answer: Proved: $∠ A P B = 18 0^{\circ} - 2 (9 0^{\circ} - ∠ O A B) = 2∠ O A B$ .
Where the marks are
- Correct figure (½ mark)
- Uses OA ⟂ AP to get ∠PAB = 90° − ∠OAB (½ mark)
- Uses PA = PB to get equal base angles and ∠APB = 180° − 2∠PAB (1 mark)
- Substitutes from (i) (½ mark)
- Concludes ∠APB = 2∠OAB (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerCircles
In the given figure, PA is the tangent to the circle with centre O such that OA = 10 cm, AB = 8 cm and AB $⊥$ OP. Find the length of PB.
Figure in the paper: Circle with centre O. External point P with tangents PA and PC touching the circle at A and C. B lies on line PO with AB perpendicular to OP (right-angle mark at B). OA = 10 cm, AB = 8 cm. PA is the tangent at A. See it in the official paper.
Try it first, then show our solutionSolution
1. In right $△ O B A$ (right angle at B):
  $O B = O A^{2} - A B^{2} = 1 0^{2} - 8^{2} = 36 = 6$ cm
  ½ mark
2. Let $∠ A O B = θ$ . In $△ O B A$ : $tan θ = \frac{A B}{O B} = \frac{8}{6} ...(i)$
  ½ mark
3. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius]. So in right $△ O A P$ :
  $tan θ = \frac{A P}{O A} = \frac{A P}{10}$
  Using (i): $\frac{A P}{10} = \frac{8}{6}$
  ½ mark
4. $\Rightarrow A P = \frac{80}{6} = \frac{40}{3}$ cm
  ½ mark
5. By Pythagoras in $△ O A P$ :
  $O P = A P^{2} + O A^{2} = \frac{1600}{9} + 100 = \frac{2500}{9} = \frac{50}{3}$ cm
  ½ mark
6. $P B = O P - O B = \frac{50}{3} - 6 = \frac{32}{3}$ cm $\approx 10.67$ cm.
  ½ mark
Answer: $P B = \frac{32}{3}$ cm ≈ 10.67 cm
Where the marks are
- Finds OB = 6 cm by Pythagoras (½ mark)
- Writes tan θ = 8/6 in triangle OBA (½ mark)
- Uses tangent ⟂ radius to write tan θ = AP/10 (½ mark)
- Gets AP = 40/3 cm (½ mark)
- Gets OP = 50/3 cm (½ mark)
- Gets PB = OP − OB = 32/3 cm (½ mark)
- Alternative: $P A^{2} = P B^{2} + 64$ and $P A^{2} = (P B + 6)^{2} - 100$ , equated to give PB = 32/3 cm (½ + ½ + 1 + 1)
Open Q31(b) on its own page

Section D

Long answer (LA)

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerStatistics
The median of the following data is 137. Find the values of x and y, given that total of frequencies is 68.
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Class f cf
  65–85 4 4
  85–105 5 9
  105–125 x 9 + x
  125–145 20 29 + x
  145–165 14 43 + x
  165–185 y 43 + x + y
  185–205 4 47 + x + y
  $n = 68$
  1½ marks
2. The median is 137, which lies in the class 125–145. So the median class is 125–145, with:
  $l = 125$
  $f = 20$
  $c f = 9 + x$
  $h = 20$
  $\frac{n}{2} = 34$
  ½ mark
3. Median $= l + (\frac{\frac{n}{2} - c f}{f}) \times h$ :
  $137 = 125 + (\frac{34 - ( 9 + x )}{20}) \times 20$
  1 mark
4. $\Rightarrow 12 = 25 - x \Rightarrow x = 13$
  1 mark
5. Total frequency: $47 + x + y = 68$
  $\Rightarrow 47 + 13 + y = 68 \Rightarrow y = 8$
  1 mark
Answer: x = 13, y = 8
Where the marks are
- Correct cumulative frequency table (1½ marks)
- Identifies the median class 125–145 (½ mark)
- Substitutes correctly in the median formula (1 mark)
- Gets x = 13 (1 mark)
- Uses total 68 to get y = 8 (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerStatistics
Find mean and mode of the following distribution :
Try it first, then show our solutionSolution
1. Step-deviation table with $a = 35$ , $h = 10$ , $u_{i} = \frac{x _{i} - 35}{10}$ :
  Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  0–10 5 3 −3 −9
  10–20 15 6 −2 −12
  20–30 25 11 −1 −11
  30–40 35 10 0 0
  40–50 45 13 1 13
  50–60 55 3 2 6
  60–70 65 4 3 12
  Total 50 −1
  2 marks
2. Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 35 + \frac{- 1}{50} \times 10 = 35 - 0.2 = 34.8$
  1 mark
3. The highest frequency is 13, so the modal class is 40–50, with:
  $l = 40$
  $f_{1} = 13$
  $f_{0} = 10$
  $f_{2} = 3$
  $h = 10$
  ½ mark
4. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 40 + (\frac{13 - 10}{26 - 10 - 3}) \times 10 = 40 + \frac{30}{13}$
  1 mark
5. $\approx 42.3$
  ½ mark
Answer: Mean = 34.8; Mode ≈ 42.3
Where the marks are
- Correct table with $\sum f_{i} = 50$ and $\sum f_{i} u_{i} = - 1$ (2 marks)
- Mean = 34.8 (1 mark)
- Identifies the modal class 40–50 (½ mark)
- Substitutes correctly in the mode formula (1 mark)
- Mode ≈ 42.3 (½ mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerTriangles
In $Δ$ ABC, AD is a median. X is a point on AD such that AX : XD = 2 : 3. BX is extended so that it intersects AC at Y. Prove that BX = 4 XY.
Figure in the paper: Triangle ABC with median AD to side BC (D is midpoint, BD = DC marked). X on AD; line BX extended meets AC at Y. See it in the official paper.
Try it first, then show our solutionSolution
1. Given: In $Δ A B C$ , AD is a median, so $B D = D C$ . X lies on AD with $A X : X D = 2 : 3$ . BX produced meets AC at Y.
  To prove: $B X = 4 X Y$ .
2. Construction: Through D, draw $D Z ∥ B Y$ meeting AC at Z.
  ½ mark
3. In $Δ C B Y$ , $D Z ∥ B Y$ . By the Basic Proportionality Theorem,
  $\frac{C D}{D B} = \frac{C Z}{Z Y}$
  Since D is the midpoint of BC, $\frac{C D}{D B} = 1$ , so $\frac{C Z}{Z Y} = 1$ and Z is the midpoint of CY.
  1 mark
4. In $Δ C B Y$ , D and Z are the midpoints of CB and CY. By the midpoint theorem, $D Z = \frac{1}{2} B Y$ , that is, $B Y = 2 D Z$ ...(i)
  ½ mark
5. In $Δ A D Z$ , $X Y ∥ D Z$ (XY is part of BY).
  So $∠ A X Y = ∠ A D Z$ and $∠ A Y X = ∠ A Z D$ [corresponding angles], with $∠ A$ common.
  Hence $Δ A X Y \sim Δ A D Z$ [AA similarity], which gives
  $\frac{A X}{A D} = \frac{X Y}{D Z}$
  1 mark
6. Since $A X : X D = 2 : 3$ , we have $A D = 2 + 3 = 5$ parts, so $\frac{A X}{A D} = \frac{2}{5}$ .
  Therefore $\frac{2}{5} = \frac{X Y}{D Z}$ , which gives $D Z = \frac{5}{2} X Y$ ...(ii)
  ½ mark
7. Substituting (ii) in (i):
  $B Y = 2 \times \frac{5}{2} X Y = 5 X Y$
  1 mark
8. Now $B X = B Y - X Y = 5 X Y - X Y = 4 X Y$ .
  Hence proved.
  ½ mark
Answer: Proved: BX = 4 XY (using DZ ∥ BY, BY = 2DZ and DZ = (5/2)XY).
Where the marks are
- Construction $D Z ∥ B Y$ (½ mark)
- Uses BPT in $Δ C B Y$ to get $C Z / Z Y = 1$ (1 mark)
- Gets $D Z = \frac{1}{2} B Y$ (½ mark)
- Shows $Δ A X Y \sim Δ A D Z$ and writes $A X / A D = X Y / D Z$ (1 mark)
- Gets $D Z = \frac{5}{2} X Y$ (½ mark)
- Combines to get $B Y = 5 X Y$ (1 mark)
- Concludes $B X = 4 X Y$ (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerPair of Linear Equations in Two Variables
Solve the following system of equations graphically : $x - 2 y = 3$ , $3 x - 8 y = 7$
Try it first, then show our solutionSolution
1. Line 1: $x - 2 y = 3$ , so $y = \frac{x - 3}{2}$ .
  x 3 5 1
  y 0 1 −1
  Plot $(3, 0)$ , $(5, 1)$ and $(1, - 1)$ on graph paper and join them to get the line $x - 2 y = 3$ .
  2 marks
2. Line 2: $3 x - 8 y = 7$ , so $y = \frac{3 x - 7}{8}$ .
  x 5 −3 1
  y 1 −2 −0.5
  Plot $(5, 1)$ , $(- 3, - 2)$ and $(1, - 0.5)$ and join them to get the line $3 x - 8 y = 7$ .
  2 marks
3. The two lines intersect at the point $(5, 1)$ .
  Check: $5 - 2 (1) = 3$ ✓ and $3 (5) - 8 (1) = 7$ ✓.
  Hence the solution is $x = 5$ , $y = 1$ .
  1 mark
Answer: x = 5, y = 1 (lines intersect at (5, 1))
Where the marks are
- Correct table and graph of $x - 2 y = 3$ (2 marks)
- Correct table and graph of $3 x - 8 y = 7$ (2 marks)
- Reads off the solution $x = 5$ (½ mark) and $y = 1$ (½ mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerPair of Linear Equations in Two Variables
Five years ago, Adil was thrice as old as Bharat. Ten years later Adil shall be twice as old as Bharat. To know the present ages of Adil and Bharat : (i) form the linear equations representing the above information. (ii) show that the system of equations is consistent with unique solution. (iii) find the present ages of Adil and Bharat.
Try it first, then show our solutionSolution
1. Let the present ages of Adil and Bharat be $x$ years and $y$ years respectively.
2. (i) Five years ago: $x - 5 = 3 (y - 5)$
  $\Rightarrow x - 5 = 3 y - 15 \Rightarrow x - 3 y = - 10$ ...(1)
  1 mark
3. Ten years later: $x + 10 = 2 (y + 10)$
  $\Rightarrow x + 10 = 2 y + 20 \Rightarrow x - 2 y = 10$ ...(2)
  1 mark
4. (ii) Write the equations as $x - 3 y + 10 = 0$ and $x - 2 y - 10 = 0$ .
  Then $\frac{a _{1}}{a _{2}} = \frac{1}{1} = 1$ and $\frac{b _{1}}{b _{2}} = \frac{- 3}{- 2} = \frac{3}{2}$ .
  Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect at one point. So the system is consistent with a unique solution.
  1 mark
5. (iii) Subtract (1) from (2):
  $(x - 2 y) - (x - 3 y) = 10 - (- 10) \Rightarrow y = 20$
  1 mark
6. Substitute $y = 20$ in (2): $x - 40 = 10 \Rightarrow x = 50$ .
  So Adil is 50 years old and Bharat is 20 years old.
  1 mark
Answer: (i) x − 3y = −10, x − 2y = 10; (ii) a₁/a₂ = 1 ≠ b₁/b₂ = 3/2, so consistent with a unique solution; (iii) Adil 50 years, Bharat 20 years
Where the marks are
- Forms $x - 3 y = - 10$ (1 mark)
- Forms $x - 2 y = 10$ (1 mark)
- Compares ratios $a_{1} / a_{2} \neq = b_{1} / b_{2}$ and concludes a unique solution (1 mark)
- Finds $y = 20$ (1 mark) and $x = 50$ (1 mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A boy standing on a horizontal plane is flying a kite with a string of length 60 m, at an angle of elevation of $3 0^{\circ}$ . Another boy standing on the roof of a 20 m high building, finds the angle of elevation of same kite to be $4 5^{\circ}$ . If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure:
  K is the kite and B is the first boy on the ground, with string $K B = 60$ m and $∠ K B P = 3 0^{\circ}$ .
  CM is the 20 m building, with the second boy at M.
  KP $⊥$ ground at P, and $M L ∥ C P$ meets KP at L, so $L P = M C = 20$ m and $M L = C P$ .
  Let $K L = h$ , $C P = M L = x$ and $P B = y$ . The angle of elevation $∠ K M L = 4 5^{\circ}$ .
  C (building) and B (first boy) lie on opposite sides of P.
  1 mark
2. In right $Δ K P B$ :
  $\frac{K P}{K B} = sin 3 0^{\circ} \Rightarrow \frac{h + 20}{60} = \frac{1}{2}$
  ½ mark
3. $\Rightarrow h + 20 = 30 \Rightarrow h = 10$ m
  ½ mark
4. Also, in $Δ K P B$ :
  $\frac{P B}{K B} = cos 3 0^{\circ} \Rightarrow \frac{y}{60} = \frac{3}{2}$
  ½ mark
5. $\Rightarrow y = 303 = 30 \times 1.73 = 51.9$ m
  ½ mark
6. In right $Δ K L M$ :
  $\frac{K L}{M L} = tan 4 5^{\circ} \Rightarrow \frac{10}{x} = 1$
  ½ mark
7. $\Rightarrow x = 10$ m
  ½ mark
8. Height of the kite from the ground: $K P = h + 20 = 10 + 20 = 30$ m.
  ½ mark
9. Distance of the first boy from the base of the building: $C B = C P + P B = 10 + 51.9 = 61.9$ m.
  ½ mark
Answer: Distance of first boy from base of building = 61.9 m; height of kite = 30 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $sin 3 0^{\circ} = \frac{h + 20}{60}$ (½ mark) and gets $h = 10$ (½ mark)
- Uses $cos 3 0^{\circ} = \frac{y}{60}$ (½ mark) and gets $y = 51.9$ m (½ mark)
- Uses $tan 4 5^{\circ} = \frac{10}{x}$ (½ mark) and gets $x = 10$ (½ mark)
- Height of kite 30 m (½ mark)
- Distance 61.9 m (½ mark)
Open Q35 on its own page

Class	f	cf
65–85	4	4
85–105	5	9
105–125	x	9 + x
125–145	20	29 + x
145–165	14	43 + x
165–185	y	43 + x + y
185–205	4	47 + x + y

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
0–10	5	3	−3	−9
10–20	15	6	−2	−12
20–30	25	11	−1	−11
30–40	35	10	0	0
40–50	45	13	1	13
50–60	55	3	2	6
60–70	65	4	3	12
Total		50		−1

x	3	5	1
y	0	1	−1

x	5	−3	1
y	1	−2	−0.5

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedPolynomials
Case Study 1: During a theatre drama, a backdrop of building arches was used. The shape of the curve shown below can be represented by the polynomial $p (x) = - x^{2} + 2 x + 8$ , where x is the length (in feet) on stage level. Based on the figure given above, answer the following questions :
1. (i)
  Determine the height of the arch.
  1 mark
2. (ii)(a)
  Find zeroes of the polynomial p(x). Which points on the graph represent the zeroes?
  2 marks
3. OR
  (ii)(b)
  Find the span of the arch on the stage floor.
  2 marks
4. (iii)
  Write the coordinates of the point of intersection of the above curve with the y-axis.
  1 mark
Figure in the paper: Left: photograph of a stage backdrop with three arches. Right: graph of an arch with axes X'OX (horizontal) and Y'OY (vertical, Y = p(x)). The arch is a curve from B on the negative x-axis to A on the positive x-axis, with its top point C(1, y) on the curve; the curve meets the y-axis above O. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The highest point of the arch is $C (1, y)$ , so the height is $p (1)$ :
  $p (1) = - (1)^{2} + 2 (1) + 8 = - 1 + 2 + 8 = 9$
  So the height of the arch is 9 feet.
  1 mark
2. (ii)(a) Factorise:
  $p (x) = - x^{2} + 2 x + 8 = - (x^{2} - 2 x - 8) = - (x - 4) (x + 2)$
  1 mark
3. (ii)(a) Setting $p (x) = 0$ gives $x = 4$ or $x = - 2$ . So the zeroes are $- 2$ and $4$ .
  ½ mark
4. (ii)(a) The zeroes are where the curve meets the x-axis, i.e. the points $B (- 2, 0)$ and $A (4, 0)$ .
  ½ mark
5. (ii)(b) Factorise: $p (x) = - (x - 4) (x + 2)$ .
  ½ mark
6. (ii)(b) The zeroes are $- 2$ and $4$ , so the arch meets the floor at $B (- 2, 0)$ and $A (4, 0)$ .
  ½ mark
7. (ii)(b) Span: $A B = 4 - (- 2) = 6$ feet.
  1 mark
8. (iii) On the y-axis, $x = 0$ :
  $p (0) = - (0)^{2} + 2 (0) + 8 = 8$
  ½ mark
9. (iii) So the curve meets the y-axis at $(0, 8)$ .
  ½ mark
Answer: (i) 9 feet; (ii)(a) zeroes −2 and 4, shown by points B and A; (ii)(b) span 6 feet; (iii) (0, 8)
Where the marks are
- (i) Evaluates $p (1) = 9$ , height 9 feet (1 mark)
- (ii)(a) Factorises $p (x) = - (x - 4) (x + 2)$ (1 mark)
- (ii)(a) States zeroes $- 2$ , $4$ (½ mark)
- (ii)(a) Identifies points B and A (½ mark)
- (ii)(b) Factorises (½ mark) and finds zeroes (½ mark)
- (ii)(b) Span $A B = 6$ feet (1 mark)
- (iii) Evaluates $p (0) = 8$ (½ mark) and writes $(0, 8)$ (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedProbability
Case Study 2: A group of friends wanted to play cards with two identical packs together. While shuffling the cards, three cards are dropped. Rest of the cards are shuffled and one card is drawn at random. Assuming that the dropped cards were a queen of hearts, a ten of spades and an ace of clubs, answer the following questions :
1. (i)
  Find the probability that the drawn card is a face card.
  1 mark
2. (ii)
  Find the probability that the drawn card is either a king or a queen.
  1 mark
3. (iii)(a)
  Do you think that the probability of getting a queen was higher if none of the cards were dropped? Justify your answer.
  2 marks
4. OR
  (iii)(b)
  Find the probability that the drawn card is a jack. Compare it with the probability when none of the cards were dropped. In which case is the probability of getting a jack higher?
  2 marks
Figure in the paper: Photograph of a wooden box with two compartments, each containing a pack of playing cards. See it in the official paper.
Try it first, then show our solutionSolution
1. Total cards in two packs $= 2 \times 52 = 104$ . After 3 cards are dropped, cards left $= 104 - 3 = 101$ .
2. (i) Face cards in two packs $= 2 \times 12 = 24$ . One queen (of hearts) was dropped, so face cards left $= 23$ .
  $P (face card) = \frac{23}{101}$
  1 mark
3. (ii) Kings left $= 8$ and queens left $= 8 - 1 = 7$ , so favourable cards $= 15$ .
  $P (king or queen) = \frac{15}{101}$
  1 mark
4. (iii)(a) Yes, the probability was higher if no cards were dropped.
  ½ mark
5. (iii)(a) With no cards dropped: $P (queen) = \frac{8}{104}$ .
  ½ mark
6. (iii)(a) With cards dropped (7 queens left): $P (queen) = \frac{7}{101}$ .
  ½ mark
7. (iii)(a) Compare by cross-multiplying: $8 \times 101 = 808$ and $7 \times 104 = 728$ .
  Since $808 > 728$ , $\frac{8}{104} > \frac{7}{101}$ . So getting a queen was more likely when no cards were dropped.
  ½ mark
8. (iii)(b) No jack was dropped, so 8 jacks remain among 101 cards: $P (jack) = \frac{8}{101}$ .
  ½ mark
9. (iii)(b) With no cards dropped: $P (jack) = \frac{8}{104}$ .
  ½ mark
10. (iii)(b) The numerators are equal and $101 < 104$ , so $\frac{8}{101} > \frac{8}{104}$ .
  The probability of a jack is higher when the 3 cards were dropped.
  1 mark
Answer: (i) 23/101; (ii) 15/101; (iii)(a) Yes, 8/104 > 7/101; (iii)(b) 8/101, which is higher than 8/104 (higher when cards were dropped)
Where the marks are
- (i) $P (face card) = 23/101$ (1 mark)
- (ii) $P (king or queen) = 15/101$ (1 mark)
- (iii)(a) Says yes (½ mark)
- (iii)(a) Finds $8/104$ (½ mark)
- (iii)(a) Finds $7/101$ (½ mark)
- (iii)(a) Correct comparison with justification (½ mark)
- (iii)(b) Finds $8/101$ (½ mark)
- (iii)(b) Finds $8/104$ (½ mark)
- (iii)(b) Compares and concludes higher when cards dropped (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Case Study 3: A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled. The diameter of spherical ball is 21 cm. Cylindrical pool - Outer diameter is 50 cm and inner diameter is 40 cm. Height of solid base is 14 cm. Height of water filled is 7 cm. Observe the figure and answer the following questions :
1. (i)
  Determine the total height of the fountain.
  1 mark
2. (ii)
  Find the volume of the ball.
  1 mark
3. (iii)(a)
  If one-third of the ball is submerged in the water, find the volume of the water filled in the pool.
  2 marks
4. OR
  (iii)(b)
  Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.
  2 marks
Figure in the paper: Photograph of a leafy spherical ball resting in a cylindrical pool. Side diagram: sphere sitting in a cylindrical pool; water level marked 7 cm above the solid base; solid base height marked 14 cm; the ball partially immersed in the water. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The ball rests on top of the solid base. Total height = height of base + diameter of ball:
  $14 + 21 = 35 cm$
  1 mark
2. (ii) Radius of ball $r = \frac{21}{2}$ cm.
  $V = \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$
  ½ mark
3. (ii) $V = 4851 cm^{3}$
  ½ mark
4. (iii)(a) The inner radius of the pool is $20$ cm and the water height is $7$ cm. One-third of the ball's volume lies in this space.
  Volume of water = inner cylinder volume − $\frac{1}{3}$ of ball volume
  $= \frac{22}{7} \times 20 \times 20 \times 7 - \frac{1}{3} \times 4851$
  1 mark
5. (iii)(a) $= 8800 - 1617 = 7183 cm^{3}$
  1 mark
6. (iii)(b) The outer radius is $25$ cm and the outer height of the cylindrical part is $14 + 7 = 21$ cm.
  Required area = $2 π R h + 4 π r^{2}$
  $= 2 \times \frac{22}{7} \times 25 \times 21 + 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$
  1 mark
7. (iii)(b) $= 3300 + 1386 = 4686 cm^{2}$
  1 mark
Answer: (i) 35 cm; (ii) 4851 cm³; (iii)(a) 7183 cm³; (iii)(b) 4686 cm²
Where the marks are
- (i) Total height $14 + 21 = 35$ cm (1 mark)
- (ii) Correct substitution in $\frac{4}{3} π r^{3}$ (½ mark) and gets 4851 cm³ (½ mark)
- (iii)(a) Inner cylinder volume minus one-third of ball volume, correctly substituted (1 mark)
- (iii)(a) Gets 7183 cm³ (1 mark)
- (iii)(b) Correct expression $2 π \times 25 \times 21 + 4 π (10.5)^{2}$ (1 mark)
- (iii)(b) Gets 4686 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.