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

CBSE Class 10 Mathematics Standard: 2025 board paper, set 30/2/1

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

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

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2025/X/041_Mathematics_Standard.zip (a zip; this paper is the file 30-2-1_Mathematics Standard.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

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

Instructions in brief: All questions compulsory; no overall choice; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 case studies of Section E (2-mark alternatives within each case study).

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQIntroduction to Trigonometry
If $7 cos^{2} θ + 3 sin^{2} θ = 4$ , then the value of $θ$ is :
1. (A) $3 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. Write $sin^{2} θ = 1 - cos^{2} θ$ . Then $7 cos^{2} θ + 3 (1 - cos^{2} θ) = 4$ , so $4 cos^{2} θ + 3 = 4$ .
  ½ mark
2. This gives $cos^{2} θ = \frac{1}{4}$ , so $cos θ = \frac{1}{2}$ and $θ = 6 0^{\circ}$ . Option (C).
  ½ mark
Answer: $θ = 6 0^{\circ}$ , option (C)
Where the marks are
- Reduces the equation to $cos^{2} θ = \frac{1}{4}$ and picks $6 0^{\circ}$ , option (C) (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQProbability
The probability of drawing an even prime number out of numbers from 1 to 30 is :
1. (A) $\frac{1}{30}$
2. (B) $\frac{4}{15}$
3. (C) $\frac{7}{30}$
4. (D) $0$
Try it first, then show our solutionSolution
1. There are 30 equally likely numbers. The only even prime number is $2$ , so there is 1 favourable outcome.
  ½ mark
2. $P = \frac{1}{30}$ . Option (A).
  ½ mark
Answer: $\frac{1}{30}$ , option (A)
Where the marks are
- Identifies 2 as the only even prime and gets $\frac{1}{30}$ , option (A) (1 mark)
Q3 · 1 mark · MCQQuadratic Equations
The quadratic equation whose roots are $7$ and $\frac{1}{7}$ is :
1. (A) $7 x^{2} - 50 x + 7 = 0$
2. (B) $7 x^{2} - 50 x + 1 = 0$
3. (C) $7 x^{2} + 50 x - 7 = 0$
4. (D) $7 x^{2} + 50 x - 1 = 0$
Try it first, then show our solutionSolution
1. Sum of roots $= 7 + \frac{1}{7} = \frac{50}{7}$ . Product of roots $= 7 \times \frac{1}{7} = 1$ .
  ½ mark
2. The equation is $x^{2} - \frac{50}{7} x + 1 = 0$ . Multiplying by 7 gives $7 x^{2} - 50 x + 7 = 0$ . Option (A).
  ½ mark
Answer: $7 x^{2} - 50 x + 7 = 0$ , option (A)
Where the marks are
- Uses $x^{2} - (sum) x + product = 0$ to reach option (A) (1 mark)
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Q4 · 1 mark · MCQReal Numbers
The least number which is a perfect square and is divisible by each of $16$ , $20$ and $50$ , is :
1. (A) $1200$
2. (B) $100$
3. (C) $3600$
4. (D) $2400$
Try it first, then show our solutionSolution
1. Factorise: $16 = 2^{4}$ , $20 = 2^{2} \times 5$ , $50 = 2 \times 5^{2}$ . So $LCM = 2^{4} \times 5^{2} = 400$ .
  ½ mark
2. $400 = 2 0^{2}$ is already a perfect square. So the least such number is $400$ , which is not among the options. Of the options, only $3600 = 6 0^{2}$ is a perfect square divisible by all three numbers, but it is not the least.
  ½ mark
Answer: 400 (no option is correct)
Where the marks are
- Finds the LCM as 400 and notes that it is a perfect square, so no option is correct. The board gives the 1 mark for any attempt (1 mark)
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Q5 · 1 mark · MCQCoordinate Geometry
The coordinates of the end points of a diameter of a circle are $(5, - 2)$ and $(5, 2)$ . The length of the radius of the circle is :
1. (A) $\pm 2$
2. (B) $\pm 4$
3. (C) $4$
4. (D) $2$
Try it first, then show our solutionSolution
1. Diameter $= (5 - 5)^{2} + (2 - (- 2))^{2} = 16 = 4$ units.
  ½ mark
2. Radius $= \frac{4}{2} = 2$ units. A length cannot be negative, so $\pm$ is ruled out. Option (D).
  ½ mark
Answer: 2, option (D)
Where the marks are
- Finds the diameter as 4 using the distance formula and halves it to get radius 2, option (D) (1 mark)
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Q6 · 1 mark · MCQCoordinate Geometry
The points $(- 5, 0)$ , $(5, 0)$ and $(0, 4)$ are the vertices of a triangle which is a/an :
1. (A)right-angled triangle
2. (B)isosceles triangle
3. (C)equilateral triangle
4. (D)scalene triangle
Try it first, then show our solutionSolution
1. Let $A (- 5, 0)$ , $B (5, 0)$ , $C (0, 4)$ . Then $A C = 25 + 16 = 41$ , $B C = 25 + 16 = 41$ and $A B = 10$ .
  ½ mark
2. Two sides are equal. Also $41 + 41 \neq = 100$ , so the triangle is not right-angled. It is an isosceles triangle. Option (B).
  ½ mark
Answer: Isosceles triangle, option (B)
Where the marks are
- Computes the sides $41, 41, 10$ and concludes isosceles, option (B) (1 mark)
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Q7 · 1 mark · MCQCircles
In the given figure, RS is the tangent to the circle at the point L and MN is the diameter. If $∠ N M L = 3 0^{\circ}$ , then $∠ R L M$ is :
1. (A) $3 0^{\circ}$
2. (B) $6 0^{\circ}$
3. (C) $9 0^{\circ}$
4. (D) $12 0^{\circ}$
Figure in the paper: A circle with centre O and horizontal diameter MN (M on the left, N on the right). Point L is on the upper part of the circle. Line RS is the tangent at L, with R on the upper-left side and S on the lower-right side. Chord LM is drawn. The angle at M between MN and ML is marked $3 0^{\circ}$ ; the angle at L between LR and LM is marked to be found. See it in the official paper.
Try it first, then show our solutionSolution
1. Join OL. Since $O L = O M$ (radii), $∠ O L M = ∠ O M L = 3 0^{\circ}$ .
  ½ mark
2. $∠ O L R = 9 0^{\circ}$ [tangent $⊥$ radius]. So $∠ R L M = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ . Option (B).
  ½ mark
Answer: $6 0^{\circ}$ , option (B)
Where the marks are
- Uses the isosceles triangle OLM and tangent $⊥$ radius to get $6 0^{\circ}$ , option (B) (1 mark)
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Q8 · 1 mark · MCQTriangles
In the given figure, $P Q ∥ B C$ . If $\frac{A P}{P B} = \frac{4}{13}$ and $A C = 20.4$ cm, then the length of AQ is :
1. (A) $2.8$ cm
2. (B) $5.8$ cm
3. (C) $3.8$ cm
4. (D) $4.8$ cm
Figure in the paper: Triangle ABC with apex A at the top, B at bottom-left and C at bottom-right. P lies on AB and Q lies on AC, with segment PQ drawn parallel to base BC. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $P Q ∥ B C$ , by the Basic Proportionality Theorem $\frac{A Q}{QC} = \frac{A P}{P B} = \frac{4}{13}$ . So $\frac{A Q}{A C} = \frac{4}{17}$ .
  ½ mark
2. $A Q = \frac{4}{17} \times 20.4 = 4.8$ cm. Option (D).
  ½ mark
Answer: 4.8 cm, option (D)
Where the marks are
- Applies BPT to get $A Q = \frac{4}{17} \times 20.4 = 4.8$ cm, option (D) (1 mark)
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Q9 · 1 mark · MCQTriangles
Which of the following statements is incorrect ?
1. (A)Two congruent figures are always similar.
2. (B)A square and a rhombus of the same area are always similar.
3. (C)Two equilateral triangles are always similar.
4. (D)Two similar triangles need not be congruent.
Try it first, then show our solutionSolution
1. (A), (C) and (D) are true facts about similarity. Congruent figures have the same shape, all equilateral triangles have equal angles of $6 0^{\circ}$ , and similar triangles may differ in size.
  ½ mark
2. A rhombus need not have right angles, so its angles may not match those of a square. Equal area does not make the two figures similar. So (B) is incorrect.
  ½ mark
Answer: Option (B)
Where the marks are
- Recognises that a rhombus's angles need not equal $9 0^{\circ}$ , so (B) is the incorrect statement (1 mark)
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Q10 · 1 mark · MCQReal Numbers
The sum of the exponents of prime factors in the prime factorisation of $4004$ is :
1. (A) $5$
2. (B) $4$
3. (C) $3$
4. (D) $2$
Try it first, then show our solutionSolution
1. $4004 = 4 \times 1001 = 2^{2} \times 7 \times 11 \times 13$ .
  ½ mark
2. Sum of exponents $= 2 + 1 + 1 + 1 = 5$ . Option (A).
  ½ mark
Answer: 5, option (A)
Where the marks are
- Writes $4004 = 2^{2} \times 7 \times 11 \times 13$ and adds the exponents to get 5, option (A) (1 mark)
Q11 · 1 mark · MCQCompetency-basedProbability
In a cricket match, a batsman hits the boundary 7 times out of the 42 balls he plays. The probability of his not hitting a boundary is :
1. (A) $\frac{1}{7}$
2. (B) $\frac{2}{7}$
3. (C) $\frac{5}{6}$
4. (D) $\frac{1}{6}$
Try it first, then show our solutionSolution
1. Balls without a boundary $= 42 - 7 = 35$ .
  ½ mark
2. $P (not a boundary) = \frac{35}{42} = \frac{5}{6}$ . Option (C).
  ½ mark
Answer: $\frac{5}{6}$ , option (C)
Where the marks are
- Gets $\frac{35}{42} = \frac{5}{6}$ , option (C) (1 mark)
Q12 · 1 mark · MCQCompetency-basedAreas Related to Circles
If a large circular pizza is divided into 5 equal sectors, then the central angle of each sector will be :
1. (A) $6 0^{\circ}$
2. (B) $9 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $7 2^{\circ}$
Try it first, then show our solutionSolution
1. The complete angle at the centre is $36 0^{\circ}$ , shared equally by 5 sectors.
2. Central angle $= \frac{36 0 ^{\circ}}{5} = 7 2^{\circ}$ , so option (D).
  1 mark
Answer: $7 2^{\circ}$ , option (D)
Where the marks are
- Divides $36 0^{\circ}$ by 5 to get $7 2^{\circ}$ (1 mark)
Q13 · 1 mark · MCQIntroduction to Trigonometry
If $sin 3 0^{\circ} tan 4 5^{\circ} = \frac{s e c 6 0 ^{\circ}}{k}$ , then the value of k is :
1. (A) $4$
2. (B) $3$
3. (C) $2$
4. (D) $1$
Try it first, then show our solutionSolution
1. $sin 3 0^{\circ} = \frac{1}{2}$ , $tan 4 5^{\circ} = 1$ , $sec 6 0^{\circ} = 2$ .
2. $\frac{1}{2} \times 1 = \frac{2}{k} \Rightarrow k = 4$ , option (A).
  1 mark
Answer: k = 4, option (A)
Where the marks are
- Substitutes standard values and gets $k = 4$ (1 mark)
Q14 · 1 mark · MCQPair of Linear Equations in Two Variables
The line represented by the equation $x - y = 0$ is :
1. (A)parallel to x-axis
2. (B)parallel to y-axis
3. (C)passing through the origin
4. (D)passing through the point $(3, 2)$
Try it first, then show our solutionSolution
1. The line is $y = x$ . The point $(0, 0)$ satisfies $0 - 0 = 0$ , so the line passes through the origin.
  ½ mark
2. It is neither horizontal nor vertical, and $(3, 2)$ gives $3 - 2 = 1 \neq = 0$ . So the answer is option (C).
  ½ mark
Answer: Passing through the origin, option (C)
Where the marks are
- Checks that $(0, 0)$ satisfies the equation and picks option (C) (1 mark)
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Q15 · 1 mark · MCQArithmetic Progressions
The $1 0^{th}$ term of the AP $5, \frac{19}{4}, \frac{9}{2}, \frac{17}{4}, \dots$ is :
1. (A) $\frac{11}{4}$
2. (B) $\frac{4}{11}$
3. (C) $\frac{13}{4}$
4. (D) $\frac{4}{13}$
Try it first, then show our solutionSolution
1. $a = 5$ and $d = \frac{19}{4} - 5 = - \frac{1}{4}$ .
  ½ mark
2. $a_{10} = a + 9 d = 5 - \frac{9}{4} = \frac{11}{4}$ . Option (A).
  ½ mark
Answer: $\frac{11}{4}$ , option (A)
Where the marks are
- Finds $d = - \frac{1}{4}$ and uses $a_{10} = a + 9 d = \frac{11}{4}$ , option (A) (1 mark)
Q16 · 1 mark · MCQPolynomials
If $- 4$ is a zero of the polynomial $p (x) = x^{2} - x - (2 + 2 k)$ , then the value of k is :
1. (A) $3$
2. (B) $9$
3. (C) $6$
4. (D) $- 9$
Try it first, then show our solutionSolution
1. $p (- 4) = 16 + 4 - 2 - 2 k = 0$ .
  ½ mark
2. $18 - 2 k = 0$ , so $k = 9$ . Option (B).
  ½ mark
Answer: $k = 9$ , option (B)
Where the marks are
- Sets $p (- 4) = 0$ and solves to get $k = 9$ , option (B) (1 mark)
Q17 · 1 mark · MCQCoordinate Geometry
The equation of a line parallel to the x-axis and at a distance of 3 units below x-axis is :
1. (A) $x = 3$
2. (B) $x = - 3$
3. (C) $y = - 3$
4. (D) $y = 3$
Try it first, then show our solutionSolution
1. A line parallel to the x-axis has every point at the same y-coordinate, so its equation is $y = c$ .
2. 3 units below the x-axis means $y = - 3$ , option (C).
  1 mark
Answer: $y = - 3$ , option (C)
Where the marks are
- Identifies the line as $y = - 3$ (1 mark)
Q18 · 1 mark · MCQReal Numbers
The HCF of $40$ , $110$ and $360$ is :
1. (A) $40$
2. (B) $110$
3. (C) $360$
4. (D) $10$
Try it first, then show our solutionSolution
1. $40 = 2^{3} \times 5$ , $110 = 2 \times 5 \times 11$ , $360 = 2^{3} \times 3^{2} \times 5$ .
  ½ mark
2. Common primes with their lowest powers: $2 \times 5 = 10$ . So HCF $= 10$ . Option (D).
  ½ mark
Answer: 10, option (D)
Where the marks are
- Prime factorises all three numbers and takes the lowest common powers to get 10, option (D) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedArithmetic Progressions
Assertion (A): Common difference of the AP : $5, 1, - 3, - 7, \dots$ is $4$ .
Reason (R): Common difference of the AP : $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ is obtained by $d = a_{n} - a_{n - 1}$ .
Select the correct answer from the codes (A), (B), (C) and (D) given below.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Assertion: $d = 1 - 5 = - 4$ , not $4$ , so (A) is false.
  ½ mark
2. Reason: in an AP, $d = a_{n} - a_{n - 1}$ (any term minus its preceding term), so (R) is true. Hence option (D).
  ½ mark
Answer: Option (D)
Where the marks are
- Finds $d = - 4$ , so Assertion false; Reason true; chooses (D) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedPair of Linear Equations in Two Variables
Assertion (A): The pair of linear equations $p x + 3 y + 59 = 0$ and $2 x + 6 y + 118 = 0$ will have infinitely many solutions if $p = 1$ .
Reason (R): If the pair of linear equations $p x + 3 y + 19 = 0$ and $2 x + 6 y + 157 = 0$ has a unique solution, then $p \neq = 1$ .
Select the correct answer from the codes (A), (B), (C) and (D) given below.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Assertion: for infinitely many solutions, $\frac{p}{2} = \frac{3}{6} = \frac{59}{118} = \frac{1}{2}$ . This gives $p = 1$ , so A is true.
  ½ mark
2. Reason: a unique solution needs $\frac{p}{2} \neq = \frac{3}{6}$ , so $p \neq = 1$ . R is true. But R is about a different pair of equations and a different condition, so it does not explain A. Option (B).
  ½ mark
Answer: Option (B)
Where the marks are
- Shows both A and R are true using the ratio conditions, and that R does not explain A, option (B) (1 mark)
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Section B

Very short answer (VSA)

Q21 · 2 marks · Very short answerPolynomials
If p and q are zeroes of the polynomial $p (y) = 21 y^{2} - y - 2$ , then find the value of $(1 - p) \cdot (1 - q)$ .
Try it first, then show our solutionSolution
1. Here $a = 21$ , $b = - 1$ , $c = - 2$ . Sum of zeroes: $p + q = - \frac{b}{a} = \frac{1}{21}$ .
  ½ mark
2. Product of zeroes: $pq = \frac{c}{a} = - \frac{2}{21}$ .
  ½ mark
3. $(1 - p) (1 - q) = 1 - (p + q) + pq = 1 - \frac{1}{21} - \frac{2}{21}$ .
  ½ mark
4. $= \frac{21 - 1 - 2}{21} = \frac{18}{21} = \frac{6}{7}$ .
  ½ mark
Answer: $(1 - p) (1 - q) = \frac{6}{7}$
Where the marks are
- Writes $p + q = \frac{1}{21}$ (½ mark)
- Writes $pq = - \frac{2}{21}$ (½ mark)
- Expands to $1 - (p + q) + pq$ and substitutes (½ mark)
- Gets $\frac{18}{21} = \frac{6}{7}$ (½ mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerCompetency-basedAreas Related to Circles
In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and $∠ A O B = 9 0^{\circ}$ . If $A O = O B = 42$ cm, then find the perimeter of the top of the table.
Figure in the paper: A major sector of a circle with centre O. Radii OA and OB form a right angle at O (marked with a square). A is at the lower-left and B at the lower-right; the large arc AB runs over the top. The sector is the reflex-angled shape of the table top. See it in the official paper.
Try it first, then show our solutionSolution
1. The table top is the major sector. Its angle is reflex $∠ A O B = 36 0^{\circ} - 9 0^{\circ} = 27 0^{\circ}$ .
  ½ mark
2. Perimeter $=$ length of major arc $+ 2 r = \frac{270}{360} \times 2 \times \frac{22}{7} \times 42 + 2 \times 42 = 198 + 84$ .
  1 mark
3. Perimeter $= 282$ cm.
  ½ mark
Answer: 282 cm
Where the marks are
- Finds the reflex angle as $27 0^{\circ}$ (½ mark)
- Uses perimeter $=$ arc length $\frac{θ}{360} \times 2 π r$ plus $2 r$ with correct substitution (1 mark)
- Gets 282 cm (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerAreas Related to Circles
In the given figure, three sectors of a circle of radius 5 cm, making angles $3 5^{\circ}$ , $5 0^{\circ}$ and $9 5^{\circ}$ at the centre are shaded. Find the area of the shaded region. [Use $π = \frac{22}{7}$ ]
Figure in the paper: A circle of radius 5 cm divided by radii into sectors. Three sectors are shaded: one labelled $9 5^{\circ}$ at the upper-left, one labelled $5 0^{\circ}$ at the right-lower side, and one labelled $3 5^{\circ}$ at the lower-left. See it in the official paper.
Try it first, then show our solutionSolution
1. Shaded area $= \frac{95}{360} π r^{2} + \frac{50}{360} π r^{2} + \frac{35}{360} π r^{2}$ , with $r = 5$ cm.
  1 mark
2. $= \frac{95 + 50 + 35}{360} \times \frac{22}{7} \times 5^{2} = \frac{180}{360} \times \frac{22}{7} \times 25$ .
  ½ mark
3. $= \frac{275}{7}$ cm $^{2} \approx 39.29$ cm $^{2}$ .
  ½ mark
Answer: $\frac{275}{7}$ cm² ≈ 39.29 cm²
Where the marks are
- Writes the sum of the three sector areas using $\frac{θ}{360} π r^{2}$ (1 mark)
- Combines the angles to $18 0^{\circ}$ (½ mark)
- Gets $\frac{275}{7}$ cm² (½ mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerIntroduction to Trigonometry
If $tan A = 3$ ; where A is an acute angle, then find the value of $\frac{s i n ^{2} A}{1 + c o s ^{2} A}$ .
Try it first, then show our solutionSolution
1. $tan A = 3 = tan 6 0^{\circ}$ , so $A = 6 0^{\circ}$ (A is acute).
  ½ mark
2. $\frac{s i n ^{2} 6 0 ^{\circ}}{1 + c o s ^{2} 6 0 ^{\circ}} = \frac{( \frac{3}{2} ) ^{2}}{1 + ( \frac{1}{2} ) ^{2}} = \frac{\frac{3}{4}}{\frac{5}{4}}$ .
  1 mark
3. $= \frac{3}{5}$ .
  ½ mark
Answer: $\frac{3}{5}$
Where the marks are
- Finds $A = 6 0^{\circ}$ (½ mark)
- Substitutes $sin 6 0^{\circ}$ and $cos 6 0^{\circ}$ correctly (1 mark)
- Gets $\frac{3}{5}$ (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerTriangles
In the given figure, D is a point on the side BC of $△ A B C$ such that $∠ A D C = ∠ B A C$ . Show that $C A^{2} = C D \cdot C B$ .
Figure in the paper: Triangle ABC with A at the top, B at bottom-left and C at bottom-right. D lies on side BC between B and C. Segment AD is drawn. The angle at A (BAC) and the angle at D (ADC) are marked. See it in the official paper.
Try it first, then show our solutionSolution
1. Compare $△ A C D$ and $△ B C A$ :
  $∠ A D C = ∠ B A C$ [given]
  $∠ A C D = ∠ B C A$ [common angle $C$ ]
  So $△ A C D \sim △ B C A$ [AA similarity criterion].
  1 mark
2. Corresponding sides of similar triangles are proportional (A↔B, C↔C, D↔A):
  $\frac{C A}{C B} = \frac{C D}{C A}$
  ½ mark
3. Cross-multiplying gives $C A^{2} = C D \cdot C B$ . Hence proved.
  ½ mark
Answer: $△ A C D \sim △ B C A$ (AA), so $\frac{C A}{C B} = \frac{C D}{C A}$ , which gives $C A^{2} = C D \cdot C B$
Where the marks are
- Proves $△ A C D \sim △ B C A$ by AA, with the given angle and the common angle $C$ (1 mark)
- Writes the correct ratio of corresponding sides $\frac{C A}{C B} = \frac{C D}{C A}$ (1/2 mark)
- Concludes $C A^{2} = C D \cdot C B$ (1/2 mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerTriangles
In the given figure, $O A \cdot O B = O C \cdot O D$ . Show that $∠ A = ∠ C$ and $∠ B = ∠ D$ .
Figure in the paper: Two line segments AB and CD intersect at point O. A is at the upper-left, D at the lower-left, C at the upper-right and B at the lower-right. Triangles OAD and OCB are formed by joining A to D and C to B. See it in the official paper.
Try it first, then show our solutionSolution
1. Given $O A \cdot O B = O C \cdot O D$ .
  Dividing both sides by $O C \cdot O B$ : $\frac{O A}{O C} = \frac{O D}{O B}$ .
  ½ mark
2. In $△ A O D$ and $△ C O B$ : $∠ A O D = ∠ C O B$ [vertically opposite angles].
  ½ mark
3. Two sides are proportional and the included angles are equal, so $△ A O D \sim △ C O B$ [SAS similarity criterion].
  ½ mark
4. Corresponding angles of similar triangles are equal (A↔C, D↔B). Therefore $∠ A = ∠ C$ and $∠ D = ∠ B$ . Hence proved.
  ½ mark
Answer: $△ A O D \sim △ C O B$ (SAS), so $∠ A = ∠ C$ and $∠ B = ∠ D$
Where the marks are
- Rewrites the given product as the ratio $\frac{O A}{O C} = \frac{O D}{O B}$ (1/2 mark)
- States $∠ A O D = ∠ C O B$ as vertically opposite angles (1/2 mark)
- Concludes $△ A O D \sim △ C O B$ by SAS (1/2 mark)
- Deduces $∠ A = ∠ C$ and $∠ B = ∠ D$ (1/2 mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerCircles
At point A on the diameter AB of a circle of radius 10 cm, tangent XAY is drawn to the circle. Find the length of the chord CD parallel to XY at a distance of 16 cm from A.
Figure in the paper: A circle with centre O and horizontal diameter BA (B on left, A on right); OA is marked 10 cm. Vertical tangent line XAY passes through A (X above, Y below). A vertical chord CD is drawn to the left of O, meeting diameter AB at point P; C is on the upper arc and D on the lower arc; OC and OD are joined. See it in the official paper.
Try it first, then show our solutionSolution
1. Chord CD meets diameter AB at P, and $A P = 16$ cm.
  $O A = 10$ cm (radius), so $O P = A P - O A = 16 - 10 = 6$ cm.
  ½ mark
2. $O A ⊥ X Y$ [tangent ⟂ radius at the point of contact]. Since $C D ∥ X Y$ , also $C D ⊥ A B$ , so $∠ C P O = 9 0^{\circ}$ .
  In right $△ O P C$ , $O C = 10$ cm (radius):
  $C P = O C^{2} - O P^{2} = 1 0^{2} - 6^{2} = 64 = 8 cm$
  1 mark
3. The perpendicular from the centre bisects the chord, so $C D = 2 \times C P = 2 \times 8 = 16$ cm.
  ½ mark
Answer: CD = 16 cm
Where the marks are
- Finds $O P = 6$ cm (1/2 mark)
- Uses $O P ⊥ C D$ and Pythagoras to get $C P = 8$ cm (1 mark)
- Doubles to get $C D = 16$ cm, as the perpendicular from the centre bisects the chord (1/2 mark)
Open Q25 on its own page

Section C

Short answer (SA)

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerCircles
Prove that the parallelogram circumscribing a circle is a rhombus.
Try it first, then show our solutionSolution
1. Figure: Draw parallelogram ABCD with all four sides touching a circle. The points of contact are P on AB, Q on BC, R on CD and S on DA.
  Given: ABCD is a parallelogram circumscribing the circle.
  To prove: ABCD is a rhombus.
  ½ mark
2. Tangents drawn from an external point to a circle are equal in length. So:
  $A P = A S, B P = B Q, C R = C Q, D R = D S$
  1 mark
3. Adding these four equations:
  $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$
  $\Rightarrow A B + C D = A D + B C$
  ½ mark
4. In a parallelogram opposite sides are equal: $A B = C D$ and $B C = A D$ .
  So $2 A B = 2 B C$ , which gives $A B = B C$ .
  ½ mark
5. Therefore $A B = B C = C D = D A$ . A parallelogram with all sides equal is a rhombus, so ABCD is a rhombus. Hence proved.
  ½ mark
Answer: AB + CD = AD + BC together with AB = CD and BC = AD gives AB = BC = CD = DA, so ABCD is a rhombus
Where the marks are
- Correct labelled figure (1/2 mark)
- Writes the four equal-tangent relations (1 mark)
- Adds them to get $A B + C D = A D + B C$ (1/2 mark)
- Uses the opposite sides of the parallelogram to get $A B = B C$ (1/2 mark)
- Concludes all sides are equal, so ABCD is a rhombus (1/2 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerCircles
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
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, and join OA, OB and AB.
  To prove: $∠ A P B + ∠ A O B = 18 0^{\circ}$ .
  1 mark
2. $O A ⊥ P A$ and $O B ⊥ P B$ [tangent ⟂ radius at the point of contact].
  So $∠ O A P = ∠ O B P = 9 0^{\circ}$ .
  1 mark
3. In quadrilateral OAPB, the angles add up to $36 0^{\circ}$ :
  $∠ A P B + ∠ O A P + ∠ O B P + ∠ A O B = 36 0^{\circ}$
  $∠ A P B + 9 0^{\circ} + 9 0^{\circ} + ∠ A O B = 36 0^{\circ}$
  ½ mark
4. So $∠ A P B + ∠ A O B = 18 0^{\circ}$ .
  Hence $∠ A P B$ and $∠ A O B$ are supplementary.
  ½ mark
Answer: $∠ A P B + ∠ A O B = 18 0^{\circ}$ , so the two angles are supplementary
Where the marks are
- Correct figure with tangents PA, PB and radii OA, OB (1 mark)
- States $∠ O A P = ∠ O B P = 9 0^{\circ}$ because tangent ⟂ radius (1 mark)
- Uses the angle sum of quadrilateral OAPB (1/2 mark)
- Concludes $∠ A P B + ∠ A O B = 18 0^{\circ}$ (1/2 mark)
Open Q26(b) on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $(1 + \frac{1}{t a n ^{2} θ}) (1 + \frac{1}{c o t ^{2} θ}) = \frac{1}{s i n ^{2} θ - s i n ^{4} θ}$
Try it first, then show our solutionSolution
1. Since $\frac{1}{t a n θ} = cot θ$ and $\frac{1}{c o t θ} = tan θ$ :
  $LHS = (1 + cot^{2} θ) (1 + tan^{2} θ)$
  1 mark
2. Using the identities $1 + cot^{2} θ = csc^{2} θ$ and $1 + tan^{2} θ = sec^{2} θ$ :
  $LHS = csc^{2} θ \cdot sec^{2} θ$
  ½ mark
3. $= \frac{1}{sin ^{2} θ} \cdot \frac{1}{cos ^{2} θ}$
  ½ mark
4. Using $cos^{2} θ = 1 - sin^{2} θ$ :
  $= \frac{1}{sin ^{2} θ ( 1 - sin ^{2} θ )}$
  ½ mark
5. $= \frac{1}{sin ^{2} θ - sin ^{4} θ} = RHS$
  Hence proved.
  ½ mark
Answer: LHS = $csc^{2} θ sec^{2} θ = \frac{1}{s i n ^{2} θ - s i n ^{4} θ}$ = RHS
Where the marks are
- Rewrites LHS as $(1 + cot^{2} θ) (1 + tan^{2} θ)$ (1 mark)
- Uses the identities to get $csc^{2} θ sec^{2} θ$ (1/2 mark)
- Converts to $\frac{1}{s i n ^{2} θ c o s ^{2} θ}$ (1/2 mark)
- Substitutes $cos^{2} θ = 1 - sin^{2} θ$ (1/2 mark)
- Simplifies to the RHS (1/2 mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{c o s e c θ - 1}{c o s e c θ + 1} + \frac{c o s e c θ + 1}{c o s e c θ - 1} = 2 sec θ$
Try it first, then show our solutionSolution
1. Take the LCM of the two square-root terms. The common denominator is $(csc θ + 1) (csc θ - 1)$ :
  $LHS = \frac{( csc θ - 1 ) + ( csc θ + 1 )}{( csc θ + 1 ) ( csc θ - 1 )}$
  1 mark
2. $= \frac{2 csc θ}{csc ^{2} θ - 1}$
  1 mark
3. Since $csc^{2} θ - 1 = cot^{2} θ$ (taking $θ$ acute):
  $= \frac{2 csc θ}{cot θ}$
  ½ mark
4. $= 2 \cdot \frac{1}{sin θ} \cdot \frac{sin θ}{cos θ} = \frac{2}{cos θ} = 2 sec θ = RHS$
  Hence proved.
  ½ mark
Answer: LHS = $\frac{2 c s c θ}{c o t θ} = 2 sec θ$ = RHS
Where the marks are
- Combines the two terms over $(csc θ + 1) (csc θ - 1)$ (1 mark)
- Simplifies to $\frac{2 c s c θ}{c s c ^{2} θ - 1}$ (1 mark)
- Uses $csc^{2} θ - 1 = cot^{2} θ$ (1/2 mark)
- Reduces to $2 sec θ$ (1/2 mark)
Open Q27(b) on its own page
Q28 · 3 marks · Short answerCoordinate Geometry
If the mid-point of the line segment joining the points $A (3, 4)$ and $B (k, 6)$ is $P (x, y)$ and $x + y - 10 = 0$ , then find the value of k.
Try it first, then show our solutionSolution
1. By the mid-point formula, $x = \frac{3 + k}{2}$ .
  ½ mark
2. $y = \frac{4 + 6}{2} = 5$ .
  ½ mark
3. P lies on $x + y - 10 = 0$ , so:
  $\frac{3 + k}{2} + 5 - 10 = 0$
  1 mark
4. $\frac{3 + k}{2} = 5 \Rightarrow 3 + k = 10 \Rightarrow k = 7$ .
  1 mark
Answer: k = 7
Where the marks are
- Finds $x = \frac{3 + k}{2}$ (1/2 mark)
- Finds $y = 5$ (1/2 mark)
- Substitutes into $x + y - 10 = 0$ (1 mark)
- Solves to get $k = 7$ (1 mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerCompetency-basedAreas Related to Circles
The length of the hour hand of a clock is 10 cm. Find the area of the minor sector swept by the hour hand of the clock between 5 a.m. to 8 a.m. Also, find the area of the major sector.
Try it first, then show our solutionSolution
1. The hour hand turns $\frac{36 0 ^{\circ}}{12} = 3 0^{\circ}$ each hour.
  From 5 a.m. to 8 a.m. is 3 hours, so the angle swept is $θ = 3 \times 3 0^{\circ} = 9 0^{\circ}$ .
  ½ mark
2. Area of minor sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{90}{360} \times \frac{22}{7} \times (10)^{2}$
  1 mark
3. $= \frac{550}{7} cm^{2} \approx 78.57 cm^{2}$
  ½ mark
4. Area of the whole circle $= \frac{22}{7} \times (10)^{2} = \frac{2200}{7} cm^{2}$
  ½ mark
5. Area of major sector $= \frac{2200}{7} - \frac{550}{7} = \frac{1650}{7} cm^{2} \approx 235.71 cm^{2}$
  ½ mark
Answer: Minor sector = 550/7 cm² ≈ 78.57 cm²; major sector = 1650/7 cm² ≈ 235.71 cm²
Where the marks are
- Finds the central angle $9 0^{\circ}$ (1/2 mark)
- Uses the sector area formula with correct values (1 mark)
- Minor sector area $\frac{550}{7}$ cm² (1/2 mark)
- Area of circle $\frac{2200}{7}$ cm² (1/2 mark)
- Major sector area $\frac{1650}{7}$ cm² (1/2 mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3$ is rational.
  Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are coprime.
  Squaring: $3 q^{2} = p^{2}$ .
  ½ mark
2. So $p^{2}$ is divisible by 3. Since 3 is prime, 3 divides $p^{2}$ means 3 divides $p$ . ... (1)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2}$ , so $q^{2} = 3 a^{2}$ .
  So $q^{2}$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)
  1 mark
4. From (1) and (2), 3 divides both $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong. Hence $3$ is irrational.
  ½ mark
Answer: Assuming $3 = p / q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $3$ is irrational
Where the marks are
- Assumes $3 = \frac{p}{q}$ with p, q coprime and gets $3 q^{2} = p^{2}$ (1/2 mark)
- Shows 3 divides p (1 mark)
- Substitutes $p = 3 a$ and shows 3 divides q (1 mark)
- States the contradiction and concludes (1/2 mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCompetency-basedArithmetic Progressions
A sum of ₹ 2,000 is invested at 7% per annum simple interest. Calculate the interests at the end of $1^{st}$ , $2^{nd}$ and $3^{rd}$ year. Do these interests form an AP ? If so, find the interest at the end of the $2 7^{th}$ year.
Try it first, then show our solutionSolution
1. Simple interest $= \frac{P \times R \times T}{100}$ .
  Interest at the end of 1st year $= \frac{2000 \times 7 \times 1}{100} = ₹140$
  ½ mark
2. Interest at the end of 2nd year $= \frac{2000 \times 7 \times 2}{100} = ₹280$
  ½ mark
3. Interest at the end of 3rd year $= \frac{2000 \times 7 \times 3}{100} = ₹420$
  ½ mark
4. The interests are 140, 280, 420, …
  $280 - 140 = 420 - 280 = 140$ . Yes, they form an AP with $a = 140$ and $d = 140$ .
  ½ mark
5. Using $a_{n} = a + (n - 1) d$ , interest at the end of the 27th year $= a_{27} = 140 + 26 \times 140$
  ½ mark
6. $= 140 + 3640 = ₹3780$
  ½ mark
Answer: ₹140, ₹280, ₹420; yes, they form an AP (d = 140); interest at the end of the 27th year = ₹3780
Where the marks are
- Interest for year 1 = ₹140 (1/2 mark)
- Interest for year 2 = ₹280 (1/2 mark)
- Interest for year 3 = ₹420 (1/2 mark)
- States they form an AP with $a = d = 140$ (1/2 mark)
- Writes $a_{27} = 140 + 26 \times 140$ (1/2 mark)
- Gets ₹3780 (1/2 mark)
Open Q31 on its own page

Section D

Long answer (LA)

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
Two ships are sailing in the sea on either side of a lighthouse. The angles of depression to the two ships as observed from the top of the lighthouse are $6 0^{\circ}$ and $4 5^{\circ}$ , respectively. If the distance between the ships is $100 (\frac{1 + 3}{3})$ m, then find the height of the lighthouse.
Figure in the paper: A vertical lighthouse with its top at the upper point. Two ships on the horizontal sea level, one on each side of the base of the lighthouse; the angles of depression from the top to each ship ( $6 0^{\circ}$ and $4 5^{\circ}$ ) are marked. See it in the official paper.
Try it first, then show our solutionSolution
1. Figure: Let AB be the lighthouse, with top A and foot B. Ships P and Q are on opposite sides of B on the sea level line PBQ, and $∠ A B P = ∠ A B Q = 9 0^{\circ}$ .
  The angles of depression from A are $6 0^{\circ}$ towards P and $4 5^{\circ}$ towards Q. By alternate angles, $∠ A P B = 6 0^{\circ}$ and $∠ A QB = 4 5^{\circ}$ .
  $P Q = 100 (\frac{1 + 3}{3})$ m.
  1 mark
2. In right $△ A B P$ :
  $\frac{A B}{P B} = tan 6 0^{\circ} = 3$
  1 mark
3. $\Rightarrow P B = \frac{A B}{3}$ ... (1)
  ½ mark
4. In right $△ A B Q$ :
  $\frac{A B}{B Q} = tan 4 5^{\circ} = 1$
  ½ mark
5. $\Rightarrow B Q = A B$ ... (2)
  ½ mark
6. Adding (1) and (2):
  $P Q = P B + B Q = \frac{A B}{3} + A B = A B (\frac{1 + 3}{3})$
  ½ mark
7. Using the given distance:
  $100 (\frac{1 + 3}{3}) = A B (\frac{1 + 3}{3})$
  ½ mark
8. $\Rightarrow A B = 100$ m.
  So the height of the lighthouse is 100 m.
  ½ mark
Answer: Height of the lighthouse = 100 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ} = \frac{A B}{P B} = 3$ (1 mark)
- Gets $P B = \frac{A B}{3}$ (1/2 mark)
- Uses $tan 4 5^{\circ} = \frac{A B}{B Q} = 1$ (1/2 mark)
- Gets $B Q = A B$ (1/2 mark)
- Writes $P Q = A B (\frac{1 + 3}{3})$ (1/2 mark)
- Equates to the given distance (1/2 mark)
- Height = 100 m (1/2 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
The angles of depression of the top and the bottom of an 8 m tall building from the top of another multistoried building are $3 0^{\circ}$ and $4 5^{\circ}$ , respectively. Find the height of the multistoried building and the distance between the two buildings.
Figure in the paper: A tall multistoried building on the left and a shorter 8 m building on the right on the same ground. From the top of the taller building, horizontal line of sight; angles of depression $3 0^{\circ}$ to the top of the 8 m building and $4 5^{\circ}$ to its bottom are marked. See it in the official paper.
Try it first, then show our solutionSolution
1. Figure: Let CD be the multistoried building, with top D and foot C. Let AB be the 8 m building, with top A and foot B. Draw AE ∥ BC meeting CD at E, so ABCE is a rectangle with $E C = A B = 8$ m and $A E = B C$ .
  The angles of depression from D are $3 0^{\circ}$ to A and $4 5^{\circ}$ to B. By alternate angles, $∠ D A E = 3 0^{\circ}$ and $∠ D B C = 4 5^{\circ}$ .
  Let $D C = h$ m. Then $D E = h - 8$ .
  1 mark
2. In right $△ D E A$ :
  $\frac{D E}{A E} = tan 3 0^{\circ} = \frac{1}{3}$
  1 mark
3. $\Rightarrow A E = 3 D E$ ... (1)
  ½ mark
4. In right $△ D C B$ :
  $\frac{D C}{B C} = tan 4 5^{\circ} = 1$
  ½ mark
5. $\Rightarrow B C = D C$ ... (2)
  ½ mark
6. Since $A E = B C$ :
  $3 (D C - 8) = D C \Rightarrow D C (3 - 1) = 83 \Rightarrow D C = \frac{8 3}{3 - 1}$
  ½ mark
7. Rationalising:
  $D C = \frac{8 3 ( 3 + 1 )}{( 3 - 1 ) ( 3 + 1 )} = \frac{8 ( 3 + 3 )}{2} = (12 + 43) m \approx 18.93 m$
  ½ mark
8. From (2), the distance between the buildings is $B C = D C = (12 + 43)$ m $\approx 18.93$ m.
  ½ mark
Answer: Height of the multistoried building = $(12 + 43)$ m ≈ 18.93 m; distance between the buildings = $(12 + 43)$ m ≈ 18.93 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 3 0^{\circ} = \frac{D E}{A E}$ (1 mark)
- Gets $A E = 3 D E$ (1/2 mark)
- Uses $tan 4 5^{\circ} = \frac{D C}{B C}$ (1/2 mark)
- Gets $B C = D C$ (1/2 mark)
- Forms $3 (D C - 8) = D C$ and gets $D C = \frac{8 3}{3 - 1}$ (1/2 mark)
- Rationalises to $12 + 43$ m (1/2 mark)
- Distance $B C = (12 + 43)$ m (1/2 mark)
Open Q32(b) on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerQuadratic Equations
The sum of the areas of two squares is $52$ cm $^{2}$ and difference of their perimeters is $8$ cm. Find the lengths of the sides of the two squares.
Try it first, then show our solutionSolution
1. Let the sides of the two squares be $x$ cm and $y$ cm, with $x > y$ .
  Sum of areas: $x^{2} + y^{2} = 52$ ...(1)
  1 mark
2. Difference of perimeters: $4 x - 4 y = 8 \Rightarrow x - y = 2 \Rightarrow x = y + 2$ ...(2)
  1 mark
3. Substitute (2) in (1): $(y + 2)^{2} + y^{2} = 52$
  $\Rightarrow 2 y^{2} + 4 y + 4 = 52 \Rightarrow 2 y^{2} + 4 y - 48 = 0 \Rightarrow y^{2} + 2 y - 24 = 0$
  1 mark
4. Factorise by splitting the middle term: $y^{2} + 6 y - 4 y - 24 = 0 \Rightarrow (y + 6) (y - 4) = 0$
  So $y = - 6$ or $y = 4$ .
  1 mark
5. A side length cannot be negative, so $y = - 6$ is rejected. Hence $y = 4$ .
  ½ mark
6. From (2), $x = 4 + 2 = 6$ .
  Check: $36 + 16 = 52$ ✓ and $24 - 16 = 8$ ✓
  The sides of the two squares are 6 cm and 4 cm.
  ½ mark
Answer: Sides are 6 cm and 4 cm
Where the marks are
- Writes the area equation $x^{2} + y^{2} = 52$ (1 mark)
- Writes the perimeter equation $4 x - 4 y = 8$ , i.e. $x - y = 2$ (1 mark)
- Forms the quadratic $y^{2} + 2 y - 24 = 0$ (1 mark)
- Factorises it as $(y + 6) (y - 4) = 0$ (1 mark)
- Rejects the negative root and gets $y = 4$ (½ mark)
- Finds $x = 6$ and states both sides (½ mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerCompetency-basedQuadratic Equations
The time taken by a person to travel an upward distance of 150 km was $2 \frac{1}{2}$ hours more than the time taken in the downward return journey. If he returned at a speed of 10 km/h more than the speed while going up, find the speeds in each direction.
Try it first, then show our solutionSolution
1. Let the speed while going up be $x$ km/h. Then the speed on the downward return journey is $(x + 10)$ km/h.
  ½ mark
2. Time = $\frac{distance}{speed}$ .
  Time going up $= \frac{150}{x}$ h and time coming down $= \frac{150}{x + 10}$ h.
  The upward trip takes $2 \frac{1}{2} = \frac{5}{2}$ h more:
  $\frac{150}{x} - \frac{150}{x + 10} = \frac{5}{2}$
  2 marks
3. $\frac{150 ( x + 10 ) - 150 x}{x ( x + 10 )} = \frac{5}{2} \Rightarrow \frac{1500}{x ^{2} + 10 x} = \frac{5}{2}$
  $\Rightarrow 5 (x^{2} + 10 x) = 3000 \Rightarrow x^{2} + 10 x - 600 = 0$
  1 mark
4. Factorise: $x^{2} + 30 x - 20 x - 600 = 0 \Rightarrow (x + 30) (x - 20) = 0$
  ½ mark
5. So $x = - 30$ or $x = 20$ . Speed cannot be negative, so $x = 20$ .
  ½ mark
6. Downward speed $= x + 10 = 30$ km/h.
  Check: $\frac{150}{20} - \frac{150}{30} = 7.5 - 5 = 2.5$ h ✓
  Speed going up = 20 km/h; speed coming down = 30 km/h.
  ½ mark
Answer: Upward speed 20 km/h, downward speed 30 km/h
Where the marks are
- Takes the upward speed as $x$ and the downward speed as $x + 10$ (½ mark)
- Forms the equation $\frac{150}{x} - \frac{150}{x + 10} = \frac{5}{2}$ (2 marks)
- Simplifies it to $x^{2} + 10 x - 600 = 0$ (1 mark)
- Factorises it as $(x + 30) (x - 20) = 0$ (½ mark)
- Gets $x = 20$ , rejecting the negative root (½ mark)
- Gets the downward speed of 30 km/h (½ mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerTriangles
Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove that $\frac{A M}{M B} = \frac{A N}{N D}$ where $L M ∥ C B$ and $L N ∥ C D$ .
Figure in the paper: Points A (left), B (top), C (right) and D (bottom). Line AC is drawn. M lies on segment AB and N lies on segment AD; L lies on AC. Segments LM (parallel to CB) and LN (parallel to CD) are drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. Theorem (Basic Proportionality Theorem).
  Given: In $△ P QR$ , a line $S T ∥ QR$ meets $P Q$ at $S$ and $P R$ at $T$ .
  To prove: $\frac{P S}{S Q} = \frac{P T}{T R}$
  Construction: Join $QT$ and $R S$ . Draw $T K ⊥ P Q$ and $S H ⊥ P R$ .
  Figure: $△ P QR$ with $P$ at the top and $QR$ as the base. Segment $S T$ is drawn parallel to $QR$ , with $S$ on $P Q$ and $T$ on $P R$ . Segments $QT$ and $R S$ are joined, and the perpendiculars $T K$ (onto $P Q$ ) and $S H$ (onto $P R$ ) are shown.
  (Letters $P, Q, R$ are used here so they do not clash with the letters of the second part.)
  1½ marks
2. Proof:
  $ar (P S T) = \frac{1}{2} \cdot P S \cdot T K$ and $ar (QS T) = \frac{1}{2} \cdot S Q \cdot T K$ , so
  $\frac{ar ( P S T )}{ar ( QS T )} = \frac{P S}{S Q} ... (1)$
  $ar (P S T) = \frac{1}{2} \cdot P T \cdot S H$ and $ar (R T S) = \frac{1}{2} \cdot T R \cdot S H$ , so
  $\frac{ar ( P S T )}{ar ( R T S )} = \frac{P T}{T R} ... (2)$
  $△ QS T$ and $△ R T S$ have the same base $S T$ and lie between the same parallels $S T$ and $QR$ . Hence $ar (QS T) = ar (R T S)$ ...(3)
  From (1), (2) and (3): $\frac{P S}{S Q} = \frac{P T}{T R}$ . Hence proved.
  1½ marks
3. Application. In $△ A B C$ , $L M ∥ C B$ , with $M$ on $A B$ and $L$ on $A C$ .
  By the BPT, $\frac{A M}{M B} = \frac{A L}{L C}$ ...(4)
  1 mark
4. In $△ A D C$ , $L N ∥ C D$ , with $N$ on $A D$ and $L$ on $A C$ .
  By the BPT, $\frac{A N}{N D} = \frac{A L}{L C}$ ...(5)
  ½ mark
5. From (4) and (5), $\frac{A M}{M B} = \frac{A N}{N D}$ . Hence proved.
  ½ mark
Answer: BPT proved by the area method; then $\frac{A M}{M B} = \frac{A L}{L C} = \frac{A N}{N D}$ , so $\frac{A M}{M B} = \frac{A N}{N D}$
Where the marks are
- Correct figure with given, to prove and construction (1½ marks)
- Correct area-ratio proof of the BPT, using equal areas of triangles on the same base between the same parallels (1½ marks)
- Applies the BPT in $△ A B C$ to get $\frac{A M}{M B} = \frac{A L}{L C}$ (1 mark)
- Applies the BPT in $△ A D C$ to get $\frac{A N}{N D} = \frac{A L}{L C}$ (½ mark)
- Concludes $\frac{A M}{M B} = \frac{A N}{N D}$ (½ mark)
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Q35 · 5 marks · Long answerStatistics
Find the Mean and Mode of the following frequency distribution :
Class Frequency
0 – 10 8
10 – 20 7
20 – 30 15
30 – 40 20
40 – 50 12
50 – 60 8
60 – 70 10
Try it first, then show our solutionSolution
1. Mean (step-deviation method) with assumed mean $a = 35$ and class size $h = 10$ , where $u_{i} = \frac{x _{i} - 35}{10}$ :
  Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
  0–10 8 5 −3 −24
  10–20 7 15 −2 −14
  20–30 15 25 −1 −15
  30–40 20 35 0 0
  40–50 12 45 1 12
  50–60 8 55 2 16
  60–70 10 65 3 30
  Total 80 5
  So $Σ f_{i} = 80$ and $Σ f_{i} u_{i} = 5$ .
  1½ marks
2. Mean $= a + \frac{Σ f _{i} u _{i}}{Σ f _{i}} \times h = 35 + \frac{5}{80} \times 10$
  1 mark
3. $= 35 + 0.625 = 35.625$
  ½ mark
4. Mode: The highest frequency is 20, so the modal class is 30–40.
  ½ mark
5. Here $l = 30$ , $f_{1} = 20$ , $f_{0} = 15$ , $f_{2} = 12$ and $h = 10$ .
  Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 30 + \frac{20 - 15}{40 - 15 - 12} \times 10$
  1 mark
6. $= 30 + \frac{50}{13} = \frac{440}{13} \approx 33.85$
  ½ mark
Answer: Mean = 35.625; Mode = 440/13 ≈ 33.85
Where the marks are
- Correct table of class marks, $u_{i}$ and $f_{i} u_{i}$ , with $Σ f_{i} = 80$ and $Σ f_{i} u_{i} = 5$ (1½ marks)
- Correct mean formula with values substituted (1 mark)
- Mean $= 35.625$ (½ mark)
- Identifies the modal class 30–40 (½ mark)
- Correct mode formula with values substituted (1 mark)
- Mode $= \frac{440}{13} \approx 33.85$ (½ mark)
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Class	Frequency
0 – 10	8
10 – 20	7
20 – 30	15
30 – 40	20
40 – 50	12
50 – 60	8
60 – 70	10

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
0–10	8	5	−3	−24
10–20	7	15	−2	−14
20–30	15	25	−1	−15
30–40	20	35	0	0
40–50	12	45	1	12
50–60	8	55	2	16
60–70	10	65	3	30
Total	80			5

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
Case Study – 1
A school is organizing a grand cultural event to show the talent of its students. To accommodate the guests, the school plans to rent chairs and tables from a local supplier. It finds that rent for each chair is ₹ 50 and for each table is ₹ 200. The school spends ₹ 30,000 for renting the chairs and tables. Also, the total number of items (chairs and tables) rented are 300.
If the school rents 'x' chairs and 'y' tables, answer the following questions :
1. (i)
  Write down the pair of linear equations representing the given information.
  1 mark
2. (ii)(a)
  Find the number of chairs and number of tables rented by the school.
  2 marks
3. OR
  (ii)(b)
  If the school wants to spend a maximum of ₹ 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent.
  2 marks
4. (iii)
  What is maximum number of tables that can be rented in ₹ 30,000 if no chairs are rented ?
  1 mark
Figure in the paper: A photograph of a classroom-style hall with rows of white tables and red and grey chairs arranged around them, with windows on the left. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Total number of items: $x + y = 300$
  ½ mark
2. (i) Total rent: $50 x + 200 y = 30000$ , i.e. $x + 4 y = 600$ (dividing by 50).
  ½ mark
3. (ii)(a) Subtract $x + y = 300$ from $x + 4 y = 600$ :
  $3 y = 300 \Rightarrow y = 100$
  1 mark
4. (ii)(a) $x = 300 - 100 = 200$
  So the school rented 200 chairs and 100 tables.
  1 mark
5. (ii)(b) Take the spending as ₹27,000 on 300 items:
  $x + y = 300$ and $50 x + 200 y = 27000$ , i.e. $x + 4 y = 540$
  1 mark
6. (ii)(b) Subtracting: $3 y = 240 \Rightarrow y = 80$ , so $x = 300 - 80 = 220$ .
  It can rent 220 chairs and 80 tables.
  1 mark
7. (iii) With no chairs, all ₹30,000 goes on tables at ₹200 each.
  Number of tables $= \frac{30000}{200} = 150$
  Maximum number of tables = 150.
  1 mark
Answer: (i) $x + y = 300$ , $50 x + 200 y = 30000$ (i.e. $x + 4 y = 600$ ); (ii)(a) 200 chairs, 100 tables; (ii)(b) 220 chairs, 80 tables; (iii) 150 tables
Where the marks are
- (i) Writes $x + y = 300$ (½ mark)
- (i) Writes $50 x + 200 y = 30000$ (½ mark)
- (ii)(a) Finds $y = 100$ (1 mark)
- (ii)(a) Finds $x = 200$ (1 mark)
- (ii)(b) Forms $x + 4 y = 540$ along with $x + y = 300$ (1 mark)
- (ii)(b) Finds $x = 220$ and $y = 80$ (½ + ½ marks)
- (iii) $30000 \div 200 = 150$ tables (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedProbability
Case Study – 2
Rahul is a lucky charm for his cricket team. He has a jar of cards with numbers from 10 to 74. Before each match, he draws a card from the jar. If the card bears an even number, the team wins. If the number is even and divisible by 5, they win by a big margin. If the number is an odd number less than 30, they win by a small margin. And if the number is a prime number between 50 and 74, they lose.
Answer the following questions if Rahul draws a card today :
1. (i)
  What is the probability that Rahul draws a card with an even number ?
  1 mark
2. (ii)
  What is the probability that Rahul draws a card with an odd number less than 30 ?
  1 mark
3. (iii)(a)
  What is the probability that Rahul draws a card with a prime number between 50 and 74 ?
  2 marks
4. OR
  (iii)(b)
  What is the probability that Rahul draws a card with an even number divisible by 5 ?
  2 marks
Figure in the paper: A photograph of a hand drawing a folded paper card from a glass jar filled with small coloured folded cards (pink, yellow, blue). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Total number of cards $= 74 - 10 + 1 = 65$
  ½ mark
2. (i) The even numbers are $10, 12, \dots, 74$ , giving $\frac{74 - 10}{2} + 1 = 33$ cards.
  $P (even) = \frac{33}{65}$
  ½ mark
3. (ii) The odd numbers less than 30 are $11, 13, \dots, 29$ , giving 10 cards.
  $P = \frac{10}{65} = \frac{2}{13}$
  1 mark
4. (iii)(a) The primes between 50 and 74 are $53, 59, 61, 67, 71, 73$ , so there are 6 favourable outcomes.
  1 mark
5. (iii)(a) $P (prime between 50 and 74) = \frac{6}{65}$
  1 mark
6. (iii)(b) The even numbers divisible by 5 are the multiples of 10: $10, 20, 30, 40, 50, 60, 70$ . There are 7 favourable outcomes.
  1 mark
7. (iii)(b) $P (even and divisible by 5) = \frac{7}{65}$
  1 mark
Answer: (i) 33/65; (ii) 2/13; (iii)(a) 6/65, (iii)(b) 7/65
Where the marks are
- (i) Total outcomes = 65 (½ mark)
- (i) $P (even) = \frac{33}{65}$ (½ mark)
- (ii) $P = \frac{10}{65} = \frac{2}{13}$ (1 mark)
- (iii)(a) Lists the 6 primes 53, 59, 61, 67, 71, 73 (1 mark)
- (iii)(a) $P = \frac{6}{65}$ (1 mark)
- (iii)(b) Lists the 7 multiples of 10 (1 mark)
- (iii)(b) $P = \frac{7}{65}$ (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Case Study – 3
A skilled carpenter decided to craft a special rolling pin for the local baker. He carefully joined three cylindrical pieces of wood – two small ones on the ends and one larger in the centre to create a perfect tool. The baker loved the rolling pin, as it rolled out the smoothest dough for breads and pastries.
The length of the bigger cylindrical part is 12 cm and diameter is 7 cm and the length of each smaller cylindrical part is 5 cm and diameter is 2·1 cm.
Based on the above information, answer the following questions :
1. (i)
  Find the volume of the bigger cylindrical part.
  1 mark
2. (ii)
  Find the curved surface area of the bigger cylindrical part.
  1 mark
3. (iii)(a)
  Find the ratio of the volume of the bigger cylindrical part to the total volume of the two smaller (identical) cylindrical parts.
  2 marks
4. OR
  (iii)(b)
  Find the sum of the curved surface areas of the two identical smaller cylindrical parts.
  2 marks
Figure in the paper: Three cylinders joined end to end, shown horizontally. The central bigger cylinder has length 12 cm and diameter 7 cm. Two identical smaller cylinders at each end have length 5 cm and diameter 2.1 cm. Dimensions are labelled on each part. A photograph of a baker rolling dough with a wooden rolling pin is shown above the diagram. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The bigger cylinder has $r = \frac{7}{2} = 3.5$ cm and $h = 12$ cm.
  Volume $= π r^{2} h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$
  ½ mark
2. (i) $= 462$ cm $^{3}$
  ½ mark
3. (ii) CSA $= 2 π r h = 2 \times \frac{22}{7} \times \frac{7}{2} \times 12$
  ½ mark
4. (ii) $= 264$ cm $^{2}$
  ½ mark
5. (iii)(a) Each smaller cylinder has $r = \frac{2.1}{2} = 1.05$ cm and $h = 5$ cm.
  Total volume of the two $= 2 \times \frac{22}{7} \times 1.05 \times 1.05 \times 5$
  ½ mark
6. (iii)(a) $= 34.65$ cm $^{3}$
  ½ mark
7. (iii)(a) Ratio $= \frac{462}{34.65} = \frac{46200}{3465} = \frac{3080}{231} = \frac{40}{3}$
  Required ratio $= 3080 : 231$ , i.e. 40 : 3 in simplest form.
  1 mark
8. (iii)(b) Sum of CSAs $= 2 \times 2 π r h = 2 \times 2 \times \frac{22}{7} \times 1.05 \times 5$
  1 mark
9. (iii)(b) $= 66$ cm $^{2}$
  1 mark
Answer: (i) 462 cm³; (ii) 264 cm²; (iii)(a) 40 : 3 (= 3080 : 231), (iii)(b) 66 cm²
Where the marks are
- (i) Uses $π r^{2} h$ with $r = 3.5$ and $h = 12$ (½ mark)
- (i) Volume = 462 cm³ (½ mark)
- (ii) Uses $2 π r h$ correctly (½ mark)
- (ii) CSA = 264 cm² (½ mark)
- (iii)(a) Volume of the two small cylinders set up correctly (½ mark)
- (iii)(a) Total = 34.65 cm³ (½ mark)
- (iii)(a) Ratio 3080 : 231, i.e. 40 : 3 (1 mark)
- (iii)(b) Sets up $2 \times 2 π r h$ with $r = 1.05$ and $h = 5$ (1 mark)
- (iii)(b) Total CSA = 66 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.