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

CBSE Class 10 Mathematics Basic: 2026 board paper, set 430/3/1

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 430-3-1_Mathematics Basic.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

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

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

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If $p = 2^{3} \times 3^{2} \times 5$ and $q = 2^{2} \times 3^{3}$ , then the LCM of p and q is:
1. (A) $2^{3} \times 3^{3}$
2. (B) $2^{2} \times 3^{2}$
3. (C) $2^{2} \times 3^{2} \times 5$
4. (D) $2^{3} \times 3^{3} \times 5$
Try it first, then show our solutionSolution
1. LCM = product of each prime factor with its highest power: $2^{3}$ , $3^{3}$ , $5^{1}$ , so LCM $= 2^{3} \times 3^{3} \times 5$ .
  1 mark
Answer: $2^{3} \times 3^{3} \times 5$ , option (D)
Where the marks are
- Takes highest powers of all primes to get $2^{3} \times 3^{3} \times 5$ (1 mark)
Q2 · 1 mark · MCQReal Numbers
$3^{n}$ , where n is a natural number, cannot end with the digit:
1. (A)3
2. (B)5
3. (C)7
4. (D)9
Try it first, then show our solutionSolution
1. The last digits of $3^{n}$ repeat in the cycle 3, 9, 7, 1.
2. To end in 5 a number needs 5 as a prime factor, but $3^{n}$ has only the prime factor 3. So $3^{n}$ cannot end in 5.
  1 mark
Answer: 5, option (B)
Where the marks are
- Identifies that 5 is not a factor of $3^{n}$ , so the last digit cannot be 5 (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQReal Numbers
A prime number has:
1. (A)exactly two prime factors
2. (B)exactly one prime factor
3. (C)at least one prime factor
4. (D)at least two prime factors
Try it first, then show our solutionSolution
1. A prime p has only the factors 1 and p, and 1 is not prime, so its only prime factor is p itself: exactly one prime factor.
  1 mark
Answer: exactly one prime factor, option (B)
Where the marks are
- Recognises a prime has only itself as a prime factor (1 mark)
Q4 · 1 mark · MCQPair of Linear Equations in Two Variables
For what value(s) of k does the system of equations $k x + 2 y = 3$ and $2 x + y = 5$ have a unique solution?
1. (A)k = a real number
2. (B) $k \neq = 8$
3. (C) $k \neq = 4$
4. (D)k = 4
Try it first, then show our solutionSolution
1. Unique solution when $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ : $\frac{k}{2} \neq = \frac{2}{1}$ , so $k \neq = 4$ .
  1 mark
Answer: $k \neq = 4$ , option (C)
Where the marks are
- Uses $a_{1} / a_{2} \neq = b_{1} / b_{2}$ to get $k \neq = 4$ (1 mark)
Q5 · 1 mark · MCQQuadratic Equations
If $x = x$ , ( $x \neq = 0$ ) is expressed as a quadratic equation in the form $a x^{2} + b x + c = 0$ , then the value of $a + b + c$ is:
1. (A)0
2. (B)1
3. (C)2
4. (D)3
Try it first, then show our solutionSolution
1. Squaring: $x^{2} = x$ , so $x^{2} - x + 0 = 0$ . Then $a = 1, b = - 1, c = 0$ .
  ½ mark
2. $a + b + c = 1 - 1 + 0 = 0$ .
  ½ mark
Answer: 0, option (A)
Where the marks are
- Forms $x^{2} - x = 0$ (½ mark)
- Gets $a + b + c = 0$ (½ mark)
Q6 · 1 mark · MCQCoordinate Geometry
If point $(1, 2)$ is the mid-point of the line segment joining the points $(3, 5)$ and $(2 a, b)$ , then $(a, b) =$
1. (A) $(- 1, - 1)$
2. (B) $(- \frac{1}{2}, - \frac{1}{2})$
3. (C) $(- \frac{1}{2}, - 1)$
4. (D) $(- 1, - \frac{1}{2})$
Try it first, then show our solutionSolution
1. Mid-point formula: $\frac{3 + 2 a}{2} = 1 \Rightarrow 2 a = - 1 \Rightarrow a = - \frac{1}{2}$ .
  ½ mark
2. $\frac{5 + b}{2} = 2 \Rightarrow b = - 1$ . So $(a, b) = (- \frac{1}{2}, - 1)$ .
  ½ mark
Answer: $(- \frac{1}{2}, - 1)$ , option (C)
Where the marks are
- Finds $a = - \frac{1}{2}$ (½ mark)
- Finds $b = - 1$ (½ mark)
Q7 · 1 mark · MCQCoordinate Geometry
The distance of point $(- 3, 4)$ from y-axis is:
1. (A) $- 3$
2. (B)3
3. (C)4
4. (D)5
Try it first, then show our solutionSolution
1. Distance from the y-axis = |x-coordinate| = $∣ - 3∣ = 3$ (distance is never negative).
  1 mark
Answer: 3, option (B)
Where the marks are
- Uses distance from y-axis = |x| = 3 (1 mark)
Q8 · 1 mark · MCQIntroduction to Trigonometry
The value of $\frac{sec ^{2} 3 0 ^{\circ} + tan ^{2} 3 0 ^{\circ}}{sin ^{2} 4 5 ^{\circ} + cos ^{2} 4 5 ^{\circ}}$ is:
1. (A)1
2. (B) $\frac{5}{3}$
3. (C) $\frac{13}{3}$
4. (D)7
Try it first, then show our solutionSolution
1. $sec^{2} 3 0^{\circ} = \frac{4}{3}$ , $tan^{2} 3 0^{\circ} = \frac{1}{3}$ , so numerator $= \frac{5}{3}$ ; denominator $= sin^{2} 4 5^{\circ} + cos^{2} 4 5^{\circ} = 1$ .
  ½ mark
2. Value $= \frac{5}{3}$ .
  ½ mark
Answer: $\frac{5}{3}$ , option (B)
Where the marks are
- Correct values of the trigonometric ratios (½ mark)
- Final value $\frac{5}{3}$ (½ mark)
Q9 · 1 mark · MCQTriangles
Which types of triangles are always similar?
1. (A)Right-angled triangles
2. (B)Acute-angled triangles
3. (C)Isosceles triangles
4. (D)Equilateral triangles
Try it first, then show our solutionSolution
1. All angles of every equilateral triangle are $6 0^{\circ}$ , so any two are similar by AAA; the other types can have different angles.
  1 mark
Answer: Equilateral triangles, option (D)
Where the marks are
- Uses AAA: equilateral triangles always have equal angles (1 mark)
Q10 · 1 mark · MCQTriangles
What values of x and y will make $△ A B C$ similar to $△ QR P$ in the figures given below?
1. (A)x = 6, y = 5
2. (B)x = 5, y = 6
3. (C)x = 6, y = 6
4. (D)x = 12, y = 3·2
Figure in the paper: Left: triangle ABC with AB = 2 cm, AC = 3 cm, BC = 2.5 cm. Right: triangle PQR with PQ = y cm, PR = x cm, QR = 4 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. For $△ A B C \sim △ QR P$ the vertices correspond as A↔Q, B↔R, C↔P. So $\frac{A B}{QR} = \frac{B C}{R P} = \frac{A C}{QP}$ .
2. $\frac{2}{4} = \frac{2.5}{x} = \frac{3}{y}$ . So $x = 2.5 \times 2 = 5$ cm and $y = 3 \times 2 = 6$ cm. The correct option is (B).
  1 mark
Answer: x = 5, y = 6, option (B)
Where the marks are
- Matches corresponding sides using A↔Q, B↔R, C↔P and obtains x = 5, y = 6, option (B) (1 mark)
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Q11 · 1 mark · MCQCircles
In the given figure, from which point can the tangent not be drawn to the circle with radius r?
1. (A)X
2. (B)P
3. (C)Q
4. (D)Y
Figure in the paper: Two concentric circles with common centre; inner circle radius r and outer circle radius R. Point X lies outside the outer circle, point P lies on the outer circle, point Q lies between the two circles, and point Y lies inside the inner circle. See it in the official paper.
Try it first, then show our solutionSolution
1. A tangent can be drawn to a circle only from a point that lies on the circle or outside it.
2. X, P and Q all lie outside the inner circle of radius r. Y lies inside it, so no tangent to that circle can be drawn from Y. The answer is option (D).
  1 mark
Answer: Y, option (D)
Where the marks are
- Identifies that a point inside the circle has no tangent, so the answer is Y (1 mark)
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Q12 · 1 mark · MCQSome Applications of Trigonometry
In the given figure, the angle of elevation of point A from point C is:
1. (A) $3 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D)Cannot be determined
Figure in the paper: Right-angled triangle ABC with right angle at B. AB = 2 cm (vertical), BC = $23$ cm (horizontal), hypotenuse AC = 4 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. $tan C = \frac{A B}{B C} = \frac{2}{2 3} = \frac{1}{3}$ , so $∠ C = 3 0^{\circ}$ .
  1 mark
Answer: $3 0^{\circ}$ , option (A)
Where the marks are
- Uses $tan C = \frac{1}{3}$ to get $3 0^{\circ}$ (1 mark)
Q13 · 1 mark · MCQAreas Related to Circles
The angle of the sector of a circle whose area is one-eighth of the area of the circle is:
1. (A) $22 \frac{1}{2}^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $\frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{1}{8} π r^{2} \Rightarrow θ = 4 5^{\circ}$ .
  1 mark
Answer: $4 5^{\circ}$ , option (B)
Where the marks are
- Sets $\frac{θ}{360} = \frac{1}{8}$ , gets $4 5^{\circ}$ (1 mark)
Q14 · 1 mark · MCQAreas Related to Circles
The perimeter of a quadrant of a circle of circumference 22 cm is:
1. (A)29 cm
2. (B)22 cm
3. (C)12·5 cm
4. (D)5·5 cm
Try it first, then show our solutionSolution
1. Circumference: $2 π r = 22$ , so $2 \times \frac{22}{7} \times r = 22$ , giving $r = 3.5$ cm.
2. Perimeter of quadrant $= 2 r + \frac{1}{4} (2 π r) = 7 + 5.5 = 12.5$ cm, which is option (C).
  1 mark
Answer: 12.5 cm, option (C)
Where the marks are
- Finds r = 3.5 cm and adds the two radii to the quarter arc to get 12.5 cm (1 mark)
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Q15 · 1 mark · MCQSurface Areas and Volumes
A cone and cylinder have same height and same radius. The volume of the cone and the volume of the cylinder are in the ratio:
1. (A)1 : 1
2. (B)1 : 3
3. (C)3 : 1
4. (D)1 : 2
Try it first, then show our solutionSolution
1. $\frac{\frac{1}{3} π r ^{2} h}{π r ^{2} h} = \frac{1}{3}$ , so ratio 1 : 3.
  1 mark
Answer: 1 : 3, option (B)
Where the marks are
- Divides the volume formulas to get 1 : 3 (1 mark)
Q16 · 1 mark · MCQStatistics
The following table shows the marks scored by 23 students of a class.
Marks 0–10 10–20 20–30 30–40 40–50
Number of Students 5 3 4 8 3
The lower limit of the modal class is:
1. (A)10
2. (B)20
3. (C)30
4. (D)40
Try it first, then show our solutionSolution
1. Highest frequency is 8, for class 30–40, so the modal class is 30–40 with lower limit 30.
  1 mark
Answer: 30, option (C)
Where the marks are
- Identifies modal class 30–40 and lower limit 30 (1 mark)
Q17 · 1 mark · MCQStatistics
For a distribution, if mean = 15 and mode = 12, then its median is:
1. (A)12
2. (B)13
3. (C)14
4. (D)15
Try it first, then show our solutionSolution
1. Empirical relation: $3 Median = Mode + 2 Mean = 12 + 30 = 42$ .
  ½ mark
2. Median $= 14$ .
  ½ mark
Answer: 14, option (C)
Where the marks are
- Uses 3 Median = Mode + 2 Mean (½ mark)
- Median = 14 (½ mark)
Q18 · 1 mark · MCQProbability
A pair of dice is thrown simultaneously. Let E denote the event that "The sum of numbers obtained on both dice is at least 9." The number of outcomes in favour of event E is:
1. (A)4
2. (B)6
3. (C)10
4. (D)26
Try it first, then show our solutionSolution
1. Sum 9: (3,6),(4,5),(5,4),(6,3) – 4; sum 10: (4,6),(5,5),(6,4) – 3; sum 11: (5,6),(6,5) – 2; sum 12: (6,6) – 1.
  ½ mark
2. Total $= 4 + 3 + 2 + 1 = 10$ .
  ½ mark
Answer: 10, option (C)
Where the marks are
- Lists outcomes for sums 9 to 12 (½ mark)
- Total 10 (½ mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Assertion (A): For $sin θ = 1$ , $cos θ$ must be 0.
Reason (R): $sin^{2} θ - cos^{2} θ = 1$ .
Select the correct answer from the codes below.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. From $sin^{2} θ + cos^{2} θ = 1$ : $1 + cos^{2} θ = 1 \Rightarrow cos θ = 0$ . A is true.
  ½ mark
2. R is false: e.g. $θ = 0^{\circ}$ gives $0 - 1 = - 1 \neq = 1$ . The correct identity is $sin^{2} θ + cos^{2} θ = 1$ .
  ½ mark
Answer: Option (C)
Where the marks are
- Shows A true (½ mark)
- Shows R false by counter-example (½ mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedQuadratic Equations
Assertion (A): Every quadratic equation has two real roots.
Reason (R): A quadratic polynomial can have at most two zeroes.
Select the correct answer from the codes below.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A is false: $x^{2} + 1 = 0$ has $D = - 4 < 0$ , so no real roots.
  ½ mark
2. R is true: a polynomial of degree 2 has at most 2 zeroes.
  ½ mark
Answer: Option (D)
Where the marks are
- Counter-example shows A false (½ mark)
- Recognises R true (½ mark)

Marks	0–10	10–20	20–30	30–40	40–50
Number of Students	5	3	4	8	3

Section B

Very short answer

Q21 · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve the following system of equations graphically: $2 x - 3 y = - 6$ and $x = 3$
Try it first, then show our solutionSolution
1. Table of values for $2 x - 3 y = - 6$ , i.e. $y = \frac{2 x + 6}{3}$ :
  $x$ $- 3$ $0$ $3$
  $y$ $0$ $2$ $4$
  The line $x = 3$ is vertical and passes through $(3, 0)$ , $(3, 2)$ , $(3, 4)$ .
2. Plot $(- 3, 0)$ , $(0, 2)$ , $(3, 4)$ and join them to get the line $2 x - 3 y = - 6$ . Draw the vertical line $x = 3$ through $(3, 0)$ , parallel to the y-axis. Both lines are drawn on the same axes and labelled.
  1½ marks
3. The two lines meet at $(3, 4)$ . So the solution is $x = 3, y = 4$ .
  ½ mark
Answer: x = 3, y = 4, i.e. the point (3, 4)
Where the marks are
- Correct table of values and correct graphs of both lines $2 x - 3 y = - 6$ and $x = 3$ (1½ marks)
- Reads the intersection point as the solution $x = 3, y = 4$ (½ mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerCompetency-basedTriangles
A vertical pole of height 10 m casts a shadow of 15 m on the ground and at the same time, a tower casts a shadow of 45 m on the ground. Find the height of the tower.
Try it first, then show our solutionSolution
1. Let the pole be AB = 10 m with shadow BC = 15 m, and the tower be PQ = $h$ m with shadow QR = 45 m. Both stand vertically, so $∠ B = ∠ Q = 9 0^{\circ}$ . The sun's rays make the same angle with the ground at the same time, so $∠ C = ∠ R$ . Hence $Δ A B C \sim Δ P QR$ [AA similarity].
  ½ mark
2. Corresponding sides are proportional:
  $\frac{A B}{P Q} = \frac{B C}{QR} \Rightarrow \frac{10}{h} = \frac{15}{45}$
  $h = \frac{10 \times 45}{15} = 30$
  Height of the tower = 30 m.
  1½ marks
Answer: Height of the tower = 30 m
Where the marks are
- Establishes that the pole-shadow triangle and the tower-shadow triangle are similar (½ mark)
- Writes the proportion $\frac{10}{h} = \frac{15}{45}$ and gets the height as 30 m (1½ marks)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerTriangles
In the given figure, $\frac{QR}{QS} = \frac{QT}{P R}$ and $∠1 = ∠2$ . Prove that $△ P QS \sim △ T QR$ .
Figure in the paper: Triangle QTR with P on side QT and S on side QR. PR is joined. Angle 1 is at R between RP and RQ; angle 2 is at Q between QP/QR region as marked. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ P QR$ , $∠1 = ∠2$ , i.e. $∠ P R Q = ∠ P QR$ . Sides opposite equal angles are equal, so $P Q = P R$ .
  ½ mark
2. Substituting $P R = P Q$ in the given ratio: $\frac{QR}{QS} = \frac{QT}{P Q} \Rightarrow \frac{QS}{QR} = \frac{QP}{QT}$ .
  ½ mark
3. In $△ P QS$ and $△ T QR$ : $\frac{QP}{QT} = \frac{QS}{QR}$ (shown above), and $∠ P QS = ∠ T QR$ (common angle $Q$ ).
  ½ mark
4. Therefore $△ P QS \sim △ T QR$ [SAS similarity criterion].
  ½ mark
Answer: $△ P QS \sim △ T QR$ by SAS similarity
Where the marks are
- Uses $∠1 = ∠2$ to get $P Q = P R$ (½ mark)
- Rewrites the ratio as $\frac{QS}{QR} = \frac{QP}{QT}$ (½ mark)
- Identifies the common included angle Q (½ mark)
- Concludes similarity by SAS (½ mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerCircles
In the given figure, TP and TQ are two tangents. If $∠ P T Q = 5 0^{\circ}$ , then find the measure of $∠ O P Q$ .
Figure in the paper: Circle with centre O, external point T with tangents TP and TQ touching the circle at P and Q. Chord PQ is drawn; angle PTQ = 50° is marked at T; angle OPQ is marked at P. See it in the official paper.
Try it first, then show our solutionSolution
1. $T P = T Q$ [tangents from an external point are equal], so in $△ T P Q$ : $∠ T P Q = ∠ T QP = \frac{18 0 ^{\circ} - 5 0 ^{\circ}}{2} = 6 5^{\circ}$ .
  1 mark
2. $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  ½ mark
3. $∠ O P Q = ∠ O P T - ∠ T P Q = 9 0^{\circ} - 6 5^{\circ} = 2 5^{\circ}$ .
  ½ mark
Answer: $∠ O P Q = 2 5^{\circ}$
Where the marks are
- Uses equal tangents to find $∠ T P Q = 6 5^{\circ}$ (1 mark)
- States $∠ O P T = 9 0^{\circ}$ with reason (½ mark)
- Obtains $∠ O P Q = 2 5^{\circ}$ (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerIntroduction to Trigonometry
If $sin 3 A = 1$ , then find the value of $cos 2 A - tan^{2} 4 5^{\circ}$ .
Try it first, then show our solutionSolution
1. $sin 3 A = 1 = sin 9 0^{\circ} \Rightarrow 3 A = 9 0^{\circ} \Rightarrow A = 3 0^{\circ}$ .
  ½ mark
2. $cos 2 A - tan^{2} 4 5^{\circ} = cos 6 0^{\circ} - (1)^{2} = \frac{1}{2} - 1 = - \frac{1}{2}$ .
  1½ marks
Answer: −1/2
Where the marks are
- Gets A = 30° from sin 3A = 1 (½ mark)
- Substitutes cos 60° = ½ and tan 45° = 1 to get −½ (1½ marks)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerIntroduction to Trigonometry
If $(sec A + tan A) (1 - sin A) = k cos A$ , then find the value of k.
Try it first, then show our solutionSolution
1. $(\frac{1}{cos A} + \frac{sin A}{cos A}) (1 - sin A) = k cos A \Rightarrow \frac{( 1 + sin A ) ( 1 - sin A )}{cos A} = k cos A$
  ½ mark
2. $1 - sin^{2} A = k cos^{2} A$
  ½ mark
3. $cos^{2} A = k cos^{2} A$ [since $1 - sin^{2} A = cos^{2} A$ ]
  ½ mark
4. $k = 1$
  ½ mark
Answer: k = 1
Where the marks are
- Writes sec A and tan A in terms of sin A and cos A (½ mark)
- Simplifies to $1 - sin^{2} A = k cos^{2} A$ (½ mark)
- Uses the identity to get $cos^{2} A = k cos^{2} A$ (½ mark)
- Concludes k = 1 (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerAreas Related to Circles
In the given figure, OPRQ is a quadrant of a circle with centre O and radius 10 cm. If OP = 10 cm and OS = 6 cm, find the area of the shaded region. (Use $π = 3.14$ )
Figure in the paper: Quadrant OPRQ of a circle with centre O; P is 10 cm to the left of O on the horizontal, Q is directly above O on the arc. S lies on OQ with OS = 6 cm. The shaded region is bounded by arc PQ and segment PS and SQ, with segment PS joined. See it in the official paper.
Try it first, then show our solutionSolution
1. Area of quadrant $O P R Q = \frac{1}{4} π r^{2} = \frac{1}{4} \times 3.14 \times 10 \times 10 = 78.5 cm^{2}$ .
  ½ mark
2. $∠ P O S = 9 0^{\circ}$ , so area of $△ O P S = \frac{1}{2} \times O P \times O S = \frac{1}{2} \times 10 \times 6 = 30 cm^{2}$ .
  ½ mark
3. Area of shaded region $=$ area of quadrant $-$ area of $△ O P S = 78.5 - 30 = 48.5 cm^{2}$ .
  1 mark
Answer: $48.5 cm^{2}$
Where the marks are
- Area of quadrant = 78.5 cm² (½ mark)
- Area of right triangle OPS = 30 cm² (½ mark)
- Shaded area = 48.5 cm² (1 mark)
Open Q25 on its own page

$x$	$- 3$	$0$	$3$
$y$	$0$	$2$	$4$

Section C

Short answer

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerReal Numbers
Prove that $2$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $2$ is rational. Then $2 = \frac{p}{q}$ , where $p$ and $q$ are co-prime integers and $q \neq = 0$ .
  ½ mark
2. Then $2 q = p$ , so $2 q^{2} = p^{2}$ . Hence 2 divides $p^{2}$ , so 2 divides $p$ [if a prime divides $a^{2}$ , it divides $a$ ]. Let $p = 2 m$ for some integer $m$ .
  1 mark
3. Then $2 q^{2} = 4 m^{2}$ , so $q^{2} = 2 m^{2}$ . Hence 2 divides $q^{2}$ , so 2 divides $q$ . Thus 2 is a common factor of $p$ and $q$ , which contradicts that $p$ and $q$ are co-prime.
  1 mark
4. So our assumption is wrong, and $2$ is irrational.
  ½ mark
Answer: √2 is irrational (proved by contradiction)
Where the marks are
- Assumes $2 = p / q$ with p, q co-prime and q ≠ 0 (½ mark)
- Shows 2 divides p (1 mark)
- Shows 2 divides q, which contradicts co-primeness (1 mark)
- Concludes √2 is irrational (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerReal Numbers
Find which among the following numbers a, b and c is/are composite numbers.
$a = 7 \times 11 \times 13 + 13$
$b = 6 \times 5 \times 4 + 4$
$c = 7 \times 13 + 6$
Try it first, then show our solutionSolution
1. $a = 7 \times 11 \times 13 + 13 = 13 (7 \times 11 + 1) = 13 \times 78 = 1014$ . It has factors other than 1 and itself, so $a$ is composite.
  1 mark
2. $b = 6 \times 5 \times 4 + 4 = 4 (30 + 1) = 4 \times 31 = 124$ . It has factors other than 1 and itself, so $b$ is composite.
  1 mark
3. $c = 7 \times 13 + 6 = 91 + 6 = 97$ . 97 is not divisible by 2, 3, 5 or 7 (and $1 1^{2} > 97$ ), so $c$ is prime, not composite.
  Hence only $a$ and $b$ are composite.
  1 mark
Answer: a and b are composite; c = 97 is prime
Where the marks are
- Shows a = 13 × 78 is composite (1 mark)
- Shows b = 4 × 31 is composite (1 mark)
- Shows c = 97 is prime and concludes only a and b are composite (1 mark)
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Q27 · 3 marks · Short answerPolynomials
Find the zeroes of the polynomial $9 s^{2} - 6 s + 1$ and verify the relationship between the zeroes and the coefficients of the given polynomial.
Try it first, then show our solutionSolution
1. $9 s^{2} - 6 s + 1 = (3 s)^{2} - 2 (3 s) (1) + 1^{2} = (3 s - 1) (3 s - 1)$ . So the zeroes are $\frac{1}{3}$ and $\frac{1}{3}$ .
  1 mark
2. Sum of zeroes $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ and $\frac{- ( coefficient of s )}{coefficient of s ^{2}} = \frac{- ( - 6 )}{9} = \frac{2}{3}$ . These are equal, so the sum relation is verified.
  1 mark
3. Product of zeroes $= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ and $\frac{constant term}{coefficient of s ^{2}} = \frac{1}{9}$ . These are equal, so the product relation is verified.
  1 mark
Answer: Zeroes: 1/3, 1/3; sum = 2/3 = −b/a, product = 1/9 = c/a (verified)
Where the marks are
- Factorises the polynomial and finds both zeroes as 1/3 (1 mark)
- Verifies sum of zeroes = −b/a (1 mark)
- Verifies product of zeroes = c/a (1 mark)
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Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerPair of Linear Equations in Two Variables
A fraction becomes $\frac{1}{3}$ , when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ , when 8 is added to its denominator. Find the fraction.
Try it first, then show our solutionSolution
1. Let the fraction be $\frac{x}{y}$ .
  ½ mark
2. $\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3 x - 3 = y \Rightarrow 3 x - y = 3 ...(i)$
  ½ mark
3. $\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4 x = y + 8 \Rightarrow 4 x - y = 8 ...(ii)$
  ½ mark
4. Subtracting (i) from (ii): $x = 5$ . From (i): $y = 3 (5) - 3 = 12$ .
  1 mark
5. The required fraction is $\frac{5}{12}$ .
  ½ mark
Answer: 5/12
Where the marks are
- Assumes the fraction as x/y (½ mark)
- Forms 3x − y = 3 (½ mark)
- Forms 4x − y = 8 (½ mark)
- Solves to get x = 5, y = 12 (1 mark)
- States the fraction 5/12 (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerPair of Linear Equations in Two Variables
Find the value of k for which the following pair of linear equations will have infinitely many solutions:
$k x + 3 y - (k - 3) = 0$ and $12 x + k y - k = 0$
Hence, find any two solutions of the given pair of equations.
Try it first, then show our solutionSolution
1. For infinitely many solutions, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ :
  $\frac{k}{12} = \frac{3}{k} = \frac{- ( k - 3 )}{- k} = \frac{k - 3}{k}$
  1 mark
2. From $\frac{k}{12} = \frac{3}{k}$ : $k^{2} = 36$ , so $k = \pm 6$ .
  From $\frac{3}{k} = \frac{k - 3}{k}$ : $3 k = k^{2} - 3 k$ , i.e. $k^{2} - 6 k = 0$ , so $k = 0$ or $6$ .
  The common value is $k = 6$ .
  1 mark
3. For $k = 6$ the equations are $6 x + 3 y - 3 = 0$ and $12 x + 6 y - 6 = 0$ . Both reduce to $2 x + y = 1$ , i.e. $y = 1 - 2 x$ .
  Solution 1: $x = 0 \Rightarrow y = 1$ , giving $(0, 1)$ .
  ½ mark
4. Solution 2: $x = 1 \Rightarrow y = - 1$ , giving $(1, - 1)$ .
  ½ mark
Answer: k = 6; two solutions: (0, 1) and (1, −1)
Where the marks are
- Writes the condition a₁/a₂ = b₁/b₂ = c₁/c₂ correctly (1 mark)
- Solves to get k = 6 (1 mark)
- Gives any two correct solutions, e.g. (0, 1) and (1, −1) (½ + ½ marks)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $\frac{1 + cosec A}{cosec A} = \frac{cos ^{2} A}{1 - sin A}$
Try it first, then show our solutionSolution
1. LHS $= \frac{1 + \frac{1}{sin A}}{\frac{1}{sin A}}$
  ½ mark
2. $= \frac{sin A + 1}{sin A} \times sin A = 1 + sin A$
  1 mark
3. $= \frac{( 1 + sin A ) ( 1 - sin A )}{1 - sin A}$
  ½ mark
4. $= \frac{1 - sin ^{2} A}{1 - sin A} = \frac{cos ^{2} A}{1 - sin A}$ = RHS [since $sin^{2} A + cos^{2} A = 1$ ]. Hence proved.
  1 mark
Answer: LHS = 1 + sin A = cos²A/(1 − sin A) = RHS (proved)
Where the marks are
- Writes cosec A as 1/sin A (½ mark)
- Simplifies LHS to 1 + sin A (1 mark)
- Multiplies and divides by (1 − sin A) (½ mark)
- Uses 1 − sin²A = cos²A to reach the RHS (1 mark)
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Q30 · 3 marks · Short answerProbability
A box contains 6 blue, 4 white and 8 red marbles. A marble is drawn at random from this box. Find the probability that the marble so drawn is:
1. (i)
  white
  1 mark
2. (ii)
  white or red
  1 mark
3. (iii)
  not red
  1 mark
Try it first, then show our solutionSolution
1. Total number of marbles $= 6 + 4 + 8 = 18$ , so there are 18 equally likely outcomes.
2. (i) P(white) $= \frac{4}{18} = \frac{2}{9}$
  1 mark
3. (ii) Marbles that are white or red $= 4 + 8 = 12$ , so P(white or red) $= \frac{12}{18} = \frac{2}{3}$
  1 mark
4. (iii) Marbles that are not red $= 6 + 4 = 10$ , so P(not red) $= \frac{10}{18} = \frac{5}{9}$
  1 mark
Answer: (i) $\frac{2}{9}$ ; (ii) $\frac{2}{3}$ ; (iii) $\frac{5}{9}$
Where the marks are
- P(white) = 2/9 (1 mark)
- P(white or red) = 2/3 (1 mark)
- P(not red) = 5/9 (1 mark)
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Q31 · 3 marks · Short answerCircles
In the given figure, a circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively. Prove that $∠ A O B + ∠ C O D = 18 0^{\circ}$ .
Figure in the paper: Quadrilateral ABCD with an inscribed circle touching AB at P, BC at Q, CD at R and DA at S. Diagonals AC and BD intersect at O; angles AOB and COD are marked at O. See it in the official paper.
Try it first, then show our solutionSolution
1. Let O be the centre of the inscribed circle. Join OA, OB, OC, OD and OP, OQ, OR, OS.
2. In $△ O A P$ and $△ O A S$ : $A P = A S$ [tangents from an external point are equal], $O P = O S$ [radii] and $O A = O A$ [common]. So $△ O A P ≅ △ O A S$ [SSS], which gives $∠ A O P = ∠ A O S$ .
  1 mark
3. Similarly, $∠ B O P = ∠ B O Q$ , $∠ C O Q = ∠ C O R$ and $∠ D O R = ∠ D O S$ .
  ½ mark
4. The angles round the point O add up to $36 0^{\circ}$ : $∠ A O P + ∠ B O P + ∠ B O Q + ∠ C O Q + ∠ C O R + ∠ D O R + ∠ D O S + ∠ A O S = 36 0^{\circ}$ , so $2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$ .
  1 mark
5. Hence $(∠ A O P + ∠ B O P) + (∠ C O R + ∠ D O R) = 18 0^{\circ}$ , i.e. $∠ A O B + ∠ C O D = 18 0^{\circ}$ .
  ½ mark
Answer: $∠ A O B + ∠ C O D = 18 0^{\circ}$ (proved)
Where the marks are
- Proves $△ O A P ≅ △ O A S$ using equal tangents, giving equal angles at O (1 mark)
- Writes the similar equal pairs at B, C and D (½ mark)
- Uses the sum of angles at O = 360° and pairs the equal angles (1 mark)
- Concludes $∠ A O B + ∠ C O D = 18 0^{\circ}$ (½ mark)
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Section D

Long answer

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerQuadratic Equations
Find two consecutive odd integers, sum of whose squares is 290.
Try it first, then show our solutionSolution
1. Let the two consecutive odd integers be $x$ and $x + 2$ .
  ½ mark
2. According to the question: $x^{2} + (x + 2)^{2} = 290$
  1½ marks
3. $x^{2} + x^{2} + 4 x + 4 = 290 \Rightarrow 2 x^{2} + 4 x - 286 = 0 \Rightarrow x^{2} + 2 x - 143 = 0$
  1½ marks
4. $x^{2} + 13 x - 11 x - 143 = 0 \Rightarrow (x - 11) (x + 13) = 0 \Rightarrow x = 11$ or $x = - 13$
  1 mark
5. The required integers are 11 and 13. Check: $121 + 169 = 290$ . (The root $x = - 13$ gives the pair $- 13, - 11$ , which also works.)
  ½ mark
Answer: 11 and 13 (also −13 and −11)
Where the marks are
- Takes the integers as x and x + 2 (½ mark)
- Forms $x^{2} + (x + 2)^{2} = 290$ (1½ marks)
- Simplifies to $x^{2} + 2 x - 143 = 0$ (1½ marks)
- Solves to get x = 11 (1 mark)
- States the integers 11 and 13 (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerQuadratic Equations
A charity trust decides to build a rectangular hall having an area of $300 m^{2}$ . The length of the hall is one metre more than twice its width. Find the length and breadth of the hall.
Try it first, then show our solutionSolution
1. Let the breadth be $x$ m. Then the length is $(2 x + 1)$ m.
  ½ mark
2. Area $= 300$ : $x (2 x + 1) = 300$
  1½ marks
3. $2 x^{2} + x - 300 = 0$
  1½ marks
4. $2 x^{2} + 25 x - 24 x - 300 = 0 \Rightarrow x (2 x + 25) - 12 (2 x + 25) = 0 \Rightarrow (x - 12) (2 x + 25) = 0$
  $x = 12$ or $x = - \frac{25}{2}$ . A breadth cannot be negative, so $x = 12$ .
  1 mark
5. Breadth = 12 m and length = $2 (12) + 1 = 25$ m. Check: $25 \times 12 = 300 m^{2}$ .
  ½ mark
Answer: Length = 25 m, breadth = 12 m
Where the marks are
- Takes breadth x and length 2x + 1 (½ mark)
- Forms x(2x + 1) = 300 (1½ marks)
- Writes the quadratic 2x² + x − 300 = 0 (1½ marks)
- Solves and rejects the negative root to get x = 12 (1 mark)
- States length 25 m and breadth 12 m (½ mark)
Open Q32(b) on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerTriangles
State and Prove "Basic Proportionality Theorem".
Try it first, then show our solutionSolution
1. Statement (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle so that it meets the other two sides in distinct points, then it divides those two sides in the same ratio.
  1 mark
2. Given: In $△ A B C$ , a line parallel to $B C$ meets $A B$ at $D$ and $A C$ at $E$ , so $D E ∥ B C$ .
  ½ mark
3. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  ½ mark
4. Construction: Join $B E$ and $C D$ . Draw $E N ⊥ A B$ (with $N$ on $A B$ ) and $D M ⊥ A C$ (with $M$ on $A C$ ).
  ½ mark
5. Figure: Draw $△ A B C$ with $A$ at the top. Mark $D$ on $A B$ and $E$ on $A C$ with $D E ∥ B C$ . Show the segments $B E$ and $C D$ , and the perpendiculars $E N$ to $A B$ and $D M$ to $A C$ .
  ½ mark
6. Proof: Take $E N$ as the height. Then $ar (△ A D E) = \frac{1}{2} \cdot A D \cdot E N$ and $ar (△ B D E) = \frac{1}{2} \cdot D B \cdot E N$ .
  So $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ ...(1)
  Now take $D M$ as the height. Then $ar (△ A D E) = \frac{1}{2} \cdot A E \cdot D M$ and $ar (△ D E C) = \frac{1}{2} \cdot E C \cdot D M$ .
  So $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ...(2)
  1 mark
7. $△ B D E$ and $△ D E C$ stand on the same base $D E$ and lie between the same parallels $D E$ and $B C$ . Hence $ar (B D E) = ar (D E C)$ ...(3)
  From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: Statement of BPT and proof that $\frac{A D}{D B} = \frac{A E}{E C}$ when $D E ∥ B C$
Where the marks are
- States the theorem correctly (1 mark)
- Writes the Given, To prove and Construction, and draws a correct labelled figure, ½ mark each (2 marks)
- Expresses the two area ratios as $A D / D B$ and $A E / E C$ using the perpendicular heights (1 mark)
- Uses ar(BDE) = ar(DEC) [same base, between same parallels] and concludes (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerTriangles
In the given figure, CM and RN are respectively, the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ , prove that:
(i) $△ A M C \sim △ P N R$
(ii) $∠ B C M = ∠ QR N$
(iii) $△ B M C \sim △ QN R$
Figure in the paper: Triangle ABC with M the midpoint of AB, CM joined; triangle PQR with N the midpoint of PQ, RN joined. Vertices: A top, B left, C bottom; P top right, Q left, R bottom. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $△ A B C \sim △ P QR$ : $∠ A = ∠ P$ , $∠ B = ∠ Q$ , $∠ A C B = ∠ P R Q$ and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R}$ ...(1)
  1 mark
2. (i) M and N are midpoints, so $A M = \frac{1}{2} A B$ and $P N = \frac{1}{2} P Q$ , giving $\frac{A M}{P N} = \frac{A B}{P Q} = \frac{A C}{P R}$ . In $△ A M C$ and $△ P N R$ we also have $∠ M A C = ∠ N P R$ . Therefore $△ A M C \sim △ P N R$ [SAS similarity].
  1½ marks
3. (ii) From (i), $∠ A C M = ∠ P R N$ [corresponding angles of similar triangles]. Also $∠ A C B = ∠ P R Q$ from (1). Subtracting, $∠ A C B - ∠ A C M = ∠ P R Q - ∠ P R N$ , i.e. $∠ B C M = ∠ QR N$ .
  1½ marks
4. (iii) In $△ B M C$ and $△ QN R$ : $∠ M B C = ∠ N QR$ (as $∠ B = ∠ Q$ ) and $∠ B C M = ∠ QR N$ (from (ii)). Therefore $△ B M C \sim △ QN R$ [AA similarity].
  1 mark
Answer: (i) $△ A M C \sim △ P N R$ (SAS); (ii) $∠ B C M = ∠ QR N$ ; (iii) $△ B M C \sim △ QN R$ (AA)
Where the marks are
- Writes the equal angles and proportional sides from the given similarity (1 mark)
- Uses the medians to get $\frac{A M}{P N} = \frac{A C}{P R}$ and proves (i) by SAS (1½ marks)
- Deduces $∠ B C M = ∠ QR N$ by subtracting equal angles (1½ marks)
- Proves (iii) by AA (1 mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A perfume bottle is in the form of a cylinder but the bottom of the bottle has a hemispherical raised portion to reduce the capacity of the bottle. The inner diameter of the bottle is 5 cm and the height of the bottle is 10 cm. Find the capacity of the perfume bottle in mL. (Use $π = 3.14$ and $1 c m^{3} = 1 mL$ )
Figure in the paper: Cylindrical bottle of height 10 cm with a hemispherical portion projecting up from the base into the inside of the bottle; the bottle's top has a small neck/cap. See it in the official paper.
Try it first, then show our solutionSolution
1. Inner radius $r = \frac{5}{2} = 2.5$ cm and height $h = 10$ cm. The hemispherical raised portion has the same radius, 2.5 cm.
  ½ mark
2. Volume of the cylinder $= π r^{2} h = 3.14 \times 2.5 \times 2.5 \times 10 = 196.25 cm^{3}$ .
  1½ marks
3. Volume of the hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 15.625 \approx 32.71 cm^{3}$ .
  1½ marks
4. Capacity $=$ volume of cylinder $-$ volume of hemisphere $= 196.25 - 32.71 = 163.54 cm^{3}$ .
  1 mark
5. Since $1 cm^{3} = 1$ mL, the capacity of the bottle is approximately $163.54$ mL.
  ½ mark
Answer: Capacity ≈ 163.54 mL
Where the marks are
- Takes radius 2.5 cm for both the cylinder and the hemisphere (½ mark)
- Volume of cylinder = 196.25 cm³ (1½ marks)
- Volume of hemisphere ≈ 32.71 cm³ (1½ marks)
- Subtracts to get ≈ 163.54 cm³ (1 mark)
- Converts to ≈ 163.54 mL (½ mark)
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Q35 · 5 marks · Long answerStatistics
The weights of 30 students of a class are given in the following distribution table:
Weight (in kg) Number of students
40 – 45 2
45 – 50 5
50 – 55 8
55 – 60 6
60 – 65 6
65 – 70 3
Find the median weight of the students.
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Weight (kg) $f$ $c f$
  40–45 2 2
  45–50 5 7
  50–55 8 15
  55–60 6 21
  60–65 6 27
  65–70 3 30
  $N = \sum f = 30$
  2 marks
2. $\frac{N}{2} = 15$ . The first $c f$ that is greater than or equal to 15 is 15, so the median class is 50–55.
  For this class: $l = 50$ , $c f = 7$ (of the preceding class), $f = 8$ , $h = 5$ .
3. Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 50 + \frac{15 - 7}{8} \times 5$
  2 marks
4. $= 50 + \frac{8}{8} \times 5 = 50 + 5 = 55$
  The median weight of the students is 55 kg.
  1 mark
Answer: Median weight = 55 kg
Where the marks are
- Correct cumulative frequency table with $N = 30$ (2 marks)
- Identifies median class 50–55, writes the median formula and substitutes $l = 50$ , $c f = 7$ , $f = 8$ , $h = 5$ (2 marks)
- Gets median = 55 kg (1 mark)
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Weight (in kg)	Number of students
40 – 45	2
45 – 50	5
50 – 55	8
55 – 60	6
60 – 65	6
65 – 70	3

Weight (kg)	$f$	$c f$
40–45	2	2
45–50	5	7
50–55	8	15
55–60	6	21
60–65	6	27
65–70	3	30

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Case Study 1
In a society, a yoga instructor was hired to train the people of the society to live a healthy lifestyle. Yoga sessions were held daily from 5 p.m. to 7 p.m. in the society park. On day one, 5 people joined the yoga session, on day two, 3 more people joined, on day three, another 3 people joined and in this manner every next day, 3 more people kept on joining.
Based on the given information, answer the following questions:
1. (i)
  On which day did 59 people join the yoga session?
  1 mark
2. (ii)
  How many people joined the yoga session on the 31st day?
  1 mark
3. (iii)
  (a) The yoga instructor was paid ₹100 for each person attending the yoga session. On which day would he earn ₹5,000?
  2 marks
4. OR
  (iii)
  (b) What was the total amount earned by the yoga instructor in 16 days?
  2 marks
Figure in the paper: Illustration of people doing yoga outdoors in a park with trees, each person on a mat in different poses. See it in the official paper.
Try it first, then show our solutionSolution
1. The numbers of people attending on successive days are 5, 8, 11, ..., an AP with $a = 5$ and $d = 3$ .
2. (i) $a_{n} = a + (n - 1) d = 59 \Rightarrow 5 + 3 (n - 1) = 59 \Rightarrow n - 1 = 18 \Rightarrow n = 19$ . So there were 59 people on the 19th day.
  1 mark
3. (ii) $a_{31} = 5 + 30 \times 3 = 95$ people.
  1 mark
4. (iii)(a) To earn ₹5,000 at ₹100 per person, the instructor needs $\frac{5000}{100} = 50$ people. Then $5 + 3 (n - 1) = 50 \Rightarrow 3 (n - 1) = 45 \Rightarrow n = 16$ .
  1½ marks
5. (iii)(a) He earns ₹5,000 on the 16th day.
  ½ mark
6. (iii)(b) Total attendance in 16 days: $S_{16} = \frac{16}{2} [2 (5) + 15 (3)] = 8 \times 55 = 440$ .
  1½ marks
7. (iii)(b) Total amount earned $= 440 \times ₹100 = ₹44, 000$ .
  ½ mark
Answer: (i) 19th day; (ii) 95 people; (iii)(a) 16th day; (iii)(b) ₹44,000
Where the marks are
- Solves $5 + 3 (n - 1) = 59$ to get n = 19 (1 mark)
- Finds $a_{31} = 95$ (1 mark)
- (iii)(a) Converts ₹5,000 into 50 people and sets up $a_{n} = 50$ (1½ marks), getting the 16th day (½ mark)
- (iii)(b) Finds $S_{16} = 440$ (1½ marks) and the amount ₹44,000 (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Case Study 2
Two motorboats A and B are waiting at the opposite banks of a river in order to reach the opposite side. From a point P on the bridge, 20 m above the river, the angles of depression of the boats are $3 0^{\circ}$ and $4 5^{\circ}$ respectively, as shown in the figure given below. Both the boats leave at the same time at the speed of 10 m/s and 5 m/s, respectively.
Based on the above information, answer the following questions:
1. (i)
  Find the distance travelled by boat A to reach point D in the river, vertically below the point P. (Use $3 = 1.73$ )
  1 mark
2. (ii)
  What is the width of the river?
  1 mark
3. (iii)
  (a) Which boat will reach point D first, and how much earlier, than the other boat?
  2 marks
4. OR
  (iii)
  (b) What is the distance between the two boats after 3 seconds?
  2 marks
Figure in the paper: Points A and B on the river's opposite banks on a horizontal line; D lies on this line vertically below P, with AD and DB along it. P is on a bridge 20 m above the river, with angle of depression 30° to A and 45° to B, marked at P. PD = 20 m is shown as a vertical dashed line. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In right $△ P D A$ , $∠ P A D = 3 0^{\circ}$ (alternate to the angle of depression), so $tan 3 0^{\circ} = \frac{P D}{A D} \Rightarrow A D = 203 = 20 \times 1.73 = 34.6$ m.
  1 mark
2. (ii) In right $△ P D B$ , $tan 4 5^{\circ} = \frac{P D}{D B} \Rightarrow D B = 20$ m. Width of the river $= A D + D B = 34.6 + 20 = 54.6$ m.
  1 mark
3. (iii)(a) Time for boat A $= \frac{34.6}{10} = 3.46$ s; time for boat B $= \frac{20}{5} = 4$ s.
  1 mark
4. (iii)(a) Boat A reaches D first, $4 - 3.46 = 0.54$ s earlier than boat B.
  1 mark
5. (iii)(b) In 3 s, boat A covers $10 \times 3 = 30$ m, so it is $34.6 - 30 = 4.6$ m from D. Boat B covers $5 \times 3 = 15$ m, so it is $20 - 15 = 5$ m from D on the other side.
  1 mark
6. (iii)(b) Distance between the boats $= 4.6 + 5 = 9.6$ m.
  1 mark
Answer: (i) 34.6 m; (ii) 54.6 m; (iii)(a) Boat A, 0.54 s earlier; (iii)(b) 9.6 m
Where the marks are
- AD = 20√3 = 34.6 m using tan 30° (1 mark)
- DB = 20 m and width = 54.6 m (1 mark)
- (iii)(a) Times 3.46 s and 4 s (1 mark); boat A first by 0.54 s (1 mark)
- (iii)(b) Remaining distances 4.6 m and 5 m (1 mark); separation 9.6 m (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Case Study 3
There is a semicircular park in Aman's society. He wishes to plant saplings along the boundary of the park. There is a borewell at the centre O of the park along the diameter AB as shown in the figure below.
Based on the above information, answer the following questions:
1. (i)
  Find the coordinates of point O.
  1 mark
2. (ii)
  Find the radius of the semicircular park.
  1 mark
3. (iii)
  (a) One sapling is kept at point C(12, y). Find the coordinates of C.
  2 marks
4. OR
  (iii)
  (b) One sapling is kept at point P along AB so that $P A = \frac{1}{3} P B$ . Find the coordinates of P.
  2 marks
Figure in the paper: Semicircular park with diameter AB along a horizontal line; A(2, 3) at left end and B(22, 3) at right end. O is the centre on AB, P is a point on AB where a sapling is planted, and C(12, y) is the lowest point of the semicircular arc. A borewell is drawn at O. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) O is the midpoint of AB: $O = (\frac{2 + 22}{2}, \frac{3 + 3}{2}) = (12, 3)$ .
  1 mark
2. (ii) $A B = (22 - 2)^{2} + (3 - 3)^{2} = 20$ , so the radius $= \frac{20}{2} = 10$ units.
  1 mark
3. (iii)(a) C lies on the boundary, so $O C = 10$ : $(12 - 12)^{2} + (y - 3)^{2} = 10 \Rightarrow (y - 3)^{2} = 100 \Rightarrow y = 13$ or $y = - 7$ .
  1 mark
4. (iii)(a) C is the lowest point of the arc, which lies below AB, so $y = - 7$ and $C = (12, - 7)$ .
  1 mark
5. (iii)(b) $P A = \frac{1}{3} P B \Rightarrow P A : P B = 1 : 3$ , so P divides AB in the ratio $1 : 3$ .
  ½ mark
6. (iii)(b) By the section formula, $P = (\frac{1 ( 22 ) + 3 ( 2 )}{1 + 3}, \frac{1 ( 3 ) + 3 ( 3 )}{1 + 3}) = (7, 3)$ .
  1½ marks
Answer: (i) O(12, 3); (ii) 10 units; (iii)(a) C(12, −7); (iii)(b) P(7, 3)
Where the marks are
- Midpoint O = (12, 3) (1 mark)
- Radius = 10 units (1 mark)
- (iii)(a) Uses OC = radius to form $(y - 3)^{2} = 100$ (1 mark) and selects C(12, −7) (1 mark)
- (iii)(b) Ratio 1 : 3 (½ mark) and P(7, 3) by the section formula (1½ marks)
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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.