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

CBSE Class 10 Mathematics Basic: 2025 board paper, set 430/3/1

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/241_Mathematics_Basic.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; internal choice in 22
C	26-31	Short Answer; internal choice in 26 and 28
D	32-35	Long Answer; internal choice in 32 and 33
E	36-38	Case study based; internal choice in the 2-mark subparts (iii) of each case

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 subparts in Section E; use of calculator not allowed; take π = 22/7 where not stated.

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 $Δ A B C$ similar to $Δ QR P$ 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, BC = 2.5 cm, AC = 3 cm. Right: triangle PQR (vertices P at top, Q bottom left, R bottom right) with PQ = y cm, PR = x cm, QR = 4 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. $Δ A B C \sim Δ QR P$ gives $\frac{A B}{QR} = \frac{B C}{R P} = \frac{A C}{QP}$ , i.e. $\frac{2}{4} = \frac{2.5}{x} = \frac{3}{y}$ .
  ½ mark
2. $x = 5$ , $y = 6$ .
  ½ mark
Answer: x = 5, y = 6, option (B)
Where the marks are
- Matches A↔Q, B↔R, C↔P and writes ratios (½ mark)
- Gets x = 5, y = 6 (½ mark)
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; the outer circle has radius R and the inner circle has radius r (marked from centre). 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. No tangent can be drawn to a circle from a point inside it. Y lies inside the circle of radius r (X, P, Q are all outside it).
  1 mark
Answer: Y, option (D)
Where the marks are
- States no tangent from an interior point; identifies Y (1 mark)
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 the right angle at B. AB = 2 cm (vertical), BC = $23$ cm (horizontal), and 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. $2 π r = 22 \Rightarrow r = \frac{22 \times 7}{2 \times 22} = 3.5$ cm.
  ½ mark
2. Perimeter $= \frac{22}{4} + 2 r = 5.5 + 7 = 12.5$ cm.
  ½ mark
Answer: 12.5 cm, option (C)
Where the marks are
- Finds r = 3.5 cm (½ mark)
- Perimeter = arc + 2 radii = 12.5 cm (½ mark)
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)
Open Q19 on its own page
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; internal choice in 22

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)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerCompetency-basedSome Applications of Trigonometry
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 $Δ P QS \sim Δ T QR$ .
Figure in the paper: Triangle QRT with P on side QT and S on side QR. Segment PR joins P to R, and segment PS joins P to S. Angle 2 is marked at vertex Q (between QS/QR and QP/QT) and angle 1 is marked at vertex R (between RP and RQ). See it in the official paper.
Try it first, then show our solutionSolution
1. In $Δ P QR$ , $∠1 = ∠2$ , i.e. $∠ P R Q = ∠ P QR$ . Hence $P Q = P R$ [sides opposite equal angles are equal].
  ½ mark
2. Given $\frac{QR}{QS} = \frac{QT}{P R}$ . Putting $P R = P Q$ gives $\frac{QR}{QS} = \frac{QT}{QP}$ , i.e. $\frac{QS}{QR} = \frac{QP}{QT}$ .
  ½ mark
3. In $Δ P QS$ and $Δ T QR$ : $∠ P QS = ∠ T QR$ , since this is $∠ Q$ , common to both.
  ½ mark
4. The sides including the equal angle are proportional, so $Δ P QS \sim Δ T QR$ [SAS similarity criterion].
  ½ mark
Answer: ΔPQS ~ ΔTQR by SAS similarity
Where the marks are
- Uses ∠1 = ∠2 to get PQ = PR (½ mark)
- Replaces PR by PQ to get $\frac{QR}{QS} = \frac{QT}{QP}$ (½ mark)
- Notes ∠Q is common (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: A circle with centre O; external point T with tangents TP and TQ touching the circle at P and Q. Chord PQ is drawn, and angle PTQ = 50° is marked at T. Angle OPQ is marked at P between PO and PQ. See it in the official paper.
Try it first, then show our solutionSolution
1. $T P = T Q$ [tangents drawn from an external point are equal]. So $∠ T P Q = ∠ T QP$ [angles opposite equal sides].
  ½ mark
2. In $Δ T P Q$ : $5 0^{\circ} + 2∠ T P Q = 18 0^{\circ}$ , so $∠ T P Q = 6 5^{\circ}$ .
  1 mark
3. $∠ O P T = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]. So $∠ O P Q = 9 0^{\circ} - 6 5^{\circ} = 2 5^{\circ}$ .
  ½ mark
Answer: ∠OPQ = 25°
Where the marks are
- Uses TP = TQ to get ∠TPQ = ∠TQP (½ mark)
- Finds ∠TPQ = 65° from the angle sum of the triangle (1 mark)
- Uses OP ⟂ TP to get ∠OPQ = 25° (½ mark)
Open Q23 on its own page
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 at bottom right; P is at the left end of the horizontal radius OP (10 cm), Q is at the top of the vertical radius OQ, and R is a point on the arc PQ. S lies on OQ with OS = 6 cm. The shaded region is bounded by the arc PRQ, the segment QS along OQ, and the segment SP. See it in the official paper.
Try it first, then show our solutionSolution
1. Area of shaded region = Area of quadrant OPRQ − Area of $Δ P O S$
  $= \frac{1}{4} \times 3.14 \times 10 \times 10 - \frac{1}{2} \times 10 \times 6$
  1 mark
2. $= 78.5 - 30 = 48.5 cm^{2}$
  1 mark
Answer: 48.5 cm²
Where the marks are
- Writes shaded area = quadrant − ΔPOS with correct substitution (1 mark)
- Gets 48.5 cm² (1 mark)
Open Q25 on its own page

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

Section C

Short Answer; internal choice in 26 and 28

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)
Open Q26(b) on its own page
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)
Open Q27 on its own page
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)
Open Q29 on its own page
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.
  (i) P(white) $= \frac{4}{18} = \frac{2}{9}$
  1 mark
2. (ii) Favourable outcomes $= 4 + 8 = 12$ . P(white or red) $= \frac{12}{18} = \frac{2}{3}$
  1 mark
3. (iii) Non-red marbles $= 6 + 4 = 10$ . P(not red) $= \frac{10}{18} = \frac{5}{9}$
  1 mark
Answer: (i) 2/9; (ii) 2/3; (iii) 5/9
Where the marks are
- (i) P(white) = 4/18 = 2/9 (1 mark)
- (ii) P(white or red) = 12/18 = 2/3 (1 mark)
- (iii) P(not red) = 10/18 = 5/9 (1 mark)
Open Q30 on its own page
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 (A bottom left, B bottom right, C top right, D top left) 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, which is inside the circle; angle AOB is marked at O, and angle COD is also marked at O on the opposite side. See it in the official paper.
Try it first, then show our solutionSolution
1. Join OP, OQ, OR, OS, where O is the centre. In $Δ O A P$ and $Δ O A S$ : $O P = O S$ [radii], $O A = O A$ [common], $∠ O P A = ∠ O S A = 9 0^{\circ}$ [tangent ⟂ radius]. So $Δ O A P ≅ Δ O A S$ [RHS], and hence $∠ A O P = ∠ A O S$ [CPCT].
  1 mark
2. Similarly, $∠ B O P = ∠ B O Q$ , $∠ C O Q = ∠ C O R$ and $∠ D O R = ∠ D O S$ .
  1 mark
3. The angles round O add up to $36 0^{\circ}$ :
  $∠ A O P + ∠ B O P + ∠ B O Q + ∠ C O Q + ∠ C O R + ∠ D O R + ∠ D O S + ∠ A O S = 36 0^{\circ}$
  $\Rightarrow 2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$
  ½ mark
4. $\Rightarrow (∠ A O P + ∠ B O P) + (∠ C O R + ∠ D O R) = 18 0^{\circ}$
  $\Rightarrow ∠ A O B + ∠ C O D = 18 0^{\circ}$ . Hence proved.
  ½ mark
Answer: ∠AOB + ∠COD = 180° (proved)
Where the marks are
- Proves ΔOAP ≅ ΔOAS and gets equal angles at O (1 mark)
- States the similar equalities for the other three pairs of angles (1 mark)
- Uses angles round O = 360° (½ mark)
- Concludes ∠AOB + ∠COD = 180° (½ mark)
Open Q31 on its own page

Section D

Long Answer; internal choice in 32 and 33

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 $Δ A B C$ and $Δ P QR$ . If $Δ A B C \sim Δ P QR$ , prove that:
(i) $Δ A M C \sim Δ P N R$
(ii) $∠ B C M = ∠ QR N$
(iii) $Δ B M C \sim Δ QN R$
Figure in the paper: Triangle ABC with M the midpoint of AB (M lies on side AB) and median CM drawn from C. Triangle PQR with N the midpoint of PQ (N lies on side QP) and median RN drawn from R. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Since $Δ A B C \sim Δ P QR$ , $\frac{A B}{P Q} = \frac{A C}{P R}$ .
  ½ mark
2. (i) M and N are midpoints of AB and PQ, so $A M = \frac{1}{2} A B$ and $P N = \frac{1}{2} P Q$ . Then $\frac{A M}{P N} = \frac{\frac{1}{2} A B}{\frac{1}{2} P Q} = \frac{A B}{P Q} = \frac{A C}{P R}$ .
  ½ mark
3. (i) Also $∠ A = ∠ P$ [corresponding angles of similar triangles].
  ½ mark
4. (i) Hence $Δ A M C \sim Δ P N R$ [SAS similarity criterion].
  ½ mark
5. (ii) From (i), $∠ A C M = ∠ P R N$ [corresponding angles]. Also $∠ A C B = ∠ P R Q$ [as $Δ A B C \sim Δ P QR$ ].
  ½ mark
6. (ii) Subtracting: $∠ A C B - ∠ A C M = ∠ P R Q - ∠ P R N$ , so $∠ B C M = ∠ QR N$ .
  ½ mark
7. (iii) Since $Δ A B C \sim Δ P QR$ , $\frac{A B}{P Q} = \frac{B C}{QR}$ .
  ½ mark
8. (iii) $B M = \frac{1}{2} A B$ and $QN = \frac{1}{2} P Q$ , so $\frac{B M}{QN} = \frac{A B}{P Q} = \frac{B C}{QR}$ .
  ½ mark
9. (iii) Also $∠ B = ∠ Q$ [corresponding angles of similar triangles].
  ½ mark
10. (iii) Hence $Δ B M C \sim Δ QN R$ [SAS similarity criterion].
  ½ mark
Answer: (i) ΔAMC ~ ΔPNR (SAS); (ii) ∠BCM = ∠QRN; (iii) ΔBMC ~ ΔQNR (SAS)
Where the marks are
- (i) Writes AB/PQ = AC/PR from the given similarity (½ mark)
- (i) Uses the medians to get AM/PN = AC/PR (½ mark)
- (i) States ∠A = ∠P (½ mark)
- (i) Concludes ΔAMC ~ ΔPNR by SAS (½ mark)
- (ii) Gets ∠ACM = ∠PRN and ∠ACB = ∠PRQ (½ mark)
- (ii) Subtracts to get ∠BCM = ∠QRN (½ mark)
- (iii) Writes AB/PQ = BC/QR (½ mark)
- (iii) Gets BM/QN = BC/QR (½ mark)
- (iii) States ∠B = ∠Q (½ mark)
- (iii) Concludes ΔBMC ~ ΔQNR by SAS (½ 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: A cylindrical bottle of height 10 cm, with a hemispherical portion raised from the base into the interior of the bottle. The height of 10 cm is marked along the outer side of the cylinder. A small nozzle sits on top of the bottle. See it in the official paper.
Try it first, then show our solutionSolution
1. Radius of cylinder = radius of hemisphere $= r = \frac{5}{2} = 2.5$ cm; height of cylinder $h = 10$ cm.
  Capacity = Volume of cylinder − Volume of hemisphere $= π r^{2} h - \frac{2}{3} π r^{3}$
  ½ mark
2. Volume of cylinder $= 3.14 \times \frac{5}{2} \times \frac{5}{2} \times 10 = 196.25 cm^{3}$
  1½ marks
3. Volume of hemisphere $= \frac{2}{3} \times 3.14 \times \frac{5}{2} \times \frac{5}{2} \times \frac{5}{2} = \frac{785}{24} \approx 32.71 cm^{3}$
  1½ marks
4. Capacity $= 196.25 - 32.71 = \frac{3925}{24} \approx 163.5 cm^{3}$
  1 mark
5. Since $1 cm^{3} = 1$ mL, capacity of the bottle ≈ 163.5 mL.
  ½ mark
Answer: ≈ 163.5 mL (3925/24 cm³)
Where the marks are
- Identifies r = 2.5 cm, h = 10 cm and capacity = cylinder − hemisphere (½ mark)
- Correct volume of the cylinder, 196.25 cm³ (1½ marks)
- Correct volume of the hemisphere, ≈ 32.71 cm³ (1½ marks)
- Subtracts to get ≈ 163.5 cm³ (1 mark)
- Converts to 163.5 mL (½ mark)
Open Q34 on its own page
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)
Open Q35 on its own page

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; internal choice in the 2-mark subparts (iii) of each case

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
Try it first, then show our solutionSolution
1. The number of people attending on successive days is 5, 8, 11, … This is an AP with $a = 5$ and $d = 3$ , so $a_{n} = 5 + (n - 1) 3$ .
2. (i) $5 + (n - 1) \times 3 = 59 \Rightarrow 3 (n - 1) = 54 \Rightarrow n - 1 = 18 \Rightarrow n = 19$ .
  So 59 people attended on the 19th day.
  1 mark
3. (ii) $a_{31} = 5 + 30 \times 3 = 5 + 90 = 95$ .
  So 95 people attended on the 31st day.
  1 mark
4. (iii)(a) To earn ₹5,000 at ₹100 per person, the number of persons needed $= \frac{5000}{100} = 50$ .
  1 mark
5. (iii)(a) $5 + (n - 1) \times 3 = 50 \Rightarrow 3 (n - 1) = 45 \Rightarrow n = 16$ .
  He earns ₹5,000 on the 16th day.
  1 mark
6. (iii)(b) $S_{16} = \frac{n}{2} [2 a + (n - 1) d] = \frac{16}{2} [2 \times 5 + 15 \times 3]$
  1 mark
7. (iii)(b) $= 8 [10 + 45] = 8 \times 55 = 440$ persons in total over 16 days.
  ½ mark
8. (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
- (i) Uses $5 + (n - 1) 3 = 59$ to get $n = 19$ (1 mark)
- (ii) $a_{31} = 95$ (1 mark)
- (iii)(a) Number of persons $= 5000/100 = 50$ (1 mark)
- (iii)(a) Solves $5 + (n - 1) 3 = 50$ to get $n = 16$ (1 mark)
- (iii)(b) Correct substitution in $S_{16} = \frac{16}{2} [10 + 15 \times 3]$ (1 mark)
- (iii)(b) $S_{16} = 440$ (½ mark)
- (iii)(b) Amount $= ₹44, 000$ (½ mark)
Open Q36 on its own page
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: Side view of a river with banks. P is a point on the bridge 20 m above the river. D is directly below P on the river surface, with a dashed vertical line PD of length 20 m. Boat A is at point A on the left, and boat B is at point B on the right, both on the water line AB with D between them. The angle of depression from P to A is 30° and from P to B is 45°. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) PD = 20 m and PD ⟂ AB. The angle of depression of A is 30°, so $∠ P A D = 30°$ [alternate angles].
  In right $△ P D A$ : $tan 30° = \frac{P D}{A D} = \frac{20}{A D}$
  ½ mark
2. (i) $\frac{1}{3} = \frac{20}{A D} \Rightarrow A D = 203 = 20 \times 1.73 = 34.6$ m.
  Boat A travels 34.6 m to reach D.
  ½ mark
3. (ii) The angle of depression of B is 45°, so $∠ P B D = 45°$ .
  In right $△ P D B$ : $tan 45° = \frac{20}{D B} \Rightarrow 1 = \frac{20}{D B} \Rightarrow D B = 20$ m.
  ½ mark
4. (ii) Width of the river $= A D + D B = 34.6 + 20 = 54.6$ m.
  ½ mark
5. (iii)(a) Time taken by boat A to reach D $= \frac{distance}{speed} = \frac{34.6}{10} = 3.46$ s.
  1 mark
6. (iii)(a) Time taken by boat B to reach D $= \frac{20}{5} = 4$ s.
  ½ mark
7. (iii)(a) Boat A reaches D first, earlier by $4 - 3.46 = 0.54$ s.
  ½ mark
8. (iii)(b) Distance covered by boat A in 3 s $= 10 \times 3 = 30$ m. This is less than 34.6 m, so A has not yet reached D.
  ½ mark
9. (iii)(b) Distance covered by boat B in 3 s $= 5 \times 3 = 15$ m. This is less than 20 m, so B has not yet reached D.
  ½ mark
10. (iii)(b) The boats move towards each other, so the distance between them $= 54.6 - (30 + 15) = 9.6$ 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
- (i) Writes $tan 30° = 20/ A D$ (½ mark)
- (i) $A D = 203 = 34.6$ m (½ mark)
- (ii) $tan 45° = 20/ D B$ , giving $D B = 20$ m (½ mark)
- (ii) Width of river $= 54.6$ m (½ mark)
- (iii)(a) Time for boat A $= 3.46$ s (1 mark)
- (iii)(a) Time for boat B $= 4$ s (½ mark)
- (iii)(a) Boat A is earlier by 0.54 s (½ mark)
- (iii)(b) A covers 30 m and B covers 15 m in 3 s (½ + ½ marks)
- (iii)(b) Distance between the boats $= 54.6 - 45 = 9.6$ m (1 mark)
Open Q37 on its own page
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: A semicircular park with diameter AB horizontal at the top, where A = (2, 3) at the left end and B = (22, 3) at the right end. The arc of the semicircle lies below AB. O is the centre on AB, with a borewell drawn at O. P is a point on AB between A and O where a sapling is planted. C(12, y) is the lowest point of the arc, where a sapling is planted. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) O is the centre, so it is the mid-point of diameter AB with A(2, 3) and B(22, 3).
  $O = (\frac{2 + 22}{2}, \frac{3 + 3}{2}) = (12, 3)$
  1 mark
2. (ii) $A B = (22 - 2)^{2} + (3 - 3)^{2} = 20$ , so the radius $= \frac{20}{2} = 10$ units.
  (Equivalently, $O A = (12 - 2)^{2} + 0^{2} = 10$ .)
  1 mark
3. (iii)(a) C lies on the boundary of the park, so $O C$ = radius = 10.
  ½ mark
4. (iii)(a) $(12 - 12)^{2} + (y - 3)^{2} = 10 \Rightarrow (y - 3)^{2} = 100 \Rightarrow y - 3 = \pm 10 \Rightarrow y = 13$ or $y = - 7$ .
  1 mark
5. (iii)(a) The coordinates of C are $(12, 13)$ or $(12, - 7)$ . In the figure the arc lies below AB, so C is $(12, - 7)$ .
  ½ mark
6. (iii)(b) $P A = \frac{1}{3} P B \Rightarrow P A : P B = 1 : 3$ , so P divides AB internally in the ratio $1 : 3$ .
  ½ mark
7. (iii)(b) By the section formula with $m_{1} = 1$ , $m_{2} = 3$ , A(2, 3) and B(22, 3):
  $P = (\frac{1 \times 22 + 3 \times 2}{1 + 3}, \frac{1 \times 3 + 3 \times 3}{1 + 3})$
  1 mark
8. (iii)(b) $= (\frac{28}{4}, \frac{12}{4}) = (7, 3)$
  ½ mark
Answer: (i) O(12, 3); (ii) radius = 10 units; (iii)(a) C(12, 13) or (12, −7), with (12, −7) as drawn; (iii)(b) P(7, 3)
Where the marks are
- (i) Mid-point of AB gives O(12, 3) (1 mark)
- (ii) Radius = 10 (1 mark)
- (iii)(a) Uses OC = radius = 10 (½ mark)
- (iii)(a) Solves $(y - 3)^{2} = 100$ to get $y = 13$ or $- 7$ (1 mark)
- (iii)(a) States C as (12, 13) or (12, −7) (½ mark)
- (iii)(b) Ratio PA : PB = 1 : 3 (½ mark)
- (iii)(b) Correct section formula substitution (1 mark)
- (iii)(b) P = (7, 3) (½ mark)
Open Q38 on its own page

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