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

CBSE Class 10 Mathematics Basic: 2025 board paper, set 430/5/1

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 430-5-1_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 (VSA)
C	26-31	Short Answer (SA)
D	32-35	Long Answer (LA)
E	36-38	Case-study based, 2-mark sub-parts carry internal choice

Instructions in brief: All questions compulsory; no overall choice, but internal choice is given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 two-mark questions in Section E; calculator not allowed; take π = 22/7 where not stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
$(2 - 53)^{2}$ is
1. (A)a negative integer
2. (B)an irrational number
3. (C)a rational number
4. (D)a positive integer
Try it first, then show our solutionSolution
1. $(2 - 53)^{2} = 4 - 203 + 75 = 79 - 203$ .
2. A rational number minus a non-zero rational multiple of $3$ is irrational, so the value is an irrational number. Option (B).
  1 mark
Answer: An irrational number, option (B)
Where the marks are
- Expands to $79 - 203$ and identifies it as irrational (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQQuadratic Equations
The value of k for which the roots of the quadratic equation $6 x^{2} + 4 k x + k = 0$ are real and equal, is
1. (A)0
2. (B) $\frac{3}{4}$
3. (C) $\frac{- 3}{2}$
4. (D) $\frac{2}{3}$
Try it first, then show our solutionSolution
1. For equal roots, $D = b^{2} - 4 a c = 0$ : $(4 k)^{2} - 4 (6) (k) = 16 k^{2} - 24 k = 0$ .
2. $8 k (2 k - 3) = 0 \Rightarrow k = 0$ or $k = \frac{3}{2}$ . Among the options only $0$ appears, so option (A).
  1 mark
Answer: k = 0, option (A)
Where the marks are
- Uses $D = 0$ to get $k = 0$ or $3/2$ and chooses 0 (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQCoordinate Geometry
The distance between the points $(2, 3)$ and $(- 2, - 3)$ is
1. (A) $413$
2. (B) $40$
3. (C) $213$
4. (D)5
Try it first, then show our solutionSolution
1. $d = (- 2 - 2)^{2} + (- 3 - 3)^{2} = 16 + 36 = 52$ .
2. $52 = 213$ . Option (C).
  1 mark
Answer: $213$ , option (C)
Where the marks are
- Applies distance formula to get $213$ (1 mark)
Q4 · 1 mark · MCQCompetency-basedPolynomials
Observe the given graph of polynomial $p (x)$ . The number of zeroes of $p (x)$ is
1. (A)0
2. (B)1
3. (C)3
4. (D)2
Figure in the paper: Cartesian axes x'Ox and y'Oy. A downward-opening parabola with its vertex above the x-axis on the y-axis; the curve cuts the x-axis at two points, one on each side of the origin. See it in the official paper.
Try it first, then show our solutionSolution
1. Number of zeroes = number of points where the graph meets the x-axis.
2. The parabola cuts the x-axis at 2 points, so $p (x)$ has 2 zeroes. Option (D).
  1 mark
Answer: 2, option (D)
Where the marks are
- Counts the 2 x-intercepts (1 mark)
Q5 · 1 mark · MCQCircles
In the given figure, AB is diameter of the circle with centre O. CD is tangent to the circle so that $∠ A C D = 4 0^{\circ}$ . The value of $∠ C B A$ is
1. (A) $5 0^{\circ}$
2. (B) $4 0^{\circ}$
3. (C) $8 0^{\circ}$
4. (D) $4 5^{\circ}$
Figure in the paper: Circle with centre O and diameter AB (A at top, B at bottom). Point C lies on the circle on the right; O is joined to C by a dashed line. Line segment CD is a tangent at C, with D above C. Chord AC is drawn and the angle ACD = 40° is marked between chord CA and tangent CD. Chord BC is also drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. $O C ⊥ C D$ [tangent ⟂ radius], so $∠ O C A = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ}$ ; $O A = O C$ gives $∠ O A C = 5 0^{\circ}$ .
2. $∠ A C B = 9 0^{\circ}$ [angle in semicircle], so $∠ C B A = 18 0^{\circ} - 9 0^{\circ} - 5 0^{\circ} = 4 0^{\circ}$ . Option (B).
  1 mark
Answer: $4 0^{\circ}$ , option (B)
Where the marks are
- Uses tangent–radius and semicircle angle to get $4 0^{\circ}$ (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQArithmetic Progressions
10th term of the A.P.: $- 12, - 19, - 26, \dots$ is
1. (A) $- 75$
2. (B) $- 65$
3. (C)51
4. (D) $- 82$
Try it first, then show our solutionSolution
1. $a = - 12$ , $d = - 7$ .
2. $a_{10} = a + 9 d = - 12 - 63 = - 75$ . Option (A).
  1 mark
Answer: $- 75$ , option (A)
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ to get $- 75$ (1 mark)
Q7 · 1 mark · MCQQuadratic Equations
The roots of the equation $x^{2} - 8 = 0$ are
1. (A)rational and distinct
2. (B)irrational and distinct
3. (C)real and equal
4. (D)not real
Try it first, then show our solutionSolution
1. $x^{2} = 8 \Rightarrow x = \pm 22$ .
2. These are two different irrational numbers: irrational and distinct. Option (B).
  1 mark
Answer: Irrational and distinct, option (B)
Where the marks are
- Finds roots $\pm 22$ (1 mark)
Q8 · 1 mark · MCQCoordinate Geometry
The point $(x, 0)$ divides the line segment joining the points $(- 4, 5)$ and $(0, - 10)$ in the ratio
1. (A)1 : 3
2. (B)2 : 1
3. (C)1 : 1
4. (D)1 : 2
Try it first, then show our solutionSolution
1. Let ratio be $k : 1$ . y-coordinate: $\frac{- 10 k + 5}{k + 1} = 0$ .
2. $k = \frac{1}{2}$ , so ratio $1 : 2$ . Option (D).
  1 mark
Answer: 1 : 2, option (D)
Where the marks are
- Sets y-coordinate of section point to 0 and gets 1:2 (1 mark)
Q9 · 1 mark · MCQProbability
A black card is lost from a deck of 52 playing cards. Rest of the cards are shuffled and one card is drawn at random from the available cards. The probability that drawn card is 'king of hearts', is
1. (A) $\frac{1}{52}$
2. (B) $\frac{1}{4}$
3. (C) $\frac{1}{51}$
4. (D) $\frac{1}{26}$
Try it first, then show our solutionSolution
1. The lost card is black, so king of hearts (red) remains; 51 cards left.
2. $P = \frac{1}{51}$ . Option (C).
  1 mark
Answer: $\frac{1}{51}$ , option (C)
Where the marks are
- Uses 51 total outcomes, 1 favourable (1 mark)
Q10 · 1 mark · MCQSurface Areas and Volumes
The largest possible cone is just fitted inside a hollow cube of edge 25 cm. The radius of the base of the cone is
1. (A)5 cm
2. (B)12.5 cm
3. (C)25 cm
4. (D)10 cm
Try it first, then show our solutionSolution
1. The base of the largest cone fits a face of the cube, so diameter = edge = 25 cm.
2. Radius $= 12.5$ cm. Option (B).
  1 mark
Answer: 12.5 cm, option (B)
Where the marks are
- Diameter equals edge, radius 12.5 cm (1 mark)
Q11 · 1 mark · MCQProbability
If E is an event such that $P (E) = 0.1$ , then $P (\overline{E})$ is equal to
1. (A)0.9
2. (B) $\frac{1}{2}$
3. (C)0.99
4. (D) $- 1$
Try it first, then show our solutionSolution
1. $P (\overline{E}) = 1 - P (E) = 1 - 0.1 = 0.9$ . Option (A).
  1 mark
Answer: 0.9, option (A)
Where the marks are
- Uses $P (\overline{E}) = 1 - P (E)$ (1 mark)
Q12 · 1 mark · MCQIntroduction to Trigonometry
If $tan A = 1$ , then $3 sin A + cos A$ is equal to
1. (A) $42$
2. (B)4
3. (C) $22$
4. (D) $4 \times 4 5^{\circ}$
Try it first, then show our solutionSolution
1. $tan A = 1 \Rightarrow A = 4 5^{\circ}$ , $sin A = cos A = \frac{1}{2}$ .
2. $3 \cdot \frac{1}{2} + \frac{1}{2} = \frac{4}{2} = 22$ . Option (C).
  1 mark
Answer: $22$ , option (C)
Where the marks are
- Finds $A = 4 5^{\circ}$ and evaluates to $22$ (1 mark)
Q13 · 1 mark · MCQPolynomials
A quadratic polynomial having only zero $(- 2)$ is
1. (A) $(x - 2)^{2}$
2. (B) $x^{2} - 2$
3. (C) $x^{2} + 2 x$
4. (D) $(x + 2)^{2}$
Try it first, then show our solutionSolution
1. Only zero $- 2$ means $- 2$ is a repeated zero, so the factor $(x + 2)$ occurs twice.
2. Polynomial $(x + 2)^{2}$ . Option (D).
  1 mark
Answer: $(x + 2)^{2}$ , option (D)
Where the marks are
- Identifies repeated zero giving $(x + 2)^{2}$ (1 mark)
Q14 · 1 mark · MCQCircles
PQ and PR are tangents to a circle with centre O such that $O Q = QP$ . The value of $∠ O P Q$ is equal to
1. (A) $4 5^{\circ}$
2. (B) $3 0^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $9 0^{\circ}$
Figure in the paper: Circle with centre O. External point P with two tangents PQ (touching at Q, upper left) and PR (touching at R, lower left). O is joined to Q and to P by a dashed line; the angle OPQ is marked at P. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius] and $O Q = QP$ , so triangle OQP is isosceles right.
2. $∠ O P Q = \frac{18 0 ^{\circ} - 9 0 ^{\circ}}{2} = 4 5^{\circ}$ . Option (A).
  1 mark
Answer: $4 5^{\circ}$ , option (A)
Where the marks are
- Uses right angle at Q and equal sides to get $4 5^{\circ}$ (1 mark)
Q15 · 1 mark · MCQStatistics
Which of the following depends on all observations of a given data?
1. (A)Median
2. (B)Mean
3. (C)Range
4. (D)Mode
Try it first, then show our solutionSolution
1. Mean = sum of all observations ÷ number of observations, so it uses every value; median, mode and range do not. Option (B).
  1 mark
Answer: Mean, option (B)
Where the marks are
- Identifies mean (1 mark)
Q16 · 1 mark · MCQCircles
PQ is tangent to the circle centred at O. If $O Q = 3$ cm, $P Q = 5$ cm, then OP is equal to
1. (A)5 cm
2. (B)4 cm
3. (C) $15$ cm
4. (D) $34$ cm
Figure in the paper: Circle centred at O with radius OQ, Q on the circle. Tangent line PQ touches the circle at Q, and P lies outside on the right. O is joined to P by a horizontal segment. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius], so $O P^{2} = O Q^{2} + P Q^{2} = 9 + 25 = 34$ .
2. $O P = 34$ cm. Option (D).
  1 mark
Answer: $34$ cm, option (D)
Where the marks are
- Applies Pythagoras with right angle at Q (1 mark)
Q17 · 1 mark · MCQProbability
Two dice are rolled together. The probability that only one die shows number 4, is
1. (A) $\frac{11}{36}$
2. (B) $\frac{1}{3}$
3. (C) $\frac{5}{18}$
4. (D) $\frac{1}{4}$
Try it first, then show our solutionSolution
1. Favourable: (4,b) with $b \neq = 4$ : 5 outcomes; (a,4) with $a \neq = 4$ : 5 outcomes; total 10 of 36.
2. $P = \frac{10}{36} = \frac{5}{18}$ . Option (C).
  1 mark
Answer: $\frac{5}{18}$ , option (C)
Where the marks are
- Counts 10 favourable outcomes out of 36 (1 mark)
Q18 · 1 mark · MCQAreas Related to Circles
An arc of length 22 cm subtends an angle of $x^{\circ}$ at the centre of the circle. If radius of circle is 36 cm, the value of $x$ is
1. (A)35
2. (B)40
3. (C)60
4. (D)30
Try it first, then show our solutionSolution
1. $\frac{x}{360} \times 2 \times \frac{22}{7} \times 36 = 22$ .
2. $x = \frac{22 \times 360 \times 7}{2 \times 22 \times 36} = 35$ . Option (A).
  1 mark
Answer: 35, option (A)
Where the marks are
- Uses arc length formula to get $x = 35$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedSurface Areas and Volumes
Assertion (A): When a hemisphere of same radius (r) is carved out from one side of a solid wooden cylinder, the total surface area of remaining solid is increased by $2 π r^{2}$ .
Reason (R): Curved surface area of hemisphere is $2 π r^{2}$ .
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. Carving removes a flat circle of area $π r^{2}$ and adds the hemisphere's curved surface $2 π r^{2}$ : net increase $= π r^{2}$ , not $2 π r^{2}$ . So A is false.
2. CSA of hemisphere $= 2 π r^{2}$ is true. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Shows increase is $π r^{2}$ , A false, R true (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Assertion (A): In a right angle triangle ABC, $∠ B = 9 0^{\circ}$ . Therefore the value of $cos (A + C)$ is equal to 0.
Reason (R): $A + B + C = 18 0^{\circ}$ and $cos 9 0^{\circ} = 0$ .
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. $A + C = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ , so $cos (A + C) = cos 9 0^{\circ} = 0$ : A is true.
2. R is true and is exactly the reason used. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Uses angle sum to get $A + C = 9 0^{\circ}$ , both true and R explains A (1 mark)

Section B

Very Short Answer (VSA)

Q21 · 2 marks · Very short answerReal Numbers
Check whether $1 5^{n} \times 2^{n}$ , $n$ being a natural number, ends with the digit zero.
Try it first, then show our solutionSolution
1. Write the number in prime factor form: $1 5^{n} \times 2^{n} = (3 \times 5)^{n} \times 2^{n} = 2^{n} \times 3^{n} \times 5^{n}$ .
  1 mark
2. A number ends with the digit zero when its prime factorisation contains both 2 and 5. Here 2 and 5 are both factors for every natural number $n$ , so $1 5^{n} \times 2^{n}$ (which equals $1 0^{n} \times 3^{n}$ ) ends with the digit zero.
  1 mark
Answer: Yes, $1 5^{n} \times 2^{n} = 2^{n} \times 3^{n} \times 5^{n}$ has both 2 and 5 as factors, so it ends with the digit zero.
Where the marks are
- Writes the prime factorisation $2^{n} \times 3^{n} \times 5^{n}$ (1 mark)
- States that 2 and 5 are both factors, so the number ends with zero (1 mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $2 tan^{2} 4 5^{\circ} + cos^{2} 3 0^{\circ} - sin^{2} 9 0^{\circ}$ .
Try it first, then show our solutionSolution
1. Substitute the standard values $tan 4 5^{\circ} = 1$ , $cos 3 0^{\circ} = \frac{3}{2}$ , $sin 9 0^{\circ} = 1$ :
  $2 (1)^{2} + (\frac{3}{2})^{2} - (1)^{2}$
  1½ marks
2. $= 2 + \frac{3}{4} - 1 = \frac{7}{4}$
  ½ mark
Answer: $\frac{7}{4}$
Where the marks are
- Substitutes correct values of $tan 4 5^{\circ}$ , $cos 3 0^{\circ}$ , $sin 9 0^{\circ}$ (1½ marks)
- Simplifies to $\frac{7}{4}$ (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerIntroduction to Trigonometry
Verify that $cos 2 A = \frac{1 - t a n ^{2} A}{1 + t a n ^{2} A}$ for $A = 3 0^{\circ}$ .
Try it first, then show our solutionSolution
1. LHS $= cos 2 A = cos 6 0^{\circ} = \frac{1}{2}$
  ½ mark
2. RHS $= \frac{1 - t a n ^{2} 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}}$ [since $tan 3 0^{\circ} = \frac{1}{3}$ ]
  1 mark
3. $= \frac{2/3}{4/3} = \frac{1}{2} =$ LHS. Hence verified.
  ½ mark
Answer: LHS = RHS = $\frac{1}{2}$ ; verified.
Where the marks are
- Finds LHS $= cos 6 0^{\circ} = \frac{1}{2}$ (½ mark)
- Substitutes $tan^{2} 3 0^{\circ} = \frac{1}{3}$ in RHS (1 mark)
- Simplifies RHS to $\frac{1}{2}$ and concludes (½ mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerCoordinate Geometry
Find the coordinates of the points of trisection of line segment joining the points $(- 4, 1)$ and $(6, 5)$ .
Try it first, then show our solutionSolution
1. Let A(−4, 1), B(6, 5) and let P, Q be the points of trisection so that AP = PQ = QB. Then P divides AB in the ratio 1 : 2.
  By section formula, $P = (\frac{1 ( 6 ) + 2 ( - 4 )}{3}, \frac{1 ( 5 ) + 2 ( 1 )}{3}) = (\frac{- 2}{3}, \frac{7}{3})$
  1 mark
2. Q divides AB in the ratio 2 : 1.
  $Q = (\frac{2 ( 6 ) + 1 ( - 4 )}{3}, \frac{2 ( 5 ) + 1 ( 1 )}{3}) = (\frac{8}{3}, \frac{11}{3})$
  1 mark
Answer: $(- \frac{2}{3}, \frac{7}{3})$ and $(\frac{8}{3}, \frac{11}{3})$
Where the marks are
- Uses ratio 1 : 2 and finds P $(- \frac{2}{3}, \frac{7}{3})$ (1 mark)
- Uses ratio 2 : 1 and finds Q $(\frac{8}{3}, \frac{11}{3})$ (1 mark)
Open Q23 on its own page
Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerProbability
A bag contains 40 marbles out of which some are white and others are black. If the probability of drawing a black marble is $\frac{3}{5}$ , then find the number of white marbles.
Try it first, then show our solutionSolution
1. Let the number of black marbles be $n$ . Then P(black) $= \frac{n}{40}$ .
  ½ mark
2. $\frac{n}{40} = \frac{3}{5} \Rightarrow n = \frac{3 \times 40}{5} = 24$
  1 mark
3. Number of white marbles $= 40 - 24 = 16$
  ½ mark
Answer: 16 white marbles
Where the marks are
- Writes P(black) $= \frac{n}{40}$ (½ mark)
- Equates to $\frac{3}{5}$ and gets 24 black marbles (1 mark)
- Finds white marbles = 16 (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerProbability
In a pre-primary class, a teacher put cards numbered 20 to 59 in a bowl. A student picked up a card at random and read the number. Find the probability that the number read was (i) a prime number (ii) a perfect square.
Try it first, then show our solutionSolution
1. Total cards numbered 20 to 59 $= 59 - 20 + 1 = 40$ .
  (i) Primes between 20 and 59: 23, 29, 31, 37, 41, 43, 47, 53, 59, i.e. 9 numbers.
  P(prime) $= \frac{9}{40}$
  1 mark
2. (ii) Perfect squares between 20 and 59: 25, 36, 49, i.e. 3 numbers.
  P(perfect square) $= \frac{3}{40}$
  1 mark
Answer: (i) $\frac{9}{40}$ ; (ii) $\frac{3}{40}$
Where the marks are
- Total outcomes 40, counts 9 primes and gets $\frac{9}{40}$ (1 mark)
- Counts 3 perfect squares and gets $\frac{3}{40}$ (1 mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerCoordinate Geometry
Using distance formula, prove that the points $(1, 5)$ , $(2, 3)$ and $(3, 1)$ are collinear.
Try it first, then show our solutionSolution
1. Let A(1, 5), B(2, 3), C(3, 1).
  $A B = (2 - 1)^{2} + (3 - 5)^{2} = 1 + 4 = 5$
  ½ mark
2. $B C = (3 - 2)^{2} + (1 - 3)^{2} = 1 + 4 = 5$
  ½ mark
3. $A C = (3 - 1)^{2} + (1 - 5)^{2} = 4 + 16 = 20 = 25$
  ½ mark
4. $A B + B C = 5 + 5 = 25 = A C$ , so A, B, C are collinear.
  ½ mark
Answer: AB = BC = $5$ , AC = $25$ ; AB + BC = AC, so the points are collinear.
Where the marks are
- Finds AB $= 5$ (½ mark)
- Finds BC $= 5$ (½ mark)
- Finds AC $= 25$ (½ mark)
- Shows AB + BC = AC and concludes collinear (½ mark)
Open Q25 on its own page

Section C

Short Answer (SA)

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerPolynomials
If $α, β$ are zeroes of the polynomial $3 x^{2} - 8 x + 4$ , then form a quadratic polynomial in $x$ whose zeroes are $\frac{1}{α}$ and $\frac{1}{β}$ .
Try it first, then show our solutionSolution
1. For $3 x^{2} - 8 x + 4$ : $α + β = - \frac{- 8}{3} = \frac{8}{3}$ and $α β = \frac{4}{3}$ .
  1 mark
2. Sum of new zeroes: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{8/3}{4/3} = 2$
  ½ mark
3. Product of new zeroes: $\frac{1}{α} \cdot \frac{1}{β} = \frac{1}{α β} = \frac{3}{4}$
  ½ mark
4. Required polynomial $= x^{2} - (sum) x + product = x^{2} - 2 x + \frac{3}{4}$ , or $k (4 x^{2} - 8 x + 3)$ for any non-zero real $k$ .
  1 mark
Answer: $x^{2} - 2 x + \frac{3}{4}$ , i.e. $k (4 x^{2} - 8 x + 3)$ , $k \neq = 0$
Where the marks are
- Writes $α + β = \frac{8}{3}$ and $α β = \frac{4}{3}$ (1 mark)
- Finds sum of reciprocals = 2 (½ mark)
- Finds product of reciprocals $= \frac{3}{4}$ (½ mark)
- Forms the polynomial $x^{2} - 2 x + \frac{3}{4}$ or $4 x^{2} - 8 x + 3$ (1 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerPolynomials
Find zeroes of the polynomial $6 x^{2} - 7 x - 3$ and verify the relationship between zeroes and its coefficients.
Try it first, then show our solutionSolution
1. Split the middle term: $6 x^{2} - 7 x - 3 = 6 x^{2} - 9 x + 2 x - 3 = 3 x (2 x - 3) + 1 (2 x - 3) = (2 x - 3) (3 x + 1)$ .
  Zeroes: $x = \frac{3}{2}$ and $x = - \frac{1}{3}$ .
  1 mark
2. Sum of zeroes $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6}$ , and $- \frac{coefficient of x}{coefficient of x ^{2}} = - \frac{- 7}{6} = \frac{7}{6}$ . Verified.
  1 mark
3. Product of zeroes $= \frac{3}{2} \times (- \frac{1}{3}) = - \frac{1}{2}$ , and $\frac{constant term}{coefficient of x ^{2}} = \frac{- 3}{6} = - \frac{1}{2}$ . Verified.
  1 mark
Answer: Zeroes $\frac{3}{2}$ and $- \frac{1}{3}$ ; sum $= \frac{7}{6}$ , product $= - \frac{1}{2}$ , relations verified.
Where the marks are
- Factorises and finds zeroes $\frac{3}{2}, - \frac{1}{3}$ (1 mark)
- Verifies sum of zeroes $= - \frac{b}{a} = \frac{7}{6}$ (1 mark)
- Verifies product of zeroes $= \frac{c}{a} = - \frac{1}{2}$ (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerQuadratic Equations
The sum of the squares of two consecutive even numbers is 452. Find the numbers.
Try it first, then show our solutionSolution
1. Let the two consecutive even numbers be $x$ and $x + 2$ .
  ½ mark
2. According to the question, $x^{2} + (x + 2)^{2} = 452 \Rightarrow 2 x^{2} + 4 x + 4 = 452 \Rightarrow x^{2} + 2 x - 224 = 0$
  1 mark
3. $x^{2} + 16 x - 14 x - 224 = 0 \Rightarrow (x + 16) (x - 14) = 0 \Rightarrow x = 14$ or $x = - 16$
  1 mark
4. Taking the positive even number $x = 14$ , the required numbers are 14 and 16. (If negative numbers are allowed, −16 and −14 also satisfy the condition.)
  ½ mark
Answer: 14 and 16
Where the marks are
- Assumes numbers as $x$ and $x + 2$ (½ mark)
- Forms the equation $x^{2} + 2 x - 224 = 0$ (1 mark)
- Solves to get $x = 14$ (1 mark)
- States the numbers 14 and 16 (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $(csc A + sin A)^{2} + (sec A + cos A)^{2} = 7 + tan^{2} A + cot^{2} A$ .
Try it first, then show our solutionSolution
1. LHS $= csc^{2} A + sin^{2} A + 2 csc A sin A + sec^{2} A + cos^{2} A + 2 sec A cos A$
  1 mark
2. Since $csc A sin A = 1$ and $sec A cos A = 1$ , and using $csc^{2} A = 1 + cot^{2} A$ , $sec^{2} A = 1 + tan^{2} A$ :
  LHS $= (sin^{2} A + cos^{2} A) + (1 + cot^{2} A) + (1 + tan^{2} A) + 2 + 2$
  1 mark
3. $= 1 + 1 + 1 + 4 + tan^{2} A + cot^{2} A = 7 + tan^{2} A + cot^{2} A =$ RHS. Hence proved.
  1 mark
Answer: LHS simplifies to $7 + tan^{2} A + cot^{2} A$ = RHS; proved.
Where the marks are
- Expands both squares correctly (1 mark)
- Uses $csc A sin A = 1$ , $sec A cos A = 1$ and the identities $csc^{2} A = 1 + cot^{2} A$ , $sec^{2} A = 1 + tan^{2} A$ (1 mark)
- Simplifies to $7 + tan^{2} A + cot^{2} A$ (1 mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerCompetency-basedReal Numbers
The traffic lights at three different road crossings change after every 45 seconds, 75 seconds and 60 seconds respectively. If they change together at 5.00 a.m., then at what time they will change together next?
Try it first, then show our solutionSolution
1. Prime factorisation: $45 = 3^{2} \times 5$ , $75 = 3 \times 5^{2}$ , $60 = 2^{2} \times 3 \times 5$
  1½ marks
2. LCM $(45, 75, 60) = 2^{2} \times 3^{2} \times 5^{2} = 900$
  ½ mark
3. 900 seconds $= \frac{900}{60} = 15$ minutes
  ½ mark
4. The lights will next change together 15 minutes after 5:00 a.m., i.e. at 5:15 a.m.
  ½ mark
Answer: 5:15 a.m.
Where the marks are
- Correct prime factorisation of 45, 75, 60 (1½ marks)
- LCM = 900 (½ mark)
- Converts 900 s to 15 minutes (½ mark)
- Next change together at 5:15 a.m. (½ mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerSurface Areas and Volumes
A spherical glass vessel has a cylindrical neck 7 cm long and 8 cm in diameter. The radius of spherical part is 10 cm. Find the volume of the vessel.
Try it first, then show our solutionSolution
1. Volume of spherical part $= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 10 \times 10 \times 10 = \frac{88000}{21} \approx 4190.48$ cm³
  1 mark
2. Neck: radius $= \frac{8}{2} = 4$ cm, height 7 cm. Volume $= π r^{2} h = \frac{22}{7} \times 4 \times 4 \times 7 = 352$ cm³
  1 mark
3. Volume of vessel $= 4190.48 + 352 = 4542.48$ cm³ (approx.)
  1 mark
Answer: ≈ 4542.48 cm³
Where the marks are
- Volume of sphere $\frac{4}{3} π (10)^{3} \approx 4190.48$ cm³ (1 mark)
- Volume of cylindrical neck = 352 cm³ (1 mark)
- Total volume ≈ 4542.48 cm³ (1 mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerSurface Areas and Volumes
From each end of a solid cylinder of height 20 cm and base radius 7 cm, a cone of base radius 2.1 cm and height 5 cm is scooped out. Find the volume of the remaining solid.
Try it first, then show our solutionSolution
1. Volume of cylinder $= π r^{2} h = \frac{22}{7} \times 7 \times 7 \times 20 = 3080$ cm³
  1 mark
2. Volume of two cones $= 2 \times \frac{1}{3} π r^{2} h = 2 \times \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 5 = 2 \times 23.1 = 46.2$ cm³
  1 mark
3. Volume of remaining solid $= 3080 - 46.2 = 3033.8$ cm³
  1 mark
Answer: 3033.8 cm³
Where the marks are
- Volume of cylinder = 3080 cm³ (1 mark)
- Volume of the two cones = 46.2 cm³ (1 mark)
- Remaining volume = 3033.8 cm³ (1 mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerTriangles
Point E lies on the extended side AD of parallelogram ABCD. BE intersects CD at F. Show that (i) $△ D F E \sim △ C F B$ (ii) $△ A E B \sim △ C B F$ .
Try it first, then show our solutionSolution
1. Figure: parallelogram ABCD with side AD produced beyond D to a point E; segment BE cuts side CD at F. Since ABCD is a parallelogram, AE ∥ BC and AB ∥ DC.
  ½ mark
2. (i) In △DFE and △CFB:
  $∠ D E F = ∠ C B F$ [alternate interior angles, DE ∥ BC, transversal EB]
  $∠ E D F = ∠ B C F$ [alternate interior angles, DE ∥ BC, transversal DC]
  1 mark
3. ∴ $△ D F E \sim △ C F B$ [AA similarity criterion]
  ½ mark
4. (ii) In △AEB and △CBF:
  $∠ A E B = ∠ C B F$ [alternate interior angles, AE ∥ BC]
  $∠ E A B = ∠ B C F$ [opposite angles of a parallelogram are equal, $∠ A = ∠ C$ ]
  ½ mark
5. ∴ $△ A E B \sim △ C B F$ [AA similarity criterion]
  ½ mark
Answer: (i) △DFE ~ △CFB by AA; (ii) △AEB ~ △CBF by AA.
Where the marks are
- Correct figure (½ mark)
- Two pairs of equal angles in △DFE and △CFB with reasons (1 mark)
- Concludes △DFE ~ △CFB by AA (½ mark)
- Two pairs of equal angles in △AEB and △CBF with reasons (½ mark)
- Concludes △AEB ~ △CBF by AA (½ mark)
Open Q31 on its own page

Section D

Long Answer (LA)

Q32 · 5 marks · Long answerPair of Linear Equations in Two Variables
Determine graphically whether the following pair of linear equations $2 x + 3 y = 12$ and $x - y = 1$ has unique solution or infinitely many solutions.
Try it first, then show our solutionSolution
1. For $2 x + 3 y = 12$ , i.e. $y = \frac{12 - 2 x}{3}$ :
  $x$ 0 3 6
  $y$ 4 2 0
  Plot (0, 4), (3, 2), (6, 0) and join them to get the first line.
  2 marks
2. For $x - y = 1$ , i.e. $y = x - 1$ :
  $x$ 0 1 3
  $y$ −1 0 2
  Plot (0, −1), (1, 0), (3, 2) on the same axes and join them to get the second line.
  2 marks
3. The two lines intersect at exactly one point (3, 2). Hence the pair of equations has a unique solution, $x = 3$ , $y = 2$ .
  1 mark
Answer: The lines intersect at (3, 2), so the pair has a unique solution ( $x = 3$ , $y = 2$ ).
Where the marks are
- Correct table and graph of $2 x + 3 y = 12$ (2 marks)
- Correct table and graph of $x - y = 1$ (2 marks)
- Identifies intersection at (3, 2) and concludes unique solution (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerTriangles
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ . Take a line parallel to $B C$ that meets $A B$ at $D$ and $A C$ at $E$ . Join $B E$ and $C D$ . Draw $E N ⊥ A B$ (N on AB) and $D M ⊥ A C$ (M on AC).
  ½ mark
2. Given: In $△ A B C$ , $D E ∥ B C$ , with $D E$ meeting $A B$ at $D$ and $A C$ at $E$ .
  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 $E N ⊥ A B$ and $D M ⊥ A C$ .
  ½ mark
5. Proof: Use area $= \frac{1}{2} \times$ base $\times$ height.
  In $△ A D E$ and $△ B D E$ , the height from $E$ is $E N$ for both:
  $\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} \cdot A D \cdot E N}{\frac{1}{2} \cdot D B \cdot E N} = \frac{A D}{D B}$ ... (1)
  1 mark
6. In $△ A D E$ and $△ D E C$ , the height from $D$ is $D M$ for both:
  $\frac{a r ( A D E )}{a r ( D E C )} = \frac{\frac{1}{2} \cdot A E \cdot D M}{\frac{1}{2} \cdot E C \cdot D M} = \frac{A E}{E C}$ ... (2)
  1 mark
7. $△ B D E$ and $△ D E C$ have the same base $D E$ and lie between the same parallels $D E$ and $B C$ . So $a r (B D E) = a r (D E C)$ ... (3)
  From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: Proved: if DE ∥ BC in △ABC, then AD/DB = AE/EC (Basic Proportionality Theorem).
Where the marks are
- Draws a correct labelled figure (½ mark)
- States what is given (½ mark)
- States what is to be proved (½ mark)
- States the construction: join BE and CD, draw EN ⊥ AB and DM ⊥ AC (½ mark)
- Expresses $a r (A D E) / a r (B D E)$ as $A D / D B$ (1 mark)
- Expresses $a r (A D E) / a r (D E C)$ as $A E / E C$ (1 mark)
- Uses equal areas of triangles on the same base between the same parallels to conclude (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerTriangles
It is given that sides AB and AC and median AD of $△ A B C$ are respectively proportional to sides PQ and PR and median PM of another $△ P QR$ . Show that $△ A B C \sim △ P QR$ .
Try it first, then show our solutionSolution
1. Given: $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ , where AD and PM are medians. To prove: △ABC ~ △PQR.
  Construction (figure): produce AD to E so that DE = AD, and join CE; produce PM to N so that MN = PM, and join RN. Mark $∠ C A D = ∠1$ , $∠ R P M = ∠2$ , $∠ B A D = ∠3$ , $∠ QP M = ∠4$ .
  1 mark
2. In △ADB and △EDC: AD = ED (construction), BD = CD (D is mid-point of BC), $∠ A D B = ∠ E D C$ (vertically opposite angles). So △ADB ≅ △EDC [SAS]. Similarly △PMQ ≅ △NMR.
  1 mark
3. ∴ AB = CE and PQ = NR [CPCT]
  ½ mark
4. From the given ratios, $\frac{C E}{N R} = \frac{A B}{P Q} = \frac{A D}{P M} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ and also $= \frac{A C}{P R}$ . So $\frac{C E}{N R} = \frac{A E}{P N} = \frac{A C}{P R}$ .
  1 mark
5. ∴ △CAE ~ △RPN [SSS similarity criterion] ⇒ $∠1 = ∠2$ (i.e. $∠ C A D = ∠ R P M$ ).
  ½ mark
6. Similarly, joining BE and QN (△ADC ≅ △EDB gives BE = AC, and PR = QN), we get △BAE ~ △QPN, so $∠3 = ∠4$ . Adding: $∠1 + ∠3 = ∠2 + ∠4$ , i.e. $∠ B A C = ∠ QP R$ .
  ½ mark
7. Now in △ABC and △PQR: $\frac{A B}{P Q} = \frac{A C}{P R}$ and included $∠ B A C = ∠ QP R$ . Hence △ABC ~ △PQR [SAS similarity criterion].
  ½ mark
Answer: △ABC ~ △PQR (proved via △CAE ~ △RPN by SSS, giving ∠BAC = ∠QPR, then SAS similarity).
Where the marks are
- Correct figure with construction AD = DE, PM = MN (1 mark)
- Proves △DAB ≅ △DEC and similarly △MPQ ≅ △MNR (1 mark)
- Gets AB = CE, PQ = NR by CPCT (½ mark)
- Establishes $\frac{C E}{N R} = \frac{A E}{P N} = \frac{A C}{P R}$ (1 mark)
- △CAE ~ △RPN by SSS (½ mark)
- Obtains ∠1 = ∠2, ∠3 = ∠4 and adds to get ∠BAC = ∠QPR (½ mark)
- Concludes △ABC ~ △PQR by SAS similarity (½ mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
The angle of elevation of the top of a building from the foot of the tower is $3 0^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $6 0^{\circ}$ . If the tower is 30 m high, find the height of the building and distance between the building and the tower. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: let CD = 30 m be the tower (foot C) and AB = $h$ m the building (foot A), with CA = $x$ m along the ground. Angle of elevation of D from A is $6 0^{\circ}$ ; angle of elevation of B from C is $3 0^{\circ}$ . $∠ D C A = ∠ B A C = 9 0^{\circ}$ .
  1 mark
2. In right △ACD: $tan 6 0^{\circ} = \frac{C D}{C A} \Rightarrow 3 = \frac{30}{x} \Rightarrow x = \frac{30}{3} = 103$ …(i)
  1½ marks
3. In right △CAB: $tan 3 0^{\circ} = \frac{A B}{C A} \Rightarrow \frac{1}{3} = \frac{h}{x} \Rightarrow x = h 3$ …(ii)
  1½ marks
4. From (i) and (ii): $h 3 = 103 \Rightarrow h = 10$ m, and $x = 10 \times 1.73 = 17.3$ m.
  Height of building = 10 m; distance between building and tower = 17.3 m.
  1 mark
Answer: Height of building = 10 m; distance = 17.3 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ}$ to get $x = 103$ (1½ marks)
- Uses $tan 3 0^{\circ}$ to get $x = h 3$ (1½ marks)
- Finds $h = 10$ m and $x = 17.3$ m (1 mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerStatistics
Find 'mean' and 'mode' of the following data:
Class 10-25 25-40 40-55 55-70 70-85 85-100
Number of Students 12 10 15 13 8 12
Try it first, then show our solutionSolution
1. Take assumed mean $a = 47.5$ , class size $h = 15$ , $u_{i} = \frac{x _{i} - 47.5}{15}$ .
  Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  10–25 17.5 12 −2 −24
  25–40 32.5 10 −1 −10
  40–55 47.5 15 0 0
  55–70 62.5 13 1 13
  70–85 77.5 8 2 16
  85–100 92.5 12 3 36
  Total 70 31
  1½ marks
2. Mean $= a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 47.5 + 15 \times \frac{31}{70} = 47.5 + 6.64 = 54.14$
  1½ marks
3. Highest frequency is 15, so modal class is 40–55: $l = 40$ , $f_{1} = 15$ , $f_{0} = 10$ , $f_{2} = 13$ , $h = 15$ .
  Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{15 - 10}{30 - 10 - 13} \times 15$
  1½ marks
4. $= 40 + \frac{5}{7} \times 15 = 40 + 10.71 = 50.71$
  ½ mark
Answer: Mean ≈ 54.14; Mode ≈ 50.71
Where the marks are
- Correct table with $x_{i}$ , $u_{i}$ , $f_{i} u_{i}$ and totals 70, 31 (1½ marks)
- Mean $= 47.5 + 15 \times \frac{31}{70} = 54.14$ (1½ marks)
- Identifies modal class 40–55 and substitutes in mode formula (1½ marks)
- Mode = 50.71 (½ mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerStatistics
The following table shows the ages of patients admitted in a hospital during a year:
Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of Patients 7 10 21 22 15 5
Find 'mode' and 'median' of the above data.
Try it first, then show our solutionSolution
1. Cumulative frequency table:
  Age (years) 5-15 15-25 25-35 35-45 45-55 55-65
  $f$ 7 10 21 22 15 5
  $c f$ 7 17 38 60 75 80
  The total is $N = 80$ .
  1 mark
2. Median: $\frac{N}{2} = 40$ . The first cf to reach 40 is 60, so the median class is $35 - 45$ .
  For this class, $l = 35$ , $c f = 38$ , $f = 22$ , $h = 10$ .
  $Median = l + \frac{\frac{N}{2} - c f}{f} \times h = 35 + \frac{40 - 38}{22} \times 10$
  1½ marks
3. $= 35 + \frac{20}{22} = 35 + 0.91 = 35.91$ years
  ½ mark
4. Mode: the highest frequency is 22, so the modal class is $35 - 45$ .
  For this class, $l = 35$ , $f_{1} = 22$ , $f_{0} = 21$ , $f_{2} = 15$ , $h = 10$ .
  $Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 35 + \frac{22 - 21}{44 - 21 - 15} \times 10$
  1½ marks
5. $= 35 + \frac{10}{8} = 35 + 1.25 = 36.25$ years
  ½ mark
Answer: Median = 35.91 years; Mode = 36.25 years
Where the marks are
- Correct cumulative frequency table with N = 80 (1 mark)
- Identifies median class 35-45 and substitutes correctly in the median formula (1½ marks)
- Median = 35.91 years (½ mark)
- Identifies modal class 35-45 and substitutes correctly in the mode formula (1½ marks)
- Mode = 36.25 years (½ mark)
Open Q35(b) on its own page

$x$	0	3	6
$y$	4	2	0

$x$	0	1	3
$y$	−1	0	2

Class	10-25	25-40	40-55	55-70	70-85	85-100
Number of Students	12	10	15	13	8	12

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
10–25	17.5	12	−2	−24
25–40	32.5	10	−1	−10
40–55	47.5	15	0	0
55–70	62.5	13	1	13
70–85	77.5	8	2	16
85–100	92.5	12	3	36
Total		70		31

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of Patients	7	10	21	22	15	5

Age (years)	5-15	15-25	25-35	35-45	45-55	55-65
$f$	7	10	21	22	15	5
$c f$	7	17	38	60	75	80

Section E

Case-study based, 2-mark sub-parts carry internal choice

Q36 · 4 marks · Case-basedCompetency-basedTriangles
In a Fine Arts class, students were asked to design triangular tiles in geometric pattern.
1. (i)
  Determine the length OB.
  1 mark
2. (ii)
  Is DE || CA? Give reason for your answer.
  1 mark
3. (iii)
  (a) Write all angles of quadrilateral OEBD and show that it is a cyclic quadrilateral.
  2 marks
4. OR
  (iii)
  (b) Find the perimeter of $△ A B C$ . (Use $3 = 1.73$ )
  2 marks
Figure in the paper: Left: a triangle with a circular design and a star-like black pattern inside a circle. Right: equilateral/isosceles triangle ABC with A at top, B and C at base. A circle with centre O touches sides AB (at E), BC (at D) and AC. OD is perpendicular to BC at D; OE is joined to E; a right-angle mark appears at the point of tangency on AC. A dashed line goes from A through O to D. Angle 30° is marked at B between BC and BO. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $O D ⊥ B C$ [radius ⟂ tangent], and $O D = 4$ cm. In right $△ O D B$ , $sin 3 0^{\circ} = \frac{O D}{O B}$ , so $\frac{1}{2} = \frac{4}{O B}$ . Hence $O B = 8$ cm.
  1 mark
2. (ii) Yes, $D E ∥ C A$ . $△ A B C$ is equilateral and $A D ⊥ B C$ , so $A D$ is also a median. Therefore D is the mid-point of BC.
  ½ mark
3. (ii) In the same way, the incircle touches AB at its mid-point, so E is the mid-point of AB. By the mid-point theorem in $△ A B C$ , $D E ∥ C A$ .
  ½ mark
4. (iii)(a) $∠ E B D = 6 0^{\circ}$ [angle of an equilateral triangle]. $∠ O E B = ∠ O D B = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact].
  ½ mark
5. (iii)(a) By the angle sum of a quadrilateral, $∠ E O D = 36 0^{\circ} - (6 0^{\circ} + 9 0^{\circ} + 9 0^{\circ}) = 12 0^{\circ}$ .
  ½ mark
6. (iii)(a) Opposite angles: $∠ O E B + ∠ O D B = 9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ . Also $∠ E B D + ∠ E O D = 6 0^{\circ} + 12 0^{\circ} = 18 0^{\circ}$ .
  ½ mark
7. (iii)(a) Since the opposite angles are supplementary, OEBD is a cyclic quadrilateral.
  ½ mark
8. (iii)(b) In right $△ O D B$ , $cos 3 0^{\circ} = \frac{B D}{O B}$ . So $B D = 8 \times \frac{3}{2} = 43 = 4 \times 1.73 = 6.92$ cm.
  1 mark
9. (iii)(b) D is the mid-point of BC, so $B C = 2 B D = 13.84$ cm. The perimeter of $△ A B C$ is $3 \times 13.84 = 41.52$ cm.
  1 mark
Answer: (i) OB = 8 cm; (ii) Yes, D and E are mid-points of BC and AB, so DE ∥ CA (mid-point theorem); (iii)(a) ∠EBD = 60°, ∠OEB = ∠ODB = 90°, ∠EOD = 120°, opposite angles sum to 180°, so cyclic; (iii)(b) 41.52 cm
Where the marks are
- (i) Uses sin 30° = 4/OB to get OB = 8 cm (1 mark)
- (ii) States Yes, and that D is the mid-point of BC as AD ⟂ BC in the equilateral triangle (½ mark)
- (ii) E is the mid-point of AB, hence DE ∥ CA by the mid-point theorem (½ mark)
- (iii)(a) ∠EBD = 60° and ∠EOD = 120° (½ mark)
- (iii)(a) ∠OEB = ∠ODB = 90° with reason (½ mark)
- (iii)(a) Sum of opposite angles = 180° (½ mark)
- (iii)(a) Concludes OEBD is cyclic (½ mark)
- (iii)(b) BD = 6.92 cm using cos 30° (1 mark)
- (iii)(b) BC = 13.84 cm and perimeter = 41.52 cm (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
A farmer has put up a decorative windmill in his farm in which there are eight blades of equal width and equally placed in a circular arrangement. A circular wire goes through them. The diagram shows two blades OAB and OPQ in a quarter circle with centre O. $∠ A O B = ∠ P O Q = 3 0^{\circ}$ , $O A = 28$ cm, $O C = 21$ cm. O is the centre of both the circles.
1. (i)
  Determine the measure of $∠ B O P$ .
  1 mark
2. (ii)
  Find length of arc CD.
  1 mark
3. (iii)
  (a) Find the area of region CABD.
  2 marks
4. OR
  (iii)
  (b) Find perimeter of region CABD.
  2 marks
Figure in the paper: Left: a windmill with eight equal blades arranged in a circle around a central hub, joined by a circular wire. Right: concentric circles with centre O. Radii OA (outer, 28 cm, pointing down) and OR (to the right). Points A, B on outer circle and C, D on inner circle along the radius OA-side; OB, OD and OP, OQ are drawn. Angles 30° are marked between OA and OB, and between OP and OQ (with OR as reference). Points on outer arc: A, B, P, Q, R; inner circle points: C, D. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $∠ A O R = 9 0^{\circ}$ (quarter circle). The blades OAB and OPQ take up $3 0^{\circ} + 3 0^{\circ} = 6 0^{\circ}$ . The blades are equally placed, so the remaining $3 0^{\circ}$ splits into two equal gaps, $∠ B O P$ and $∠ QO R$ . Hence $∠ B O P = \frac{1}{2} (9 0^{\circ} - 6 0^{\circ}) = 1 5^{\circ}$ .
  1 mark
2. (ii) Arc CD lies on the inner circle ( $r = 21$ cm) and subtends $3 0^{\circ}$ at O. Length of arc $C D = \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{30}{360} \times 2 \times \frac{22}{7} \times 21 = 11$ cm.
  1 mark
3. (iii)(a) Area of region CABD = area of sector OAB − area of sector OCD $= \frac{30}{360} \times \frac{22}{7} \times (2 8^{2} - 2 1^{2})$
  1 mark
4. (iii)(a) $= \frac{1}{12} \times \frac{22}{7} \times (784 - 441) = \frac{1}{12} \times \frac{22}{7} \times 343 = \frac{1078}{12} \approx 89.83 cm^{2}$
  1 mark
5. (iii)(b) Length of arc $A B = \frac{30}{360} \times 2 \times \frac{22}{7} \times 28 = \frac{44}{3} \approx 14.67$ cm.
  1 mark
6. (iii)(b) Perimeter of CABD = arc AB + arc CD + CA + DB. Here $C A = D B = 28 - 21 = 7$ cm, so the perimeter $= 14.67 + 11 + 7 + 7 = 39.67$ cm.
  1 mark
Answer: (i) 15°; (ii) 11 cm; (iii)(a) ≈ 89.83 cm²; (iii)(b) ≈ 39.67 cm
Where the marks are
- (i) ∠BOP = ½(90° − 60°) = 15° (1 mark)
- (ii) Arc CD = (30/360) × 2 × (22/7) × 21 = 11 cm (1 mark)
- (iii)(a) Writes area as difference of two sectors: (30/360) × (22/7) × (28² − 21²) (1 mark)
- (iii)(a) Area ≈ 89.8 cm² (1 mark)
- (iii)(b) Arc AB = 44/3 ≈ 14.67 cm (1 mark)
- (iii)(b) Perimeter = 14.67 + 11 + 2 × 7 = 39.67 cm (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
A tent house owner provides furniture on rent. He stacks chairs in his shop to save space. In the diagram, the height of seat of chair from ground is represented by $h_{1}, h_{2}, h_{3}, \dots$ . The height of first seat is 44 cm from ground level and gap between every two seats is 10 cm.
1. (i)
  Write the values of $h_{1}, h_{2}, h_{3}, h_{4}$ and $h_{5}$ in this order only.
  1 mark
2. (ii)
  Show that the above values form an A.P. Write its first term and common difference.
  1 mark
3. (iii)
  (a) If chairs can be stacked up to the maximum height of 160 cm, then find the maximum number of chairs in a stack.
  2 marks
4. OR
  (iii)
  (b) Is it possible to stack 15 chairs if maximum height of the stack can not be more than 180 cm? Justify your answer.
  2 marks
Figure in the paper: Left: a stack of chairs. Right: five stacked chair seats drawn as trapezoids, with the bottom seat's height from the ground marked as 44 cm and the vertical gap between successive seats marked as 10 cm. Heights from ground are labelled h1 (lowest), h2, h3, h4, h5 (highest) with arrows from the ground line. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Each seat is 10 cm above the one below it. So $h_{1} = 44$ cm, $h_{2} = 54$ cm, $h_{3} = 64$ cm, $h_{4} = 74$ cm and $h_{5} = 84$ cm.
  1 mark
2. (ii) $h_{2} - h_{1} = h_{3} - h_{2} = h_{4} - h_{3} = h_{5} - h_{4} = 10$ . The difference between consecutive terms is constant, so the heights form an A.P.
  ½ mark
3. (ii) First term $a = 44$ cm and common difference $d = 10$ cm.
  ½ mark
4. (iii)(a) The seat height of the $n$ th chair is $a_{n} = a + (n - 1) d$ . For the maximum height of 160 cm: $44 + (n - 1) \times 10 = 160$ .
  1 mark
5. (iii)(a) This gives $(n - 1) \times 10 = 116$ , so $n - 1 = 11.6$ and $n = 12.6$ .
  ½ mark
6. (iii)(a) $n$ must be a whole number and the height must not exceed 160 cm, so at most 12 chairs can be stacked.
  ½ mark
7. (iii)(b) $h_{15} = a + 14 d = 44 + 14 \times 10 = 184$ cm.
  1 mark
8. (iii)(b) $184$ cm $> 180$ cm.
  ½ mark
9. (iii)(b) Hence it is not possible to stack 15 chairs.
  ½ mark
Answer: (i) 44, 54, 64, 74, 84 cm; (ii) common difference 10, so an A.P. with a = 44, d = 10; (iii)(a) 12 chairs; (iii)(b) No, since h₁₅ = 184 cm > 180 cm
Where the marks are
- (i) Writes 44, 54, 64, 74, 84 (1 mark)
- (ii) Shows the common difference is constant, hence an A.P. (½ mark)
- (ii) a = 44, d = 10 (½ mark)
- (iii)(a) Forms 44 + (n − 1) × 10 = 160 (1 mark)
- (iii)(a) Gets n = 12.6 (½ mark)
- (iii)(a) Concludes maximum 12 chairs (½ mark)
- (iii)(b) Finds h₁₅ = 44 + 14 × 10 = 184 cm (1 mark)
- (iii)(b) Compares 184 > 180 (½ mark)
- (iii)(b) Concludes 15 chairs cannot be stacked (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.