MathsCBSE Class 10 · Standard · Board exam 2024 (session 2023-24)

CBSE Class 10 Mathematics Standard: 2024 board paper, set 30/1/1

The question paper set in the CBSE Class 10 board exam of 2024, 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/2024/X/MATHEMATICS_STANDARD.zip (a zip; this paper is the file 30-1-1(Mathematics Standard).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 23 and 25
C	26-31	Short answer; internal choice in 26 and 30
D	32-35	Long answer; internal choice in 33 and 34
E	36-38	Case-study based; internal choice in 2-mark subparts of each case

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choice is given in 2 marks questions of Section B, 2 questions of Section C, 2 questions of Section D and 3 two-mark subparts of Section E. Use of calculators is not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPolynomials
If the sum of zeroes of the polynomial $p (x) = 2 x^{2} - k 2 x + 1$ is $2$ , then value of $k$ is:
1. (A) $2$
2. (B)2
3. (C) $22$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. For $a x^{2} + b x + c$ , the sum of zeroes is $- \frac{b}{a}$ . Here $a = 2$ and $b = - k 2$ , so the sum of zeroes is $\frac{k 2}{2}$ .
2. Given that the sum is $2$ : $\frac{k 2}{2} = 2 \Rightarrow k = 2$ . So the answer is option (B).
  1 mark
Answer: k = 2, option (B)
Where the marks are
- Uses sum of zeroes $= - b / a$ and gets $k = 2$ , option (B) (1 mark)
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Q2 · 1 mark · MCQProbability
If the probability of a player winning a game is 0.79, then the probability of his losing the same game is:
1. (A)1.79
2. (B)0.31
3. (C)0.21%
4. (D)0.21
Try it first, then show our solutionSolution
1. Winning and losing are complementary events, so $P (lose) = 1 - P (win)$ .
2. $P (lose) = 1 - 0.79 = 0.21$ . Option (C) is wrong because $0.21%$ means $0.0021$ . So the answer is option (D).
  1 mark
Answer: 0.21, option (D)
Where the marks are
- Uses $P (\overset{ˉ}{E}) = 1 - P (E)$ to get 0.21, option (D) (1 mark)
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Q3 · 1 mark · MCQQuadratic Equations
If the roots of equation $a x^{2} + b x + c = 0$ , $a \neq = 0$ are real and equal, then which of the following relation is true?
1. (A) $a = \frac{b ^{2}}{c}$
2. (B) $b^{2} = a c$
3. (C) $a c = \frac{b ^{2}}{4}$
4. (D) $c = \frac{b ^{2}}{a}$
Try it first, then show our solutionSolution
1. The roots are real and equal when the discriminant is zero: $b^{2} - 4 a c = 0$ .
2. So $b^{2} = 4 a c$ , which gives $a c = \frac{b ^{2}}{4}$ . The answer is option (C).
  1 mark
Answer: $a c = \frac{b ^{2}}{4}$ , option (C)
Where the marks are
- Uses $D = b^{2} - 4 a c = 0$ to get $a c = b^{2} /4$ , option (C) (1 mark)
Q4 · 1 mark · MCQArithmetic Progressions
In an A.P., if the first term $a = 7$ , $n$ th term $a_{n} = 84$ and the sum of first $n$ terms $s_{n} = \frac{2093}{2}$ , then $n$ is equal to:
1. (A)22
2. (B)24
3. (C)23
4. (D)26
Try it first, then show our solutionSolution
1. Use $S_{n} = \frac{n}{2} (a + a_{n}) = \frac{n}{2} (7 + 84) = \frac{91 n}{2}$ .
2. $\frac{91 n}{2} = \frac{2093}{2} \Rightarrow n = \frac{2093}{91} = 23$ . The answer is option (C).
  1 mark
Answer: n = 23, option (C)
Where the marks are
- Uses $S_{n} = \frac{n}{2} (a + l)$ and gets $n = 23$ , option (C) (1 mark)
Q5 · 1 mark · MCQReal Numbers
If two positive integers $p$ and $q$ can be expressed as $p = 18 a^{2} b^{4}$ and $q = 20 a^{3} b^{2}$ , where $a$ and $b$ are prime numbers, then LCM $(p, q)$ is:
1. (A) $2 a^{2} b^{2}$
2. (B) $180 a^{2} b^{2}$
3. (C) $12 a^{2} b^{2}$
4. (D) $180 a^{3} b^{4}$
Try it first, then show our solutionSolution
1. Write the prime factorisations: $p = 2 \times 3^{2} \times a^{2} b^{4}$ and $q = 2^{2} \times 5 \times a^{3} b^{2}$ .
2. The LCM is the product of the highest powers: $2^{2} \times 3^{2} \times 5 \times a^{3} b^{4} = 180 a^{3} b^{4}$ . The answer is option (D).
  1 mark
Answer: $180 a^{3} b^{4}$ , option (D)
Where the marks are
- Takes the highest powers of all prime factors to get $180 a^{3} b^{4}$ , option (D) (1 mark)
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Q6 · 1 mark · MCQCoordinate Geometry
AD is a median of $Δ A B C$ with vertices A(5, -6), B(6, 4) and C(0, 0). Length AD is equal to:
1. (A) $68$ units
2. (B) $215$ units
3. (C) $101$ units
4. (D)10 units
Try it first, then show our solutionSolution
1. D is the mid-point of BC: $D = (\frac{6 + 0}{2}, \frac{4 + 0}{2}) = (3, 2)$ .
2. $A D = (5 - 3)^{2} + (- 6 - 2)^{2} = 4 + 64 = 68$ units. The answer is option (A).
  1 mark
Answer: $68$ units, option (A)
Where the marks are
- Finds the mid-point $D (3, 2)$ and the distance $A D = 68$ , option (A) (1 mark)
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Q7 · 1 mark · MCQIntroduction to Trigonometry
If $sec θ - tan θ = m$ , then the value of $sec θ + tan θ$ is:
1. (A) $1 - \frac{1}{m}$
2. (B) $m^{2} - 1$
3. (C) $\frac{1}{m}$
4. (D) $- m$
Try it first, then show our solutionSolution
1. Use the identity $sec^{2} θ - tan^{2} θ = 1$ , that is, $(sec θ - tan θ) (sec θ + tan θ) = 1$ .
2. So $m (sec θ + tan θ) = 1$ , which gives $sec θ + tan θ = \frac{1}{m}$ . The answer is option (C).
  1 mark
Answer: $\frac{1}{m}$ , option (C)
Where the marks are
- Factorises $sec^{2} θ - tan^{2} θ = 1$ to get $\frac{1}{m}$ , option (C) (1 mark)
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Q8 · 1 mark · MCQProbability
From the data 1, 4, 7, 9, 16, 21, 25, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is:
1. (A) $\frac{2}{5}$
2. (B) $\frac{1}{5}$
3. (C) $\frac{1}{7}$
4. (D) $\frac{2}{7}$
Try it first, then show our solutionSolution
1. Remove the even numbers 4 and 16. The remaining data are 1, 7, 9, 21, 25, which is 5 numbers.
2. The only prime among them is 7 (1 is not prime; 9, 21 and 25 are composite). So $P = \frac{1}{5}$ . The answer is option (B).
  1 mark
Answer: $\frac{1}{5}$ , option (B)
Where the marks are
- Identifies the 5 remaining numbers with one prime, giving $\frac{1}{5}$ , option (B) (1 mark)
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Q9 · 1 mark · MCQStatistics
For some data $x_{1}, x_{2}, \dots, x_{n}$ with respective frequencies $f_{1}, f_{2}, \dots, f_{n}$ , the value of $\sum_{i = 1}^{n} f_{i} (x_{i} - \overset{x}{ˉ})$ is equal to:
1. (A) $n \overset{x}{ˉ}$
2. (B)1
3. (C) $\sum f_{i}$
4. (D)0
Try it first, then show our solutionSolution
1. $\sum f_{i} (x_{i} - \overset{x}{ˉ}) = \sum f_{i} x_{i} - \overset{x}{ˉ} \sum f_{i}$ .
2. Since $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , we have $\overset{x}{ˉ} \sum f_{i} = \sum f_{i} x_{i}$ , so the expression equals $0$ . The answer is option (D).
  1 mark
Answer: 0, option (D)
Where the marks are
- Uses $\sum f_{i} x_{i} = \overset{x}{ˉ} \sum f_{i}$ to get 0, option (D) (1 mark)
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Q10 · 1 mark · MCQPolynomials
The zeroes of a polynomial $x^{2} + p x + q$ are twice the zeroes of the polynomial $4 x^{2} - 5 x - 6$ . The value of $p$ is:
1. (A) $- \frac{5}{2}$
2. (B) $\frac{5}{2}$
3. (C) $- 5$
4. (D)10
Try it first, then show our solutionSolution
1. For $4 x^{2} - 5 x - 6$ , the sum of zeroes is $α + β = \frac{5}{4}$ .
2. The zeroes of $x^{2} + p x + q$ are $2 α$ and $2 β$ , so their sum is $\frac{5}{2} = - p$ . Hence $p = - \frac{5}{2}$ . The answer is option (A).
  1 mark
Answer: $p = - \frac{5}{2}$ , option (A)
Where the marks are
- Doubles the sum of zeroes and equates it to $- p$ , giving $p = - \frac{5}{2}$ , option (A) (1 mark)
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Q11 · 1 mark · MCQCoordinate Geometry
If the distance between the points $(3, - 5)$ and $(x, - 5)$ is 15 units, then the values of $x$ are:
1. (A)12, -18
2. (B)-12, 18
3. (C)18, 5
4. (D)-9, -12
Try it first, then show our solutionSolution
1. $(x - 3)^{2} + (- 5 + 5)^{2} = 15 \Rightarrow (x - 3)^{2} = 225$ .
2. So $x - 3 = \pm 15$ , which gives $x = 18$ or $x = - 12$ . The answer is option (B).
  1 mark
Answer: x = -12, 18, option (B)
Where the marks are
- Uses the distance formula to get $x - 3 = \pm 15$ , so $x = 18, - 12$ , option (B) (1 mark)
Q12 · 1 mark · MCQIntroduction to Trigonometry
If $cos (α + β) = 0$ , then value of $cos (\frac{α + β}{2})$ is equal to:
1. (A) $\frac{1}{2}$
2. (B) $\frac{1}{2}$
3. (C)0
4. (D) $2$
Try it first, then show our solutionSolution
1. $cos (α + β) = 0 = cos 9 0^{\circ} \Rightarrow α + β = 9 0^{\circ}$ .
2. So $cos (\frac{α + β}{2}) = cos 4 5^{\circ} = \frac{1}{2}$ . The answer is option (A).
  1 mark
Answer: $\frac{1}{2}$ , option (A)
Where the marks are
- Gets $α + β = 9 0^{\circ}$ and $cos 4 5^{\circ} = \frac{1}{2}$ , option (A) (1 mark)
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Q13 · 1 mark · MCQSurface Areas and Volumes
A solid sphere is cut into two hemispheres. The ratio of the surface areas of sphere to that of two hemispheres taken together, is:
1. (A)1 : 1
2. (B)1 : 4
3. (C)2 : 3
4. (D)3 : 2
Try it first, then show our solutionSolution
1. Surface area of the sphere $= 4 π r^{2}$ . Each hemisphere has total surface area $2 π r^{2} + π r^{2} = 3 π r^{2}$ , so the two together have $6 π r^{2}$ .
2. Ratio $= 4 π r^{2} : 6 π r^{2} = 2 : 3$ . The answer is option (C).
  1 mark
Answer: 2 : 3, option (C)
Where the marks are
- Uses TSA of a hemisphere $= 3 π r^{2}$ to get the ratio $2 : 3$ , option (C) (1 mark)
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Q14 · 1 mark · MCQStatistics
The middle most observation of every data arranged in order is called:
1. (A)mode
2. (B)median
3. (C)mean
4. (D)deviation
Try it first, then show our solutionSolution
1. When data are arranged in order, the value that divides them into two equal halves, the middle-most observation, is called the median. The answer is option (B).
  1 mark
Answer: median, option (B)
Where the marks are
- Identifies the median, option (B) (1 mark)
Q15 · 1 mark · MCQSurface Areas and Volumes
The volume of the largest right circular cone that can be carved out from a solid cube of edge 2 cm is:
1. (A) $\frac{4 π}{3}$ cu cm
2. (B) $\frac{5 π}{3}$ cu cm
3. (C) $\frac{8 π}{3}$ cu cm
4. (D) $\frac{2 π}{3}$ cu cm
Try it first, then show our solutionSolution
1. The largest cone has its base diameter equal to the edge, so $r = 1$ cm, and its height equal to the edge, so $h = 2$ cm.
2. $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (1)^{2} (2) = \frac{2 π}{3}$ cu cm. The answer is option (D).
  1 mark
Answer: $\frac{2 π}{3}$ cu cm, option (D)
Where the marks are
- Takes $r = 1$ , $h = 2$ and computes $\frac{2 π}{3}$ , option (D) (1 mark)
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Q16 · 1 mark · MCQProbability
Two dice are rolled together. The probability of getting sum of numbers on the two dice as 2, 3 or 5, is:
1. (A) $\frac{7}{36}$
2. (B) $\frac{11}{36}$
3. (C) $\frac{5}{36}$
4. (D) $\frac{4}{9}$
Try it first, then show our solutionSolution
1. There are 36 equally likely outcomes. Sum 2: (1,1), which is 1 outcome. Sum 3: (1,2), (2,1), which is 2 outcomes. Sum 5: (1,4), (4,1), (2,3), (3,2), which is 4 outcomes.
2. Favourable outcomes $= 1 + 2 + 4 = 7$ , so $P = \frac{7}{36}$ . The answer is option (A).
  1 mark
Answer: $\frac{7}{36}$ , option (A)
Where the marks are
- Counts 7 favourable outcomes out of 36, option (A) (1 mark)
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Q17 · 1 mark · MCQCoordinate Geometry
The centre of a circle is at (2, 0). If one end of a diameter is at (6, 0), then the other end is at:
1. (A)(0, 0)
2. (B)(4, 0)
3. (C)(-2, 0)
4. (D)(-6, 0)
Try it first, then show our solutionSolution
1. The centre is the mid-point of the diameter. Let the other end be $(x, y)$ . Then $\frac{6 + x}{2} = 2$ and $\frac{0 + y}{2} = 0$ .
2. So $x = - 2$ and $y = 0$ , and the other end is $(- 2, 0)$ . The answer is option (C).
  1 mark
Answer: (-2, 0), option (C)
Where the marks are
- Uses the mid-point formula to get $(- 2, 0)$ , option (C) (1 mark)
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Q18 · 1 mark · MCQCompetency-basedPair of Linear Equations in Two Variables
In the given figure, graphs of two linear equations are shown. The pair of these linear equations is:
1. (A)consistent with unique solution.
2. (B)consistent with infinitely many solutions.
3. (C)inconsistent.
4. (D)inconsistent but can be made consistent by extending these lines.
Figure in the paper: Coordinate axes X-X1 and Y-Y1 meeting at O. Two straight lines are drawn, both sloping upward to the right, and they appear to be non-intersecting within the drawn region, i.e. parallel lines. See it in the official paper.
Try it first, then show our solutionSolution
1. In the figure the two lines have different slopes (one is steeper than the other), so they are not parallel. Lines extend endlessly, as the arrows show.
2. Two non-parallel lines meet at exactly one point. So the pair is consistent with a unique solution, which is option (A).
  1 mark
Answer: Consistent with unique solution, option (A)
Where the marks are
- Recognises the lines are intersecting (different slopes), so the pair is consistent with a unique solution, option (A) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedCircles
Assertion (A): The tangents drawn at the end points of a diameter of a circle, are parallel.
Reason (R): Diameter of a circle is the longest chord.
1. (A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
2. (B)Both, Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation for Assertion (A).
3. (C)Assertion (A) is true but Reason (R) is false.
4. (D)Assertion (A) is false but Reason (R) is true.
Try it first, then show our solutionSolution
1. Assertion: each tangent is perpendicular to the radius at its point of contact, so both tangents are perpendicular to the same diameter. Hence they are parallel, and A is true.
2. Reason: the diameter is indeed the longest chord, so R is true. However, the parallelism comes from 'tangent ⟂ radius', not from the diameter being longest. So R does not explain A, and the answer is option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Judges both statements true, with R not explaining A, option (B) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedPolynomials
Assertion (A): If the graph of a polynomial touches x-axis at only one point, then the polynomial cannot be a quadratic polynomial.
Reason (R): A polynomial of degree $n (n > 1)$ can have at most $n$ zeroes.
1. (A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
2. (B)Both, Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation for Assertion (A).
3. (C)Assertion (A) is true but Reason (R) is false.
4. (D)Assertion (A) is false but Reason (R) is true.
Try it first, then show our solutionSolution
1. Assertion: $p (x) = (x - 1)^{2} = x^{2} - 2 x + 1$ is quadratic, and its graph touches the x-axis only at $(1, 0)$ . So A is false.
2. Reason: a polynomial of degree $n$ has at most $n$ zeroes, so R is true. Hence the answer is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Gives a counter-example such as $(x - 1)^{2}$ to show A is false and R is true, option (D) (1 mark)
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Section B

Very short answer; internal choice in 23 and 25

Q21 · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve the following system of linear equations $7 x - 2 y = 5$ and $8 x + 7 y = 15$ and verify your answer.
Try it first, then show our solutionSolution
1. $7 x - 2 y = 5$ …(i), $8 x + 7 y = 15$ …(ii). Multiply (i) by 7 and (ii) by 2: $49 x - 14 y = 35$ and $16 x + 14 y = 30$ . Adding gives $65 x = 65$ .
  1 mark
2. So $x = 1$ . Substituting in (i): $7 - 2 y = 5 \Rightarrow y = 1$ . Hence $x = 1, y = 1$ .
  ½ mark
3. Verification: in (i), $7 (1) - 2 (1) = 5$ ✓; in (ii), $8 (1) + 7 (1) = 15$ ✓. Both equations are satisfied.
  ½ mark
Answer: x = 1, y = 1
Where the marks are
- Eliminates a variable correctly to reach $65 x = 65$ (1 mark)
- Obtains $x = 1, y = 1$ (½ mark)
- Verifies the solution in both equations (½ mark)
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Q22 · 2 marks · Very short answerProbability
In a pack of 52 playing cards one card is lost. From the remaining cards, a card is drawn at random. Find the probability that the drawn card is queen of heart, if the lost card is a black card.
Try it first, then show our solutionSolution
1. After one card is lost, the number of remaining cards is $52 - 1 = 51$ , so there are 51 possible outcomes.
  1 mark
2. The lost card is black, so the queen of hearts (a red card) is still in the pack: 1 favourable outcome. $P (queen of hearts) = \frac{1}{51}$ .
  1 mark
Answer: $\frac{1}{51}$
Where the marks are
- Takes the total number of outcomes as 51 (1 mark)
- Gets the probability $\frac{1}{51}$ (1 mark)
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Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $22 cos 4 5^{\circ} sin 3 0^{\circ} + 23 cos 3 0^{\circ}$
Try it first, then show our solutionSolution
1. Substitute $cos 4 5^{\circ} = \frac{1}{2}$ , $sin 3 0^{\circ} = \frac{1}{2}$ and $cos 3 0^{\circ} = \frac{3}{2}$ .
2. $22 \times \frac{1}{2} \times \frac{1}{2} + 23 \times \frac{3}{2}$
  1½ marks
3. $= 1 + 3 = 4$
  ½ mark
Answer: 4
Where the marks are
- Substitutes the correct values of $cos 4 5^{\circ}$ , $sin 3 0^{\circ}$ and $cos 3 0^{\circ}$ (½ mark each, 1½ marks)
- Simplifies to 4 (½ mark)
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OR
Q23(B) · 2 marks · Very short answerIntroduction to Trigonometry
If $A = 6 0^{\circ}$ and $B = 3 0^{\circ}$ , verify that: $sin (A + B) = sin A cos B + cos A sin B$
Try it first, then show our solutionSolution
1. LHS $= sin (6 0^{\circ} + 3 0^{\circ}) = sin 9 0^{\circ} = 1$
  1 mark
2. RHS $= sin 6 0^{\circ} cos 3 0^{\circ} + cos 6 0^{\circ} sin 3 0^{\circ} = \frac{3}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1$ . Hence LHS = RHS, which is verified.
  1 mark
Answer: LHS = RHS = 1, verified
Where the marks are
- Evaluates LHS $= sin 9 0^{\circ} = 1$ (1 mark)
- Evaluates RHS $= \frac{3}{4} + \frac{1}{4} = 1$ and concludes LHS = RHS (1 mark)
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Q24 · 2 marks · Very short answerTriangles
In the given figure, ABCD is a quadrilateral. Diagonal BD bisects $∠ B$ and $∠ D$ both. Prove that:
1. (i)
  $Δ A B D \sim Δ C B D$
  1 mark
2. (ii)
  $A B = B C$
  1 mark
Figure in the paper: Quadrilateral ABCD with vertices A (bottom left), B (bottom right), C (top right), D (top left). Diagonal BD is drawn, joining B and D. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Name the angles. Since BD bisects $∠ B$ , $∠ A B D = ∠ C B D$ . Since BD bisects $∠ D$ , $∠ A D B = ∠ C D B$ .
2. (i) In $Δ A B D$ and $Δ C B D$ :
  $∠ A B D = ∠ C B D$ [BD bisects $∠ B$ ]
  $∠ A D B = ∠ C D B$ [BD bisects $∠ D$ ]
  $∴ Δ A B D \sim Δ C B D$ [AA similarity criterion].
  1 mark
3. (ii) From the similarity, corresponding sides are proportional: $\frac{A B}{C B} = \frac{B D}{B D} = 1$ .
  (Equivalently, the two equal angle pairs and the common side BD give $Δ A B D ≅ Δ C B D$ by ASA.)
4. (ii) Hence $A B = B C$ . Hence proved.
  1 mark
Answer: (i) $Δ A B D \sim Δ C B D$ by AA; (ii) $A B = B C$ , since the ratio of sides is $B D / B D = 1$ (or by ASA congruence).
Where the marks are
- States the two pairs of equal angles with reasons and concludes $Δ A B D \sim Δ C B D$ by AA (1 mark)
- Uses the common side BD, through the side ratio or ASA congruence, to conclude $A B = B C$ (1 mark)
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Answer one of these 2 alternatives.
Q25(A) · 2 marks · Very short answerReal Numbers
Prove that $5 - 23$ is an irrational number. It is given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $5 - 23$ is rational. Then $5 - 23 = \frac{a}{b}$ , where $a, b$ are integers and $b \neq = 0$ .
  ½ mark
2. Rearranging: $23 = 5 - \frac{a}{b} = \frac{5 b - a}{b}$ , so $3 = \frac{5 b - a}{2 b}$ .
  ½ mark
3. Since $a$ and $b$ are integers, $\frac{5 b - a}{2 b}$ is rational. That would make $3$ rational, which contradicts the given fact that $3$ is irrational.
  ½ mark
4. So our assumption is wrong. Hence $5 - 23$ is an irrational number.
  ½ mark
Answer: $5 - 23$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 - 23 = a / b$ with integers $a, b$ and $b \neq = 0$ (½ mark)
- Isolates $3 = \frac{5 b - a}{2 b}$ (½ mark)
- Notes that the RHS is rational while $3$ is irrational, which is a contradiction (½ mark)
- Concludes that $5 - 23$ is irrational (½ mark)
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OR
Q25(B) · 2 marks · Very short answerReal Numbers
Show that the number $5 \times 11 \times 17 + 3 \times 11$ is a composite number.
Try it first, then show our solutionSolution
1. Take 11 out as a common factor: $5 \times 11 \times 17 + 3 \times 11 = 11 \times (5 \times 17 + 3)$ .
  1 mark
2. Simplify the bracket: $= 11 \times (85 + 3) = 11 \times 88 = 11 \times 11 \times 2^{3}$ $(= 968)$ .
  ½ mark
3. The number has factors other than 1 and itself (for example 11 and 88). Hence it is a composite number.
  ½ mark
Answer: $5 \times 11 \times 17 + 3 \times 11 = 11 \times 88 = 2^{3} \times 1 1^{2}$ , so it is composite.
Where the marks are
- Takes 11 out as a common factor (1 mark)
- Simplifies to $11 \times 88$ or $2^{3} \times 1 1^{2}$ (½ mark)
- Concludes the number is composite because it has factors other than 1 and itself (½ mark)
Open Q25(B) on its own page

Section C

Short answer; internal choice in 26 and 30

Answer one of these 2 alternatives.
Q26(A) · 3 marks · Short answerCoordinate Geometry
Find the ratio in which the point $(\frac{8}{5}, y)$ divides the line segment joining the points (1, 2) and (2, 3). Also, find the value of $y$ .
Try it first, then show our solutionSolution
1. Let $P (\frac{8}{5}, y)$ divide AB, with $A (1, 2)$ and $B (2, 3)$ , in the ratio $k : 1$ .
  By the section formula, the x-coordinate gives $\frac{k \cdot 2 + 1 \cdot 1}{k + 1} = \frac{8}{5}$ .
  1 mark
2. Solve: $5 (2 k + 1) = 8 (k + 1) \Rightarrow 10 k + 5 = 8 k + 8 \Rightarrow 2 k = 3 \Rightarrow k = \frac{3}{2}$ .
  ½ mark
3. So the required ratio is $\frac{3}{2} : 1 = 3 : 2$ .
  ½ mark
4. For the y-coordinate, use $m : n = 3 : 2$ :
  $y = \frac{3 \times 3 + 2 \times 2}{3 + 2} = \frac{13}{5}$ .
  1 mark
Answer: Ratio $3 : 2$ ; $y = \frac{13}{5}$
Where the marks are
- Applies the section formula to the x-coordinate with ratio $k : 1$ (1 mark)
- Obtains $k = 3/2$ (½ mark)
- States the ratio as $3 : 2$ (½ mark)
- Finds $y = 13/5$ (1 mark)
Open Q26(A) on its own page
OR
Q26(B) · 3 marks · Short answerCoordinate Geometry
ABCD is a rectangle formed by the points A (-1, -1), B (-1, 6), C (3, 6) and D (3, -1). P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.
Try it first, then show our solutionSolution
1. Use the mid-point formula $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ .
  P, the mid-point of AB: $(\frac{- 1 - 1}{2}, \frac{- 1 + 6}{2}) = (- 1, \frac{5}{2})$ .
  ½ mark
2. Q, the mid-point of BC: $(\frac{- 1 + 3}{2}, \frac{6 + 6}{2}) = (1, 6)$ .
  ½ mark
3. R, the mid-point of CD: $(\frac{3 + 3}{2}, \frac{6 - 1}{2}) = (3, \frac{5}{2})$ .
  ½ mark
4. S, the mid-point of DA: $(\frac{3 - 1}{2}, \frac{- 1 - 1}{2}) = (1, - 1)$ .
  ½ mark
5. Mid-point of diagonal PR: $(\frac{- 1 + 3}{2}, \frac{\frac{5}{2} + \frac{5}{2}}{2}) = (1, \frac{5}{2})$ .
  ½ mark
6. Mid-point of diagonal QS: $(\frac{1 + 1}{2}, \frac{6 - 1}{2}) = (1, \frac{5}{2})$ .
  Both diagonals have the same mid-point, so the diagonals PR and QS of PQRS bisect each other.
  ½ mark
Answer: P(−1, 5/2), Q(1, 6), R(3, 5/2), S(1, −1). Both diagonals PR and QS have mid-point (1, 5/2), so they bisect each other.
Where the marks are
- Coordinates of P (½ mark)
- Coordinates of Q (½ mark)
- Coordinates of R (½ mark)
- Coordinates of S (½ mark)
- Mid-point of PR is (1, 5/2) (½ mark)
- Mid-point of QS is (1, 5/2), with the conclusion (½ mark)
Open Q26(B) on its own page
Q27 · 3 marks · Short answerCompetency-basedReal Numbers
In a teachers' workshop, the number of teachers teaching French, Hindi and English are 48, 80 and 144 respectively. Find the minimum number of rooms required if in each room the same number of teachers are seated and all of them are of the same subject.
Try it first, then show our solutionSolution
1. To use the fewest rooms, each room must hold the largest possible equal number of teachers of one subject. That number is HCF(48, 80, 144).
  Prime factorise: $48 = 2^{4} \times 3$
  ½ mark
2. $80 = 2^{4} \times 5$
  ½ mark
3. $144 = 2^{4} \times 3^{2}$
  ½ mark
4. HCF $(48, 80, 144) = 2^{4} = 16$ , so 16 teachers sit in each room.
  ½ mark
5. Minimum number of rooms $= \frac{48}{16} + \frac{80}{16} + \frac{144}{16} = 3 + 5 + 9 = 17$ .
  1 mark
Answer: 17 rooms
Where the marks are
- Prime factorisation of 48 (½ mark)
- Prime factorisation of 80 (½ mark)
- Prime factorisation of 144 (½ mark)
- HCF = 16 (½ mark)
- Number of rooms = 3 + 5 + 9 = 17 (1 mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $\frac{t a n θ}{1 - c o t θ} + \frac{c o t θ}{1 - t a n θ} = 1 + sec θ cosec θ$
Try it first, then show our solutionSolution
1. Write everything in sines and cosines:
  LHS $= \frac{\frac{s i n θ}{c o s θ}}{\frac{s i n θ - c o s θ}{s i n θ}} + \frac{\frac{c o s θ}{s i n θ}}{\frac{c o s θ - s i n θ}{c o s θ}}$
  ½ mark
2. Simplify each fraction:
  $= \frac{sin ^{2} θ}{cos θ ( sin θ - cos θ )} - \frac{cos ^{2} θ}{sin θ ( sin θ - cos θ )} = \frac{1}{sin θ - cos θ} [\frac{sin ^{2} θ}{cos θ} - \frac{cos ^{2} θ}{sin θ}]$
  1 mark
3. Take the LCM and use $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ :
  $= \frac{1}{sin θ - cos θ} \cdot \frac{sin ^{3} θ - cos ^{3} θ}{sin θ cos θ} = \frac{( sin θ - cos θ ) ( sin ^{2} θ + cos ^{2} θ + sin θ cos θ )}{( sin θ - cos θ ) sin θ cos θ}$
  1 mark
4. Cancel $(sin θ - cos θ)$ and use $sin^{2} θ + cos^{2} θ = 1$ :
  $= \frac{1 + sin θ cos θ}{sin θ cos θ} = \frac{1}{sin θ cos θ} + 1 = 1 + sec θ cosec θ =$ RHS. Hence proved.
  ½ mark
Answer: LHS simplifies to $1 + sec θ cosec θ$ = RHS. Hence proved.
Where the marks are
- Converts tan and cot into sin and cos (½ mark)
- Simplifies to $\frac{1}{s i n θ - c o s θ} [\frac{s i n ^{2} θ}{c o s θ} - \frac{c o s ^{2} θ}{s i n θ}]$ (1 mark)
- Factorises $sin^{3} θ - cos^{3} θ$ (1 mark)
- Uses $sin^{2} θ + cos^{2} θ = 1$ to reach the RHS (½ mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now?
Try it first, then show our solutionSolution
1. Let the present ages of Rashmi and Nazma be $x$ years and $y$ years.
  Three years ago: $x - 3 = 3 (y - 3) \Rightarrow x - 3 y + 6 = 0$ ...(1)
  1 mark
2. Ten years later: $x + 10 = 2 (y + 10) \Rightarrow x - 2 y - 10 = 0$ ...(2)
  1 mark
3. Subtract (1) from (2): $(x - 2 y - 10) - (x - 3 y + 6) = 0 \Rightarrow y - 16 = 0 \Rightarrow y = 16$ .
  Then from (2): $x = 2 (16) + 10 = 42$ .
  Check: $39 = 3 \times 13$ ✓ and $52 = 2 \times 26$ ✓.
  So Rashmi is 42 years old and Nazma is 16 years old.
  1 mark
Answer: Rashmi: 42 years; Nazma: 16 years
Where the marks are
- Forms $x - 3 = 3 (y - 3)$ (1 mark)
- Forms $x + 10 = 2 (y + 10)$ (1 mark)
- Solves to get $x = 42$ , $y = 16$ (1 mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(A) · 3 marks · Short answerCircles
In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that $∠ P O Q = 9 0^{\circ}$ .
Figure in the paper: A circle with centre O and diameter AB vertical. Tangent at A and tangent at B are horizontal lines. Point Q lies on the upper tangent to the right of A, and point P lies on the lower tangent to the right of B. Tangent PQ joins P and Q, and OQ and OP are drawn with the angle at O marked. See it in the official paper.
Try it first, then show our solutionSolution
1. Let the tangent PQ touch the circle at R. Join OR.
  ½ mark
2. In $Δ O A Q$ and $Δ O R Q$ :
  $O A = O R$ [radii]
  $Q A = QR$ [tangents from an external point are equal]
  $O Q = O Q$ [common]
  $∴ Δ O A Q ≅ Δ O R Q$ [SSS], so $∠ A O Q = ∠ R O Q$ [CPCT] ...(1)
  1 mark
3. In the same way, $Δ O B P ≅ Δ O R P$ (OB = OR, PB = PR, OP common), so $∠ B O P = ∠ R O P$ ...(2)
  ½ mark
4. AOB is a diameter, so it is a straight line: $∠ A O R + ∠ R O B = 18 0^{\circ}$ .
  Using (1) and (2): $2∠ QO R + 2∠ R O P = 18 0^{\circ}$ .
  ½ mark
5. $\Rightarrow ∠ QO R + ∠ R O P = 9 0^{\circ}$ , that is, $∠ P O Q = 9 0^{\circ}$ . Hence proved.
  ½ mark
Answer: $∠ P O Q = 9 0^{\circ}$ (proved)
Where the marks are
- Takes R, the point of contact of PQ, and joins OR (½ mark)
- Proves $Δ O A Q ≅ Δ O R Q$ and gets $∠ A O Q = ∠ R O Q$ (1 mark)
- Proves $Δ O B P ≅ Δ O R P$ and gets $∠ B O P = ∠ R O P$ (½ mark)
- Uses $∠ A O R + ∠ R O B = 18 0^{\circ}$ (½ mark)
- Concludes $∠ P O Q = 9 0^{\circ}$ (½ mark)
Open Q30(A) on its own page
OR
Q30(B) · 3 marks · Short answerCircles
A circle with centre O and radius 8 cm is inscribed in a quadrilateral ABCD in which P, Q, R, S are the points of contact as shown. If AD is perpendicular to DC, BC = 30 cm and BS = 24 cm, then find the length DC.
Figure in the paper: Quadrilateral ABCD with the circle inscribed. A top-left, B top-right, C bottom-right, D bottom-left with a right angle at D. S is the touching point on side AB, P on side AD, Q on side DC, R on side BC. See it in the official paper.
Try it first, then show our solutionSolution
1. Join OP and OQ.
  $B R = B S = 24$ cm [tangents from the external point B are equal].
  ½ mark
2. $C R = B C - B R = 30 - 24 = 6$ cm.
  ½ mark
3. $C Q = C R = 6$ cm [tangents from C are equal].
  ½ mark
4. In quadrilateral OPDQ: $∠ D = 9 0^{\circ}$ and $∠ O P D = ∠ O Q D = 9 0^{\circ}$ [tangent ⟂ radius]. Also OP = OQ = 8 cm, so OPDQ is a square.
  $∴ D Q = O P = 8$ cm.
  ½ mark
5. $D C = D Q + QC = 8 + 6 = 14$ cm.
  1 mark
Answer: DC = 14 cm
Where the marks are
- $B R = B S = 24$ cm (½ mark)
- $C R = 6$ cm (½ mark)
- $C Q = 6$ cm (½ mark)
- $D Q = 8$ cm, since OPDQ is a square (½ mark)
- $D C = 14$ cm (1 mark)
Open Q30(B) on its own page
Q31 · 3 marks · Short answerSurface Areas and Volumes
The difference between the outer and inner radii of a hollow right circular cylinder of length 14 cm is 1 cm. If the volume of the metal used in making the cylinder is 176 cm $^{3}$ , find the outer and inner radii of the cylinder.
Try it first, then show our solutionSolution
1. Let the outer radius be $R$ cm and the inner radius be $r$ cm. Then $R - r = 1$ ...(1)
  ½ mark
2. Volume of metal $= π h (R^{2} - r^{2})$ :
  $\frac{22}{7} \times 14 \times (R^{2} - r^{2}) = 176 \Rightarrow 44 (R^{2} - r^{2}) = 176 \Rightarrow R^{2} - r^{2} = 4$ .
  1 mark
3. $(R + r) (R - r) = 4$ . Using (1), $R + r = 4$ ...(2)
  ½ mark
4. Add (1) and (2): $2 R = 5 \Rightarrow R = 2.5$ cm. Then $r = 1.5$ cm.
  Outer radius = 2.5 cm, inner radius = 1.5 cm.
  1 mark
Answer: Outer radius 2.5 cm, inner radius 1.5 cm
Where the marks are
- Writes $R - r = 1$ (½ mark)
- Sets up $\frac{22}{7} \times 14 \times (R^{2} - r^{2}) = 176$ (1 mark)
- Obtains $R + r = 4$ (½ mark)
- Solves to get $R = 2.5$ cm and $r = 1.5$ cm (1 mark)
Open Q31 on its own page

Section D

Long answer; internal choice in 33 and 34

Q32 · 5 marks · Long answerAreas Related to Circles
An arc of a circle of radius 21 cm subtends an angle of $6 0^{\circ}$ at the centre. Find:
1. (i)
  the length of the arc.
  2 marks
2. (ii)
  the area of the minor segment of the circle made by the corresponding chord.
  3 marks
Try it first, then show our solutionSolution
1. Figure: a circle with centre O, chord AB, $∠ A O B = 6 0^{\circ}$ and $r = 21$ cm. The minor segment lies between chord AB and arc ALB.
  (i) Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$
  1½ marks
2. (i) $= \frac{1}{6} \times 132 = 22$ cm.
  ½ mark
3. (ii) Area of sector OALB $= \frac{60}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times 1386 = 231$ cm².
  1 mark
4. (ii) In $Δ O A B$ , $O A = O B = 21$ cm and $∠ A O B = 6 0^{\circ}$ . The base angles are equal and add to $12 0^{\circ}$ , so each is $6 0^{\circ}$ and the triangle is equilateral.
  Area $= \frac{3}{4} \times 2 1^{2} = \frac{441 3}{4}$ cm² $\approx 190.95$ cm².
  1½ marks
5. (ii) Area of minor segment = area of sector − area of triangle $= (231 - \frac{441 3}{4})$ cm² $\approx 231 - 190.95 = 40.05$ cm².
  ½ mark
Answer: (i) 22 cm; (ii) $(231 - \frac{441 3}{4})$ cm² ≈ 40.05 cm²
Where the marks are
- Uses the arc length formula with correct values (1½ marks)
- Arc length = 22 cm (½ mark)
- Area of sector = 231 cm² (1 mark)
- Area of the equilateral triangle OAB = $\frac{441 3}{4}$ cm² (1½ marks)
- Area of minor segment = $231 - \frac{441 3}{4}$ ≈ 40.05 cm² (½ mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerArithmetic Progressions
The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.
Try it first, then show our solutionSolution
1. Let the first term be $a$ and the common difference be $d$ , so $a_{8} = a + 7 d$ .
  Sum: $a + (a + 7 d) = 32 \Rightarrow 2 a + 7 d = 32$ ...(1)
  1 mark
2. Product: $a (a + 7 d) = 60$ ...(2)
  1 mark
3. From (1), $a + 7 d = 32 - a$ . Substituting in (2): $a (32 - a) = 60 \Rightarrow a^{2} - 32 a + 60 = 0 \Rightarrow (a - 2) (a - 30) = 0$ .
  So $a = 2$ or $a = 30$ .
  If $a = 2$ : $7 d = 32 - 4 = 28 \Rightarrow d = 4$ .
  If $a = 30$ : $7 d = 32 - 60 = - 28 \Rightarrow d = - 4$ .
  So the first term and common difference are 2 and 4, or 30 and −4.
  2 marks
4. $S_{20} = \frac{20}{2} [2 a + 19 d]$ .
  For $a = 2$ , $d = 4$ : $S_{20} = 10 (4 + 76) = 800$ .
  ½ mark
5. For $a = 30$ , $d = - 4$ : $S_{20} = 10 (60 - 76) = - 160$ .
  ½ mark
Answer: $a = 2, d = 4$ giving $S_{20} = 800$ ; or $a = 30, d = - 4$ giving $S_{20} = - 160$
Where the marks are
- Forms $2 a + 7 d = 32$ (1 mark)
- Forms $a (a + 7 d) = 60$ (1 mark)
- Solves to get $a = 2, d = 4$ or $a = 30, d = - 4$ (2 marks)
- $S_{20} = 800$ for the first case (½ mark)
- $S_{20} = - 160$ for the second case (½ mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerArithmetic Progressions
In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.
Try it first, then show our solutionSolution
1. Let the first term be $a$ and the common difference be $d$ . The A.P. has $n = 40$ terms.
2. Sum of the first 9 terms: $S_{9} = \frac{9}{2} [2 a + 8 d] = 153$
  $\Rightarrow 2 a + 8 d = 34 \Rightarrow a + 4 d = 17$ ... (i)
  1 mark
3. The last 6 terms are $a_{35}, a_{36}, \dots, a_{40}$ , so their sum is $S_{40} - S_{34} = 687$ .
  Using $a_{n} = a + (n - 1) d$ :
  $(a + 34 d) + (a + 35 d) + (a + 36 d) + (a + 37 d) + (a + 38 d) + (a + 39 d) = 687$
  1 mark
4. $\Rightarrow 6 a + 219 d = 687 \Rightarrow 2 a + 73 d = 229$ ... (ii)
  1 mark
5. From (i), $2 a + 8 d = 34$ . Subtract this from (ii):
  $65 d = 195 \Rightarrow d = 3$
  Then $a = 17 - 4 (3) = 5$ .
  1 mark
6. Sum of all 40 terms: $S_{40} = \frac{40}{2} [2 (5) + 39 (3)] = 20 (10 + 117) = 20 \times 127 = 2540$
  So first term $= 5$ , common difference $= 3$ , and the sum of all terms $= 2540$ .
  1 mark
Answer: a = 5, d = 3, S₄₀ = 2540
Where the marks are
- Uses $S_{9} = 153$ to get $a + 4 d = 17$ (1 mark)
- Writes the sum of the last 6 terms as $a_{35}$ to $a_{40}$ and simplifies to $6 a + 219 d = 687$ , i.e. $2 a + 73 d = 229$ (2 marks)
- Solves the two equations to get $a = 5$ , $d = 3$ (1 mark)
- Finds $S_{40} = 2540$ (1 mark)
Open Q33(B) on its own page
Answer one of these 2 alternatives.
Q34(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 Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerTriangles
In the given figure PA, QB and RC are each perpendicular to AC. If AP = x, BQ = y and CR = z, then prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ .
Figure in the paper: Points A and C lie on a horizontal base AC. B lies between A and C on the base. P is above A with PA perpendicular to AC (length x). Q lies on the segment PC with QB perpendicular to AC (length y). R is above C with RC perpendicular to AC (length z). Line AR passes through Q, and line PC also passes through Q... (Q is the intersection of PC and AR). See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ P A C$ and $△ QB C$ :
  $∠ P A C = ∠ QB C = 9 0^{\circ}$ [given perpendiculars]
  $∠ P C A = ∠ QC B$ [common angle]
  So $△ P A C \sim △ QB C$ [AA similarity].
  1 mark
2. Corresponding sides are proportional:
  $\frac{QB}{P A} = \frac{B C}{A C} \Rightarrow \frac{y}{x} = \frac{B C}{A C}$ ... (i)
  1 mark
3. In $△ R C A$ and $△ QB A$ :
  $∠ R C A = ∠ QB A = 9 0^{\circ}$
  $∠ R A C = ∠ Q A B$ [common angle]
  So $△ R C A \sim △ QB A$ [AA similarity].
  1 mark
4. Corresponding sides are proportional:
  $\frac{QB}{R C} = \frac{A B}{A C} \Rightarrow \frac{y}{z} = \frac{A B}{A C}$ ... (ii)
  1 mark
5. Add (i) and (ii):
  $\frac{y}{x} + \frac{y}{z} = \frac{B C + A B}{A C} = \frac{A C}{A C} = 1$ [since B lies on AC]
  ½ mark
6. Divide both sides by $y$ :
  $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ . Hence proved.
  ½ mark
Answer: Proved: 1/x + 1/z = 1/y
Where the marks are
- Proves $△ P A C \sim △ QB C$ (1 mark)
- Obtains $y / x = B C / A C$ (1 mark)
- Proves $△ R C A \sim △ QB A$ (1 mark)
- Obtains $y / z = A B / A C$ (1 mark)
- Adds the two ratios and uses $A B + B C = A C$ (½ mark)
- Concludes $1/ x + 1/ z = 1/ y$ (½ mark)
Open Q34(B) on its own page
Q35 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A pole 6 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is $6 0^{\circ}$ and the angle of depression of the point P from the top of the tower is $4 5^{\circ}$ . Find the height of the tower and the distance of point P from the foot of the tower. (Use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: Draw a vertical line $A C$ with $A$ on the ground. $A B$ is the tower of height $h$ m and $B C$ is the pole of 6 m on top of it. $P$ is a point on the ground with $A P = x$ m. Join $P B$ and $P C$ . Mark $∠ C P A = 6 0^{\circ}$ , and the angle of depression of $P$ from $B$ as $4 5^{\circ}$ , so $∠ B P A = 4 5^{\circ}$ [alternate angles]. $∠ B A P = 9 0^{\circ}$ .
  1 mark
2. In right $△ B A P$ :
  $tan 4 5^{\circ} = \frac{A B}{A P} \Rightarrow 1 = \frac{h}{x}$
  1 mark
3. $\Rightarrow h = x$ ... (i)
  ½ mark
4. In right $△ C A P$ :
  $tan 6 0^{\circ} = \frac{A C}{A P} \Rightarrow 3 = \frac{h + 6}{x}$
  1 mark
5. $\Rightarrow h + 6 = 3 x$ ... (ii)
  ½ mark
6. Substitute (i) in (ii):
  $h + 6 = 3 h \Rightarrow h = \frac{6}{3 - 1} = \frac{6 ( 3 + 1 )}{2} = 3 (3 + 1)$
  $= 3 (2.73) = 8.19$ m
  ½ mark
7. $x = h = 8.19$ m.
  Height of the tower $= 8.19$ m. Distance of P from the foot of the tower $= 8.19$ m.
  ½ mark
Answer: Height of tower = 3(√3+1) ≈ 8.19 m; distance of P from foot of tower ≈ 8.19 m
Where the marks are
- Draws a correct figure with the tower, pole and both angles marked (1 mark)
- Uses $tan 4 5^{\circ} = h / x$ (1 mark)
- Gets $h = x$ (½ mark)
- Uses $tan 6 0^{\circ} = (h + 6) / x$ (1 mark)
- Gets $h + 6 = 3 x$ (½ mark)
- Finds $h = 3 (3 + 1) \approx 8.19$ m (½ mark)
- Finds distance $x \approx 8.19$ m (½ mark)
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Section E

Case-study based; internal choice in 2-mark subparts of each case

Q36 · 4 marks · Case-basedCompetency-basedQuadratic Equations
A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(Figure: photograph of a floor covered with grey rectangular tiles.)
1. (i)
  Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information.
  1 mark
2. (ii)
  Write the corresponding quadratic equation in standard form.
  1 mark
3. (iii)
  (a) Find the value of $x$ , the length of side of a tile, by factorisation.
  2 marks
4. OR
  (iii)
  (b) Solve the quadratic equation for $x$ , using quadratic formula.
  2 marks
Figure in the paper: Photograph of a floor laid with grey rectangular tiles in a staggered pattern. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Area of one original tile $= x^{2}$ , so floor area $= 200 x^{2}$ .
  New side $= (x + 1)$ , so floor area $= 128 (x + 1)^{2}$ .
  The floor is the same, so:
  $200 x^{2} = 128 (x + 1)^{2}$
  1 mark
2. (ii) Divide both sides by 8:
  $25 x^{2} = 16 (x^{2} + 2 x + 1) = 16 x^{2} + 32 x + 16$
  $\Rightarrow 9 x^{2} - 32 x - 16 = 0$
  1 mark
3. (iii)(a) Split the middle term ( $9 \times (- 16) = - 144 = - 36 \times 4$ ):
  $9 x^{2} - 36 x + 4 x - 16 = 0$
  $9 x (x - 4) + 4 (x - 4) = 0$
  $(9 x + 4) (x - 4) = 0$
  1 mark
4. (iii)(a) So $x = 4$ or $x = - \frac{4}{9}$ .
  A length cannot be negative, so $x = 4$ units.
  1 mark
5. (iii)(b) For $9 x^{2} - 32 x - 16 = 0$ : $a = 9$ , $b = - 32$ , $c = - 16$ .
  $D = b^{2} - 4 a c = 1024 + 576 = 1600$
  $x = \frac{32 \pm 1600}{18} = \frac{32 \pm 40}{18}$
  1 mark
6. (iii)(b) So $x = \frac{72}{18} = 4$ or $x = \frac{- 8}{18} = - \frac{4}{9}$ .
  Since a length cannot be negative, $x = 4$ units.
  1 mark
Answer: (i) 200x² = 128(x+1)²; (ii) 9x² − 32x − 16 = 0; (iii)(a) x = 4 units; (iii)(b) x = 4 units (rejecting −4/9)
Where the marks are
- (i) Equates the two floor areas: $200 x^{2} = 128 (x + 1)^{2}$ (1 mark)
- (ii) Simplifies to $9 x^{2} - 32 x - 16 = 0$ (1 mark)
- (iii)(a) Factorises as $(9 x + 4) (x - 4) = 0$ (1 mark)
- (iii)(a) Rejects the negative root and gets $x = 4$ (1 mark)
- (iii)(b) Substitutes correctly in the quadratic formula: $x = (32 \pm 40) /18$ (1 mark)
- (iii)(b) Gets $x = 4$ , rejecting $- 4/9$ (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedStatistics
BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.
The table given below shows the data of one such game where 48 balls were used before Tara said 'BINGO'.
Numbers announced Number of times
0-15 8
15-30 9
30-45 10
45-60 12
60-75 9
Based on the above information, answer the following:
1. (i)
  Write the median class.
  1 mark
2. (ii)
  When first ball was picked up, what was the probability of calling out an even number?
  1 mark
3. (iii)
  (a) Find median of the given data.
  2 marks
4. OR
  (iii)
  (b) Find mode of the given data.
  2 marks
Figure in the paper: Photograph-style image of BINGO balls labelled B, I, N, G, O over a background of numbered bingo cards. See it in the official paper.
Try it first, then show our solutionSolution
1. Cumulative frequencies:
  Class 0–15 15–30 30–45 45–60 60–75
  f 8 9 10 12 9
  cf 8 17 27 39 48
  $N = 48$
2. (i) $\frac{N}{2} = 24$ . The first cf greater than 24 is 27, in class 30–45.
  So the median class is 30–45.
  1 mark
3. (ii) Balls are numbered 1 to 75, so there are 75 equally likely outcomes.
  Even numbers are $2, 4, \dots, 74$ , which is 37 numbers.
  $P (even) = \frac{37}{75}$
  1 mark
4. (iii)(a) For the median class 30–45: $l = 30$ , $c f = 17$ , $f = 10$ , $h = 15$ .
  Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 30 + \frac{24 - 17}{10} \times 15$
  1 mark
5. (iii)(a) $= 30 + 10.5 = 40.5$
  1 mark
6. (iii)(b) The highest frequency is 12, so the modal class is 45–60.
  Here $l = 45$ , $f_{1} = 12$ , $f_{0} = 10$ , $f_{2} = 9$ , $h = 15$ .
  ½ mark
7. (iii)(b) Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 45 + \frac{12 - 10}{24 - 10 - 9} \times 15$
  1 mark
8. (iii)(b) $= 45 + \frac{2}{5} \times 15 = 45 + 6 = 51$
  ½ mark
Answer: (i) 30–45; (ii) 37/75; (iii)(a) 40.5; (iii)(b) 51
Where the marks are
- (i) Uses $N /2 = 24$ to identify the median class 30–45 (1 mark)
- (ii) Counts 37 even numbers out of 75 and gets probability $37/75$ (1 mark)
- (iii)(a) Substitutes correctly in the median formula (1 mark)
- (iii)(a) Gets median $= 40.5$ (1 mark)
- (iii)(b) Identifies the modal class 45–60 (½ mark)
- (iii)(b) Substitutes correctly in the mode formula (1 mark)
- (iii)(b) Gets mode $= 51$ (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedCircles
A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC = 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = $x$ m.
1. (i)
  Find the length of AR in terms of $x$ .
  1 mark
2. (ii)
  Write the type of quadrilateral BQOR.
  1 mark
3. (iii)
  (a) Find the length PC in terms of $x$ and hence find the value of $x$ .
  2 marks
4. OR
  (iii)
  (b) Find $x$ and hence find the radius $r$ of circle.
  2 marks
Figure in the paper: Right triangle ABC with right angle at B; AB = 7 m vertical, BC = 15 m horizontal. A circle with centre O and radius r is inscribed, touching AC at P, BC at Q and AB at R. OR, OQ are drawn to the touching points. Labels: 7 m on AB, 15 m on BC. A photograph of a triangular backyard with a circular pit is shown alongside. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Tangents drawn from an external point to a circle are equal. So from A: $A R = A P = x$ m.
  1 mark
2. (ii) In BQOR:
  $∠ B = 9 0^{\circ}$ [given]
  $∠ O R B = ∠ O QB = 9 0^{\circ}$ [tangent ⟂ radius]
  So all four angles are $9 0^{\circ}$ . Also $O R = O Q = r$ [radii], and $B R = B Q$ [tangents from B].
  Hence BQOR is a square.
  1 mark
3. (iii)(a) $B R = A B - A R = 7 - x$ . Since BQOR is a square, $B Q = B R = 7 - x$ .
  $C Q = B C - B Q = 15 - (7 - x) = 8 + x$
  $P C = C Q$ [tangents from C], so $P C = 8 + x$ .
  ½ mark
4. (iii)(a) $A C = A P + P C = x + 8 + x = 8 + 2 x$ .
  By Pythagoras in $△ A B C$ :
  $A C^{2} = A B^{2} + B C^{2} = 49 + 225 = 274$
  $\Rightarrow (8 + 2 x)^{2} = 274 \Rightarrow 8 + 2 x = 274$
  1 mark
5. (iii)(a) $x = \frac{274 - 8}{2} \approx \frac{16.55 - 8}{2} \approx 4.28$ m
  ½ mark
6. (iii)(b) As in (iii)(a): $A R = x$ , $B Q = B R = 7 - x$ , and $P C = C Q = 8 + x$ , so $A C = 8 + 2 x$ .
  By Pythagoras: $(8 + 2 x)^{2} = 7^{2} + 1 5^{2} = 274$
  $\Rightarrow 8 + 2 x = 274$
  1 mark
7. (iii)(b) $x = \frac{274 - 8}{2} \approx 4.28$ m
  ½ mark
8. (iii)(b) BQOR is a square, so $r = B R = 7 - x$ :
  $r = 7 - (\frac{274}{2} - 4) = 11 - \frac{274}{2} \approx 2.72$ m
  ½ mark
Answer: (i) AR = x m; (ii) square; (iii)(a) PC = (8 + x) m, x = (√274 − 8)/2 ≈ 4.28 m; (iii)(b) x ≈ 4.28 m, r = 11 − √274/2 ≈ 2.72 m
Where the marks are
- (i) Uses equal tangents from A: $A R = x$ m (1 mark)
- (ii) Identifies BQOR as a square (1 mark)
- (iii)(a) Gets $P C = 8 + x$ (½ mark)
- (iii)(a) Uses Pythagoras: $(8 + 2 x)^{2} = 274$ (1 mark)
- (iii)(a) Finds $x = (274 - 8) /2 \approx 4.28$ (½ mark)
- (iii)(b) Forms $(8 + 2 x)^{2} = 274$ (1 mark)
- (iii)(b) Finds $x \approx 4.28$ (½ mark)
- (iii)(b) Finds $r = 7 - x \approx 2.72$ m (½ mark)
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Numbers announced	Number of times
0-15	8
15-30	9
30-45	10
45-60	12
60-75	9

Class	0–15	15–30	30–45	45–60	60–75
f	8	9	10	12	9
cf	8	17	27	39	48

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