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

CBSE Class 10 Mathematics Basic: 2024 board paper, set 430/1/1

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_BASIC.zip (a zip; this paper is the file 430-1-1MATHEMATICS (BASIC).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

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

Instructions in brief: All 38 questions are compulsory with no overall choice; internal choice is given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E; calculators are not allowed; take π = 22/7 where not stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPolynomials
For what value of $k$ , the product of zeroes of the polynomial $k x^{2} - 4 x - 7$ is $2$ ?
1. (A) $- \frac{1}{14}$
2. (B) $- \frac{7}{2}$
3. (C) $\frac{7}{2}$
4. (D) $- \frac{2}{7}$
Try it first, then show our solutionSolution
1. Product of zeroes $= \frac{c}{a} = \frac{- 7}{k}$ .
2. $\frac{- 7}{k} = 2 \Rightarrow k = - \frac{7}{2}$ , option (B).
  1 mark
Answer: $k = - \frac{7}{2}$ , option (B)
Where the marks are
- Uses product $= c / a$ to get $k = - 7/2$ (1 mark)
Q2 · 1 mark · MCQArithmetic Progressions
In an A.P., if $a = 8$ and $a_{10} = - 19$ , then value of $d$ is :
1. (A)3
2. (B) $- \frac{11}{9}$
3. (C) $- \frac{27}{10}$
4. (D) $- 3$
Try it first, then show our solutionSolution
1. $a_{10} = a + 9 d \Rightarrow - 19 = 8 + 9 d$ .
2. $9 d = - 27 \Rightarrow d = - 3$ , option (D).
  1 mark
Answer: $d = - 3$ , option (D)
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ to get $d = - 3$ (1 mark)
Q3 · 1 mark · MCQCoordinate Geometry
The mid-point of the line segment joining the points $(- 1, 3)$ and $(8, \frac{3}{2})$ is :
1. (A) $(\frac{7}{2}, - \frac{3}{4})$
2. (B) $(\frac{7}{2}, \frac{9}{2})$
3. (C) $(\frac{9}{2}, - \frac{3}{4})$
4. (D) $(\frac{7}{2}, \frac{9}{4})$
Try it first, then show our solutionSolution
1. Mid-point $= (\frac{- 1 + 8}{2}, \frac{3 + \frac{3}{2}}{2})$ .
2. $= (\frac{7}{2}, \frac{9}{4})$ , option (D).
  1 mark
Answer: $(\frac{7}{2}, \frac{9}{4})$ , option (D)
Where the marks are
- Applies mid-point formula correctly (1 mark)
Q4 · 1 mark · MCQIntroduction to Trigonometry
If $sin θ = \frac{1}{3}$ , then $sec θ$ is equal to :
1. (A) $\frac{2 2}{3}$
2. (B) $\frac{3}{2 2}$
3. (C) $3$
4. (D) $\frac{1}{3}$
Try it first, then show our solutionSolution
1. $cos θ = 1 - \frac{1}{9} = \frac{2 2}{3}$ .
2. $sec θ = \frac{1}{c o s θ} = \frac{3}{2 2}$ , option (B).
  1 mark
Answer: $\frac{3}{2 2}$ , option (B)
Where the marks are
- Finds $cos θ$ and takes reciprocal (1 mark)
Q5 · 1 mark · MCQReal Numbers
HCF $(132, 77)$ is :
1. (A)11
2. (B)77
3. (C)22
4. (D)44
Try it first, then show our solutionSolution
1. $132 = 2^{2} \times 3 \times 11$ , $77 = 7 \times 11$ .
2. Common factor: $11$ , so HCF $= 11$ , option (A).
  1 mark
Answer: 11, option (A)
Where the marks are
- Prime factorisation giving HCF 11 (1 mark)
Q6 · 1 mark · MCQQuadratic Equations
If the roots of quadratic equation $4 x^{2} - 5 x + k = 0$ are real and equal, then value of $k$ is :
1. (A) $\frac{5}{4}$
2. (B) $\frac{25}{16}$
3. (C) $- \frac{5}{4}$
4. (D) $- \frac{25}{16}$
Try it first, then show our solutionSolution
1. Equal roots: $D = b^{2} - 4 a c = 0 \Rightarrow 25 - 16 k = 0$ .
2. $k = \frac{25}{16}$ , option (B).
  1 mark
Answer: $k = \frac{25}{16}$ , option (B)
Where the marks are
- Sets discriminant zero and solves (1 mark)
Q7 · 1 mark · MCQProbability
If probability of winning a game is $p$ , then probability of losing the game is :
1. (A) $1 + p$
2. (B) $- p$
3. (C) $p - 1$
4. (D) $1 - p$
Try it first, then show our solutionSolution
1. Winning and losing are complementary events, so $P (lose) = 1 - P (win) = 1 - p$ , option (D).
  1 mark
Answer: $1 - p$ , option (D)
Where the marks are
- Uses $P (E) + P (\overset{ˉ}{E}) = 1$ (1 mark)
Q8 · 1 mark · MCQCoordinate Geometry
The distance between the points $(2, - 3)$ and $(- 2, 3)$ is :
1. (A) $213$ units
2. (B)5 units
3. (C) $132$ units
4. (D)10 units
Try it first, then show our solutionSolution
1. $d = (- 2 - 2)^{2} + (3 + 3)^{2} = 16 + 36 = 52$ .
2. $= 213$ units, option (A).
  1 mark
Answer: $213$ units, option (A)
Where the marks are
- Applies distance formula correctly (1 mark)
Q9 · 1 mark · MCQIntroduction to Trigonometry
For what value of $θ$ , $sin^{2} θ + sin θ + cos^{2} θ$ is equal to $2$ ?
1. (A) $4 5^{\circ}$
2. (B) $0^{\circ}$
3. (C) $9 0^{\circ}$
4. (D) $3 0^{\circ}$
Try it first, then show our solutionSolution
1. $sin^{2} θ + cos^{2} θ = 1$ , so expression $= 1 + sin θ = 2 \Rightarrow sin θ = 1$ .
2. $θ = 9 0^{\circ}$ , option (C).
  1 mark
Answer: $9 0^{\circ}$ , option (C)
Where the marks are
- Uses identity to get $sin θ = 1$ (1 mark)
Q10 · 1 mark · MCQProbability
A card is drawn from a well shuffled deck of 52 playing cards. The probability that drawn card is a red queen, is :
1. (A) $\frac{1}{13}$
2. (B) $\frac{2}{13}$
3. (C) $\frac{1}{52}$
4. (D) $\frac{1}{26}$
Try it first, then show our solutionSolution
1. Red queens: queen of hearts and queen of diamonds = 2.
2. $P = \frac{2}{52} = \frac{1}{26}$ , option (D).
  1 mark
Answer: $\frac{1}{26}$ , option (D)
Where the marks are
- Counts 2 favourable outcomes out of 52 (1 mark)
Q11 · 1 mark · MCQStatistics
If a certain variable $x$ divides a statistical data arranged in order into two equal parts; then the value of $x$ is called the :
1. (A)mean
2. (B)median
3. (C)mode
4. (D)range
Try it first, then show our solutionSolution
1. The middle value that splits ordered data into two equal halves is the median, option (B).
  1 mark
Answer: median, option (B)
Where the marks are
- Identifies median (1 mark)
Q12 · 1 mark · MCQSurface Areas and Volumes
The radius of a sphere is $\frac{7}{2}$ cm. The volume of the sphere is :
1. (A) $\frac{231}{3}$ cu cm
2. (B) $\frac{539}{12}$ cu cm
3. (C) $\frac{539}{3}$ cu cm
4. (D)154 cu cm
Try it first, then show our solutionSolution
1. $V = \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8}$ .
2. $= \frac{539}{3}$ cu cm, option (C).
  1 mark
Answer: $\frac{539}{3}$ cu cm, option (C)
Where the marks are
- Applies $\frac{4}{3} π r^{3}$ correctly (1 mark)
Q13 · 1 mark · MCQStatistics
The mean and median of a statistical data are 21 and 23 respectively. The mode of the data is :
1. (A)27
2. (B)22
3. (C)17
4. (D)23
Try it first, then show our solutionSolution
1. Empirical relation: Mode $= 3$ Median $- 2$ Mean.
2. $= 3 (23) - 2 (21) = 69 - 42 = 27$ , option (A).
  1 mark
Answer: 27, option (A)
Where the marks are
- Uses empirical relation to get 27 (1 mark)
Q14 · 1 mark · MCQSurface Areas and Volumes
The height and radius of a right circular cone are 24 cm and 7 cm respectively. The slant height of the cone is :
1. (A)24 cm
2. (B)31 cm
3. (C)26 cm
4. (D)25 cm
Try it first, then show our solutionSolution
1. $l = h^{2} + r^{2} = 576 + 49 = 625$ .
2. $l = 25$ cm, option (D).
  1 mark
Answer: 25 cm, option (D)
Where the marks are
- Uses $l = h^{2} + r^{2}$ (1 mark)
Q15 · 1 mark · MCQPolynomials
If one of the zeroes of the quadratic polynomial $(α - 1) x^{2} + α x + 1$ is $- 3$ , then the value of $α$ is :
1. (A) $- \frac{2}{3}$
2. (B) $\frac{2}{3}$
3. (C) $\frac{4}{3}$
4. (D) $\frac{3}{4}$
Try it first, then show our solutionSolution
1. Put $x = - 3$ : $9 (α - 1) - 3 α + 1 = 0 \Rightarrow 6 α - 8 = 0$ .
2. $α = \frac{4}{3}$ , option (C).
  1 mark
Answer: $α = \frac{4}{3}$ , option (C)
Where the marks are
- Substitutes zero and solves (1 mark)
Q16 · 1 mark · MCQCoordinate Geometry
The diameter of a circle is of length 6 cm. If one end of the diameter is $(- 4, 0)$ , the other end on x-axis is at :
1. (A) $(0, 2)$
2. (B) $(6, 0)$
3. (C) $(2, 0)$
4. (D) $(4, 0)$
Try it first, then show our solutionSolution
1. Other end is $(x, 0)$ with distance $∣ x + 4∣ = 6 \Rightarrow x = 2$ or $x = - 10$ .
2. Among the options, $(2, 0)$ , option (C).
  1 mark
Answer: $(2, 0)$ , option (C)
Where the marks are
- Uses distance 6 along x-axis to get $(2, 0)$ (1 mark)
Q17 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of $k$ for which the pair of linear equations $5 x + 2 y - 7 = 0$ and $2 x + k y + 1 = 0$ don't have a solution, is :
1. (A)5
2. (B) $\frac{4}{5}$
3. (C) $\frac{5}{4}$
4. (D) $\frac{5}{2}$
Try it first, then show our solutionSolution
1. No solution: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , i.e. $\frac{5}{2} = \frac{2}{k}$ .
2. $k = \frac{4}{5}$ (and $\frac{5}{2} \neq = \frac{- 7}{1}$ ), option (B).
  1 mark
Answer: $k = \frac{4}{5}$ , option (B)
Where the marks are
- Uses parallel-lines condition to get $k = 4/5$ (1 mark)
Q18 · 1 mark · MCQProbability
Two dice are rolled together. The probability of getting a doublet is :
1. (A) $\frac{2}{36}$
2. (B) $\frac{1}{36}$
3. (C) $\frac{1}{6}$
4. (D) $\frac{5}{6}$
Try it first, then show our solutionSolution
1. Doublets: (1,1),…,(6,6) = 6 out of 36 outcomes.
2. $P = \frac{6}{36} = \frac{1}{6}$ , option (C).
  1 mark
Answer: $\frac{1}{6}$ , option (C)
Where the marks are
- Counts 6 doublets out of 36 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedCircles
Directions: In Q. No. 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Select the correct option from the following options.
Assertion (A): If the PA and PB are tangents drawn to a circle with centre O from an external point P, then the quadrilateral OAPB is a cyclic quadrilateral.
Reason (R): In a cyclic quadrilateral, opposite angles are equal.
1. (A)Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely.
2. (B)Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain Assertion (A).
3. (C)Assertion (A) is true but Reason (R) is false.
4. (D)Assertion (A) is false but Reason (R) is true.
Figure in the paper: Circle with centre O. Two tangents from external point P touch the circle at A (upper) and B (lower). Radii OA and OB are drawn, forming quadrilateral OAPB with P on the right. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A P = ∠ O B P = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ O A P + ∠ O B P = 18 0^{\circ}$ ; opposite angles supplementary ⇒ OAPB is cyclic. A is true.
2. In a cyclic quadrilateral opposite angles are supplementary, not necessarily equal, so R is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- A true via supplementary opposite angles, R false (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedPolynomials
Assertion (A): Zeroes of a polynomial $p (x) = x^{2} - 2 x - 3$ are $- 1$ and $3$ .
Reason (R): The graph of polynomial $p (x) = x^{2} - 2 x - 3$ intersects x-axis at $(- 1, 0)$ and $(3, 0)$ .
1. (A)Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely.
2. (B)Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain 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. $x^{2} - 2 x - 3 = (x - 3) (x + 1)$ , zeroes $3, - 1$ : A true.
2. Zeroes are the x-coordinates where the graph meets the x-axis, so R is true and explains A. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both true and R explains A (1 mark)

Section B

Very short answer

Q21 · 2 marks · Very short answerTriangles
D is a point on the side BC of $Δ$ ABC such that $∠$ ADC = $∠$ BAC. Show that $A C^{2} = B C \times D C$ .
Figure in the paper: Triangle ABC with vertex A at top, base BC horizontal. Point D lies on BC, and segment AD is drawn from A to D. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ and $△ D A C$ : $∠ B A C = ∠ A D C$ [given]
  ½ mark
2. $∠ A C B = ∠ D C A$ [common angle at C]
  ½ mark
3. $∴ △ A B C \sim △ D A C$ [AA similarity criterion]
  ½ mark
4. Corresponding sides of similar triangles are proportional, so $\frac{A C}{D C} = \frac{B C}{A C}$ . Cross-multiplying gives $A C^{2} = B C \times D C$ . Hence proved.
  ½ mark
Answer: $△ A B C \sim △ D A C$ (AA), so $A C^{2} = B C \times D C$
Where the marks are
- States $∠ B A C = ∠ A D C$ (given) (½ mark)
- States $∠ C$ is common (½ mark)
- Concludes $△ A B C \sim △ D A C$ by AA (½ mark)
- Writes the ratio $\frac{A C}{D C} = \frac{B C}{A C}$ and obtains $A C^{2} = B C \times D C$ (½ mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(A) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve the following pair of linear equations for $x$ and $y$ algebraically : $x + 2 y = 9$ and $y - 2 x = 2$
Try it first, then show our solutionSolution
1. $x + 2 y = 9$ ...(i), $y - 2 x = 2$ ...(ii). From (ii), $y = 2 x + 2$ . Substituting in (i): $x + 2 (2 x + 2) = 9 \Rightarrow 5 x + 4 = 9 \Rightarrow x = 1$
  1 mark
2. Then $y = 2 (1) + 2 = 4$ . Check in (i): $1 + 8 = 9$ ✓. So $x = 1, y = 4$ .
  1 mark
Answer: $x = 1, y = 4$
Where the marks are
- Correct elimination/substitution leading to $x = 1$ (1 mark)
- Finds $y = 4$ (1 mark)
Open Q22(A) on its own page
OR
Q22(B) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Check whether the point $(- 4, 3)$ lies on both the lines represented by the linear equations $x + y + 1 = 0$ and $x - y = 1$ .
Try it first, then show our solutionSolution
1. For $x + y + 1 = 0$ : put $x = - 4, y = 3$ : LHS $= - 4 + 3 + 1 = 0 =$ RHS. So $(- 4, 3)$ lies on this line.
  1 mark
2. For $x - y = 1$ : LHS $= - 4 - 3 = - 7 \neq = 1$ . So $(- 4, 3)$ does not lie on this line. Hence the point does not lie on both lines.
  1 mark
Answer: $(- 4, 3)$ lies on $x + y + 1 = 0$ but not on $x - y = 1$ ; so it does not lie on both lines.
Where the marks are
- Substitutes in $x + y + 1 = 0$ and shows it is satisfied (1 mark)
- Substitutes in $x - y = 1$ , shows it is not satisfied and concludes (1 mark)
Open Q22(B) on its own page
Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerReal Numbers
Prove that $6 - 45$ is an irrational number, given that $5$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $6 - 45 = x$ is rational. Then $45 = 6 - x \Rightarrow 5 = \frac{6 - x}{4}$ .
  1 mark
2. Since $x$ is rational, $\frac{6 - x}{4}$ is rational, so $5$ would be rational. This contradicts the fact that $5$ is irrational. Hence our assumption is wrong and $6 - 45$ is irrational.
  1 mark
Answer: $6 - 45$ is irrational (by contradiction).
Where the marks are
- Assumes the number rational and expresses $5 = \frac{6 - x}{4}$ (1 mark)
- Points out RHS rational, LHS irrational — contradiction, conclusion (1 mark)
Open Q23(A) on its own page
OR
Q23(B) · 2 marks · Very short answerReal Numbers
Show that $11 \times 19 \times 23 + 3 \times 11$ is not a prime number.
Try it first, then show our solutionSolution
1. $11 \times 19 \times 23 + 3 \times 11 = 11 \times (19 \times 23 + 3) = 11 \times (437 + 3) = 11 \times 440$
  1 mark
2. So the number has factors other than 1 and itself (e.g. 11 and 440); it is a composite number, not a prime.
  1 mark
Answer: $= 11 \times 440 = 4840$ , composite, hence not prime.
Where the marks are
- Takes 11 common: $11 \times (19 \times 23 + 3)$ (1 mark)
- States it has more than two factors, hence not prime (1 mark)
Open Q23(B) on its own page
Q24 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $sin A cos B + cos A sin B$ ; if $A = 3 0^{\circ}$ and $B = 4 5^{\circ}$ .
Try it first, then show our solutionSolution
1. $sin 3 0^{\circ} cos 4 5^{\circ} + cos 3 0^{\circ} sin 4 5^{\circ} = \frac{1}{2} \times \frac{1}{2} + \frac{3}{2} \times \frac{1}{2}$
  1½ marks
2. $= \frac{1 + 3}{2 2}$ $(= \frac{6 + 2}{4})$
  ½ mark
Answer: $\frac{3 + 1}{2 2}$
Where the marks are
- Substitutes correct values of $sin 3 0^{\circ}, cos 4 5^{\circ}, cos 3 0^{\circ}, sin 4 5^{\circ}$ (1½ marks)
- Simplifies to $\frac{3 + 1}{2 2}$ (½ mark)
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Q25 · 2 marks · Very short answerProbability
A bag contains 4 red, 5 white and some yellow balls. If probability of drawing a red ball at random is $\frac{1}{5}$ , then find the probability of drawing a yellow ball at random.
Try it first, then show our solutionSolution
1. Let the number of yellow balls be $n$ . Total balls $= 4 + 5 + n = 9 + n$ .
2. $P (red) = \frac{4}{9 + n} = \frac{1}{5} \Rightarrow 9 + n = 20 \Rightarrow n = 11$
  1½ marks
3. $P (yellow) = \frac{11}{20}$
  ½ mark
Answer: $\frac{11}{20}$
Where the marks are
- Forms $\frac{4}{9 + n} = \frac{1}{5}$ and gets $n = 11$ (1½ marks)
- $P (yellow) = \frac{11}{20}$ (½ mark)
Open Q25 on its own page

Section C

Short answer

Q26 · 3 marks · Short answerReal Numbers
Two alarm clocks ring their alarms at regular intervals of 20 minutes and 25 minutes respectively. If they first beep together at 12 noon, at what time will they beep again together next time ?
Try it first, then show our solutionSolution
1. They beep together again after LCM(20, 25) minutes. $20 = 2^{2} \times 5$ , $25 = 5^{2}$ , so LCM $= 2^{2} \times 5^{2} = 100$ minutes.
  2 marks
2. 100 minutes = 1 hour 40 minutes. So next time together = 12:00 noon + 1 h 40 min = 1:40 pm.
  1 mark
Answer: 1:40 pm
Where the marks are
- Finds LCM(20, 25) = 100 (2 marks)
- Converts to 1 h 40 min and states 1:40 pm (1 mark)
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Q27 · 3 marks · Short answerPair of Linear Equations in Two Variables
The greater of two supplementary angles exceeds the smaller by $1 8^{\circ}$ . Find measures of these two angles.
Try it first, then show our solutionSolution
1. Let the greater angle be $x^{\circ}$ and the smaller $y^{\circ}$ . Supplementary: $x + y = 180$ ...(i)
  1 mark
2. Given: $x - y = 18$ ...(ii)
  1 mark
3. Adding (i) and (ii): $2 x = 198 \Rightarrow x = 99$
  ½ mark
4. $y = 180 - 99 = 81$ . The angles are $9 9^{\circ}$ and $8 1^{\circ}$ .
  ½ mark
Answer: $9 9^{\circ}$ and $8 1^{\circ}$
Where the marks are
- Forms $x + y = 180$ (1 mark)
- Forms $x - y = 18$ (1 mark)
- $x = 9 9^{\circ}$ (½ mark)
- $y = 8 1^{\circ}$ (½ mark)
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Q28 · 3 marks · Short answerCoordinate Geometry
Find the co-ordinates of the points of trisection of the line segment joining the points $(- 2, 2)$ and $(7, - 4)$ .
Try it first, then show our solutionSolution
1. Let $A (- 2, 2)$ , $B (7, - 4)$ and let P, Q be the points of trisection, so $A P : P B = 1 : 2$ and $A Q : QB = 2 : 1$ . Section formula: $(\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})$ .
2. P (1:2): $(\frac{1 ( 7 ) + 2 ( - 2 )}{3}, \frac{1 ( - 4 ) + 2 ( 2 )}{3}) = (\frac{3}{3}, \frac{0}{3}) = (1, 0)$
  1½ marks
3. Q (2:1): $(\frac{2 ( 7 ) + 1 ( - 2 )}{3}, \frac{2 ( - 4 ) + 1 ( 2 )}{3}) = (\frac{12}{3}, \frac{- 6}{3}) = (4, - 2)$
  1½ marks
Answer: $(1, 0)$ and $(4, - 2)$
Where the marks are
- Uses ratio 1:2 with section formula and gets $P (1, 0)$ (1½ marks)
- Uses ratio 2:1 and gets $Q (4, - 2)$ (1½ marks)
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Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerCircles
In two concentric circles, the radii are $O A = r$ cm and $O Q = 6$ cm, as shown in the figure. Chord CD of larger circle is a tangent to smaller circle at Q. PA is tangent to larger circle. If $P A = 16$ cm and $O P = 20$ cm, find the length CD.
Figure in the paper: Two concentric circles with centre O. A is a point on the outer circle, with tangent PA drawn from external point P, where OP = 20 cm; OA is joined. Chord CD of the outer circle is horizontal, touching the inner circle at Q, with OQ perpendicular to CD. D lies on the outer circle to the right of Q and OD is joined. See it in the official paper.
Try it first, then show our solutionSolution
1. $O A ⊥ P A$ [tangent ⟂ radius]. In right $△ O A P$ : $O A^{2} = O P^{2} - P A^{2} = 2 0^{2} - 1 6^{2} = 144 \Rightarrow r = O A = 12$ cm. Also $O D = 12$ cm (radius of larger circle).
  1 mark
2. $O Q ⊥ C D$ [tangent CD ⟂ radius OQ of smaller circle]. In right $△ O Q D$ : $Q D^{2} = O D^{2} - O Q^{2} = 1 2^{2} - 6^{2} = 108 \Rightarrow Q D = 63$ cm
  1 mark
3. The perpendicular from the centre to a chord bisects it, so $C D = 2 Q D = 123$ cm $(\approx 20.78$ cm $)$ .
  1 mark
Answer: $C D = 123$ cm
Where the marks are
- Uses tangent ⟂ radius and Pythagoras to get $r = 12$ cm (1 mark)
- Finds $Q D = 63$ cm in $△ O Q D$ (1 mark)
- Uses perpendicular bisects chord to get $C D = 123$ cm (1 mark)
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerCircles
In given figure, two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $∠$ PTQ = $2∠$ OPQ.
Figure in the paper: Circle with centre O. External point T with tangents TP and TQ touching the circle at P (upper) and Q (lower). Chord PQ and radius OP are drawn; the angle at T is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. Let $∠ P T Q = θ$ . $T P = T Q$ [tangents from an external point are equal], so $∠ T P Q = ∠ T QP$ . In $△ T P Q$ : $∠ T P Q + ∠ T QP + θ = 18 0^{\circ} \Rightarrow ∠ T P Q = 9 0^{\circ} - \frac{θ}{2}$
  1½ marks
2. $O P ⊥ P T$ [tangent ⟂ radius], so $∠ O P Q + ∠ T P Q = 9 0^{\circ} \Rightarrow ∠ O P Q = 9 0^{\circ} - (9 0^{\circ} - \frac{θ}{2}) = \frac{θ}{2}$
  1 mark
3. Hence $∠ P T Q = θ = 2∠ O P Q$ . Proved.
  ½ mark
Answer: $∠ O P Q = \frac{θ}{2}$ , so $∠ P T Q = 2∠ O P Q$
Where the marks are
- Uses TP = TQ and angle sum to get $∠ T P Q = 9 0^{\circ} - \frac{θ}{2}$ (1½ marks)
- Uses OP ⟂ PT to get $∠ O P Q = \frac{θ}{2}$ (1 mark)
- Concludes $∠ P T Q = 2∠ O P Q$ (½ mark)
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Answer one of these 2 alternatives.
Q30(A) · 3 marks · Short answerSurface Areas and Volumes
A solid is in the form of a cylinder with hemi-spherical ends of same radii. The total height of the solid is 20 cm and the diameter of the cylinder is 14 cm. Find the surface area of the solid.
Try it first, then show our solutionSolution
1. Radius of cylinder = radius of each hemisphere $= r = \frac{14}{2} = 7$ cm
  ½ mark
2. Height of cylinder $h = 20 - 2 \times 7 = 6$ cm
  ½ mark
3. Surface area = CSA of cylinder + CSA of two hemispheres $= 2 π r h + 2 (2 π r^{2}) = 2 π r (h + 2 r) = 2 \times \frac{22}{7} \times 7 \times (6 + 14) = 44 \times 20$
  1½ marks
4. $= 880$ cm²
  ½ mark
Answer: 880 cm²
Where the marks are
- r = 7 cm (½ mark)
- Cylinder height 6 cm (½ mark)
- Correct formula $2 π r h + 4 π r^{2}$ and substitution (1½ marks)
- 880 cm² (½ mark)
Open Q30(A) on its own page
OR
Q30(B) · 3 marks · Short answerSurface Areas and Volumes
A juice glass is cylindrical in shape with hemi-spherical raised up portion at the bottom. The inner diameter of glass is 10 cm and its height is 14 cm. Find the capacity of the glass. (use $π = 3.14$ )
Try it first, then show our solutionSolution
1. Radius of glass (and of hemisphere) $r = \frac{10}{2} = 5$ cm, $h = 14$ cm
  ½ mark
2. Capacity = volume of cylinder − volume of hemisphere $= π r^{2} h - \frac{2}{3} π r^{3} = 3.14 \times 5 \times 5 \times 14 - \frac{2}{3} \times 3.14 \times 5 \times 5 \times 5 = 1099 - 261.67$
  2 marks
3. $= \frac{2512}{3} \approx 837.33$ cm³
  ½ mark
Answer: $\frac{2512}{3}$ cm³ ≈ 837.33 cm³
Where the marks are
- r = 5 cm (½ mark)
- Capacity = cylinder − hemisphere with correct substitution (2 marks)
- ≈ 837.33 cm³ (½ mark)
Open Q30(B) on its own page
Q31 · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $(cot θ - cosec θ)^{2} = \frac{1 - c o s θ}{1 + c o s θ}$ .
Try it first, then show our solutionSolution
1. LHS $= (\frac{cos θ}{sin θ} - \frac{1}{sin θ})^{2}$
  1 mark
2. $= \frac{( cos θ - 1 ) ^{2}}{sin ^{2} θ} = \frac{( 1 - cos θ ) ^{2}}{1 - cos ^{2} θ}$ [since $sin^{2} θ = 1 - cos^{2} θ$ ]
  1 mark
3. $= \frac{( 1 - cos θ ) ^{2}}{( 1 - cos θ ) ( 1 + cos θ )}$
  ½ mark
4. $= \frac{1 - cos θ}{1 + cos θ} =$ RHS. Proved.
  ½ mark
Answer: LHS simplifies to $\frac{1 - c o s θ}{1 + c o s θ}$ = RHS
Where the marks are
- Writes cot and cosec in terms of sin and cos (1 mark)
- Combines to $\frac{( c o s θ - 1 ) ^{2}}{s i n ^{2} θ}$ (1 mark)
- Uses $sin^{2} θ = (1 - cos θ) (1 + cos θ)$ (½ mark)
- Cancels to reach RHS (½ mark)
Open Q31 on its own page

Section D

Long answer

Answer one of these 2 alternatives.
Q32(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 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 Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerTriangles
Sides AB and BC and median AD of a $Δ$ ABC are respectively proportional to sides PQ and PR and median PM of $Δ$ PQR. Show that $Δ$ ABC $\sim$ $Δ$ PQR.
Figure in the paper: Left: triangle ABC with A at top, base BC, and median AD drawn to midpoint D of BC. Right: triangle PQR with P at top, base QR, and median PM drawn to midpoint M of QR. See it in the official paper.
Try it first, then show our solutionSolution
1. Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ , AD and PM medians. Construction: produce AD to E with $D E = A D$ , join CE; produce PM to N with $M N = P M$ , join RN. Mark $∠ B A D = ∠3$ , $∠ D A C = ∠1$ , $∠ QP M = ∠4$ , $∠ M P R = ∠2$ .
  1 mark
2. In $△ A D B$ and $△ E D C$ : $B D = D C$ [D midpoint], $A D = D E$ [construction], $∠ A D B = ∠ E D C$ [vertically opposite]. So $△ A D B ≅ △ E D C$ (SAS) $\Rightarrow A B = C E$ and $∠3 = ∠ C E D$ . Similarly $△ P M Q ≅ △ N M R \Rightarrow P Q = R N$ and $∠4 = ∠ R N M$ .
  1 mark
3. Now $\frac{C E}{R N} = \frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ . Hence $△ A E C \sim △ P N R$ [SSS similarity].
  1 mark
4. So $∠1 = ∠2$ (i.e. $∠ C A E = ∠ R P N$ ) and $∠ A E C = ∠ P N R$ , giving $∠3 = ∠ C E D = ∠ R N M = ∠4$ . Adding: $∠1 + ∠3 = ∠2 + ∠4 \Rightarrow ∠ B A C = ∠ QP R$ .
  1 mark
5. In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{A C}{P R}$ [given] and $∠ B A C = ∠ QP R$ . Therefore $△ A B C \sim △ P QR$ [SAS similarity].
  1 mark
Answer: $△ A B C \sim △ P QR$ (SAS similarity, after showing $∠ B A C = ∠ QP R$ )
Where the marks are
- Construction: produce medians to double length (1 mark)
- Congruence $△ A D B ≅ △ E D C$ giving AB = CE, similarly PQ = RN (1 mark)
- Shows $△ A E C \sim △ P N R$ by SSS (1 mark)
- Deduces $∠ B A C = ∠ QP R$ (1 mark)
- Concludes $△ A B C \sim △ P QR$ by SAS (1 mark)
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Q33 · 5 marks · Long answerArithmetic Progressions
How many terms of the A.P. $27, 24, 21, \dots$ must be taken so that their sum is $105$ ? Which term of the A.P. is zero?
Try it first, then show our solutionSolution
1. Here $a = 27$ , $d = 24 - 27 = - 3$ . Let n terms be required.
  ½ mark
2. $S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow 105 = \frac{n}{2} [54 + (n - 1) (- 3)] = \frac{n}{2} (57 - 3 n)$
  1½ marks
3. $210 = 57 n - 3 n^{2} \Rightarrow 3 n^{2} - 57 n + 210 = 0 \Rightarrow n^{2} - 19 n + 70 = 0$
  1 mark
4. $(n - 5) (n - 14) = 0 \Rightarrow n = 5$ or $n = 14$ . Both are valid: the terms from the 6th to the 14th $(12, 9, \dots, - 12)$ add up to zero, so 5 terms or 14 terms give sum 105.
  1 mark
5. $a_{n} = 0 \Rightarrow 27 + (n - 1) (- 3) = 0 \Rightarrow n - 1 = 9 \Rightarrow n = 10$ . The 10th term is zero.
  1 mark
Answer: 5 terms or 14 terms; the 10th term is zero
Where the marks are
- d = −3 (½ mark)
- Sets up $105 = \frac{n}{2} [54 + (n - 1) (- 3)]$ (1½ marks)
- Simplifies to $n^{2} - 19 n + 70 = 0$ (1 mark)
- n = 5 or 14 (1 mark)
- $a_{10} = 0$ (1 mark)
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Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerSome Applications of Trigonometry
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is $3 0^{\circ}$ than when it was $6 0^{\circ}$ . Find the height of the tower and the length of original shadow. (use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: AB = h m is the tower (A at the foot). AC = x m is the shadow when altitude is $6 0^{\circ}$ ( $∠ B C A = 6 0^{\circ}$ ); AD = (x + 40) m is the shadow when altitude is $3 0^{\circ}$ ( $∠ B D A = 3 0^{\circ}$ ), with CD = 40 m, A, C, D collinear on the ground.
  1 mark
2. In right $△ B A D$ : $tan 3 0^{\circ} = \frac{A B}{A D} \Rightarrow \frac{1}{3} = \frac{h}{x + 40}$
  1 mark
3. $\Rightarrow x + 40 = h 3$ ...(i)
  ½ mark
4. In right $△ B A C$ : $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow 3 = \frac{h}{x} \Rightarrow h = x 3$ ...(ii)
  1 mark
5. Putting $x = \frac{h}{3}$ in (i): $\frac{h}{3} + 40 = h 3 \Rightarrow h (3 - \frac{1}{3}) = 40 \Rightarrow \frac{2 h}{3} = 40 \Rightarrow h = 203 = 20 \times 1.73 = 34.6$ m
  1 mark
6. $x = \frac{h}{3} = 20$ m. Height of tower = 34.6 m, original shadow = 20 m.
  ½ mark
Answer: Height = $203$ = 34.6 m; original shadow = 20 m
Where the marks are
- Correct labelled figure (1 mark)
- $tan 3 0^{\circ} = \frac{h}{x + 40}$ (1 mark)
- $x + 40 = h 3$ (½ mark)
- $tan 6 0^{\circ}$ gives $h = x 3$ (1 mark)
- h = $203$ = 34.6 m (1 mark)
- x = 20 m (½ mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerSome Applications of Trigonometry
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (use $3 = 1.73$ )
Try it first, then show our solutionSolution
1. Figure: AB = h m is the multi-storeyed building, CD = 8 m the small building, BD = x m the distance between them. Draw CE ⟂ AB, so EB = CD = 8 m, CE = BD = x and AE = (h − 8) m. Angle of depression of C is $3 0^{\circ}$ , so $∠ A C E = 3 0^{\circ}$ ; of D is $4 5^{\circ}$ , so $∠ A D B = 4 5^{\circ}$ [alternate angles].
  1 mark
2. In right $△ A E C$ : $tan 3 0^{\circ} = \frac{A E}{C E} = \frac{h - 8}{x} \Rightarrow h - 8 = \frac{x}{3}$ ...(i)
  1½ marks
3. In right $△ A B D$ : $tan 4 5^{\circ} = \frac{h}{x} \Rightarrow h = x$ ...(ii)
  1 mark
4. From (i) and (ii): $x - \frac{x}{3} = 8 \Rightarrow x (3 - 1) = 83 \Rightarrow x = \frac{8 3 ( 3 + 1 )}{2} = 12 + 43 = 12 + 6.92 = 18.92$ m. So $h = x = 18.92$ m.
  1½ marks
Answer: Height of multi-storeyed building = distance between buildings = $12 + 43$ = 18.92 m
Where the marks are
- Correct labelled figure (1 mark)
- $tan 3 0^{\circ} = \frac{h - 8}{x}$ giving $h - 8 = \frac{x}{3}$ (1½ marks)
- $tan 4 5^{\circ}$ gives $h = x$ (1 mark)
- Solves to $h = x = 12 + 43 = 18.92$ m (1½ marks)
Open Q34(B) on its own page
Q35 · 5 marks · Long answerAreas Related to Circles
A chord of a circle of radius 14 cm subtends an angle of $9 0^{\circ}$ at the centre. Find the area of the corresponding minor and major segments of the circle.
Try it first, then show our solutionSolution
1. Here $r = 14$ cm and $θ = 9 0^{\circ}$ . Area of minor sector $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{90}{360} \times \frac{22}{7} \times 14 \times 14 = 154 cm^{2}$ .
  1 mark
2. The chord and the two radii form a right triangle at the centre, since $θ = 9 0^{\circ}$ . Its area $= \frac{1}{2} r^{2} = \frac{1}{2} \times 14 \times 14 = 98 cm^{2}$ .
  1 mark
3. Area of minor segment $=$ area of sector $-$ area of triangle $= 154 - 98 = 56 cm^{2}$ .
  1½ marks
4. Area of major segment $=$ area of circle $-$ area of minor segment $= \frac{22}{7} \times 14 \times 14 - 56 = 616 - 56 = 560 cm^{2}$ .
  1½ marks
Answer: Minor segment = 56 cm²; major segment = 560 cm²
Where the marks are
- Area of sector $\frac{1}{4} \times \frac{22}{7} \times 14 \times 14$ (1 mark)
- Area of triangle $\frac{1}{2} \times 14 \times 14$ (1 mark)
- Minor segment $= 154 - 98 = 56$ cm² (1½ marks)
- Major segment $= 616 - 56 = 560$ cm² (1½ marks)
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Section E

Case-based

Q36 · 4 marks · Case-basedCompetency-basedQuadratic Equations
To keep the lawn green and cool, Sadhna uses water sprinklers which rotate in circular shape and cover a particular area.
The diagram below shows the circular areas covered by two sprinklers:
Two circles touch externally. The sum of their areas is $130 π$ sq m and the distance between their centres is 14 m.
Based on above information, answer the following questions:
1. (i)
  Obtain a quadratic equation involving R and r from above.
  1 mark
2. (ii)
  Write a quadratic equation involving only r.
  1 mark
3. (iii)
  (a) Find the radius r and the corresponding area irrigated.
  2 marks
4. OR
  (iii)
  (b) Find the radius R and the corresponding area irrigated.
  2 marks
Figure in the paper: Two circles touching externally: larger circle with radius R and centre marked with a dot joined by a radius; smaller circle with radius r and centre marked with a dot joined by a radius. Caption (R > r). A photo of lawn sprinklers spraying water is also shown. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The sum of the areas is $π R^{2} + π r^{2} = 130 π$ , so $R^{2} + r^{2} = 130$ .
  1 mark
2. (ii) The circles touch externally, so the distance between the centres is $R + r = 14$ , giving $R = 14 - r$ . Substituting: $(14 - r)^{2} + r^{2} = 130 \Rightarrow 2 r^{2} - 28 r + 66 = 0 \Rightarrow r^{2} - 14 r + 33 = 0$ .
  1 mark
3. (iii)(a) Factorising: $r^{2} - 14 r + 33 = 0 \Rightarrow (r - 11) (r - 3) = 0 \Rightarrow r = 3$ or $11$ . If $r = 11$ , then $R = 3$ , which contradicts $R > r$ . Hence $r = 3$ m.
  1 mark
4. (iii)(a) Area irrigated by the smaller sprinkler $= π r^{2} = 9 π m^{2}$ .
  1 mark
5. (iii)(b) Put $r = 14 - R$ in $R^{2} + r^{2} = 130$ . This gives $R^{2} - 14 R + 33 = 0 \Rightarrow (R - 11) (R - 3) = 0$ . Since $R > r$ , $R = 11$ m and not 3 m.
  1 mark
6. (iii)(b) Area irrigated by the larger sprinkler $= π R^{2} = 121 π m^{2}$ .
  1 mark
Answer: (i) $R^{2} + r^{2} = 130$ ; (ii) $r^{2} - 14 r + 33 = 0$ ; (iii)(a) r = 3 m, area = 9π m²; (iii)(b) R = 11 m, area = 121π m²
Where the marks are
- Writes $R^{2} + r^{2} = 130$ (1 mark)
- Uses $R + r = 14$ to get $r^{2} - 14 r + 33 = 0$ (1 mark)
- (iii)(a) Solves to get r = 3 m, rejecting 11 because r < R (1 mark)
- (iii)(a) Area $= 9 π$ m² (1 mark)
- (iii)(b) Solves to get R = 11 m, rejecting 3 because R > r (1 mark)
- (iii)(b) Area $= 121 π$ m² (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedStatistics
Gurpreet is very fond of doing research on plants. She collected some leaves from different plants and measured their lengths in mm.
The data obtained is represented in the following table:
Length (in mm) 70-80 80-90 90-100 100-110 110-120 120-130 130-140
Number of leaves 3 5 9 12 5 4 2
Based on the above information, answer the following questions:
1. (i)
  Write the median class of the data.
  1 mark
2. (ii)
  How many leaves are of length equal to or more than 10 cm ?
  1 mark
3. (iii)
  (a) Find median of the data.
  2 marks
4. OR
  (iii)
  (b) Write the modal class and find the mode of the data.
  2 marks
Figure in the paper: Photograph of many leaves of different shapes and sizes spread out on a surface, collected from various plants. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The cumulative frequencies are 3, 8, 17, 29, 34, 38, 40, so $N = 40$ and $\frac{N}{2} = 20$ . The cf first exceeds 20 in the class 100–110, so the median class is 100–110.
  1 mark
2. (ii) 10 cm = 100 mm. Leaves of length ≥ 100 mm $= 12 + 5 + 4 + 2 = 23$ .
  1 mark
3. (iii)(a) Cumulative frequency table: 70–80: f = 3, cf = 3; 80–90: f = 5, cf = 8; 90–100: f = 9, cf = 17; 100–110: f = 12, cf = 29; 110–120: f = 5, cf = 34; 120–130: f = 4, cf = 38; 130–140: f = 2, cf = 40.
  ½ mark
4. (iii)(a) For the median class, $l = 100$ , $c f = 17$ , $f = 12$ and $h = 10$ . Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 100 + \frac{20 - 17}{12} \times 10 = 100 + 2.5 = 102.5$ mm.
  1½ marks
5. (iii)(b) The highest frequency is 12, so the modal class is 100–110.
  ½ mark
6. (iii)(b) Here $l = 100$ , $f_{1} = 12$ , $f_{0} = 9$ , $f_{2} = 5$ and $h = 10$ . Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 100 + \frac{12 - 9}{24 - 9 - 5} \times 10 = 100 + 3 = 103$ mm.
  1½ marks
Answer: (i) 100–110; (ii) 23; (iii)(a) median = 102.5 mm; (iii)(b) modal class 100–110, mode = 103 mm
Where the marks are
- Identifies the median class 100–110 (1 mark)
- Counts 23 leaves of length ≥ 100 mm (1 mark)
- (iii)(a) Correct cumulative frequency table (½ mark)
- (iii)(a) Substitutes in the median formula and gets 102.5 (1½ marks)
- (iii)(b) Modal class 100–110 (½ mark)
- (iii)(b) Substitutes in the mode formula and gets 103 (1½ marks)
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Q38 · 4 marks · Case-basedCompetency-basedCircles
The picture given below shows a circular mirror hanging on the wall with a cord. The diagram represents the mirror as a circle with centre O. AP and AQ are tangents to the circle at P and Q respectively such that $A P = 30$ cm and $∠$ PAQ = $6 0^{\circ}$ .
Based on the above information; answer the following questions:
1. (i)
  Find the length PQ.
  1 mark
2. (ii)
  Find m $∠$ POQ.
  1 mark
3. (iii)
  (a) Find the length OA.
  2 marks
4. OR
  (iii)
  (b) Find the radius of the mirror.
  2 marks
Figure in the paper: Geometric figure: external point A at top, with tangent lines AP (left) and AQ (right) touching a circle with centre O at P and Q. Chord PQ is drawn horizontally, and line AO is drawn from A through the centre to O. A photo of a round mirror hanging from a wall by a cord is also shown. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $A P = A Q$ [tangents from an external point are equal], so $∠ A P Q = ∠ A QP = \frac{18 0 ^{\circ} - 6 0 ^{\circ}}{2} = 6 0^{\circ}$ . Hence $△ A P Q$ is equilateral.
  ½ mark
2. (i) Therefore $P Q = A P = 30$ cm.
  ½ mark
3. (ii) $∠ O P A = ∠ O Q A = 9 0^{\circ}$ [tangent ⟂ radius]. In quadrilateral APOQ, $∠ P O Q = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
  1 mark
4. (iii)(a) OA bisects $∠ P A Q$ , so $∠ P A O = 3 0^{\circ}$ .
  ½ mark
5. (iii)(a) In right $△ O P A$ : $cos 3 0^{\circ} = \frac{A P}{O A} \Rightarrow \frac{3}{2} = \frac{30}{O A}$ .
  1 mark
6. (iii)(a) $O A = \frac{60}{3} = 203$ cm.
  ½ mark
7. (iii)(b) $∠ P A O = 3 0^{\circ}$ [OA bisects ∠PAQ].
  ½ mark
8. (iii)(b) In right $△ O P A$ : $tan 3 0^{\circ} = \frac{O P}{A P} \Rightarrow \frac{1}{3} = \frac{O P}{30}$ .
  1 mark
9. (iii)(b) Radius $O P = \frac{30}{3} = 103$ cm.
  ½ mark
Answer: (i) PQ = 30 cm; (ii) ∠POQ = 120°; (iii)(a) OA = 20√3 cm; (iii)(b) radius = 10√3 cm
Where the marks are
- Shows △APQ is equilateral (½ mark)
- PQ = 30 cm (½ mark)
- ∠POQ = 120° (1 mark)
- (iii)(a) ∠PAO = 30° (½ mark)
- (iii)(a) Uses cos 30° = AP/OA (1 mark)
- (iii)(a) OA = 20√3 cm (½ mark)
- (iii)(b) ∠PAO = 30° (½ mark)
- (iii)(b) Uses tan 30° = OP/AP (1 mark)
- (iii)(b) Radius = 10√3 cm (½ mark)
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Length (in mm)	70-80	80-90	90-100	100-110	110-120	120-130	130-140
Number of leaves	3	5	9	12	5	4	2

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