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

CBSE Class 10 Mathematics Basic: 2024 board paper, set 430/2/1

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 430-2-1MATHEMATICS (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 study based

Instructions in brief: No overall choice; internal choice (OR) in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 of Section E (2-mark sub-parts in each case study). Use of calculators not allowed; take π = 22/7 unless stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The HCF of smallest 2-digit number and the smallest composite number is:
1. (A)2
2. (B)20
3. (C)40
4. (D)4
Try it first, then show our solutionSolution
1. Smallest 2-digit number $= 10 = 2 \times 5$ ; smallest composite number $= 4 = 2^{2}$ .
2. Common prime factor with least power: $2$ , so HCF $= 2$ . Option (A).
  1 mark
Answer: HCF = 2, option (A)
Where the marks are
- Identifies 10 and 4 and gets HCF 2 (1 mark)
Q2 · 1 mark · MCQPair of Linear Equations in Two Variables
The value of $k$ for which the pair of linear equations $x + y - 4 = 0$ and $2 x + k y - 8 = 0$ has infinitely many solutions, is
1. (A) $k \neq = 2$
2. (B) $k \neq = - 2$
3. (C) $k = 2$
4. (D) $k = - 2$
Try it first, then show our solutionSolution
1. For infinitely many solutions: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , i.e. $\frac{1}{2} = \frac{1}{k} = \frac{- 4}{- 8}$ .
2. So $k = 2$ . Option (C).
  1 mark
Answer: k = 2, option (C)
Where the marks are
- Uses equal-ratio condition to get k = 2 (1 mark)
Q3 · 1 mark · MCQQuadratic Equations
Which of the following equations has 2 as a root?
1. (A) $x^{2} - 4 x + 5 = 0$
2. (B) $x^{2} + 3 x - 12 = 0$
3. (C) $2 x^{2} - 7 x + 6 = 0$
4. (D) $3 x^{2} - 6 x - 2 = 0$
Try it first, then show our solutionSolution
1. Substitute $x = 2$ in (C): $2 (4) - 7 (2) + 6 = 8 - 14 + 6 = 0$ .
2. Others give $1, - 2, - 2$ , not zero. So 2 is a root of (C).
  1 mark
Answer: Option (C)
Where the marks are
- Verifies x = 2 satisfies $2 x^{2} - 7 x + 6 = 0$ (1 mark)
Q4 · 1 mark · MCQArithmetic Progressions
In an A.P., if $d = - 4$ and $a_{7} = 4$ , then the first term $a$ is equal to
1. (A)6
2. (B)7
3. (C)20
4. (D)28
Try it first, then show our solutionSolution
1. $a_{7} = a + 6 d \Rightarrow 4 = a + 6 (- 4)$ .
2. $a = 4 + 24 = 28$ . Option (D).
  1 mark
Answer: a = 28, option (D)
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ to get a = 28 (1 mark)
Q5 · 1 mark · MCQCoordinate Geometry
The distance of the point $(5, 4)$ from the origin is
1. (A) $41$
2. (B) $41$
3. (C)3
4. (D)9
Try it first, then show our solutionSolution
1. Distance from origin $= x^{2} + y^{2} = 25 + 16 = 41$ . Option (B).
  1 mark
Answer: $41$ , option (B)
Where the marks are
- Applies distance formula to get $41$ (1 mark)
Q6 · 1 mark · MCQIntroduction to Trigonometry
If $sin A = \frac{3}{5}$ , then value of $cot A$ is:
1. (A) $\frac{3}{4}$
2. (B) $\frac{4}{3}$
3. (C) $\frac{4}{5}$
4. (D) $\frac{5}{4}$
Try it first, then show our solutionSolution
1. $cos A = 1 - \frac{9}{25} = \frac{4}{5}$ .
2. $cot A = \frac{c o s A}{s i n A} = \frac{4/5}{3/5} = \frac{4}{3}$ . Option (B).
  1 mark
Answer: $\frac{4}{3}$ , option (B)
Where the marks are
- Finds cos A = 4/5 and cot A = 4/3 (1 mark)
Q7 · 1 mark · MCQIntroduction to Trigonometry
$\frac{1 + t a n ^{2} A}{1 + c o t ^{2} A}$ is equal to
1. (A) $sec^{2} A$
2. (B) $- 1$
3. (C) $cot^{2} A$
4. (D) $tan^{2} A$
Try it first, then show our solutionSolution
1. $\frac{1 + t a n ^{2} A}{1 + c o t ^{2} A} = \frac{s e c ^{2} A}{c s c ^{2} A} = \frac{s i n ^{2} A}{c o s ^{2} A} = tan^{2} A$ . Option (D).
  1 mark
Answer: $tan^{2} A$ , option (D)
Where the marks are
- Uses identities to simplify to $tan^{2} A$ (1 mark)
Q8 · 1 mark · MCQIntroduction to Trigonometry
$\frac{2 t a n 3 0 ^{\circ}}{1 - t a n ^{2} 3 0 ^{\circ}}$ is equal to
1. (A) $cos 6 0^{\circ}$
2. (B) $sin 6 0^{\circ}$
3. (C) $tan 6 0^{\circ}$
4. (D) $sin 3 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan 3 0^{\circ} = \frac{1}{3}$ : $\frac{2/ 3}{1 - 1/3} = \frac{2}{3} \times \frac{3}{2} = 3$ .
2. $3 = tan 6 0^{\circ}$ . Option (C).
  1 mark
Answer: $tan 6 0^{\circ}$ , option (C)
Where the marks are
- Evaluates to $3 = tan 6 0^{\circ}$ (1 mark)
Q9 · 1 mark · MCQPolynomials
A quadratic polynomial, the sum of whose zeroes is $- 5$ and their product is $6$ , is
1. (A) $x^{2} + 5 x + 6$
2. (B) $x^{2} - 5 x + 6$
3. (C) $x^{2} - 5 x - 6$
4. (D) $- x^{2} + 5 x + 6$
Try it first, then show our solutionSolution
1. Polynomial $= x^{2} - (sum) x + product = x^{2} - (- 5) x + 6 = x^{2} + 5 x + 6$ . Option (A).
  1 mark
Answer: $x^{2} + 5 x + 6$ , option (A)
Where the marks are
- Uses $x^{2} - S x + P$ form (1 mark)
Q10 · 1 mark · MCQPolynomials
The zeroes of the polynomial $3 x^{2} + 11 x - 4$ are:
1. (A) $\frac{1}{3}, 4$
2. (B) $\frac{- 1}{3}, - 4$
3. (C) $\frac{1}{3}, - 4$
4. (D) $\frac{- 1}{3}, 4$
Try it first, then show our solutionSolution
1. $3 x^{2} + 12 x - x - 4 = 3 x (x + 4) - 1 (x + 4) = (3 x - 1) (x + 4)$ .
2. Zeroes: $\frac{1}{3}, - 4$ . Option (C).
  1 mark
Answer: $\frac{1}{3}, - 4$ , option (C)
Where the marks are
- Factorises and finds zeroes 1/3 and −4 (1 mark)
Q11 · 1 mark · MCQStatistics
The annual rainfall record of a city for 66 days is given in the following table:
Rainfall (in cm) 0–10 10–20 20–30 30–40 40–50 50–60
Number of days 22 10 8 15 5 6
The difference of upper limits of modal and median classes is:
1. (A)10
2. (B)15
3. (C)20
4. (D)30
Try it first, then show our solutionSolution
1. Modal class (highest frequency 22): 0–10, upper limit 10.
2. $N /2 = 33$ ; cumulative frequencies 22, 32, 40… so median class is 20–30, upper limit 30.
3. Difference $= 30 - 10 = 20$ . Option (C).
  1 mark
Answer: 20, option (C)
Where the marks are
- Identifies modal class 0–10 and median class 20–30, difference 20 (1 mark)
Open Q11 on its own page
Q12 · 1 mark · MCQProbability
If $P (A)$ denotes the probability of an event A, then
1. (A) $P (A) < 0$
2. (B) $P (A) > 1$
3. (C) $0 \leq P (A) \leq 1$
4. (D) $- 1 \leq P (A) \leq 1$
Try it first, then show our solutionSolution
1. Probability of an impossible event is 0 and of a sure event is 1, so $0 \leq P (A) \leq 1$ . Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- States $0 \leq P (A) \leq 1$ (1 mark)
Q13 · 1 mark · MCQSurface Areas and Volumes
The total surface area of a solid hemisphere of radius 7 cm is:
1. (A) $98 π$ cm $^{2}$
2. (B) $147 π$ cm $^{2}$
3. (C) $196 π$ cm $^{2}$
4. (D) $228 \frac{2}{3} π$ cm $^{2}$
Try it first, then show our solutionSolution
1. TSA of hemisphere $= 3 π r^{2} = 3 π (49) = 147 π$ cm $^{2}$ . Option (B).
  1 mark
Answer: $147 π$ cm², option (B)
Where the marks are
- Uses $3 π r^{2}$ (1 mark)
Q14 · 1 mark · MCQAreas Related to Circles
The difference of the areas of a minor sector of angle $12 0^{\circ}$ and its corresponding major sector of a circle of radius 21 cm, is
1. (A)231 cm $^{2}$
2. (B)462 cm $^{2}$
3. (C)346.5 cm $^{2}$
4. (D)693 cm $^{2}$
Try it first, then show our solutionSolution
1. Circle area $= \frac{22}{7} \times 21 \times 21 = 1386$ cm $^{2}$ ; minor sector $= \frac{120}{360} \times 1386 = 462$ ; major $= 924$ .
2. Difference $= 924 - 462 = 462$ cm $^{2}$ . Option (B).
  1 mark
Answer: 462 cm², option (B)
Where the marks are
- Finds both sector areas and difference 462 cm² (1 mark)
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Q15 · 1 mark · MCQPair of Linear Equations in Two Variables
The graph of a pair of linear equations $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ in two variables $x$ and $y$ represents parallel lines, if
1. (A) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$
2. (B) $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
3. (C) $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$
4. (D) $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$
Try it first, then show our solutionSolution
1. Parallel lines have no solution, which occurs when $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ . Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- States inconsistency condition (1 mark)
Q16 · 1 mark · MCQCircles
In the given figure, tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of $8 0^{\circ}$ . $∠ A B O$ is equal to
1. (A) $4 0^{\circ}$
2. (B) $8 0^{\circ}$
3. (C) $10 0^{\circ}$
4. (D) $5 0^{\circ}$
Figure in the paper: A circle with centre O. External point P on the left with two tangents PA (upper) and PB (lower) touching the circle at A and B. The angle at P between the tangents is marked $8 0^{\circ}$ . Chord AB is drawn, joining A and B, and radius OB is drawn. The angle ABO (between chord BA and radius BO at B) is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. $P A = P B$ so $∠ P B A = \frac{18 0 ^{\circ} - 8 0 ^{\circ}}{2} = 5 0^{\circ}$ .
2. $∠ O B P = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ A B O = 9 0^{\circ} - 5 0^{\circ} = 4 0^{\circ}$ . Option (A).
  1 mark
Answer: $4 0^{\circ}$ , option (A)
Where the marks are
- Uses equal tangents and radius ⟂ tangent to get 40° (1 mark)
Open Q16 on its own page
Q17 · 1 mark · MCQCircles
A line intersecting a circle in two distinct points is called a
1. (A)secant
2. (B)chord
3. (C)diameter
4. (D)tangent
Try it first, then show our solutionSolution
1. A line (not a segment) meeting a circle in two distinct points is a secant. Option (A).
  1 mark
Answer: secant, option (A)
Where the marks are
- Identifies secant (1 mark)
Q18 · 1 mark · MCQSome Applications of Trigonometry
If a pole 6 m high casts a shadow $23$ m long on the ground, then the sun's elevation is
1. (A) $3 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan θ = \frac{6}{2 3} = 3 \Rightarrow θ = 6 0^{\circ}$ . Option (C).
  1 mark
Answer: $6 0^{\circ}$ , option (C)
Where the marks are
- Uses tan θ = height/shadow = √3 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedTriangles
Directions: In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option: (A) Both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are correct but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): A line drawn parallel to any one side of a triangle intersects the other two sides in the same ratio.
Reason (R): Parallel lines cannot be drawn to any side of a triangle.
1. (A)Both (A) and (R) are correct and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are correct but (R) is not the correct explanation of (A).
3. (C)(A) is true, but (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. Assertion is the Basic Proportionality Theorem — true.
2. A line parallel to any side can always be drawn, so Reason is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- A true (BPT), R false (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedCoordinate Geometry
Assertion (A): The point $(0, 4)$ lies on $y$ -axis.
Reason (R): The $x$ -coordinate of a point, lying on $y$ -axis, is zero.
1. (A)Both (A) and (R) are correct and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are correct but (R) is not the correct explanation of (A).
3. (C)(A) is true, but (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. Every point on the y-axis has x-coordinate 0 (R true).
2. $(0, 4)$ has x-coordinate 0, so it lies on y-axis; R explains A. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both true and R explains A (1 mark)

Rainfall (in cm)	0–10	10–20	20–30	30–40	40–50	50–60
Number of days	22	10	8	15	5	6

Section B

Very short answer

Q21 · 2 marks · Very short answerReal Numbers
Find the HCF of 84 and 144 by prime factorisation method.
Try it first, then show our solutionSolution
1. Write 84 as a product of primes: $84 = 2 \times 2 \times 3 \times 7 = 2^{2} \times 3 \times 7$
  ½ mark
2. Write 144 as a product of primes: $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^{4} \times 3^{2}$
  ½ mark
3. HCF is the product of the smallest powers of the common prime factors: $HCF = 2^{2} \times 3 = 12$
  1 mark
Answer: HCF(84, 144) = 12
Where the marks are
- Correct prime factorisation of 84 (½ mark)
- Correct prime factorisation of 144 (½ mark)
- HCF = $2^{2} \times 3 = 12$ (1 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
The sum of two natural numbers is 70 and their difference is 10. Find the natural numbers.
Try it first, then show our solutionSolution
1. Let the larger number be $x$ and the smaller be $y$ . Their sum is 70: $x + y = 70$ ...(1)
  ½ mark
2. Their difference is 10: $x - y = 10$ ...(2)
  ½ mark
3. Adding (1) and (2): $2 x = 80 \Rightarrow x = 40$
  ½ mark
4. Substituting in (1): $40 + y = 70 \Rightarrow y = 30$ . The numbers are 40 and 30.
  ½ mark
Answer: The numbers are 40 and 30
Where the marks are
- Forms $x + y = 70$ (½ mark)
- Forms $x - y = 10$ (½ mark)
- Gets $x = 40$ (½ mark)
- Gets $y = 30$ (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve for $x$ and $y$ :
$x - 3 y = 7$
$3 x - 3 y = 5$
Try it first, then show our solutionSolution
1. Subtract $x - 3 y = 7$ from $3 x - 3 y = 5$ : $2 x = - 2 \Rightarrow x = - 1$
  1 mark
2. Substitute in $x - 3 y = 7$ : $- 1 - 3 y = 7 \Rightarrow - 3 y = 8 \Rightarrow y = - \frac{8}{3}$
  1 mark
Answer: $x = - 1$ , $y = - \frac{8}{3}$
Where the marks are
- Eliminates $y$ and finds $x = - 1$ (1 mark)
- Finds $y = - \frac{8}{3}$ (1 mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerProbability
15 defective pens are accidentally mixed with 145 good ones. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Try it first, then show our solutionSolution
1. Total number of pens $= 145 + 15 = 160$ . These are the equally likely outcomes.
  1 mark
2. Number of good pens (favourable outcomes) $= 145$ . $P (good pen) = \frac{145}{160} = \frac{29}{32}$
  1 mark
Answer: $\frac{29}{32}$
Where the marks are
- Total outcomes = 160 (1 mark)
- $P (good) = \frac{145}{160} = \frac{29}{32}$ (1 mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerTriangles
In the given figure, $O A \cdot O B = O C \cdot O D$ . Prove that $△ A O D \sim △ C O B$ .
Figure in the paper: Two line segments AB and CD intersect at point O inside the figure. A and D are on one side (A upper-left, D lower-left), and C and B are on the other side (C upper-right, B lower-right). Segments AD and CB are drawn to form the two triangles AOD and COB. See it in the official paper.
Try it first, then show our solutionSolution
1. Given $O A \cdot O B = O C \cdot O D$ . Dividing both sides by $O C \cdot O B$ : $\frac{O A}{O C} = \frac{O D}{O B}$ ...(i)
  ½ mark
2. In $△ A O D$ and $△ C O B$ : $\frac{O A}{O C} = \frac{O D}{O B}$ [from (i)], and $∠ A O D = ∠ C O B$ [vertically opposite angles]
  1 mark
3. Therefore $△ A O D \sim △ C O B$ [SAS similarity criterion].
  ½ mark
Answer: $△ A O D \sim △ C O B$ (by SAS similarity)
Where the marks are
- Rewrites the given product as the ratio $\frac{O A}{O C} = \frac{O D}{O B}$ (½ mark)
- Uses the ratio and the vertically opposite angles $∠ A O D = ∠ C O B$ (1 mark)
- Concludes similarity by SAS (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerTriangles
In the given figure, $∠ D = ∠ E$ and $\frac{A D}{D B} = \frac{A E}{E C}$ . Prove that $△ A B C$ is isosceles.
Figure in the paper: Triangle ABC with apex A at the top and base BC horizontal at the bottom. Point D lies on side AB and point E lies on side AC. Segment DE is drawn. Angles at D (between DA and DE) and at E (between ED and EA) are marked as equal. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ , $\frac{A D}{D B} = \frac{A E}{E C}$ (given), so $D E ∥ B C$ [converse of Basic Proportionality Theorem].
  ½ mark
2. Hence $∠ A D E = ∠ A B C$ and $∠ A E D = ∠ A C B$ [corresponding angles, $D E ∥ B C$ ].
  ½ mark
3. But $∠ A D E = ∠ A E D$ (given), so $∠ B = ∠ C$ .
  ½ mark
4. Therefore $A B = A C$ [sides opposite equal angles of a triangle are equal], so $△ A B C$ is isosceles.
  ½ mark
Answer: $A B = A C$ , so $△ A B C$ is isosceles
Where the marks are
- Uses the converse of BPT to get $D E ∥ B C$ (½ mark)
- States the corresponding angles $∠ D = ∠ B$ and $∠ E = ∠ C$ (½ mark)
- Concludes $∠ B = ∠ C$ (½ mark)
- Gets $A B = A C$ , so the triangle is isosceles (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerCircles
Prove that the tangents drawn at the ends of a diameter of a circle are parallel to each other.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O and diameter AB. Tangent $l$ is drawn at A and tangent $m$ at B. Let $∠1$ be the angle between OA and $l$ , and $∠2$ be the angle between OB and $m$ , on opposite sides of AB. To prove: $l ∥ m$ .
  ½ mark
2. $∠1 = 9 0^{\circ}$ and $∠2 = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]. So $∠1 = ∠2$ .
  1 mark
3. $∠1$ and $∠2$ are alternate interior angles made by the transversal AB with lines $l$ and $m$ . Since they are equal, $l ∥ m$ .
  ½ mark
Answer: The tangents at the ends of a diameter are parallel (alternate angles, each 90°, are equal)
Where the marks are
- Correct figure with given and to prove (½ mark)
- Both angles are $9 0^{\circ}$ because the tangent is perpendicular to the radius (1 mark)
- Equal alternate interior angles, so the tangents are parallel (½ mark)
Open Q25 on its own page

Section C

Short answer

Q26 · 3 marks · Short answerProbability
Two dice are tossed simultaneously. Find the probability of getting
1. (a)
  an even number on both the dice.
  0 marks
2. (b)
  the sum of two numbers more than 9.
  0 marks
Try it first, then show our solutionSolution
1. When two dice are tossed, total outcomes $= 6 \times 6 = 36$ .
2. (a) Even on both: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6). Favourable outcomes $= 9$ .
  1 mark
3. (a) $P (even on both) = \frac{9}{36} = \frac{1}{4}$
  ½ mark
4. (b) Sum more than 9 (that is, 10, 11 or 12): (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). Favourable outcomes $= 6$ .
  1 mark
5. (b) $P (sum > 9) = \frac{6}{36} = \frac{1}{6}$
  ½ mark
Answer: (a) $\frac{1}{4}$ ; (b) $\frac{1}{6}$
Where the marks are
- Lists the 9 favourable outcomes for even on both dice (1 mark)
- Probability $\frac{9}{36} = \frac{1}{4}$ (½ mark)
- Lists the 6 outcomes with sum 10, 11 or 12 (1 mark)
- Probability $\frac{6}{36} = \frac{1}{6}$ (½ mark)
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Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerCircles
In two concentric circles, a chord of length 24 cm of larger circle touches the smaller circle, whose radius is 5 cm. Find the radius of the larger circle.
Try it first, then show our solutionSolution
1. Figure: concentric circles with centre O. Chord AB of the larger circle touches the smaller circle at P. Join OP and OA, with $O P = 5$ cm.
  ½ mark
2. $O P ⊥ A B$ [tangent ⟂ radius], and the perpendicular from the centre bisects the chord, so $A P = \frac{1}{2} (24) = 12$ cm.
  1 mark
3. In right $△ O P A$ : $O A^{2} = O P^{2} + A P^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169$ [Pythagoras theorem]
  1 mark
4. $O A = 13$ cm, so the radius of the larger circle is 13 cm.
  ½ mark
Answer: 13 cm
Where the marks are
- Correct figure (½ mark)
- $O P ⊥ A B$ and $A P = 12$ cm (1 mark)
- Applies Pythagoras: $O A^{2} = 5^{2} + 1 2^{2}$ (1 mark)
- Radius = 13 cm (½ mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerCircles
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O. PQ and PR are tangents from an external point P touching the circle at Q and R. Join OQ and OR. To prove: $∠ QP R + ∠ QO R = 18 0^{\circ}$ .
  ½ mark
2. $∠ O QP = 9 0^{\circ}$ and $∠ O R P = 9 0^{\circ}$ [radius ⟂ tangent at the point of contact].
  1 mark
3. In quadrilateral OQPR, the angle sum is $36 0^{\circ}$ : $∠ O QP + ∠ QP R + ∠ P R O + ∠ R O Q = 36 0^{\circ}$ , so $9 0^{\circ} + ∠ QP R + 9 0^{\circ} + ∠ QO R = 36 0^{\circ}$ .
  1 mark
4. Hence $∠ QP R + ∠ QO R = 18 0^{\circ}$ . The angle between the tangents is supplementary to the angle subtended by QR at the centre.
  ½ mark
Answer: $∠ QP R + ∠ QO R = 18 0^{\circ}$ (proved)
Where the marks are
- Figure with given and to prove (½ mark)
- Both angles between radius and tangent are $9 0^{\circ}$ (1 mark)
- Uses the quadrilateral angle sum of $36 0^{\circ}$ (1 mark)
- Concludes $∠ QP R + ∠ QO R = 18 0^{\circ}$ (½ mark)
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Q28 · 3 marks · Short answerReal Numbers
Prove that $7 - 35$ is an irrational number, given that $5$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $7 - 35$ is rational. Then $7 - 35 = x$ for some rational number $x$ .
  1 mark
2. Then $35 = 7 - x$ , so $5 = \frac{7 - x}{3}$ .
  1 mark
3. Since $x$ is rational, $\frac{7 - x}{3}$ is rational, so $5$ would be rational. This contradicts the given fact that $5$ is irrational. So our assumption is wrong, and $7 - 35$ is irrational.
  1 mark
Answer: $7 - 35$ is irrational (proved by contradiction)
Where the marks are
- Assumes $7 - 35$ is rational (1 mark)
- Expresses $5 = \frac{7 - x}{3}$ (1 mark)
- States the contradiction and concludes the number is irrational (1 mark)
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Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerPolynomials
Zeroes of the quadratic polynomial $x^{2} - 3 x + 2$ are $α$ and $β$ . Construct a quadratic polynomial whose zeroes are $2 α + 1$ and $2 β + 1$ .
Try it first, then show our solutionSolution
1. For $x^{2} - 3 x + 2$ : $α + β = - \frac{b}{a} = 3$
  ½ mark
2. $α β = \frac{c}{a} = 2$
  ½ mark
3. Sum of the new zeroes: $(2 α + 1) + (2 β + 1) = 2 (α + β) + 2 = 2 (3) + 2 = 8$
  ½ mark
4. Product of the new zeroes: $(2 α + 1) (2 β + 1) = 4 α β + 2 (α + β) + 1 = 4 (2) + 2 (3) + 1 = 15$
  1 mark
5. Required polynomial is $x^{2} - (sum) x + product = x^{2} - 8 x + 15$ , or $k (x^{2} - 8 x + 15)$ for any non-zero constant $k$ .
  ½ mark
Answer: $x^{2} - 8 x + 15$ (or $k (x^{2} - 8 x + 15)$ , $k \neq = 0$ )
Where the marks are
- $α + β = 3$ (½ mark)
- $α β = 2$ (½ mark)
- New sum = 8 (½ mark)
- New product = 15 (1 mark)
- Writes the polynomial $x^{2} - 8 x + 15$ (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerPolynomials
Find the zeroes of the polynomial $4 x^{2} - 4 x + 1$ and verify the relationship between the zeroes and the coefficients.
Try it first, then show our solutionSolution
1. $4 x^{2} - 4 x + 1 = 4 x^{2} - 2 x - 2 x + 1 = 2 x (2 x - 1) - 1 (2 x - 1) = (2 x - 1) (2 x - 1)$
  1 mark
2. Setting $(2 x - 1)^{2} = 0$ gives the zeroes $\frac{1}{2}$ and $\frac{1}{2}$ .
  1 mark
3. Sum of zeroes $= \frac{1}{2} + \frac{1}{2} = 1$ , and $- \frac{coefficient of x}{coefficient of x ^{2}} = - \frac{( - 4 )}{4} = 1$ . Verified.
  ½ mark
4. Product of zeroes $= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ , and $\frac{constant term}{coefficient of x ^{2}} = \frac{1}{4}$ . Verified.
  ½ mark
Answer: Zeroes: $\frac{1}{2}, \frac{1}{2}$ ; sum = 1 = $- b / a$ , product = $\frac{1}{4}$ = $c / a$ (verified)
Where the marks are
- Factorises as $(2 x - 1) (2 x - 1)$ (1 mark)
- Zeroes $\frac{1}{2}, \frac{1}{2}$ (1 mark)
- Verifies the sum of zeroes (½ mark)
- Verifies the product of zeroes (½ mark)
Open Q29(b) on its own page
Q30 · 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 $cot θ = \frac{1}{tan θ}$ : LHS $= \frac{tan θ}{1 - \frac{1}{t a n θ}} + \frac{\frac{1}{t a n θ}}{1 - tan θ} = \frac{tan ^{2} θ}{tan θ - 1} + \frac{1}{tan θ ( 1 - tan θ )}$
  ½ mark
2. $= \frac{tan ^{2} θ}{tan θ - 1} - \frac{1}{tan θ ( tan θ - 1 )} = \frac{tan ^{3} θ - 1}{tan θ ( tan θ - 1 )}$
  ½ mark
3. Using $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ : $= \frac{( tan θ - 1 ) ( tan ^{2} θ + tan θ + 1 )}{tan θ ( tan θ - 1 )}$
  ½ mark
4. Cancel $(tan θ - 1)$ and use $1 + tan^{2} θ = sec^{2} θ$ : $= \frac{sec ^{2} θ + tan θ}{tan θ}$
  ½ mark
5. $= \frac{sec ^{2} θ}{tan θ} + 1$
  ½ mark
6. $\frac{sec ^{2} θ}{tan θ} = \frac{1}{cos ^{2} θ} \cdot \frac{cos θ}{sin θ} = \frac{1}{sin θ cos θ} = sec θ cosec θ$ . So LHS $= 1 + sec θ cosec θ =$ RHS.
  ½ mark
Answer: LHS = $1 + sec θ cosec θ$ = RHS (proved)
Where the marks are
- Converts $cot θ$ to $\frac{1}{t a n θ}$ and simplifies each term (½ mark)
- Combines into $\frac{t a n ^{3} θ - 1}{t a n θ ( t a n θ - 1 )}$ (½ mark)
- Factorises using the difference of cubes (½ mark)
- Cancels and uses $1 + tan^{2} θ = sec^{2} θ$ (½ mark)
- Splits as $\frac{s e c ^{2} θ}{t a n θ} + 1$ (½ mark)
- Shows this equals $sec θ cosec θ + 1$ (½ mark)
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Q31 · 3 marks · Short answerAreas Related to Circles
A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of $12 0^{\circ}$ . Find the area cleaned at each sweep of the blades.
Try it first, then show our solutionSolution
1. Each wiper sweeps a sector with $r = 21$ cm and $θ = 12 0^{\circ}$ .
  ½ mark
2. Area swept by one wiper $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21$
  1½ marks
3. $= \frac{1}{3} \times 22 \times 3 \times 21 = 462$ cm²
  ½ mark
4. The wipers do not overlap, so the total area cleaned at each sweep $= 2 \times 462 = 924$ cm².
  ½ mark
Answer: 924 cm² (462 cm² by each wiper)
Where the marks are
- Identifies $r = 21$ cm and $θ = 12 0^{\circ}$ (½ mark)
- Uses the sector area formula with correct substitution (1½ marks)
- Area for one wiper = 462 cm² (½ mark)
- Total area for both wipers = 924 cm² (½ mark)
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Section D

Long answer

Q32 · 5 marks · Long answerQuadratic Equations
A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. Find the total number of toys produced on that day.
Try it first, then show our solutionSolution
1. Let the number of toys produced that day be $x$ . Then the cost of each toy is ₹ $(55 - x)$ .
  ½ mark
2. Total cost $=$ number of toys $\times$ cost per toy: $x (55 - x) = 750$
  1 mark
3. $55 x - x^{2} = 750 \Rightarrow x^{2} - 55 x + 750 = 0$
  1 mark
4. Split the middle term, since $25 \times 30 = 750$ and $25 + 30 = 55$ : $x^{2} - 25 x - 30 x + 750 = 0 \Rightarrow x (x - 25) - 30 (x - 25) = 0 \Rightarrow (x - 25) (x - 30) = 0$
  1½ marks
5. $x = 25$ or $x = 30$ . Both values give a total cost of ₹750 (25 × ₹30 or 30 × ₹25). So 25 or 30 toys were produced that day.
  1 mark
Answer: 25 toys or 30 toys
Where the marks are
- Takes the number of toys as $x$ and the cost per toy as $55 - x$ (½ mark)
- Forms $x (55 - x) = 750$ (1 mark)
- Gets $x^{2} - 55 x + 750 = 0$ (1 mark)
- Factorises as $(x - 25) (x - 30) = 0$ (1½ marks)
- Gets $x = 25$ or $30$ (1 mark)
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Q33 · 5 marks · Long answerSome Applications of Trigonometry
A TV tower stands vertically on a bank of a canal. From a point on the other bank exactly opposite the tower, the angle of elevation of the top of the tower is $6 0^{\circ}$ . From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $3 0^{\circ}$ . Find the height of the tower and the width of the canal. [Use $3 = 1.732$ ]
Try it first, then show our solutionSolution
1. Figure: AB is the tower of height $h$ m, with foot A and top B. C is the point on the opposite bank, with width of canal $C A = x$ m and $∠ B C A = 6 0^{\circ}$ . D is 20 m beyond C on line AC, so $D A = x + 20$ and $∠ B D A = 3 0^{\circ}$ . Angle A $= 9 0^{\circ}$ .
  1 mark
2. In right $△ B A C$ : $tan 6 0^{\circ} = \frac{A B}{C A} = \frac{h}{x}$
  1 mark
3. $3 = \frac{h}{x} \Rightarrow h = 3 x$ ...(i)
  ½ mark
4. In right $△ B A D$ : $tan 3 0^{\circ} = \frac{h}{x + 20} = \frac{1}{3}$ , so $3 h = x + 20$ ...(ii)
  1 mark
5. Substituting (i) in (ii): $3 (3 x) = x + 20 \Rightarrow 3 x = x + 20 \Rightarrow x = 10$
  1 mark
6. $h = 3 \times 10 = 1.732 \times 10 = 17.32$ m. Height of the tower = 17.32 m and width of the canal = 10 m.
  ½ mark
Answer: Height of tower = 17.32 m; width of canal = 10 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ} = \frac{h}{x}$ (1 mark)
- Gets $h = 3 x$ (½ mark)
- Uses $tan 3 0^{\circ} = \frac{h}{x + 20}$ (1 mark)
- Solves to get $x = 10$ (1 mark)
- Height = 17.32 m, width = 10 m (½ mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerTriangles
In the given figure, altitudes CE and AD of $△$ ABC intersect each other at the point P. Show that:
1. (i)
  $△$ AEP $\sim$ $△$ CDP
  1 mark
2. (ii)
  $△$ ABD $\sim$ $△$ CBE
  2 marks
3. (iii)
  $△$ AEP $\sim$ $△$ ADB
  2 marks
Figure in the paper: Triangle ABC with A at bottom-left, B at bottom-right, C at top. Altitude CE drops from C to base AB at foot E (right angle marked at E). Altitude AD drops from A to side BC at foot D (right angle marked at D). The two altitudes intersect at point P inside the triangle. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In $△ A E P$ and $△ C D P$ : $∠ A P E = ∠ C P D$ [vertically opposite angles]
  ½ mark
2. (i) $∠ A E P = ∠ C D P = 9 0^{\circ}$ [CE and AD are altitudes]. So $△ A E P \sim △ C D P$ [AA similarity].
  ½ mark
3. (ii) In $△ A B D$ and $△ C B E$ : $∠ A B D = ∠ C B E$ [common angle B]
  ½ mark
4. (ii) $∠ A D B = ∠ C E B = 9 0^{\circ}$ [AD ⟂ BC, CE ⟂ AB]
  1 mark
5. (ii) So $△ A B D \sim △ C B E$ [AA similarity].
  ½ mark
6. (iii) In $△ A E P$ and $△ A D B$ : $∠ P A E = ∠ B A D$ [common angle A, since P lies on AD and E lies on AB]
  ½ mark
7. (iii) $∠ A E P = ∠ A D B = 9 0^{\circ}$ [altitudes]
  1 mark
8. (iii) So $△ A E P \sim △ A D B$ [AA similarity].
  ½ mark
Answer: (i) $△ A E P \sim △ C D P$ ; (ii) $△ A B D \sim △ C B E$ ; (iii) $△ A E P \sim △ A D B$ , all by AA similarity
Where the marks are
- (i) Vertically opposite angles at P (½ mark)
- (i) Right angles at E and D, so similar by AA (½ mark)
- (ii) Common angle B (½ mark)
- (ii) $∠ A D B = ∠ C E B = 9 0^{\circ}$ (1 mark)
- (ii) Concludes similarity by AA (½ mark)
- (iii) Common angle A (½ mark)
- (iii) $∠ A E P = ∠ A D B = 9 0^{\circ}$ (1 mark)
- (iii) Concludes similarity by AA (½ mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerTriangles
AD and PM are medians of triangles ABC and PQR, respectively, where $△$ ABC $\sim$ $△$ PQR. Prove that $\frac{A B}{P Q} = \frac{A D}{P M}$ .
Try it first, then show our solutionSolution
1. Figure: $△ A B C$ with median AD (D the mid-point of BC) and $△ P QR$ with median PM (M the mid-point of QR). Given: $△ A B C \sim △ P QR$ . To prove: $\frac{A B}{P Q} = \frac{A D}{P M}$ .
  1 mark
2. Since $△ A B C \sim △ P QR$ , corresponding sides are proportional: $\frac{A B}{P Q} = \frac{B C}{QR}$
  1 mark
3. $B C = 2 B D$ and $QR = 2 QM$ [D and M are mid-points], so $\frac{A B}{P Q} = \frac{2 B D}{2 QM} = \frac{B D}{QM}$
  1 mark
4. In $△ A B D$ and $△ P QM$ : $\frac{A B}{P Q} = \frac{B D}{QM}$ [proved above]
  ½ mark
5. $∠ B = ∠ Q$ [corresponding angles of similar triangles ABC and PQR]
  ½ mark
6. Therefore $△ A B D \sim △ P QM$ [SAS similarity].
  ½ mark
7. Hence corresponding sides are proportional: $\frac{A B}{P Q} = \frac{A D}{P M}$ .
  ½ mark
Answer: $△ A B D \sim △ P QM$ (SAS), hence $\frac{A B}{P Q} = \frac{A D}{P M}$ (proved)
Where the marks are
- Figure with given and to prove (1 mark)
- $\frac{A B}{P Q} = \frac{B C}{QR}$ from the similar triangles (1 mark)
- Uses the mid-points to get $\frac{A B}{P Q} = \frac{B D}{QM}$ (1 mark)
- Ratio stated for triangles ABD and PQM (½ mark)
- $∠ B = ∠ Q$ (½ mark)
- $△ A B D \sim △ P QM$ by SAS (½ mark)
- Concludes $\frac{A B}{P Q} = \frac{A D}{P M}$ (½ mark)
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Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerSurface Areas and Volumes
A textile industry runs in a shed. This shed is in the shape of a cuboid surmounted by a half cylinder. If the base of the industry is of dimensions 14 m × 20 m and the height of the cuboidal portion is 7 m, find the volume of air that the industry can hold. Further, suppose the machinery in the industry occupies a total space of 400 m $^{3}$ . Then, how much space is left in the industry?
Figure in the paper: A 3D shape: a cuboid with a semicircular (half-cylinder) roof on top. Front face width is 14 m (the semicircle spans this width, radius 7 m) and height of the cuboidal part is 7 m. The length along the shed is 20 m, marked along the top edge. See it in the official paper.
Try it first, then show our solutionSolution
1. Cuboidal part: length $= 20$ m, breadth $= 14$ m, height $= 7$ m.
  Volume of cuboid $= l \times b \times h = 20 \times 14 \times 7 = 1960 m^{3}$
  1 mark
2. Half-cylinder roof: its diameter is the width $14$ m, so $r = \frac{14}{2} = 7$ m. Its length (height of the cylinder) is $h = 20$ m.
  1 mark
3. Volume of half cylinder $= \frac{1}{2} π r^{2} h = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 \times 20 = 1540 m^{3}$
  2 marks
4. Volume of air the shed can hold $= 1960 + 1540 = 3500 m^{3}$
  ½ mark
5. The machinery occupies $400 m^{3}$ .
  Space left $= 3500 - 400 = 3100 m^{3}$
  ½ mark
Answer: Volume of air = 3500 m³; space left = 3100 m³
Where the marks are
- Volume of cuboid $= 1960 m^{3}$ (1 mark)
- Identifies $r = 7$ m and $h = 20$ m for the half cylinder (1 mark)
- Volume of half cylinder $= \frac{1}{2} π r^{2} h = 1540 m^{3}$ (2 marks)
- Total volume of air $= 3500 m^{3}$ (½ mark)
- Space left $= 3100 m^{3}$ (½ mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerSurface Areas and Volumes
From a solid cylinder of height 8 cm and radius 6 cm, a conical cavity of the same height and same radius is carved out. Find the total surface area of the remaining solid. (Take $π = 3.14$ )
Try it first, then show our solutionSolution
1. For the cone: $r = 6$ cm, $h = 8$ cm.
  Slant height $l = r^{2} + h^{2} = 6^{2} + 8^{2} = 100 = 10$ cm
  1 mark
2. The remaining solid is bounded by the curved surface of the cylinder, its uncarved circular base and the curved surface of the conical cavity.
  TSA $= 2 π r h + π r^{2} + π r l = 2 π (6) (8) + π (6)^{2} + π (6) (10)$
  2½ marks
3. $= 96 π + 36 π + 60 π = 192 π$
  1 mark
4. $= 192 \times 3.14 = 602.88 cm^{2}$
  ½ mark
Answer: Total surface area of the remaining solid = 192π = 602.88 cm²
Where the marks are
- Finds slant height $l = 10$ cm (1 mark)
- Writes TSA $= 2 π r h + π r^{2} + π r l$ with correct substitution (2½ marks)
- Simplifies to $192 π$ (1 mark)
- Final value $602.88 cm^{2}$ (½ mark)
Open Q35(b) on its own page

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Saving money is a good habit and it should be inculcated in children right from the beginning. Rehan's mother brought a piggy bank for Rehan and puts one ₹ 5 coin of her savings in the piggy bank on the first day. She increases his savings by one ₹ 5 coin daily.
Based on the above information, answer the following questions:
1. (i)
  How many coins were added to the piggy bank on 8th day?
  1 mark
2. (ii)
  How much money will be there in the piggy bank after 8 days?
  1 mark
3. (iii) (a)
  If the piggy bank can hold one hundred twenty ₹ 5 coins in all, find the number of days she can contribute to put ₹ 5 coins into it.
  2 marks
4. OR
  (iii) (b)
  Find the total money saved, when the piggy bank is full.
  2 marks
Figure in the paper: Photograph of a pink piggy bank on the left, with a hand dropping a coin into it; to the right, stacks of coins of decreasing height, representing the savings increasing day by day (the stacks shown get progressively shorter from left to right). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) She puts 1 coin on day 1 and one more coin each day, so coins added on successive days are $1, 2, 3, \dots$ (AP with $a = 1, d = 1$ ).
  Coins on 8th day $= a_{8} = 1 + 7 (1) = 8$ coins
  1 mark
2. (ii) Money added day-wise: ₹5, ₹10, ₹15, ₹20, … (AP with $a = 5, d = 5$ ).
  $S_{8} = \frac{8}{2} [2 (5) + 7 (5)] = 4 \times 45 = ₹180$
  1 mark
3. (iii)(a) Total coins in $n$ days $= \frac{n}{2} [2 (1) + (n - 1) (1)]$ . The bank holds 120 coins, so
  $\frac{n}{2} [2 + (n - 1)] = 120$
  1 mark
4. $\Rightarrow n (n + 1) = 240 \Rightarrow n^{2} + n - 240 = 0$
  ½ mark
5. $\Rightarrow (n + 16) (n - 15) = 0 \Rightarrow n = 15$ (since $n \neq = - 16$ , days cannot be negative).
  She can contribute for 15 days.
  ½ mark
6. (iii)(b) When full, the bank holds 120 coins of ₹5 each.
  Total money saved $= 120 \times 5$
  1 mark
7. $= ₹600$
  1 mark
Answer: (i) 8 coins; (ii) ₹180; (iii)(a) 15 days; (iii)(b) ₹600
Where the marks are
- (i) 8 coins on the 8th day (1 mark)
- (ii) Money after 8 days = ₹180 (1 mark)
- (iii)(a) Sets up $\frac{n}{2} [2 + (n - 1)] = 120$ (1 mark)
- (iii)(a) Obtains $n^{2} + n - 240 = 0$ (½ mark)
- (iii)(a) Solves to get $n = 15$ , rejecting $- 16$ (½ mark)
- (iii)(b) Total money $= 120 \times 5 = ₹600$ (1 + 1 marks)
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Q37 · 4 marks · Case-basedCompetency-basedStatistics
Heart Rate: The heart rate is one of the 'vital signs' of health in the human body. It measures the number of times per minute that the heart contracts or beats. While a normal heart rate does not guarantee that a person is free of health problems, it is a useful benchmark for identifying a range of health issues.
Thirty women were examined by doctors of AIIMS and the number of heart beats per minute were recorded and summarized as follows:
Number of heart beats per minute Number of Women
65–68 2
68–71 4
71–74 3
74–77 8
77–80 7
80–83 4
83–86 2
Based on the above information, answer the following questions:
1. (i)
  How many women are having heart beat in the range 68 – 77?
  1 mark
2. (ii)
  What is the median class of heart beats per minute for these women?
  1 mark
3. (iii) (a)
  Find the modal value of heart beats per minute for these women.
  2 marks
4. OR
  (iii) (b)
  Find the median value of heart beats per minute for these women.
  2 marks
Figure in the paper: Two ECG-style heart rhythm traces drawn on a grid: each shows repeated sharp spikes (QRS complexes) with smaller bumps before and after, one trace above the other. The graph is illustrative only. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The range 68–77 covers the classes 68–71, 71–74 and 74–77.
  Number of women $= 4 + 3 + 8 = 15$
  1 mark
2. (ii) $N = 30$ , so $\frac{N}{2} = 15$ . Cumulative frequencies: 2, 6, 9, 17, 24, 28, 30. The first cf greater than 15 is 17, so the median class is 74–77.
  1 mark
3. (iii)(a) Highest frequency is 8, so modal class is 74–77: $l = 74$ , $f_{1} = 8$ , $f_{0} = 3$ , $f_{2} = 7$ , $h = 3$ .
  Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 74 + (\frac{8 - 3}{16 - 3 - 7}) \times 3$
  1½ marks
4. $= 74 + \frac{5}{6} \times 3 = 74 + 2.5 = 76.5$ beats per minute
  ½ mark
5. (iii)(b) Cumulative frequency table:
  Class f cf
  65–68 2 2
  68–71 4 6
  71–74 3 9
  74–77 8 17
  77–80 7 24
  80–83 4 28
  83–86 2 30
  $\frac{N}{2} = 15$ , median class 74–77: $l = 74$ , $c f = 9$ , $f = 8$ , $h = 3$ .
  ½ mark
6. Median $= l + (\frac{\frac{N}{2} - c f}{f}) \times h = 74 + \frac{15 - 9}{8} \times 3$
  1 mark
7. $= 74 + 2.25 = 76.25$ beats per minute
  ½ mark
Answer: (i) 15 women; (ii) 74–77; (iii)(a) Mode = 76.5; (iii)(b) Median = 76.25
Where the marks are
- (i) $4 + 3 + 8 = 15$ women (1 mark)
- (ii) Median class 74–77 (1 mark)
- (iii)(a) Correct values substituted in the mode formula (1½ marks)
- (iii)(a) Mode $= 76.5$ (½ mark)
- (iii)(b) Correct cumulative frequency table (½ mark)
- (iii)(b) Correct substitution in the median formula (1 mark)
- (iii)(b) Median $= 76.25$ (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
The top of a table is hexagonal in shape. The hexagonal top is shown as a 3D sketch, and its plan is drawn on a coordinate grid (x-axis 0 to 13, y-axis 0 to 13). The points are: A(1, 9), B(5, 13), C(9, 13), D(13, 9), E(13, 5), F(9, 1), G(5, 1), H(1, 5) forming the outer octagon; inner points M(5, 11), N(9, 11), O(11, 9), P(11, 5), Q(9, 3), R(5, 3), S(3, 5), T(3, 9) forming an inner octagon; the dashed lines from M, N, O, P, Q, R, S, T all pass through the centre (7, 7).
On the basis of the information given above, answer the following questions:
1. (i)
  Write the coordinates of A and B.
  1 mark
2. (ii)
  Write the coordinates of the mid-point of line segment joining C and D.
  1 mark
3. (iii) (a)
  Find the distance between M and Q.
  2 marks
4. OR
  (iii) (b)
  Find the coordinates of the point which divides the line segment joining M and N in the ratio 1:3 internally.
  2 marks
Figure in the paper: Coordinate grid with x-axis (0 to 13) and y-axis (0 to 13). An outer hexagon-like octagon with vertices A(1,9), B(5,13), C(9,13), D(13,9), E(13,5), F(9,1), G(5,1), H(1,5), and an inner octagon with vertices M(5,11), N(9,11), O(11,9), P(11,5), Q(9,3), R(5,3), S(3,5), T(3,9). Dashed lines join M, N, O, P, Q, R, S, T to the centre at (7,7). A small 3D sketch of a hexagonal table top is shown above the grid. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Reading from the grid: $A (1, 9)$ and $B (5, 13)$ .
  1 mark
2. (ii) $C (9, 13)$ and $D (13, 9)$ .
  Mid-point $= (\frac{9 + 13}{2}, \frac{13 + 9}{2}) = (11, 11)$
  1 mark
3. (iii)(a) From the grid: $M (5, 11)$ and $Q (9, 3)$ .
  1 mark
4. $M Q = (9 - 5)^{2} + (3 - 11)^{2} = 16 + 64 = 80 = 45$ units
  1 mark
5. (iii)(b) From the grid: $M (5, 11)$ and $N (9, 11)$ ; ratio $m_{1} : m_{2} = 1 : 3$ .
  1 mark
6. By the section formula, the point is $(\frac{1 \times 9 + 3 \times 5}{1 + 3}, \frac{1 \times 11 + 3 \times 11}{1 + 3}) = (\frac{24}{4}, \frac{44}{4}) = (6, 11)$
  1 mark
Answer: (i) A(1, 9), B(5, 13); (ii) (11, 11); (iii)(a) MQ = √80 = 4√5 units; (iii)(b) (6, 11)
Where the marks are
- (i) A(1, 9) and B(5, 13) (½ + ½ marks)
- (ii) Reads C(9, 13), D(13, 9) (½ mark) and gets mid-point (11, 11) (½ mark)
- (iii)(a) Coordinates M(5, 11), Q(9, 3) (1 mark)
- (iii)(a) $M Q = 80 = 45$ (1 mark)
- (iii)(b) Coordinates M(5, 11), N(9, 11) (1 mark)
- (iii)(b) Section formula gives (6, 11) (1 mark)
Open Q38 on its own page

Number of heart beats per minute	Number of Women
65–68	2
68–71	4
71–74	3
74–77	8
77–80	7
80–83	4
83–86	2

Class	f	cf
65–68	2	2
68–71	4	6
71–74	3	9
74–77	8	17
77–80	7	24
80–83	4	28
83–86	2	30

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