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

CBSE Class 10 Mathematics Basic: 2024 board paper, set 430/4/1

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 430-4-1MATHEMATICS (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 (VSA)
C	26-31	Short Answer (SA)
D	32-35	Long Answer (LA)
E	36-38	Case study based

Instructions in brief: No overall choice. Internal choice in 2 questions of Section B, 2 in Section C, 2 in Section D and 3 in Section E (2-mark sub-parts of each case study); use of calculator not allowed; take π = 22/7 where required, if not stated.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQCoordinate Geometry
The distance between the points $(2, - 1)$ and $(- 1, - 5)$ is :
1. (A)15 units
2. (B)5 units
3. (C)25 units
4. (D)41 units
Try it first, then show our solutionSolution
1. Distance $= (- 1 - 2)^{2} + (- 5 + 1)^{2} = 9 + 16 = 25 = 5$ units, so option (B).
  1 mark
Answer: 5 units, option (B)
Where the marks are
- Applies the distance formula to get 5 units (1 mark)
Q2 · 1 mark · MCQCoordinate Geometry
If $C (1, - 1)$ is the mid-point of the line segment AB joining points $A (4, x)$ and $B (- 2, 4)$ , then value of $x$ is :
1. (A)5
2. (B)-5
3. (C)6
4. (D)-6
Try it first, then show our solutionSolution
1. The y-coordinate of the mid-point gives $\frac{x + 4}{2} = - 1$ , so $x + 4 = - 2$ and $x = - 6$ . This is option (D).
  1 mark
Answer: x = −6, option (D)
Where the marks are
- Equates the mid-point y-coordinate to −1 and gets x = −6 (1 mark)
Q3 · 1 mark · MCQProbability
Which of the following relationship is correct ?
1. (A) $P (E) = 1 + P (\overline{E})$
2. (B) $P (\overline{E}) - P (E) = 1$
3. (C) $P (E) + P (\overline{E}) = 1$
4. (D) $P (E) = 2 P (\overline{E})$
Try it first, then show our solutionSolution
1. An event and its complement together cover all outcomes, so $P (E) + P (\overline{E}) = 1$ . This is option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Identifies $P (E) + P (\overline{E}) = 1$ (1 mark)
Q4 · 1 mark · MCQStatistics
The following distribution gives the daily income of 50 workers of a factory :
Income (in ₹) 400 – 424 425 – 449 450 – 474 475 – 499 500 – 524
Number of workers 12 14 8 6 10
The lower limit of the modal class is :
1. (A)425
2. (B)449
3. (C)424·5
4. (D)425·5
Try it first, then show our solutionSolution
1. The highest frequency is 14, so the modal class is 425–449.
2. The classes are not continuous, so make them continuous by subtracting 0.5. The class becomes 424.5–449.5, and its lower limit is 424.5. This is option (C).
  1 mark
Answer: 424·5, option (C)
Where the marks are
- Finds the modal class 425–449 and its true lower limit 424.5 (1 mark)
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Q5 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A lamp post 9 m high casts a shadow $33$ m long on the ground. The Sun's elevation at this moment is :
1. (A) $6 0^{\circ}$
2. (B) $9 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $3 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan θ = \frac{9}{3 3} = 3$ , so $θ = 6 0^{\circ}$ . This is option (A).
  1 mark
Answer: 60°, option (A)
Where the marks are
- Uses $tan θ$ = height/shadow $= 3$ to get 60° (1 mark)
Q6 · 1 mark · MCQPolynomials
If one zero of a quadratic polynomial $k x^{2} + 4 x + k$ is 1, then the value of $k$ is :
1. (A)2
2. (B)-2
3. (C)4
4. (D)-4
Try it first, then show our solutionSolution
1. Put $x = 1$ : $k + 4 + k = 0$ , so $2 k = - 4$ and $k = - 2$ . This is option (B).
  1 mark
Answer: k = −2, option (B)
Where the marks are
- Substitutes x = 1 and solves to get k = −2 (1 mark)
Q7 · 1 mark · MCQPolynomials
The number of quadratic polynomials having zeroes $- 1$ and $3$ is :
1. (A)1
2. (B)2
3. (C)3
4. (D)more than 3
Try it first, then show our solutionSolution
1. Every polynomial $k (x^{2} - 2 x - 3)$ with $k \neq = 0$ has zeroes −1 and 3. There are infinitely many such polynomials, so the answer is more than 3, option (D).
  1 mark
Answer: More than 3, option (D)
Where the marks are
- Recognises that $k (x^{2} - 2 x - 3)$ gives infinitely many polynomials (1 mark)
Q8 · 1 mark · MCQQuadratic Equations
The roots of the quadratic equation $x^{2} - 4 = 0$ is/are :
1. (A)2 only
2. (B)-2, 2
3. (C)4 only
4. (D)-4, 4
Try it first, then show our solutionSolution
1. $x^{2} = 4$ , so $x = \pm 2$ . This is option (B).
  1 mark
Answer: −2, 2, option (B)
Where the marks are
- Solves $x^{2} = 4$ to get both roots ±2 (1 mark)
Q9 · 1 mark · MCQQuadratic Equations
Which of the following is not a quadratic equation ?
1. (A) $(x - 2)^{2} + 1 = 2 x - 3$
2. (B) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
3. (C) $x (x + 1) + 8 = (x + 2) (x - 2)$
4. (D) $2 x + \frac{3}{x} = 5$
Try it first, then show our solutionSolution
1. In (C), $x^{2} + x + 8 = x^{2} - 4$ , which reduces to $x + 12 = 0$ . This is linear, not quadratic.
  1 mark
2. The other options reduce to equations of degree 2, for example (D) becomes $2 x^{2} - 5 x + 3 = 0$ . So the answer is option (C).
Answer: Option (C)
Where the marks are
- Simplifies (C) and shows the $x^{2}$ terms cancel (1 mark)
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Q10 · 1 mark · MCQArithmetic Progressions
The common difference of an A.P., if $a_{23} - a_{19} = 32$ , is :
1. (A)8
2. (B)-8
3. (C)-4
4. (D)4
Try it first, then show our solutionSolution
1. $a_{23} - a_{19} = (a + 22 d) - (a + 18 d) = 4 d = 32$ , so $d = 8$ . This is option (A).
  1 mark
Answer: d = 8, option (A)
Where the marks are
- Gets 4d = 32 and hence d = 8 (1 mark)
Q11 · 1 mark · MCQIntroduction to Trigonometry
$(\frac{1}{t a n ^{2} θ} - \frac{1}{s i n ^{2} θ})$ is equal to :
1. (A)1
2. (B)-1
3. (C) $sec^{2} θ$
4. (D) $sin^{2} θ$
Try it first, then show our solutionSolution
1. $\frac{1}{t a n ^{2} θ} - \frac{1}{s i n ^{2} θ} = cot^{2} θ - csc^{2} θ = - 1$ , using $csc^{2} θ - cot^{2} θ = 1$ . This is option (B).
  1 mark
Answer: −1, option (B)
Where the marks are
- Uses the identity $csc^{2} θ - cot^{2} θ = 1$ to get −1 (1 mark)
Q12 · 1 mark · MCQCircles
The region between a chord and either of the two arcs of a circle is called :
1. (A)an arc
2. (B)a sector
3. (C)a segment
4. (D)a semicircle
Try it first, then show our solutionSolution
1. The region bounded by a chord and an arc is called a segment, so the answer is option (C).
  1 mark
Answer: A segment, option (C)
Where the marks are
- Identifies the region as a segment (1 mark)
Q13 · 1 mark · MCQReal Numbers
If $1080 = 2^{x} \times 3^{y} \times 5$ , then $(x - y)$ is equal to :
1. (A)6
2. (B)-1
3. (C)1
4. (D)0
Try it first, then show our solutionSolution
1. $1080 = 2^{3} \times 3^{3} \times 5$ , so $x = 3$ and $y = 3$ , which gives $x - y = 0$ . This is option (D).
  1 mark
Answer: 0, option (D)
Where the marks are
- Prime factorises 1080 to find x = y = 3 and x − y = 0 (1 mark)
Q14 · 1 mark · MCQCircles
In the given figure, PA is a tangent from an external point P to a circle with centre O. If $∠ A O P = 7 0^{\circ}$ , then the measure of $∠ A P O$ is :
1. (A) $7 0^{\circ}$
2. (B) $9 0^{\circ}$
3. (C) $11 0^{\circ}$
4. (D) $2 0^{\circ}$
Figure in the paper: Circle with centre O. A point P outside the circle with tangent PA touching the circle at A. A straight line through A and O continues to a point B on the opposite side of the circle (AB is a diameter). Segment OP is drawn from O to P. The angle AOP at the centre is marked as 70°. The angle at P (APO) is marked by an arc. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius].
2. In $△ O A P$ , $∠ A P O = 18 0^{\circ} - 9 0^{\circ} - 7 0^{\circ} = 2 0^{\circ}$ . This is option (D).
  1 mark
Answer: 20°, option (D)
Where the marks are
- Uses tangent ⟂ radius and the angle sum of a triangle to get 20° (1 mark)
Q15 · 1 mark · MCQStatistics
The median group in the following frequency distribution is :
Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 5 8 20 15 7 5
1. (A)10 – 20
2. (B)20 – 30
3. (C)30 – 40
4. (D)40 – 50
Try it first, then show our solutionSolution
1. $N = 60$ , so $\frac{N}{2} = 30$ . The cumulative frequencies are 5, 13, 33, …
2. The first cumulative frequency greater than 30 is 33, so the median class is 20–30, option (B).
  1 mark
Answer: 20 – 30, option (B)
Where the marks are
- Uses N/2 = 30 and cumulative frequency to find the median class 20–30 (1 mark)
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Q16 · 1 mark · MCQAreas Related to Circles
In a circle of radius 21 cm, if an arc subtends an angle of $6 0^{\circ}$ at the centre of the circle, then the length of the arc is :
1. (A)11 cm
2. (B)44 cm
3. (C) $\frac{22}{7}$ cm
4. (D)22 cm
Try it first, then show our solutionSolution
1. Length of arc $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22$ cm. This is option (D).
  1 mark
Answer: 22 cm, option (D)
Where the marks are
- Applies the arc length formula to get 22 cm (1 mark)
Q17 · 1 mark · MCQCircles
A tangent to a circle is a line that touches the circle at :
1. (A)one point only
2. (B)two points
3. (C)three points
4. (D)infinite number of points
Try it first, then show our solutionSolution
1. A tangent meets the circle at exactly one point, so the answer is option (A).
  1 mark
Answer: One point only, option (A)
Where the marks are
- States that a tangent touches the circle at exactly one point (1 mark)
Q18 · 1 mark · MCQTriangles
In the given figure, if $△ A B C \sim △ QP R$ , then the value of $x$ is :
1. (A)5·3 cm
2. (B)4·6 cm
3. (C)2·3 cm
4. (D)4 cm
Figure in the paper: Two triangles. Triangle ABC has AC = 6 cm, BC = 4·6 cm, and AB as the base. Triangle QPR has QR = 3 cm and side PR labelled x. Vertices: R at the top, Q at the bottom left, P at the bottom right. See it in the official paper.
Try it first, then show our solutionSolution
1. From $△ A B C \sim △ QP R$ , the correspondence is A↔Q, B↔P, C↔R. So $\frac{A C}{QR} = \frac{B C}{P R}$ .
2. $\frac{6}{3} = \frac{4.6}{x}$ , so $x = 2.3$ cm. This is option (C).
  1 mark
Answer: 2·3 cm, option (C)
Where the marks are
- Uses the ratio of corresponding sides to get x = 2.3 cm (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedPair of Linear Equations in Two Variables
Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) given below.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, 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): The pair of linear equations $5 x + 2 y + 6 = 0$ and $7 x + 6 y + 18 = 0$ have infinitely many solutions.
Reason (R): The pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ have infinitely many solutions, if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $\frac{a _{1}}{a _{2}} = \frac{5}{7}$ and $\frac{b _{1}}{b _{2}} = \frac{2}{6} = \frac{1}{3}$ . These ratios are not equal, so the pair has a unique solution and the Assertion is false.
  1 mark
2. The Reason states the correct condition for infinitely many solutions, so it is true. The answer is option (D).
Answer: Option (D)
Where the marks are
- Compares the ratios to show A is false while R is true (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedProbability
Questions number 19 and 20 are Assertion and Reason based questions. Select the correct answer from the codes (A), (B), (C) and (D) given below.
Assertion (A): The probability of getting number 8 on rolling a die is zero (0).
Reason (R): The probability of an impossible event is zero (0).
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A die shows only the numbers 1 to 6, so getting 8 is an impossible event. The probability of an impossible event is 0, so both statements are true and R explains A. The answer is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises that getting 8 is an impossible event, so R explains A (1 mark)

Income (in ₹)	400 – 424	425 – 449	450 – 474	475 – 499	500 – 524
Number of workers	12	14	8	6	10

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	5	8	20	15	7	5

Section B

Very Short Answer (VSA)

Q21 · 2 marks · Very short answerReal Numbers
Given that HCF $(306, 1314) = 18$ , find LCM of $(306, 1314)$ .
Try it first, then show our solutionSolution
1. For any two positive integers, HCF × LCM = product of the numbers, so $18 \times LCM = 306 \times 1314$ .
  1 mark
2. $LCM = \frac{306 \times 1314}{18} = 17 \times 1314 = 22338$ .
  1 mark
Answer: LCM (306, 1314) = 22338
Where the marks are
- Uses the relation HCF × LCM = product of the two numbers (1 mark)
- Correctly computes LCM = 22338 (1 mark)
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Q22 · 2 marks · Very short answerCircles
XY and PQ are two tangents drawn at the end points of the diameter AB of a circle. Prove that $X Y ∥ P Q$ .
Try it first, then show our solutionSolution
1. Figure: a circle with centre O and diameter AB. Tangent XY touches the circle at A and tangent PQ touches it at B, with X, P on the same side of AB. Given: XY and PQ are tangents at the ends of diameter AB. To prove: $X Y ∥ P Q$ .
  ½ mark
2. OA is a radius and XY is the tangent at A, so $O A ⊥ X Y$ [tangent ⟂ radius at the point of contact]. Hence $∠ O A Y = 9 0^{\circ}$ , i.e. $∠ B A Y = 9 0^{\circ}$ .
  ½ mark
3. OB is a radius and PQ is the tangent at B, so $O B ⊥ P Q$ [tangent ⟂ radius]. Hence $∠ O B P = 9 0^{\circ}$ , i.e. $∠ A B P = 9 0^{\circ}$ .
  ½ mark
4. AB is a transversal of XY and PQ, and $∠ B A Y$ and $∠ A B P$ are alternate interior angles. They are equal (each $9 0^{\circ}$ ), so $X Y ∥ P Q$ .
  ½ mark
Answer: XY ∥ PQ, because the alternate angles ∠OAY and ∠OBP are both 90°.
Where the marks are
- Draws a correct figure with the tangents at A and B (½ mark)
- ∠OAY = 90° with reason: tangent ⟂ radius (½ mark)
- ∠OBP = 90° with reason: tangent ⟂ radius (½ mark)
- Equal alternate interior angles, therefore XY ∥ PQ (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerTriangles
In the given figure, $P Q ∥ R S$ . Prove that $O P \times O R = O Q \times O S$ .
Figure in the paper: Two lines PR and QS intersect at O. P and Q are joined by segment PQ, and R and S by segment RS, with PQ parallel to RS (arrow marks). P is on one arm and R on the other arm of the crossing at O; Q and S lie on the other pair of arms. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ P O Q$ and $△ S O R$ : $∠ P O Q = ∠ S O R$ [vertically opposite angles].
  ½ mark
2. $∠ O P Q = ∠ O S R$ [alternate interior angles, since $P Q ∥ R S$ and PS is a transversal].
  ½ mark
3. Therefore $△ P O Q \sim △ S O R$ [AA similarity criterion].
  ½ mark
4. Corresponding sides are proportional, so $\frac{O P}{O S} = \frac{O Q}{O R}$ . Cross-multiplying gives $O P \times O R = O Q \times O S$ . Hence proved.
  ½ mark
Answer: △POQ ~ △SOR (AA), so OP/OS = OQ/OR, i.e. OP × OR = OQ × OS.
Where the marks are
- Vertically opposite angles are equal (½ mark)
- Alternate interior angles are equal because PQ ∥ RS (½ mark)
- Concludes △POQ ~ △SOR by AA (½ mark)
- Writes the ratio of corresponding sides and cross-multiplies (½ mark)
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OR
Q23(b) · 2 marks · Very short answerTriangles
In the given figure, $L M ∥ C B$ and $L N ∥ C D$ . Prove that $\frac{A M}{A B} = \frac{A N}{A D}$ .
Figure in the paper: Quadrilateral ABCD with diagonal AC. Point L lies on AC. M lies on side AB and N lies on side AD. LM is drawn parallel to CB and LN parallel to CD (arrow marks). See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ A B C$ , $L M ∥ C B$ , so by the Basic Proportionality Theorem, $\frac{A M}{A B} = \frac{A L}{A C}$ ...(i)
  0.75 marks
2. In $△ A D C$ , $L N ∥ C D$ , so by BPT, $\frac{A N}{A D} = \frac{A L}{A C}$ ...(ii)
  0.75 marks
3. From (i) and (ii), $\frac{A M}{A B} = \frac{A N}{A D}$ . Hence proved.
  ½ mark
Answer: AM/AB = AL/AC = AN/AD, hence AM/AB = AN/AD.
Where the marks are
- Applies BPT in △ABC to get AM/AB = AL/AC, and in △ADC to get AN/AD = AL/AC (1½ marks)
- Equates the two ratios to conclude AM/AB = AN/AD (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerPolynomials
If $α, β$ are zeroes of the polynomial $8 x^{2} + 14 x + 3$ , then find the value of $(\frac{1}{α} + \frac{1}{β})$ .
Try it first, then show our solutionSolution
1. For $8 x^{2} + 14 x + 3$ : $a = 8, b = 14, c = 3$ . Sum of zeroes: $α + β = - \frac{b}{a} = - \frac{14}{8}$ .
  ½ mark
2. Product of zeroes: $α β = \frac{c}{a} = \frac{3}{8}$ .
  ½ mark
3. $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$
  ½ mark
4. $= \frac{- 14/8}{3/8} = - \frac{14}{3}$
  ½ mark
Answer: $\frac{1}{α} + \frac{1}{β} = - \frac{14}{3}$
Where the marks are
- Sum of zeroes = −14/8 (½ mark)
- Product of zeroes = 3/8 (½ mark)
- Writes 1/α + 1/β = (α + β)/(αβ) (½ mark)
- Gets the value −14/3 (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerPolynomials
Find a quadratic polynomial whose zeroes are $- 9$ and $6$ .
Try it first, then show our solutionSolution
1. Sum of zeroes $= - 9 + 6 = - 3$ .
  ½ mark
2. Product of zeroes $= (- 9) (6) = - 54$ .
  ½ mark
3. The required polynomial is $k [x^{2} - (sum) x + product] = k (x^{2} + 3 x - 54)$ , where $k$ is any non-zero real. Taking $k = 1$ , the polynomial is $x^{2} + 3 x - 54$ .
  1 mark
Answer: $x^{2} + 3 x - 54$ (or $k (x^{2} + 3 x - 54)$ , $k \neq = 0$ )
Where the marks are
- Sum of zeroes = −3 (½ mark)
- Product of zeroes = −54 (½ mark)
- Forms the polynomial x² + 3x − 54 (1 mark)
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Q25 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $sin 3 0^{\circ} cos 6 0^{\circ} + cos 3 0^{\circ} sin 6 0^{\circ} - cot 4 5^{\circ}$
Try it first, then show our solutionSolution
1. Substitute the standard values: $sin 3 0^{\circ} = \frac{1}{2}, cos 6 0^{\circ} = \frac{1}{2}, cos 3 0^{\circ} = \frac{3}{2}, sin 6 0^{\circ} = \frac{3}{2}, cot 4 5^{\circ} = 1$ . The expression becomes $\frac{1}{2} \times \frac{1}{2} + \frac{3}{2} \times \frac{3}{2} - 1$ .
  1½ marks
2. $= \frac{1}{4} + \frac{3}{4} - 1 = 0$
  ½ mark
Answer: 0
Where the marks are
- Substitutes all five trigonometric values correctly (1½ marks)
- Simplifies to 0 (½ mark)
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Section C

Short Answer (SA)

Q26 · 3 marks · Short answerAreas Related to Circles
A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding (i) minor sector (ii) major sector. (Use $π = 3 \cdot 14$ )
Try it first, then show our solutionSolution
1. (i) The angle of the minor sector is $θ = 9 0^{\circ}$ and $r = 10$ cm. Area of minor sector $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{90}{360} \times 3.14 \times 10 \times 10$
  1 mark
2. $= \frac{314}{4} = 78.5 cm^{2}$
  ½ mark
3. (ii) Area of the circle $= π r^{2} = 3.14 \times 10 \times 10 = 314 cm^{2}$
  1 mark
4. Area of major sector $= 314 - 78.5 = 235.5 cm^{2}$
  ½ mark
Answer: (i) 78.5 cm²; (ii) 235.5 cm²
Where the marks are
- Correct sector formula with substitution (1 mark)
- Minor sector area = 78.5 cm² (½ mark)
- Area of circle = 314 cm² (1 mark)
- Major sector area = 235.5 cm² (½ mark)
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Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerCircles
Prove that opposite sides of a quadrilateral circumscribing a circle subtends supplementary angles at the centre of the circle.
Try it first, then show our solutionSolution
1. Figure: quadrilateral ABCD circumscribes a circle with centre O. The sides AB, BC, CD, DA touch the circle at P, Q, R, S. Join OA, OB, OC, OD, OP, OQ, OR, OS. Mark $∠ A O P = ∠1$ , $∠ A O S = ∠2$ , $∠ D O S = ∠3$ , $∠ D O R = ∠4$ , $∠ C O R = ∠5$ , $∠ C O Q = ∠6$ , $∠ B O Q = ∠7$ , $∠ B O P = ∠8$ . To prove: $∠ A O B + ∠ C O D = 18 0^{\circ}$ and $∠ B O C + ∠ A O D = 18 0^{\circ}$ .
  ½ mark
2. In $△ A O P$ and $△ A O S$ : $O A = O A$ [common], $O P = O S$ [radii], $A P = A S$ [tangents from an external point are equal]. So $△ A O P ≅ △ A O S$ [SSS], giving $∠1 = ∠2$ [CPCT].
  1 mark
3. In the same way, $∠3 = ∠4$ , $∠5 = ∠6$ and $∠7 = ∠8$ .
  ½ mark
4. The angles round O add up to $36 0^{\circ}$ : $∠1 + ∠2 + \dots + ∠8 = 36 0^{\circ}$ . So $2 (∠1 + ∠8 + ∠4 + ∠5) = 36 0^{\circ}$ , i.e. $(∠1 + ∠8) + (∠4 + ∠5) = 18 0^{\circ}$ .
  ½ mark
5. Since $∠1 + ∠8 = ∠ A O B$ and $∠4 + ∠5 = ∠ C O D$ , we get $∠ A O B + ∠ C O D = 18 0^{\circ}$ . Similarly, $(∠6 + ∠7) + (∠2 + ∠3) = 18 0^{\circ}$ , i.e. $∠ B O C + ∠ A O D = 18 0^{\circ}$ . Hence opposite sides subtend supplementary angles at the centre.
  ½ mark
Answer: ∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°.
Where the marks are
- Labelled figure with given and to prove (½ mark)
- △AOP ≅ △AOS by SSS, so ∠1 = ∠2 (1 mark)
- States the other pairs of equal angles (½ mark)
- Uses the 360° sum round O to get (∠1+∠8)+(∠4+∠5) = 180° (½ mark)
- Concludes both pairs are supplementary (½ mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerCircles
If O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of $4 0^{\circ}$ with PQ, then find the measure of $∠ P O Q$ .
Figure in the paper: Circle with centre O. Chord PQ drawn with P on the circle at the top and Q on the right side of the circle. Radii OP and OQ are drawn. At P, a tangent line PR extends horizontally to the right with an arrow; the angle between tangent PR and chord PQ is marked 40°. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O P R = 9 0^{\circ}$ [radius ⟂ tangent at the point of contact], and $∠ R P Q = 4 0^{\circ}$ (given). So $∠ O P Q = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ}$ .
  1 mark
2. In $△ P O Q$ , $O P = O Q$ [radii], so $∠ O QP = ∠ O P Q = 5 0^{\circ}$ [angles opposite equal sides].
  1 mark
3. By the angle sum property, $∠ P O Q = 18 0^{\circ} - 5 0^{\circ} - 5 0^{\circ} = 8 0^{\circ}$ .
  1 mark
Answer: ∠POQ = 80°
Where the marks are
- ∠OPQ = 50° using radius ⟂ tangent (1 mark)
- ∠OQP = 50° because OP = OQ (1 mark)
- ∠POQ = 80° by angle sum (1 mark)
Open Q27(b) on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerQuadratic Equations
Using quadratic formula, find the real roots of the equation $2 x^{2} + 2 x + 9 = 0$ , if they exist.
Try it first, then show our solutionSolution
1. Compare $2 x^{2} + 2 x + 9 = 0$ with $a x^{2} + b x + c = 0$ : $a = 2, b = 2, c = 9$ .
  ½ mark
2. Discriminant $D = b^{2} - 4 a c = 2^{2} - 4 (2) (9) = 4 - 72 = - 68$ .
  1½ marks
3. By the quadratic formula $x = \frac{- b \pm D}{2 a}$ , but $- 68$ is not a real number. Since $D < 0$ , the equation has no real roots.
  1 mark
Answer: D = −68 < 0, so no real roots exist.
Where the marks are
- Identifies a, b, c and computes D = −68 (2 marks)
- Concludes there are no real roots since D < 0 (1 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerQuadratic Equations
Find the values of 'k' for which the quadratic equation $k x^{2} - 2 k x + 6 = 0$ has real and equal roots. Also, find the roots.
Try it first, then show our solutionSolution
1. For equal roots, $D = 0$ . Here $a = k, b = - 2 k, c = 6$ , so $D = (- 2 k)^{2} - 4 (k) (6) = 4 k^{2} - 24 k = 0$ .
  1 mark
2. $4 k (k - 6) = 0 \Rightarrow k = 0$ or $k = 6$ . Since $k = 0$ makes the equation non-quadratic, $k = 6$ .
  1 mark
3. With $k = 6$ : $6 x^{2} - 12 x + 6 = 0 \Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow (x - 1)^{2} = 0 \Rightarrow x = 1, 1$ .
  1 mark
Answer: k = 6; roots are 1, 1
Where the marks are
- Sets D = 0 to get 4k² − 24k = 0 (1 mark)
- k = 6, rejecting k = 0 (1 mark)
- Finds the roots x = 1, 1 (1 mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerProbability
One card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability that the card drawn is :
(i) a red king.
(ii) not a black card.
(iii) an ace of hearts.
Try it first, then show our solutionSolution
1. Total possible outcomes = 52. (i) There are 2 red kings (hearts and diamonds), so $P (red king) = \frac{2}{52} = \frac{1}{26}$ .
  1 mark
2. (ii) The cards that are not black are the 26 red cards, so $P (not black) = \frac{26}{52} = \frac{1}{2}$ .
  1 mark
3. (iii) There is 1 ace of hearts, so $P (ace of hearts) = \frac{1}{52}$ .
  1 mark
Answer: (i) 1/26; (ii) 1/2; (iii) 1/52
Where the marks are
- P(red king) = 2/52 = 1/26 (1 mark)
- P(not black) = 26/52 = 1/2 (1 mark)
- P(ace of hearts) = 1/52 (1 mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerReal Numbers
Prove that $2 + 53$ is an irrational number, if it is given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $x = 2 + 53$ is a rational number.
  1 mark
2. Then $53 = x - 2$ , so $3 = \frac{x - 2}{5}$ .
  1 mark
3. Since $x$ is rational, $\frac{x - 2}{5}$ is rational, which would make $3$ rational. This contradicts the given fact that $3$ is irrational. So the assumption is wrong, and $2 + 53$ is irrational.
  1 mark
Answer: 2 + 5√3 is irrational (proved by contradiction).
Where the marks are
- Assumes 2 + 5√3 is rational (1 mark)
- Expresses √3 = (x − 2)/5 (1 mark)
- Identifies the contradiction and concludes (1 mark)
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Q31 · 3 marks · Short answerIntroduction to Trigonometry
Prove that :
$(tan A + sec A)^{2} + (tan A - sec A)^{2} = 2 (\frac{1 + sin ^{2} A}{1 - sin ^{2} A})$
Try it first, then show our solutionSolution
1. LHS $= (tan^{2} A + sec^{2} A + 2 sec A tan A) + (tan^{2} A + sec^{2} A - 2 sec A tan A) = 2 (tan^{2} A + sec^{2} A)$
  1 mark
2. $= 2 (\frac{sin ^{2} A}{cos ^{2} A} + \frac{1}{cos ^{2} A}) = 2 (\frac{sin ^{2} A + 1}{cos ^{2} A})$
  1½ marks
3. Using $cos^{2} A = 1 - sin^{2} A$ : $= 2 (\frac{1 + sin ^{2} A}{1 - sin ^{2} A}) =$ RHS. Hence proved.
  ½ mark
Answer: LHS = 2(1 + sin²A)/(1 − sin²A) = RHS
Where the marks are
- Expands both squares to get 2(tan²A + sec²A) (1 mark)
- Converts to sine and cosine to get 2(sin²A + 1)/cos²A (1½ marks)
- Uses cos²A = 1 − sin²A to reach the RHS (½ mark)
Open Q31 on its own page

Section D

Long Answer (LA)

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerSurface Areas and Volumes
Two cubes each of volume $125 c m^{3}$ are joined end to end. Find the volume and the surface area of the resulting cuboid.
Try it first, then show our solutionSolution
1. Side of each cube: $a^{3} = 125 \Rightarrow a = 5$ cm.
  1 mark
2. Volume of the cuboid = volume of 2 cubes $= 2 \times 125 = 250 cm^{3}$ .
  1 mark
3. Dimensions of the cuboid: $l = 5 + 5 = 10$ cm, $b = 5$ cm, $h = 5$ cm.
  1½ marks
4. Surface area $= 2 (l b + bh + h l) = 2 (10 \times 5 + 5 \times 5 + 5 \times 10)$
  1 mark
5. $= 2 (50 + 25 + 50) = 250 cm^{2}$
  ½ mark
Answer: Volume = 250 cm³; Surface area = 250 cm²
Where the marks are
- Side of cube = 5 cm (1 mark)
- Volume of cuboid = 250 cm³ (1 mark)
- Dimensions 10 cm × 5 cm × 5 cm (1½ marks)
- Surface area formula with substitution (1 mark)
- Surface area = 250 cm² (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerSurface Areas and Volumes
A solid is in the shape of a cone surmounted on a hemisphere with both their diameters being equal to 7 cm and the height of the cone is equal to its radius. Find the volume of the solid.
Try it first, then show our solutionSolution
1. Radius of hemisphere = radius of cone $= r = \frac{7}{2}$ cm.
  1 mark
2. Height of cone $h = r = \frac{7}{2}$ cm.
  ½ mark
3. Volume of solid = volume of hemisphere + volume of cone $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (2 r + h)$
  1½ marks
4. $= \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times (2 \times \frac{7}{2} + \frac{7}{2}) = \frac{1}{3} \times \frac{77}{2} \times \frac{21}{2}$
  1½ marks
5. $= \frac{539}{4} = 134.75 cm^{3}$
  ½ mark
Answer: Volume = 539/4 cm³ = 134.75 cm³
Where the marks are
- Radius = 7/2 cm (1 mark)
- Height of cone = 7/2 cm (½ mark)
- Volume = hemisphere + cone, with correct formulas and substitution (3 marks)
- Final volume 134.75 cm³ (½ mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A contractor plans to install two slides for the children to play in a park. For the children below the age of 6 years, he prefers to have a slide whose top is at a height of $2 \cdot 0$ m and is inclined at an angle of $3 0^{\circ}$ to the ground, whereas for older children, he wants to have a steep slide at a height of $4 \cdot 0$ m and inclined at an angle of $6 0^{\circ}$ to the ground. What would be the length of the slide in each case ?
Try it first, then show our solutionSolution
1. Figure: two right triangles. In $△ A B C$ (right-angled at B), BC = 2 m is the height, AC is the slide and $∠ A = 3 0^{\circ}$ . In $△ P QR$ (right-angled at Q), QR = 4 m is the height, PR is the slide and $∠ P = 6 0^{\circ}$ .
  1 mark
2. In $△ A B C$ : $sin 3 0^{\circ} = \frac{B C}{A C}$ .
  1 mark
3. $\frac{2}{A C} = \frac{1}{2}$
  ½ mark
4. $\Rightarrow A C = 4$ m. So the slide for children below 6 years is 4 m long.
  ½ mark
5. In $△ P QR$ : $sin 6 0^{\circ} = \frac{QR}{P R}$ .
  1 mark
6. $\frac{4}{P R} = \frac{3}{2}$
  ½ mark
7. $\Rightarrow P R = \frac{8}{3} = \frac{8 3}{3} \approx 4.62$ m. So the slide for older children is about 4.62 m long.
  ½ mark
Answer: 4 m for children below 6 years; 8√3/3 m ≈ 4.62 m for older children
Where the marks are
- Correct labelled figures (1 mark)
- sin 30° = BC/AC (1 mark)
- Substitution 2/AC = 1/2 (½ mark)
- AC = 4 m (½ mark)
- sin 60° = QR/PR (1 mark)
- Substitution 4/PR = √3/2 (½ mark)
- PR = 8√3/3 ≈ 4.62 m (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerTriangles
If BD and QM are medians of triangles ABC and PQR, respectively, where $△ A B C \sim △ P QR$ , prove that $\frac{A B}{P Q} = \frac{B D}{QM}$ .
Try it first, then show our solutionSolution
1. Figure: $△ A B C$ with median BD to side AC (D is the midpoint of AC), and $△ P QR$ with median QM to side PR (M is the midpoint of PR). Given: $△ A B C \sim △ P QR$ . To prove: $\frac{A B}{P Q} = \frac{B D}{QM}$ .
  1 mark
2. Since $△ A B C \sim △ P QR$ , corresponding sides are proportional: $\frac{A B}{P Q} = \frac{A C}{P R}$ .
  1 mark
3. As D and M are midpoints, $A C = 2 A D$ and $P R = 2 P M$ . So $\frac{A B}{P Q} = \frac{2 A D}{2 P M} = \frac{A D}{P M}$ .
  1 mark
4. In $△ A B D$ and $△ P QM$ : $\frac{A B}{P Q} = \frac{A D}{P M}$ (proved above) and $∠ A = ∠ P$ [corresponding angles of similar triangles]. So $△ A B D \sim △ P QM$ [SAS similarity].
  1½ marks
5. Corresponding sides of similar triangles are proportional, so $\frac{A B}{P Q} = \frac{B D}{QM}$ . Hence proved.
  ½ mark
Answer: △ABD ~ △PQM (SAS), hence AB/PQ = BD/QM.
Where the marks are
- Correct figure with given and to prove (1 mark)
- AB/PQ = AC/PR from the similarity (1 mark)
- Uses the midpoints to get AB/PQ = AD/PM (1 mark)
- Proves △ABD ~ △PQM by SAS using ∠A = ∠P (1½ marks)
- Concludes AB/PQ = BD/QM (½ mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerTriangles
CD and GH are respectively the bisectors of $∠ A C B$ and $∠ E GF$ such that D and H lie on sides AB and FE of $△ A B C$ and $△ F E G$ respectively. If $△ A B C \sim △ F E G$ , show that :
(i) $\frac{C D}{G H} = \frac{A C}{F G}$
(ii) $△ D C B \sim △ H GE$
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ with CD bisecting $∠ A C B$ (D on AB). Mark $∠ A C D = ∠1$ and $∠ D C B = ∠2$ . Draw $△ F E G$ with GH bisecting $∠ E GF$ (H on FE). Mark $∠ F G H = ∠3$ and $∠ H GE = ∠4$ .
  Given: $△ A B C \sim △ F E G$ , and CD, GH are the bisectors of $∠ A C B$ and $∠ E GF$ .
  1 mark
2. (i) Since $△ A B C \sim △ F E G$ , corresponding angles are equal, so $∠ A C B = ∠ F GE$ .
  ½ mark
3. Halving both angles gives $\frac{1}{2} ∠ A C B = \frac{1}{2} ∠ F GE$ [CD and GH are bisectors].
  Hence $∠1 = ∠3$ and $∠2 = ∠4$ .
  1 mark
4. In $△ A C D$ and $△ F G H$ :
  $∠ A = ∠ F$ [corresponding angles of similar triangles]
  $∠1 = ∠3$ [proved above]
  So $△ A C D \sim △ F G H$ [AA similarity criterion].
  1 mark
5. Corresponding sides of similar triangles are proportional, so $\frac{C D}{G H} = \frac{A C}{F G}$ .
  ½ mark
6. (ii) In $△ D C B$ and $△ H GE$ :
  $∠ B = ∠ E$ [corresponding angles, as $△ A B C \sim △ F E G$ ]
  $∠2 = ∠4$ [proved above]
  So $△ D C B \sim △ H GE$ [AA similarity criterion].
  1 mark
Answer: (i) $△ A C D \sim △ F G H$ (AA), so $\frac{C D}{G H} = \frac{A C}{F G}$ ; (ii) $△ D C B \sim △ H GE$ (AA).
Where the marks are
- Correct labelled figure with given data (1 mark)
- States $∠ A C B = ∠ F GE$ from the similarity (½ mark)
- Deduces $∠1 = ∠3$ and $∠2 = ∠4$ using the bisectors (1 mark)
- Proves $△ A C D \sim △ F G H$ by AA (1 mark)
- Writes the side ratio $\frac{C D}{G H} = \frac{A C}{F G}$ (½ mark)
- Proves $△ D C B \sim △ H GE$ by AA (1 mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerCompetency-basedArithmetic Progressions
A manufacturer of TV sets produced 720 TV sets in the fourth year and 880 TV sets in the eighth year. Assuming that the production increases uniformly by a fixed number every year, find the production in the tenth year and the total production in the first seven years.
Try it first, then show our solutionSolution
1. The production increases by a fixed number every year, so the yearly productions form an A.P. Let the first term be $a$ and the common difference be $d$ .
  Fourth year: $a_{4} = a + 3 d = 720$ ...(i)
  1 mark
2. Eighth year: $a_{8} = a + 7 d = 880$ ...(ii)
  1 mark
3. Subtracting (i) from (ii): $4 d = 160$ , so $d = 40$ .
  Substituting in (i): $a = 720 - 120 = 600$ .
  1 mark
4. Production in the tenth year:
  $a_{10} = a + 9 d = 600 + 9 \times 40 = 960$ TV sets.
  1 mark
5. Total production in the first seven years:
  $S_{7} = \frac{7}{2} [2 a + 6 d] = \frac{7}{2} [1200 + 240] = \frac{7}{2} \times 1440 = 5040$ TV sets.
  1 mark
Answer: Tenth year: 960 TV sets; first seven years: 5040 TV sets
Where the marks are
- Forms $a + 3 d = 720$ (1 mark)
- Forms $a + 7 d = 880$ (1 mark)
- Solves to get $d = 40$ and $a = 600$ (½ + ½ marks)
- Finds $a_{10} = 960$ (1 mark)
- Finds $S_{7} = 5040$ (1 mark)
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Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedStatistics
Case Study – 1
Mutual Fund : A mutual fund is a type of investment vehicle that pools money from multiple investors to invest in securities like stocks, bonds or other securities. Mutual funds are operated by professional money managers, who allocate the fund's assets and attempt to produce capital gains or income for the fund's investors.
Net Asset Value (NAV) represents a fund's per share market value. It is the price at which the investors buy fund shares from a fund company and sell them to a fund company.
The following table shows the Net Asset Value (NAV) per unit of mutual fund of ICICI mutual funds :
NAV (in ₹) 0 – 5 5 – 10 10 – 15 15 – 20 20 – 25
Number of mutual funds 13 16 22 18 11
Based on the above information, answer the following questions :
1. (i)
  What is the upper limit of modal class of the data ?
  1 mark
2. (ii)
  What is the median class of the data ?
  1 mark
3. (iii)(a)
  What is the mode NAV of mutual funds ?
  2 marks
4. OR
  (iii)(b)
  What is the median NAV of mutual funds ?
  2 marks
Try it first, then show our solutionSolution
1. (i) The highest frequency is 22, in the class 10 – 15, so 10 – 15 is the modal class.
  Its upper limit is 15.
  1 mark
2. (ii) Total frequency $N = 13 + 16 + 22 + 18 + 11 = 80$ , so $\frac{N}{2} = 40$ .
  The cumulative frequencies are 13, 29, 51, 69, 80. The first cf greater than 40 is 51, so the median class is 10 – 15.
  1 mark
3. (iii)(a) For the modal class 10 – 15: $l = 10$ , $f_{1} = 22$ , $f_{0} = 16$ , $f_{2} = 18$ , $h = 5$ .
  $Mode = l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) h = 10 + (\frac{22 - 16}{44 - 16 - 18}) \times 5$
  1½ marks
4. $= 10 + \frac{6}{10} \times 5 = 10 + 3 = 13$
  The mode NAV is ₹ 13.
  ½ mark
5. (iii)(b) Cumulative frequency table:
  NAV (₹) f cf
  0 – 5 13 13
  5 – 10 16 29
  10 – 15 22 51
  15 – 20 18 69
  20 – 25 11 80
  ½ mark
6. For the median class 10 – 15: $l = 10$ , $\frac{N}{2} = 40$ , $c f = 29$ , $f = 22$ , $h = 5$ .
  $Median = l + (\frac{\frac{N}{2} - c f}{f}) h = 10 + (\frac{40 - 29}{22}) \times 5$
  1 mark
7. $= 10 + \frac{11}{22} \times 5 = 10 + 2.5 = 12.5$
  The median NAV is ₹ 12.5.
  ½ mark
Answer: (i) 15; (ii) 10 – 15; (iii)(a) Mode = ₹ 13; (iii)(b) Median = ₹ 12.5
Where the marks are
- (i) Upper limit of the modal class is 15 (1 mark)
- (ii) Median class is 10 – 15 (1 mark)
- (iii)(a) Correct values substituted in the mode formula (1½ marks)
- (iii)(a) Mode = 13 (½ mark)
- (iii)(b) Correct cumulative frequency table (½ mark)
- (iii)(b) Correct values substituted in the median formula (1 mark)
- (iii)(b) Median = 12.5 (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Case Study – 2
Resident Welfare Association (RWA) of Gulmohar Society in Delhi, have installed three electric poles A, B and C in the society's common park. Despite these three poles, some parts of the park are still in the dark. So, RWA decides to have one more electric pole D in the park. The park can be modelled as a coordinate system given below.
On the basis of the above information, answer the following questions :
1. (i)
  What is the position of the pole C ?
  1 mark
2. (ii)
  What is the distance of the pole B from the corner O of the park ?
  1 mark
3. (iii)(a)
  Find the position of the fourth pole D so that the four points A, B, C and D form a parallelogram ABCD.
  2 marks
4. OR
  (iii)(b)
  Find the distance between poles A and C.
  2 marks
Figure in the paper: Coordinate graph with X-axis and Y-axis meeting at origin O, grid marked 1 to 6 on X and 1 to 10 on Y. Three poles drawn as dots with upward arrows: A at (2, 7) with its arrow going up to about 8; B at (6, 6) with arrow going up to about 7; C at (5, 4) with arrow going up to about 5. The park is the first-quadrant region shown. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Reading from the graph, pole C is at the point (5, 4).
  1 mark
2. (ii) Pole B is at (6, 6) and O is at (0, 0).
  ½ mark
3. By the distance formula, $B O = (6 - 0)^{2} + (6 - 0)^{2} = 36 + 36 = 72 = 62$ units.
  ½ mark
4. (iii)(a) The poles are at A(2, 7), B(6, 6) and C(5, 4). Let D be $(x, y)$ .
  The diagonals of a parallelogram bisect each other, so the mid-point of AC equals the mid-point of BD:
  $(\frac{2 + 5}{2}, \frac{7 + 4}{2}) = (\frac{6 + x}{2}, \frac{6 + y}{2})$
  1 mark
5. Comparing x-coordinates: $\frac{7}{2} = \frac{6 + x}{2} \Rightarrow x = 1$
  ½ mark
6. Comparing y-coordinates: $\frac{11}{2} = \frac{6 + y}{2} \Rightarrow y = 5$
  So the fourth pole D is at (1, 5).
  ½ mark
7. (iii)(b) The poles are at A(2, 7) and C(5, 4).
  $A C = (5 - 2)^{2} + (4 - 7)^{2} = 9 + 9$
  1½ marks
8. $= 18 = 32$ units
  ½ mark
Answer: (i) C(5, 4); (ii) $62$ units; (iii)(a) D(1, 5); (iii)(b) $32$ units
Where the marks are
- (i) Writes C as (5, 4) (1 mark)
- (ii) Takes B(6, 6) and applies the distance formula from O (½ mark)
- (ii) BO = $72 = 62$ (½ mark)
- (iii)(a) Equates the mid-points of the diagonals AC and BD (1 mark)
- (iii)(a) Finds x = 1 (½ mark)
- (iii)(a) Finds y = 5, so D(1, 5) (½ mark)
- (iii)(b) Substitutes correctly in the distance formula for AC (1½ marks)
- (iii)(b) AC = $32$ (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
Case Study – 3
Deepankar bought 3 notebooks and 2 pens for ₹ 80 and his friend Suryansh bought 4 notebooks and 3 pens for ₹ 110 from the school bookshop.
Based on the above information, answer the following questions.
1. (i)
  If the price of one notebook be ₹ x and the price of one pen be ₹ y, write the given situation algebraically.
  1 mark
2. (ii)(a)
  What is the price of one notebook ?
  2 marks
3. OR
  (ii)(b)
  What is the price of one pen ?
  2 marks
4. (iii)
  What is the total amount to be paid by Suryansh, if he purchases 6 notebooks and 3 pens ?
  1 mark
Try it first, then show our solutionSolution
1. (i) Deepankar: $3 x + 2 y = 80$ ...(1)
  ½ mark
2. Suryansh: $4 x + 3 y = 110$ ...(2)
  ½ mark
3. (ii)(a) Multiply (1) by 3 and (2) by 2:
  $9 x + 6 y = 240$ and $8 x + 6 y = 220$
  1 mark
4. Subtracting the second from the first gives $x = 20$ .
  The price of one notebook is ₹ 20.
  1 mark
5. (ii)(b) Multiply (1) by 4 and (2) by 3:
  $12 x + 8 y = 320$ and $12 x + 9 y = 330$
  1 mark
6. Subtracting the first from the second gives $y = 10$ .
  The price of one pen is ₹ 10.
  1 mark
7. (iii) Amount for 6 notebooks and 3 pens $= 6 \times 20 + 3 \times 10 = 120 + 30 = ₹ 150$
  1 mark
Answer: (i) $3 x + 2 y = 80$ , $4 x + 3 y = 110$ ; (ii)(a) ₹ 20; (ii)(b) ₹ 10; (iii) ₹ 150
Where the marks are
- (i) Writes $3 x + 2 y = 80$ (½ mark)
- (i) Writes $4 x + 3 y = 110$ (½ mark)
- (ii)(a) Solves the pair of equations correctly to get x = 20 (2 marks)
- (ii)(b) Solves the pair of equations correctly to get y = 10 (2 marks)
- (iii) Total amount = ₹ 150 (1 mark)
Open Q38 on its own page

NAV (in ₹)	0 – 5	5 – 10	10 – 15	15 – 20	20 – 25
Number of mutual funds	13	16	22	18	11

NAV (₹)	f	cf
0 – 5	13	13
5 – 10	16	29
10 – 15	22	51
15 – 20	18	69
20 – 25	11	80

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