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

CBSE Class 10 Mathematics Standard: 2024 board paper, set 30/5/1

The question paper set in the CBSE Class 10 board exam of 2024, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2024/X/MATHEMATICS_STANDARD.zip (a zip; this paper is the file 30-5-1(Mathematics Standard).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 21 and 24
C	26-31	Short answer; internal choice in 26 and 27
D	32-35	Long answer; internal choice in 32 and 34
E	36-38	Case study based; internal choice in the 2-mark sub-part of each case

Instructions in brief: All 38 questions are compulsory; no overall choice, but internal choice is given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 case studies in Section E.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQArithmetic Progressions
The next (4th) term of the A.P. $18, 50, 98, \dots$ is :
1. (A) $128$
2. (B) $140$
3. (C) $162$
4. (D) $200$
Try it first, then show our solutionSolution
1. Simplify: $18 = 32$ , $50 = 52$ , $98 = 72$ , so $d = 22$ .
2. $a_{4} = 72 + 22 = 92 = 162$ , option (C).
  1 mark
Answer: $162$ , option (C)
Where the marks are
- Finds $a_{4} = 92 = 162$ (1 mark)
Q2 · 1 mark · MCQIntroduction to Trigonometry
If $\frac{x}{3} = 2 sin A$ , $\frac{y}{3} = 2 cos A$ , then the value of $x^{2} + y^{2}$ is :
1. (A)36
2. (B)9
3. (C)6
4. (D)18
Try it first, then show our solutionSolution
1. $x = 6 sin A$ , $y = 6 cos A$ .
2. $x^{2} + y^{2} = 36 (sin^{2} A + cos^{2} A) = 36$ , option (A).
  1 mark
Answer: 36, option (A)
Where the marks are
- Uses $sin^{2} A + cos^{2} A = 1$ to get 36 (1 mark)
Q3 · 1 mark · MCQIntroduction to Trigonometry
If $4 sec θ - 5 = 0$ , then the value of $cot θ$ is :
1. (A) $\frac{3}{4}$
2. (B) $\frac{4}{5}$
3. (C) $\frac{5}{3}$
4. (D) $\frac{4}{3}$
Try it first, then show our solutionSolution
1. $sec θ = \frac{5}{4}$ , so $tan^{2} θ = sec^{2} θ - 1 = \frac{9}{16}$ , $tan θ = \frac{3}{4}$ .
2. $cot θ = \frac{4}{3}$ , option (D).
  1 mark
Answer: $\frac{4}{3}$ , option (D)
Where the marks are
- Obtains $cot θ = 4/3$ (1 mark)
Q4 · 1 mark · MCQPair of Linear Equations in Two Variables
Which out of the following type of straight lines will be represented by the system of equations $3 x + 4 y = 5$ and $6 x + 8 y = 7$ ?
1. (A)Parallel
2. (B)Intersecting
3. (C)Coincident
4. (D)Perpendicular to each other
Try it first, then show our solutionSolution
1. $\frac{a _{1}}{a _{2}} = \frac{3}{6} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{4}{8} = \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{5}{7}$ .
2. Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , lines are parallel, option (A).
  1 mark
Answer: Parallel, option (A)
Where the marks are
- Compares ratios and concludes parallel (1 mark)
Open Q4 on its own page
Q5 · 1 mark · MCQQuadratic Equations
The ratio of the sum and product of the roots of the quadratic equation $5 x^{2} - 6 x + 21 = 0$ is :
1. (A) $5 : 21$
2. (B) $2 : 7$
3. (C) $21 : 5$
4. (D) $7 : 2$
Try it first, then show our solutionSolution
1. Sum $= \frac{6}{5}$ , product $= \frac{21}{5}$ .
2. Ratio $= 6 : 21 = 2 : 7$ , option (B).
  1 mark
Answer: 2 : 7, option (B)
Where the marks are
- Finds sum and product, ratio 2:7 (1 mark)
Q6 · 1 mark · MCQStatistics
For the data $2, 9, x + 6, 2 x + 3, 5, 10, 5$ ; if the mean is $7$ , then the value of $x$ is :
1. (A)9
2. (B)6
3. (C)5
4. (D)3
Try it first, then show our solutionSolution
1. Sum $= 2 + 9 + x + 6 + 2 x + 3 + 5 + 10 + 5 = 3 x + 40$ ; mean $= \frac{3 x + 40}{7} = 7$ .
2. $3 x + 40 = 49 \Rightarrow x = 3$ , option (D).
  1 mark
Answer: x = 3, option (D)
Where the marks are
- Forms $\frac{3 x + 40}{7} = 7$ and gets $x = 3$ (1 mark)
Q7 · 1 mark · MCQProbability
One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 7 is :
1. (A) $\frac{1}{7}$
2. (B) $\frac{1}{8}$
3. (C) $\frac{1}{5}$
4. (D) $\frac{7}{40}$
Try it first, then show our solutionSolution
1. Multiples of 7 up to 40: 7, 14, 21, 28, 35 — 5 outcomes out of 40.
2. $P = \frac{5}{40} = \frac{1}{8}$ , option (B).
  1 mark
Answer: $\frac{1}{8}$ , option (B)
Where the marks are
- Counts 5 favourable outcomes, P = 1/8 (1 mark)
Q8 · 1 mark · MCQAreas Related to Circles
The perimeter of the sector of a circle of radius 21 cm which subtends an angle of $6 0^{\circ}$ at the centre of circle, is :
1. (A)22 cm
2. (B)43 cm
3. (C)64 cm
4. (D)462 cm
Try it first, then show our solutionSolution
1. Arc $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22$ cm.
2. Perimeter $= 22 + 2 (21) = 64$ cm, option (C).
  1 mark
Answer: 64 cm, option (C)
Where the marks are
- Arc 22 cm plus two radii gives 64 cm (1 mark)
Q9 · 1 mark · MCQAreas Related to Circles
The length of an arc of a circle with radius 12 cm is $10 π$ cm. The angle subtended by the arc at the centre of the circle, is :
1. (A) $12 0^{\circ}$
2. (B) $6^{\circ}$
3. (C) $7 5^{\circ}$
4. (D) $15 0^{\circ}$
Try it first, then show our solutionSolution
1. $\frac{θ}{360} \times 2 π (12) = 10 π \Rightarrow \frac{24 θ}{360} = 10$ .
2. $θ = 15 0^{\circ}$ , option (D).
  1 mark
Answer: $15 0^{\circ}$ , option (D)
Where the marks are
- Uses arc length formula to get 150° (1 mark)
Q10 · 1 mark · MCQReal Numbers
The greatest number which divides 281 and 1249, leaving remainder 5 and 7 respectively, is :
1. (A)23
2. (B)276
3. (C)138
4. (D)69
Try it first, then show our solutionSolution
1. Required number = HCF $(281 - 5, 1249 - 7)$ = HCF $(276, 1242)$ .
2. $276 = 2^{2} \cdot 3 \cdot 23$ , $1242 = 2 \cdot 3^{3} \cdot 23$ ; HCF $= 2 \cdot 3 \cdot 23 = 138$ , option (C).
  1 mark
Answer: 138, option (C)
Where the marks are
- Finds HCF(276, 1242) = 138 (1 mark)
Q11 · 1 mark · MCQArithmetic Progressions
The number of terms in the A.P. $3, 6, 9, 12, \dots, 111$ is :
1. (A)36
2. (B)40
3. (C)37
4. (D)30
Try it first, then show our solutionSolution
1. $a = 3, d = 3$ : $3 + (n - 1) 3 = 111$ .
2. $3 n = 111 \Rightarrow n = 37$ , option (C).
  1 mark
Answer: 37, option (C)
Where the marks are
- Uses $a_{n} = a + (n - 1) d$ to get n = 37 (1 mark)
Q12 · 1 mark · MCQCircles
A chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is :
1. (A) $52$
2. (B) $102$
3. (C) $\frac{5}{2}$
4. (D)5
Figure in the paper: A circle with centre O. Chord AB is drawn; OA and OB are radii, each labelled 10 cm. The angle AOB at the centre is marked as a right angle. See it in the official paper.
Try it first, then show our solutionSolution
1. In right △AOB (right angle at O), $A B^{2} = 1 0^{2} + 1 0^{2} = 200$ [Pythagoras].
2. $A B = 102$ cm, option (B).
  1 mark
Answer: $102$ cm, option (B)
Where the marks are
- Applies Pythagoras to get $102$ (1 mark)
Q13 · 1 mark · MCQReal Numbers
The LCM of three numbers 28, 44, 132 is :
1. (A)258
2. (B)231
3. (C)462
4. (D)924
Try it first, then show our solutionSolution
1. $28 = 2^{2} \cdot 7$ , $44 = 2^{2} \cdot 11$ , $132 = 2^{2} \cdot 3 \cdot 11$ .
2. LCM $= 2^{2} \cdot 3 \cdot 7 \cdot 11 = 924$ , option (D).
  1 mark
Answer: 924, option (D)
Where the marks are
- Prime factorises and gets LCM 924 (1 mark)
Q14 · 1 mark · MCQReal Numbers
If the product of two co-prime numbers is 553, then their HCF is :
1. (A)1
2. (B)553
3. (C)7
4. (D)79
Try it first, then show our solutionSolution
1. Co-prime numbers have no common factor other than 1, so HCF = 1 regardless of the product, option (A).
  1 mark
Answer: 1, option (A)
Where the marks are
- States HCF of co-prime numbers is 1 (1 mark)
Q15 · 1 mark · MCQPolynomials
If $α$ and $β$ are the zeroes of the polynomial $p (x) = k x^{2} - 30 x + 45 k$ and $α + β = α β$ , then the value of $k$ is :
1. (A) $- \frac{2}{3}$
2. (B) $- \frac{3}{2}$
3. (C) $\frac{3}{2}$
4. (D) $\frac{2}{3}$
Try it first, then show our solutionSolution
1. $α + β = \frac{30}{k}$ , $α β = \frac{45 k}{k} = 45$ .
2. $\frac{30}{k} = 45 \Rightarrow k = \frac{2}{3}$ , option (D).
  1 mark
Answer: $k = \frac{2}{3}$ , option (D)
Where the marks are
- Equates sum and product to get k = 2/3 (1 mark)
Q16 · 1 mark · MCQCircles
In the given figure, RJ and RL are two tangents to the circle. If $∠ R J L = 4 2^{\circ}$ , then the measure of $∠ J O L$ is :
1. (A) $4 2^{\circ}$
2. (B) $8 4^{\circ}$
3. (C) $9 6^{\circ}$
4. (D) $13 8^{\circ}$
Figure in the paper: A circle with centre O. Two tangents RJ and RL meet at external point R, touching the circle at J and L. Chord JL is drawn; the angle RJL = 42° is marked at J. OJ and OL are radii joined to O. See it in the official paper.
Try it first, then show our solutionSolution
1. RJ = RL [tangents from external point], so $∠ R L J = 4 2^{\circ}$ and $∠ J R L = 18 0^{\circ} - 8 4^{\circ} = 9 6^{\circ}$ .
2. In quadrilateral OJRL, $∠ O J R = ∠ O L R = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ J O L = 18 0^{\circ} - 9 6^{\circ} = 8 4^{\circ}$ , option (B).
  1 mark
Answer: $8 4^{\circ}$ , option (B)
Where the marks are
- Uses equal tangents and tangent ⟂ radius to get 84° (1 mark)
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Q17 · 1 mark · MCQTriangles
In the given figure, in $△ A B C$ , $D E ∥ B C$ . If $A D = 2 \cdot 4$ cm, $D B = 4$ cm and $A E = 2$ cm, then the length of AC is :
1. (A) $\frac{10}{3}$ cm
2. (B) $\frac{3}{10}$ cm
3. (C) $\frac{16}{3}$ cm
4. (D) $1 \cdot 2$ cm
Figure in the paper: Triangle ABC with apex A at top, base BC horizontal. D lies on AB and E lies on AC, with DE drawn parallel to BC. See it in the official paper.
Try it first, then show our solutionSolution
1. By BPT, $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{2.4}{4} = \frac{2}{E C} \Rightarrow E C = \frac{10}{3}$ cm.
2. $A C = 2 + \frac{10}{3} = \frac{16}{3}$ cm, option (C).
  1 mark
Answer: $\frac{16}{3}$ cm, option (C)
Where the marks are
- Applies BPT and adds AE + EC = 16/3 cm (1 mark)
Q18 · 1 mark · MCQSome Applications of Trigonometry
If a vertical pole of length $7 \cdot 5$ m casts a shadow 5 m long on the ground and at the same time, a tower casts a shadow 24 m long, then the height of the tower is :
1. (A)20 m
2. (B)40 m
3. (C)60 m
4. (D)36 m
Try it first, then show our solutionSolution
1. Same sun elevation gives similar triangles: $\frac{h}{24} = \frac{7.5}{5}$ .
2. $h = 36$ m, option (D).
  1 mark
Answer: 36 m, option (D)
Where the marks are
- Uses similar triangles ratio to get 36 m (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedTriangles
Assertion (A): ABCD is a trapezium with DC $∥$ AB. E and F are points on AD and BC respectively, such that EF $∥$ AB. Then $\frac{A E}{E D} = \frac{B F}{F C}$ .
Reason (R): Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
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. R is a standard result (join AC; apply BPT in △ADC and △ABC), hence true.
2. A follows directly from R with EF ∥ AB ∥ DC, so both true and R explains A: option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises both true and R explains A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedPolynomials
Assertion (A): Degree of a zero polynomial is not defined.
Reason (R): Degree of a non-zero constant polynomial is 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 is true: zero polynomial has no nonzero term, so its degree is not defined. R is true: $c = c x^{0}$ , $c \neq = 0$ , has degree 0.
2. R concerns non-zero constants, so it does not explain A: option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Both true, R not the explanation (1 mark)
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Section B

Very short answer; internal choice in 21 and 24

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerCircles
If two tangents inclined at an angle of $6 0^{\circ}$ are drawn to a circle of radius 3 cm, then find the length of each tangent.
Try it first, then show our solutionSolution
1. Figure: Draw a circle with centre O and radius OA = OB = 3 cm. From an external point P draw tangents PA and PB with $∠ A P B = 6 0^{\circ}$ . Join OA, OB and OP.
  ½ mark
2. $O A ⊥ P A$ [tangent ⟂ radius]. OP bisects $∠ A P B$ [tangents from an external point are equally inclined to OP], so $∠ A P O = 3 0^{\circ}$ .
  ½ mark
3. In right $△ O A P$ : $tan 3 0^{\circ} = \frac{O A}{A P} \Rightarrow \frac{1}{3} = \frac{3}{A P}$
  ½ mark
4. $A P = 33$ cm. Since tangents from an external point are equal, $P A = P B = 33$ cm.
  ½ mark
Answer: Each tangent = $33$ cm
Where the marks are
- Correct labelled figure (½ mark)
- States $∠ A P O = 3 0^{\circ}$ (½ mark)
- Writes $tan 3 0^{\circ} = 3/ A P$ (½ mark)
- Gets $A P = 33$ cm (½ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerCircles
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Try it first, then show our solutionSolution
1. Figure: a circle with centre O and diameter AB. Line PQ is the tangent at A (P and Q on either side of A), and line RS is the tangent at B (R above P, S above Q).
  ½ mark
2. $∠ O A P = 9 0^{\circ}$ and $∠ O B R = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]
  ½ mark
3. $∠ O A P + ∠ O B R = 9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ . These are co-interior angles made by the transversal AB with the lines PQ and RS, and they are supplementary.
  ½ mark
4. Therefore $P Q ∥ R S$ , i.e. the tangents at the ends of a diameter are parallel.
  ½ mark
Answer: The co-interior angles at A and B add to $18 0^{\circ}$ , so the tangents PQ and RS are parallel.
Where the marks are
- Correct figure (½ mark)
- Both angles $= 9 0^{\circ}$ with reason (½ mark)
- Sum $= 18 0^{\circ}$ , co-interior angles supplementary (½ mark)
- Concludes $P Q ∥ R S$ (½ mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $\frac{2 t a n 3 0 ^{\circ} \cdot s e c 6 0 ^{\circ} \cdot t a n 4 5 ^{\circ}}{1 - s i n ^{2} 6 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. Substitute $tan 3 0^{\circ} = \frac{1}{3}$ , $sec 6 0^{\circ} = 2$ , $tan 4 5^{\circ} = 1$ , $sin 6 0^{\circ} = \frac{3}{2}$ :
  $\frac{2 \times \frac{1}{3} \times 2 \times 1}{1 - \frac{3}{4}} = \frac{\frac{4}{3}}{\frac{1}{4}}$
  1½ marks
2. $= \frac{16}{3} = \frac{16 3}{3}$
  ½ mark
Answer: $\frac{16}{3}$ or $\frac{16 3}{3}$
Where the marks are
- Correct substitution of all trigonometric values (1½ marks)
- Correct simplification to $16/ 3$ (½ mark)
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Q23 · 2 marks · Very short answerPolynomials
If $α, β$ are zeroes of the polynomial $p (x) = 5 x^{2} - 6 x + 1$ , then find the value of $α + β + α β$ .
Try it first, then show our solutionSolution
1. Here $a = 5, b = - 6, c = 1$ . Sum of zeroes $α + β = - \frac{b}{a} = \frac{6}{5}$
  ½ mark
2. Product of zeroes $α β = \frac{c}{a} = \frac{1}{5}$
  ½ mark
3. $α + β + α β = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$
  1 mark
Answer: $\frac{7}{5}$
Where the marks are
- $α + β = 6/5$ (½ mark)
- $α β = 1/5$ (½ mark)
- Adds to get $7/5$ (1 mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerCoordinate Geometry
Find the ratio in which the point $P (- 4, 6)$ divides the line segment joining the points $A (- 6, 10)$ and $B (3, - 8)$ .
Try it first, then show our solutionSolution
1. Let P divide AB in the ratio $k : 1$ . By the section formula, the x-coordinate of P is
  $- 4 = \frac{3 k + ( - 6 )}{k + 1} = \frac{3 k - 6}{k + 1}$
  1 mark
2. $- 4 k - 4 = 3 k - 6 \Rightarrow 7 k = 2 \Rightarrow k = \frac{2}{7}$ (check: y-coordinate $\frac{- 8 \cdot \frac{2}{7} + 10}{\frac{9}{7}} = 6$ ✓)
  ½ mark
3. Required ratio is $2 : 7$ .
  ½ mark
Answer: 2 : 7
Where the marks are
- Applies section formula with ratio $k : 1$ correctly (1 mark)
- Solves $k = 2/7$ (½ mark)
- States ratio 2 : 7 (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerCoordinate Geometry
Prove that the points $(3, 0)$ , $(6, 4)$ and $(- 1, 3)$ are the vertices of an isosceles triangle.
Try it first, then show our solutionSolution
1. Let $A (3, 0)$ , $B (6, 4)$ , $C (- 1, 3)$ . By distance formula, $A B = (6 - 3)^{2} + (4 - 0)^{2} = 9 + 16 = 5$
  ½ mark
2. $B C = (6 + 1)^{2} + (4 - 3)^{2} = 49 + 1 = 50$
  ½ mark
3. $C A = (3 + 1)^{2} + (0 - 3)^{2} = 16 + 9 = 5$
  ½ mark
4. Since $A B = C A = 5$ (two sides equal), $△ A B C$ is an isosceles triangle.
  ½ mark
Answer: AB = CA = 5, BC = √50, so the triangle is isosceles.
Where the marks are
- AB = 5 (½ mark)
- BC = √50 (½ mark)
- CA = 5 (½ mark)
- Concludes isosceles since AB = AC (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerProbability
A carton consists of 60 shirts of which 48 are good, 8 have major defects and 4 have minor defects. Nigam, a trader, will accept the shirts which are good but Anmol, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. Find the probability that it is acceptable to Anmol.
Try it first, then show our solutionSolution
1. Anmol rejects only shirts with major defects, so he accepts good shirts and those with minor defects: $48 + 4 = 52$ (i.e. $60 - 8 = 52$ ) favourable outcomes.
  ½ mark
2. P(shirt acceptable to Anmol) $= \frac{52}{60} = \frac{13}{15}$
  1½ marks
Answer: $\frac{13}{15}$
Where the marks are
- Number of shirts acceptable to Anmol = 52 (½ mark)
- Probability $= 52/60 = 13/15$ (1½ marks)
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Section C

Short answer; internal choice in 26 and 27

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3$ is rational.
  Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are coprime.
  Squaring: $3 q^{2} = p^{2}$ .
  ½ mark
2. So $p^{2}$ is divisible by 3. Since 3 is prime, 3 divides $p^{2}$ means 3 divides $p$ . ... (1)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2}$ , so $q^{2} = 3 a^{2}$ .
  So $q^{2}$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)
  1 mark
4. From (1) and (2), 3 divides both $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong. Hence $3$ is irrational.
  ½ mark
Answer: Assuming $3 = p / q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $3$ is irrational
Where the marks are
- Assumes $3 = \frac{p}{q}$ with p, q coprime and gets $3 q^{2} = p^{2}$ (1/2 mark)
- Shows 3 divides p (1 mark)
- Substitutes $p = 3 a$ and shows 3 divides q (1 mark)
- States the contradiction and concludes (1/2 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerReal Numbers
Prove that $(2 + 3)^{2}$ is an irrational number, given that $6$ is an irrational number.
Try it first, then show our solutionSolution
1. $(2 + 3)^{2} = 2 + 3 + 26 = 5 + 26$
  1 mark
2. Assume, to the contrary, that $5 + 26$ is rational. Then $5 + 26 = \frac{a}{b}$ , where $a, b$ are integers, $b \neq = 0$ .
  ½ mark
3. $6 = \frac{a - 5 b}{2 b}$
  ½ mark
4. Since $a, b$ are integers, $\frac{a - 5 b}{2 b}$ is rational, so $6$ would be rational. But $6$ is irrational (given) — a contradiction.
  ½ mark
5. Hence our assumption is wrong, and $5 + 26 = (2 + 3)^{2}$ is irrational.
  ½ mark
Answer: $(2 + 3)^{2} = 5 + 26$ is irrational.
Where the marks are
- Expands to $5 + 26$ (1 mark)
- Assumes it is rational, $= a / b$ (½ mark)
- Isolates $6 = (a - 5 b) /2 b$ (½ mark)
- Identifies contradiction: rational RHS vs irrational $6$ (½ mark)
- Concludes irrational (½ mark)
Open Q26(b) on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerArithmetic Progressions
If the sum of the first 14 terms of an A.P. is 1050 and the first term is 10, then find the 20th term and the $n^{t h}$ term.
Try it first, then show our solutionSolution
1. $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with $a = 10$ , $n = 14$ : $\frac{14}{2} (20 + 13 d) = 1050$
  1 mark
2. $20 + 13 d = 150 \Rightarrow 13 d = 130 \Rightarrow d = 10$
  1 mark
3. $a_{20} = a + 19 d = 10 + 19 \times 10 = 200$
  ½ mark
4. $a_{n} = a + (n - 1) d = 10 + (n - 1) 10 = 10 n$
  ½ mark
Answer: $a_{20} = 200$ , $a_{n} = 10 n$
Where the marks are
- Forms $\frac{14}{2} (20 + 13 d) = 1050$ (1 mark)
- Finds $d = 10$ (1 mark)
- $a_{20} = 200$ (½ mark)
- $a_{n} = 10 n$ (½ mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerArithmetic Progressions
The first term of an A.P. is 5, the last term is 45 and the sum of all the terms is 400. Find the number of terms and the common difference of the A.P.
Try it first, then show our solutionSolution
1. $a = 5$ , $l = a_{n} = 45$ , $S_{n} = 400$ . Using $S_{n} = \frac{n}{2} (a + l)$ : $\frac{n}{2} (5 + 45) = 400$
  1 mark
2. $25 n = 400 \Rightarrow n = 16$
  1 mark
3. $a_{16} = a + 15 d$ : $5 + 15 d = 45$
  ½ mark
4. $15 d = 40 \Rightarrow d = \frac{40}{15} = \frac{8}{3}$
  ½ mark
Answer: n = 16, d = 8/3
Where the marks are
- Uses $S_{n} = \frac{n}{2} (a + l)$ correctly (1 mark)
- n = 16 (1 mark)
- Forms $5 + 15 d = 45$ (½ mark)
- d = 8/3 (½ mark)
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Q28 · 3 marks · Short answerCircles
Prove that the parallelogram circumscribing a circle is a rhombus.
Try it first, then show our solutionSolution
1. Figure: Draw parallelogram ABCD with all four sides touching a circle. The points of contact are P on AB, Q on BC, R on CD and S on DA.
  Given: ABCD is a parallelogram circumscribing the circle.
  To prove: ABCD is a rhombus.
  ½ mark
2. Tangents drawn from an external point to a circle are equal in length. So:
  $A P = A S, B P = B Q, C R = C Q, D R = D S$
  1 mark
3. Adding these four equations:
  $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$
  $\Rightarrow A B + C D = A D + B C$
  ½ mark
4. In a parallelogram opposite sides are equal: $A B = C D$ and $B C = A D$ .
  So $2 A B = 2 B C$ , which gives $A B = B C$ .
  ½ mark
5. Therefore $A B = B C = C D = D A$ . A parallelogram with all sides equal is a rhombus, so ABCD is a rhombus. Hence proved.
  ½ mark
Answer: AB + CD = AD + BC together with AB = CD and BC = AD gives AB = BC = CD = DA, so ABCD is a rhombus
Where the marks are
- Correct labelled figure (1/2 mark)
- Writes the four equal-tangent relations (1 mark)
- Adds them to get $A B + C D = A D + B C$ (1/2 mark)
- Uses the opposite sides of the parallelogram to get $A B = B C$ (1/2 mark)
- Concludes all sides are equal, so ABCD is a rhombus (1/2 mark)
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Q29 · 3 marks · Short answerIntroduction to Trigonometry
Prove that : $\frac{t a n A}{1 - c o t A} + \frac{c o t A}{1 - t a n A} = 1 + sec A csc A$
Try it first, then show our solutionSolution
1. LHS $= \frac{\frac{s i n A}{c o s A}}{\frac{s i n A - c o s A}{s i n A}} + \frac{\frac{c o s A}{s i n A}}{\frac{c o s A - s i n A}{c o s A}}$
  ½ mark
2. $= \frac{sin ^{2} A}{cos A ( sin A - cos A )} - \frac{cos ^{2} A}{sin A ( sin A - cos A )} = \frac{1}{sin A - cos A} [\frac{sin ^{2} A}{cos A} - \frac{cos ^{2} A}{sin A}] = \frac{sin ^{3} A - cos ^{3} A}{sin A cos A ( sin A - cos A )}$
  1 mark
3. Using $a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + ab)$ : $= \frac{( sin A - cos A ) ( sin ^{2} A + cos ^{2} A + sin A cos A )}{sin A cos A ( sin A - cos A )} = \frac{1 + sin A cos A}{sin A cos A}$ [since $sin^{2} A + cos^{2} A = 1$ ]
  1 mark
4. $= \frac{1}{sin A cos A} + 1 = 1 + sec A csc A =$ RHS. Hence proved.
  ½ mark
Answer: LHS simplifies to $1 + sec A csc A$ = RHS; hence proved.
Where the marks are
- Writes tan and cot in terms of sin and cos (½ mark)
- Combines into a single expression with common factor $(sin A - cos A)$ (1 mark)
- Factorises $sin^{3} A - cos^{3} A$ and uses $sin^{2} A + cos^{2} A = 1$ (1 mark)
- Reaches $1 + sec A csc A$ (½ mark)
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Q30 · 3 marks · Short answerProbability
Three unbiased coins are tossed simultaneously. Find the probability of getting :
1. (i)
  at least one head.
  1 mark
2. (ii)
  exactly one tail.
  1 mark
3. (iii)
  two heads and one tail.
  1 mark
Try it first, then show our solutionSolution
1. Sample space: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}; total outcomes = 8.
  (i) At least one head: all except TTT, i.e. 7 outcomes. P $= \frac{7}{8}$
  1 mark
2. (ii) Exactly one tail: HHT, HTH, THH — 3 outcomes. P $= \frac{3}{8}$
  1 mark
3. (iii) Two heads and one tail: HHT, HTH, THH — 3 outcomes. P $= \frac{3}{8}$
  1 mark
Answer: (i) 7/8; (ii) 3/8; (iii) 3/8
Where the marks are
- (i) P(at least one head) = 7/8 (1 mark)
- (ii) P(exactly one tail) = 3/8 (1 mark)
- (iii) P(two heads and one tail) = 3/8 (1 mark)
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Q31 · 3 marks · Short answerAreas Related to Circles
An arc of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding major sector. (Use $π = 3 \cdot 14$ )
Try it first, then show our solutionSolution
1. Area of circle $= π r^{2} = 3.14 \times 10 \times 10 = 314$ cm²
  1 mark
2. Area of minor sector (θ = 90°) $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{90}{360} \times 3.14 \times 10 \times 10 = 78.5$ cm²
  1 mark
3. Area of major sector $= 314 - 78.5 = 235.5$ cm²
  1 mark
Answer: 235.5 cm²
Where the marks are
- Area of circle = 314 cm² (1 mark)
- Area of minor sector = 78.5 cm² (1 mark)
- Area of major sector = 235.5 cm² (1 mark)
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Section D

Long answer; internal choice in 32 and 34

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerQuadratic Equations
Find the value of 'k' for which the quadratic equation $(k + 1) x^{2} - 6 (k + 1) x + 3 (k + 9) = 0$ , $k \neq = - 1$ has real and equal roots.
Try it first, then show our solutionSolution
1. Here $a = k + 1$ , $b = - 6 (k + 1)$ , $c = 3 (k + 9)$ . For real and equal roots, $D = b^{2} - 4 a c = 0$ :
  $36 (k + 1)^{2} - 4 (k + 1) \cdot 3 (k + 9) = 0$
  2 marks
2. $36 k^{2} + 72 k + 36 - 12 k^{2} - 120 k - 108 = 0 \Rightarrow 24 k^{2} - 48 k - 72 = 0 \Rightarrow k^{2} - 2 k - 3 = 0$
  1½ marks
3. $(k - 3) (k + 1) = 0 \Rightarrow k = 3$ or $k = - 1$
  1 mark
4. Since $k \neq = - 1$ , $k = 3$ .
  ½ mark
Answer: k = 3
Where the marks are
- Uses D = 0 and substitutes correctly: $36 (k + 1)^{2} - 12 (k + 1) (k + 9) = 0$ (2 marks)
- Simplifies to $k^{2} - 2 k - 3 = 0$ (1½ marks)
- Factorises $(k - 3) (k + 1) = 0$ (1 mark)
- Rejects $k = - 1$ and gives $k = 3$ (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerQuadratic Equations
The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find their present ages.
Try it first, then show our solutionSolution
1. Let the present age of the son be $x$ years. Then the present age of the man $= 2 x^{2}$ years.
  1 mark
2. After 8 years: man's age $= 2 x^{2} + 8$ , son's age $= x + 8$ . According to the question,
  $2 x^{2} + 8 = 3 (x + 8) + 4$
  1 mark
3. $2 x^{2} + 8 = 3 x + 28 \Rightarrow 2 x^{2} - 3 x - 20 = 0$
  1 mark
4. $2 x^{2} - 8 x + 5 x - 20 = 0 \Rightarrow (2 x + 5) (x - 4) = 0 \Rightarrow x = 4$ or $x = - \frac{5}{2}$
  1 mark
5. Age cannot be negative, so $x \neq = - \frac{5}{2}$ ; hence $x = 4$ .
  ½ mark
6. Present age of son = 4 years; present age of man $= 2 \times 4^{2} = 32$ years. (Check: after 8 years, 40 = 3 × 12 + 4 ✓)
  ½ mark
Answer: Son: 4 years; Man: 32 years
Where the marks are
- Takes son's age x and man's age $2 x^{2}$ (1 mark)
- Forms equation $2 x^{2} + 8 = 3 (x + 8) + 4$ (1 mark)
- Gets $2 x^{2} - 3 x - 20 = 0$ (1 mark)
- Factorises $(2 x + 5) (x - 4) = 0$ (1 mark)
- Rejects negative root, x = 4 (½ mark)
- States ages 4 years and 32 years (½ mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerSome Applications of Trigonometry
From a point on a bridge across the river, the angles of depressions of the banks on opposite sides of the river are $3 0^{\circ}$ and $6 0^{\circ}$ respectively. If the bridge is at a height of 4 m from the banks, find the width of the river.
Try it first, then show our solutionSolution
1. Figure: P is the point on the bridge, Q is the foot of the perpendicular from P on the line of the banks, PQ = 4 m. A and B are the banks on opposite sides of Q. Angles of depression from P are 30° to A and 60° to B, so $∠ P A Q = 3 0^{\circ}$ and $∠ P B Q = 6 0^{\circ}$ [alternate angles]. Let AQ = x, QB = y; width AB = x + y.
  1 mark
2. In right $△ P Q A$ : $tan 3 0^{\circ} = \frac{P Q}{A Q} \Rightarrow \frac{1}{3} = \frac{4}{x}$
  1 mark
3. $x = 43$ m
  ½ mark
4. In right $△ P QB$ : $tan 6 0^{\circ} = \frac{P Q}{QB} \Rightarrow 3 = \frac{4}{y}$
  1 mark
5. $y = \frac{4}{3}$ m
  ½ mark
6. Width of river $= x + y = 43 + \frac{4}{3} = \frac{12 + 4}{3} = \frac{16}{3} = \frac{16 3}{3}$ m ( $\approx 9.24$ m)
  1 mark
Answer: Width of river $= \frac{16 3}{3}$ m (≈ 9.24 m)
Where the marks are
- Correct labelled figure (1 mark)
- $tan 3 0^{\circ} = 4/ x$ (1 mark)
- $x = 43$ (½ mark)
- $tan 6 0^{\circ} = 4/ y$ (1 mark)
- $y = 4/ 3$ (½ mark)
- Width $= 163 /3$ m (1 mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerTriangles
In the given figure, $△ F E C ≅ △ G D B$ and $∠1 = ∠2$ . Prove that $△ A D E \sim △ A B C$ .
Figure in the paper: Triangle ABC with apex A at top. D lies on AB and E lies on AC, with DE joined. Angles 1 (at D, between AD and DE) and 2 (at E, between DE and AE) are marked. Line FG is the base line; B and C lie on it between F and G. Segments FE and GD cross the figure; F is left of B, G is right of C, and FE and GD are drawn as lines from F to E and from G to D. Angles 3 and 4 are marked at B and C. See it in the official paper.
Try it first, then show our solutionSolution
1. $△ F E C ≅ △ G D B$ (given), with F↔G, E↔D, C↔B. So $∠ E C F = ∠ D B G$ [CPCT]. Since F, B, C, G are collinear, $∠ E C F = ∠4$ and $∠ D B G = ∠3$ . Hence $∠3 = ∠4$ .
  1 mark
2. In $△ A B C$ , $∠3 = ∠4 \Rightarrow A B = A C$ ...(i) [sides opposite equal angles]
  1 mark
3. In $△ A D E$ , $∠1 = ∠2 \Rightarrow A D = A E$ ...(ii) [sides opposite equal angles]
  1 mark
4. Dividing (ii) by (i): $\frac{A D}{A B} = \frac{A E}{A C}$ , so D and E divide AB and AC in the same ratio; hence $D E ∥ B C$ [converse of BPT].
  1 mark
5. As $D E ∥ B C$ : $∠1 = ∠3$ and $∠2 = ∠4$ [corresponding angles]. Therefore $△ A D E \sim △ A B C$ [AA similarity].
  1 mark
Answer: $△ A D E \sim △ A B C$ (by AA similarity), hence proved.
Where the marks are
- Uses congruence (CPCT) to get $∠3 = ∠4$ (1 mark)
- Deduces AB = AC (1 mark)
- Deduces AD = AE from $∠1 = ∠2$ (1 mark)
- Gets $A D / A B = A E / A C$ and $D E ∥ B C$ (1 mark)
- Corresponding angles equal and concludes similarity (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerTriangles
Sides AB and AC and median AD of a $△ A B C$ are respectively proportional to sides PQ and PR and median PM of another $△ P QR$ . Show that $△ A B C \sim △ P QR$ .
Try it first, then show our solutionSolution
1. Figure: $△ A B C$ with median AD (D midpoint of BC) and $△ P QR$ with median PM (M midpoint of QR). Mark $∠ B A D = ∠1$ , $∠ D A C = ∠2$ , $∠ QP M = ∠3$ , $∠ M P R = ∠4$ . Given: $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ .
  1 mark
2. Construction: Produce AD to E such that AD = DE and join EC. Produce PM to L such that PM = ML and join LR.
  ½ mark
3. In $△ A B D$ and $△ E C D$ : BD = DC, AD = DE, $∠ A D B = ∠ E D C$ [vertically opposite]. So $△ A B D ≅ △ E C D$ [SAS] $\Rightarrow A B = E C$ and $∠ B A D = ∠ C E D$ [CPCT]. Similarly, $△ P QM ≅ △ L R M$ , so $P Q = L R$ and $∠ QP M = ∠ R L M$ .
  1 mark
4. Now $\frac{E C}{L R} = \frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M} = \frac{2 A D}{2 P M} = \frac{A E}{P L}$ . Hence $△ A E C \sim △ P L R$ [SSS similarity].
  1 mark
5. Therefore $∠ C A E = ∠ R P L$ , i.e. $∠2 = ∠4$ , and $∠ A E C = ∠ P L R$ .
  ½ mark
6. Also $∠1 = ∠ B A D = ∠ C E D = ∠ R L M = ∠ QP M = ∠3$ . Adding, $∠1 + ∠2 = ∠3 + ∠4$ , i.e. $∠ B A C = ∠ QP R$ .
  ½ mark
7. In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{A C}{P R}$ and included $∠ B A C = ∠ QP R$ . Hence $△ A B C \sim △ P QR$ [SAS similarity].
  ½ mark
Answer: $△ A B C \sim △ P QR$ (by SAS similarity), hence proved.
Where the marks are
- Correct figure with given data (1 mark)
- Construction: produce medians to double length (½ mark)
- Proves $△ A B D ≅ △ E C D$ , so AB = EC (and PQ = LR) (1 mark)
- Shows $△ A E C \sim △ P L R$ by SSS (1 mark)
- Gets $∠2 = ∠4$ (½ mark)
- Gets $∠1 = ∠3$ and adds to get $∠ B A C = ∠ QP R$ (½ mark)
- Concludes $△ A B C \sim △ P QR$ by SAS (½ mark)
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Q35 · 5 marks · Long answerSurface Areas and Volumes
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is $5 \cdot 8$ cm and its base is of radius $2 \cdot 1$ cm, find the total surface area of the article.
Figure in the paper: A solid cylinder standing upright with a hemispherical cavity scooped out of the top end and another hemispherical cavity scooped out of the bottom end. The cavities are shown as shaded hemispherical bowls at both ends. See it in the official paper.
Try it first, then show our solutionSolution
1. TSA of article = CSA of cylinder + CSA of two hemispherical cavities (r = 2.1 cm, h = 5.8 cm).
  CSA of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 2.1 \times 5.8$
  1 mark
2. $= 13.2 \times 5.8 = 76.56$ cm²
  1 mark
3. CSA of two hemispheres $= 2 \times 2 π r^{2} = 4 \times \frac{22}{7} \times 2.1 \times 2.1$
  1 mark
4. $= 4 \times 13.86 = 55.44$ cm²
  1 mark
5. Total surface area of article $= 76.56 + 55.44 = 132$ cm²
  1 mark
Answer: 132 cm²
Where the marks are
- Writes CSA of cylinder $2 π r h$ with values (1 mark)
- CSA of cylinder = 76.56 cm² (1 mark)
- Writes CSA of two hemispheres $4 π r^{2}$ with values (1 mark)
- = 55.44 cm² (1 mark)
- TSA = 132 cm² (1 mark)
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Section E

Case study based; internal choice in the 2-mark sub-part of each case

Q36 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
Case Study – 1
Essel World is one of India's largest amusement parks that offers a diverse range of thrilling rides, water attractions and entertainment options for visitors of all ages. The park is known for its iconic "Water Kingdom" section, making it a popular destination for family outings and fun-filled adventure. The ticket charges for the park are ₹ 150 per child and ₹ 250 per adult.
On a day, the cashier of the park found that 300 tickets were sold and an amount of ₹ 55,000 was collected.
Based on the above, answer the following questions :
1. (i)
  If the number of children visited be $x$ and the number of adults visited be $y$ , then write the given situation algebraically.
  1 mark
2. (ii)(a)
  How many children visited the amusement park that day ?
  2 marks
3. OR
  (ii)(b)
  How many adults visited the amusement park that day ?
  2 marks
4. (iii)
  How much amount will be collected if 250 children and 100 adults visit the amusement park ?
  1 mark
Try it first, then show our solutionSolution
1. (i) Total number of tickets sold is 300: $x + y = 300$ ...(1)
  ½ mark
2. (i) Total amount collected at ₹150 per child and ₹250 per adult: $150 x + 250 y = 55000$ , i.e. $3 x + 5 y = 1100$ ...(2)
  ½ mark
3. (ii)(a) From (1), $x = 300 - y$ . Substituting in (2): $3 (300 - y) + 5 y = 1100 \Rightarrow 900 + 2 y = 1100 \Rightarrow y = 100$ .
  1 mark
4. (ii)(a) So $x = 300 - 100 = 200$ . 200 children visited the park that day.
  1 mark
5. (ii)(b) From (1), $x = 300 - y$ . Substituting in (2): $900 - 3 y + 5 y = 1100 \Rightarrow 2 y = 200$ .
  1 mark
6. (ii)(b) So $y = 100$ . 100 adults visited the park that day. (Check: $150 (200) + 250 (100) = 30000 + 25000 = 55000$ ✓)
  1 mark
7. (iii) Amount collected $= 250 \times 150 + 100 \times 250 = 37500 + 25000 =$ ₹62,500.
  1 mark
Answer: (i) $x + y = 300$ , $150 x + 250 y = 55000$ ; (ii)(a) 200 children; (ii)(b) 100 adults; (iii) ₹62,500
Where the marks are
- Writes $x + y = 300$ (½ mark)
- Writes $150 x + 250 y = 55000$ (½ mark)
- (ii)(a) Solves the pair of equations correctly to get 200 children (2 marks)
- (ii)(b) Solves the pair of equations correctly to get 100 adults (2 marks)
- (iii) Computes $250 \times 150 + 100 \times 250 = ₹62500$ (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Case Study – 2
A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at the distance of 1 m from each other. He wants to decorate the garden with rose plants. He chose a triangular region inside the garden to grow rose plants. In the above situation, the gardener took help from the students of class 10. They made a chart for it which looks like the given figure.
Based on the above, answer the following questions :
1. (i)
  If A is taken as origin, what are the coordinates of the vertices of $△$ PQR ?
  1 mark
2. (ii)(a)
  Find distances PQ and QR.
  2 marks
3. OR
  (ii)(b)
  Find the coordinates of the point which divides the line segment joining points P and R in the ratio 2 : 1 internally.
  2 marks
4. (iii)
  Find out if $△$ PQR is an isosceles triangle.
  1 mark
Figure in the paper: A square garden ABCD with A at bottom-left, B at top-left, C at top-right and D at bottom-right. Ashoka saplings are along the boundary at 1 m intervals. A grid of 1 m squares is drawn inside. A shaded triangle PQR is inside the garden: P is above and to the right of Q, R is to the right of P at the same level as the grid line near P's lower height, with Q at the bottom. Labels P, Q and R mark the vertices. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Taking A as origin, AD along the x-axis and AB along the y-axis (each grid square = 1 m), the vertices are $P (4, 6)$ , $Q (3, 2)$ , $R (6, 5)$ .
  1 mark
2. (ii)(a) By the distance formula, $P Q = (4 - 3)^{2} + (6 - 2)^{2} = 1 + 16 = 17$ m.
  1 mark
3. (ii)(a) $QR = (6 - 3)^{2} + (5 - 2)^{2} = 9 + 9 = 18 = 32$ m.
  1 mark
4. (ii)(b) By the section formula with $m : n = 2 : 1$ , $P (4, 6)$ , $R (6, 5)$ , the point is $(\frac{2 \times 6 + 1 \times 4}{3}, \frac{2 \times 5 + 1 \times 6}{3})$ .
  1 mark
5. (ii)(b) $= (\frac{16}{3}, \frac{16}{3})$ .
  1 mark
6. (iii) $P R = (6 - 4)^{2} + (5 - 6)^{2} = 4 + 1 = 5$ m, and from (ii)(a) $P Q = 17$ m, $QR = 18$ m.
  ½ mark
7. (iii) Since $P Q \neq = QR \neq = P R$ , no two sides are equal, so $△ P QR$ is not an isosceles triangle.
  ½ mark
Answer: (i) P(4, 6), Q(3, 2), R(6, 5); (ii)(a) PQ = √17 m, QR = √18 = 3√2 m; (ii)(b) (16/3, 16/3); (iii) PR = √5, all sides unequal, so not isosceles
Where the marks are
- (i) Writes the correct coordinates of P, Q and R (1 mark)
- (ii)(a) $P Q = 17$ (1 mark)
- (ii)(a) $QR = 18$ (1 mark)
- (ii)(b) Correct substitution in the section formula (1 mark)
- (ii)(b) Gets $(16/3, 16/3)$ (1 mark)
- (iii) Finds $P R = 5$ (½ mark)
- (iii) Concludes the sides are unequal, so the triangle is not isosceles (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedStatistics
Case Study – 3
Activities like running or cycling reduce stress and the risk of mental disorders like depression. Running helps build endurance. Children develop stronger bones and muscles and are less prone to gain weight. The physical education teacher of a school has decided to conduct an inter school running tournament in his school premises. The time taken by a group of students to run 100 m, was noted as follows :
Time (in seconds) 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
Number of students 8 10 13 6 3
Based on the above, answer the following questions :
1. (i)
  What is the median class of the above given data ?
  1 mark
2. (ii)(a)
  Find the mean time taken by the students to finish the race.
  2 marks
3. OR
  (ii)(b)
  Find the mode of the above given data.
  2 marks
4. (iii)
  How many students took time less than 60 seconds ?
  1 mark
Figure in the paper: A photograph, taken from above, of runners starting a 100 m race on a red athletics track with white lane lines and a starting line; the runners are in their starting positions. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The cumulative frequencies are 8, 18, 31, 37, 40, so $N = 40$ and $\frac{N}{2} = 20$ .
  ½ mark
2. (i) The first cf greater than 20 is 31, for class 40–60. So the median class is 40–60.
  ½ mark
3. (ii)(a) Class marks $x_{i}$ are 10, 30, 50, 70, 90. The products $f_{i} x_{i}$ are $8 \times 10 = 80$ , $10 \times 30 = 300$ , $13 \times 50 = 650$ , $6 \times 70 = 420$ and $3 \times 90 = 270$ . So $\sum f_{i} = 40$ and $\sum f_{i} x_{i} = 1720$ .
  1½ marks
4. (ii)(a) Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{1720}{40} = 43$ seconds.
  ½ mark
5. (ii)(b) The highest frequency is 13, so the modal class is 40–60. Here $l = 40$ , $f_{1} = 13$ , $f_{0} = 10$ , $f_{2} = 6$ , $h = 20$ .
  ½ mark
6. (ii)(b) Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{13 - 10}{26 - 10 - 6} \times 20$
  1 mark
7. (ii)(b) $= 40 + \frac{3}{10} \times 20 = 46$ seconds.
  ½ mark
8. (iii) Students taking less than 60 s $= 8 + 10 + 13 = 31$ (the cf up to 60). So 31 students.
  1 mark
Answer: (i) 40–60; (ii)(a) Mean = 43 s; (ii)(b) Mode = 46 s; (iii) 31 students
Where the marks are
- (i) Correct cumulative frequencies (½ mark)
- (i) Median class 40–60 (½ mark)
- (ii)(a) Correct table of $x_{i}$ and $f_{i} x_{i}$ with $\sum f_{i} x_{i} = 1720$ (1½ marks)
- (ii)(a) Mean $= 1720/40 = 43$ (½ mark)
- (ii)(b) Modal class 40–60 (½ mark)
- (ii)(b) Correct substitution in the mode formula (1 mark)
- (ii)(b) Mode = 46 (½ mark)
- (iii) 31 students (1 mark)
Open Q38 on its own page

Time (in seconds)	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Number of students	8	10	13	6	3

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