MathsCBSE Class 10 · Standard · Sample paper, session 2025-26

CBSE Class 10 Mathematics Standard: sample paper 2025-26

The sample question paper CBSE published for the 2025-26 session, 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 sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassX_2025_26/MathsStandard-SQP.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	VSA
C	26-31	SA
D	32-35	LA
E	36-38	Case study, sub parts 1, 1 and 2 marks

Instructions in brief: All questions compulsory; no overall choice. Internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D, and in all the 2-mark parts of Section E. Take pi = 22/7 where required if not stated. Calculators not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If $a = 2^{2} \times 3^{x}$ , $b = 2^{2} \times 3 \times 5$ , $c = 2^{2} \times 3 \times 7$ and LCM $(a, b, c) = 3780$ , then $x$ is equal to
1. (A)1
2. (B)2
3. (C)3
4. (D)0
Try it first, then show our solutionSolution
1. The LCM takes the highest power of each prime. Assuming $x \geq 1$ : LCM $(a, b, c) = 2^{2} \times 3^{x} \times 5 \times 7 = 140 \times 3^{x}$
2. Set $140 \times 3^{x} = 3780$ . Then $3^{x} = \frac{3780}{140} = 27 = 3^{3}$ , so $x = 3$ .
  1 mark
3. Hence the answer is option (C).
Answer: x = 3, option (C)
Where the marks are
- Writes the LCM as $140 \times 3^{x}$ , equates it to 3780 and gets $x = 3$ , option (C) (1 mark)
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Q2 · 1 mark · MCQCoordinate Geometry
The shortest distance (in units) of the point (2,3) from y-axis is
1. (A)2
2. (B)3
3. (C)5
4. (D)1
Try it first, then show our solutionSolution
1. The shortest distance from a point to the y-axis is the perpendicular distance, which equals $∣ x -coordinate ∣$ .
2. For $(2, 3)$ this distance is $∣2∣ = 2$ units, so the answer is option (A).
  1 mark
Answer: 2 units, option (A)
Where the marks are
- Uses the fact that the distance from the y-axis is |x| and gets 2, option (A) (1 mark)
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Q3 · 1 mark · MCQPair of Linear Equations in Two Variables
If the lines given by $3 x + 2 k y = 2$ and $2 x + 5 y + 1 = 0$ are not parallel, then k has to be
1. (A) $\frac{15}{4}$
2. (B) $\neq = \frac{15}{4}$
3. (C)any rational number
4. (D)any rational number having 4 as denominator
Try it first, then show our solutionSolution
1. The lines are not parallel exactly when $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ . Here $a_{1} = 3, b_{1} = 2 k, a_{2} = 2, b_{2} = 5$ .
2. So we need $\frac{3}{2} \neq = \frac{2 k}{5}$ , which gives $4 k \neq = 15$ , i.e. $k \neq = \frac{15}{4}$ . The answer is option (B).
  1 mark
Answer: k ≠ 15/4, option (B)
Where the marks are
- Applies the condition $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ and gets $k \neq = \frac{15}{4}$ , option (B) (1 mark)
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Q4 · 1 mark · MCQCircles
A quadrilateral ABCD is drawn to circumscribe a circle. If BC=7cm, CD=4cm and AD=3cm, then the length of AB is
1. (A)3cm
2. (B)4cm
3. (C)6cm
4. (D)7cm
Try it first, then show our solutionSolution
1. For a quadrilateral circumscribing a circle, $A B + C D = A D + B C$ , because tangents from an external point are equal.
2. Substituting: $A B + 4 = 3 + 7$ , so $A B = 6$ cm. The answer is option (C).
  1 mark
Answer: AB = 6 cm, option (C)
Where the marks are
- Uses $A B + C D = A D + B C$ and gets AB = 6 cm, option (C) (1 mark)
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Q5 · 1 mark · MCQIntroduction to Trigonometry
If $sec θ + tan θ = x$ , then $sec θ - tan θ$ will be
1. (A) $x$
2. (B) $x^{2}$
3. (C) $\frac{2}{x}$
4. (D) $\frac{1}{x}$
Try it first, then show our solutionSolution
1. Use the identity $sec^{2} θ - tan^{2} θ = 1$ , which factorises as $(sec θ + tan θ) (sec θ - tan θ) = 1$ .
2. So $x (sec θ - tan θ) = 1$ , giving $sec θ - tan θ = \frac{1}{x}$ . The answer is option (D).
  1 mark
Answer: 1/x, option (D)
Where the marks are
- Uses $sec^{2} θ - tan^{2} θ = 1$ to get $\frac{1}{x}$ , option (D) (1 mark)
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Q6 · 1 mark · MCQQuadratic Equations
Which one of the following is not a quadratic equation?
1. (A) $(x + 2)^{2} = 2 (x + 3)$
2. (B) $x^{2} + 3 x = (- 1) (1 - 3 x)^{2}$
3. (C) $x^{3} - x^{2} + 2 x + 1 = (x + 1)^{3}$
4. (D) $(x + 2) (x + 1) = x^{2} + 2 x + 3$
Try it first, then show our solutionSolution
1. Simplify each option. (A) $x^{2} + 4 x + 4 = 2 x + 6$ gives $x^{2} + 2 x - 2 = 0$ , which is quadratic. (B) $x^{2} + 3 x = - (1 - 6 x + 9 x^{2})$ gives $10 x^{2} - 3 x + 1 = 0$ , which is quadratic. (C) $x^{3} - x^{2} + 2 x + 1 = x^{3} + 3 x^{2} + 3 x + 1$ gives $4 x^{2} + x = 0$ , which is quadratic.
2. (D) $x^{2} + 3 x + 2 = x^{2} + 2 x + 3$ gives $x - 1 = 0$ . This has degree 1, so it is not a quadratic equation. The answer is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Simplifies (D) to the linear equation $x - 1 = 0$ and concludes it is not quadratic, option (D) (1 mark)
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Q7 · 1 mark · MCQCompetency-basedAreas Related to Circles
Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1cm. Total area of all the dotted regions (assuming the thickness of the rings to be negligible) is
For Visually Impaired candidates: The area of the circle that can be inscribed in a square of 6 cm is (A) $36 π$ cm $^{2}$ (B) $18 π$ cm $^{2}$ (C) $12 π$ cm $^{2}$ (D) $9 π$ cm $^{2}$
1. (A) $4 [\frac{π}{12} - \frac{3}{4}]$ cm $^{2}$
2. (B) $[\frac{π}{6} - \frac{3}{4}]$ cm $^{2}$
3. (C) $4 [\frac{π}{6} - \frac{3}{4}]$ cm $^{2}$
4. (D) $8 [\frac{π}{6} - \frac{3}{4}]$ cm $^{2}$
Figure in the paper: Five interlocking Olympic rings (three on top row: blue, black, red; two below: yellow, green). Adjacent rings intersect; the lens-shaped overlapping regions are dotted. Points A and B are marked at the intersection points of the blue and yellow rings, with chord AB drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. Adjacent rings overlap in 4 places (blue–yellow, yellow–black, black–green, green–red). Each lens-shaped overlap is made of 2 circular segments, so there are 8 segments in all.
2. The chord AB = 1 cm equals the radius. So triangle OAB is equilateral and the central angle is $6 0^{\circ}$ .
3. Area of one segment $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} π (1)^{2} - \frac{3}{4} (1)^{2} = \frac{π}{6} - \frac{3}{4}$ . Total area $= 8 [\frac{π}{6} - \frac{3}{4}]$ cm $^{2}$ , which is option (D).
  1 mark
4. For Visually Impaired candidates: the circle inscribed in a square of side 6 cm has radius 3 cm. Its area is $π (3)^{2} = 9 π$ cm $^{2}$ , which is option (D).
Answer: $8 [\frac{π}{6} - \frac{3}{4}]$ cm², option (D) (VI: 9π cm², option (D))
Where the marks are
- Finds the 60° segment area, multiplies by 8 segments and selects option (D) (1 mark)
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Q8 · 1 mark · MCQProbability
A pair of dice is tossed. The probability of not getting the sum eight is
1. (A) $\frac{5}{36}$
2. (B) $\frac{31}{36}$
3. (C) $\frac{5}{18}$
4. (D) $\frac{5}{9}$
Try it first, then show our solutionSolution
1. Tossing two dice gives 36 equally likely outcomes. A sum of 8 comes from (2,6), (3,5), (4,4), (5,3) and (6,2), which is 5 outcomes. So P(sum 8) $= \frac{5}{36}$ .
2. P(not sum 8) $= 1 - \frac{5}{36} = \frac{31}{36}$ . The answer is option (B).
  1 mark
Answer: 31/36, option (B)
Where the marks are
- Counts 5 favourable outcomes and uses the complement to get $\frac{31}{36}$ , option (B) (1 mark)
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Q9 · 1 mark · MCQIntroduction to Trigonometry
If $2 sin 5 x = 3$ , $0^{\circ} \leq x \leq 9 0^{\circ}$ , then $x$ is equal to
1. (A) $1 0^{\circ}$
2. (B) $1 2^{\circ}$
3. (C) $2 0^{\circ}$
4. (D) $5 0^{\circ}$
Try it first, then show our solutionSolution
1. $2 sin 5 x = 3$ gives $sin 5 x = \frac{3}{2} = sin 6 0^{\circ}$ .
2. So $5 x = 6 0^{\circ}$ and $x = 1 2^{\circ}$ . The answer is option (B).
  1 mark
Answer: x = 12°, option (B)
Where the marks are
- Identifies $sin 6 0^{\circ} = \frac{3}{2}$ , solves $5 x = 6 0^{\circ}$ to get $1 2^{\circ}$ , option (B) (1 mark)
Q10 · 1 mark · MCQReal Numbers
The sum of two numbers is 1215 and their HCF is 81, then the possible pairs of such numbers are
1. (A)2
2. (B)3
3. (C)4
4. (D)5
Try it first, then show our solutionSolution
1. Since the HCF is 81, write the numbers as $81 x$ and $81 y$ , where $x$ and $y$ are co-prime. Then $81 x + 81 y = 1215$ , so $x + y = 15$ .
2. The co-prime pairs with sum 15 are (1,14), (2,13), (4,11) and (7,8). The pairs (3,12), (5,10) and (6,9) are rejected because they share a common factor. So there are 4 pairs, which is option (C).
  1 mark
Answer: 4 pairs, option (C)
Where the marks are
- Reduces to $x + y = 15$ with x, y co-prime and counts 4 pairs, option (C) (1 mark)
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Q11 · 1 mark · MCQSurface Areas and Volumes
If the area of the base of a right circular cone is 51cm $^{2}$ and it's volume is 85cm $^{2}$ , then the height of the cone is given as
1. (A) $\frac{5}{6}$ cm
2. (B) $\frac{5}{3}$ cm
3. (C) $\frac{5}{2}$ cm
4. (D)5cm
Try it first, then show our solutionSolution
1. $V = \frac{1}{3} \times (base area) \times h$ , so $85 = \frac{1}{3} \times 51 \times h$ .
2. $h = \frac{3 \times 85}{51} = 5$ cm. The answer is option (D). (The volume should be read as 85 cm $^{3}$ .)
  1 mark
Answer: h = 5 cm, option (D)
Where the marks are
- Uses $V = \frac{1}{3} π r^{2} h$ with $π r^{2} = 51$ and gets h = 5 cm, option (D) (1 mark)
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Q12 · 1 mark · MCQPolynomials
If zeroes of the quadratic polynomial $a x^{2} + b x + c$ (a, c ≠0) are equal, then
1. (A)c and b must have opposite signs
2. (B)c and a must have opposite signs
3. (C)c and b must have same signs
4. (D)c and a must have same signs
Try it first, then show our solutionSolution
1. Equal zeroes means the discriminant is zero: $b^{2} - 4 a c = 0$ , so $a c = \frac{b ^{2}}{4}$ .
2. Since $a, c \neq = 0$ , $b \neq = 0$ , so $b^{2} > 0$ and hence $a c > 0$ . Therefore a and c have the same sign, which is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Uses $b^{2} = 4 a c$ to deduce $a c > 0$ , so a and c have the same sign, option (D) (1 mark)
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Q13 · 1 mark · MCQAreas Related to Circles
The area (in cm $^{2}$ ) of a sector of a circle of radius 21cm cut off by an arc of length 22cm is
1. (A)441
2. (B)321
3. (C)231
4. (D)221
Try it first, then show our solutionSolution
1. Area of a sector $= \frac{1}{2} \times l \times r$ , where $l$ is the arc length.
2. $= \frac{1}{2} \times 22 \times 21 = 231$ cm $^{2}$ . The answer is option (C).
  1 mark
Answer: 231 cm², option (C)
Where the marks are
- Uses area $= \frac{1}{2} l r$ to get 231, option (C) (1 mark)
Q14 · 1 mark · MCQTriangles
If $△ A B C \sim △ D E F$ , AB=6cm, DE=9cm, EF=6cm and FD=12cm, then the perimeter of $△ A B C$ is
1. (A)28cm
2. (B)28.5cm
3. (C)18cm
4. (D)23cm
Try it first, then show our solutionSolution
1. For similar triangles, the ratio of perimeters equals the ratio of corresponding sides: $\frac{A B}{D E} = \frac{6}{9} = \frac{2}{3}$ .
2. Perimeter of DEF $= 9 + 6 + 12 = 27$ cm. So the perimeter of ABC $= \frac{2}{3} \times 27 = 18$ cm, which is option (C).
  1 mark
Answer: 18 cm, option (C)
Where the marks are
- Uses ratio of perimeters = ratio of sides and gets 18 cm, option (C) (1 mark)
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Q15 · 1 mark · MCQProbability
If the probability of the letter chosen at random from the letters of the word “Mathematics” to be a vowel is $\frac{2}{2 x + 1}$ , then $x$ is equal to
1. (A) $\frac{4}{11}$
2. (B) $\frac{9}{4}$
3. (C) $\frac{11}{4}$
4. (D) $\frac{4}{9}$
Try it first, then show our solutionSolution
1. MATHEMATICS has 11 letters, and the vowels are A, E, A, I (4 letters). So P(vowel) $= \frac{4}{11}$ .
2. $\frac{2}{2 x + 1} = \frac{4}{11}$ gives $8 x + 4 = 22$ , so $x = \frac{9}{4}$ . The answer is option (B).
  1 mark
Answer: x = 9/4, option (B)
Where the marks are
- Finds P(vowel) $= \frac{4}{11}$ , equates it to $\frac{2}{2 x + 1}$ and gets $x = \frac{9}{4}$ , option (B) (1 mark)
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Q16 · 1 mark · MCQCoordinate Geometry
The points A(9,0), B(9,-6), C(-9,0) and D(-9,6) are the vertices of a
1. (A)Square
2. (B)Rectangle
3. (C)Parallelogram
4. (D)Trapezium
Try it first, then show our solutionSolution
1. AB joins (9,0) and (9,−6), and CD joins (−9,0) and (−9,6). Both are vertical with length 6, so AB ∥ CD and AB = CD.
2. Slope of BC $= \frac{0 - ( - 6 )}{- 9 - 9} = - \frac{1}{3}$ and slope of DA $= \frac{0 - 6}{9 - ( - 9 )} = - \frac{1}{3}$ , so BC ∥ DA. Both pairs of opposite sides are parallel.
3. Diagonal AC $= 18$ and diagonal BD $= 1 8^{2} + 1 2^{2} = 468$ . These are unequal, so ABCD is not a rectangle or a square. It is a parallelogram, option (C).
  1 mark
Answer: Parallelogram, option (C)
Where the marks are
- Shows both pairs of opposite sides are parallel and the diagonals are unequal, so ABCD is a parallelogram, option (C) (1 mark)
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Q17 · 1 mark · MCQStatistics
The median of a set of 9 distinct observation is 20.5. If each of the observations of a set is increased by 2, then the median of a new set
1. (A)is increased by 2
2. (B)is decreased by 2
3. (C)is two times the original number
4. (D)Remains same as that of original observations
Try it first, then show our solutionSolution
1. Adding 2 to every observation does not change their order, so the same middle (5th) observation stays the median.
2. That observation has also increased by 2, so the new median is $20.5 + 2 = 22.5$ . The median increases by 2, which is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Reasons that the order is unchanged, so the median increases by 2, option (A) (1 mark)
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Q18 · 1 mark · MCQCircles
The length of a tangent drawn to a circle of radius 9 cm from a point at a distance of 41cm from the centre of the circle is
1. (A)40cm
2. (B)9cm
3. (C)41cm
4. (D)50cm
Try it first, then show our solutionSolution
1. The tangent is perpendicular to the radius at the point of contact T, so triangle OTP is right-angled at T.
2. $T P = 4 1^{2} - 9^{2} = 1681 - 81 = 1600 = 40$ cm. The answer is option (A).
  1 mark
Answer: 40 cm, option (A)
Where the marks are
- Uses tangent ⟂ radius and Pythagoras to get 40 cm, option (A) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
Assertion (A): The number $5^{n}$ cannot end with the digit 0, where n is a natural number
Reason (R): A number ends with 0, if its prime factorization contains both 2 and 5
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 and 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. Reason (R) is true: a number ends in 0 only if it is divisible by 10 = 2 × 5, so its prime factorisation must contain both 2 and 5.
2. $5^{n}$ has only the prime 5 in its factorisation and no 2, so by uniqueness of prime factorisation it cannot end in 0. Hence (A) is true and (R) explains it, which is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises that $5^{n}$ has no factor 2, so (A) is true and explained by (R), option (A) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Assertion (A): If $cos A + cos^{2} A = 1$ , then $sin^{2} A + sin^{4} A = 1$
Reason (R): $sin^{2} A + cos^{2} A = 1$
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 and 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. From $cos A + cos^{2} A = 1$ we get $cos A = 1 - cos^{2} A = sin^{2} A$ , using (R).
2. Then $sin^{2} A + sin^{4} A = cos A + (cos A)^{2} = 1$ . So (A) is true, (R) is true, and (R) is the identity used to prove it. The answer is option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Shows $cos A = sin^{2} A$ using (R) and substitutes to prove (A), option (A) (1 mark)
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Section B

VSA

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerArithmetic Progressions
The A.P 8, 10, 12,……. has 60 terms. Find the sum of last 10 terms.
Try it first, then show our solutionSolution
1. Here $a = 8$ , $d = 10 - 8 = 2$ and $n = 60$ .
  ½ mark
2. Last term: $t_{60} = a + 59 d = 8 + 59 (2) = 126$ .
  ½ mark
3. The last 10 terms are $t_{51}$ to $t_{60}$ , with $t_{51} = 8 + 50 (2) = 108$ . Their sum $= \frac{10}{2} (t_{51} + t_{60}) = 5 (108 + 126) = 1170$ .
  1 mark
Answer: 1170
Where the marks are
- Identifies a = 8, d = 2 (½ mark)
- Finds $t_{60} = 126$ (½ mark)
- Finds $t_{51} = 108$ and the sum of the last 10 terms = 1170 (1 mark)
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OR
Q21(B) · 2 marks · Very short answerArithmetic Progressions
Find the middle term of A.P 6,13, 20, ……., 230
Try it first, then show our solutionSolution
1. $a = 6$ , $d = 7$ and the last term is 230. Using $230 = 6 + (n - 1) 7$ : $(n - 1) = 32$ , so $n = 33$ terms.
  1 mark
2. With 33 terms (an odd number), the middle term is the $\frac{33 + 1}{2} = 17$ th term: $t_{17} = 6 + 16 (7) = 118$ .
  1 mark
Answer: Middle term = 118
Where the marks are
- Finds the number of terms n = 33 (1 mark)
- Finds the middle term $t_{17} = 118$ (1 mark)
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Q22 · 2 marks · Very short answerIntroduction to Trigonometry
If $sin (A + B) = 1$ and $cos (A - B) = \frac{3}{2}$ , $0^{\circ} < A, B < 9 0^{\circ}$ , find the measure of angles $A$ and $B$ .
Try it first, then show our solutionSolution
1. $sin (A + B) = 1 = sin 9 0^{\circ}$ , so $A + B = 9 0^{\circ}$ . Also $cos (A - B) = \frac{3}{2} = cos 3 0^{\circ}$ , so $A - B = 3 0^{\circ}$ (taking $A > B$ ).
  1 mark
2. Adding the two equations: $2 A = 12 0^{\circ}$ , so $A = 6 0^{\circ}$ . Then $B = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ .
  1 mark
Answer: A = 60°, B = 30°
Where the marks are
- Forms $A + B = 9 0^{\circ}$ and $A - B = 3 0^{\circ}$ (1 mark)
- Solves to get $A = 6 0^{\circ}$ , $B = 3 0^{\circ}$ (1 mark)
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Q23 · 2 marks · Very short answerTriangles
If AP and DQ are medians of triangles ABC and DEF respectively, where $△ A B C \sim △ D E F$ , then prove that $\frac{A B}{D E} = \frac{A P}{D Q}$
Try it first, then show our solutionSolution
1. Figure: triangles ABC and DEF, with P the midpoint of BC and Q the midpoint of EF; AP and DQ are drawn. Since $△ A B C \sim △ D E F$ , $\frac{A B}{D E} = \frac{B C}{E F}$ [corresponding sides of similar triangles are proportional].
  ½ mark
2. AP and DQ are medians, so $B C = 2 B P$ and $E F = 2 E Q$ . Hence $\frac{A B}{D E} = \frac{2 B P}{2 E Q} = \frac{B P}{E Q}$ .
  ½ mark
3. In $△ A B P$ and $△ D E Q$ : $\frac{A B}{D E} = \frac{B P}{E Q}$ , and $∠ B = ∠ E$ [corresponding angles of similar triangles]. So $△ A B P \sim △ D E Q$ [SAS similarity].
  ½ mark
4. Corresponding sides of the similar triangles ABP and DEQ are proportional. Therefore $\frac{A B}{D E} = \frac{A P}{D Q}$ . Hence proved.
  ½ mark
Answer: Proved: AB/DE = AP/DQ (via △ABP ~ △DEQ by SAS)
Where the marks are
- Writes $\frac{A B}{D E} = \frac{B C}{E F}$ from the similarity (½ mark)
- Uses the medians to get $\frac{A B}{D E} = \frac{B P}{E Q}$ (½ mark)
- Proves $△ A B P \sim △ D E Q$ by SAS using $∠ B = ∠ E$ (½ mark)
- Concludes $\frac{A B}{D E} = \frac{A P}{D Q}$ (½ mark)
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Answer one of these 2 alternatives.
Q24(A) · 2 marks · Very short answerCompetency-basedAreas Related to Circles
A horse, a cow and a goat are tied, each by ropes of length 14m, at the corners A, B and C respectively, of a grassy triangular field ABC with sides of lengths 35m, 40m and 50 m. Find the area of grass field that can be grazed by them.
Try it first, then show our solutionSolution
1. Each animal grazes a sector of radius $r = 14$ m whose central angle is the angle of the triangle at its corner. The sides are 35 m, 40 m and 50 m, each longer than $2 \times 14 = 28$ m, so the three sectors do not overlap. Let the angles at A, B, C be $θ_{1}, θ_{2}, θ_{3}$ .
2. Total grazed area $= \frac{θ _{1}}{36 0 ^{\circ}} π r^{2} + \frac{θ _{2}}{36 0 ^{\circ}} π r^{2} + \frac{θ _{3}}{36 0 ^{\circ}} π r^{2} = \frac{π r ^{2}}{36 0 ^{\circ}} (θ_{1} + θ_{2} + θ_{3})$ . Since $θ_{1} + θ_{2} + θ_{3} = 18 0^{\circ}$ [angle sum property of a triangle], this is $\frac{π r ^{2}}{36 0 ^{\circ}} \times 18 0^{\circ} = \frac{π r ^{2}}{2}$ .
  1 mark
3. $= \frac{1}{2} \times \frac{22}{7} \times 14 \times 14 = 308$ m $^{2}$ . So the three animals can graze 308 m $^{2}$ of the field.
  1 mark
Answer: 308 m²
Where the marks are
- Adds the three sectors and uses the angle sum $18 0^{\circ}$ to get $\frac{18 0 ^{\circ}}{36 0 ^{\circ}} π r^{2}$ (1 mark)
- Calculates the area as 308 m² (1 mark)
Open Q24(A) on its own page
OR
Q24(B) · 2 marks · Very short answerAreas Related to Circles
Find the area of the major segment (in terms of $π$ ) of a circle of radius 5cm, formed by a chord subtending an angle of 90° at the centre.
Try it first, then show our solutionSolution
1. Area of minor segment = area of sector − area of $△$ formed by the two radii and the chord. With $θ = 9 0^{\circ}$ the triangle is right-angled at the centre:
  $= \frac{9 0 ^{\circ}}{36 0 ^{\circ}} π (5)^{2} - \frac{1}{2} \times 5 \times 5 = (\frac{25 π}{4} - \frac{25}{2})$ cm $^{2}$ .
  1 mark
2. Area of major segment = area of circle − area of minor segment
  $= 25 π - (\frac{25 π}{4} - \frac{25}{2}) = (\frac{75 π}{4} + \frac{25}{2})$ cm $^{2}$ .
  1 mark
Answer: $(\frac{75 π}{4} + \frac{25}{2})$ cm²
Where the marks are
- Finds the minor segment area $\frac{25 π}{4} - \frac{25}{2}$ cm² (1 mark)
- Subtracts it from the circle's area to get the major segment $\frac{75 π}{4} + \frac{25}{2}$ cm² (1 mark)
Open Q24(B) on its own page
Q25 · 2 marks · Very short answerCircles
A $△$ ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that ar( $△$ ABC) = 90cm $^{2}$
For Visually Impaired candidates: A circle is inscribed in a right-angled triangle ABC, right angled at B. If BC=7cm and AB=24cm, find the radius of the circle
Figure in the paper: Triangle ABC with an inscribed circle with centre O touching BC at D, CA at E and AB at F. A is at top, C bottom-left, B bottom-right. Lines OA, OB, OC, OD, OE, OF are drawn from centre O. See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents drawn from an external point are equal. So $B F = B D = 10$ cm and $C E = C D = 8$ cm. Let $A E = A F = x$ cm.
  Then $A B = x + 10$ , $A C = x + 8$ and $B C = 18$ cm.
  ½ mark
2. The radii OD, OE, OF are perpendicular to the sides [tangent ⟂ radius]. Hence
  ar( $△ A B C$ ) = ar( $△ A O B$ ) + ar( $△ B O C$ ) + ar( $△ A O C$ ) $= \frac{1}{2} r (A B + B C + A C)$ .
  ½ mark
3. $90 = \frac{1}{2} \times 4 (x + 10 + 18 + x + 8) = 2 (2 x + 36) = 4 x + 72$ , so $4 x = 18$ and $x = 4.5$ cm.
  ½ mark
4. $A B = 4.5 + 10 = 14.5$ cm and $A C = 4.5 + 8 = 12.5$ cm.
  ½ mark
5. For Visually Impaired candidates (alternative question): In right $△ A B C$ with $∠ B = 9 0^{\circ}$ , $A C^{2} = A B^{2} + B C^{2} = 2 4^{2} + 7^{2} = 625$ , so $A C = 25$ cm.
6. (VI) Area $= \frac{1}{2} \times 7 \times 24 = 84$ cm $^{2}$ … (i)
7. (VI) Joining the centre to the vertices, area $= \frac{1}{2} r (24 + 25 + 7) = 28 r$ … (ii)
8. (VI) From (i) and (ii), $28 r = 84$ , so $r = 3$ cm. (This version carries ½ mark for each of these four steps.)
Answer: AB = 14.5 cm, AC = 12.5 cm (VI version: r = 3 cm)
Where the marks are
- Uses equal tangents: BF = 10, CE = 8, AE = AF = x (½ mark)
- Writes the area of ABC as the sum of areas of AOB, BOC and AOC (½ mark)
- Solves $90 = 2 (2 x + 36)$ to get x = 4.5 cm (½ mark)
- States AB = 14.5 cm and AC = 12.5 cm (½ mark)
- VI: AC = 25 cm (½), area 84 cm² (½), area = 28r (½), r = 3 cm (½)
Open Q25 on its own page

Section C

Q26 · 3 marks · Short answerCircles
In Figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that $∠ A O B = 9 0^{\circ}$
For Visually Impaired candidates: Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that $∠ A P B = 2 (∠ O A B)$
Figure in the paper: Circle with centre O between two parallel horizontal tangent lines XY (top, touching at P) and X′Y′ (bottom, touching at Q). A third tangent AB touches the circle at C, meeting XY at A and X′Y′ at B. OA, OB, OC, OP, OQ are joined. See it in the official paper.
Try it first, then show our solutionSolution
1. Join OC. In $△ A P O$ and $△ A C O$ : $A P = A C$ [tangents from external point A], $O P = O C$ [radii], $O A = O A$ [common]. So $△ A P O ≅ △ A C O$ [SSS], and hence $∠ P O A = ∠ C O A$ [CPCT].
  In the same way, $△ B QO ≅ △ B C O$ , so $∠ QO B = ∠ C O B$ .
  1 mark
2. $O P ⊥ X Y$ and $O Q ⊥ X^{'} Y^{'}$ [tangent ⟂ radius]. Since $X Y ∥ X^{'} Y^{'}$ , the points P, O, Q are collinear, so PQ is a diameter. Therefore
  $∠ P O A + ∠ C O A + ∠ C O B + ∠ QO B = 18 0^{\circ}$ .
  1 mark
3. Hence $2∠ C O A + 2∠ C O B = 18 0^{\circ}$ , so $∠ C O A + ∠ C O B = 9 0^{\circ}$ , i.e. $∠ A O B = 9 0^{\circ}$ . Hence proved.
  1 mark
4. For Visually Impaired candidates (alternative): $P A = P B$ [tangents from external point], so $∠ P A B = ∠ P B A = x$ [angles opposite equal sides]. In $△ P A B$ , $∠ A P B = 18 0^{\circ} - 2 x$ … (i). (½ mark in VI version)
5. (VI) $∠ O A P = 9 0^{\circ}$ [radius ⟂ tangent], so $x + ∠ O A B = 9 0^{\circ}$ , giving $x = 9 0^{\circ} - ∠ O A B$ … (ii). (1 mark)
6. (VI) Putting (ii) in (i): $∠ A P B = 18 0^{\circ} - 2 (9 0^{\circ} - ∠ O A B) = 2∠ O A B$ . Hence proved. (1 + ½ marks)
Answer: ∠AOB = 90° (proved); VI version: ∠APB = 2∠OAB (proved)
Where the marks are
- Proves $△ A P O ≅ △ A C O$ (and similarly for the triangles at B), getting equal angles (1 mark)
- Uses PQ as a diameter so that the four angles at O add to 180° (1 mark)
- Concludes $∠ A O B = 9 0^{\circ}$ (1 mark)
- VI: equal tangents and base angles (½), angle APB = 180° − 2x (1), x = 90° − ∠OAB using radius ⟂ tangent (1), conclusion (½)
Open Q26 on its own page
Q27 · 3 marks · Short answerCompetency-basedReal Numbers
In a workshop, the number of teachers of English, Hindi and Science are 36, 60 and 84 respectively. Find the minimum number of rooms required, if in each room the same number of teachers are to be seated and all of them being of the same subject.
Try it first, then show our solutionSolution
1. For the fewest rooms, each room must hold the greatest possible number of teachers, so we need the HCF of 36, 60 and 84.
  $36 = 2^{2} \times 3^{2}$ , $60 = 2^{2} \times 3 \times 5$ , $84 = 2^{2} \times 3 \times 7$ .
  HCF $= 2^{2} \times 3 = 12$ .
  1½ marks
2. Number of rooms $= \frac{36}{12} + \frac{60}{12} + \frac{84}{12} = 3 + 5 + 7$
  1 mark
3. $= 15$ . So the minimum number of rooms required is 15.
  ½ mark
Answer: 15 rooms
Where the marks are
- Finds HCF(36, 60, 84) = 12 (1½ marks)
- Divides each number of teachers by 12 to count the rooms (1 mark)
- States the minimum number of rooms as 15 (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerPolynomials
Find the zeroes of the quadratic polynomial $2 x^{2} - (1 + 22) x + 2$ and verify the relationship between the zeroes and coefficents of the polynomial.
Try it first, then show our solutionSolution
1. Split the middle term: $2 x^{2} - (1 + 22) x + 2 = 2 x^{2} - x - 22 x + 2$ .
  1 mark
2. $= x (2 x - 1) - 2 (2 x - 1) = (2 x - 1) (x - 2)$ .
  So the zeroes are $\frac{1}{2}$ and $2$ .
  1 mark
3. Verification ( $a = 2$ , $b = - (1 + 22)$ , $c = 2$ ):
  Sum of zeroes $= \frac{1}{2} + 2 = \frac{1 + 2 2}{2} = \frac{- b}{a}$ .
  Product of zeroes $= \frac{1}{2} \times 2 = \frac{2}{2} = \frac{c}{a}$ .
  Hence the relationship is verified.
  1 mark
Answer: Zeroes are 1/2 and √2; sum = −b/a and product = c/a verified
Where the marks are
- Splits the middle term correctly (1 mark)
- Factorises and gets the zeroes 1/2 and √2 (1 mark)
- Verifies sum = −b/a and product = c/a (1 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerIntroduction to Trigonometry
If $sin θ + cos θ = 3$ , then prove that $tan θ + cot θ = 1$
Try it first, then show our solutionSolution
1. Squaring both sides: $(sin θ + cos θ)^{2} = 3$ , so $sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 3$ , which gives $1 + 2 sin θ cos θ = 3$ .
  1 mark
2. So $2 sin θ cos θ = 2$ , i.e. $sin θ cos θ = 1$ .
  1 mark
3. $tan θ + cot θ = \frac{sin θ}{cos θ} + \frac{cos θ}{sin θ} = \frac{sin ^{2} θ + cos ^{2} θ}{sin θ cos θ} = \frac{1}{sin θ cos θ} = \frac{1}{1} = 1$ . Hence proved.
  1 mark
Answer: tan θ + cot θ = 1 (proved)
Where the marks are
- Squares the given relation and uses $sin^{2} θ + cos^{2} θ = 1$ (1 mark)
- Gets $sin θ cos θ = 1$ (1 mark)
- Writes $tan θ + cot θ = 1/ (sin θ cos θ) = 1$ (1 mark)
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{c o s A - s i n A + 1}{c o s A + s i n A - 1} = cosec A + cot A$
Try it first, then show our solutionSolution
1. Multiply the numerator and the denominator by $(cos A + sin A + 1)$ :
  LHS $= \frac{( cos A + 1 - sin A ) ( cos A + 1 + sin A )}{( cos A + sin A - 1 ) ( cos A + sin A + 1 )}$ .
  1 mark
2. $= \frac{( cos A + 1 ) ^{2} - sin ^{2} A}{( cos A + sin A ) ^{2} - 1} = \frac{cos ^{2} A + 2 cos A + 1 - sin ^{2} A}{2 sin A cos A}$ [using $sin^{2} A + cos^{2} A = 1$ in the denominator].
  1 mark
3. In the numerator, $1 - sin^{2} A = cos^{2} A$ , so it becomes $2 cos^{2} A + 2 cos A = 2 cos A (1 + cos A)$ .
  LHS $= \frac{2 cos A ( 1 + cos A )}{2 sin A cos A} = \frac{1 + cos A}{sin A} = \frac{1}{sin A} + \frac{cos A}{sin A} = cosec A + cot A$ = RHS. Hence proved.
  1 mark
Answer: LHS = cosec A + cot A (proved)
Where the marks are
- Multiplies the numerator and denominator by $(cos A + sin A + 1)$ (1 mark)
- Simplifies to $\frac{c o s ^{2} A + 2 c o s A + 1 - s i n ^{2} A}{2 s i n A c o s A}$ (1 mark)
- Reduces this to $\frac{1 + c o s A}{s i n A} = cosec A + cot A$ (1 mark)
Open Q29(B) on its own page
Q30 · 3 marks · Short answerCompetency-basedProbability
On a particular day, Vidhi and Unnati couldn’t decide on who would get to drive the car. They had one coin each and flipped their coin exactly three times. The following was agreed upon:
1. If Vidhi gets two heads in a row, she would drive the car
2. If Unnati gets a head immediately followed by a tail, she would drive the car.
Who has greater probability to drive the car that day? Justify your answer.
Try it first, then show our solutionSolution
1. Three flips of a coin give $2^{3} = 8$ equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
2. Vidhi needs two heads in a row: HHH, HHT, THH, which is 3 outcomes.
  $P (Vidhi drives) = \frac{3}{8}$ .
  1 mark
3. Unnati needs a head immediately followed by a tail: HHT, HTH, HTT, THT, which is 4 outcomes.
  $P (Unnati drives) = \frac{4}{8} = \frac{1}{2}$ .
  1 mark
4. Since $\frac{4}{8} > \frac{3}{8}$ , Unnati has the greater probability of driving the car that day.
  1 mark
Answer: Unnati (4/8 > 3/8)
Where the marks are
- Finds P(Vidhi) = 3/8 with the favourable outcomes listed (1 mark)
- Finds P(Unnati) = 4/8 with the favourable outcomes listed (1 mark)
- Compares the two and concludes that Unnati is more likely to drive (1 mark)
Open Q30 on its own page
Answer one of these 2 alternatives.
Q31(A) · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
The monthly income of Aryan and Babban are in the ratio 3:4 and their monthly expenditures are in ratio 5:7. If each saves ₹ 15,000 per month, find their monthly incomes.
Try it first, then show our solutionSolution
1. Let the monthly incomes of Aryan and Babban be ₹ $3 x$ and ₹ $4 x$ , and their expenditures be ₹ $5 y$ and ₹ $7 y$ .
  1 mark
2. Savings = income − expenditure, so
  $3 x - 5 y = 15000$ … (1)
  $4 x - 7 y = 15000$ … (2)
  1 mark
3. (1)×4: $12 x - 20 y = 60000$ . (2)×3: $12 x - 21 y = 45000$ . Subtracting gives $y = 15000$ . Then $3 x = 15000 + 75000 = 90000$ , so $x = 30000$ .
  Monthly incomes: Aryan ₹ $3 \times 30000$ = ₹90,000 and Babban ₹ $4 \times 30000$ = ₹1,20,000.
  1 mark
Answer: Aryan ₹90,000; Babban ₹1,20,000
Where the marks are
- Takes incomes as 3x, 4x and expenditures as 5y, 7y (1 mark)
- Forms the equations $3 x - 5 y = 15000$ and $4 x - 7 y = 15000$ (1 mark)
- Solves to x = 30000 and gives the incomes ₹90,000 and ₹1,20,000 (1 mark)
Open Q31(A) on its own page
OR
Q31(B) · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve the following system of equations graphically:
$2 x + y = 6$ , $2 x - y - 2 = 0$ . Find the area of the triangle so formed by two lines and $x$ -axis.
For Visually Impaired candidates: Five years hence, fathers age will be three times the age of son. Five years ago, father was seven times as old as his son. Find their present ages.
Try it first, then show our solutionSolution
1. Table for $2 x + y = 6$ (i.e. $y = 6 - 2 x$ ): points $(0, 6)$ , $(1, 4)$ , $(3, 0)$ .
  Table for $2 x - y - 2 = 0$ (i.e. $y = 2 x - 2$ ): points $(0, - 2)$ , $(1, 0)$ , $(2, 2)$ .
  1 mark
2. Plot these points and draw both lines on graph paper. The first line cuts the x-axis at C(3, 0), the second at B(1, 0), and the two lines meet at A(2, 2).
  1 mark
3. From the graph, the lines intersect at $(2, 2)$ , so the solution is $x = 2$ , $y = 2$ .
  ½ mark
4. The triangle formed with the x-axis has vertices A(2, 2), B(1, 0), C(3, 0). Base $B C = 2$ units and height = 2 units (the y-coordinate of A).
  Area $= \frac{1}{2} \times 2 \times 2 = 2$ sq. units.
  ½ mark
5. For Visually Impaired candidates (alternative): Let the present ages of the father and son be $x$ and $y$ years. $x + 5 = 3 (y + 5)$ gives $x - 3 y = 10$ . (1 mark in VI version)
6. (VI) $x - 5 = 7 (y - 5)$ gives $x - 7 y = - 30$ . (1 mark)
7. (VI) Subtracting: $4 y = 40$ , so $y = 10$ and $x = 40$ . The father is 40 years old and the son is 10 years old. (1 mark)
Answer: x = 2, y = 2; area = 2 sq. units (VI version: father 40 years, son 10 years)
Where the marks are
- Draws both lines correctly from tables of values (2 marks)
- Reads the solution x = 2, y = 2 (½ mark)
- Finds the area of the triangle with the x-axis as 2 sq. units (½ mark)
- VI: forms $x - 3 y = 10$ (1), forms $x - 7 y = - 30$ (1), solves to 40 and 10 years (1)
Open Q31(B) on its own page

Section D

Q32 · 5 marks · Long answerCompetency-basedQuadratic Equations
A train travels at a certain average speed for a distance of 63km and then travels at a distance of 72km at an average speed of 6km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?
Try it first, then show our solutionSolution
1. Let the original average speed be $x$ km/h. Time taken for 63 km is $t_{1} = \frac{63}{x}$ h.
  1 mark
2. The new speed is $(x + 6)$ km/h, so the time for 72 km is $t_{2} = \frac{72}{x + 6}$ h. Total time: $t_{1} + t_{2} = 3$ .
  1 mark
3. $\frac{63}{x} + \frac{72}{x + 6} = 3$ , so $63 (x + 6) + 72 x = 3 x (x + 6)$ .
  1 mark
4. $135 x + 378 = 3 x^{2} + 18 x$ , which gives $3 x^{2} - 117 x - 378 = 0$ , i.e. $x^{2} - 39 x - 126 = 0$ .
  1 mark
5. $x^{2} - 42 x + 3 x - 126 = 0$ , so $(x - 42) (x + 3) = 0$ and $x = 42$ or $x = - 3$ . Speed cannot be negative, so $x = 42$ .
  Check: $\frac{63}{42} + \frac{72}{48} = 1.5 + 1.5 = 3$ h ✓. The original average speed is 42 km/h.
  1 mark
Answer: 42 km/h
Where the marks are
- Writes the time for the first part as 63/x (1 mark)
- Writes the time for the second part as 72/(x+6) and uses total = 3 h (1 mark)
- Forms the equation $\frac{63}{x} + \frac{72}{x + 6} = 3$ (1 mark)
- Reduces it to $x^{2} - 39 x - 126 = 0$ (1 mark)
- Solves it, rejects −3 and states 42 km/h (1 mark)
Open Q32 on its own page
Q33 · 5 marks · Long answerTriangles
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Hence in $Δ$ PQR, prove that a line $ℓ$ intersects the sides PQ and PR of a $Δ$ PQR at L and M respectively such that LM II QR. If PL = 5.7cm, PQ=15.2cm and MR=5.5cm, then find the length of PM (in cm)
Try it first, then show our solutionSolution
1. Given: In $△ A B C$ , a line DE ∥ BC meets AB at D and AC at E.
  To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
  Figure: Draw triangle ABC with D on AB and E on AC, with DE parallel to BC.
  Construction: Join BE and CD. Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  2 marks
2. Proof: $ar (A D E) = \frac{1}{2} A D \cdot E N$ and $ar (B D E) = \frac{1}{2} D B \cdot E N$ , so $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ … (1).
  Also $ar (A D E) = \frac{1}{2} A E \cdot D M$ and $ar (D E C) = \frac{1}{2} E C \cdot D M$ , so $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ … (2).
  1 mark
3. $△ B D E$ and $△ D E C$ have the same base DE and lie between the same parallels DE and BC, so $ar (B D E) = ar (D E C)$ … (3).
  From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
4. Application: In $△ P QR$ , LM ∥ QR. Let $P M = x$ cm, so $P R = x + 5.5$ cm. By the theorem (in the equivalent form), $\frac{P L}{P Q} = \frac{P M}{P R}$ , which gives $\frac{5.7}{15.2} = \frac{x}{x + 5.5}$ .
  ½ mark
5. $5.7 x + 31.35 = 15.2 x$ , so $9.5 x = 31.35$ and $x = 3.3$ .
  (Check: $L Q = 15.2 - 5.7 = 9.5$ , and $\frac{5.7}{9.5} = \frac{3.3}{5.5} = 0.6$ ✓.) So $P M = 3.3$ cm.
  ½ mark
Answer: Theorem (BPT) proved; PM = 3.3 cm
Where the marks are
- Writes the given, to prove, figure and construction correctly (2 marks)
- Proves the result using area ratios and equal areas of triangles BDE and DEC (2 marks)
- Sets up $\frac{5.7}{15.2} = \frac{x}{x + 5.5}$ (½ mark)
- Finds PM = 3.3 cm (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerSurface Areas and Volumes
From a solid right circular cone, whose height is 6cm and radius of base is 12cm, a right circular cylindrical cavity of height 3cm and radius 4cm is hollowed out such that bases of cone and cylinder form concentric circles. Find the surface area of the remaining solid in terms of $π$ .
Try it first, then show our solutionSolution
1. Given: cone with radius $R = 12$ cm and height $H = 6$ cm. Cylindrical cavity with radius $r = 4$ cm and height $h = 3$ cm, hollowed out from the base. The two bases are concentric circles.
  Slant height of the cone: $L = R^{2} + H^{2} = 1 2^{2} + 6^{2} = 180 = 65$ cm
  ½ mark
2. Curved surface area of the cone $= π R L = π \times 12 \times 65 = 725 π$ cm²
  1 mark
3. Area of the base. The exposed base is a ring: the outer circle minus the inner circle. The top face of the cavity, a circle of radius 4 cm, is also exposed. The two circles of radius 4 cm cancel, so:
  $π R^{2} - π r^{2} + π r^{2} = π R^{2} = π (12)^{2} = 144 π$ cm²
  1 mark
4. Inner curved surface of the cylindrical cavity $= 2 π r h = 2 π \times 4 \times 3 = 24 π$ cm²
  1 mark
5. Total surface area of the remaining solid
  $=$ CSA of cone $+$ base area $+$ CSA of cylinder
  $= 725 π + 144 π + 24 π$
  1 mark
6. $= (168 + 725) π$ cm²
  This is the same as $(168 + 3620) π$ cm².
  ½ mark
Answer: $(168 + 725) π$ cm² (same as $(168 + 3620) π$ cm²)
Where the marks are
- Finds slant height $65$ cm (½ mark)
- Finds CSA of cone $725 π$ cm² (1 mark)
- Finds base area $144 π$ cm² after allowing for the cavity's top face (1 mark)
- Finds CSA of cylinder $24 π$ cm² (1 mark)
- Adds the correct three parts for the total surface area (1 mark)
- Final answer $(168 + 725) π$ cm² (½ mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerSurface Areas and Volumes
An empty cone of radius 3cm and height 12cm is filled with ice-cream such that the lower part of the cone which is $(\frac{1}{6})^{t h}$ of the volume of the cone is unfilled (empty) but a hemisphere is formed on the top. Find the volume of the ice-cream.
Try it first, then show our solutionSolution
1. Volume of the cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} π \times 3 \times 3 \times 12 = 36 π$ cm³
  The lower $\frac{1}{6}$ of the cone is empty, so the ice-cream fills $\frac{5}{6}$ of it:
  $\frac{5}{6} \times 36 π = 30 π$ cm³
  2 marks
2. Hemisphere on top. Its radius equals the cone's radius, $r = 3$ cm.
  Volume $= \frac{2}{3} π r^{3} = \frac{2}{3} π \times 27 = 18 π$ cm³
  1½ marks
3. Total volume of ice-cream $= 30 π + 18 π = 48 π$ cm³
  $= 48 \times \frac{22}{7} \approx 150.86$ cm³
  1½ marks
Answer: $48 π$ cm³ ≈ 150.86 cm³
Where the marks are
- Finds cone volume $36 π$ cm³ and ice-cream inside the cone $30 π$ cm³ (2 marks)
- Finds hemisphere volume $18 π$ cm³ (1½ marks)
- Adds to get the total $48 π \approx 150.86$ cm³ (1½ marks)
Open Q34(B) on its own page
Answer one of these 2 alternatives.
Q35(A) · 5 marks · Long answerStatistics
If the mode of the following distribution is 55, then find the value of $x$ . Hence, find the mean.
Class Interval 0 − 15 15 − 30 30 − 45 45 − 60 60 − 75 75 − 90
Frequency 10 7 $x$ 15 10 12
Try it first, then show our solutionSolution
1. The mode is 55, which lies in the class 45–60. So the modal class is 45–60.
  $l = 45$ , $h = 15$ , $f_{1} = 15$ , $f_{0} = x$ , $f_{2} = 10$
  ½ mark
2. Mode formula: $Mode = l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h$
  Substituting: $55 = 45 + \frac{15 - x}{30 - x - 10} \times 15$
  1 mark
3. $10 = \frac{15 ( 15 - x )}{20 - x}$
  $\Rightarrow 200 - 10 x = 225 - 15 x$
  $\Rightarrow 5 x = 25$
  $\Rightarrow x = 5$
  1 mark
4. Table for the mean (using $x = 5$ ):
  C.I. $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  0–15 10 7.5 75
  15–30 7 22.5 157.5
  30–45 5 37.5 187.5
  45–60 15 52.5 787.5
  60–75 10 67.5 675
  75–90 12 82.5 990
  Total 59 2872.5
  1½ marks
5. Mean $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2872.5}{59} \approx 48.69$
  1 mark
Answer: $x = 5$ ; mean ≈ 48.69 (48.68 if truncated)
Where the marks are
- Identifies modal class 45–60 with $l = 45$ , $h = 15$ (½ mark)
- Substitutes correctly in the mode formula (1 mark)
- Solves to get $x = 5$ (1 mark)
- Correct table with $\sum f_{i} = 59$ and $\sum f_{i} x_{i} = 2872.5$ (1½ marks)
- Mean ≈ 48.69 (1 mark)
Open Q35(A) on its own page
OR
Q35(B) · 5 marks · Long answerStatistics
A survey regarding heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
Heights (in cm) Number of girls
less than 140 04
less than 145 11
less than 150 29
less than 155 40
less than 160 46
less than 165 51
Find the median height of girls. If mode of the above distribution is 148.05, find the mean using empirical formula.
Try it first, then show our solutionSolution
1. Convert the cumulative ('less than') data into class frequencies:
  C.I. $f$ c.f.
  135–140 4 4
  140–145 7 11
  145–150 18 29
  150–155 11 40
  155–160 6 46
  160–165 5 51
  1 mark
2. $N = 51$ , so $\frac{N}{2} = 25.5$ .
  The first c.f. greater than 25.5 is 29, so the median class is 145–150.
  $l = 145$ , $c f = 11$ , $f = 18$ , $h = 5$
  Median formula: $Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h$
  1 mark
3. $Median = 145 + \frac{25.5 - 11}{18} \times 5 = 145 + \frac{72.5}{18} \approx 145 + 4.03 = 149.03$
  1 mark
4. Median height ≈ 149.03 cm
  1 mark
5. Empirical formula: $3 Median = Mode + 2 Mean$
  $3 (149.03) = 148.05 + 2 Mean$
  $\Rightarrow 2 Mean = 447.09 - 148.05 = 299.04$
  $\Rightarrow Mean = 149.52$ cm
  1 mark
Answer: Median ≈ 149.03 cm; Mean ≈ 149.52 cm
Where the marks are
- Converts the cumulative data into class intervals with frequencies (1 mark)
- Identifies the median class 145–150 and writes the median formula (1 mark)
- Substitutes correctly (1 mark)
- Median ≈ 149.03 cm (1 mark)
- Uses $3 Median = Mode + 2 Mean$ to get Mean ≈ 149.52 cm (1 mark)
Open Q35(B) on its own page

Class Interval	0 − 15	15 − 30	30 − 45	45 − 60	60 − 75	75 − 90
Frequency	10	7	$x$	15	10	12

C.I.	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0–15	10	7.5	75
15–30	7	22.5	157.5
30–45	5	37.5	187.5
45–60	15	52.5	787.5
60–75	10	67.5	675
75–90	12	82.5	990
Total	59		2872.5

Heights (in cm)	Number of girls
less than 140	04
less than 145	11
less than 150	29
less than 155	40
less than 160	46
less than 165	51

C.I.	$f$	c.f.
135–140	4	4
140–145	7	11
145–150	18	29
150–155	11	40
155–160	6	46
160–165	5	51

Section E

Case study, sub parts 1, 1 and 2 marks

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
In a class, the teacher asks every student to write an example of A.P. Two boys Aryan and Roshan writes the progression as −5, −2, 1, 4, …… and 187, 184, 181, …. respectively. Now the teacher asks his various students the following questions on progression.
Help the students to find answers for the following:
1. i.
  Find the sum of the common difference of two progressions.
  1 mark
2. ii.
  Find the 34th term of progression written by Roshan.
  1 mark
3. iii.
  (A) Find the sum of first 10 terms of the progression written by Aryan.
  2 marks
4. OR
  iii.
  (B) Which term of the progressions will have the same value?
  2 marks
Try it first, then show our solutionSolution
1. (i) Aryan's AP: $d_{1} = - 2 - (- 5) = 3$ .
  Roshan's AP: $d_{2} = 184 - 187 = - 3$ .
  Sum of the common differences $= 3 + (- 3) = 0$ .
  1 mark
2. (ii) For Roshan's AP, $a = 187$ and $d = - 3$ .
  $t_{34} = a + 33 d = 187 + 33 (- 3) = 187 - 99 = 88$
  1 mark
3. (iii)(a) For Aryan's AP, $a = - 5$ , $d = 3$ , $n = 10$ .
  $S_{10} = \frac{n}{2} [2 a + (n - 1) d] = \frac{10}{2} [2 (- 5) + 9 (3)]$
  1 mark
4. (iii)(a) $S_{10} = 5 (- 10 + 27) = 5 \times 17 = 85$
  1 mark
5. (iii)(b) Let the $n$ th terms be equal:
  $- 5 + (n - 1) 3 = 187 + (n - 1) (- 3)$
  1 mark
6. (iii)(b) $6 (n - 1) = 192$
  $\Rightarrow n - 1 = 32$
  $\Rightarrow n = 33$
  So the 33rd terms are equal. Check: both equal $- 5 + 96 = 187 - 96 = 91$ .
  1 mark
Answer: (i) 0; (ii) 88; (iii)(a) 85, (iii)(b) 33rd term (value 91)
Where the marks are
- (i) Finds $d = 3$ and $d = - 3$ , sum 0 (1 mark)
- (ii) $t_{34} = 88$ (1 mark)
- (iii)(a) Correct substitution in the $S_{n}$ formula (1 mark); $S_{10} = 85$ (1 mark)
- (iii)(b) Equates the two $n$ th terms (1 mark); $n = 33$ (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
A group of class X students goes to picnic during winter holidays. The position of three friends Aman, Kirti and Chahat are shown by the points P, Q and R.
For Visually Impaired Candidates: The co-ordinates of P (2,5), Q (4,4) and R (8,3) are given.
1. (i)
  Find the distance between P and R.
  1 mark
2. (ii)
  Is Q, the midpoint of PR? Justify by finding midpoint of PR.
  1 mark
3. (iii)
  (A) Find the point on x-axis which is equidistant from P and Q.
  2 marks
4. OR
  (iii)
  (B) Let S be a point which divides the line joining PQ in ratio 2:3. Find the coordinates of S.
  2 marks
Figure in the paper: Coordinate grid with x-axis marked 0 to 9 and y-axis 1 to 6. Points P(2,5), Q(4,4), R(8,3) are marked with dotted lines to the axes, and P, Q, R are joined by a line segment. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) With $P (2, 5)$ and $R (8, 3)$ , by the distance formula:
  $P R = (8 - 2)^{2} + (3 - 5)^{2} = 36 + 4 = 40 = 210$ units
  1 mark
2. (ii) From the figure, $Q = (4, 4)$ .
  Midpoint of $P R = (\frac{2 + 8}{2}, \frac{5 + 3}{2}) = (5, 4)$
  ½ mark
3. (ii) Since $(5, 4) \neq = (4, 4)$ , Q is not the midpoint of PR.
  ½ mark
4. (iii)(a) Let the point on the x-axis be $A (x, 0)$ . Set $A P = A Q$ , so $A P^{2} = A Q^{2}$ :
  $(x - 2)^{2} + 25 = (x - 4)^{2} + 16$
  1 mark
5. (iii)(a) $x^{2} - 4 x + 29 = x^{2} - 8 x + 32$
  $\Rightarrow 4 x = 3$
  $\Rightarrow x = \frac{3}{4}$
  The required point is $(\frac{3}{4}, 0)$ .
  1 mark
6. (iii)(b) S divides $P (2, 5)$ and $Q (4, 4)$ in the ratio $m : n = 2 : 3$ . By the section formula:
  $S = (\frac{2 \times 4 + 3 \times 2}{2 + 3}, \frac{2 \times 4 + 3 \times 5}{2 + 3})$
  1 mark
7. (iii)(b) $S = (\frac{14}{5}, \frac{23}{5})$
  1 mark
Answer: (i) $210$ units; (ii) No, midpoint of PR is (5,4) ≠ Q(4,4); (iii)(a) $(\frac{3}{4}, 0)$ , (iii)(b) $(\frac{14}{5}, \frac{23}{5})$
Where the marks are
- (i) $P R = 210$ (1 mark)
- (ii) Midpoint (5,4) found (½ mark); concludes Q is not the midpoint (½ mark)
- (iii)(a) Equates squared distances from $(x, 0)$ (1 mark); gets $(\frac{3}{4}, 0)$ (1 mark)
- (iii)(b) Correct section formula substitution (1 mark); $S = (\frac{14}{5}, \frac{23}{5})$ (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
India gate (formerly known as All India war memorial) is located near Karthavya path (formerly Rajpath) at New Delhi. It stands as a memorial to 74187 soldiers of Indian Army, who gave their life in the first world war. This 42m tall structure was designed by Sir Edwin Lutyens in the style of Roman triumphal arches. A student Shreya of height 1 m visited India Gate as a part of her study tour.
1. i.
  What is the angle of elevation from Shreya’s eye to the top of India Gate, if she is standing at a distance of 41m away from the India Gate?
  1 mark
2. ii.
  If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60°, then how far is the she standing from the base of the India Gate?
  1 mark
3. iii.
  (A) If the angle of elevation from Shreya’s eye changes from 45° to 30°, when she moves some distance back from the original position. Find the distance she moves back.
  2 marks
4. OR
  iii.
  (B) If Shreya moves to a point which is at a distance of $\frac{41}{3}$ m from the India Gate, then find the angle of elevation made by her eye to the top of India Gate.
  2 marks
Figure in the paper: Photograph of India Gate, a tall arched memorial structure on a road. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Shreya's eye is 1 m above the ground, so the height of the top above her eye is $42 - 1 = 41$ m. Her horizontal distance from the gate is 41 m.
  $tan θ = \frac{41}{41} = 1$
  ½ mark
2. (i) $tan θ = 1$ , so $θ = 4 5^{\circ}$ .
  ½ mark
3. (ii) Let the distance be $x$ m. The height above her eye is still 41 m.
  $tan 6 0^{\circ} = \frac{41}{x}$
  ½ mark
4. (ii) $3 = \frac{41}{x}$
  $\Rightarrow x = \frac{41}{3} = \frac{41 3}{3}$ m $\approx 23.67$ m
  ½ mark
5. (iii)(a) At 45° she stands 41 m away (from part (i)). Let her move back $x$ m, so her new distance is $(41 + x)$ m.
  $tan 3 0^{\circ} = \frac{41}{41 + x}$
  $\Rightarrow \frac{1}{3} = \frac{41}{41 + x}$
  1 mark
6. (iii)(a) $41 + x = 413$
  $\Rightarrow x = 41 (3 - 1)$ m $\approx 30.01$ m
  1 mark
7. (iii)(b) Height above her eye is 41 m and distance is $\frac{41}{3}$ m.
  $tan θ = \frac{41}{41/ 3} = 3$
  1 mark
8. (iii)(b) $tan θ = 3$ , so $θ = 6 0^{\circ}$ .
  1 mark
Answer: (i) 45°; (ii) $\frac{41 3}{3}$ m (≈ 23.67 m); (iii)(a) $41 (3 - 1)$ m (≈ 30.01 m), (iii)(b) 60°
Where the marks are
- (i) Uses height 41 m above eye level with $tan θ = 1$ (½ mark); $θ = 4 5^{\circ}$ (½ mark)
- (ii) Writes $tan 6 0^{\circ} = 41/ x$ (½ mark); $x = \frac{41 3}{3}$ m (½ mark)
- (iii)(a) Writes $tan 3 0^{\circ} = \frac{41}{41 + x}$ (1 mark); $x = 41 (3 - 1)$ m (1 mark)
- (iii)(b) Gets $tan θ = 3$ (1 mark); $θ = 6 0^{\circ}$ (1 mark)
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Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.