MathsCBSE Class 10 · Standard · Sample paper, session 2026-27

CBSE Class 10 Mathematics Standard: sample paper 2026-27

The sample question paper CBSE published for the 2026-27 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_2026_27/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	Very Short Answer (VSA)
C	26-31	Short Answer (SA)
D	32-35	Long Answer (LA)
E	36-38	Case study based, sub-parts (i), (ii), (iii) of marks 1, 1, 2

Instructions in brief: No overall choice; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D, and in the 2-mark sub-part (iii) of every Section E case study.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
The greatest number which divides both 134 and 188, leaving remainders 4 and 6 respectively, is:
1. (A)13
2. (B)26
3. (C)39
4. (D)65
Try it first, then show our solutionSolution
1. Remove the remainders. The required number must exactly divide $134 - 4 = 130$ and $188 - 6 = 182$ .
2. $130 = 2 \times 5 \times 13$ and $182 = 2 \times 7 \times 13$ , so $HCF (130, 182) = 2 \times 13 = 26$ .
  1 mark
3. The greatest such number is 26, which is option (B).
Answer: 26, option (B)
Where the marks are
- Subtracts the remainders and finds HCF(130, 182) = 26 (1 mark)
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Q2 · 1 mark · MCQPolynomials
If $f (x) = p x^{2} + q x + r$ , $p \neq = 0$ and $p + r = q$ , then one of the zeroes of $f (x)$ is:
1. (A) $\frac{q}{p}$
2. (B) $\frac{r}{p}$
3. (C) $- \frac{r}{p}$
4. (D) $- \frac{q}{p}$
Try it first, then show our solutionSolution
1. Put $q = p + r$ into the polynomial: $f (x) = p x^{2} + (p + r) x + r$ .
2. Factorise: $p x^{2} + p x + r x + r = p x (x + 1) + r (x + 1) = (x + 1) (p x + r)$ .
  1 mark
3. The zeroes are $- 1$ and $- \frac{r}{p}$ , so the answer is option (C).
Answer: $- \frac{r}{p}$ , option (C)
Where the marks are
- Substitutes q = p + r and factorises to get the zero $- \frac{r}{p}$ (1 mark)
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Q3 · 1 mark · MCQPair of Linear Equations in Two Variables
Tarun correctly solved a pair of linear equations in two variables and found their only point of intersection as $(5, - 1)$ . One of the lines was $x - y = 6$ . Which of the following could have been the other line?
I: $3 x - 3 y = 18$
II: $2 x - 3 y = 13$
III: $2 x - 3 y = 16$
1. (A)I only
2. (B)II only
3. (C)I and II
4. (D)II and III
Try it first, then show our solutionSolution
1. I: $3 x - 3 y = 18$ is just $3 (x - y) = 3 \times 6$ . It is the same line as $x - y = 6$ , so the pair would have infinitely many solutions, not only one. I is rejected.
2. II: at $(5, - 1)$ , $2 (5) - 3 (- 1) = 13$ , so the point lies on it. Also $\frac{1}{2} \neq = \frac{- 1}{- 3}$ , so the lines intersect at exactly one point. II works.
  1 mark
3. III: $2 (5) - 3 (- 1) = 13 \neq = 16$ , so $(5, - 1)$ does not lie on it. III is rejected.
4. Only II is possible, which is option (B).
Answer: II only, option (B)
Where the marks are
- Checks each line for passing through (5, −1) with a unique intersection and selects II only (1 mark)
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Q4 · 1 mark · MCQQuadratic Equations
If $(1 - p)$ is a root of the quadratic equation $x^{2} + p x + 1 - p = 0$ , then its roots are:
1. (A) $0, 1$
2. (B) $- 1, 1$
3. (C) $0, - 1$
4. (D) $- 1, 2$
Try it first, then show our solutionSolution
1. Put $x = 1 - p$ : $(1 - p)^{2} + p (1 - p) + (1 - p) = 0 \Rightarrow (1 - p) (1 - p + p + 1) = 0 \Rightarrow 2 (1 - p) = 0 \Rightarrow p = 1$ .
  1 mark
2. The equation becomes $x^{2} + x = 0$ , so $x (x + 1) = 0$ and the roots are $0$ and $- 1$ . This is option (C).
Answer: 0, −1, option (C)
Where the marks are
- Finds p = 1 and solves x² + x = 0 to get roots 0, −1 (1 mark)
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Q5 · 1 mark · MCQArithmetic Progressions
The middle term of the A.P.: $10, 7, 4, \dots, - 62$ is:
1. (A) $- 26$
2. (B) $- 29$
3. (C) $- 32$
4. (D) $- 35$
Try it first, then show our solutionSolution
1. Here $a = 10$ and $d = - 3$ . Use $a_{n} = a + (n - 1) d$ : $- 62 = 10 + (n - 1) (- 3)$ , so $n - 1 = 24$ and $n = 25$ .
2. With 25 terms, the middle term is the $\frac{25 + 1}{2} = 1 3^{t h}$ term: $a_{13} = 10 + 12 (- 3) = - 26$ . This is option (A).
  1 mark
Answer: −26, option (A)
Where the marks are
- Finds n = 25 and computes $a_{13} = - 26$ (1 mark)
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Q6 · 1 mark · MCQTriangles
The perimeters of two similar triangles are 56 cm and 70 cm respectively. If one side of the first triangle is 14 cm, then the corresponding side of the second triangle (in cm) is:
1. (A)5
2. (B)7.5
3. (C)10
4. (D)17.5
Try it first, then show our solutionSolution
1. For similar triangles, the ratio of perimeters equals the ratio of corresponding sides.
2. $\frac{56}{70} = \frac{14}{x} \Rightarrow x = \frac{14 \times 70}{56} = 17.5$ cm. This is option (D).
  1 mark
Answer: 17.5 cm, option (D)
Where the marks are
- Uses ratio of perimeters = ratio of corresponding sides to get 17.5 cm (1 mark)
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Q7 · 1 mark · MCQCoordinate Geometry
The point which lies on the perpendicular bisector of the line segment joining the points $A (- 3, - 4)$ and $B (3, 4)$ is:
1. (A) $(0, 0)$
2. (B) $(0, 3)$
3. (C) $(3, 0)$
4. (D) $(0, 4)$
Try it first, then show our solutionSolution
1. The perpendicular bisector of AB passes through the mid-point of AB.
2. Mid-point $= (\frac{- 3 + 3}{2}, \frac{- 4 + 4}{2}) = (0, 0)$ . So $(0, 0)$ lies on it, which is option (A). (The other options are not equidistant from A and B.)
  1 mark
Answer: (0, 0), option (A)
Where the marks are
- Identifies that the mid-point lies on the perpendicular bisector and finds (0, 0) (1 mark)
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Q8 · 1 mark · MCQIntroduction to Trigonometry
If $tan (A + B) = 3$ and $tan (A - B) = \frac{1}{3}$ , $0^{\circ} < A + B < 9 0^{\circ}$ , $A > B$ , then the value of A and B respectively are:
1. (A) $6 0^{\circ}, 3 0^{\circ}$
2. (B) $6 0^{\circ}, 4 5^{\circ}$
3. (C) $4 5^{\circ}, 1 5^{\circ}$
4. (D) $6 0^{\circ}, 1 5^{\circ}$
Try it first, then show our solutionSolution
1. $tan (A + B) = 3 \Rightarrow A + B = 6 0^{\circ}$ ...(i) and $tan (A - B) = \frac{1}{3} \Rightarrow A - B = 3 0^{\circ}$ ...(ii).
2. Add (i) and (ii): $2 A = 9 0^{\circ}$ , so $A = 4 5^{\circ}$ and $B = 1 5^{\circ}$ . This is option (C).
  1 mark
Answer: A = 45°, B = 15°, option (C)
Where the marks are
- Forms A + B = 60° and A − B = 30° and solves them (1 mark)
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Q9 · 1 mark · MCQProbability
T-shirts marked with numbers 4 to 99 are placed in a box. Gunika is fond of numbers. She randomly takes out a T-shirt from this box. The probability that she gets a T-shirt marked with a number that is either a perfect square or a perfect cube is:
1. (A) $\frac{1}{12}$
2. (B) $\frac{1}{32}$
3. (C) $\frac{11}{96}$
4. (D) $\frac{5}{48}$
Try it first, then show our solutionSolution
1. Numbers 4 to 99 give $99 - 4 + 1 = 96$ outcomes.
2. Perfect squares are 4, 9, 16, 25, 36, 49, 64, 81 (8 numbers). Perfect cubes are 8, 27, 64 (3 numbers). The number 64 is in both lists, so the favourable outcomes are $8 + 3 - 1 = 10$ .
3. $P = \frac{10}{96} = \frac{5}{48}$ , which is option (D).
  1 mark
Answer: $\frac{5}{48}$ , option (D)
Where the marks are
- Counts 96 outcomes and 10 favourable outcomes, getting 5/48 (1 mark)
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Q10 · 1 mark · MCQCompetency-basedAreas Related to Circles
The diameter of a car wheel is 21 cm. The number of complete revolutions it will make in moving 66 km is:
1. (A) $1 0^{4}$
2. (B) $1 0^{5}$
3. (C) $1 0^{6}$
4. (D) $1 0^{7}$
Try it first, then show our solutionSolution
1. Circumference $= π d = \frac{22}{7} \times 21 = 66$ cm. This is the distance covered in one revolution.
2. $66$ km $= 66 \times 1 0^{5}$ cm. Number of revolutions $= \frac{66 \times 1 0 ^{5}}{66} = 1 0^{5}$ , which is option (B).
  1 mark
Answer: $1 0^{5}$ , option (B)
Where the marks are
- Finds circumference 66 cm and divides 66 × 10⁵ cm by it to get 10⁵ (1 mark)
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Q11 · 1 mark · MCQSurface Areas and Volumes
Two cubes each of volume $64 cm^{3}$ are joined end to end to form a solid. The surface area of the resultant cuboid is:
1. (A) $192 cm^{2}$
2. (B) $160 cm^{2}$
3. (C) $96 cm^{2}$
4. (D) $80 cm^{2}$
Try it first, then show our solutionSolution
1. Side of each cube: $a^{3} = 64 \Rightarrow a = 4$ cm. The cuboid formed is $8 cm \times 4 cm \times 4 cm$ .
2. TSA $= 2 (l b + bh + h l) = 2 (32 + 16 + 32) = 160 cm^{2}$ , which is option (B).
  1 mark
Answer: 160 cm², option (B)
Where the marks are
- Finds cuboid 8 × 4 × 4 and its surface area 160 cm² (1 mark)
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Q12 · 1 mark · MCQStatistics
The mean age of a combined group of men and women is 35 years. If the mean ages of the men and women are 38 years and 30 years respectively, then the percentage of women in the group is:
1. (A)15
2. (B)25.5
3. (C)35
4. (D)37.5
Try it first, then show our solutionSolution
1. Let there be $x$ men and $y$ women. Then $\frac{38 x + 30 y}{x + y} = 35 \Rightarrow 3 x = 5 y \Rightarrow x : y = 5 : 3$ .
2. Women make up $\frac{3}{5 + 3} \times 100 = 37.5%$ of the group, which is option (D).
  1 mark
Answer: 37.5%, option (D)
Where the marks are
- Uses the combined mean to get x : y = 5 : 3 and finds 37.5% (1 mark)
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Q13 · 1 mark · MCQProbability
Two dice are rolled simultaneously. The probability of getting number less than 4 on each die is:
1. (A) $\frac{1}{4}$
2. (B) $\frac{1}{9}$
3. (C) $\frac{1}{36}$
4. (D) $\frac{1}{6}$
Try it first, then show our solutionSolution
1. Total outcomes $= 36$ . Each die must show 1, 2 or 3, giving $3 \times 3 = 9$ favourable outcomes: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3).
2. $P = \frac{9}{36} = \frac{1}{4}$ , which is option (A).
  1 mark
Answer: $\frac{1}{4}$ , option (A)
Where the marks are
- Counts 9 favourable outcomes out of 36 and gets 1/4 (1 mark)
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Q14 · 1 mark · MCQCoordinate Geometry
PM is a median of $Δ$ PQR with vertices $P (5, - 6)$ , $Q (6, 4)$ and $R (0, 0)$ . The length of PM is:
1. (A) $217$ units
2. (B) $215$ units
3. (C) $101$ units
4. (D)10 units
Try it first, then show our solutionSolution
1. M is the mid-point of QR: $M = (\frac{6 + 0}{2}, \frac{4 + 0}{2}) = (3, 2)$ .
2. $P M = (5 - 3)^{2} + (- 6 - 2)^{2} = 4 + 64 = 68 = 217$ units, which is option (A).
  1 mark
Answer: $217$ units, option (A)
Where the marks are
- Finds M(3, 2) and uses the distance formula to get $217$ (1 mark)
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Q15 · 1 mark · MCQAreas Related to Circles
The arc of a circle is of length $6 π$ cm and the sector it bounds has an area of $24 π cm^{2}$ . The radius of the circle is:
1. (A)4 cm
2. (B)8 cm
3. (C)16 cm
4. (D)18 cm
Try it first, then show our solutionSolution
1. Area of sector $= \frac{1}{2} \times l \times r$ , where $l$ is the arc length.
2. $24 π = \frac{1}{2} \times 6 π \times r \Rightarrow r = 8$ cm, which is option (B).
  1 mark
Answer: 8 cm, option (B)
Where the marks are
- Uses area = ½ × l × r to get r = 8 cm (1 mark)
Q16 · 1 mark · MCQStatistics
The mean and median of the data are 35.5 and 32 respectively. The value of mode for this data is:
1. (A)23
2. (B)26
3. (C)25
4. (D)30
Try it first, then show our solutionSolution
1. Use the empirical relation between the three measures of central tendency: $Mode = 3 Median - 2 Mean$ .
2. Substitute Median $= 32$ and Mean $= 35.5$ : $Mode = 3 (32) - 2 (35.5) = 96 - 71 = 25$ . Hence the correct option is (C).
  1 mark
Answer: Mode = 25, option (C)
Where the marks are
- Uses Mode = 3 Median − 2 Mean and gets 25, option (C) (1 mark)
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Q17 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A flying kite is tied to a point on the ground with the help of a string. The string makes an angle $θ$ with the ground level such that $tan θ = \frac{12}{5}$ . If the length of the string is 52 m, then the height of the kite above the ground is:
1. (A)40 m
2. (B)45.5 m
3. (C)48 m
4. (D)50 m
Try it first, then show our solutionSolution
1. $tan θ = \frac{12}{5}$ , so the sides are in the ratio 12 : 5 and the hypotenuse is $1 2^{2} + 5^{2} = 13$ . This gives $sin θ = \frac{12}{13}$ .
2. The string is the hypotenuse: $\frac{h}{52} = \frac{12}{13} \Rightarrow h = 48$ m, which is option (C).
  1 mark
Answer: 48 m, option (C)
Where the marks are
- Finds sin θ = 12/13 and gets h = 48 m (1 mark)
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Q18 · 1 mark · MCQProbability
2 cards of diamonds and 4 cards of spades are missing from a pack of 52 cards. A card is drawn at random from this pack. The probability of getting a card of heart is:
1. (A) $\frac{13}{52}$
2. (B) $\frac{13}{46}$
3. (C) $\frac{11}{52}$
4. (D) $\frac{11}{46}$
Try it first, then show our solutionSolution
1. Cards left $= 52 - (2 + 4) = 46$ . No heart is missing, so all 13 hearts are still there.
2. $P (heart) = \frac{13}{46}$ , which is option (B).
  1 mark
Answer: $\frac{13}{46}$ , option (B)
Where the marks are
- Uses 46 remaining cards and 13 hearts to get 13/46 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedCoordinate Geometry
Assertion (A): If three vertices of a parallelogram taken in order are $(- 1, - 6)$ , $(2, - 5)$ and $(7, 2)$ , then its fourth vertex is $(4, 1)$ .
Reason (R): Diagonals of a parallelogram bisect each other.
Options: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
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. Let the vertices be $A (- 1, - 6)$ , $B (2, - 5)$ , $C (7, 2)$ , $D (x, y)$ . Diagonals AC and BD bisect each other, so they share a mid-point. Mid-point of AC $= (3, - 2)$ .
2. $\frac{2 + x}{2} = 3 \Rightarrow x = 4$ and $\frac{- 5 + y}{2} = - 2 \Rightarrow y = 1$ . So $D = (4, 1)$ and A is true.
3. R is a true property, and it is exactly what was used to find D. So R correctly explains A: option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Uses mid-points of the diagonals to get (4, 1) and recognises R as the explanation (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedPolynomials
Assertion (A): If zeroes of the polynomial $(2 k - 1) x^{2} + 4 x - 3$ are reciprocal of each other, then $k = - 1$ .
Reason (R): If $a = c$ , then zeroes of the polynomial $a x^{2} + b x + c$ , $a \neq = 0$ are reciprocal of each other.
Options: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
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. If the zeroes are reciprocals, their product is 1: $\frac{c}{a} = \frac{- 3}{2 k - 1} = 1 \Rightarrow 2 k - 1 = - 3 \Rightarrow k = - 1$ . A is true.
2. R: if $a = c$ , the product of zeroes $= \frac{c}{a} = 1$ , so the zeroes are reciprocals. R is true, and it gives the condition $2 k - 1 = - 3$ used in A. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Uses product of zeroes = 1 to get k = −1 and links it to R (1 mark)
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Section B

Very Short Answer (VSA)

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerCompetency-basedReal Numbers
A hall has a length of 9.75 m, breadth of 6.75 m and height of 5.25 m. What is the length of the longest unmarked ruler that can exactly measure the dimensions of the hall?
Try it first, then show our solutionSolution
1. Convert to cm: length $= 975$ cm, breadth $= 675$ cm, height $= 525$ cm. The longest ruler that measures all three exactly has length $HCF (975, 675, 525)$ .
  ½ mark
2. Prime factorisation: $975 = 3 \times 5^{2} \times 13$ , $675 = 3^{3} \times 5^{2}$ , $525 = 3 \times 5^{2} \times 7$ .
  1 mark
3. $HCF = 3 \times 5^{2} = 75$ . The longest ruler is 75 cm, that is 0.75 m.
  ½ mark
Answer: 75 cm (0.75 m)
Where the marks are
- Recognises that the HCF of the three dimensions (in cm) is required (½ mark)
- Correct prime factorisation of 975, 675 and 525 (1 mark)
- HCF = 75 cm (½ mark)
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OR
Q21(B) · 2 marks · Very short answerCompetency-basedReal Numbers
There are three bells placed at different swings in a park, which toll at intervals of 5, 6 and 8 minutes, respectively. They all toll together when the park opens for the visitors at 10:00 a.m. How many more times do they all toll together till the park is closed at 7:00 p.m.?
Try it first, then show our solutionSolution
1. $5 = 5$ , $6 = 2 \times 3$ , $8 = 2^{3}$ , so $LCM (5, 6, 8) = 2^{3} \times 3 \times 5 = 120$ . The bells toll together every 120 minutes, that is every 2 hours.
  1 mark
2. Starting at 10:00 a.m., they toll together again at 12:00 noon, 2:00 p.m., 4:00 p.m. and 6:00 p.m. The next time would be 8:00 p.m., which is after closing. So they toll together 4 more times.
  1 mark
Answer: 4 more times
Where the marks are
- LCM of 5, 6, 8 = 120 min = 2 hours (1 mark)
- Lists the times up to 7 p.m. and concludes 4 more times (1 mark)
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Answer one of these 2 alternatives.
Q22(A) · 2 marks · Very short answerTriangles
ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect each other at the point O. Show that $\frac{A O}{B O} = \frac{C O}{D O}$ .
Try it first, then show our solutionSolution
1. In $Δ A O B$ and $Δ C O D$ : $∠ O A B = ∠ O C D$ and $∠ O B A = ∠ O D C$ [alternate interior angles, $A B ∥ D C$ with transversals AC and BD].
  1 mark
2. $∴ Δ A O B \sim Δ C O D$ [AA similarity criterion].
  ½ mark
3. Corresponding sides are proportional, so $\frac{A O}{C O} = \frac{B O}{D O}$ . Hence $\frac{A O}{B O} = \frac{C O}{D O}$ . Hence proved.
  ½ mark
Answer: $Δ A O B \sim Δ C O D$ (AA), hence $\frac{A O}{B O} = \frac{C O}{D O}$
Where the marks are
- Two pairs of equal alternate angles with reason (1 mark)
- States ΔAOB ~ ΔCOD by AA (½ mark)
- Writes the proportion and rearranges to the result (½ mark)
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OR
Q22(B) · 2 marks · Very short answerTriangles
In the given figure, if $∠ P QR = ∠ QS P$ , PQ = 6 cm and PS = 3 cm, then find the length of PR.
For visually impaired candidates: In a right-angled triangle ABC, right-angled at B, a perpendicular BD is drawn to the hypotenuse AC. Prove that the triangle BCD is similar to the triangle ACB.
Figure in the paper: Triangle PQR with a point S on side PR; segment QS joins Q to S. Angle PQR (at Q) and angle QSP (at S) are marked as equal. PQ = 6 cm and PS = 3 cm are given; PR is to be found. See it in the official paper.
Try it first, then show our solutionSolution
1. In $Δ P QR$ and $Δ P S Q$ : $∠ P QR = ∠ P S Q$ (given) and $∠ QP R = ∠ S P Q$ (common).
  ½ mark
2. $∴ Δ P QR \sim Δ P S Q$ [AA criterion], with $P \leftrightarrow P$ , $Q \leftrightarrow S$ , $R \leftrightarrow Q$ .
  ½ mark
3. Corresponding sides are proportional: $\frac{P Q}{P S} = \frac{P R}{P Q} \Rightarrow \frac{6}{3} = \frac{P R}{6}$ .
  ½ mark
4. $P R = 12$ cm.
  ½ mark
5. For visually impaired candidates (this alternative carries the same 2 marks: ½ + ½ + 1). In $Δ B C D$ and $Δ A C B$ : $∠ B D C = ∠ A B C = 9 0^{\circ}$ [ $B D ⊥ A C$ , and the triangle is right-angled at B].
6. $∠ B C D = ∠ A C B$ (common angle C).
7. $∴ Δ B C D \sim Δ A C B$ [AA criterion]. Hence proved.
Answer: PR = 12 cm (VI version: ΔBCD ~ ΔACB by AA)
Where the marks are
- Equal angle pair given and common angle P (½ mark)
- Concludes ΔPQR ~ ΔPSQ by AA (½ mark)
- Writes PQ/PS = PR/PQ with values (½ mark)
- PR = 12 cm (½ mark)
- VI alternative: right angles equal (½ mark), common angle C (½ mark), AA similarity concluded (1 mark)
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Q23 · 2 marks · Very short answerIntroduction to Trigonometry
If $tan θ = \frac{1}{5}$ , then find the value of $\frac{cosec ^{2} θ - s e c ^{2} θ}{cosec ^{2} θ + s e c ^{2} θ}$ .
Try it first, then show our solutionSolution
1. Given $tan θ = \frac{1}{5}$ , so $cot θ = \frac{1}{t a n θ} = 5$ .
2. Using $sec^{2} θ = 1 + tan^{2} θ$ : $sec^{2} θ = 1 + \frac{1}{5} = \frac{6}{5}$ .
  ½ mark
3. Using $cosec^{2} θ = 1 + cot^{2} θ$ : $cosec^{2} θ = 1 + 5 = 6$ .
  ½ mark
4. Substitute: $\frac{cosec ^{2} θ - sec ^{2} θ}{cosec ^{2} θ + sec ^{2} θ} = \frac{6 - \frac{6}{5}}{6 + \frac{6}{5}} = \frac{\frac{24}{5}}{\frac{36}{5}} = \frac{24}{36} = \frac{2}{3}$ .
  1 mark
Answer: $\frac{2}{3}$
Where the marks are
- Finds $sec^{2} θ = \frac{6}{5}$ (½ mark)
- Finds $cosec^{2} θ = 6$ (½ mark)
- Substitutes and simplifies to $\frac{2}{3}$ (1 mark)
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Q24 · 2 marks · Very short answerAreas Related to Circles
Find the area of a sector of a circle with diameter 28 cm, if the length of the corresponding arc is 22 cm.
Try it first, then show our solutionSolution
1. Radius $r = \frac{28}{2} = 14$ cm.
  ½ mark
2. Area of a sector in terms of its arc length $l$ : $A = \frac{1}{2} \times l \times r = \frac{1}{2} \times 22 \times 14$ .
  1 mark
3. $A = 154 cm^{2}$ .
  ½ mark
Answer: 154 cm²
Where the marks are
- Finds radius = 14 cm (½ mark)
- Uses $A = \frac{1}{2} l r$ with correct values (1 mark)
- Gets area $= 154 cm^{2}$ (½ mark)
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Q25 · 2 marks · Very short answerProbability
Two dice are thrown simultaneously. Find the probability that the product of the numbers appearing on them is a prime number.
Try it first, then show our solutionSolution
1. When two dice are thrown, the total number of outcomes is $6 \times 6 = 36$ .
  ½ mark
2. A product of two natural numbers is prime only when one number is 1 and the other is prime (2, 3 or 5).
  Favourable outcomes: $(1, 2), (2, 1), (1, 3), (3, 1), (1, 5), (5, 1)$ . That is 6 outcomes.
  ½ mark
3. $P (product is prime) = \frac{6}{36} = \frac{1}{6}$ .
  1 mark
Answer: $\frac{1}{6}$
Where the marks are
- States total outcomes = 36 (½ mark)
- Lists the 6 favourable outcomes (½ mark)
- Gets probability $= \frac{1}{6}$ (1 mark)
Open Q25 on its own page

Section C

Short Answer (SA)

Q26 · 3 marks · Short answerReal Numbers
Given that $3$ is an irrational number, prove that $(2 + 33)$ is an irrational number.
Try it first, then show our solutionSolution
1. Suppose, on the contrary, that $2 + 33$ is rational.
  Then $2 + 33 = \frac{a}{b}$ , where $a, b$ are integers and $b \neq = 0$ .
  So $33 = \frac{a}{b} - 2$ , which gives $3 = \frac{a - 2 b}{3 b}$ .
  1 mark
2. Since $a$ and $b$ are integers, $a - 2 b$ and $3 b$ are integers, with $3 b \neq = 0$ .
  So $\frac{a - 2 b}{3 b}$ is rational, and therefore $3$ would be rational.
  1 mark
3. This contradicts the given fact that $3$ is irrational. So our assumption is false.
  Hence $2 + 33$ is irrational.
  1 mark
Answer: Proved: $2 + 33$ is irrational.
Where the marks are
- Assumes the number is rational, writes it as $\frac{a}{b}$ and isolates $3 = \frac{a - 2 b}{3 b}$ (1 mark)
- Argues that $\frac{a - 2 b}{3 b}$ is rational, so $3$ would be rational (1 mark)
- States the contradiction and concludes the number is irrational (1 mark)
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Q27 · 3 marks · Short answerPolynomials
If $α$ and $β$ are the zeroes of the quadratic polynomial $2 x^{2} - 8 x + 5$ , then find the value of $(α + \frac{1}{β}) \times (β + \frac{1}{α})$ .
Try it first, then show our solutionSolution
1. For $2 x^{2} - 8 x + 5$ : $α + β = - \frac{- 8}{2} = 4$ and $α β = \frac{5}{2}$ .
  1 mark
2. Combine each bracket into one fraction:
  $(α + \frac{1}{β}) (β + \frac{1}{α}) = \frac{α β + 1}{β} \times \frac{α β + 1}{α} = \frac{( α β + 1 ) ^{2}}{α β}$ .
  ½ mark
3. Substitute $α β = \frac{5}{2}$ : $\frac{( \frac{5}{2} + 1 ) ^{2}}{\frac{5}{2}} = \frac{\frac{49}{4}}{\frac{5}{2}}$ .
  1 mark
4. $= \frac{49}{4} \times \frac{2}{5} = \frac{49}{10}$ .
  ½ mark
Answer: $\frac{49}{10}$
Where the marks are
- Finds sum of zeroes = 4 and product = $\frac{5}{2}$ (1 mark)
- Rewrites the expression as $\frac{( α β + 1 ) ^{2}}{α β}$ (½ mark)
- Substitutes the value of $α β$ (1 mark)
- Simplifies to $\frac{49}{10}$ (½ mark)
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Answer one of these 2 alternatives.
Q28(A) · 3 marks · Short answerArithmetic Progressions
Ridhi drew a polygon with n sides. The smallest exterior angle is $8^{\circ}$ and each subsequent exterior angle is $4^{\circ}$ more than the previous exterior angle. Find the number of sides of the polygon that Ridhi had drawn.
Try it first, then show our solutionSolution
1. The exterior angles form an A.P. with first term $a = 8$ and common difference $d = 4$ .
  The exterior angles of any polygon add up to $36 0^{\circ}$ , so $S_{n} = 360$ :
  $\frac{n}{2} [2 \times 8 + (n - 1) \times 4] = 360$ .
  1 mark
2. Simplify:
  $\frac{n}{2} (4 n + 12) = 360$
  $\Rightarrow 2 n^{2} + 6 n - 360 = 0$
  $\Rightarrow n^{2} + 3 n - 180 = 0$
  $\Rightarrow (n + 15) (n - 12) = 0$ .
  1 mark
3. So $n = - 15$ or $n = 12$ .
  ½ mark
4. The number of sides cannot be negative, so $n = 12$ .
  The polygon has 12 sides.
  ½ mark
Answer: 12 sides
Where the marks are
- Uses sum of exterior angles = 360° to form $\frac{n}{2} [16 + 4 (n - 1)] = 360$ (1 mark)
- Reduces to $n^{2} + 3 n - 180 = 0$ and factorises (1 mark)
- Gets $n = - 15$ or $12$ (½ mark)
- Rejects the negative value and concludes 12 sides (½ mark)
Open Q28(A) on its own page
OR
Q28(B) · 3 marks · Short answerArithmetic Progressions
Find the sum of integers between 1 and 400 that are multiples of 4 as well as of 5.
Try it first, then show our solutionSolution
1. A number that is a multiple of both 4 and 5 is a multiple of LCM(4, 5) = 20.
  Such numbers between 1 and 400 form the A.P. $20, 40, 60, \dots, 380$ . (400 is not included, since we need numbers between 1 and 400.)
  1 mark
2. Here $a = 20$ and $d = 20$ . Let $a_{n} = 380$ :
  $20 + (n - 1) \times 20 = 380$ .
  ½ mark
3. $\Rightarrow (n - 1) \times 20 = 360 \Rightarrow n = 19$ .
  ½ mark
4. Using $S_{n} = \frac{n}{2} (a + l)$ : $S_{19} = \frac{19}{2} (20 + 380)$ .
  ½ mark
5. $S_{19} = \frac{19}{2} \times 400 = 3800$ .
  ½ mark
Answer: 3800
Where the marks are
- Identifies the A.P. $20, 40, \dots, 380$ of multiples of 20 (1 mark)
- Sets up $a_{n} = 20 + (n - 1) 20 = 380$ (½ mark)
- Finds $n = 19$ (½ mark)
- Applies the sum formula $S_{19} = \frac{19}{2} (20 + 380)$ (½ mark)
- Gets 3800 (½ mark)
Open Q28(B) on its own page
Q29 · 3 marks · Short answerCircles
Prove that the parallelogram circumscribing a circle is a rhombus.
Try it first, then show our solutionSolution
1. Figure: ABCD is a parallelogram whose sides AB, BC, CD and DA touch a circle with centre O at S, P, Q and R respectively.
  Tangents drawn from an external point to a circle are equal in length, so:
  $A S = A R$ …(i)
  $B S = B P$ …(ii)
  $C Q = C P$ …(iii)
  $D Q = D R$ …(iv)
  1 mark
2. Adding (i), (ii), (iii) and (iv):
  $(A S + B S) + (C Q + D Q) = (A R + D R) + (B P + C P)$
  $\Rightarrow A B + C D = A D + B C$ .
  1 mark
3. In a parallelogram, opposite sides are equal, so $C D = A B$ and $B C = A D$ .
  $\Rightarrow 2 A B = 2 A D \Rightarrow A B = A D$ .
  So the adjacent sides of parallelogram ABCD are equal, and all four sides are equal. Hence ABCD is a rhombus.
  1 mark
Answer: Proved: $A B = A D$ , so the parallelogram ABCD is a rhombus.
Where the marks are
- Writes the four pairs of equal tangent lengths with the reason (1 mark)
- Adds them to get $A B + C D = A D + B C$ (1 mark)
- Uses opposite sides of a parallelogram being equal to get $A B = A D$ and concludes it is a rhombus (1 mark)
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Q30 · 3 marks · Short answerCoordinate Geometry
In what ratio does the x-axis divide the line segment joining the points $(- 4, - 6)$ and $(- 1, 7)$ ? Also, find the coordinates of the point of division.
Try it first, then show our solutionSolution
1. Let the x-axis divide the segment joining $A (- 4, - 6)$ and $B (- 1, 7)$ at $P (x, 0)$ in the ratio $k : 1$ .
  By the section formula:
  $P = (\frac{k ( - 1 ) + 1 ( - 4 )}{k + 1}, \frac{k ( 7 ) + 1 ( - 6 )}{k + 1}) = (\frac{- k - 4}{k + 1}, \frac{7 k - 6}{k + 1})$ .
  1 mark
2. P lies on the x-axis, so its y-coordinate is 0:
  $\frac{7 k - 6}{k + 1} = 0 \Rightarrow k = \frac{6}{7}$ .
  ½ mark
3. So the x-axis divides the segment in the ratio $6 : 7$ .
  ½ mark
4. x-coordinate of P:
  $x = \frac{- \frac{6}{7} - 4}{\frac{6}{7} + 1} = \frac{- \frac{34}{7}}{\frac{13}{7}} = - \frac{34}{13}$ .
  Point of division is $(- \frac{34}{13}, 0)$ .
  1 mark
Answer: Ratio 6 : 7; point of division $(- \frac{34}{13}, 0)$
Where the marks are
- Writes the coordinates of P using the section formula with ratio $k : 1$ (1 mark)
- Puts y-coordinate = 0 and gets $k = \frac{6}{7}$ (½ mark)
- States the ratio 6 : 7 (½ mark)
- Finds the point $(- \frac{34}{13}, 0)$ (1 mark)
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Answer one of these 2 alternatives.
Q31(A) · 3 marks · Short answerIntroduction to Trigonometry
During a math class, Ms. Isha wrote the expression below on the board and asked the students to simplify it.
$\frac{c o s θ}{1 - s i n θ} + \frac{1 - s i n θ}{c o s θ}$
Jyoti solved it in her note book as follows:
$= \frac{c o s ^{2} θ + ( 1 - s i n θ ) ^{2}}{( 1 - s i n θ ) \times c o s θ}$ .... (step 1)
$= \frac{c o s ^{2} θ + c o s ^{2} θ}{( 1 - s i n θ ) \times c o s θ}$ .... (step 2)
$= \frac{2 c o s ^{2} θ}{( 1 - s i n θ ) \times c o s θ}$ .... (step 3)
$= \frac{2 c o s θ}{( 1 - s i n θ )}$ .... (step 4)
Identify the step in which Jyoti has made error(s), if any. Rectify the same and find the correct answer.
Try it first, then show our solutionSolution
1. Jyoti's error is in step 2. She replaced $(1 - sin θ)^{2}$ by $cos^{2} θ$ . But $cos^{2} θ = 1 - sin^{2} θ$ , which is not the same as $(1 - sin θ)^{2}$ .
  1 mark
2. Correct working:
  $\frac{cos θ}{1 - sin θ} + \frac{1 - sin θ}{cos θ} = \frac{cos ^{2} θ + ( 1 - sin θ ) ^{2}}{( 1 - sin θ ) cos θ}$
  $= \frac{cos ^{2} θ + 1 - 2 sin θ + sin ^{2} θ}{( 1 - sin θ ) cos θ}$
  1 mark
3. Using $sin^{2} θ + cos^{2} θ = 1$ :
  $= \frac{2 - 2 sin θ}{( 1 - sin θ ) cos θ}$
  ½ mark
4. $= \frac{2 ( 1 - sin θ )}{( 1 - sin θ ) cos θ} = \frac{2}{cos θ} = 2 sec θ$ .
  ½ mark
Answer: Error in step 2; correct answer $2 sec θ$
Where the marks are
- Identifies step 2 as the wrong step (1 mark)
- Expands $(1 - sin θ)^{2}$ correctly in the numerator (1 mark)
- Uses $sin^{2} θ + cos^{2} θ = 1$ to get $2 - 2 sin θ$ (½ mark)
- Cancels and gets $2 sec θ$ (½ mark)
Open Q31(A) on its own page
OR
Q31(B) · 3 marks · Short answerIntroduction to Trigonometry
If $\frac{1}{s i n x - c o s x} = \frac{cosec x}{2}$ , then prove that $(\frac{1}{s i n x + c o s x})^{2} = \frac{s e c ^{2} x}{2}$ .
Try it first, then show our solutionSolution
1. Given $\frac{1}{sin x - cos x} = \frac{cosec x}{2} = \frac{1}{2 sin x}$ .
  Taking reciprocals: $sin x - cos x = 2 sin x$ .
  ½ mark
2. Squaring both sides:
  $sin^{2} x + cos^{2} x - 2 sin x cos x = 2 sin^{2} x$
  $\Rightarrow 1 - 2 sin x cos x = 2 sin^{2} x$
  $\Rightarrow 2 sin x cos x = 1 - 2 sin^{2} x$ …(i)
  1 mark
3. LHS $= (\frac{1}{sin x + cos x})^{2} = \frac{1}{sin ^{2} x + cos ^{2} x + 2 sin x cos x} = \frac{1}{1 + 2 sin x cos x}$ .
  ½ mark
4. Using (i): $= \frac{1}{1 + 1 - 2 sin ^{2} x} = \frac{1}{2 ( 1 - sin ^{2} x )} = \frac{1}{2 cos ^{2} x}$ .
  ½ mark
5. $= \frac{sec ^{2} x}{2} =$ RHS. Hence proved.
  ½ mark
Answer: Proved: $(\frac{1}{s i n x + c o s x})^{2} = \frac{s e c ^{2} x}{2}$
Where the marks are
- Simplifies the given condition to $sin x - cos x = 2 sin x$ (½ mark)
- Squares to get $2 sin x cos x = 1 - 2 sin^{2} x$ (1 mark)
- Writes LHS as $\frac{1}{1 + 2 s i n x c o s x}$ (½ mark)
- Substitutes to get $\frac{1}{2 c o s ^{2} x}$ (½ mark)
- Concludes $\frac{s e c ^{2} x}{2}$ = RHS (½ mark)
Open Q31(B) on its own page

Section D

Long Answer (LA)

Q32 · 5 marks · Long answerCompetency-basedQuadratic Equations
In a rectangular park measuring 50 m $\times$ 40 m, the gram panchayat plans to construct a rectangular swimming pool in the middle, surrounded by a uniform-width grass strip throughout the park. Find the dimensions of the swimming pool, if the area of the grassy strip is $1184 m^{2}$ .
Try it first, then show our solutionSolution
1. Figure: ABCD is the park, 50 m by 40 m. PQRS is the pool in the middle, with a grass strip of width $x$ m on all four sides.
  Length of pool $= (50 - 2 x)$ m and breadth of pool $= (40 - 2 x)$ m.
  ½ mark
2. Area of grass strip = Area of park − Area of pool:
  $50 \times 40 - (50 - 2 x) (40 - 2 x) = 1184$ .
  1 mark
3. $2000 - (2000 - 180 x + 4 x^{2}) = 1184$
  $\Rightarrow 180 x - 4 x^{2} = 1184$
  $\Rightarrow 4 x^{2} - 180 x + 1184 = 0$
  $\Rightarrow x^{2} - 45 x + 296 = 0$ .
  1 mark
4. $(x - 8) (x - 37) = 0 \Rightarrow x = 8$ or $x = 37$ .
  ½ mark
5. If $x = 37$ , the breadth $40 - 74 = - 34$ m, which is negative. So reject $x = 37$ .
  Hence $x = 8$ .
  1 mark
6. Length of pool $= 50 - 2 (8) = 34$ m.
  ½ mark
7. Breadth of pool $= 40 - 2 (8) = 24$ m.
  (Check: $2000 - 34 \times 24 = 2000 - 816 = 1184 m^{2}$ .)
  The swimming pool is 34 m × 24 m.
  ½ mark
Answer: Swimming pool: 34 m × 24 m (strip width 8 m)
Where the marks are
- Writes the pool's dimensions as $(50 - 2 x)$ m and $(40 - 2 x)$ m (½ mark)
- Forms the area equation $2000 - (50 - 2 x) (40 - 2 x) = 1184$ (1 mark)
- Reduces to $x^{2} - 45 x + 296 = 0$ (1 mark)
- Factorises to get $x = 8, 37$ (½ mark)
- Rejects $x = 37$ with a reason and takes $x = 8$ (1 mark)
- Finds length 34 m (½ mark)
- Finds breadth 24 m (½ mark)
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Q33 · 5 marks · Long answerTriangles
Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points, divides the other two sides in the same ratio.
Using the above theorem solve the following: PQRS is a trapezium with PQ $∥$ SR. X and Y are points on non-parallel sides PS and QR respectively such that XY $∥$ PQ. Show that $\frac{P X}{X S} = \frac{Q Y}{Y R}$ .
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ with a line DE parallel to BC, meeting AB at D and AC at E.
  Given: In $△ A B C$ , $D E ∥ B C$ , with D on AB and E on AC.
  To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
  Construction: Join BE and CD. Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  1 mark
2. Proof: Use area $= \frac{1}{2} \times$ base $\times$ height, with $E N$ as the common height:
  $\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{\frac{1}{2} \times A D \times E N}{\frac{1}{2} \times D B \times E N} = \frac{A D}{D B}$ …(1)
  ½ mark
3. Similarly, with $D M$ as the common height:
  $\frac{a r ( △ A D E )}{a r ( △ D E C )} = \frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times E C \times D M} = \frac{A E}{E C}$ …(2)
  ½ mark
4. $△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC.
  So $a r (△ B D E) = a r (△ D E C)$ …(3)
  ½ mark
5. From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence the theorem is proved.
  ½ mark
6. Application.
  Figure: Draw trapezium PQRS with $P Q ∥ S R$ , X on PS and Y on QR, with $X Y ∥ P Q$ .
  Join the diagonal PR. Let it cut XY at M.
  Since $X Y ∥ P Q$ and $P Q ∥ S R$ , we get $X Y ∥ S R$ , so $X M ∥ S R$ .
  ½ mark
7. In $△ P S R$ , $X M ∥ S R$ . By the theorem above (BPT):
  $\frac{P X}{X S} = \frac{P M}{M R}$ …(i)
  ½ mark
8. In $△ R P Q$ , $M Y ∥ P Q$ . By BPT:
  $\frac{R M}{M P} = \frac{R Y}{Y Q}$ , that is, $\frac{P M}{M R} = \frac{Q Y}{Y R}$ …(ii)
  ½ mark
9. From (i) and (ii): $\frac{P X}{X S} = \frac{Q Y}{Y R}$ . Hence shown.
  ½ mark
Answer: Theorem (BPT) proved; using it, $\frac{P X}{X S} = \frac{Q Y}{Y R}$ is shown.
Where the marks are
- Correct figure, given, to prove and construction for the theorem (1 mark)
- Writes area ratios of $△ A D E$ to $△ B D E$ and $△ D E C$ as $\frac{A D}{D B}$ and $\frac{A E}{E C}$ (1 mark)
- Uses equal areas of $△ B D E$ and $△ D E C$ (same base, same parallels) and concludes the theorem (1 mark)
- Joins PR meeting XY at M (½ mark)
- Applies BPT in $△ P S R$ to get $\frac{P X}{X S} = \frac{P M}{M R}$ (½ mark)
- Applies BPT in $△ P QR$ to get $\frac{Q Y}{Y R} = \frac{P M}{M R}$ (½ mark)
- Concludes $\frac{P X}{X S} = \frac{Q Y}{Y R}$ (½ mark)
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Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, then find the area of the canvas required for making the tent, keeping a provision of $26 m^{2}$ of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of $₹ 250$ per $m^{2}$ .
Try it first, then show our solutionSolution
1. Figure (in words): a cylinder of radius $r = 14$ m and height $3$ m, with a cone of the same radius on top. The total height is $13.5$ m.
  Height of the conical part: $h = 13.5 - 3 = 10.5$ m
  ½ mark
2. Slant height of the cone: $l = r^{2} + h^{2} = 1 4^{2} + 10. 5^{2} = 196 + 110.25 = 306.25 = 17.5$ m
  1 mark
3. Curved surface area of the cone: $π r l = \frac{22}{7} \times 14 \times 17.5 = 770 m^{2}$
  1 mark
4. Curved surface area of the cylinder: $2 π r h = 2 \times \frac{22}{7} \times 14 \times 3 = 264 m^{2}$
  The base of the tent is open ground, so no canvas is needed for it.
  1 mark
5. Canvas required = CSA of cone + CSA of cylinder + wastage
  $= 770 + 264 + 26 = 1060 m^{2}$
  ½ mark
6. Cost of canvas $= 1060 \times 250 = ₹ 2, 65, 000$
  1 mark
Answer: Canvas required = 1060 m²; cost = ₹ 2,65,000
Where the marks are
- Finds height of cone = 10.5 m (½ mark)
- Finds slant height = 17.5 m (1 mark)
- Finds CSA of cone = 770 m² (1 mark)
- Finds CSA of cylinder = 264 m² (1 mark)
- Adds wastage to get total canvas = 1060 m² (½ mark)
- Finds cost ₹ 265000 (1 mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes
A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of hemisphere is 3.5 cm and the total wood used in making the toy is $166 \frac{5}{6} cm^{3}$ . Find the height of the conical part. Also, find the cost of painting the hemispherical part of the toy at the rate of $₹ 15$ per $cm^{2}$ .
Try it first, then show our solutionSolution
1. Figure (in words): a hemisphere of radius $r = 3.5$ cm with a cone of the same radius and height $h$ on its flat face.
  Volume of toy = volume of cone + volume of hemisphere:
  $166 \frac{5}{6} = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3}$
  $\Rightarrow \frac{1001}{6} = \frac{1}{3} \times \frac{22}{7} \times (3.5)^{2} \times h + \frac{2}{3} \times \frac{22}{7} \times (3.5)^{3}$
  2 marks
2. Here $π r^{2} = \frac{22}{7} \times 12.25 = 38.5$ . So:
  $\frac{38.5 h}{3} + \frac{539}{6} = \frac{1001}{6}$
  $\Rightarrow \frac{38.5 h}{3} = \frac{462}{6} = 77$
  $\Rightarrow h = \frac{77 \times 3}{38.5} = 6$ cm
  1 mark
3. CSA of the hemispherical part: $2 π r^{2} = 2 \times \frac{22}{7} \times (3.5)^{2} = 77 cm^{2}$
  1 mark
4. Cost of painting $= 77 \times 15 = ₹ 1155$
  1 mark
Answer: Height of cone = 6 cm; cost of painting = ₹ 1155
Where the marks are
- Writes the volume equation (cone + hemisphere = 1001/6) with correct substitution (2 marks)
- Solves to get h = 6 cm (1 mark)
- Finds CSA of hemisphere = 77 cm² (1 mark)
- Finds cost ₹ 1155 (1 mark)
Open Q34(B) on its own page
Answer one of these 2 alternatives.
Q35(A) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
Manish is standing on level ground and observes a kite flying 200 m away from him at an angle of elevation of $3 0^{\circ}$ . Mahesh, standing on the roof of a 50 m high building on the opposite side of the kite, observes the same kite at an angle of elevation of $4 5^{\circ}$ . Find the distance between the kite and Mahesh.
Try it first, then show our solutionSolution
1. Figure (in words):
  Manish stands at A on the ground.
  The kite is at B, with BE ⟂ ground and E on the ground.
  The building is DC = 50 m, with Mahesh at D on its roof, on the other side of E.
  R is the point on BE with DR ∥ CE, so RE = DC = 50 m.
  AB = 200 m and ∠BAE = 30°.
  ∠BDR = 45°.
  1 mark
2. In right $△ A E B$ : $\frac{B E}{A B} = sin 3 0^{\circ} = \frac{1}{2}$
  1 mark
3. $\Rightarrow B E = \frac{200}{2} = 100$ m
  ½ mark
4. $B R = B E - R E = B E - D C = 100 - 50 = 50$ m
  1 mark
5. In right $△ B R D$ : $\frac{B R}{B D} = sin 4 5^{\circ} \Rightarrow \frac{50}{B D} = \frac{1}{2}$
  1 mark
6. $\Rightarrow B D = 502$ m $\approx 70.7$ m
  So the distance between the kite and Mahesh is $502$ m.
  ½ mark
Answer: $502$ m (≈ 70.7 m)
Where the marks are
- Draws a correct labelled figure (1 mark)
- Uses sin 30° in triangle AEB (1 mark)
- Gets BE = 100 m (½ mark)
- Finds BR = 50 m by subtracting the building height (1 mark)
- Uses sin 45° in triangle BRD (1 mark)
- Gets BD = 50√2 m (½ mark)
Open Q35(A) on its own page
OR
Q35(B) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be $4 5^{\circ}$ and $3 0^{\circ}$ . If the ships are 200 m apart and one ship is exactly behind the other, then find the height of the lighthouse.
Try it first, then show our solutionSolution
1. Figure (in words):
  AB = $h$ is the lighthouse, with B at its foot.
  C and D are the ships, on the same line through B, with D beyond C and CD = 200 m.
  The angles of depression from A are 45° (to C) and 30° (to D).
  Hence $∠ A C B = 4 5^{\circ}$ and $∠ A D B = 3 0^{\circ}$ [alternate angles].
  1 mark
2. In right $△ A B C$ : $\frac{h}{B C} = tan 4 5^{\circ} = 1$
  1 mark
3. $\Rightarrow B C = h$ ...(i)
  ½ mark
4. In right $△ A B D$ : $\frac{h}{B C + C D} = tan 3 0^{\circ} \Rightarrow \frac{h}{h + 200} = \frac{1}{3}$
  1 mark
5. $3 h = h + 200$
  $\Rightarrow h = \frac{200}{3 - 1} \times \frac{3 + 1}{3 + 1}$
  ½ mark
6. $h = \frac{200 ( 3 + 1 )}{2} = 100 (3 + 1)$ m $\approx 273.2$ m
  So the height of the lighthouse is $100 (3 + 1)$ m.
  1 mark
Answer: $100 (3 + 1)$ m (≈ 273.2 m)
Where the marks are
- Draws a correct labelled figure (1 mark)
- Uses tan 45° in triangle ABC (1 mark)
- Gets BC = h (½ mark)
- Uses tan 30° with base h + 200 (1 mark)
- Rationalises the denominator (½ mark)
- Gets h = 100(√3 + 1) m (1 mark)
Open Q35(B) on its own page

Section E

Case study based, sub-parts (i), (ii), (iii) of marks 1, 1, 2

Q36 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
A student entrepreneur started a company that manufactures sanitizers in two sizes – small and large. The cost of a small bottle of sanitizer is ₹10 and that of a large bottle is ₹15. In June, the company sold 1000 bottles and recorded a total sale of ₹12,750. Seeing the increased demand, the company decided to increase the price of both sanitizer bottles by ₹2 each. In the next month, the company sold 2500 bottles and recorded a total sale of ₹34,250.
Based on the above information, answer the following questions:
1. (i)
  Form a linear equation in two variables representing the sale for June.
  1 mark
2. (ii)
  Form a linear equation in two variables representing the sale for July.
  1 mark
3. (iii)
  (A) How many sanitizer bottles of each type were actually sold in June?
  2 marks
4. OR
  (iii)
  (B) How many sanitizer bottles of each type were sold in July?
  2 marks
Try it first, then show our solutionSolution
1. Let $x$ = number of small bottles and $y$ = number of large bottles.
  (i) June prices are ₹10 and ₹15, and total sale is ₹12750:
  $10 x + 15 y = 12750$ , i.e. $2 x + 3 y = 2550$
  1 mark
2. (ii) July prices are ₹12 and ₹17, and total sale is ₹34250:
  $12 x + 17 y = 34250$
  1 mark
3. (iii)(a) For June:
  $x + y = 1000$ ...(1)
  $10 x + 15 y = 12750$ ...(2)
  1 mark
4. From (1), $x = 1000 - y$ . Substituting in (2):
  $10 (1000 - y) + 15 y = 12750$
  $\Rightarrow 5 y = 2750 \Rightarrow y = 550$ , so $x = 450$
  In June, 450 small and 550 large bottles were sold.
  1 mark
5. (iii)(b) For July:
  $x + y = 2500$ ...(1)
  $12 x + 17 y = 34250$ ...(2)
  1 mark
6. From (1), $x = 2500 - y$ . Substituting in (2):
  $12 (2500 - y) + 17 y = 34250$
  $\Rightarrow 5 y = 4250 \Rightarrow y = 850$ , so $x = 1650$
  In July, 1650 small and 850 large bottles were sold.
  1 mark
Answer: (i) 10x + 15y = 12750; (ii) 12x + 17y = 34250; (iii)(a) 450 small, 550 large; (iii)(b) 1650 small, 850 large
Where the marks are
- (i) Forms 10x + 15y = 12750 (1 mark)
- (ii) Forms 12x + 17y = 34250 (1 mark)
- (iii)(a) Writes the pair x + y = 1000 and 10x + 15y = 12750 (1 mark)
- (iii)(a) Solves to x = 450, y = 550 (1 mark)
- (iii)(b) Writes the pair x + y = 2500 and 12x + 17y = 34250 (1 mark)
- (iii)(b) Solves to x = 1650, y = 850 (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedCircles
In a technology park in Hyderabad, a company installed a circular meditation garden for its employees as shown in the figure. Arjun, a designer, stood at a point P outside the garden and drew two tangents PA and PB to the circular boundary. The radius of the garden is 9 m, and the distance of point P from the centre O is 18 m. For preparing safety guidelines, Arjun needed the lengths of the tangents. During a demonstration to interns, he asked them to compute the angle formed between the two tangents using the properties of circle.
Based on the above information, answer the following questions:
1. (i)
  What is the measure of $∠ O A P$ ?
  1 mark
2. (ii)
  Calculate the length of the tangent PA.
  1 mark
3. (iii)
  (A) Calculate the measure of $∠ A O B$ .
  2 marks
4. OR
  (iii)
  (B) In the above design, if PA and PB were inclined at $6 0^{\circ}$ , then what would be their length?
  2 marks
Figure in the paper: A circle with centre O and points A and B on the circle, with AB a chord drawn vertically through the figure. Point P lies outside the circle to the right; tangents PA and PB are drawn from P touching the circle at A and B. Radius OA and OB are drawn to the points of contact; OP is the line from the centre to P (OP = 18 m, radius = 9 m). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius at the point of contact]
  1 mark
2. (ii) In right $△ O A P$ : $O P^{2} = O A^{2} + A P^{2} \Rightarrow 1 8^{2} = 9^{2} + A P^{2}$
  ½ mark
3. $A P^{2} = 324 - 81 = 243 \Rightarrow A P = 243 = 93$ m $\approx 15.6$ m
  ½ mark
4. (iii)(a) Let $∠ A O P = θ$ . In right $△ O A P$ :
  $cos θ = \frac{O A}{O P} = \frac{9}{18} = \frac{1}{2} = cos 6 0^{\circ}$
  1 mark
5. $\Rightarrow θ = 6 0^{\circ}$
  ½ mark
6. OP bisects $∠ A O B$ [ $△ O A P ≅ △ O B P$ by RHS].
  So $∠ A O B = 2 θ = 12 0^{\circ}$ .
  ½ mark
7. (iii)(b) If $∠ A P B = 6 0^{\circ}$ , then OP bisects it [tangents from an external point are equally inclined to OP].
  So $∠ O P A = \frac{1}{2} \times 6 0^{\circ} = 3 0^{\circ}$ .
  ½ mark
8. In right $△ O A P$ : $\frac{O A}{A P} = tan 3 0^{\circ} \Rightarrow \frac{9}{A P} = \frac{1}{3}$
  1 mark
9. $\Rightarrow A P = 93$ m, and $P B = P A = 93$ m [tangents from an external point are equal]
  ½ mark
Answer: (i) 90°; (ii) 9√3 m; (iii)(a) 120°; (iii)(b) 9√3 m each
Where the marks are
- (i) States ∠OAP = 90° with reason (1 mark)
- (ii) Applies Pythagoras theorem in triangle OAP (½ mark)
- (ii) Gets AP = 9√3 m (½ mark)
- (iii)(a) Uses cos θ = 9/18 = 1/2 (1 mark)
- (iii)(a) Gets θ = 60° (½ mark)
- (iii)(a) Gets ∠AOB = 120° (½ mark)
- (iii)(b) Gets ∠OPA = 30° (½ mark)
- (iii)(b) Uses tan 30° = 9/AP (1 mark)
- (iii)(b) Gets AP = 9√3 m (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedStatistics
A school conducted a weekly test for Class X students, before the commencement of the pre-board examination and recorded their scores (out of 50). To analyse performance patterns, the academic coordinator grouped the marks into intervals. The grouped frequency distribution is as below:
Marks Obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Number of students 3 6 12 15 14
The academic coordinator computed the central tendencies to judge overall learning level, the most common performance range and to understand consistency across the batch for planning the remedial sessions.
Based on the above information, answer the following questions:
1. (i)
  Identify the class with the most common performance range.
  1 mark
2. (ii)
  Find the class interval containing the median.
  1 mark
3. (iii)
  (A) What is the average performance of the students?
  2 marks
4. OR
  (iii)
  (B) Find the mode of the data.
  2 marks
Try it first, then show our solutionSolution
1. (i) The most common performance range is the class with the highest frequency (15).
  So the modal class is 30 – 40.
  1 mark
2. (ii) Cumulative frequencies are 3, 9, 21, 36, 50, so $N = 50$ and $\frac{N}{2} = 25$ .
  The first cf greater than 25 is 36, which belongs to class 30 – 40. So the median class is 30 – 40.
  1 mark
3. (iii)(a) Table using class marks $x_{i}$ :
  Class $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  0 – 10 3 5 15
  10 – 20 6 15 90
  20 – 30 12 25 300
  30 – 40 15 35 525
  40 – 50 14 45 630
  Total 50 1560
  1 mark
4. Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{1560}{50} = 31.2$ marks
  1 mark
5. (iii)(b) Modal class is 30 – 40, with:
  $l = 30$ , $h = 10$
  $f_{1} = 15$ , $f_{0} = 12$ , $f_{2} = 14$
  ½ mark
6. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 30 + (\frac{15 - 12}{30 - 12 - 14}) \times 10$
  1 mark
7. $= 30 + \frac{3}{4} \times 10 = 37.5$ marks
  ½ mark
Answer: (i) 30 – 40; (ii) 30 – 40; (iii)(a) 31.2; (iii)(b) 37.5
Where the marks are
- (i) Identifies modal class 30 – 40 (1 mark)
- (ii) Uses cumulative frequency and N/2 = 25 to get median class 30 – 40 (1 mark)
- (iii)(a) Correct table with class marks and fᵢxᵢ, total 1560 (1 mark)
- (iii)(a) Mean = 31.2 (1 mark)
- (iii)(b) Correct l, f₀, f₁, f₂ (½ mark)
- (iii)(b) Correct substitution in the mode formula (1 mark)
- (iii)(b) Mode = 37.5 (½ mark)
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Marks Obtained	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Number of students	3	6	12	15	14

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0 – 10	3	5	15
10 – 20	6	15	90
20 – 30	12	25	300
30 – 40	15	35	525
40 – 50	14	45	630
Total	50		1560

Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.