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

CBSE Class 10 Mathematics Basic: 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/MathsBasic-SQP.pdf. Questions are transcribed from that PDF; figures are described in words, so check the official paper for them. Questions printed only for visually impaired candidates are left out (3).

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 21 and 25
C	26-31	Short answer; internal choice in 27 and 30
D	32-35	Long answer; internal choice in 33 and 35
E	36-38	Case study based; sub-parts 1, 1, 2 marks; internal choice in sub-part (iii) of each

Instructions in brief: There is no overall choice. Internal choice is provided in 2 questions of Section B (21, 25), 2 questions of Section C (27, 30) and 2 questions of Section D (33, 35), and in all 2-mark sub-parts of Section E (sub-part (iii) of 36, 37 and 38). Visually Impaired alternatives are given for Q24, Q27 and Q36. Use of calculators is not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
If $HCF (2520, 6600) = 40$ and $LCM (2520, 6600) = 252 \times k$ , then the value of $k$ is
1. (A)165
2. (B)1600
3. (C)1625
4. (D)1650
Try it first, then show our solutionSolution
1. Use HCF × LCM = product of the numbers: $40 \times 252 k = 2520 \times 6600$ .
2. $k = \frac{2520 \times 6600}{40 \times 252} = 10 \times 165 = 1650$ . Option (D).
  1 mark
Answer: k = 1650, option (D)
Where the marks are
- Uses HCF × LCM = product and gets k = 1650 (1 mark)
Q2 · 1 mark · MCQPolynomials
If $α$ and $β$ are the Zeroes of the polynomial $2 x^{2} - 4 x - 5$ , the value of $(α - β)^{2}$ is
1. (A)4
2. (B)6
3. (C)14
4. (D)56
Try it first, then show our solutionSolution
1. $α + β = \frac{4}{2} = 2$ and $α β = - \frac{5}{2}$ .
2. $(α - β)^{2} = (α + β)^{2} - 4 α β = 4 + 10 = 14$ . Option (C).
  1 mark
Answer: 14, option (C)
Where the marks are
- Uses the sum and product of zeroes to get 14 (1 mark)
Q3 · 1 mark · MCQPair of Linear Equations in Two Variables
For what value of $p$ does the pair of Linear equations $4 x + p y + 8 = 0$ and $2 x + 2 y + 2 = 0$ has a unique solution
1. (A) $p = 4$ only
2. (B) $p = 2$ only
3. (C) $p \neq = 4$
4. (D) $p \neq = 2$
Try it first, then show our solutionSolution
1. A pair of linear equations has a unique solution when $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ .
2. $\frac{4}{2} \neq = \frac{p}{2} \Rightarrow p \neq = 4$ . Option (C).
  1 mark
Answer: p ≠ 4, option (C)
Where the marks are
- Applies $a_{1} / a_{2} \neq = b_{1} / b_{2}$ to get p ≠ 4 (1 mark)
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Q4 · 1 mark · MCQPair of Linear Equations in Two Variables
The sum of the numerator and denominator of a fraction is 11. If the denominator is increased by 1, the fraction becomes $\frac{1}{2}$ , then the fraction is
1. (A) $\frac{2}{9}$
2. (B) $\frac{3}{8}$
3. (C) $\frac{4}{7}$
4. (D) $\frac{5}{6}$
Try it first, then show our solutionSolution
1. Let the fraction be $\frac{x}{y}$ . Then $x + y = 11$ and $\frac{x}{y + 1} = \frac{1}{2}$ , so $2 x = y + 1$ .
2. Substituting gives $3 x = 12$ , so $x = 4$ and $y = 7$ . The fraction is $\frac{4}{7}$ . Option (C).
  1 mark
Answer: 4/7, option (C)
Where the marks are
- Forms both equations and solves to get 4/7 (1 mark)
Open Q4 on its own page
Q5 · 1 mark · MCQQuadratic Equations
If one root of the quadratic equation $a x^{2} + b x + c = 0$ is the reciprocal of the other, then
1. (A) $b = c$
2. (B) $a = b$
3. (C) $a c = 1$
4. (D) $a = c$
Try it first, then show our solutionSolution
1. Let the roots be $α$ and $\frac{1}{α}$ . Their product is $1$ .
2. Product of roots $= \frac{c}{a} = 1 \Rightarrow a = c$ . Option (D).
  1 mark
Answer: a = c, option (D)
Where the marks are
- Uses product of roots = c/a = 1 (1 mark)
Q6 · 1 mark · MCQArithmetic Progressions
The first term of AP is $p$ and the common difference is $q$ , then its $1 0^{t h}$ term is
1. (A) $q + 10 p$
2. (B) $p - 9 q$
3. (C) $p + 9 q$
4. (D) $p + 10 q$
Try it first, then show our solutionSolution
1. The nth term of an AP is $a_{n} = a + (n - 1) d$ .
2. $a_{10} = p + 9 q$ . Option (C).
  1 mark
Answer: p + 9q, option (C)
Where the marks are
- Applies $a_{n} = a + (n - 1) d$ (1 mark)
Q7 · 1 mark · MCQArithmetic Progressions
Which term of the AP: 21, 42, 63, 84 ... is 210?
1. (A) $9^{t h}$
2. (B) $1 0^{t h}$
3. (C) $1 1^{t h}$
4. (D) $1 2^{t h}$
Try it first, then show our solutionSolution
1. Here $a = 21$ and $d = 21$ .
2. $210 = 21 + (n - 1) 21 \Rightarrow n - 1 = 9 \Rightarrow n = 10$ . So 210 is the $1 0^{t h}$ term. Option (B).
  1 mark
Answer: 10th term, option (B)
Where the marks are
- Solves $21 + (n - 1) 21 = 210$ to get n = 10 (1 mark)
Q8 · 1 mark · MCQCoordinate Geometry
If the point P (5, 2) divides the line segment joining A( 8, 5) and B( 4, $y$ ) in the ratio 3 : 1, then the value of $y$ is
1. (A)4
2. (B)3
3. (C)2
4. (D)1
Try it first, then show our solutionSolution
1. By the section formula with ratio 3 : 1, the y-coordinate of P is $\frac{3 y + 1 ( 5 )}{3 + 1}$ .
2. $\frac{3 y + 5}{4} = 2 \Rightarrow 3 y = 3 \Rightarrow y = 1$ . Option (D).
  1 mark
Answer: y = 1, option (D)
Where the marks are
- Applies the section formula to the y-coordinate to get y = 1 (1 mark)
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Q9 · 1 mark · MCQProbability
A letter from the word INDEPENDENCE is selected at random. What is the probability that the letter selected is a vowel which occurs the maximum number of times in the given word?
1. (A) $\frac{1}{12}$
2. (B) $\frac{1}{4}$
3. (C) $\frac{1}{3}$
4. (D) $\frac{5}{12}$
Try it first, then show our solutionSolution
1. INDEPENDENCE has 12 letters. Its vowels are I (once) and E (4 times), so E is the most frequent vowel.
2. P(E) $= \frac{4}{12} = \frac{1}{3}$ . Option (C).
  1 mark
Answer: 1/3, option (C)
Where the marks are
- Counts 4 E's out of 12 letters to get 1/3 (1 mark)
Q10 · 1 mark · MCQStatistics
For the following distribution
Class 0-5 5-10 10-15 15-20 20-25
Frequency 10 15 12 20 9
The lower limit of the median class is
1. (A)5
2. (B)10
3. (C)15
4. (D)20
Try it first, then show our solutionSolution
1. Total frequency $N = 66$ , so $\frac{N}{2} = 33$ . The cumulative frequencies are 10, 25, 37, 57, 66.
2. The first cumulative frequency exceeding 33 is 37, so the median class is 10–15 and its lower limit is 10. Option (B).
  1 mark
Answer: 10, option (B)
Where the marks are
- Finds the median class 10–15 from the cumulative frequencies (1 mark)
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Q11 · 1 mark · MCQCircles
The length of a tangent drawn from a point at a distance of 10 cm from the centre of the circle is 8 cm. The radius of the circle is
1. (A)4 cm
2. (B)5 cm
3. (C)6 cm
4. (D)7 cm
Try it first, then show our solutionSolution
1. The radius is perpendicular to the tangent at the point of contact, so $1 0^{2} = 8^{2} + r^{2}$ .
2. $r = 100 - 64 = 6$ cm. Option (C).
  1 mark
Answer: 6 cm, option (C)
Where the marks are
- Applies Pythagoras with radius ⟂ tangent to get 6 cm (1 mark)
Q12 · 1 mark · MCQStatistics
Mean and median of certain data are 32 and 30 respectively. Using empirical formula, the value of mode is
1. (A)36
2. (B)26
3. (C)30
4. (D)20
Try it first, then show our solutionSolution
1. Empirical formula: Mode = 3 Median − 2 Mean.
2. Mode $= 3 (30) - 2 (32) = 90 - 64 = 26$ . Option (B).
  1 mark
Answer: 26, option (B)
Where the marks are
- Uses Mode = 3 Median − 2 Mean to get 26 (1 mark)
Q13 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
A ladder 15 m long reaches a window 12 m above the ground. The distance of the foot of the ladder from the base of the wall is
1. (A)8 m
2. (B)9 m
3. (C)10 m
4. (D)13 m
Try it first, then show our solutionSolution
1. The ladder is the hypotenuse, so $1 5^{2} = 1 2^{2} + d^{2}$ .
2. $d = 225 - 144 = 81 = 9$ m. Option (B).
  1 mark
Answer: 9 m, option (B)
Where the marks are
- Applies Pythagoras' theorem to get 9 m (1 mark)
Q14 · 1 mark · MCQIntroduction to Trigonometry
The value of $sin^{2} 6 0^{\circ} - 2 tan^{2} 4 5^{\circ} - cos^{2} 3 0^{\circ}$ is
1. (A) $- 2$
2. (B) $- 1$
3. (C)1
4. (D)2
Try it first, then show our solutionSolution
1. $sin 6 0^{\circ} = cos 3 0^{\circ} = \frac{3}{2}$ and $tan 4 5^{\circ} = 1$ .
2. $\frac{3}{4} - 2 (1) - \frac{3}{4} = - 2$ . Option (A).
  1 mark
Answer: −2, option (A)
Where the marks are
- Substitutes standard values to get −2 (1 mark)
Q15 · 1 mark · MCQTriangles
The perimeter of two similar triangles is 28 cm and 35 cm respectively. If one side of the first triangle is 8 cm, then the corresponding side of the second triangle is
1. (A)10 cm
2. (B)12 cm
3. (C)14 cm
4. (D)16 cm
Try it first, then show our solutionSolution
1. For similar triangles, the ratio of perimeters equals the ratio of corresponding sides: $\frac{28}{35} = \frac{8}{x}$ .
2. $x = \frac{8 \times 35}{28} = 10$ cm. Option (A).
  1 mark
Answer: 10 cm, option (A)
Where the marks are
- Uses ratio of perimeters = ratio of sides to get 10 cm (1 mark)
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Q16 · 1 mark · MCQTriangles
If in $△ A B C$ and $△ P QR$ , $∠ B = ∠ Q$ , $∠ R = ∠ C$ and $A B = 2 P Q$ , then the two triangles are
1. (A)Congruent but not similar
2. (B)Similar but not congruent
3. (C)Neither congruent nor similar
4. (D)Congruent as well as similar
Try it first, then show our solutionSolution
1. Two pairs of angles are equal, so $△ A B C \sim △ P QR$ by AA similarity.
2. Since $A B = 2 P Q$ , corresponding sides are not equal, so the triangles are not congruent. Option (B).
  1 mark
Answer: Similar but not congruent, option (B)
Where the marks are
- States AA similarity and notes AB ≠ PQ, so not congruent (1 mark)
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Q17 · 1 mark · MCQCircles
If tangents PA and PB from a point P to a circle with centre O and respective point of contacts as A and B, are inclined to each other at angle of $8 0^{\circ}$ , then $∠ A O B$ is equal to
1. (A) $6 0^{\circ}$
2. (B) $7 0^{\circ}$
3. (C) $8 0^{\circ}$
4. (D) $10 0^{\circ}$
Try it first, then show our solutionSolution
1. $∠ O A P = ∠ O B P = 9 0^{\circ}$ [tangent ⟂ radius]. The angles of quadrilateral OAPB add up to $36 0^{\circ}$ .
2. $∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 8 0^{\circ} = 10 0^{\circ}$ . Option (D).
  1 mark
Answer: 100°, option (D)
Where the marks are
- Uses radius ⟂ tangent and the angle sum of a quadrilateral to get 100° (1 mark)
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Q18 · 1 mark · MCQSurface Areas and Volumes
Two cubes each of volume $64 cm^{3}$ are joined end to end to form a cuboid. The total surface area of the resulting cuboid is
1. (A) $128 cm^{2}$
2. (B) $160 cm^{2}$
3. (C) $176 cm^{2}$
4. (D) $192 cm^{2}$
Try it first, then show our solutionSolution
1. $a^{3} = 64 \Rightarrow a = 4$ cm. The cuboid formed is 8 cm × 4 cm × 4 cm.
2. TSA $= 2 (32 + 16 + 32) = 160 cm^{2}$ . Option (B).
  1 mark
Answer: 160 cm², option (B)
Where the marks are
- Finds the cuboid dimensions and uses TSA = 2(lb + bh + hl) to get 160 cm² (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Directions: Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the options given below.
ASSERTION (A): The probability of getting number 8 on rolling a die is zero (0).
REASON (R): The probability of an impossible event is zero (0).
1. (A)Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true but Reason (R) is false.
4. (D)Assertion (A) is false but Reason (R) is true.
Try it first, then show our solutionSolution
1. A die shows only the numbers 1 to 6, so getting 8 is an impossible event. Hence A is true.
2. R is true: an impossible event has probability 0. R explains why A is true. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Identifies both statements as true and R as the correct explanation of A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedTriangles
Directions: Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the options given below.
ASSERTION (A): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
REASON (R): Line drawn from midpoint of one side of triangle parallel to another will bisect the third side.
1. (A)Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true but Reason (R) is false.
4. (D)Assertion (A) is false but Reason (R) is true.
Try it first, then show our solutionSolution
1. A is the converse of the Basic Proportionality Theorem, so it is true. R is the mid-point theorem result, also true.
2. R deals with a special case about bisecting a side; it does not explain the general converse of BPT. Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Recognises both statements as true and R as not explaining A (1 mark)
Open Q20 on its own page

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

Section B

Very short answer; internal choice in 21 and 25

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerReal Numbers
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Try it first, then show our solutionSolution
1. The required number increased by 17 must be a common multiple of 520 and 468. For the smallest such number, we use the LCM.
  Prime factorisation: $520 = 2^{3} \times 5 \times 13$ and $468 = 2^{2} \times 3^{2} \times 13$
  1 mark
2. $LCM (520, 468) = 2^{3} \times 3^{2} \times 5 \times 13 = 4680$
  ½ mark
3. Required number $= 4680 - 17 = 4663$
  ½ mark
Answer: 4663
Where the marks are
- Correct prime factorisation of 520 and 468 (1 mark)
- LCM = 4680 (½ mark)
- Required number = 4680 − 17 = 4663 (½ mark)
Open Q21(A) on its own page
OR
Q21(B) · 2 marks · Very short answerReal Numbers
Show that $(25)^{n}$ will never end with digit zero for any natural number $n$ .
Try it first, then show our solutionSolution
1. $(25)^{n} = (5^{2})^{n} = 5^{2 n}$
  ½ mark
2. So the prime factorisation of $(25)^{n}$ contains only the prime 5. By the Fundamental Theorem of Arithmetic, this factorisation is unique.
  ½ mark
3. A number ends with digit 0 only if it is divisible by 10, that is, its prime factorisation must contain both 2 and 5. Here 2 is not a factor of $(25)^{n}$ .
  ½ mark
4. Hence $(25)^{n}$ can never end with the digit zero for any natural number $n$ .
  ½ mark
Answer: $(25)^{n} = 5^{2 n}$ has no factor 2, so it is not divisible by 10 and never ends in 0.
Where the marks are
- Writes $(25)^{n} = 5^{2 n}$ (½ mark)
- States that the factorisation contains only 5 and is unique (½ mark)
- States that ending in 0 needs both 2 and 5 as factors, and 2 is absent (½ mark)
- Concludes that it never ends in 0 (½ mark)
Open Q21(B) on its own page
Q22 · 2 marks · Very short answerIntroduction to Trigonometry
If $3 sin θ = 4 cos θ$ , find the value of $sin θ + cos^{2} θ - 1$ ; $0^{\circ} < θ < 9 0^{\circ}$
Try it first, then show our solutionSolution
1. $3 sin θ = 4 cos θ \Rightarrow \frac{sin θ}{cos θ} = \frac{4}{3} \Rightarrow tan θ = \frac{4}{3}$
  ½ mark
2. Take a right triangle with perpendicular 4 and base 3. Then the hypotenuse is $3^{2} + 4^{2} = 5$ .
  So $sin θ = \frac{4}{5}$ and $cos θ = \frac{3}{5}$ .
  ½ mark
3. $sin θ + cos^{2} θ - 1 = \frac{4}{5} + \frac{9}{25} - 1 = \frac{20 + 9 - 25}{25} = \frac{4}{25}$
  1 mark
Answer: $\frac{4}{25}$
Where the marks are
- Obtains $tan θ = \frac{4}{3}$ (½ mark)
- Finds $sin θ = \frac{4}{5}$ and $cos θ = \frac{3}{5}$ (½ mark)
- Substitutes and gets $\frac{4}{25}$ (1 mark)
Open Q22 on its own page
Q23 · 2 marks · Very short answerPolynomials
If one zero of the quadratic polynomial $2 x^{2} - 3 x + p$ is 3, find the value of $p$ . Also, find the other zero.
Try it first, then show our solutionSolution
1. Since 3 is a zero, $p (3) = 0$ : $2 (3)^{2} - 3 (3) + p = 0$
  ½ mark
2. $18 - 9 + p = 0 \Rightarrow p = - 9$
  ½ mark
3. Let the other zero be $β$ . Sum of zeroes $= - \frac{b}{a}$ , so $3 + β = \frac{- ( - 3 )}{2} = \frac{3}{2}$
  ½ mark
4. $β = \frac{3}{2} - 3 = - \frac{3}{2}$
  ½ mark
Answer: $p = - 9$ ; other zero $= - \frac{3}{2}$
Where the marks are
- Substitutes $x = 3$ and equates to 0 (½ mark)
- Gets $p = - 9$ (½ mark)
- Uses sum of zeroes $= - b / a$ (½ mark)
- Other zero $= - \frac{3}{2}$ (½ mark)
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Q24 · 2 marks · Very short answerCircles
In the given figure, PA is a common tangent and QB and PC are the tangents from Q and P to the smaller and larger circle respectively. If QB = 5 cm and PC = 9 cm, then evaluate length of PQ.
Figure in the paper: Two circles with common centre O, the smaller inside the larger, touching at point A on the bottom. A straight line PA is a common tangent touching both circles at A (P lies to the right, on the line through A). Q lies on segment AP between A and P and is the point from which tangent QB touches the smaller circle at B (QB = 5 cm, labelled on the segment from Q to B). From P a tangent PC touches the larger circle at C (PC = 9 cm, labelled along the segment from C to P). The points C, B are on the upper part of the circles and the tangent lines CP and QB are drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. From Q, QA and QB are tangents to the smaller circle. [Lengths of tangents from an external point are equal], so $Q A = QB = 5$ cm.
  ½ mark
2. From P, PA and PC are tangents to the larger circle. [Lengths of tangents from an external point are equal], so $P A = P C = 9$ cm.
  ½ mark
3. Q lies on AP, so $P Q = A P - A Q = 9 - 5 = 4$ cm.
  1 mark
Answer: PQ = 4 cm
Where the marks are
- QA = QB = 5 cm with the equal-tangents reason (½ mark)
- PA = PC = 9 cm (½ mark)
- PQ = AP − AQ = 4 cm (1 mark)
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Answer one of these 2 alternatives.
Q25(A) · 2 marks · Very short answerCoordinate Geometry
Points A (3, 1), B (5, 1), C (a, b) and D (4, 3) are vertices of a parallelogram ABCD. Find the values of $a$ and $b$ .
Try it first, then show our solutionSolution
1. The diagonals of a parallelogram bisect each other, so the midpoint of AC = the midpoint of BD.
  ½ mark
2. $(\frac{3 + a}{2}, \frac{1 + b}{2}) = (\frac{5 + 4}{2}, \frac{1 + 3}{2})$ , so $\frac{3 + a}{2} = \frac{9}{2}$ and $\frac{1 + b}{2} = 2$
  1 mark
3. $a = 6, b = 3$
  ½ mark
Answer: $a = 6$ , $b = 3$
Where the marks are
- Uses the property that the diagonals bisect each other (½ mark)
- Equates the midpoints correctly (1 mark)
- $a = 6$ , $b = 3$ (½ mark)
Open Q25(A) on its own page
OR
Q25(B) · 2 marks · Very short answerCoordinate Geometry
Find a linear relation between $x$ and $y$ such that P $(x, y)$ is equidistant from the points A (1, 4) and B ( $- 1$ , 2).
Try it first, then show our solutionSolution
1. $P A = P B \Rightarrow (x - 1)^{2} + (y - 4)^{2} = (x + 1)^{2} + (y - 2)^{2}$
  ½ mark
2. Squaring both sides: $x^{2} - 2 x + 1 + y^{2} - 8 y + 16 = x^{2} + 2 x + 1 + y^{2} - 4 y + 4$ , which simplifies to $4 x + 4 y = 12$
  1 mark
3. $x + y = 3$
  ½ mark
Answer: $x + y = 3$
Where the marks are
- Writes PA = PB using the distance formula (½ mark)
- Squares and simplifies correctly (1 mark)
- $x + y = 3$ (½ mark)
Open Q25(B) on its own page

Section C

Short answer; internal choice in 27 and 30

Q26 · 3 marks · Short answerReal Numbers
Given that $5$ is irrational, prove that $2 + 35$ is irrational.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $2 + 35$ is rational. Then $2 + 35 = \frac{p}{q}$ , where $p$ and $q$ are integers and $q \neq = 0$ .
  ½ mark
2. $35 = \frac{p}{q} - 2 \Rightarrow 5 = \frac{p - 2 q}{3 q}$
  1 mark
3. Since $p$ and $q$ are integers, $\frac{p - 2 q}{3 q}$ is rational, so $5$ would be rational.
  ½ mark
4. But $5$ is irrational (given), which is a contradiction.
  ½ mark
5. Hence our assumption is wrong, and $2 + 35$ is irrational.
  ½ mark
Answer: $2 + 35$ is irrational (proved by contradiction).
Where the marks are
- Assumes $2 + 35 = p / q$ with $q \neq = 0$ (½ mark)
- Expresses $5 = \frac{p - 2 q}{3 q}$ (1 mark)
- States that the RHS is rational (½ mark)
- Identifies the contradiction with $5$ being irrational (½ mark)
- Conclusion (½ mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(A) · 3 marks · Short answerCircles
In the given figure, AX, AY and BC are tangents to the circle with P, R and Q respective point of contacts. Prove that $A P = \frac{1}{2} (A B + B C + C A)$
Figure in the paper: A circle with three tangents forming a triangle ABC. A is the external apex; two tangent lines AX and AY meet at A, with AX going down-left and AY going down-right. A third tangent BC touches the circle at Q, cutting across the top of the circle, with B on line AX and C on line AY. The points of contact on the circle are P (on AX), R (on AY) and Q (on BC). Labels: A at top, B and C near the top of the circle, X and Y at the ends of the tangent lines, P, Q and R on the circle. See it in the official paper.
Try it first, then show our solutionSolution
1. Perimeter: $A B + B C + C A = A B + (B Q + QC) + C A$ [Q lies on BC]
  1 mark
2. $B Q = B P$ and $C Q = C R$ [lengths of tangents from an external point to a circle are equal]
  So $A B + B C + C A = (A B + B P) + (C R + A C)$
  ½ mark
3. $= A P + A R$ (since B lies on AP and C lies on AR)
  $= 2 A P$ [ $A P = A R$ , tangents from A]
  1 mark
4. Hence $A P = \frac{1}{2} (A B + B C + C A)$
  ½ mark
Answer: $A P = \frac{1}{2} (A B + B C + C A)$ , proved
Where the marks are
- Splits BC as BQ + QC in the perimeter (1 mark)
- Uses BQ = BP and CQ = CR with the reason (½ mark)
- Rewrites the perimeter as AP + AR = 2AP using AP = AR (1 mark)
- Final conclusion (½ mark)
Open Q27(A) on its own page
OR
Q27(B) · 3 marks · Short answerCircles
Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that $∠ A P B = 2∠ O A B$
Try it first, then show our solutionSolution
1. Figure: Draw a circle with centre O and an external point P. Draw tangents PA and PB touching the circle at A and B. Join OA and AB.
  ½ mark
2. $O A ⊥ A P$ [tangent ⟂ radius at the point of contact], so $∠ O A P = 9 0^{\circ}$ .
  Therefore $∠ P A B = 9 0^{\circ} - ∠ O A B ...(i)$
  ½ mark
3. $P A = P B$ [tangents from an external point are equal], so $∠ P A B = ∠ P B A$ [angles opposite equal sides].
  In $△ P A B$ , by the angle sum property:
  $∠ A P B = 18 0^{\circ} - 2∠ P A B$
  1 mark
4. Using (i): $∠ A P B = 18 0^{\circ} - 2 (9 0^{\circ} - ∠ O A B)$
  ½ mark
5. $\Rightarrow ∠ A P B = 2∠ O A B$ . Hence proved.
  ½ mark
Answer: Proved: $∠ A P B = 18 0^{\circ} - 2 (9 0^{\circ} - ∠ O A B) = 2∠ O A B$ .
Where the marks are
- Correct figure (½ mark)
- Uses OA ⟂ AP to get ∠PAB = 90° − ∠OAB (½ mark)
- Uses PA = PB to get equal base angles and ∠APB = 180° − 2∠PAB (1 mark)
- Substitutes from (i) (½ mark)
- Concludes ∠APB = 2∠OAB (½ mark)
Open Q27(B) on its own page
Q28 · 3 marks · Short answerCoordinate Geometry
Determine the ratio in which the point $(- 6, y)$ divides the line segment joining the points A $(- 3, - 1)$ and B $(- 8, 9)$ . Also, find the value of $y$ .
Try it first, then show our solutionSolution
1. Let the point $(- 6, y)$ divide AB in the ratio $k : 1$ .
  ½ mark
2. By the section formula, the point is $(\frac{- 8 k - 3}{k + 1}, \frac{9 k - 1}{k + 1})$
  ½ mark
3. $\frac{- 8 k - 3}{k + 1} = - 6 \Rightarrow - 8 k - 3 = - 6 k - 6 \Rightarrow k = \frac{3}{2}$ , so the ratio is $3 : 2$
  1 mark
4. $y = \frac{9 \cdot \frac{3}{2} - 1}{\frac{3}{2} + 1} = \frac{\frac{25}{2}}{\frac{5}{2}} = 5$
  1 mark
Answer: Ratio 3 : 2 (internally); $y = 5$
Where the marks are
- Takes the ratio as k : 1 (½ mark)
- Writes the section formula coordinates (½ mark)
- Finds k = 3/2, i.e. 3 : 2 (1 mark)
- Finds y = 5 (1 mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerIntroduction to Trigonometry
Consider $△ A C B$ , right-angled at C, in which AB = 29 units, BC = 21 units and $∠ A B C = θ$ . Determine the values of
1. i.
  $1 + tan^{2} θ$
  0 marks
2. ii.
  $cos^{2} θ - sin^{2} θ$
  0 marks
Try it first, then show our solutionSolution
1. By Pythagoras theorem, $A C = A B^{2} - B C^{2} = 2 9^{2} - 2 1^{2} = 400 = 20$ units
  ½ mark
2. For $∠ B = θ$ : $sin θ = \frac{A C}{A B} = \frac{20}{29}$ , $cos θ = \frac{B C}{A B} = \frac{21}{29}$ , $tan θ = \frac{20}{21}$
  ½ mark
3. (i) $1 + tan^{2} θ = 1 + \frac{400}{441} = \frac{841}{441}$
  1 mark
4. (ii) $cos^{2} θ - sin^{2} θ = \frac{441}{841} - \frac{400}{841} = \frac{41}{841}$
  1 mark
Answer: (i) $\frac{841}{441}$ ; (ii) $\frac{41}{841}$
Where the marks are
- AC = 20 units (½ mark)
- Correct sinθ and cosθ (½ mark)
- (i) 841/441 (1 mark)
- (ii) 41/841 (1 mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(A) · 3 marks · Short answerStatistics
The mean of the following distribution is 48 and sum of all the frequencies is 50. Find the missing frequencies $x$ and $y$ .
Class 20-30 30-40 40-50 50-60 60-70
Frequency 8 6 $x$ 11 $y$
Try it first, then show our solutionSolution
1. Step-deviation table with $a = 45$ , $h = 10$ , $u_{i} = \frac{x _{i} - 45}{10}$ :
  Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
  20-30 8 25 −2 −16
  30-40 6 35 −1 −6
  40-50 x 45 0 0
  50-60 11 55 1 11
  60-70 y 65 2 2y
  Total 25 + x + y 2y − 11
  1 mark
2. $\sum f_{i} = 50 \Rightarrow 25 + x + y = 50 \Rightarrow x + y = 25$ ...(1)
  ½ mark
3. Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h$ , so $48 = 45 + \frac{2 y - 11}{50} \times 10$
  ½ mark
4. $3 = \frac{2 y - 11}{5} \Rightarrow 2 y = 26 \Rightarrow y = 13$
  ½ mark
5. From (1), $x = 25 - 13 = 12$
  ½ mark
Answer: $x = 12$ , $y = 13$
Where the marks are
- Correct table with $f_{i} u_{i}$ total $2 y - 11$ (1 mark)
- Forms $x + y = 25$ (½ mark)
- Substitutes in the mean formula (½ mark)
- y = 13 (½ mark)
- x = 12 (½ mark)
Open Q30(A) on its own page
OR
Q30(B) · 3 marks · Short answerStatistics
The distribution below gives the weight of 50 students of class X. Find the Modal weight of the students.
Weight (in kg) 35-45 45-55 55-65 65-75 75-85
Number of students 5 10 20 12 3
Try it first, then show our solutionSolution
1. The maximum frequency is 20, so the modal class is 55-65.
  $f_{1} = 20$ , $f_{0} = 10$ (class 45-55), $f_{2} = 12$ (class 65-75)
  1 mark
2. $l = 55$ , $h = 10$
  ½ mark
3. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) h = 55 + \frac{20 - 10}{40 - 10 - 12} \times 10 = 55 + \frac{100}{18} = 60.555 \dots$
  1 mark
4. Modal weight $\approx 60.56$ kg
  ½ mark
Answer: Modal weight ≈ 60.56 kg
Where the marks are
- Identifies the modal class 55-65 with $f_{0}$ , $f_{1}$ , $f_{2}$ (1 mark)
- Writes l = 55 and h = 10 (½ mark)
- Substitutes correctly in the mode formula (1 mark)
- Mode ≈ 60.56 kg (½ mark)
Open Q30(B) on its own page
Q31 · 3 marks · Short answerPair of Linear Equations in Two Variables
The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Try it first, then show our solutionSolution
1. Let the tens digit be $x$ and the units digit be $y$ . Then the number is $10 x + y$ and the reversed number is $10 y + x$ .
  ½ mark
2. $x + y = 9$ ...(1)
  ½ mark
3. $9 (10 x + y) = 2 (10 y + x) \Rightarrow 90 x + 9 y = 20 y + 2 x \Rightarrow 88 x = 11 y \Rightarrow y = 8 x$ ...(2)
  ½ mark
4. Substituting (2) in (1): $x + 8 x = 9 \Rightarrow x = 1$ , so $y = 8$
  1 mark
5. Required number $= 10 (1) + 8 = 18$
  ½ mark
Answer: 18
Where the marks are
- Expresses the number and its reverse in terms of the digits (½ mark)
- Forms x + y = 9 (½ mark)
- Forms 9(10x + y) = 2(10y + x) (½ mark)
- Solves to get x = 1, y = 8 (1 mark)
- Number = 18 (½ mark)
Open Q31 on its own page

Class	20-30	30-40	40-50	50-60	60-70
Frequency	8	6	$x$	11	$y$

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
20-30	8	25	−2	−16
30-40	6	35	−1	−6
40-50	x	45	0	0
50-60	11	55	1	11
60-70	y	65	2	2y
Total	25 + x + y			2y − 11

Weight (in kg)	35-45	45-55	55-65	65-75	75-85
Number of students	5	10	20	12	3

Section D

Long answer; internal choice in 33 and 35

Q32 · 5 marks · Long answerTriangles
Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
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. Join BE and CD, and draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  ½ mark
2. Given: In $△ A B C$ , $D E ∥ B C$ , and DE meets AB at D and AC at E.
  ½ mark
3. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  ½ mark
4. Construction: Join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  ½ mark
5. Proof: $a r (A D E) = \frac{1}{2} \times A D \times E N$ and $a r (B D E) = \frac{1}{2} \times D B \times E N$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ ...(1)
  1 mark
6. Similarly, $a r (A D E) = \frac{1}{2} \times A E \times D M$ and $a r (D E C) = \frac{1}{2} \times E C \times D M$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ ...(2)
  1 mark
7. $△ B D E$ and $△ D E C$ lie 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)
  From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
Answer: $\frac{A D}{D B} = \frac{A E}{E C}$ (Basic Proportionality Theorem proved)
Where the marks are
- Correct figure (½ mark)
- Given (½ mark)
- To prove (½ mark)
- Construction (½ mark)
- Ratio of areas ADE : BDE = AD : DB (1 mark)
- Ratio of areas ADE : DEC = AE : EC (1 mark)
- Equal areas of BDE and DEC (same base, same parallels) and conclusion (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerCompetency-basedQuadratic Equations
A train travels a distance of 360 km at a uniform speed. If the speed had been 5 km/h more, then it would have taken 1 hour less to cover the same distance. Find the speed of the train.
Try it first, then show our solutionSolution
1. Let the uniform speed of the train be $x$ km/h; the increased speed is $(x + 5)$ km/h.
  ½ mark
2. Time = distance ÷ speed, so $\frac{360}{x} - \frac{360}{x + 5} = 1$
  1 mark
3. $360 (x + 5) - 360 x = x (x + 5) \Rightarrow 1800 = x^{2} + 5 x \Rightarrow x^{2} + 5 x - 1800 = 0$
  1 mark
4. $x = \frac{- 5 \pm 25 + 7200}{2} = \frac{- 5 \pm 85}{2}$
  1 mark
5. $x = 40$ or $x = - 45$
  1 mark
6. Speed cannot be negative, so $x = 40$ . The speed of the train is 40 km/h.
  ½ mark
Answer: 40 km/h
Where the marks are
- Defines the speed variable (½ mark)
- Forms the time equation (1 mark)
- Simplifies to $x^{2} + 5 x - 1800 = 0$ (1 mark)
- Applies the quadratic formula or factorisation (1 mark)
- Gets roots 40 and −45 (1 mark)
- Rejects the negative root; speed = 40 km/h (½ mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerQuadratic Equations
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find the number of marbles they had with them in the beginning.
Try it first, then show our solutionSolution
1. Let John have $x$ marbles at the beginning; then Jivanti had $45 - x$ .
  ½ mark
2. After losing 5 each, they have $x - 5$ and $40 - x$ marbles, so $(x - 5) (40 - x) = 124$
  1 mark
3. $40 x - x^{2} - 200 + 5 x = 124 \Rightarrow x^{2} - 45 x + 324 = 0$
  ½ mark
4. $x = \frac{45 \pm 2025 - 1296}{2} = \frac{45 \pm 27}{2}$
  1 mark
5. $x = 36$ or $x = 9$
  1 mark
6. Case 1: $x = 36$ , so John had 36 and Jivanti had $45 - 36 = 9$ .
  ½ mark
7. Case 2: $x = 9$ , so John had 9 and Jivanti had 36.
  ½ mark
Answer: 36 and 9 marbles (John 36, Jivanti 9, or John 9, Jivanti 36)
Where the marks are
- Defines the variables (½ mark)
- Forms $(x - 5) (40 - x) = 124$ (1 mark)
- Gets $x^{2} - 45 x + 324 = 0$ (½ mark)
- Applies the quadratic formula correctly (1 mark)
- Gets roots 36 and 9 (1 mark)
- Case 1 interpretation (½ mark)
- Case 2 interpretation (½ mark)
Open Q33(B) on its own page
Q34 · 5 marks · Long answerSurface Areas and Volumes
The largest possible hemisphere is drilled out of a wooden cubical block of side 21 cm such that the base of the hemisphere is on one of the faces of the cube. Then, find
1. i.
  the volume of the wood left in the block.
  0 marks
2. ii.
  the total surface area of the remaining solid.
  0 marks
Try it first, then show our solutionSolution
1. Side of cube $a = 21$ cm. The largest hemisphere has diameter equal to the side, so $r = \frac{21}{2}$ cm.
  ½ mark
2. (i) Volume of wood left = Volume of cube − Volume of hemisphere $= a^{3} - \frac{2}{3} π r^{3}$
  ½ mark
3. $= 2 1^{3} - \frac{2}{3} \times \frac{22}{7} \times \frac{21 \times 21 \times 21}{2 \times 2 \times 2} = 9261 - 2425.5$
  1 mark
4. $= 6835.5 cm^{3}$
  ½ mark
5. (ii) TSA of remaining solid = TSA of cube − area of circular top removed + curved surface of hemisphere
  $= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$
  ½ mark
6. $= 6 (21)^{2} + \frac{22}{7} \times (\frac{21}{2})^{2} = 2646 + 346.5$
  1 mark
7. $= 2992.5 cm^{2}$
  1 mark
Answer: (i) 6835.5 cm³; (ii) 2992.5 cm²
Where the marks are
- r = 21/2 cm (½ mark)
- Volume formula: cube − hemisphere (½ mark)
- Correct substitution (1 mark)
- Volume = 6835.5 cm³ (½ mark)
- TSA formula $6 a^{2} - π r^{2} + 2 π r^{2}$ (½ mark)
- Correct substitution (1 mark)
- TSA = 2992.5 cm² (1 mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(A) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
To assist an ambulance, a drone is hovering over an accident site at a constant height on a highway. The angle of depression of the ambulance, approaching the accident site, observed from drone is $3 0^{\circ}$ . Twelve minutes later, the angle of depression of the ambulance is found to be $6 0^{\circ}$ . If the ambulance is moving at a constant speed then, find the total time taken by the ambulance to reach the site of accident.
Try it first, then show our solutionSolution
1. Figure: Let A be the drone and B the accident site directly below it, with AB = $h$ m. D is the first position of the ambulance (angle of depression 30°, so $∠ A D B = 3 0^{\circ}$ ). C is its position 12 minutes later (angle of depression 60°, so $∠ A C B = 6 0^{\circ}$ ). D, C and B lie on a straight line, with $∠ A B C = 9 0^{\circ}$ .
  1 mark
2. Let the speed of the ambulance be $x$ m/min, so $C D = 12 x$ m.
  In right $△ A B C$ : $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow 3 = \frac{h}{B C} \Rightarrow B C = \frac{h}{3}$
  1 mark
3. In right $△ A B D$ : $tan 3 0^{\circ} = \frac{A B}{B D} \Rightarrow \frac{1}{3} = \frac{h}{B D} \Rightarrow B D = h 3$
  1 mark
4. $B D = B C + C D \Rightarrow h 3 = \frac{h}{3} + 12 x \Rightarrow \frac{2 h}{3} = 12 x \Rightarrow h = 63 x$
  So $B C = \frac{6 3 x}{3} = 6 x$
  1 mark
5. Time from C to B $= \frac{6 x}{x} = 6$ minutes
  ½ mark
6. Total time taken $= 12 + 6 = 18$ minutes
  ½ mark
Answer: 18 minutes
Where the marks are
- Correct labelled figure (1 mark)
- BC = h/√3 from tan 60° (1 mark)
- BD = h√3 from tan 30° (1 mark)
- Relates using CD = 12x to get h = 6√3x and BC = 6x (1 mark)
- Time from C to B = 6 min (½ mark)
- Total time = 18 min (½ mark)
Open Q35(A) on its own page
OR
Q35(B) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
A statue standing on a pedestal, is 1.6 m tall. From a point on the ground, the angles of elevation of the top and bottom of the statue are $6 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the pedestal.
Try it first, then show our solutionSolution
1. Figure: Let A be the point on the ground, BC the pedestal of height $h$ m, and CD the statue of height 1.6 m, with B the foot of the pedestal and $∠ A B C = 9 0^{\circ}$ . The angle of elevation of C (bottom of the statue) from A is $∠ B A C = 4 5^{\circ}$ . The angle of elevation of D (top of the statue) from A is $∠ B A D = 6 0^{\circ}$ . Then $B D = h + 1.6$ m.
  1 mark
2. In right $△ A B C$ : $tan 4 5^{\circ} = \frac{B C}{A B} \Rightarrow 1 = \frac{h}{A B} \Rightarrow A B = h$ ...(i)
  1 mark
3. In right $△ A B D$ : $tan 6 0^{\circ} = \frac{B D}{A B} \Rightarrow 3 = \frac{h + 1.6}{A B} \Rightarrow A B = \frac{h + 1.6}{3}$ ...(ii)
  1 mark
4. Equating (i) and (ii): $h = \frac{h + 1.6}{3} \Rightarrow 3 h = h + 1.6$
  ½ mark
5. $(3 - 1) h = 1.6 \Rightarrow h = \frac{1.6}{3 - 1} \times \frac{3 + 1}{3 + 1} = \frac{1.6 ( 3 + 1 )}{2} = 0.8 (3 + 1)$
  1 mark
6. Hence the height of the pedestal is $0.8 (3 + 1)$ m $\approx 2.19$ m.
  ½ mark
Answer: Height of pedestal = $0.8 (3 + 1)$ m ≈ 2.19 m
Where the marks are
- Correct labelled figure with both angles and heights (1 mark)
- Uses $tan 4 5^{\circ}$ to get $A B = h$ (1 mark)
- Uses $tan 6 0^{\circ}$ to get $A B = \frac{h + 1.6}{3}$ (1 mark)
- Equates the two expressions for AB (½ mark)
- Solves to get $h = \frac{1.6}{3 - 1} = 0.8 (3 + 1)$ (1 mark)
- States the height of the pedestal with unit (½ mark)
Open Q35(B) on its own page

Section E

Case study based; sub-parts 1, 1, 2 marks; internal choice in sub-part (iii) of each

Q36 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
An Art and craft teacher prepared a fan which looked like the sector of a circle using ice-cream sticks and black sheet as shown in the given figure:
1. (i)
  The angle subtended at the centre of the sector is $6 0^{\circ}$ . Find the area of one face of the fan.
  1 mark
2. (ii)
  With the given sector angle of $6 0^{\circ}$ at the centre, find area of black sheet used in the fan.
  1 mark
3. (iii)(A)
  The boundary of the black sheet region on one face is to be covered by a designer tape. What length of tape is required?
  2 marks
4. OR
  (iii)(B)
  If 29.6 cm of designer tape is required to cover the boundary of the black sheet region on one face of the fan, then evaluate the central angle of the sector.
  2 marks
Figure in the paper: A folding fan shaped as a sector of a circle. The black sheet forms the outer sector region (fan face) and white ice-cream sticks form the narrow part near the pivot at the bottom. A straight edge is marked 10 cm (the outer radius of the black sheet) and the shorter edge near the pivot is marked 4 cm (the radius of the ice-cream stick portion). Arrows label 'Black sheet' pointing to the dark fabric region and 'Ice-cream sticks' pointing to the white stick region at the bottom. See it in the official paper.
Try it first, then show our solutionSolution
1. Given: central angle $θ = 6 0^{\circ}$ , outer radius $R = 10$ cm, radius of the stick portion $r = 4$ cm.
2. (i) Area of one face $= \frac{θ}{36 0 ^{\circ}} π R^{2} = \frac{60}{360} \times \frac{22}{7} \times 1 0^{2}$
  ½ mark
3. (i) $= \frac{1100}{21} \approx 52.38$ cm²
  ½ mark
4. (ii) The black sheet is the region between the two arcs, so its area $= \frac{θ}{36 0 ^{\circ}} π (R^{2} - r^{2}) = \frac{60}{360} \times \frac{22}{7} \times (100 - 16)$
  ½ mark
5. (ii) $= \frac{1}{6} \times \frac{22}{7} \times 84 = 44$ cm²
  ½ mark
6. (iii)(A) The boundary is made of two straight edges of length $(R - r)$ each, the outer arc and the inner arc. Length $= 2 (10 - 4) + \frac{60}{360} \times 2 \times \frac{22}{7} \times (10 + 4)$
  1 mark
7. (iii)(A) $= 12 + \frac{1}{6} \times 88 = 12 + \frac{44}{3} = \frac{80}{3}$ cm $\approx 26.67$ cm of tape
  1 mark
8. (iii)(B) Using the same boundary: $2 (10 - 4) + \frac{θ}{360} \times 2 \times \frac{22}{7} \times (10 + 4) = 29.6$
  ½ mark
9. (iii)(B) $12 + \frac{θ}{360} \times 88 = 29.6 \Rightarrow \frac{θ}{360} \times 88 = 17.6$
  1 mark
10. (iii)(B) $θ = \frac{17.6 \times 360}{88} = 7 2^{\circ}$
  ½ mark
Answer: (i) 52.38 cm²; (ii) 44 cm²; (iii)(A) 80/3 cm ≈ 26.67 cm; (iii)(B) 72°
Where the marks are
- (i) Sector area formula with R = 10 cm substituted (½ mark) and result 52.38 cm² (½ mark)
- (ii) Area of black sheet as difference of sectors (½ mark) and result 44 cm² (½ mark)
- (iii)(A) Boundary expression: two straight edges plus two arcs (1 mark); length 80/3 cm (1 mark)
- (iii)(B) Forms equation for the boundary equal to 29.6 (½ mark); simplifies to θ/360 × 88 = 17.6 (1 mark); θ = 72° (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Delhi's pollution is a severe, often seasonal issue caused by a combination of factors like high vehicular and industrial emissions, dust from construction, and crop burning in surrounding states. In a school, students of classes I to XII thought of planting trees in and around the school to reduce air pollution.
It was decided that the number of trees that each section of each class will plant be three more than the double of the class in which they are studying, e.g., a section of class IV will plant 11 trees. If in the given school, there are two sections of each class, then answer the following questions:
1. (i)
  If number of trees planted by students class-wise follow arithmetic progression, then find the common difference.
  1 mark
2. (ii)
  What is the number of trees planted by students of class VI?
  1 mark
3. (iii)(A)
  What is the total number of trees planted by the students of given school?
  2 marks
4. OR
  (iii)(B)
  Identify the class that planted 34 trees.
  2 marks
Figure in the paper: Two illustrations side by side. Left: a grey image of a field with crop burning, showing flames and smoke rising over dry stalks. Right: an illustration of a factory with several smoke-emitting chimneys, a car on the left, a tree on the right, and a group of people standing in between. See it in the official paper.
Try it first, then show our solutionSolution
1. Each section of class $n$ plants $2 n + 3$ trees. With 2 sections, class $n$ plants $2 (2 n + 3)$ trees. Class I: $2 (5) = 10$ ; Class II: $2 (7) = 14$ ; Class III: $2 (9) = 18$ ; ... This gives the AP 10, 14, 18, 22, ...
2. (i) Common difference $d = 14 - 10 = 4$
  1 mark
3. (ii) Class VI: $2 {2 (6) + 3} = 2 \times 15 = 30$ trees
  1 mark
4. (iii)(A) There are 12 classes, so $n = 12$ , $a = 10$ , $d = 4$ . Then $S_{12} = \frac{n}{2} {2 a + (n - 1) d} = \frac{12}{2} {2 (10) + 11 (4)}$
  1 mark
5. (iii)(A) $= 6 \times (20 + 44) = 6 \times 64 = 384$ trees
  1 mark
6. (iii)(B) Let class $n$ plant 34 trees: $2 (2 n + 3) = 34$
  1 mark
7. (iii)(B) $2 n + 3 = 17$
  ½ mark
8. (iii)(B) $n = 7$ , i.e. Class VII planted 34 trees.
  ½ mark
Answer: (i) 4; (ii) 30 trees; (iii)(A) 384 trees; (iii)(B) Class VII
Where the marks are
- (i) Finds the common difference 4 (1 mark)
- (ii) Class VI plants 2(2×6+3) = 30 trees (1 mark)
- (iii)(A) Uses $S_{n}$ with n = 12, a = 10, d = 4 (1 mark); total 384 (1 mark)
- (iii)(B) Forms 2(2n+3) = 34 (1 mark); 2n+3 = 17 (½ mark); n = 7, Class VII (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedProbability
A labourer prints 400 T-shirts in a day. The supervisor checked the T-shirts and found that 312 prints were good, 54 prints were with minor defects and rest of the prints were of major defects.
Harish, a customer will buy a T-shirt only if it is good but a trader will buy, if it has no major defect. If a random T-shirt is picked, then
1. (i)
  find the probability that Harish will buy a T-shirt.
  1 mark
2. (ii)
  find the probability that a trader will buy the T-shirt.
  1 mark
3. (iii)(A)
  find the probability that the T-shirt is not good.
  2 marks
4. OR
  (iii)(B)
  find the probability that neither Harish nor trader buy the T-shirt.
  2 marks
Figure in the paper: A photograph of a garment factory floor: several workers in dark clothing sit at long work tables with machines and fabric, in a large industrial hall. See it in the official paper.
Try it first, then show our solutionSolution
1. Total T-shirts = 400. Good = 312, minor defect = 54, major defect = $400 - 312 - 54 = 34$ .
2. (i) Harish buys only a good T-shirt: $P = \frac{312}{400} = \frac{39}{50}$
  1 mark
3. (ii) The trader buys if there is no major defect (good or minor defect): $P = \frac{312 + 54}{400} = \frac{366}{400} = \frac{183}{200}$
  1 mark
4. (iii)(A) $P (not good) = 1 - P (good) = 1 - \frac{312}{400}$
  1 mark
5. (iii)(A) $= \frac{88}{400} = \frac{11}{50}$
  1 mark
6. (iii)(B) Neither buys only if the T-shirt has a major defect. Favourable outcomes $= 400 - 312 - 54 = 34$ .
  1 mark
7. (iii)(B) $P = \frac{34}{400} = \frac{17}{200}$
  1 mark
Answer: (i) 39/50; (ii) 183/200; (iii)(A) 11/50; (iii)(B) 17/200
Where the marks are
- (i) P(Harish buys) = 312/400 = 39/50 (1 mark)
- (ii) P(trader buys) = 366/400 = 183/200 (1 mark)
- (iii)(A) Uses the complement 1 − 312/400 (1 mark); gets 88/400 = 11/50 (1 mark)
- (iii)(B) Identifies 34 T-shirts with major defect (1 mark); probability 34/400 = 17/200 (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.