MathsCBSE Class 10 · Basic · Board exam 2026 (session 2025-26)

CBSE Class 10 Mathematics Basic: 2026 board paper, set 430/5/1

The question paper set in the CBSE Class 10 board exam of 2026, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

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

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2026/X/Mathematics_Standard.zip (a zip; this paper is the file 1159-1_430-5-1 Mathematics (Basic).pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	Very short answer; internal choice in 21 and 24
C	26-31	Short answer; internal choice in 28 and 30
D	32-35	Long answer; internal choice in 32 and 34
E	36-38	Case study; internal choice in 2-mark subparts of 36, 37, 38

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

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
$7 \times 11 \times 13 + 5$ is
1. (A)a prime number.
2. (B)an odd number.
3. (C)a composite number.
4. (D)a multiple of 5.
Try it first, then show our solutionSolution
1. $7 \times 11 \times 13 + 5 = 1001 + 5 = 1006 = 2 \times 503$ .
2. Since 1006 has a factor 2 other than 1 and itself, it is composite (it is even, so not odd; it ends in 6, so not a multiple of 5).
  1 mark
Answer: Composite number, option (C)
Where the marks are
- Shows the number is 1006, which is divisible by 2, hence composite (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQQuadratic Equations
The roots of the quadratic equation $x^{2} + 9 = 0$ are
1. (A)real and equal
2. (B)not real
3. (C)real and negative of each other
4. (D)rational numbers
Try it first, then show our solutionSolution
1. Here $a = 1, b = 0, c = 9$ , so $D = b^{2} - 4 a c = 0 - 36 = - 36 < 0$ .
2. Since $D < 0$ , the equation has no real roots.
  1 mark
Answer: Not real, option (B)
Where the marks are
- Finds $D = - 36 < 0$ and concludes the roots are not real (1 mark)
Q3 · 1 mark · MCQCoordinate Geometry
The distance between the points $(- 2, 5)$ and $(5, - 2)$ is
1. (A) $72$
2. (B)14
3. (C) $27$
4. (D)7
Try it first, then show our solutionSolution
1. $d = (5 - (- 2))^{2} + (- 2 - 5)^{2} = 49 + 49 = 98 = 72$ .
  1 mark
Answer: $72$ , option (A)
Where the marks are
- Applies the distance formula to get $72$ (1 mark)
Q4 · 1 mark · MCQSurface Areas and Volumes
A cylinder of radius $r$ is surmounted on a hemisphere of same radius. If total height of the object is 13 cm, then its inner surface area is
1. (A) $2 π r (r + 13)$
2. (B) $13 π r$
3. (C) $2 π (13 + r)^{2}$
4. (D) $26 π r$
Figure in the paper: A solid object: a cylinder of radius r standing on a hemisphere of the same radius (hemisphere at the bottom, flat face up, with the radius r marked across the flat circular face). A vertical double-headed arrow on the right side is labelled 13 cm, spanning the total height from the bottom of the hemisphere to the top of the cylinder. See it in the official paper.
Try it first, then show our solutionSolution
1. The cylinder's height is $h = 13 - r$ . Inner surface area = CSA of cylinder + CSA of hemisphere $= 2 π r (13 - r) + 2 π r^{2}$ .
2. $= 26 π r - 2 π r^{2} + 2 π r^{2} = 26 π r$ .
  1 mark
Answer: $26 π r$ , option (D)
Where the marks are
- Uses height $13 - r$ and adds the curved areas to get $26 π r$ (1 mark)
Q5 · 1 mark · MCQTriangles
Which of the following statements is not always true?
1. (A)Two circles are similar.
2. (B)Two isosceles right triangles are similar.
3. (C)Two rectangles are similar.
4. (D)Two equilateral triangles are similar.
Try it first, then show our solutionSolution
1. All circles, all equilateral triangles and all isosceles right triangles have the same shape. Two rectangles have equal angles, but their sides need not be proportional (e.g. $2 \times 1$ and $3 \times 1$ ).
  1 mark
Answer: Two rectangles are similar, option (C)
Where the marks are
- Identifies rectangles as not always similar because their sides need not be proportional (1 mark)
Q6 · 1 mark · MCQIntroduction to Trigonometry
If value of $cot θ$ is $5$ , then $sin θ$ equals
1. (A) $\frac{1}{6}$
2. (B) $6$
3. (C) $\frac{5}{6}$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. $csc^{2} θ = 1 + cot^{2} θ = 1 + 5 = 6$ , so $csc θ = 6$ .
2. $sin θ = \frac{1}{6}$ .
  1 mark
Answer: $\frac{1}{6}$ , option (A)
Where the marks are
- Uses $1 + cot^{2} θ = csc^{2} θ$ to get $sin θ = 1/ 6$ (1 mark)
Q7 · 1 mark · MCQCircles
A chord QR subtends an angle of $10 5^{\circ}$ at the centre O of the circle. The measure of $∠$ RQP is
1. (A) $\frac{7 5 ^{\circ}}{2}$
2. (B) $\frac{10 5 ^{\circ}}{2}$
3. (C) $7 5^{\circ}$
4. (D) $1 5^{\circ}$
Figure in the paper: A circle with centre O. A chord QR is drawn with Q and R on the circle; O is joined to Q and R forming the central angle. A ray from Q extends outward through point P, lying outside the circle, so that P, Q are on a line beyond Q away from the circle's interior. See it in the official paper.
Try it first, then show our solutionSolution
1. QP is a tangent at Q, so $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius]. In isosceles $△ O QR$ , $∠ O QR = \frac{18 0 ^{\circ} - 10 5 ^{\circ}}{2} = 37. 5^{\circ}$ .
2. $∠ R QP = 9 0^{\circ} - 37. 5^{\circ} = 52. 5^{\circ} = \frac{10 5 ^{\circ}}{2}$ .
  1 mark
Answer: $\frac{10 5 ^{\circ}}{2}$ , option (B)
Where the marks are
- Uses tangent ⟂ radius and the isosceles triangle to get $10 5^{\circ} /2$ (1 mark)
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Q8 · 1 mark · MCQProbability
The probability of getting sum greater than 10, when two dice are rolled together, is
1. (A) $\frac{1}{9}$
2. (B) $\frac{1}{18}$
3. (C) $\frac{1}{12}$
4. (D)1
Try it first, then show our solutionSolution
1. Total outcomes = 36. Sum > 10: (5,6), (6,5), (6,6) — 3 outcomes.
2. $P = \frac{3}{36} = \frac{1}{12}$ .
  1 mark
Answer: $\frac{1}{12}$ , option (C)
Where the marks are
- Lists 3 favourable outcomes out of 36 to get $1/12$ (1 mark)
Q9 · 1 mark · MCQCompetency-basedPolynomials
The graph of a polynomial $p (x)$ is shown here. The number of zeroes of the polynomial $p (x)$ is
1. (A)5
2. (B)1
3. (C)0
4. (D)4
Figure in the paper: Graph of $y = p (x)$ on axes X'OX and YOY'. The curve starts at the left above the x-axis, dips below to a trough, rises above the axis to a peak, comes down to a lower trough below the axis near the origin, and rises again above the axis to a peak on the right. It crosses the x-axis at several points. See it in the official paper.
Try it first, then show our solutionSolution
1. The number of zeroes equals the number of points where the graph meets the x-axis. The curve crosses the x-axis 4 times.
  1 mark
Answer: 4, option (D)
Where the marks are
- Counts the 4 intersections with the x-axis (1 mark)
Q10 · 1 mark · MCQArithmetic Progressions
$n^{t h}$ term of the A.P.: $\frac{- 1}{3}, \frac{4}{3}, 3, \dots$ is
1. (A) $\frac{5 n - 9}{3}$
2. (B) $\frac{5 n - 6}{3}$
3. (C) $\frac{3 n - 4}{3}$
4. (D) $\frac{3 n + 2}{3}$
Try it first, then show our solutionSolution
1. $a = - \frac{1}{3}$ , $d = \frac{4}{3} + \frac{1}{3} = \frac{5}{3}$ .
2. $a_{n} = - \frac{1}{3} + (n - 1) \frac{5}{3} = \frac{5 n - 6}{3}$ .
  1 mark
Answer: $\frac{5 n - 6}{3}$ , option (B)
Where the marks are
- Finds $d = 5/3$ and $a_{n} = (5 n - 6) /3$ (1 mark)
Q11 · 1 mark · MCQTriangles
In the given figure, DE $∥$ BC. If AD : AB = 1 : 3 and AE = 2.5 cm, then AC equals
1. (A)7.5 cm
2. (B)5 cm
3. (C)10 cm
4. (D)2.5 cm
Figure in the paper: Triangle ABC with vertex A at the top. D lies on side AB and E lies on side AC, with DE parallel to BC (arrow marks on DE and BC). See it in the official paper.
Try it first, then show our solutionSolution
1. Since DE ∥ BC, $△ A D E \sim △ A B C$ , so $\frac{A E}{A C} = \frac{A D}{A B} = \frac{1}{3}$ .
2. $A C = 3 \times 2.5 = 7.5$ cm.
  1 mark
Answer: 7.5 cm, option (A)
Where the marks are
- Uses $A E / A C = A D / A B$ to get AC = 7.5 cm (1 mark)
Q12 · 1 mark · MCQPolynomials
The value of k for which sum of the zeroes of the polynomial $p (x) = 3 x^{2} - k x + 6$ is 2, is
1. (A)2
2. (B)-6
3. (C)-2
4. (D)6
Try it first, then show our solutionSolution
1. Sum of zeroes $= - \frac{b}{a} = \frac{k}{3} = 2$ .
2. $k = 6$ .
  1 mark
Answer: k = 6, option (D)
Where the marks are
- Uses sum $= - b / a = k /3 = 2$ to get k = 6 (1 mark)
Q13 · 1 mark · MCQProbability
A bag contains some red and some white balls. A ball is drawn at random from the bag. If the probability of getting a red ball is $\frac{2}{7}$ , then the probability of getting a white ball is
1. (A) $\frac{1}{14}$
2. (B) $\frac{5}{7}$
3. (C) $\frac{1}{7}$
4. (D) $\frac{2}{7}$
Try it first, then show our solutionSolution
1. The balls are only red or white, so $P (white) = 1 - \frac{2}{7} = \frac{5}{7}$ .
  1 mark
Answer: $\frac{5}{7}$ , option (B)
Where the marks are
- Uses complementary events to get $5/7$ (1 mark)
Q14 · 1 mark · MCQArithmetic Progressions
If $- 26, x, 2$ are in A.P., then the value of $x$ is
1. (A)14
2. (B)-13
3. (C)-12
4. (D)-14
Try it first, then show our solutionSolution
1. For an A.P., $2 x = - 26 + 2 = - 24$ , so $x = - 12$ .
  1 mark
Answer: x = −12, option (C)
Where the marks are
- Uses the middle term as the average to get $x = - 12$ (1 mark)
Q15 · 1 mark · MCQCircles
PQ is tangent to a circle at a point P on the circle. The number of tangents which can be drawn to the circle parallel to PQ, is
1. (A)2
2. (B)1
3. (C)many
4. (D)zero
Try it first, then show our solutionSolution
1. Tangents parallel to PQ touch the circle at the ends of the diameter through P. One is PQ itself; the only other one is at the opposite end of that diameter.
  1 mark
Answer: 1, option (B)
Where the marks are
- Recognises only one more parallel tangent, at the other end of the diameter (1 mark)
Q16 · 1 mark · MCQProbability
A card is drawn from a well-shuffled deck of 52 playing cards. The probability of getting a queen of spade is
1. (A) $\frac{1}{26}$
2. (B) $\frac{1}{52}$
3. (C)0
4. (D) $\frac{1}{4}$
Try it first, then show our solutionSolution
1. There is exactly one queen of spades among 52 cards, so $P = \frac{1}{52}$ .
  1 mark
Answer: $\frac{1}{52}$ , option (B)
Where the marks are
- One favourable card out of 52 (1 mark)
Q17 · 1 mark · MCQCompetency-basedAreas Related to Circles
The length of a pendulum is 70 cm and it describes an arc of length 88 cm when swings. The angle subtended by the arc at the centre is
1. (A) $3 6^{\circ}$
2. (B) $7 0^{\circ}$
3. (C) $7 2^{\circ}$
4. (D) $8 0^{\circ}$
Try it first, then show our solutionSolution
1. Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r$ : $88 = \frac{θ}{360} \times 2 \times \frac{22}{7} \times 70 = \frac{θ}{360} \times 440$ .
2. $θ = \frac{88 \times 360}{440} = 7 2^{\circ}$ .
  1 mark
Answer: $7 2^{\circ}$ , option (C)
Where the marks are
- Applies arc-length formula with r = 70 cm to get $7 2^{\circ}$ (1 mark)
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Q18 · 1 mark · MCQSurface Areas and Volumes
The total surface area of a solid cone of radius 7 cm and slant height 25 cm, is
1. (A) $724 cm^{2}$
2. (B) $704 cm^{2}$
3. (C) $550 cm^{2}$
4. (D) $616 cm^{2}$
Try it first, then show our solutionSolution
1. TSA $= π r (l + r) = \frac{22}{7} \times 7 \times (25 + 7) = 22 \times 32 = 704$ cm².
  1 mark
Answer: 704 cm², option (B)
Where the marks are
- Uses $π r (l + r)$ to get 704 cm² (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Assertion (A): For an acute angle $θ$ , $cos θ$ is always less than 1.
Reason (R): In a right-angled triangle, hypotenuse is the longest side and $cos θ = \frac{Base}{Hypotenuse}$ .
1. (A)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. R is true: the hypotenuse is the longest side and $cos θ =$ base/hypotenuse.
2. Since base < hypotenuse for an acute angle, $cos θ < 1$ ; so A is true and R explains it.
  1 mark
Answer: Option (A)
Where the marks are
- Both true and R correctly explains A (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedStatistics
Assertion (A): Median of a data is the value of $\frac{N}{2}$ , where N represents sum of all frequencies.
Reason (R): Median divides the whole distribution in two equal parts.
1. (A)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $\frac{N}{2}$ only locates the position of the median (median class); it is not the median value itself, so A is false.
2. R is true: the median divides the data into two equal halves.
  1 mark
Answer: Option (D)
Where the marks are
- A false (N/2 is the position, not the value), R true (1 mark)
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Section B

Very short answer; internal choice in 21 and 24

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerIntroduction to Trigonometry
If $sec A = 2$ and $tan B = 3$ , then find the value of $2 sin A cos B$ .
Try it first, then show our solutionSolution
1. $sec A = 2 = sec 4 5^{\circ} \Rightarrow A = 4 5^{\circ}$
  ½ mark
2. $tan B = 3 = tan 6 0^{\circ} \Rightarrow B = 6 0^{\circ}$
  ½ mark
3. $2 sin A cos B = 2 sin 4 5^{\circ} cos 6 0^{\circ} = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} = \frac{2}{2}$
  1 mark
Answer: $2 sin A cos B = \frac{1}{2}$ (or $\frac{2}{2}$ )
Where the marks are
- Finds $A = 4 5^{\circ}$ (½ mark)
- Finds $B = 6 0^{\circ}$ (½ mark)
- Substitutes and gets $\frac{1}{2}$ (1 mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate: $\frac{4 cos ^{3} 6 0 ^{\circ} + cosec 3 0 ^{\circ}}{tan ^{2} 3 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. Put $cos 6 0^{\circ} = \frac{1}{2}$ , $cosec 3 0^{\circ} = 2$ , $tan 3 0^{\circ} = \frac{1}{3}$ :
  $\frac{4 ( \frac{1}{2} ) ^{3} + 2}{( \frac{1}{3} ) ^{2}} = \frac{\frac{1}{2} + 2}{\frac{1}{3}}$
  1½ marks
2. $= \frac{5}{2} \times 3 = \frac{15}{2}$
  ½ mark
Answer: $\frac{15}{2}$
Where the marks are
- Correct substitution of all three standard values (1½ marks)
- Simplifies to $\frac{15}{2}$ (½ mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerReal Numbers
Find the H.C.F. and L.C.M. of 1530 and 2040.
Try it first, then show our solutionSolution
1. $1530 = 2 \times 3^{2} \times 5 \times 17$
  ½ mark
2. $2040 = 2^{3} \times 3 \times 5 \times 17$
  ½ mark
3. HCF = product of smallest powers of common primes $= 2 \times 3 \times 5 \times 17 = 510$
  ½ mark
4. LCM = product of greatest powers of all primes $= 2^{3} \times 3^{2} \times 5 \times 17 = 6120$
  ½ mark
Answer: HCF = 510, LCM = 6120
Where the marks are
- Prime factorisation of 1530 (½ mark)
- Prime factorisation of 2040 (½ mark)
- HCF = 510 (½ mark)
- LCM = 6120 (½ mark)
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Q23 · 2 marks · Very short answerCoordinate Geometry
If A(a, 0), B(1, 1) and C(0, b) form a triangle, right angled at B when joined, then establish a relation between a and b.
Try it first, then show our solutionSolution
1. Since $∠ B = 9 0^{\circ}$ , by Pythagoras theorem $A C^{2} = A B^{2} + B C^{2}$ .
  Using distance formula: $A B^{2} = (a - 1)^{2} + 1$ , $B C^{2} = 1 + (b - 1)^{2}$ , $A C^{2} = a^{2} + b^{2}$
  $\Rightarrow a^{2} + b^{2} = (a - 1)^{2} + 1 + 1 + (b - 1)^{2}$
  1 mark
2. $a^{2} + b^{2} = a^{2} - 2 a + b^{2} - 2 b + 4 \Rightarrow 2 a + 2 b = 4 \Rightarrow a + b = 2$
  1 mark
Answer: $a + b = 2$
Where the marks are
- Uses $A C^{2} = A B^{2} + B C^{2}$ with distance formula correctly (1 mark)
- Simplifies to $a + b = 2$ (1 mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerProbability
Find the probability that a number selected at random from the numbers $30, 31, 32, 33, \dots, 60$ is (i) a prime number (ii) a multiple of 6.
Try it first, then show our solutionSolution
1. Total numbers from 30 to 60 $= 60 - 30 + 1 = 31$ .
  (i) Primes: 31, 37, 41, 43, 47, 53, 59 → 7 outcomes. $P (prime) = \frac{7}{31}$
  1 mark
2. (ii) Multiples of 6: 30, 36, 42, 48, 54, 60 → 6 outcomes. $P (multiple of 6) = \frac{6}{31}$
  1 mark
Answer: (i) $\frac{7}{31}$ ; (ii) $\frac{6}{31}$
Where the marks are
- Lists 7 primes and gets $\frac{7}{31}$ (1 mark)
- Lists 6 multiples of 6 and gets $\frac{6}{31}$ (1 mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerProbability
Slips of letters of the word 'BACKGROUND' are put in a bowl and thoroughly mixed. One slip is picked up at random. Find the probability that picked up slip's letter is (i) a vowel (ii) present in the word 'BALL'.
Try it first, then show our solutionSolution
1. BACKGROUND has 10 letters: B, A, C, K, G, R, O, U, N, D.
  (i) Vowels: A, O, U → 3. $P (vowel) = \frac{3}{10}$
  1 mark
2. (ii) Letters of 'BALL' are B, A, L; of these only B and A occur in BACKGROUND → 2 slips. $P = \frac{2}{10} = \frac{1}{5}$
  1 mark
Answer: (i) $\frac{3}{10}$ ; (ii) $\frac{1}{5}$
Where the marks are
- P(vowel) = $\frac{3}{10}$ (1 mark)
- P(letter in BALL) = $\frac{2}{10} = \frac{1}{5}$ (1 mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerTriangles
In the given figure, AB $∥$ DC. If OB = 3OD and CD = 1.8 cm, then find the length AB.
Figure in the paper: Two parallel horizontal lines: the upper line XY contains points D and C (D left, C right); the lower line X'Y' contains points A (left) and B (right). Segments AC and BD are drawn and intersect at O, which lies between the two lines. Given OB = 3OD and CD = 1.8 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ O D C$ and $△ O B A$ : $∠ O D C = ∠ O B A$ and $∠ O C D = ∠ O A B$ [alternate angles, $D C ∥ A B$ ]. So $△ O D C \sim △ O B A$ (AA similarity).
  ½ mark
2. $\frac{O D}{O B} = \frac{C D}{A B}$ [corresponding sides of similar triangles]
  ½ mark
3. $O B = 3 O D \Rightarrow \frac{1}{3} = \frac{1.8}{A B}$
  ½ mark
4. $A B = 3 \times 1.8 = 5.4$ cm
  ½ mark
Answer: AB = 5.4 cm
Where the marks are
- Proves $△ O C D \sim △ O A B$ by AA (½ mark)
- Writes ratio of corresponding sides (½ mark)
- Substitutes $\frac{1}{3} = \frac{1.8}{A B}$ (½ mark)
- AB = 5.4 cm (½ mark)
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Section C

Short answer; internal choice in 28 and 30

Q26 · 3 marks · Short answerReal Numbers
Prove that $2$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $2$ is rational. Then $2 = \frac{p}{q}$ , where $p$ and $q$ are co-prime integers and $q \neq = 0$ .
  ½ mark
2. Then $2 q = p$ , so $2 q^{2} = p^{2}$ . Hence 2 divides $p^{2}$ , so 2 divides $p$ [if a prime divides $a^{2}$ , it divides $a$ ]. Let $p = 2 m$ for some integer $m$ .
  1 mark
3. Then $2 q^{2} = 4 m^{2}$ , so $q^{2} = 2 m^{2}$ . Hence 2 divides $q^{2}$ , so 2 divides $q$ . Thus 2 is a common factor of $p$ and $q$ , which contradicts that $p$ and $q$ are co-prime.
  1 mark
4. So our assumption is wrong, and $2$ is irrational.
  ½ mark
Answer: √2 is irrational (proved by contradiction)
Where the marks are
- Assumes $2 = p / q$ with p, q co-prime and q ≠ 0 (½ mark)
- Shows 2 divides p (1 mark)
- Shows 2 divides q, which contradicts co-primeness (1 mark)
- Concludes √2 is irrational (½ mark)
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Q27 · 3 marks · Short answerCoordinate Geometry
Points P(6, 0), Q(2, 8) and R(-2, 4) are vertices of $△$ PQR. It is given that MN $∥$ QR such that $\frac{P M}{M Q} = \frac{1}{3}$ . Using distance formula and ratio formula, show that $\frac{M N}{QR} = \frac{1}{4}$ .
Figure in the paper: Triangle PQR with P(6, 0) at the top, Q(2, 8) at bottom left and R(-2, 4) at bottom right. Point M lies on side PQ and point N lies on side PR, with MN drawn parallel to QR (arrows on MN and QR). See it in the official paper.
Try it first, then show our solutionSolution
1. $M N ∥ QR$ , so by BPT $\frac{P M}{M Q} = \frac{P N}{N R} = \frac{1}{3}$ ; M and N divide PQ and PR in ratio $1 : 3$ .
  ½ mark
2. $M = (\frac{1 \times 2 + 3 \times 6}{1 + 3}, \frac{1 \times 8 + 3 \times 0}{1 + 3}) = (\frac{20}{4}, \frac{8}{4}) = (5, 2)$
  ½ mark
3. $N = (\frac{1 \times ( - 2 ) + 3 \times 6}{4}, \frac{1 \times 4 + 3 \times 0}{4}) = (\frac{16}{4}, \frac{4}{4}) = (4, 1)$
  ½ mark
4. $M N = (5 - 4)^{2} + (2 - 1)^{2} = 2$
  ½ mark
5. $QR = (2 + 2)^{2} + (8 - 4)^{2} = 32 = 42$
  ½ mark
6. $\frac{M N}{QR} = \frac{2}{4 2} = \frac{1}{4}$ . Hence shown.
  ½ mark
Answer: M(5, 2), N(4, 1), MN = $2$ , QR = $42$ , so $\frac{M N}{QR} = \frac{1}{4}$
Where the marks are
- Uses BPT: $\frac{P N}{N R} = \frac{1}{3}$ (½ mark)
- M = (5, 2) by section formula (½ mark)
- N = (4, 1) by section formula (½ mark)
- MN = $2$ (½ mark)
- QR = $42$ (½ mark)
- Concludes ratio $\frac{1}{4}$ (½ mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerArithmetic Progressions
In an A.P., it is given that $a = 2$ , $d = 8$ and $S_{n} = 90$ . Find the value of $n$ .
Try it first, then show our solutionSolution
1. $S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow \frac{n}{2} [4 + (n - 1) 8] = 90$
  1 mark
2. $n (4 n - 2) = 90 \Rightarrow 4 n^{2} - 2 n - 90 = 0 \Rightarrow 2 n^{2} - n - 45 = 0$
  1 mark
3. $(n - 5) (2 n + 9) = 0 \Rightarrow n = 5$ or $n = - \frac{9}{2}$ (rejected, as $n$ is a natural number). So $n = 5$ .
  1 mark
Answer: $n = 5$
Where the marks are
- Correct substitution in $S_{n}$ formula (1 mark)
- Forms $2 n^{2} - n - 45 = 0$ (1 mark)
- Solves, rejects $- \frac{9}{2}$ , gets $n = 5$ (1 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerArithmetic Progressions
How many 4-digit numbers are divisible by 7?
Try it first, then show our solutionSolution
1. Smallest 4-digit multiple of 7 is $1001$ ( $7 \times 143$ ); largest is $9996$ ( $7 \times 1428$ ). The numbers $1001, 1008, 1015, \dots, 9996$ form an AP with $a = 1001$ , $d = 7$ , $a_{n} = 9996$ .
  1½ marks
2. $a_{n} = a + (n - 1) d \Rightarrow 9996 = 1001 + (n - 1) \times 7$
  1 mark
3. $(n - 1) = \frac{8995}{7} = 1285 \Rightarrow n = 1286$
  ½ mark
Answer: 1286 four-digit numbers
Where the marks are
- Identifies AP 1001, 1008, …, 9996 with a, d, $a_{n}$ (1½ marks)
- Uses $a_{n} = a + (n - 1) d$ correctly (1 mark)
- n = 1286 (½ mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{1 - sin A}{1 + sin A} = \frac{1}{sec A + tan A}$ .
Try it first, then show our solutionSolution
1. LHS $= \frac{1 - sin A}{1 + sin A} \times \frac{1 - sin A}{1 - sin A} = \frac{( 1 - sin A ) ^{2}}{1 - sin ^{2} A} = \frac{( 1 - sin A ) ^{2}}{cos ^{2} A} = \frac{1 - sin A}{cos A}$
  1 mark
2. $= \frac{1}{cos A} - \frac{sin A}{cos A} = sec A - tan A$
  ½ mark
3. $= (sec A - tan A) \times \frac{sec A + tan A}{sec A + tan A}$
  ½ mark
4. $= \frac{sec ^{2} A - tan ^{2} A}{sec A + tan A}$
  ½ mark
5. $= \frac{1}{sec A + tan A}$ [ $sec^{2} A - tan^{2} A = 1$ ] $=$ RHS. Hence proved.
  ½ mark
Answer: LHS = $sec A - tan A = \frac{1}{s e c A + t a n A}$ = RHS (proved)
Where the marks are
- Rationalises with $(1 - sin A)$ and gets $\frac{1 - s i n A}{c o s A}$ (1 mark)
- Writes as $sec A - tan A$ (½ mark)
- Multiplies and divides by $sec A + tan A$ (½ mark)
- Gets $\frac{s e c ^{2} A - t a n ^{2} A}{s e c A + t a n A}$ (½ mark)
- Uses identity to reach RHS (½ mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerPolynomials
If $α, β$ are zeroes of the polynomial $p (x) = 5 x^{2} - 7 x - 3$ , then form a quadratic polynomial whose zeroes are $\frac{2}{α}$ and $\frac{2}{β}$ .
Try it first, then show our solutionSolution
1. For $5 x^{2} - 7 x - 3$ : $α + β = - \frac{- 7}{5} = \frac{7}{5}$ , $α β = \frac{- 3}{5}$
  1 mark
2. Sum of new zeroes $= \frac{2}{α} + \frac{2}{β} = \frac{2 ( α + β )}{α β} = \frac{2 \times \frac{7}{5}}{- \frac{3}{5}} = - \frac{14}{3}$
  ½ mark
3. Product of new zeroes $= \frac{4}{α β} = \frac{4}{- \frac{3}{5}} = - \frac{20}{3}$
  ½ mark
4. Required polynomial $= x^{2} - (sum) x + product = x^{2} + \frac{14}{3} x - \frac{20}{3}$ , or $k (3 x^{2} + 14 x - 20)$ , $k \neq = 0$ .
  1 mark
Answer: $x^{2} + \frac{14}{3} x - \frac{20}{3}$ , i.e. $k (3 x^{2} + 14 x - 20)$
Where the marks are
- $α + β = \frac{7}{5}$ and $α β = - \frac{3}{5}$ (½ + ½ mark)
- Sum of new zeroes $- \frac{14}{3}$ (½ mark)
- Product of new zeroes $- \frac{20}{3}$ (½ mark)
- Writes the polynomial (1 mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerPolynomials
Find the zeroes of the polynomial $p (x) = 3 x^{2} + 7 x - 20$ and verify the relationship between its zeroes and the coefficients.
Try it first, then show our solutionSolution
1. $3 x^{2} + 7 x - 20 = 3 x^{2} + 12 x - 5 x - 20 = 3 x (x + 4) - 5 (x + 4) = (3 x - 5) (x + 4)$ . Zeroes: $x = \frac{5}{3}$ and $x = - 4$ .
  1 mark
2. Sum of zeroes $= - 4 + \frac{5}{3} = - \frac{7}{3}$ , and $- \frac{coefficient of x}{coefficient of x ^{2}} = - \frac{7}{3}$ . Verified.
  1 mark
3. Product of zeroes $= - 4 \times \frac{5}{3} = - \frac{20}{3}$ , and $\frac{constant term}{coefficient of x ^{2}} = \frac{- 20}{3}$ . Verified.
  1 mark
Answer: Zeroes $- 4$ and $\frac{5}{3}$ ; sum $- \frac{7}{3}$ , product $- \frac{20}{3}$ (verified)
Where the marks are
- Finds zeroes $- 4, \frac{5}{3}$ (1 mark)
- Verifies sum relation (1 mark)
- Verifies product relation (1 mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerAreas Related to Circles
Chord AB of a circle subtends an angle of $12 0^{\circ}$ at the centre O of the circle. Find the length of arc AB, if radius of the circle is 21 cm.
Figure in the paper: A circle with centre O. Chord AB is drawn with A and B on the circle; O is joined to A and to B, and the angle AOB at the centre is marked 120°. See it in the official paper.
Try it first, then show our solutionSolution
1. Length of arc $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{120}{360} \times 2 \times \frac{22}{7} \times 21$
  2 marks
2. $= \frac{1}{3} \times 132 = 44$ cm
  1 mark
Answer: Length of arc AB = 44 cm
Where the marks are
- Correct formula with substitution (2 marks)
- Arc length = 44 cm (1 mark)
Open Q31 on its own page

Section D

Long answer; internal choice in 32 and 34

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerTriangles
In the given figure, $△$ ABC is right angled triangle with $∠ A = 9 0^{\circ}$ . AD is perpendicular to BC. Prove that: (i) $△$ DBA $\sim$ $△$ DAC (ii) $D A^{2} = D B \times D C$ (iii) Find the area of $△$ ABC when DB = 9 cm and DC = 16 cm.
Figure in the paper: Right triangle ABC with the right angle at A (marked). A is at the top, B at the right and C at the lower left on the base. D lies on BC with AD drawn perpendicular to BC (right-angle mark at D). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In $△ D B A$ and $△ D A C$ : $∠ B D A = ∠ A D C = 9 0^{\circ}$ [ $A D ⊥ B C$ ]
  1 mark
2. In $△ A B D$ , $∠ B A D = 9 0^{\circ} - ∠ B$ ; in $△ A B C$ , $∠ A C B = 9 0^{\circ} - ∠ B$ [ $∠ A = 9 0^{\circ}$ ]. So $∠ D A B = ∠ D C A$ . Hence $△ D B A \sim △ D A C$ (AA similarity).
  1 mark
3. (ii) Corresponding sides of similar triangles are proportional: $\frac{D B}{D A} = \frac{D A}{D C} \Rightarrow D A^{2} = D B \times D C$
  1 mark
4. (iii) $D A^{2} = 9 \times 16 = 144 \Rightarrow D A = 12$ cm
  1 mark
5. $B C = D B + D C = 25$ cm. Area of $△ A B C = \frac{1}{2} \times B C \times A D = \frac{1}{2} \times 25 \times 12 = 150 cm^{2}$
  1 mark
Answer: (i), (ii) proved; (iii) Area of $△ A B C$ = 150 cm²
Where the marks are
- Right angles at D equal (1 mark)
- Second pair of equal angles and AA similarity (1 mark)
- Proportion of sides giving $D A^{2} = D B \times D C$ (1 mark)
- DA = 12 cm (1 mark)
- Area = 150 cm² (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerTriangles
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
Try it first, then show our solutionSolution
1. Figure: Draw $△ A B C$ . Take a line parallel to $B C$ that meets $A B$ at $D$ and $A C$ at $E$ . Join $B E$ and $C D$ . Draw $E N ⊥ A B$ (N on AB) and $D M ⊥ A C$ (M on AC).
  ½ mark
2. Given: In $△ A B C$ , $D E ∥ B C$ , with $D E$ meeting $A B$ at $D$ and $A C$ at $E$ .
  ½ mark
3. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
  ½ mark
4. Construction: Join $B E$ and $C D$ . Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
  ½ mark
5. Proof: Use area $= \frac{1}{2} \times$ base $\times$ height.
  In $△ A D E$ and $△ B D E$ , the height from $E$ is $E N$ for both:
  $\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} \cdot A D \cdot E N}{\frac{1}{2} \cdot D B \cdot E N} = \frac{A D}{D B}$ ... (1)
  1 mark
6. In $△ A D E$ and $△ D E C$ , the height from $D$ is $D M$ for both:
  $\frac{a r ( A D E )}{a r ( D E C )} = \frac{\frac{1}{2} \cdot A E \cdot D M}{\frac{1}{2} \cdot E C \cdot D M} = \frac{A E}{E C}$ ... (2)
  1 mark
7. $△ B D E$ and $△ D E C$ have the same base $D E$ and lie between the same parallels $D E$ and $B C$ . 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: Proved: if DE ∥ BC in △ABC, then AD/DB = AE/EC (Basic Proportionality Theorem).
Where the marks are
- Draws a correct labelled figure (½ mark)
- States what is given (½ mark)
- States what is to be proved (½ mark)
- States the construction: join BE and CD, draw EN ⊥ AB and DM ⊥ AC (½ mark)
- Expresses $a r (A D E) / a r (B D E)$ as $A D / D B$ (1 mark)
- Expresses $a r (A D E) / a r (D E C)$ as $A E / E C$ (1 mark)
- Uses equal areas of triangles on the same base between the same parallels to conclude (1 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ respectively. Find the height of the poles and the distance of the point from the poles.
Try it first, then show our solutionSolution
1. Figure: poles AB and CD, each of height $h$ m, at the two ends of road AC = 80 m, with right angles at A and C. O is the point on AC with AO = $x$ m, OC = $(80 - x)$ m; $∠ A O B = 6 0^{\circ}$ , $∠ C O D = 3 0^{\circ}$ .
  1 mark
2. In right $△ D C O$ : $tan 3 0^{\circ} = \frac{C D}{C O} = \frac{h}{80 - x}$
  1 mark
3. $\Rightarrow \frac{1}{3} = \frac{h}{80 - x} \Rightarrow 80 - x = h 3$ …(i)
  ½ mark
4. In right $△ B A O$ : $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow h = x 3$ …(ii)
  1 mark
5. Substituting (ii) in (i): $80 - x = 3 x \Rightarrow x = 20$ , so AO = 20 m and $h = 203$ m ( $\approx 34.64$ m).
  1 mark
6. CO = $80 - 20 = 60$ m. Height of each pole = $203$ m; the point is 20 m from one pole and 60 m from the other.
  ½ mark
Answer: Height = $203$ m; distances 20 m and 60 m from the poles
Where the marks are
- Correct labelled figure (1 mark)
- $tan 3 0^{\circ} = \frac{h}{80 - x}$ (1 mark)
- Simplifies to $80 - x = h 3$ (½ mark)
- $tan 6 0^{\circ} = \frac{h}{x}$ giving $h = x 3$ (1 mark)
- x = 20 m and $h = 203$ m (½ + ½ mark)
- Other distance 60 m (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerQuadratic Equations
ABCD is a rectangle of dimensions 80 cm $\times$ 60 cm. Another rectangle PQRS is drawn inside ABCD leaving space of equal width $x$ cm along the edges of ABCD. If area PQRS is half of the area ABCD, then find the value of $x$ .
Try it first, then show our solutionSolution
1. Leaving a strip of width $x$ cm on each side: $P Q = (80 - 2 x)$ cm, $QR = (60 - 2 x)$ cm.
  2 marks
2. Area PQRS $= \frac{1}{2}$ Area ABCD $\Rightarrow (80 - 2 x) (60 - 2 x) = \frac{1}{2} \times 80 \times 60$ , i.e. $2 (80 - 2 x) (60 - 2 x) = 4800$
  1 mark
3. $4800 - 280 x + 4 x^{2} = 2400 \Rightarrow 4 x^{2} - 280 x + 2400 = 0 \Rightarrow x^{2} - 70 x + 600 = 0$
  1 mark
4. $(x - 10) (x - 60) = 0 \Rightarrow x = 10$ or $60$ . $x = 60$ is rejected since then $60 - 2 x < 0$ . So $x = 10$ cm.
  1 mark
Answer: $x = 10$ cm
Where the marks are
- Dimensions of PQRS: $80 - 2 x$ and $60 - 2 x$ (2 marks)
- Forms equation from area condition (1 mark)
- Gets $x^{2} - 70 x + 600 = 0$ (1 mark)
- Solves and rejects 60; $x = 10$ cm (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerQuadratic Equations
A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the same journey. Find the original speed of the train.
Try it first, then show our solutionSolution
1. Let the original speed be $x$ km/h. Time at original speed $= \frac{90}{x}$ h, at increased speed $= \frac{90}{x + 15}$ h. Difference is 30 min $= \frac{1}{2}$ h:
  $\frac{90}{x} - \frac{90}{x + 15} = \frac{30}{60}$
  2 marks
2. $\frac{90 ( x + 15 ) - 90 x}{x ( x + 15 )} = \frac{1}{2} \Rightarrow 2700 = x^{2} + 15 x \Rightarrow x^{2} + 15 x - 2700 = 0$
  2 marks
3. $(x + 60) (x - 45) = 0 \Rightarrow x = 45$ or $x = - 60$ (rejected, speed cannot be negative). Original speed = 45 km/h.
  1 mark
Answer: Original speed = 45 km/h
Where the marks are
- Forms the time equation $\frac{90}{x} - \frac{90}{x + 15} = \frac{1}{2}$ (2 marks)
- Reduces to $x^{2} + 15 x - 2700 = 0$ (2 marks)
- Solves, rejects −60, speed 45 km/h (1 mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerStatistics
Find mean and mode of the following data:
Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 5 4 10 13 12 10 6
Try it first, then show our solutionSolution
1. Take assumed mean $a = 45$ , $h = 10$ , $u_{i} = \frac{x _{i} - 45}{10}$ :
  Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
  10-20 5 15 −3 −15
  20-30 4 25 −2 −8
  30-40 10 35 −1 −10
  40-50 13 45 0 0
  50-60 12 55 1 12
  60-70 10 65 2 20
  70-80 6 75 3 18
  Total 60 17
  1½ marks
2. Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 45 + \frac{17}{60} \times 10$
  1 mark
3. $= 45 + 2.83 = 47.83$ (approx.)
  ½ mark
4. Highest frequency is 13, so modal class is 40-50: $l = 40$ , $f_{1} = 13$ , $f_{0} = 10$ , $f_{2} = 12$ , $h = 10$ .
  Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{13 - 10}{26 - 10 - 12} \times 10$
  1½ marks
5. $= 40 + \frac{3}{4} \times 10 = 47.5$
  ½ mark
Answer: Mean ≈ 47.83; Mode = 47.5
Where the marks are
- Correct table with $\sum f_{i} = 60$ , $\sum f_{i} u_{i} = 17$ (1½ marks)
- Correct mean formula with substitution (1 mark)
- Mean ≈ 47.83 (½ mark)
- Modal class 40-50 and correct mode formula substitution (1½ marks)
- Mode = 47.5 (½ mark)
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Class	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	4	10	13	12	10	6

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
10-20	5	15	−3	−15
20-30	4	25	−2	−8
30-40	10	35	−1	−10
40-50	13	45	0	0
50-60	12	55	1	12
60-70	10	65	2	20
70-80	6	75	3	18
Total	60			17

Section E

Case study; internal choice in 2-mark subparts of 36, 37, 38

Q36 · 4 marks · Case-basedCompetency-basedPair of Linear Equations in Two Variables
Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories.
To find the number of calories burned per minute on each machine, answer the following:
1. (i)
  Represent the above situation in terms of a pair of linear equations in two variables.
  1 mark
2. (ii)
  Show that the equations have unique solution.
  1 mark
3. (iii)(a)
  Solve both equations to find the values of the variables using elimination method.
  2 marks
4. OR
  (iii)(b)
  Solve both equations to find the values of the variables using substitution method.
  2 marks
Figure in the paper: Photograph of outdoor exercise machines in a park, including an exercise bicycle and a double cross walker, set on a paved surface. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let Seema burn $x$ calories per minute on the exercise bicycle and $y$ calories per minute on the double cross walker. The two situations give:
  $15 x + 30 y = 435$ ...(1)
  $30 x + 40 y = 690$ ...(2)
  1 mark
2. (ii) Here $a_{1} = 15, b_{1} = 30, a_{2} = 30, b_{2} = 40$ .
  $\frac{a _{1}}{a _{2}} = \frac{15}{30} = \frac{1}{2}$ and $\frac{b _{1}}{b _{2}} = \frac{30}{40} = \frac{3}{4}$ .
  Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect at exactly one point, so the pair of equations has a unique solution.
  1 mark
3. (iii)(a) Elimination method: dividing (1) by 15 gives $x + 2 y = 29$ ...(3). Dividing (2) by 10 gives $3 x + 4 y = 69$ ...(4).
  Multiply (3) by 2: $2 x + 4 y = 58$ ...(5).
  Subtract (5) from (4): $(3 x + 4 y) - (2 x + 4 y) = 69 - 58$ , so $x = 11$ .
  1 mark
4. Put $x = 11$ in (3): $11 + 2 y = 29 \Rightarrow 2 y = 18 \Rightarrow y = 9$ .
  So 11 calories per minute are burned on the bicycle and 9 calories per minute on the walker. Check: $15 (11) + 30 (9) = 165 + 270 = 435$ ✓.
  1 mark
5. (iii)(b) Substitution method: from (1), $x + 2 y = 29 \Rightarrow x = 29 - 2 y$ ...(6).
  Substitute in (2) written as $3 x + 4 y = 69$ : $3 (29 - 2 y) + 4 y = 69 \Rightarrow 87 - 6 y + 4 y = 69 \Rightarrow - 2 y = - 18 \Rightarrow y = 9$ .
  1 mark
6. From (6): $x = 29 - 2 (9) = 11$ .
  So $x = 11$ and $y = 9$ . The bicycle burns 11 calories per minute and the walker burns 9 calories per minute. Check: $30 (11) + 40 (9) = 330 + 360 = 690$ ✓.
  1 mark
Answer: (i) $15 x + 30 y = 435$ , $30 x + 40 y = 690$ ; (ii) $\frac{15}{30} \neq = \frac{30}{40}$ , so the solution is unique; (iii) $x = 11$ , $y = 9$ (11 cal/min on the bicycle, 9 cal/min on the walker) by either method
Where the marks are
- Defines the variables and writes both equations $15 x + 30 y = 435$ and $30 x + 40 y = 690$ (1 mark)
- Compares $a_{1} / a_{2}$ with $b_{1} / b_{2}$ , shows they are unequal and concludes the solution is unique (1 mark)
- (iii)(a) Eliminates one variable correctly to get $x = 11$ (1 mark)
- (iii)(a) Back-substitutes to get $y = 9$ (1 mark)
- (iii)(b) Expresses one variable in terms of the other and substitutes to get $y = 9$ (1 mark)
- (iii)(b) Finds $x = 11$ (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
There are many varieties of mushrooms available in the world. One such mushroom 'Amanita muscaria' has a upper part which is like red cap (hemispherical) and lower part is like white stem (cylinderical).
The hemispherical cap's radius = 3 cm and cylindrical stem is 2 cm high with diameter 1.4 cm. Considering mushroom a solid object, answer the following questions:
1. (i)
  What is the total height of a mushroom?
  1 mark
2. (ii)
  Find the volume of the stem.
  1 mark
3. (iii)(a)
  Determine the volume of 7 such mushrooms.
  2 marks
4. OR
  (iii)(b)
  Find the total surface area of 7 such mushrooms.
  2 marks
Figure in the paper: Left: photograph of an Amanita muscaria mushroom with a red-looking hemispherical cap spotted white and a white cylindrical stem. Right: a line diagram of the same mushroom, a semicircular dome on top of a rectangular stem. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Total height = height of stem + radius of hemispherical cap $= 2 + 3 = 5$ cm.
  1 mark
2. (ii) Radius of stem $r = \frac{1.4}{2} = 0.7$ cm and $h = 2$ cm.
  Volume of stem $= π r^{2} h = \frac{22}{7} \times 0.7 \times 0.7 \times 2 = 3.08 cm^{3}$ .
  1 mark
3. (iii)(a) Volume of one mushroom = volume of hemisphere + volume of cylinder $= \frac{2}{3} π R^{3} + π r^{2} h$ , with $R = 3$ cm, $r = 0.7$ cm and $h = 2$ cm.
  Volume of 7 mushrooms $= 7 (\frac{2}{3} \times \frac{22}{7} \times 3^{3} + \frac{22}{7} \times (0.7)^{2} \times 2)$ .
  1 mark
4. $= 7 (\frac{396}{7} + 3.08) = 396 + 21.56 = 417.56 cm^{3}$ .
  1 mark
5. (iii)(b) Surface of one solid mushroom = curved surface of hemisphere ( $2 π R^{2}$ ) + exposed ring of the flat base of the cap ( $π R^{2} - π r^{2}$ ) + curved surface of stem ( $2 π r h$ ) + bottom of stem ( $π r^{2}$ ) $= 3 π R^{2} + 2 π r h$ .
  TSA of 7 mushrooms $= 7 (3 \times \frac{22}{7} \times 3 \times 3 + 2 \times \frac{22}{7} \times 0.7 \times 2)$ .
  1 mark
6. $= 7 (\frac{594}{7} + 8.8) = 594 + 61.6 = 655.6 cm^{2}$ .
  1 mark
Answer: (i) 5 cm; (ii) 3.08 cm³; (iii)(a) 417.56 cm³; (iii)(b) 655.6 cm²
Where the marks are
- Total height $2 + 3 = 5$ cm (1 mark)
- Volume of stem $π r^{2} h = 3.08$ cm³ (1 mark)
- (iii)(a) Correct expression $7 (\frac{2}{3} π R^{3} + π r^{2} h)$ with values substituted (1 mark)
- (iii)(a) Volume of 7 mushrooms $= 417.56$ cm³ (1 mark)
- (iii)(b) Correct surface area expression $7 (3 π R^{2} + 2 π r h)$ with values substituted (1 mark)
- (iii)(b) TSA $= 655.6$ cm² (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
In a circular museum hall of radius 14 m, some statues are displayed. Statues are kept inside the inner concentric circle of radius 7 m. One such statue lying in sector OAB, is fenced along line segments OA, AP, PB and BO where P is a point on outer circle.
Based on above information, answer the following questions:
1. (i)
  Find $m ∠ A O P$ .
  1 mark
2. (ii)
  Prove that $△ O A P ≅ △ O B P$ .
  1 mark
3. (iii)(a)
  Find the length of fencing required to protect the statue. (Take $3 = 1.73$ )
  2 marks
4. OR
  (iii)(b)
  Find area of quadrilateral OAPB. (Take $3 = 1.73$ )
  2 marks
Figure in the paper: Photograph of a circular museum hall with statues, and a figure: an outer circle of radius 14 m and an inner concentric circle of radius 7 m, both centred at O. A and B lie on the inner circle; P lies on the outer circle. Segments OA, AP, PB and BO are drawn, with OP as a dashed line from O through the inner region to P. The region OAPB is the fenced area. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Here $O A = 7$ m (radius of the inner circle) and $O P = 14$ m (radius of the outer circle). PA is a tangent to the inner circle at A, so $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius].
  In right $△ O A P$ : $cos ∠ A O P = \frac{O A}{O P} = \frac{7}{14} = \frac{1}{2}$ , so $m ∠ A O P = 6 0^{\circ}$ .
  1 mark
2. (ii) In $△ O A P$ and $△ O B P$ :
  $∠ O A P = ∠ O B P = 9 0^{\circ}$ [radius ⟂ tangent at the point of contact]
  $O P = O P$ [common hypotenuse]
  $O A = O B$ [radii of the inner circle]
  Therefore $△ O A P ≅ △ O B P$ [RHS congruence].
  1 mark
3. (iii)(a) In right $△ O A P$ : $sin 6 0^{\circ} = \frac{A P}{O P} \Rightarrow \frac{3}{2} = \frac{A P}{14} \Rightarrow A P = 73 = 7 \times 1.73 = 12.11$ m.
  1 mark
4. $P B = A P$ (CPCT) and $O B = O A$ .
  Length of fencing $= O A + A P + P B + B O = 2 (O A + A P) = 2 (7 + 12.11) = 38.22$ m.
  1 mark
5. (iii)(b) As in (iii)(a), $sin 6 0^{\circ} = \frac{A P}{14} \Rightarrow A P = 73 = 12.11$ m.
  1 mark
6. Area of OAPB $= 2 \times ar (△ O A P)$ [the triangles are congruent] $= 2 \times \frac{1}{2} \times O A \times A P = 7 \times 12.11 = 84.77 m^{2}$ .
  1 mark
Answer: (i) $6 0^{\circ}$ ; (ii) $△ O A P ≅ △ O B P$ by RHS; (iii)(a) 38.22 m; (iii)(b) 84.77 m²
Where the marks are
- Uses $cos ∠ A O P = 7/14$ to get $∠ A O P = 6 0^{\circ}$ (1 mark)
- Proves the congruence by RHS with correct reasons (right angle, common hypotenuse, equal radii) (1 mark)
- (iii)(a) Finds $A P = 73 = 12.11$ m (1 mark)
- (iii)(a) Fencing $= 2 (O A + A P) = 38.22$ m (1 mark)
- (iii)(b) Finds $A P = 12.11$ m (1 mark)
- (iii)(b) Area $= 2 \times \frac{1}{2} \times O A \times A P = 84.77$ m² (1 mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.