MathsCBSE Class 10 · Basic · Sample paper, session 2024-25

CBSE Class 10 Mathematics Basic: sample paper 2024-25

The sample question paper CBSE published for the 2024-25 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_2024_25/MathsBasic-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 of 1, 1 and 2 marks

Instructions in brief: All questions compulsory; internal choice in 2 questions of Section B, 2 of Section C, 2 of Section D, and in all 2-mark sub-parts of Section E.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
HCF OF $(3^{3} \times 5^{2} \times 2)$ , $(3^{2} \times 5^{3} \times 2^{2})$ and $(3^{4} \times 5 \times 2^{3})$ is
1. (A)450
2. (B)90
3. (C)180
4. (D)630
Try it first, then show our solutionSolution
1. HCF takes the smallest power of each common prime: $3^{2}$ , $5^{1}$ , $2^{1}$ .
2. HCF $= 9 \times 5 \times 2 = 90$ , option (B).
  1 mark
Answer: 90, option (B)
Where the marks are
- Takes lowest powers of common primes to get 90 (1 mark)
Q2 · 1 mark · MCQPair of Linear Equations in Two Variables
The system of linear equations represented by the lines l and m is
1. (A)consistent with unique solution
2. (B)inconsistent
3. (C)consistent with three solutions
4. (D)consistent with many solutions
Figure in the paper: Coordinate axes X and Y with origin O. Two lines l and m are drawn. Line l passes through (0,2) on the Y-axis and (-1,0) on the X-axis. Line m passes through (0,2) and (3/2,0) on the X-axis. The two lines intersect at (0,2). Labels (0,2), (-1,0), (3/2,0) and O are marked. See it in the official paper.
Try it first, then show our solutionSolution
1. Lines l and m cross at exactly one point, (0, 2).
2. Intersecting lines give a unique solution, so the system is consistent with a unique solution: option (A).
  1 mark
Answer: Consistent with unique solution, option (A)
Where the marks are
- Recognises the lines intersect at one point, so there is a unique solution (1 mark)
Q3 · 1 mark · MCQQuadratic Equations
The value of k for which the quadratic equation $k x^{2} - 5 x + 1 = 0$ does not have a real solution, is
1. (A)0
2. (B) $\frac{25}{4}$
3. (C) $\frac{4}{25}$
4. (D)7
Try it first, then show our solutionSolution
1. There are no real roots when $D < 0$ : $25 - 4 k < 0 \Rightarrow k > \frac{25}{4} = 6.25$ .
2. Among the options only $k = 7$ satisfies this, so the answer is option (D).
  1 mark
Answer: k = 7, option (D)
Where the marks are
- Uses $b^{2} - 4 a c < 0$ to get $k > 25/4$ and picks 7 (1 mark)
Q4 · 1 mark · MCQCoordinate Geometry
The distance between the points $(a, b)$ and $(- a, - b)$ is
1. (A) $a^{2} + b^{2}$
2. (B) $a^{2} + b^{2}$
3. (C) $2 a^{2} + b^{2}$
4. (D) $4 a^{2} + b^{2}$
Try it first, then show our solutionSolution
1. $d = (2 a)^{2} + (2 b)^{2} = 4 a^{2} + 4 b^{2} = 2 a^{2} + b^{2}$ , option (C).
  1 mark
Answer: $2 a^{2} + b^{2}$ , option (C)
Where the marks are
- Applies the distance formula correctly (1 mark)
Q5 · 1 mark · MCQCircles
In the given figure, PQ and PR are tangents to a circle centred at O. If $∠ QP R = 3 5^{\circ}$ then $∠ QO R$ is equal to
1. (A) $7 0^{\circ}$
2. (B) $9 0^{\circ}$
3. (C) $13 5^{\circ}$
4. (D) $14 5^{\circ}$
Figure in the paper: A circle with centre O. External point P with two tangents PQ and PR touching the circle at Q and R. Angle QPR = 35° is marked at P. Radii OQ and OR are drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ O QP = ∠ O R P = 9 0^{\circ}$ [tangent ⟂ radius], so in quadrilateral OQPR, $∠ QO R = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 3 5^{\circ} = 14 5^{\circ}$ , option (D).
  1 mark
Answer: 145°, option (D)
Where the marks are
- Uses $∠ QO R = 18 0^{\circ} - ∠ QP R$ to get 145° (1 mark)
Q6 · 1 mark · MCQTriangles
If $△ A B C \sim △ P QR$ such that $3 A B = 2 P Q$ and $B C = 10$ cm, then length QR is equal to
1. (A)10 cm
2. (B)15 cm
3. (C) $\frac{20}{3}$ cm
4. (D)30 cm
Try it first, then show our solutionSolution
1. $\frac{A B}{P Q} = \frac{2}{3} = \frac{B C}{QR}$ , so $QR = \frac{3 \times 10}{2} = 15$ cm, option (B).
  1 mark
Answer: 15 cm, option (B)
Where the marks are
- Uses the ratio of corresponding sides, 2:3, to get QR = 15 cm (1 mark)
Q7 · 1 mark · MCQIntroduction to Trigonometry
If $3 cot A = 4$ , where $0^{\circ} < A < 9 0^{\circ}$ , then $sec A$ is equal to
1. (A) $\frac{5}{4}$
2. (B) $\frac{4}{3}$
3. (C) $\frac{5}{3}$
4. (D) $\frac{3}{4}$
Try it first, then show our solutionSolution
1. $cot A = \frac{4}{3} \Rightarrow tan A = \frac{3}{4}$ .
2. $sec A = 1 + \frac{9}{16} = \frac{5}{4}$ , option (A).
  1 mark
Answer: 5/4, option (A)
Where the marks are
- Uses $sec^{2} A = 1 + tan^{2} A$ to get 5/4 (1 mark)
Q8 · 1 mark · MCQTriangles
In the given figure, $△ B A C$ is similar to
1. (A) $△ A E D$
2. (B) $△ E A D$
3. (C) $△ A C B$
4. (D) $△ B C A$
Figure in the paper: A right-angled triangle ABC with the right angle at C (marked). Point D lies on AB extended beyond... on segment AB with a right angle marked at D, and a vertical segment DE from D up to E such that A, C, E are collinear and A, D, B are collinear, with BC drawn. Points A, B, C, D, E are labelled. See it in the official paper.
Try it first, then show our solutionSolution
1. In $△ B A C$ and $△ E A D$ , $∠ A$ is common and $∠ C = ∠ D = 9 0^{\circ}$ .
2. So by AA similarity, B↔E, A↔A, C↔D, which gives $△ B A C \sim △ E A D$ , option (B).
  1 mark
Answer: △EAD, option (B)
Where the marks are
- Uses AA similarity with the correct correspondence of vertices (1 mark)
Open Q8 on its own page
Q9 · 1 mark · MCQReal Numbers
If H.C.F(420,189) = 21 then L.C.M(420,189) is
1. (A)420
2. (B)1890
3. (C)3780
4. (D)3680
Try it first, then show our solutionSolution
1. LCM $= \frac{420 \times 189}{21} = 20 \times 189 = 3780$ , option (C).
  1 mark
Answer: 3780, option (C)
Where the marks are
- Uses HCF × LCM = product of the two numbers (1 mark)
Q10 · 1 mark · MCQArithmetic Progressions
The $4^{t h}$ term from the end of the A.P $- 8, - 5, - 2, \dots, 49$ is
1. (A)37
2. (B)40
3. (C)1
4. (D)43
Try it first, then show our solutionSolution
1. Here $d = 3$ . Counting back from the end, the terms are 49, 46, 43, 40.
2. So the 4th term from the end is $49 - 3 \times 3 = 40$ , option (B).
  1 mark
Answer: 40, option (B)
Where the marks are
- Uses $l - (n - 1) d$ to get 40 (1 mark)
Q11 · 1 mark · MCQTriangles
In the given figure, if $△ O C A \sim △ O B D$ then $∠ O A C$ is equal to
1. (A) $5 8^{\circ}$
2. (B) $5 5^{\circ}$
3. (C) $12 8^{\circ}$
4. (D) $5 2^{\circ}$
Figure in the paper: Two lines BD and AC intersect at O. B and D lie on one line (top), A and C on another (bottom), with segments AD and BC crossing at O. Angle at D between the top line and segment DO is marked 58°. Angle BOA (at O) is marked 110°. See it in the official paper.
Try it first, then show our solutionSolution
1. $∠ B O D = 18 0^{\circ} - 11 0^{\circ} = 7 0^{\circ}$ [linear pair]. In $△ O B D$ , $∠ O B D = 18 0^{\circ} - 7 0^{\circ} - 5 8^{\circ} = 5 2^{\circ}$ .
2. Since $△ O C A \sim △ O B D$ , $∠ O A C = ∠ O B D = 5 2^{\circ}$ , option (D).
  1 mark
Answer: 52°, option (D)
Where the marks are
- Finds $∠ O B D = 5 2^{\circ}$ and uses the similarity correspondence (1 mark)
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Q12 · 1 mark · MCQCircles
If perimeter of given triangle is 38 cm, then length AP is equal to
1. (A)19 cm
2. (B)5 cm
3. (C)10 cm
4. (D)8 cm
Figure in the paper: Triangle ABC with an incircle touching AB at P, AC at Q and BC at R. BR = 6 cm and RC = 8 cm are marked on side BC. See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents from an external point are equal, so BP = BR = 6, CQ = CR = 8 and AP = AQ = x.
2. Perimeter: $2 x + 12 + 16 = 38 \Rightarrow x = 5$ cm, option (B).
  1 mark
Answer: 5 cm, option (B)
Where the marks are
- Uses equal tangent lengths and the perimeter to get AP = 5 cm (1 mark)
Q13 · 1 mark · MCQIntroduction to Trigonometry
$\frac{1 - t a n ^{2} 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}}$ is equal to
1. (A) $cos 6 0^{\circ}$
2. (B) $sin 6 0^{\circ}$
3. (C)1
4. (D) $t a n^{2} 6 0^{\circ}$
Try it first, then show our solutionSolution
1. $\frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{2/3}{4/3} = \frac{1}{2} = cos 6 0^{\circ}$ , option (A).
  1 mark
Answer: cos 60°, option (A)
Where the marks are
- Substitutes $tan^{2} 3 0^{\circ} = 1/3$ to get 1/2 (1 mark)
Q14 · 1 mark · MCQSurface Areas and Volumes
The total surface area of solid hemisphere of radius $r$ is
1. (A) $π r^{2}$
2. (B) $2 π r^{2}$
3. (C) $3 π r^{2}$
4. (D) $4 π r^{2}$
Try it first, then show our solutionSolution
1. TSA = curved surface + base $= 2 π r^{2} + π r^{2} = 3 π r^{2}$ , option (C).
  1 mark
Answer: $3 π r^{2}$ , option (C)
Where the marks are
- Adds the curved surface and the flat circular base (1 mark)
Q15 · 1 mark · MCQProbability
Which of the following cannot be the probability of an event?
1. (A)0.4
2. (B)4%
3. (C)0.04%
4. (D)4
Try it first, then show our solutionSolution
1. A probability must satisfy $0 \leq P \leq 1$ . Since 4 > 1, it cannot be a probability: option (D).
  1 mark
Answer: 4, option (D)
Where the marks are
- Uses $0 \leq P (E) \leq 1$ (1 mark)
Q16 · 1 mark · MCQQuadratic Equations
The roots of quadratic equation $3 x^{2} - 43 x + 4 = 0$ are
1. (A)not real
2. (B)real and equal
3. (C)rational and distinct
4. (D)irrational and distinct
Try it first, then show our solutionSolution
1. $D = (43)^{2} - 4 (3) (4) = 48 - 48 = 0$ , so the roots are real and equal: option (B).
  1 mark
Answer: Real and equal, option (B)
Where the marks are
- Computes discriminant D = 0 (1 mark)
Q17 · 1 mark · MCQStatistics
The following distribution shows the marks distribution of 80 students.
Marks Below 10 Below 20 Below 30 Below 40 Below 50 Below 60
No. of students 2 12 28 56 76 80
The median class is
1. (A)20-30
2. (B)40-50
3. (C)30-40
4. (D)10-20
Try it first, then show our solutionSolution
1. $n /2 = 40$ . The first cumulative frequency greater than 40 is 56 (below 40), so the median class is 30-40, option (C).
  1 mark
Answer: 30-40, option (C)
Where the marks are
- Locates the class containing n/2 = 40 (1 mark)
Q18 · 1 mark · MCQPolynomials
A quadratic polynomial whose zeroes are $\frac{2}{5}$ and $\frac{- 1}{5}$ is
1. (A) $25 x^{2} + 5 x - 2$
2. (B) $5 x^{2} - 2 x + 1$
3. (C) $5 x^{2} + 2 x - 1$
4. (D) $25 x^{2} - 5 x - 2$
Try it first, then show our solutionSolution
1. Sum of zeroes $= \frac{1}{5}$ , product of zeroes $= - \frac{2}{25}$ .
2. $p (x) = x^{2} - \frac{1}{5} x - \frac{2}{25}$ , i.e. $25 x^{2} - 5 x - 2$ , option (D).
  1 mark
Answer: $25 x^{2} - 5 x - 2$ , option (D)
Where the marks are
- Uses $x^{2} - (sum) x + product$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedArithmetic Progressions
DIRECTION: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion(A): The sequence $- 1, - 1 - 1, \dots, - 1$ is an AP.
Reason(R): In an AP, $a_{n} - a_{n - 1}$ is constant where $n \geq 2$ and $n \in N$
1. (A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
2. (B)Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
3. (C)Assertion (A) is true but reason (R) is false.
4. (D)Assertion (A) is false but reason (R) is true.
Try it first, then show our solutionSolution
1. The terms are constant (each is -1), so every difference is 0, which is constant. Hence the sequence is an AP and A is true.
2. R is the definition of an AP, so it is true and correctly explains A: option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Recognises common difference 0 and that R is the definition of an AP (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedReal Numbers
DIRECTION: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion(A): $(2 + 3) 3$ is an irrational number.
Reason(R): Product of two irrational numbers is always irrational.
1. (A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
2. (B)Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
3. (C)Assertion (A) is true but reason (R) is false.
4. (D)Assertion (A) is false but reason (R) is true.
Try it first, then show our solutionSolution
1. $(2 + 3) 3 = 23 + 3$ , which is irrational, so A is true.
2. R is false, since for example $2 \times 2 = 2$ is rational. So the answer is option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows A is true and gives a counterexample to R (1 mark)

Marks	Below 10	Below 20	Below 30	Below 40	Below 50	Below 60
No. of students	2	12	28	56	76	80

Section B

Very short answer (VSA)

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerCoordinate Geometry
$P (x, y)$ is a point equidistant from the points $A (4, 3)$ and $B (3, 4)$ . Prove that $x - y = 0$ .
Try it first, then show our solutionSolution
1. Since $P (x, y)$ is equidistant from $A (4, 3)$ and $B (3, 4)$ , $P A = P B$ , so $P A^{2} = P B^{2}$ . By the distance formula:
  $(x - 4)^{2} + (y - 3)^{2} = (x - 3)^{2} + (y - 4)^{2}$
  1 mark
2. Expanding: $x^{2} - 8 x + 16 + y^{2} - 6 y + 9 = x^{2} - 6 x + 9 + y^{2} - 8 y + 16$ .
  This gives $- 8 x - 6 y = - 6 x - 8 y$ , so $- 2 x + 2 y = 0$ .
  Hence $x = y$ , i.e. $x - y = 0$ . Proved.
  1 mark
Answer: $x - y = 0$ (proved)
Where the marks are
- Writes $P A^{2} = P B^{2}$ using the distance formula (1 mark)
- Simplifies to $x - y = 0$ (1 mark)
Open Q21(A) on its own page
OR
Q21(B) · 2 marks · Very short answerCoordinate Geometry
In the given figure, $△ A B C$ is an equilateral triangle. Coordinates of vertices A and B are $(0, 3)$ and $(0, - 3)$ respectively. Find the coordinates of points C.
Figure in the paper: Coordinate axes with X'OX horizontal and YOY' vertical. A is on the positive Y-axis at (0,3), B is on the negative Y-axis at (0,-3). C lies on the positive X-axis, forming equilateral triangle ABC with AB as the base on the Y-axis. See it in the official paper.
Try it first, then show our solutionSolution
1. $A B = 3 - (- 3) = 6$ units. Since $△ A B C$ is equilateral, $A C = A B = 6$ units.
  ½ mark
2. $C$ lies on the positive x-axis, so the origin $O$ is the midpoint of $A B$ and $O A = 3$ .
  In right $△ A O C$ : $O C = A C^{2} - O A^{2} = 36 - 9 = 27 = 33$ units.
  1 mark
3. Hence the coordinates of C are $(33, 0)$ .
  ½ mark
Answer: $C (33, 0)$
Where the marks are
- Finds $A B = A C = 6$ (½ mark)
- Uses Pythagoras to get $O C = 33$ (1 mark)
- States $C = (33, 0)$ (½ mark)
Open Q21(B) on its own page
Q22 · 2 marks · Very short answerCircles
In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm, then find the radius of the smaller circle.
Try it first, then show our solutionSolution
1. Figure: two concentric circles with centre $O$ . Chord $A B$ of the larger circle touches the smaller circle at $M$ . Join $O A$ and $O M$ .
  $O M ⊥ A B$ [tangent ⟂ radius at the point of contact].
  ½ mark
2. The perpendicular from the centre bisects the chord, so $A M = \frac{8}{2} = 4$ cm.
  ½ mark
3. In right $△ O M A$ : $O M = O A^{2} - A M^{2} = 5^{2} - 4^{2} = 9 = 3$ cm.
  So the radius of the smaller circle is 3 cm.
  1 mark
Answer: 3 cm
Where the marks are
- Correct figure with $O M ⊥ A B$ (½ mark)
- $A M = 4$ cm (½ mark)
- $O M = 25 - 16 = 3$ cm (1 mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerArithmetic Progressions
The sum of the first 12 terms of an A.P. is 900. If its first term is 20 then find the common difference and $1 2^{t h}$ term.
Try it first, then show our solutionSolution
1. Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with $n = 12$ , $a = 20$ :
  $\frac{12}{2} [2 \times 20 + 11 d] = 900$
  ½ mark
2. $6 (40 + 11 d) = 900 \Rightarrow 40 + 11 d = 150 \Rightarrow 11 d = 110 \Rightarrow d = 10$
  1 mark
3. $a_{12} = a + 11 d = 20 + 11 \times 10 = 130$
  ½ mark
Answer: $d = 10$ , $a_{12} = 130$
Where the marks are
- Correct substitution in the $S_{n}$ formula (½ mark)
- $d = 10$ (1 mark)
- $a_{12} = 130$ (½ mark)
Open Q23(A) on its own page
OR
Q23(B) · 2 marks · Very short answerArithmetic Progressions
The sum of first $n$ terms of an A.P. is represented by $S_{n} = 6 n - n^{2}$ . Find the common difference.
Try it first, then show our solutionSolution
1. Put $n = 1$ : $S_{1} = a = 6 (1) - 1^{2} = 5$ …(i)
  ½ mark
2. Put $n = 2$ : $S_{2} = a_{1} + a_{2} = 2 a + d = 6 (2) - 2^{2} = 8$ …(ii)
  1 mark
3. From (i) and (ii): $10 + d = 8 \Rightarrow d = - 2$
  ½ mark
Answer: $d = - 2$
Where the marks are
- $S_{1} = a = 5$ (½ mark)
- $S_{2} = 2 a + d = 8$ (1 mark)
- $d = - 2$ (½ mark)
Open Q23(B) on its own page
Q24 · 2 marks · Very short answerIntroduction to Trigonometry
If $s in (A - B) = \frac{1}{2}$ and $cos (A + B) = \frac{1}{2}$ , $0^{\circ} < A + B < 9 0^{\circ}$ and $A > B$ , then find the values of $A$ and $B$ .
Try it first, then show our solutionSolution
1. $sin (A - B) = \frac{1}{2} = sin 3 0^{\circ} \Rightarrow A - B = 3 0^{\circ}$ …(i)
  ½ mark
2. $cos (A + B) = \frac{1}{2} = cos 6 0^{\circ} \Rightarrow A + B = 6 0^{\circ}$ …(ii)
  ½ mark
3. Adding (i) and (ii): $2 A = 9 0^{\circ} \Rightarrow A = 4 5^{\circ}$
  ½ mark
4. Substituting in (ii): $B = 6 0^{\circ} - 4 5^{\circ} = 1 5^{\circ}$
  ½ mark
Answer: $A = 4 5^{\circ}$ , $B = 1 5^{\circ}$
Where the marks are
- $A - B = 3 0^{\circ}$ (½ mark)
- $A + B = 6 0^{\circ}$ (½ mark)
- $A = 4 5^{\circ}$ (½ mark)
- $B = 1 5^{\circ}$ (½ mark)
Open Q24 on its own page
Q25 · 2 marks · Very short answerStatistics
Calculate mode of the following distribution:
Class 5-10 10-15 15-20 20-25 25-30 30-35
Frequency 5 6 15 10 5 4
Try it first, then show our solutionSolution
1. The highest frequency is 15, so the modal class is 15–20.
  Here $l = 15$ , $h = 5$ , $f_{1} = 15$ , $f_{0} = 6$ , $f_{2} = 10$ .
  ½ mark
2. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 15 + (\frac{15 - 6}{30 - 6 - 10}) \times 5$
  1 mark
3. $= 15 + \frac{45}{14} = 15 + 3.21 = 18.21$ (approx.)
  ½ mark
Answer: Mode ≈ 18.21
Where the marks are
- Identifies the modal class 15–20 (½ mark)
- Correct substitution in the mode formula (1 mark)
- Mode ≈ 18.21 (½ mark)
Open Q25 on its own page

Class	5-10	10-15	15-20	20-25	25-30	30-35
Frequency	5	6	15	10	5	4

Section C

Short answer (SA)

Q26 · 3 marks · Short answerReal Numbers
Prove that $5$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $5$ is rational.
  Then $5 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are co-prime.
  ½ mark
2. Squaring gives $5 q^{2} = p^{2}$ . So $p^{2}$ is divisible by 5.
  Since 5 is prime, $p$ is divisible by 5 [if a prime divides $p^{2}$ , it divides $p$ ].
  So $p = 5 a$ for some integer $a$ .
  1 mark
3. Substitute: $25 a^{2} = 5 q^{2} \Rightarrow q^{2} = 5 a^{2}$ . So $q^{2}$ is divisible by 5, and hence $q$ is divisible by 5.
  ½ mark
4. Now 5 is a common factor of $p$ and $q$ . This contradicts the assumption that $p$ and $q$ are co-prime.
  Hence our assumption is wrong, and $5$ is irrational.
  1 mark
Answer: $5$ is irrational (proved by contradiction).
Where the marks are
- Assumes $5 = p / q$ with $p, q$ co-prime (½ mark)
- Shows 5 divides $p$ , writes $p = 5 a$ (1 mark)
- Shows 5 divides $q$ (½ mark)
- States the contradiction and concludes (1 mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(A) · 3 marks · Short answerCoordinate Geometry
Find the ratio in which the y-axis divides the line segment joining the points $(4, - 5)$ and $(- 1, 2)$ . Also find the point of intersection.
Try it first, then show our solutionSolution
1. Let the y-axis meet the segment joining $A (4, - 5)$ and $B (- 1, 2)$ at $P (0, y)$ . Let $A P : P B = k : 1$ .
  ½ mark
2. By the section formula, the x-coordinate of P is $\frac{k ( - 1 ) + 1 ( 4 )}{k + 1} = 0$ .
  1 mark
3. $- k + 4 = 0 \Rightarrow k = 4$ . So the required ratio is $4 : 1$ .
  ½ mark
4. $y = \frac{4 ( 2 ) + 1 ( - 5 )}{4 + 1} = \frac{8 - 5}{5} = \frac{3}{5}$
  ½ mark
5. Hence the point of intersection is $(0, \frac{3}{5})$ .
  ½ mark
Answer: Ratio $4 : 1$ ; point $(0, \frac{3}{5})$
Where the marks are
- Takes the point as $(0, y)$ and the ratio as $k : 1$ (½ mark)
- Sets the x-coordinate from the section formula equal to 0 (1 mark)
- $k = 4$ , ratio $4 : 1$ (½ mark)
- $y = \frac{3}{5}$ (½ mark)
- Point $(0, \frac{3}{5})$ (½ mark)
Open Q27(A) on its own page
OR
Q27(B) · 3 marks · Short answerCoordinate Geometry
Line $4 x + y = 4$ divides the line segment joining the points $(- 2, - 1)$ and $(3, 5)$ in a certain ratio. Find the ratio.
Try it first, then show our solutionSolution
1. Let the line $4 x + y = 4$ meet the segment joining $A (- 2, - 1)$ and $B (3, 5)$ at $P (x_{1}, y_{1})$ , with $A P : P B = k : 1$ . By the section formula:
  $x_{1} = \frac{3 k - 2}{k + 1}$ , $y_{1} = \frac{5 k - 1}{k + 1}$
  1 mark
2. Since P lies on $4 x + y = 4$ :
  $4 (\frac{3 k - 2}{k + 1}) + \frac{5 k - 1}{k + 1} = 4$
  ½ mark
3. $12 k - 8 + 5 k - 1 = 4 k + 4 \Rightarrow 13 k = 13 \Rightarrow k = 1$
  1 mark
4. Hence the required ratio is $1 : 1$ .
  ½ mark
Answer: $1 : 1$
Where the marks are
- Coordinates of P by the section formula in ratio $k : 1$ (1 mark)
- Substitutes P into $4 x + y = 4$ (½ mark)
- Solves to get $k = 1$ (1 mark)
- States the ratio $1 : 1$ (½ mark)
Open Q27(B) on its own page
Q28 · 3 marks · Short answerIntroduction to Trigonometry
Prove that: $(cosec A - s in A) (sec A - cos A) = \frac{1}{t an A + co t A}$ .
Try it first, then show our solutionSolution
1. LHS $= (\frac{1}{s i n A} - sin A) (\frac{1}{c o s A} - cos A)$
  ½ mark
2. $= \frac{1 - s i n ^{2} A}{s i n A} \times \frac{1 - c o s ^{2} A}{c o s A} = \frac{c o s ^{2} A}{s i n A} \times \frac{s i n ^{2} A}{c o s A}$ [using $sin^{2} A + cos^{2} A = 1$ ]
  1 mark
3. $= sin A cos A$
  ½ mark
4. RHS $= \frac{1}{\frac{s i n A}{c o s A} + \frac{c o s A}{s i n A}} = \frac{s i n A c o s A}{s i n ^{2} A + c o s ^{2} A} = sin A cos A$ .
  Hence LHS = RHS. Proved.
  1 mark
Answer: Both sides equal $sin A cos A$ ; hence proved.
Where the marks are
- Writes cosec A and sec A in terms of sin A and cos A (½ mark)
- Uses $1 - sin^{2} A = cos^{2} A$ and $1 - cos^{2} A = sin^{2} A$ (1 mark)
- Simplifies LHS to $sin A cos A$ (½ mark)
- Simplifies RHS to $sin A cos A$ (1 mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerStatistics
Find the mean using the step deviation method.
Class 0-10 10-20 20-30 30-40 40-50
Frequency 6 10 15 9 10
Try it first, then show our solutionSolution
1. Take assumed mean $a = 25$ and class size $h = 10$ , with $u_{i} = \frac{x _{i} - 25}{10}$ .
  Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  0-10 5 6 −2 −12
  10-20 15 10 −1 −10
  20-30 25 15 0 0
  30-40 35 9 1 9
  40-50 45 10 2 20
  Total $Σ f_{i} = 50$ $Σ f_{i} u_{i} = 7$
  1½ marks
2. Mean $= a + h \times \frac{Σ f _{i} u _{i}}{Σ f _{i}} = 25 + 10 \times \frac{7}{50}$
  1 mark
3. $= 25 + 1.4 = 26.4$
  ½ mark
Answer: Mean = 26.4
Where the marks are
- Correct table with $x_{i}$ , $u_{i}$ , $f_{i} u_{i}$ , $Σ f_{i} = 50$ and $Σ f_{i} u_{i} = 7$ (1½ marks)
- Correct substitution in the step-deviation formula (1 mark)
- Mean = 26.4 (½ mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(A) · 3 marks · Short answerCircles
In the given figure, PA and PB are tangents to a circle centred at O. Prove that (i) OP bisects $∠ A P B$ (ii) OP is the right bisector of AB.
Figure in the paper: A circle with centre O. External point P with tangents PA and PB touching the circle at A and B. Chord AB is drawn, and the line OP meets AB at Q. Radii OA and OB are drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In $△ O A P$ and $△ O B P$ :
  $O A = O B$ [radii]
  $P A = P B$ [tangents from an external point are equal]
  $O P = O P$ [common]
  So $△ O A P ≅ △ O B P$ [SSS]. Hence $∠ A P O = ∠ B P O$ [CPCT], i.e. OP bisects $∠ A P B$ .
  1 mark
2. (ii) In $△ A QP$ and $△ B QP$ :
  $P A = P B$ [equal tangents]
  $∠ A P Q = ∠ B P Q$ [from (i)]
  $P Q = P Q$ [common]
  So $△ A QP ≅ △ B QP$ [SAS]. Hence $A Q = QB$ and $∠ A QP = ∠ B QP$ [CPCT].
  1 mark
3. Since AQB is a straight line, $∠ A QP + ∠ B QP = 18 0^{\circ}$ , so $∠ A QP = ∠ B QP = 9 0^{\circ}$ .
  Thus OP is perpendicular to AB and bisects it at Q, so OP is the right (perpendicular) bisector of AB.
  1 mark
Answer: OP bisects ∠APB, and OP is the perpendicular bisector of AB (proved).
Where the marks are
- $△ O A P ≅ △ O B P$ , giving $∠ A P O = ∠ B P O$ (1 mark)
- $△ A QP ≅ △ B QP$ , giving $A Q = QB$ and equal angles at Q (1 mark)
- Linear pair gives $9 0^{\circ}$ , so OP is the right bisector of AB (1 mark)
Open Q30(A) on its own page
OR
Q30(B) · 3 marks · Short answerCircles
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Try it first, then show our solutionSolution
1. Figure: circle with centre O; from external point P draw tangents PQ and PR touching the circle at Q and R; join OP, OQ, OR.
  1 mark
2. Given: PQ and PR are tangents from external point P to a circle with centre O. To prove: $P Q = P R$ . Construction: join OP, OQ, OR.
  ½ mark
3. In $△ O QP$ and $△ O R P$ : $O Q = O R$ [radii]; $O P = O P$ [common]; $∠ O QP = ∠ O R P = 9 0^{\circ}$ [tangent ⟂ radius at point of contact]. So $△ O QP ≅ △ O R P$ [RHS congruence].
  1 mark
4. Hence $P Q = P R$ [CPCT].
  ½ mark
Answer: PQ = PR (proved)
Where the marks are
- Correct labelled figure (1 mark)
- Given, To prove, Construction (½ mark)
- Proves △OQP ≅ △ORP by RHS with reasons (1 mark)
- Concludes PQ = PR by CPCT (½ mark)
Open Q30(B) on its own page
Q31 · 3 marks · Short answerPair of Linear Equations in Two Variables
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If ten's digit is 3 more than the unit's digit, then find the number.
Try it first, then show our solutionSolution
1. Let the ten's digit be $x$ and the unit's digit be $y$ . Then the number is $10 x + y$ .
  ½ mark
2. The reversed number is $10 y + x$ , so $(10 x + y) + (10 y + x) = 99$ .
  ½ mark
3. $11 x + 11 y = 99 \Rightarrow x + y = 9$ …(i)
  ½ mark
4. The ten's digit is 3 more than the unit's digit: $x = y + 3$ …(ii)
  ½ mark
5. Substituting (ii) in (i): $2 y + 3 = 9 \Rightarrow y = 3$ , so $x = 6$ .
  ½ mark
6. The required number is $10 (6) + 3 = 63$ .
  ½ mark
Answer: 63
Where the marks are
- Represents the number as $10 x + y$ (½ mark)
- Forms the equation for number + reversed number = 99 (½ mark)
- Simplifies to $x + y = 9$ (½ mark)
- Writes $x = y + 3$ (½ mark)
- Solves to get $x = 6$ , $y = 3$ (½ mark)
- States the number 63 (½ mark)
Open Q31 on its own page

Class	0-10	10-20	20-30	30-40	40-50
Frequency	6	10	15	9	10

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
0-10	5	6	−2	−12
10-20	15	10	−1	−10
20-30	25	15	0	0
30-40	35	9	1	9
40-50	45	10	2	20
Total		$Σ f_{i} = 50$		$Σ f_{i} u_{i} = 7$

Section D

Long answer (LA)

Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerQuadratic Equations
Amita buys some books for ₹1920. If she had bought 4 more books for the same amount each book would cost her ₹ 24 less. How many books did she buy? What was the initial price of one book?
Try it first, then show our solutionSolution
1. Let the number of books bought be $x$ . Then the cost of one book is $₹ \frac{1920}{x}$ .
  1 mark
2. With $x + 4$ books, each costs $₹ \frac{1920}{x + 4}$ . So:
  $\frac{1920}{x} - \frac{1920}{x + 4} = 24$
  1 mark
3. $1920 \times \frac{( x + 4 ) - x}{x ( x + 4 )} = 24 \Rightarrow 7680 = 24 x (x + 4) \Rightarrow x^{2} + 4 x - 320 = 0$
  $(x + 20) (x - 16) = 0$
  1 mark
4. $x = 16$ or $x = - 20$ . The number of books cannot be negative, so $x = 16$ . Amita bought 16 books.
  1 mark
5. Initial price of one book $= \frac{1920}{16} = ₹120$ .
  1 mark
Answer: 16 books; ₹120 per book
Where the marks are
- Writes the cost of one book as $\frac{1920}{x}$ (1 mark)
- Forms the equation $\frac{1920}{x} - \frac{1920}{x + 4} = 24$ (1 mark)
- Reduces to $x^{2} + 4 x - 320 = 0$ and factorises (1 mark)
- Rejects $- 20$ and gets 16 books (1 mark)
- Price of one book = ₹120 (1 mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerQuadratic Equations
A train travels at a certain average speed for a distance of 132 km and then travels a distance of 140 km at an average speed of 4 km/h more than the initial speed. If it takes 4 hours to complete the whole journey, what was the initial average speed? Determine the time taken by train to cover the distances separately.
Try it first, then show our solutionSolution
1. Let the initial average speed be $x$ km/h. Using time = distance ÷ speed:
  $\frac{132}{x} + \frac{140}{x + 4} = 4$
  1 mark
2. $132 (x + 4) + 140 x = 4 x (x + 4) \Rightarrow 272 x + 528 = 4 x^{2} + 16 x \Rightarrow 4 x^{2} - 256 x - 528 = 0$
  1 mark
3. Dividing by 4: $x^{2} - 64 x - 132 = 0 \Rightarrow (x - 66) (x + 2) = 0$
  1 mark
4. $x = 66$ or $x = - 2$ . Speed cannot be negative, so the initial average speed is 66 km/h.
  1 mark
5. Time for 132 km $= \frac{132}{66} = 2$ h.
  Time for 140 km $= \frac{140}{70} = 2$ h.
  1 mark
Answer: Initial speed 66 km/h; 2 hours for each part
Where the marks are
- Forms the equation $\frac{132}{x} + \frac{140}{x + 4} = 4$ (1 mark)
- Simplifies to $4 x^{2} - 256 x - 528 = 0$ (1 mark)
- Gets $x^{2} - 64 x - 132 = 0$ and factorises (1 mark)
- Rejects $- 2$ , so speed = 66 km/h (1 mark)
- Finds both times: 2 h and 2 h (1 mark)
Open Q32(B) on its own page
Q33 · 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 Q33 on its own page
Q34 · 5 marks · Long answerAreas Related to Circles
The perimeter of sector OACB of the circle centred at O and of radius 24, is 73.12 cm.
(i) Find the central angle $∠ A O B$ .
(ii) Find the area of the minor segment ACB. (Use $π = 3.14$ and $3 = 1.73$ )
Figure in the paper: A circle with centre O and radius 24. Points A and B on the circle are joined to O, forming sector OACB with arc ACB (C on the minor arc). Chord AB is drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Perimeter of the sector $= 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r$ , so:
  $2 (24) + \frac{2 \times 3.14 \times 24 \times θ}{360} = 73.12$
  1 mark
2. $\frac{150.72 θ}{360} = 73.12 - 48 = 25.12 \Rightarrow θ = \frac{25.12 \times 360}{150.72} = 6 0^{\circ}$ .
  So $∠ A O B = 6 0^{\circ}$ .
  1 mark
3. (ii) Since $O A = O B$ and $∠ A O B = 6 0^{\circ}$ , $△ O A B$ is equilateral with side 24 cm.
  Area of minor segment = area of sector − area of $△ O A B$
  $= (\frac{60}{360} \times 3.14 \times 24 \times 24 - \frac{1.73}{4} \times 24 \times 24)$ cm²
  2 marks
4. $= (301.44 - 249.12)$ cm² $= 52.32$ cm²
  1 mark
Answer: (i) $6 0^{\circ}$ ; (ii) 52.32 cm²
Where the marks are
- Writes the perimeter equation $2 r + \frac{2 π r θ}{360} = 73.12$ (1 mark)
- Solves to get $θ = 6 0^{\circ}$ (1 mark)
- Segment area = sector area − equilateral triangle area, with correct substitution (2 marks)
- Area = 52.32 cm² (1 mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(A) · 5 marks · Long answerSome Applications of Trigonometry
From the top of a 9 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and angle of depression of its foot is $4 5^{\circ}$ . Determine the height of the tower and distance between building and tower. (Use $3 = 1.732$ )
Try it first, then show our solutionSolution
1. Figure: AB is the building with $A B = 9$ m, and B is its top. CD is the tower with foot C and top D. Draw $B E ⊥ C D$ , so $E C = A B = 9$ m and $B E = C A = x$ . Let $D E = h$ .
  Angle of elevation of D from B is $∠ D B E = 6 0^{\circ}$ . Angle of depression of C from B is $∠ E B C = 4 5^{\circ} = ∠ B C A$ .
  1 mark
2. In right $△ D E B$ : $tan 6 0^{\circ} = \frac{D E}{B E} \Rightarrow 3 = \frac{h}{x}$
  1 mark
3. $\Rightarrow h = x 3$ …(i)
  ½ mark
4. In right $△ B A C$ : $tan 4 5^{\circ} = \frac{A B}{A C} \Rightarrow 1 = \frac{9}{x}$
  1 mark
5. $\Rightarrow x = 9$ m …(ii). This is the distance between the building and the tower.
  ½ mark
6. From (i) and (ii): $h = 9 \times 1.732 = 15.588$ m
  ½ mark
7. Height of tower $C D = h + 9 = 15.588 + 9 = 24.588$ m
  ½ mark
Answer: Height of tower = 24.588 m; distance between building and tower = 9 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 6 0^{\circ} = \frac{h}{x}$ (1 mark)
- Gets $h = x 3$ (½ mark)
- Uses $tan 4 5^{\circ} = \frac{9}{x}$ (1 mark)
- Gets $x = 9$ m (½ mark)
- $h = 15.588$ m (½ mark)
- Tower height = 24.588 m (½ mark)
Open Q35(A) on its own page
OR
Q35(B) · 5 marks · Long answerSome Applications of Trigonometry
As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are $3 0^{\circ}$ and $4 5^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships (Use $3 = 1.732$ )
Try it first, then show our solutionSolution
1. Figure: AB is the lighthouse with $A B = 75$ m, and B is its top. D and C are the two ships on the same side of A, with C farther away. Let $A D = y$ and $D C = x$ .
  The angles of depression equal the alternate angles at the ships: $∠ B C A = 3 0^{\circ}$ and $∠ B D A = 4 5^{\circ}$ .
  1 mark
2. In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{3} = \frac{75}{x + y}$
  1 mark
3. $\Rightarrow x + y = 753$ …(i)
  ½ mark
4. In right $△ B A D$ : $tan 4 5^{\circ} = \frac{A B}{A D} \Rightarrow 1 = \frac{75}{y}$
  1 mark
5. $\Rightarrow y = 75$ m …(ii)
  ½ mark
6. From (i) and (ii): $x = 753 - 75 = 75 (3 - 1) = 75 \times 0.732 = 54.9$ m.
  So the distance between the two ships is 54.9 m.
  1 mark
Answer: 54.9 m
Where the marks are
- Correct labelled figure (1 mark)
- Uses $tan 3 0^{\circ} = \frac{75}{x + y}$ (1 mark)
- Gets $x + y = 753$ (½ mark)
- Uses $tan 4 5^{\circ} = \frac{75}{y}$ (1 mark)
- Gets $y = 75$ (½ mark)
- Finds $x = 75 (3 - 1) = 54.9$ m (1 mark)
Open Q35(B) on its own page

Section E

Case study based, sub-parts of 1, 1 and 2 marks

Q36 · 4 marks · Case-basedCompetency-basedProbability
A group of students conducted a survey to find out about the preferred mode of transportation to school among their classmates. They surveyed 200 students from their school. The results of the survey are as follows:
120 students preferred to walk to school.
25% of the students preferred to use bicycles.
10% of the students preferred to take the bus.
Remaining students preferred to be dropped off by car.
Based on the above information, answer the following questions:
1. (i)
  What is the probability that a randomly selected student does not prefer to walk to school?
  1 mark
2. (ii)
  Find the probability of a randomly selected student who prefers to walk or use a bicycle.
  1 mark
3. (iii)
  (A) One day 50% of walking students decided to come by bicycle. What is the probability that a randomly selected student comes to school using a bicycle on that day?
  2 marks
4. OR
  (iii)
  (B) What is the probability that a randomly selected student prefers to be dropped off by car?
  2 marks
Try it first, then show our solutionSolution
1. First find the number of students in each group. Walk = 120. Bicycle = $25%$ of $200 = 50$ . Bus = $10%$ of $200 = 20$ . Car = $200 - (120 + 50 + 20) = 10$ .
2. (i) Students who do not prefer to walk $= 200 - 120 = 80$ .
  ½ mark
3. (i) $P (not walk) = \frac{80}{200} = \frac{2}{5}$ .
  ½ mark
4. (ii) Students who prefer to walk or use a bicycle $= 120 + 50 = 170$ .
  ½ mark
5. (ii) $P (walk or bicycle) = \frac{170}{200} = \frac{17}{20}$ .
  ½ mark
6. (iii)(a) Walking students who switch to bicycle $= 50%$ of $120 = 60$ .
  ½ mark
7. (iii)(a) Students already using a bicycle $= 50$ , so bicycle users that day $= 60 + 50 = 110$ .
  ½ mark
8. (iii)(a) $P (bicycle that day) = \frac{110}{200} = \frac{11}{20}$ .
  1 mark
9. (iii)(b) Students dropped by car $= 200 - (120 + 50 + 20) = 10$ .
  1 mark
10. (iii)(b) $P (car) = \frac{10}{200} = \frac{1}{20}$ .
  1 mark
Answer: (i) 2/5; (ii) 17/20; (iii)(A) 11/20; (iii)(B) 1/20
Where the marks are
- Finds 80 students who do not walk (½ mark)
- Probability $\frac{80}{200} = \frac{2}{5}$ (½ mark)
- Finds 170 students who walk or cycle (½ mark)
- Probability $\frac{17}{20}$ (½ mark)
- (iii)(A): 60 walkers switch to bicycle (½ mark)
- (iii)(A): adds the existing 50 to get 110 (½ mark)
- (iii)(A): probability $\frac{11}{20}$ (1 mark)
- (iii)(B): car students $= 10$ (1 mark)
- (iii)(B): probability $\frac{1}{20}$ (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedPolynomials
Radha, an aspiring landscape designer, is tasked with creating a visually captivating pool design that incorporates a unique arrangement of fountains. The challenge entails arranging the fountains in such a way that when water is thrown upwards, it forms the shape of a parabola. The graph of one such parabola is given below.
The height of each fountain rod above water level is 10 cm. The equation of the downward-facing parabola representing the water fountain is given by $p (x) = - x^{2} + 5 x - 4$ .
Based on the above information, answer the following questions:
1. (i)
  Find the zeroes of the polynomial p(x) from the graph.
  1 mark
2. (ii)
  Find the value of x at which water attains maximum height.
  1 mark
3. (iii)
  (A) If h is the maximum height attained by the water stream from the water level of the pool, then find the value of h.
  2 marks
4. OR
  (iii)
  (B) At what point(s) on x-axis, the height of water above x-axis is 2 m?
  2 marks
Figure in the paper: Graph of a downward-facing green parabola on axes X'OX and YY'. On the Y-axis 1 unit = 1 m; on the X-axis 1 unit = 0.5 m. The curve crosses the X-axis at (1,0) and (4,0), peaks at roughly (2.5, 2.25) and extends below the axis. A vertical black rod is drawn at x = 1 from the water level line, which is a horizontal line at the level of the X-axis labelled 'Water level'. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The parabola cuts the x-axis at $(1, 0)$ and $(4, 0)$ , so the zeroes of $p (x)$ are $1$ and $4$ . Check: $p (x) = - (x - 1) (x - 4)$ .
  1 mark
2. (ii) A parabola is symmetric about the vertical line midway between its zeroes. So the maximum occurs at $x = \frac{1 + 4}{2} = \frac{5}{2}$ .
  1 mark
3. (iii)(a) Substitute $x = \frac{5}{2}$ : $p (\frac{5}{2}) = - \frac{25}{4} + \frac{25}{2} - 4 = \frac{- 25 + 50 - 16}{4} = \frac{9}{4} = 2.25$ m.
  1 mark
4. (iii)(a) Add the rod height above water, $10 cm = 0.10$ m: $h = 0.10 + 2.25 = 2.35$ m.
  1 mark
5. (iii)(b) Set $p (x) = 2$ : $- x^{2} + 5 x - 4 = 2$ .
  ½ mark
6. (iii)(b) Rearrange to get $x^{2} - 5 x + 6 = 0$ .
  ½ mark
7. (iii)(b) Factorise: $(x - 2) (x - 3) = 0$ , so $x = 2$ or $x = 3$ .
  ½ mark
8. (iii)(b) The required points on the x-axis are $(2, 0)$ and $(3, 0)$ .
  ½ mark
Answer: (i) 1 and 4; (ii) x = 5/2; (iii)(A) h = 2.35 m; (iii)(B) (2, 0) and (3, 0)
Where the marks are
- Reads the zeroes 1 and 4 from the graph (1 mark)
- Finds $x = \frac{5}{2}$ for the maximum height (1 mark)
- (iii)(A): finds $p (5/2) = 2.25$ (1 mark)
- (iii)(A): adds 0.10 m to get $h = 2.35$ m (1 mark)
- (iii)(B): forms $- x^{2} + 5 x - 4 = 2$ (½ mark)
- (iii)(B): simplifies to $x^{2} - 5 x + 6 = 0$ (½ mark)
- (iii)(B): solves to get $x = 2, 3$ (½ mark)
- (iii)(B): states the points $(2, 0)$ and $(3, 0)$ (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Rinku was very happy to receive a fancy jumbo pencil from his best friend Rohan on his birthday. Pencil is a basic writing tool, when sharpened its shape is a combination of cylinder & cone as given in the picture.
Cylindrical pencil with conical head is a common shape worldwide since ages. Commonly pencils are made up of wood & plastic but we should promote pencils made up of eco-friendly material (many options available in the market these days) to save environment.
The dimensions of Rinku's pencil are given as follows: Length of cylindrical portion is 21cm. Diameter of the base is 1 cm and height of the conical portion is 1.2 cm.
Based on the above information, answer the following questions:
1. (i)
  Find the slant height of the sharpened part.
  1 mark
2. (ii)
  Find curved surface area of sharpened part (in terms of $π$ ).
  1 mark
3. (iii)
  (A) Find the total surface area of the pencil (in terms of $π$ ).
  2 marks
4. OR
  (iii)
  (B) The pencil's total height decreases by 8.2 cm after sharpening it many times, what is the volume of the cylindrical part of the shortened pencil (in terms of $π$ )?
  2 marks
Figure in the paper: A photograph of a jumbo pencil with a cylindrical body and a sharpened conical tip at one end. Dimensions given in text: cylindrical portion length 21 cm, base diameter 1 cm, conical portion height 1.2 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. Given: radius $r = \frac{1}{2} = 0.5$ cm, cone height $h = 1.2$ cm, cylinder length $H = 21$ cm.
2. (i) Slant height: $l^{2} = h^{2} + r^{2} = (1.2)^{2} + (0.5)^{2} = 1.44 + 0.25 = 1.69$ .
  ½ mark
3. (i) $l = 1.69 = 1.3$ cm.
  ½ mark
4. (ii) Curved surface area of the cone: $π r l = π \times 0.5 \times 1.3$ .
  ½ mark
5. (ii) This gives $0.65 π$ cm².
  ½ mark
6. (iii)(a) TSA of pencil = CSA of cylinder + CSA of cone + area of the flat circular base $= 2 π r H + π r l + π r^{2}$ .
  ½ mark
7. (iii)(a) Substitute: $2 π (0.5) (21) + 0.65 π + π (0.5)^{2} = 21 π + 0.65 π + 0.25 π$ .
  1 mark
8. (iii)(a) TSA $= 21.9 π$ cm².
  ½ mark
9. (iii)(b) The conical tip stays the same after sharpening, so the cylindrical part shortens by 8.2 cm. New length $= 21 - 8.2 = 12.8$ cm.
  ½ mark
10. (iii)(b) Volume of the cylinder: $π r^{2} H^{'} = π \times 0.5 \times 0.5 \times 12.8$ .
  1 mark
11. (iii)(b) This gives $3.2 π$ cm³.
  ½ mark
Answer: (i) 1.3 cm; (ii) 0.65π cm²; (iii)(A) 21.9π cm²; (iii)(B) 3.2π cm³
Where the marks are
- Uses $l^{2} = h^{2} + r^{2}$ with the correct values (½ mark)
- Gets $l = 1.3$ cm (½ mark)
- Uses $π r l$ for the cone (½ mark)
- Gets $0.65 π$ cm² (½ mark)
- (iii)(A): writes TSA as cylinder CSA + cone CSA + base area (½ mark)
- (iii)(A): substitutes correctly, with cylinder CSA $= 21 π$ (1 mark)
- (iii)(A): gets $21.9 π$ cm² (½ mark)
- (iii)(B): finds the new cylinder length 12.8 cm (½ mark)
- (iii)(B): substitutes into $π r^{2} H$ (1 mark)
- (iii)(B): gets $3.2 π$ cm³ (½ mark)
Open Q38 on its own page

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