MathsCBSE Class 10 · Standard · Board exam 2025 (session 2024-25)

CBSE Class 10 Mathematics Standard: 2025 board paper, set 30/5/1

The question paper set in the CBSE Class 10 board exam of 2025, 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/2025/X/041_Mathematics_Standard.zip (a zip; this paper is the file 30-5-1_Mathematics Standard.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
C	26-31	Short answer
D	32-35	Long answer
E	36-38	Case study based

Instructions in brief: No overall choice; internal choice given in 2 questions of Section B, 2 of Section C, 2 of Section D and 3 two-mark subparts in Section E (one alternative per case-study sub-part pair).

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQReal Numbers
$(3 + 2)^{2} + (3 - 2)^{2}$ is a/an
1. (A)positive rational number
2. (B)negative rational number
3. (C)positive irrational number
4. (D)negative irrational number
Try it first, then show our solutionSolution
1. $(3 + 2)^{2} = 7 + 43$ and $(3 - 2)^{2} = 7 - 43$ .
2. Sum $= 14$ , which is a positive rational number. Option (A).
  1 mark
Answer: 14, positive rational number, option (A)
Where the marks are
- Expands both squares and adds to get 14 (1 mark)
Q2 · 1 mark · MCQReal Numbers
Let $x = a^{2} b^{3} c^{n}$ and $y = a^{3} b^{m} c^{2}$ , where a, b, c are prime numbers. If LCM of x and y is $a^{3} b^{4} c^{3}$ , then the value of m + n is
1. (A)10
2. (B)7
3. (C)6
4. (D)5
Try it first, then show our solutionSolution
1. LCM takes the highest power of each prime: $b$ : $max (3, m) = 4 \Rightarrow m = 4$ ; $c$ : $max (n, 2) = 3 \Rightarrow n = 3$ .
2. $m + n = 7$ . Option (B).
  1 mark
Answer: m + n = 7, option (B)
Where the marks are
- Finds m = 4, n = 3, so m + n = 7 (1 mark)
Q3 · 1 mark · MCQReal Numbers
For any prime number p, if p divides $a^{2}$ , where a is any real number then p also divides
1. (A) $a$
2. (B) $a^{\frac{1}{2}}$
3. (C) $a^{\frac{3}{2}}$
4. (D) $a^{\frac{1}{8}}$
Try it first, then show our solutionSolution
1. By the standard theorem: if a prime $p$ divides $a^{2}$ , then $p$ divides $a$ . Option (A).
  1 mark
Answer: a, option (A)
Where the marks are
- Recalls theorem: p | a² ⇒ p | a (1 mark)
Q4 · 1 mark · MCQQuadratic Equations
Which of the following equations is a quadratic equation?
1. (A) $x^{2} + 1 = (x - 1)^{2}$
2. (B) $(x + x)^{2} = 2 x x$
3. (C) $x^{3} + 3 x^{2} = (x + 1)^{3}$
4. (D) $(x + 1) (x - 1) = (x + 1)^{2}$
Try it first, then show our solutionSolution
1. (A) reduces to $2 x = 0$ ; (C) to $3 x + 1 = 0$ ; (D) to $2 x + 2 = 0$ : all linear.
2. (B): $x^{2} + 2 x x + x = 2 x x \Rightarrow x^{2} + x = 0$ , a quadratic. Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Simplifies (B) to $x^{2} + x = 0$ (1 mark)
Q5 · 1 mark · MCQQuadratic Equations
If $x^{2} + b x + b = 0$ has two real and distinct roots, then the value of b can be
1. (A)0
2. (B)4
3. (C)3
4. (D) $- 3$
Try it first, then show our solutionSolution
1. Need $D = b^{2} - 4 b > 0$ , i.e. $b (b - 4) > 0$ , so $b < 0$ or $b > 4$ .
2. Of the options only $b = - 3$ works ( $9 + 12 = 21 > 0$ ). Option (D).
  1 mark
Answer: b = −3, option (D)
Where the marks are
- Uses $b^{2} - 4 b > 0$ and selects −3 (1 mark)
Q6 · 1 mark · MCQCoordinate Geometry
In the figure given below, points P, Q, R divide the line segment AB in four equal parts. The point Q divides PB in the ratio
1. (A)1 : 3
2. (B)2 : 3
3. (C)1 : 2
4. (D)1 : 1
Figure in the paper: A straight line segment AB with five equally spaced marked points in order A, P, Q, R, B (dots on the line). See it in the official paper.
Try it first, then show our solutionSolution
1. Each part has length $k$ : $P Q = k$ , $QB = 2 k$ .
2. $P Q : QB = 1 : 2$ . Option (C).
  1 mark
Answer: 1 : 2, option (C)
Where the marks are
- Finds PQ = 1 part, QB = 2 parts (1 mark)
Q7 · 1 mark · MCQProbability
A bag contains red balls and black balls in the ratio 3 : 7. A ball is drawn at random. The probability that ball so drawn is black in colour, is
1. (A) $\frac{3}{7}$
2. (B)0.3
3. (C)0.7
4. (D) $\frac{1}{7}$
Try it first, then show our solutionSolution
1. Let red $= 3 k$ , black $= 7 k$ ; total $10 k$ .
2. $P (black) = \frac{7 k}{10 k} = 0.7$ . Option (C).
  1 mark
Answer: 0.7, option (C)
Where the marks are
- P(black) = 7/10 = 0.7 (1 mark)
Q8 · 1 mark · MCQTriangles
Which of the following statements is false?
1. (A)Two right triangles are always similar.
2. (B)Two squares are always similar.
3. (C)Two equilateral triangles are always similar.
4. (D)Two circles are always similar.
Try it first, then show our solutionSolution
1. Squares, equilateral triangles and circles all have the same shape, so they are always similar.
2. Two right triangles may have different acute angles (e.g. 30° and 45°), so they need not be similar. Statement (A) is false.
  1 mark
Answer: Option (A)
Where the marks are
- Identifies right triangles need not be similar (1 mark)
Open Q8 on its own page
Q9 · 1 mark · MCQTriangles
In the adjoining figure, ABCD is a trapezium in which $X Y ∥ A B ∥ C D$ . If $A X = \frac{2}{3} A D$ , then CY : YB =
1. (A)2 : 3
2. (B)3 : 2
3. (C)1 : 3
4. (D)1 : 2
Figure in the paper: Trapezium ABCD with AB at bottom (longer) and DC at top (shorter). X lies on AD and Y lies on BC, with segment XY drawn parallel to AB and CD. See it in the official paper.
Try it first, then show our solutionSolution
1. $A X = \frac{2}{3} A D \Rightarrow D X : X A = 1 : 2$ .
2. Lines parallel to the bases divide the legs proportionally, so $C Y : Y B = D X : X A = 1 : 2$ . Option (D).
  1 mark
Answer: 1 : 2, option (D)
Where the marks are
- Uses proportional division by parallel lines: CY:YB = DX:XA = 1:2 (1 mark)
Q10 · 1 mark · MCQCircles
Which of the following statements is false?
1. (A)Infinite number of tangents can be drawn to a circle.
2. (B)Infinite number of tangents can be drawn to a circle from a point outside the circle.
3. (C)Infinite number of secants can be drawn to a circle from a point outside the circle.
4. (D)Angle between tangent and diameter at point of contact is $9 0^{\circ}$ .
Try it first, then show our solutionSolution
1. From an external point exactly two tangents can be drawn to a circle, so (B) is false; others are true.
  1 mark
Answer: Option (B)
Where the marks are
- States only two tangents from an external point (1 mark)
Q11 · 1 mark · MCQCircles
In the adjoining figure, PA and PB are tangents to a circle with centre O. The measure of angle APB is
1. (A) $21 0^{\circ}$
2. (B) $15 0^{\circ}$
3. (C) $10 5^{\circ}$
4. (D) $3 0^{\circ}$
Figure in the paper: A circle with centre O. External point P with two tangents touching the circle at A and B. The reflex angle AOB at the centre, measured through the major arc side, is marked $21 0^{\circ}$ . See it in the official paper.
Try it first, then show our solutionSolution
1. Non-reflex $∠ A O B = 36 0^{\circ} - 21 0^{\circ} = 15 0^{\circ}$ .
2. In quadrilateral OAPB, $∠ O A P = ∠ O B P = 9 0^{\circ}$ [tangent ⟂ radius], so $∠ A P B = 18 0^{\circ} - 15 0^{\circ} = 3 0^{\circ}$ . Option (D).
  1 mark
Answer: 30°, option (D)
Where the marks are
- Finds ∠AOB = 150° and ∠APB = 30° (1 mark)
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Q12 · 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) $sin 6 0^{\circ}$
2. (B) $cos 6 0^{\circ}$
3. (C) $tan 6 0^{\circ}$
4. (D) $sec 6 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan^{2} 3 0^{\circ} = \frac{1}{3}$ , so value $= \frac{2/3}{4/3} = \frac{1}{2}$ .
2. $\frac{1}{2} = cos 6 0^{\circ}$ . Option (B).
  1 mark
Answer: cos 60°, option (B)
Where the marks are
- Evaluates to 1/2 = cos 60° (1 mark)
Q13 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
An observer 1.8 m tall stands away from a chimney at a distance of 38.2 m along the ground. The angle of elevation of top of chimney from the eyes of observer is $4 5^{\circ}$ . The height of chimney above the ground is
1. (A)38.2 m
2. (B)36.4 m
3. (C)40 m
4. (D) $(38.2) 2$ m
Try it first, then show our solutionSolution
1. Height above eye level $= 38.2 tan 4 5^{\circ} = 38.2$ m.
2. Chimney height $= 38.2 + 1.8 = 40$ m. Option (C).
  1 mark
Answer: 40 m, option (C)
Where the marks are
- Adds observer height to 38.2 m to get 40 m (1 mark)
Q14 · 1 mark · MCQCircles
In the adjoining figure, the sum of radii of two concentric circles is 16 cm. The length of chord AB which touches the inner circle at P is 16 cm. The difference of the radii of the given circles is
1. (A)8 cm
2. (B)4 cm
3. (C)2 cm
4. (D)3 cm
Figure in the paper: Two concentric circles with centre O. Chord AB of the outer circle touches the inner circle at P, with P lying on the chord between A and B; A and B are at the ends of the chord on the outer circle. See it in the official paper.
Try it first, then show our solutionSolution
1. OP ⟂ AB [tangent ⟂ radius], so P bisects AB: $A P = 8$ . With $R$ outer, $r$ inner: $R^{2} - r^{2} = 64$ .
2. $(R + r) (R - r) = 64 \Rightarrow 16 (R - r) = 64 \Rightarrow R - r = 4$ cm. Option (B).
  1 mark
Answer: 4 cm, option (B)
Where the marks are
- Uses $R^{2} - r^{2} = A P^{2}$ and factorises to get 4 cm (1 mark)
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Q15 · 1 mark · MCQSurface Areas and Volumes
A cone of height 12 cm and slant height 13 cm is surmounted on a hemisphere having radius equal to that of cone. The entire height of the solid is
1. (A)17 cm
2. (B)18 cm
3. (C)22 cm
4. (D)23 cm
Try it first, then show our solutionSolution
1. $r = 1 3^{2} - 1 2^{2} = 5$ cm.
2. Total height $= 12 + 5 = 17$ cm. Option (A).
  1 mark
Answer: 17 cm, option (A)
Where the marks are
- Finds r = 5 and height 12 + 5 = 17 cm (1 mark)
Q16 · 1 mark · MCQStatistics
If $x$ median $+$ $y$ mean $=$ $z$ mode; is the empirical relationship between mean, median and mode, then the value of $x + y + z$ is
1. (A)6
2. (B)3
3. (C)2
4. (D)1
Try it first, then show our solutionSolution
1. Empirical relation: $3 median = mode + 2 mean$ , i.e. $3 median - 2 mean = mode$ , so $x = 3, y = - 2, z = 1$ .
2. $x + y + z = 2$ . Option (C).
  1 mark
Answer: 2, option (C)
Where the marks are
- Writes relation as 3 median − 2 mean = mode and sums 2 (1 mark)
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Q17 · 1 mark · MCQStatistics
Following data shows the marks obtained by 100 students in a class test:
Marks obtained 20 29 28 33 42 38 43 25
Number of students 6 28 24 15 2 4 1 20
The median will be the average of which two observations?
1. (A)29 and 33
2. (B)25 and 28
3. (C)28 and 29
4. (D)33 and 38
Try it first, then show our solutionSolution
1. Ascending: 20(6), 25(20), 28(24), 29(28),...; cumulative: 6, 26, 50, 78,...
2. n = 100, median = average of 50th and 51st values = 28 and 29. Option (C).
  1 mark
Answer: 28 and 29, option (C)
Where the marks are
- Arranges data and identifies 50th = 28, 51st = 29 (1 mark)
Q18 · 1 mark · MCQProbability
The probability of getting a composite number greater than 3 on throwing a die is
1. (A) $\frac{1}{6}$
2. (B) $\frac{1}{3}$
3. (C) $\frac{1}{2}$
4. (D) $\frac{2}{3}$
Try it first, then show our solutionSolution
1. Composite numbers >3 on a die: 4, 6 → 2 outcomes.
2. $P = \frac{2}{6} = \frac{1}{3}$ . Option (B).
  1 mark
Answer: 1/3, option (B)
Where the marks are
- Identifies 4, 6 and gets 1/3 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedIntroduction to Trigonometry
Assertion (A): For an acute angle $θ$ , $sin θ = \frac{3}{5} \Rightarrow cos θ = - \frac{4}{5}$ .
Reason (R): For any value of $θ$ , $(0^{\circ} \leq θ \leq 9 0^{\circ})$ , $sin^{2} θ + cos^{2} θ = 1$ .
Directions: In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from following:
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of (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 the identity $sin^{2} θ + cos^{2} θ = 1$ : true.
2. For acute θ, cos θ > 0, so $cos θ = \frac{4}{5}$ , not $- \frac{4}{5}$ ; A is false. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Notes cos θ positive for acute θ, A false, R true (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedArithmetic Progressions
Assertion (A): For an A.P., $3, 6, 9, \dots, 198$ , $1 0^{th}$ term from the end is $168$ .
Reason (R): If $a$ and $l$ are the first term and last term of an A.P. with common difference $d$ , then $n^{th}$ term from the end of the given A.P. is $l - (n - 1) d$ .
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of (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 the standard formula: true.
2. 10th term from end $= 198 - 9 \times 3 = 171 \neq = 168$ , so A is false. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Computes 171 using l − (n−1)d; A false, R true (1 mark)

Marks obtained	20	29	28	33	42	38	43	25
Number of students	6	28	24	15	2	4	1	20

Section B

Very short answer

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerCompetency-basedPair of Linear Equations in Two Variables
The cost of 2 kg apples and 1 kg of grapes on a day was found to be ₹ 320. The cost of 4 kg apples and 2 kg grapes was found to be ₹ 600. If cost of 1 kg of apples and 1 kg of grapes is ₹ $x$ and ₹ $y$ respectively, represent the given situation algebraically as a system of equations and check whether the system so obtained is consistent or not.
Try it first, then show our solutionSolution
1. Cost of 2 kg apples and 1 kg grapes is ₹320, so $2 x + y = 320$ ...(1)
  ½ mark
2. Cost of 4 kg apples and 2 kg grapes is ₹600, so $4 x + 2 y = 600$ ...(2)
  ½ mark
3. Comparing coefficients: $\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{1}{2}$ and $\frac{c _{1}}{c _{2}} = \frac{320}{600} = \frac{8}{15}$
  ½ mark
4. Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the lines are parallel and have no common solution. Hence the system is not consistent (inconsistent).
  ½ mark
Answer: $2 x + y = 320$ , $4 x + 2 y = 600$ ; the system is inconsistent.
Where the marks are
- Writes $2 x + y = 320$ (½ mark)
- Writes $4 x + 2 y = 600$ (½ mark)
- Finds the three coefficient ratios correctly (½ mark)
- Concludes the system is not consistent since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ (½ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerPair of Linear Equations in Two Variables
Solve for $x$ and $y$ :
$2 x + 3 y = 5$ and $3 x - 8 y = - 6$
Try it first, then show our solutionSolution
1. Multiply the first equation by $3$ : $6 x + 3 y = 53$ ...(1)
  ½ mark
2. Multiply the second equation by $2$ , using $8 \cdot 2 = 4$ and $6 \cdot 2 = 23$ : $6 x - 4 y = - 23$ ...(2)
  ½ mark
3. Subtract (2) from (1): $7 y = 73$ , so $y = 3$ .
  ½ mark
4. Substitute in the first equation: $2 x + 3 = 5$ , so $2 x = 2$ and $x = 2$ .
  ½ mark
Answer: $x = 2$ , $y = 3$
Where the marks are
- Makes the coefficient of $x$ equal in the first equation ( $\times 3$ ) (½ mark)
- Does the same in the second equation ( $\times 2$ ) (½ mark)
- Obtains $y = 3$ (½ mark)
- Obtains $x = 2$ (½ mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerCoordinate Geometry
The coordinates of the end points of the line segment AB are A $(- 2, - 2)$ and B $(2, - 4)$ . P is the point on AB such that $B P = \frac{4}{7} A B$ . Find the coordinates of point P.
Try it first, then show our solutionSolution
1. $B P = \frac{4}{7} A B$ , so $A P = \frac{3}{7} A B$ . Hence P divides AB internally in the ratio $A P : P B = 3 : 4$ .
  ½ mark
2. By the section formula with $m_{1} = 3$ , $m_{2} = 4$ , A $(- 2, - 2)$ and B $(2, - 4)$ : $x = \frac{3 ( 2 ) + 4 ( - 2 )}{3 + 4} = \frac{- 2}{7}$
  ½ mark
3. $y = \frac{3 ( - 4 ) + 4 ( - 2 )}{3 + 4} = \frac{- 20}{7}$
  ½ mark
4. Hence the coordinates of P are $(- \frac{2}{7}, - \frac{20}{7})$ .
  ½ mark
Answer: $P (- \frac{2}{7}, - \frac{20}{7})$
Where the marks are
- Finds the ratio $A P : P B = 3 : 4$ (½ mark)
- Finds the x-coordinate $- \frac{2}{7}$ (½ mark)
- Finds the y-coordinate $- \frac{20}{7}$ (½ mark)
- States P correctly (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerIntroduction to Trigonometry
It is given that $sin (A - B) = sin A cos B - cos A sin B$ . Use it to find the value of $sin 1 5^{\circ}$ .
Try it first, then show our solutionSolution
1. $sin 1 5^{\circ} = sin (4 5^{\circ} - 3 0^{\circ}) = sin 4 5^{\circ} cos 3 0^{\circ} - cos 4 5^{\circ} sin 3 0^{\circ}$
  ½ mark
2. $= \frac{1}{2} \times \frac{3}{2} - \frac{1}{2} \times \frac{1}{2}$
  1 mark
3. $= \frac{3 - 1}{2 2} = \frac{6 - 2}{4}$
  ½ mark
Answer: $sin 1 5^{\circ} = \frac{3 - 1}{2 2} = \frac{6 - 2}{4}$
Where the marks are
- Writes $1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}$ and applies the formula (½ mark)
- Substitutes the correct standard values (1 mark)
- Simplifies to $\frac{3 - 1}{2 2}$ (½ mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerIntroduction to Trigonometry
If $sin A = y$ , then express $cos A$ and $tan A$ in terms of $y$ .
Try it first, then show our solutionSolution
1. Since $sin^{2} A + cos^{2} A = 1$ : $cos A = 1 - sin^{2} A = 1 - y^{2}$
  1 mark
2. $tan A = \frac{s i n A}{c o s A} = \frac{y}{1 - y ^{2}}$
  1 mark
Answer: $cos A = 1 - y^{2}$ , $tan A = \frac{y}{1 - y ^{2}}$
Where the marks are
- Obtains $cos A = 1 - y^{2}$ using the identity (1 mark)
- Obtains $tan A = \frac{y}{1 - y ^{2}}$ (1 mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerTriangles
AD and PS are medians of triangles ABC and PQR respectively such that $△ A B D \sim △ P QS$ . Prove that $△ A B C \sim △ P QR$ .
Try it first, then show our solutionSolution
1. Figure: draw $△ A B C$ with median AD (D the midpoint of BC) and $△ P QR$ with median PS (S the midpoint of QR).
  ½ mark
2. Since $△ A B D \sim △ P QS$ , $\frac{A B}{P Q} = \frac{B D}{QS}$ . As $B D = \frac{1}{2} B C$ and $QS = \frac{1}{2} QR$ [D, S are midpoints]: $\frac{A B}{P Q} = \frac{\frac{1}{2} B C}{\frac{1}{2} QR}$
3. $\Rightarrow \frac{A B}{P Q} = \frac{B C}{QR}$
  ½ mark
4. Also $∠ B = ∠ Q$ [corresponding angles of similar triangles ABD and PQS].
  ½ mark
5. Hence $△ A B C \sim △ P QR$ [SAS similarity criterion].
  ½ mark
Answer: $△ A B C \sim △ P QR$ by SAS similarity.
Where the marks are
- Draws a correct figure (½ mark)
- Deduces $\frac{A B}{P Q} = \frac{B C}{QR}$ using $B D = \frac{B C}{2}$ and $QS = \frac{QR}{2}$ (½ mark)
- States $∠ B = ∠ Q$ (½ mark)
- Concludes similarity by SAS (½ mark)
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Q25 · 2 marks · Very short answerProbability
While shuffling a pack of 52 cards, one card was accidentally dropped. Find the probability that the dropped card
1. (i)
  is not a face card.
  1 mark
2. (ii)
  is a black king.
  1 mark
Try it first, then show our solutionSolution
1. (i) Total outcomes = 52. There are 12 face cards (J, Q, K in each of 4 suits), so 52 − 12 = 40 cards are not face cards. P(not a face card) $= \frac{40}{52} = \frac{10}{13}$
  1 mark
2. (ii) There are 2 black kings (spades and clubs). P(black king) $= \frac{2}{52} = \frac{1}{26}$
  1 mark
Answer: (i) $\frac{10}{13}$ ; (ii) $\frac{1}{26}$
Where the marks are
- P(not a face card) $= \frac{40}{52} = \frac{10}{13}$ (1 mark)
- P(black king) $= \frac{2}{52} = \frac{1}{26}$ (1 mark)
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Section C

Short answer

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerReal Numbers
Prove that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, on the contrary, that $3$ is rational.
  Then $3 = \frac{p}{q}$ , where $p, q$ are integers, $q \neq = 0$ , and $p, q$ are coprime.
  Squaring: $3 q^{2} = p^{2}$ .
  ½ mark
2. So $p^{2}$ is divisible by 3. Since 3 is prime, 3 divides $p^{2}$ means 3 divides $p$ . ... (1)
  1 mark
3. Let $p = 3 a$ for some integer $a$ . Then $9 a^{2} = 3 q^{2}$ , so $q^{2} = 3 a^{2}$ .
  So $q^{2}$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)
  1 mark
4. From (1) and (2), 3 divides both $p$ and $q$ . This contradicts the fact that $p$ and $q$ are coprime.
  So our assumption is wrong. Hence $3$ is irrational.
  ½ mark
Answer: Assuming $3 = p / q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $3$ is irrational
Where the marks are
- Assumes $3 = \frac{p}{q}$ with p, q coprime and gets $3 q^{2} = p^{2}$ (1/2 mark)
- Shows 3 divides p (1 mark)
- Substitutes $p = 3 a$ and shows 3 divides q (1 mark)
- States the contradiction and concludes (1/2 mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerReal Numbers
State true or false for each of the following statements and justify in each case:
(i) $2 \times 3 \times 5 \times 7 + 7$ is a composite number.
(ii) $2 \times 3 \times 5 \times 7 + 1$ is a composite number.
Try it first, then show our solutionSolution
1. (i) True.
  1 mark
2. $2 \times 3 \times 5 \times 7 + 7 = 7 \times (2 \times 3 \times 5 + 1) = 7 \times 31 = 217$ . It has factors 1, 7, 31 and 217, i.e. more than two factors, so it is composite.
  ½ mark
3. (ii) False.
  1 mark
4. $2 \times 3 \times 5 \times 7 + 1 = 211$ . It is not divisible by any prime up to $211$ (2, 3, 5, 7, 11, 13). So its only factors are 1 and 211, i.e. it is prime, not composite.
  ½ mark
Answer: (i) True (= 7 × 31); (ii) False (211 is prime)
Where the marks are
- (i) States True (1 mark)
- (i) Justifies by showing 7 is a common factor: $7 \times 31$ (½ mark)
- (ii) States False (1 mark)
- (ii) Justifies that 211 has only two factors (½ mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerPolynomials
Obtain the zeroes of the polynomial $7 x^{2} + 18 x - 9$ . Hence, write a polynomial each of whose zeroes is twice the zeroes of given polynomial.
Try it first, then show our solutionSolution
1. Split the middle term: $7 x^{2} + 18 x - 9 = 7 x^{2} + 21 x - 3 x - 9 = 7 x (x + 3) - 3 (x + 3) = (7 x - 3) (x + 3)$ . So the zeroes are $\frac{3}{7}$ and $- 3$ .
  1 mark
2. Twice the zeroes: $\frac{6}{7}$ and $- 6$
  ½ mark
3. Sum of the new zeroes $= \frac{6}{7} + (- 6) = - \frac{36}{7}$
  ½ mark
4. Product of the new zeroes $= \frac{6}{7} \times (- 6) = - \frac{36}{7}$
  ½ mark
5. Required polynomial $= x^{2} - (sum) x + product = x^{2} + \frac{36}{7} x - \frac{36}{7}$ , or $7 x^{2} + 36 x - 36$ (or any non-zero multiple of it).
  ½ mark
Answer: Zeroes are $\frac{3}{7}$ and $- 3$ ; required polynomial $7 x^{2} + 36 x - 36$
Where the marks are
- Factorises and finds the zeroes $\frac{3}{7}$ , $- 3$ (1 mark)
- Writes the new zeroes $\frac{6}{7}$ , $- 6$ (½ mark)
- Sum $= - \frac{36}{7}$ (½ mark)
- Product $= - \frac{36}{7}$ (½ mark)
- Forms the polynomial $7 x^{2} + 36 x - 36$ (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerPair of Linear Equations in Two Variables
Solve the following system of equations graphically:
$2 x - y - 2 = 0$
$- 4 x + y + 4 = 0$
Also, find the absolute difference between the ordinates of the points where given lines cut y-axis.
Try it first, then show our solutionSolution
1. For $2 x - y - 2 = 0$ , i.e. $y = 2 x - 2$ : plot the points $(0, - 2)$ , $(1, 0)$ , $(2, 2)$ and join them to get line 1.
  1 mark
2. For $- 4 x + y + 4 = 0$ , i.e. $y = 4 x - 4$ : plot the points $(0, - 4)$ , $(1, 0)$ , $(2, 4)$ and join them to get line 2, on the same axes.
  1 mark
3. The two lines intersect at $(1, 0)$ , so the solution is $x = 1$ , $y = 0$ .
  ½ mark
4. Line 1 cuts the y-axis at $(0, - 2)$ and line 2 at $(0, - 4)$ . Absolute difference of ordinates $= ∣ - 2 - (- 4) ∣ = 2$ .
  ½ mark
Answer: $x = 1$ , $y = 0$ ; absolute difference = 2
Where the marks are
- Table of points and correct graph of both lines (2 marks)
- Solution $x = 1$ , $y = 0$ read from the intersection (½ mark)
- Absolute difference of y-intercepts = 2 (½ mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerCoordinate Geometry
Find a relation between $x$ and $y$ such that $P (x, y)$ is equidistant from the points A $(3, 5)$ and B $(7, 1)$ . Hence, write the coordinates of the points on x-axis and y-axis which are equidistant from points A and B.
Try it first, then show our solutionSolution
1. P is equidistant from A and B, so $P A = P B \Rightarrow P A^{2} = P B^{2}$ .
  ½ mark
2. $(x - 3)^{2} + (y - 5)^{2} = (x - 7)^{2} + (y - 1)^{2}$
3. $x^{2} - 6 x + 9 + y^{2} - 10 y + 25 = x^{2} - 14 x + 49 + y^{2} - 2 y + 1 \Rightarrow 8 x - 8 y = 16$
  1 mark
4. $\Rightarrow x - y = 2$ is the required relation.
  ½ mark
5. On the x-axis $y = 0$ , so $x = 2$ : the point is $(2, 0)$ .
  ½ mark
6. On the y-axis $x = 0$ , so $y = - 2$ : the point is $(0, - 2)$ .
  ½ mark
Answer: $x - y = 2$ ; points (2, 0) and (0, −2)
Where the marks are
- Uses $P A^{2} = P B^{2}$ (½ mark)
- Applies the distance formula and expands correctly (1 mark)
- Obtains $x - y = 2$ (½ mark)
- Point on x-axis (2, 0) (½ mark)
- Point on y-axis (0, −2) (½ mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerIntroduction to Trigonometry
Prove the following trigonometric identity:
$\frac{1 + cosec A}{cosec A} = \frac{c o s ^{2} A}{1 - s i n A}$
Try it first, then show our solutionSolution
1. LHS $= \frac{1 + \frac{1}{s i n A}}{\frac{1}{s i n A}}$ [using $cosec A = \frac{1}{s i n A}$ ]
  ½ mark
2. $= \frac{\frac{s i n A + 1}{s i n A}}{\frac{1}{s i n A}} = 1 + sin A$
  1 mark
3. $= \frac{( 1 + s i n A ) ( 1 - s i n A )}{1 - s i n A}$ [multiplying and dividing by $1 - sin A$ ]
  ½ mark
4. $= \frac{1 - s i n ^{2} A}{1 - s i n A}$
  ½ mark
5. $= \frac{c o s ^{2} A}{1 - s i n A}$ = RHS [since $1 - sin^{2} A = cos^{2} A$ ]. Hence proved.
  ½ mark
Answer: LHS = $1 + sin A$ = $\frac{c o s ^{2} A}{1 - s i n A}$ = RHS. Hence proved.
Where the marks are
- Writes cosec A as $\frac{1}{s i n A}$ (½ mark)
- Simplifies LHS to $1 + sin A$ (1 mark)
- Multiplies and divides by $1 - sin A$ (½ mark)
- Gets $\frac{1 - s i n ^{2} A}{1 - s i n A}$ (½ mark)
- Uses the identity to reach RHS (½ mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerIntroduction to Trigonometry
Let $2 A + B$ and $A + 2 B$ be acute angles such that $sin (2 A + B) = \frac{3}{2}$ and $tan (A + 2 B) = 1$ . Find the value of $cot (4 A - 7 B)$ .
Try it first, then show our solutionSolution
1. $sin (2 A + B) = \frac{3}{2} = sin 6 0^{\circ} \Rightarrow 2 A + B = 6 0^{\circ}$ ...(1)
  ½ mark
2. $tan (A + 2 B) = 1 = tan 4 5^{\circ} \Rightarrow A + 2 B = 4 5^{\circ}$ ...(2)
  ½ mark
3. Adding (1) and (2): $3 A + 3 B = 10 5^{\circ} \Rightarrow A + B = 3 5^{\circ}$ . Subtracting (2) from (1): $A - B = 1 5^{\circ}$ . Hence $A = 2 5^{\circ}$ .
  ½ mark
4. $B = 1 0^{\circ}$
  ½ mark
5. $4 A - 7 B = 10 0^{\circ} - 7 0^{\circ} = 3 0^{\circ}$ , so $cot (4 A - 7 B) = cot 3 0^{\circ}$
  ½ mark
6. $= 3$
  ½ mark
Answer: $A = 2 5^{\circ}$ , $B = 1 0^{\circ}$ ; $cot (4 A - 7 B) = 3$
Where the marks are
- Obtains $2 A + B = 6 0^{\circ}$ (½ mark)
- Obtains $A + 2 B = 4 5^{\circ}$ (½ mark)
- Finds $A = 2 5^{\circ}$ (½ mark) and $B = 1 0^{\circ}$ (½ mark)
- Gets $4 A - 7 B = 3 0^{\circ}$ (½ mark)
- Final value $3$ (½ mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerCircles
In the adjoining figure, XY and X′Y′ are parallel tangents to a circle with centre O. Another tangent AB touches the circle at C intersecting XY at A and X′Y′ at B. Prove that AB subtends right angle at the centre of the circle; or $∠ A O B = 9 0^{\circ}$ .
Figure in the paper: A circle with centre O. Two horizontal parallel tangent lines XY (top, touching at P) and X′Y′ (bottom, touching at Q), with PQ a vertical diameter through O. Points A on the top tangent and B on the bottom tangent; the slanted tangent AB touches the circle at C between A and B. Radii OA and OB are drawn and angle AOB is marked at O. See it in the official paper.
Try it first, then show our solutionSolution
1. Join OC, OP and OQ. Since $O P ⊥ X Y$ and $O Q ⊥ X^{'} Y^{'}$ [tangent ⟂ radius] and $X Y ∥ X^{'} Y^{'}$ , the points P, O, Q are collinear, so POQ is a diameter.
  ½ mark
2. In $△ P O A$ and $△ C O A$ : $O P = O C$ [radii], $A P = A C$ [tangents from an external point are equal] and OA is common. So $△ P O A ≅ △ C O A$ [SSS].
  ½ mark
3. $∴ ∠ P O A = ∠ C O A$ [CPCT] ...(1)
  ½ mark
4. Similarly, $△ QO B ≅ △ C O B$ , so $∠ QO B = ∠ C O B$ ...(2)
  ½ mark
5. POQ is a straight line, so $∠ P O A + ∠ C O A + ∠ C O B + ∠ QO B = 18 0^{\circ}$ .
  ½ mark
6. Using (1) and (2): $2∠ C O A + 2∠ C O B = 18 0^{\circ} \Rightarrow ∠ C O A + ∠ C O B = 9 0^{\circ}$ . Hence $∠ A O B = 9 0^{\circ}$ .
  ½ mark
Answer: $∠ A O B = 9 0^{\circ}$ (proved)
Where the marks are
- Joins OC (with PQ as diameter) (½ mark)
- Proves $△ P O A ≅ △ C O A$ with reasons (½ mark)
- Gets $∠ P O A = ∠ C O A$ (½ mark)
- Similarly gets $∠ QO B = ∠ C O B$ (½ mark)
- Sum of the four angles at O on line PQ = 180° (½ mark)
- Concludes $∠ A O B = 9 0^{\circ}$ (½ mark)
Open Q31 on its own page

Section D

Long answer

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerQuadratic Equations
A 2-digit number is seven times the sum of its digits and two (2) more than five times the product of its digits. Find the number.
Try it first, then show our solutionSolution
1. Let the unit's digit be $x$ and the ten's digit be $y$ . Then the number $= 10 y + x$ .
  1 mark
2. Given $10 y + x = 7 (x + y) \Rightarrow 3 y = 6 x \Rightarrow y = 2 x$ ...(1)
  1 mark
3. Also $10 y + x = 5 x y + 2$ ...(2)
  ½ mark
4. Substituting (1) in (2): $20 x + x = 10 x^{2} + 2 \Rightarrow 10 x^{2} - 21 x + 2 = 0$
  1 mark
5. $(x - 2) (10 x - 1) = 0 \Rightarrow x = 2$ or $x = \frac{1}{10}$ . A digit must be a whole number, so $x = 2$ .
  ½ mark
6. $y = 2 x = 4$
  ½ mark
7. The required number is 42. (Check: $7 (4 + 2) = 42$ and $5 \times 4 \times 2 + 2 = 42$ .)
  ½ mark
Answer: 42
Where the marks are
- Sets up the digits and the number $10 y + x$ (1 mark)
- Forms $10 y + x = 7 (x + y)$ and gets $y = 2 x$ (1 mark)
- Forms $10 y + x = 5 x y + 2$ (½ mark)
- Obtains $10 x^{2} - 21 x + 2 = 0$ (1 mark)
- Solves to get $x = 2$ , rejecting $\frac{1}{10}$ (½ mark)
- Finds $y = 4$ (½ mark)
- States the number 42 (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerQuadratic Equations
Find the value(s) of p for which the quadratic equation given as $(p + 4) x^{2} - (p + 1) x + 1 = 0$ has real and equal roots. Also, find the roots of the equation(s) so obtained.
Try it first, then show our solutionSolution
1. For real and equal roots, the discriminant $D = b^{2} - 4 a c = 0$ .
  ½ mark
2. Here $a = p + 4$ , $b = - (p + 1)$ , $c = 1$ , so $[- (p + 1)]^{2} - 4 (p + 4) (1) = 0$ .
  ½ mark
3. $p^{2} + 2 p + 1 - 4 p - 16 = 0 \Rightarrow p^{2} - 2 p - 15 = 0 \Rightarrow (p - 5) (p + 3) = 0$
  1 mark
4. $∴ p = 5$ or $p = - 3$ (both make $p + 4 \neq = 0$ ).
  1 mark
5. For $p = 5$ : $9 x^{2} - 6 x + 1 = 0 \Rightarrow (3 x - 1)^{2} = 0 \Rightarrow x = \frac{1}{3}, \frac{1}{3}$
  1 mark
6. For $p = - 3$ : $x^{2} + 2 x + 1 = 0 \Rightarrow (x + 1)^{2} = 0 \Rightarrow x = - 1, - 1$
  1 mark
Answer: p = 5 (roots $\frac{1}{3}, \frac{1}{3}$ ) and p = −3 (roots −1, −1)
Where the marks are
- States the condition D = 0 (½ mark)
- Substitutes correctly into $b^{2} - 4 a c$ (½ mark)
- Forms and factorises $p^{2} - 2 p - 15 = 0$ (1 mark)
- Gets p = 5, −3 (1 mark)
- Roots $\frac{1}{3}, \frac{1}{3}$ for p = 5 (1 mark)
- Roots −1, −1 for p = −3 (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 intersecting the other two sides in distinct points then it divides the two sides in the same ratio. Prove it.
Also, state the converse of the above statement.
Try it first, then show our solutionSolution
1. Given: In $△ A B C$ , a line $D E ∥ B C$ meets AB at D and AC at E. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ . Construction: Join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ . (Figure: triangle ABC with D on AB, E on AC, DE parallel to BC, segments BE and CD, perpendiculars EN and DM.)
  2 marks
2. Proof: $ar (A D E) = \frac{1}{2} \cdot A D \cdot E N$ and $ar (B D E) = \frac{1}{2} \cdot D B \cdot E N$ , so $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ ...(1). Also $ar (A D E) = \frac{1}{2} \cdot A E \cdot D M$ and $ar (D E C) = \frac{1}{2} \cdot E C \cdot D M$ , so $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ...(2)
  1 mark
3. $△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so $ar (B D E) = ar (D E C)$ ...(3). From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
  1 mark
4. Converse: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
  1 mark
Answer: $\frac{A D}{D B} = \frac{A E}{E C}$ (proved). Converse: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
Where the marks are
- Correct figure, given, to prove and construction (2 marks)
- Area ratios $\frac{A D}{D B}$ and $\frac{A E}{E C}$ (1 mark)
- Uses $ar (B D E) = ar (D E C)$ to conclude (1 mark)
- Correct statement of the converse (1 mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerSurface Areas and Volumes
From one of the faces of a solid wooden cube of side 14 cm, maximum number of hemispheres of diameter 1.4 cm are scooped out. Find the total number of hemispheres that can be scooped out. Also, find the total surface area of the remaining solid.
Try it first, then show our solutionSolution
1. Each hemisphere of diameter 1.4 cm occupies a $1.4 \times 1.4$ square on the face. Number of hemispheres $= \frac{14 \times 14}{1.4 \times 1.4}$
  1 mark
2. $= 10 \times 10 = 100$
  1 mark
3. TSA of remaining solid = TSA of cube + CSA of 100 hemispheres − area of 100 circular tops, with $r = 0.7$ cm: $= 6 \times 14 \times 14 + 100 \times 2 \times \frac{22}{7} \times 0.7 \times 0.7 - 100 \times \frac{22}{7} \times 0.7 \times 0.7 = 1176 + 308 - 154$
  2 marks
4. $= 1330 cm^{2}$
  1 mark
Answer: 100 hemispheres; TSA = 1330 cm²
Where the marks are
- Sets up the count $\frac{14 \times 14}{1.4 \times 1.4}$ (1 mark)
- Gets 100 hemispheres (1 mark)
- Correct TSA expression: cube + hemisphere CSA − circles (2 marks)
- TSA = 1330 cm² (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerSurface Areas and Volumes
From a solid cylinder of height 24 cm and radius 5 cm, two cones of height 12 cm and radius 5 cm are hollowed out. Find the volume and surface area of the remaining solid.
Try it first, then show our solutionSolution
1. Volume of remaining solid = volume of cylinder − volume of 2 cones $= \frac{22}{7} \times 5 \times 5 \times 24 - 2 \times \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 = \frac{13200}{7} - \frac{4400}{7}$
  1 mark
2. $= \frac{8800}{7} \approx 1257.14 cm^{3}$
  1 mark
3. Slant height of each cone: $l = 1 2^{2} + 5^{2} = 169 = 13$ cm
  1 mark
4. The two cones are hollowed from the two ends, so both flat ends are removed. Surface area = CSA of cylinder + CSA of 2 cones $= 2 \times \frac{22}{7} \times 5 \times 24 + 2 \times \frac{22}{7} \times 5 \times 13 = \frac{5280}{7} + \frac{2860}{7}$
  1 mark
5. $= \frac{8140}{7} \approx 1162.86 cm^{2}$
  1 mark
Answer: Volume = $\frac{8800}{7} \approx 1257.14$ cm³; surface area = $\frac{8140}{7} \approx 1162.86$ cm²
Where the marks are
- Correct volume expression: cylinder − 2 cones (1 mark)
- Volume $= \frac{8800}{7}$ cm³ (1 mark)
- Slant height l = 13 cm (1 mark)
- Surface area expression: CSA of cylinder + CSA of 2 cones (1 mark)
- Surface area $= \frac{8140}{7}$ cm² (1 mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerStatistics
Medical check-up was carried out for 35 students of a class and their weights were recorded as follows:
Weight (in kg) 38-40 40-42 42-44 44-46 46-48 48-50 50-52
Number of Students 3 2 4 5 14 4 3
Find the difference between the mean weight and the median weight.
Try it first, then show our solutionSolution
1. Take assumed mean $a = 45$ and class width $h = 2$ , so $u_{i} = \frac{x _{i} - 45}{2}$ .
  Weight (kg) $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$ c.f.
  38–40 3 39 −3 −9 3
  40–42 2 41 −2 −4 5
  42–44 4 43 −1 −4 9
  44–46 5 45 0 0 14
  46–48 14 47 1 14 28
  48–50 4 49 2 8 32
  50–52 3 51 3 9 35
  Total 35 14
  1½ marks
2. Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 45 + \frac{14}{35} \times 2$
  1 mark
3. $= 45 + 0.8 = 45.8$ kg
  ½ mark
4. $\frac{n}{2} = \frac{35}{2} = 17.5$ . The first c.f. greater than 17.5 is 28, so the median class is $46$ – $48$ , with $l = 46$ , $c f = 14$ , $f = 14$ , $h = 2$ .
  ½ mark
5. Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 46 + \frac{17.5 - 14}{14} \times 2$
  ½ mark
6. $= 46 + \frac{3.5}{14} \times 2 = 46 + 0.5 = 46.5$ kg
  ½ mark
7. Difference between the median and the mean $= 46.5 - 45.8 = 0.7$ kg
  ½ mark
Answer: Mean = 45.8 kg, Median = 46.5 kg, Difference = 0.7 kg
Where the marks are
- Correct table with class marks, $u_{i}$ , $f_{i} u_{i}$ and cumulative frequencies (1½ marks)
- Correct substitution in the mean formula (1 mark)
- Mean = 45.8 kg (½ mark)
- Identifies the median class as 46–48 (½ mark)
- Correct substitution in the median formula (½ mark)
- Median = 46.5 kg (½ mark)
- Difference = 0.7 kg (½ mark)
Open Q35 on its own page

Weight (in kg)	38-40	40-42	42-44	44-46	46-48	48-50	50-52
Number of Students	3	2	4	5	14	4	3

Weight (kg)	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$	c.f.
38–40	3	39	−3	−9	3
40–42	2	41	−2	−4	5
42–44	4	43	−1	−4	9
44–46	5	45	0	0	14
46–48	14	47	1	14	28
48–50	4	49	2	8	32
50–52	3	51	3	9	35
Total	35			14

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
A farmer has a circular piece of land. He wishes to construct his house in the form of largest possible square within the land as shown below.
The radius of circular piece of land is 35 m.
Based on given information, answer the following questions:
1. (i)
  Find the length of wire needed to fence the entire land.
  1 mark
2. (ii)
  Find the length of each side of the square land on which house will be constructed.
  1 mark
3. (iii)
  (a) The farmer wishes to grow grass on the shaded region around the house. Find the cost of growing the grass at the rate of ₹ 50 per square metre.
  2 marks
4. OR
  (iii)
  (b) Find the ratio of area of land on which house is built to remaining area of circular piece of land.
  2 marks
Figure in the paper: Left: a circle showing a house with two gabled sections and a tree below, drawn within the circular land. Right: a circle with a largest inscribed square (house floor plan with rooms), with the four circular segments outside the square shaded grey. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The length of wire for fencing equals the circumference: $2 π r = 2 \times \frac{22}{7} \times 35$
  ½ mark
2. (i) $= 220$ m
  ½ mark
3. (ii) The largest square inside a circle has its diagonal equal to the diameter. So diagonal $= 2 \times 35 = 70$ m.
  ½ mark
4. (ii) Side $= \frac{diagonal}{2} = \frac{70}{2} = 352$ m ( $\approx 49.49$ m)
  ½ mark
5. (iii)(a) The shaded region consists of two segments, each cut off by a side of the square. Each side subtends $9 0^{\circ}$ at the centre. Area $= 2 [\frac{90}{360} \times \frac{22}{7} \times 35 \times 35 - \frac{1}{2} \times 35 \times 35]$
  1 mark
6. (iii)(a) $= 2 [962.5 - 612.5] = 2 \times 350 = 700 m^{2}$
  ½ mark
7. (iii)(a) Cost $= 700 \times ₹50 = ₹35000$
  ½ mark
8. (iii)(b) Area of square $= (352)^{2} = 2450 m^{2}$ . Area of circle $= \frac{22}{7} \times 35 \times 35 = 3850 m^{2}$ . Remaining area $= 3850 - 2450 = 1400 m^{2}$ .
  1½ marks
9. (iii)(b) Required ratio $= \frac{2450}{1400} = \frac{7}{4}$ , i.e. $7 : 4$
  ½ mark
Answer: (i) 220 m; (ii) $352$ m; (iii)(a) ₹35000; (iii)(b) 7 : 4
Where the marks are
- (i) Uses circumference $2 π r$ (½ mark) and gets 220 m (½ mark)
- (ii) Diagonal of square = diameter = 70 m (½ mark); side $= 352$ m (½ mark)
- (iii)(a) Correct expression for the area of two segments (1 mark); 700 m² (½ mark); cost ₹35000 (½ mark)
- (iii)(b) Correct areas of square and remaining region, 2450 and 1400 (1½ marks); ratio 7 : 4 (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
In an equilateral triangle of side 10 cm, equilateral triangles of side 1 cm are formed as shown in the figure below, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row and so on.
Based on given information, answer the following questions using Arithmetic Progression.
1. (i)
  How many triangles will be there in bottom most row?
  1 mark
2. (ii)
  How many triangles will be there in fourth row from the bottom?
  1 mark
3. (iii)
  (a) Find the total number of triangles of side 1 cm each till $8^{th}$ row.
  2 marks
4. OR
  (iii)
  (b) How many more number of triangles are there from $5^{th}$ row to $1 0^{th}$ row than in first 4 rows? Show working.
  2 marks
Figure in the paper: A large equilateral triangle subdivided into small unit equilateral triangles, alternately shaded grey and white. Rows are labelled from the top: 1st Row (1 triangle), 2nd Row (3), 3rd Row (5), continuing down to the 10th Row at the base, which has its bottom edge with 5 visible unit triangles across the base in the picture. See it in the official paper.
Try it first, then show our solutionSolution
1. The numbers of triangles in successive rows are $1, 3, 5, \dots$ , an A.P. with $a = 1$ , $d = 2$ and 10 rows (side 10 cm, each small side 1 cm).
  (i) Bottom row: $a_{10} = a + 9 d = 1 + 9 \times 2 = 19$
  1 mark
2. (ii) Counting upward from the bottom, the rows form an A.P. with first term 19 and $d = - 2$ . The fourth row from the bottom has $19 + 3 \times (- 2) = 13$ triangles (this is the 7th row from the top: $1 + 6 \times 2 = 13$ ).
  1 mark
3. (iii)(a) $S_{8} = \frac{8}{2} [2 \times 1 + 7 \times 2]$
  1 mark
4. (iii)(a) $= 4 \times 16 = 64$ triangles
  1 mark
5. (iii)(b) Triangles from the 5th to the 10th row $= S_{10} - S_{4} = \frac{10}{2} [2 + 18] - \frac{4}{2} [2 + 6] = 100 - 16 = 84$
  1 mark
6. (iii)(b) Triangles in the first 4 rows: $S_{4} = \frac{4}{2} [2 \times 1 + 3 \times 2] = 16$
  ½ mark
7. (iii)(b) Required difference $= 84 - 16 = 68$ more triangles
  ½ mark
Answer: (i) 19; (ii) 13; (iii)(a) 64; (iii)(b) 68 more triangles
Where the marks are
- (i) $a_{10} = 1 + 9 \times 2 = 19$ (1 mark)
- (ii) Fourth row from the bottom = 13 (1 mark)
- (iii)(a) Correct use of $S_{n}$ formula for $n = 8$ (1 mark); $S_{8} = 64$ (1 mark)
- (iii)(b) $S_{10} - S_{4} = 84$ (1 mark); $S_{4} = 16$ (½ mark); difference 68 (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry
Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a mobile means to travel between the aircraft doors and the ground. Larger aircraft have door sills 5 to 20 feet (1 foot = 30 cm) high. Stairs facilitate safe boarding and de-boarding.
An aircraft has a door sill at a height of 15 feet above the ground. A stair car is placed at a horizontal distance of 15 feet from the plane.
Based on given information, answer the questions given in part (i) and (ii).
1. (i)
  Find the angle at which stairs are inclined to reach the door sill 15 feet high above the ground.
  1 mark
2. (ii)
  Find the length of stairs used to reach the door sill.
  1 mark
3. (iii)
  (a) If the 20 feet long stairs is inclined at an angle of $6 0^{\circ}$ to reach the door sill, then find the height of the door sill above the ground. (use $3 = 1.732$ )
  2 marks
4. OR
  (iii)
  (b) What should be the shortest possible length of stairs to reach the door sill of the plane 20 feet above the ground, if the angle of elevation cannot exceed $3 0^{\circ}$ ? Also, find the horizontal distance of base of stair car from the plane.
  2 marks
Figure in the paper: Left: an aircraft with its front door open, a passenger with luggage walking down a set of boarding stairs with railings to the tarmac. Right: a line drawing of a boarding stair car labelled 'STAIRS', with a sloped staircase with railings and a base platform on wheels at ground level. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Draw a right triangle: vertical side (door sill height) 15 ft, horizontal side 15 ft, stairs as hypotenuse $l$ , with inclination $θ$ at the ground. Then $tan θ = \frac{15}{15} = 1$
  ½ mark
2. (i) $\Rightarrow θ = 4 5^{\circ}$
  ½ mark
3. (ii) $sin 4 5^{\circ} = \frac{15}{l} \Rightarrow \frac{1}{2} = \frac{15}{l}$
  ½ mark
4. (ii) $l = 152 \approx 21.21$ ft
  ½ mark
5. (iii)(a) The stairs (20 ft) form the hypotenuse and $h$ is the height. $sin 6 0^{\circ} = \frac{h}{20} = \frac{3}{2}$
  1 mark
6. (iii)(a) $h = 103$
  ½ mark
7. (iii)(a) $h = 10 \times 1.732 = 17.32$ ft
  ½ mark
8. (iii)(b) The height is 20 ft. For a fixed height, the stair length $l = \frac{20}{sin θ}$ is shortest when $θ$ is largest, i.e. $θ = 3 0^{\circ}$ . Then $sin 3 0^{\circ} = \frac{20}{l} = \frac{1}{2}$
  ½ mark
9. (iii)(b) $l = 40$ ft
  ½ mark
10. (iii)(b) Let $x$ be the horizontal distance. $tan 3 0^{\circ} = \frac{20}{x} = \frac{1}{3}$
  ½ mark
11. (iii)(b) $x = 203 \approx 34.64$ ft
  ½ mark
Answer: (i) 45°; (ii) $152$ ft ≈ 21.21 ft; (iii)(a) 17.32 ft; (iii)(b) shortest length 40 ft, horizontal distance $203$ ft ≈ 34.64 ft
Where the marks are
- (i) $tan θ = 1$ (½ mark); $θ = 4 5^{\circ}$ (½ mark)
- (ii) Uses $sin 4 5^{\circ} = 15/ l$ (½ mark); $l = 152$ ft (½ mark)
- (iii)(a) $sin 6 0^{\circ} = h /20$ (1 mark); $h = 103$ (½ mark); 17.32 ft (½ mark)
- (iii)(b) $sin 3 0^{\circ} = 20/ l$ (½ mark); $l = 40$ ft (½ mark); $tan 3 0^{\circ} = 20/ x$ (½ mark); $x = 203$ ft (½ mark)
Open Q38 on its own page

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.