MathsCBSE Class 10 · Standard · Board exam 2024 (session 2023-24)

CBSE Class 10 Mathematics Standard: 2024 board paper, set 30/4/1

The question paper set in the CBSE Class 10 board exam of 2024, 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/2024/X/MATHEMATICS_STANDARD.zip (a zip; this paper is the file 30-4-1(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; internal choice in 22 and 23
C	26-31	Short answer; internal choice in 27 and 29
D	32-35	Long answer; internal choice in 33 and 34
E	36-38	Case study based; internal choice in each case study (sub-part (ii) in 36, 37, 38)

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

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQPair of Linear Equations in Two Variables
If $a x + b y = a^{2} - b^{2}$ and $b x + a y = 0$ , then the value of $x + y$ is :
1. (A) $a^{2} - b^{2}$
2. (B) $a + b$
3. (C) $a - b$
4. (D) $a^{2} + b^{2}$
Try it first, then show our solutionSolution
1. Add the two equations: $(a + b) x + (a + b) y = a^{2} - b^{2} = (a + b) (a - b)$ .
2. Divide by $(a + b)$ : $x + y = a - b$ . Option (C).
  1 mark
Answer: $x + y = a - b$ , option (C)
Where the marks are
- Adds the equations and simplifies to $a - b$ (1 mark)
Q2 · 1 mark · MCQReal Numbers
The HCF of two numbers 65 and 104 is 13. If LCM of 65 and 104 is $40 x$ , then the value of $x$ is :
1. (A)5
2. (B)13
3. (C)40
4. (D)8
Try it first, then show our solutionSolution
1. HCF × LCM = product of the numbers: $LCM = \frac{65 \times 104}{13} = 520$ .
2. $40 x = 520 \Rightarrow x = 13$ . Option (B).
  1 mark
Answer: $x = 13$ , option (B)
Where the marks are
- Uses HCF × LCM = product to get LCM 520 and $x = 13$ (1 mark)
Q3 · 1 mark · MCQPolynomials
If a polynomial $p (x)$ is given by $p (x) = x^{2} -5 x + 6$ , then the value of $p (1) + p (4)$ is :
1. (A)0
2. (B)4
3. (C)2
4. (D) $- 4$
Try it first, then show our solutionSolution
1. $p (1) = 1 - 5 + 6 = 2$ , $p (4) = 16 - 20 + 6 = 2$ .
2. $p (1) + p (4) = 4$ . Option (B).
  1 mark
Answer: 4, option (B)
Where the marks are
- Evaluates both values and adds to get 4 (1 mark)
Q4 · 1 mark · MCQQuadratic Equations
If the discriminant of the quadratic equation $3 x^{2} -2 x + c = 0$ is 16, then the value of $c$ is :
1. (A)1
2. (B)0
3. (C) $- 1$
4. (D) $2$
Try it first, then show our solutionSolution
1. $D = b^{2} - 4 a c = (- 2)^{2} - 4 (3) (c) = 4 - 12 c$ .
2. $4 - 12 c = 16 \Rightarrow c = - 1$ . Option (C).
  1 mark
Answer: $c = - 1$ , option (C)
Where the marks are
- Sets $4 - 12 c = 16$ and gets $c = - 1$ (1 mark)
Q5 · 1 mark · MCQAreas Related to Circles
If an arc subtends an angle of $9 0^{\circ}$ at the centre of a circle, then the ratio of its length to the circumference of the circle is :
1. (A)2 : 3
2. (B)1 : 4
3. (C)4 : 1
4. (D)1 : 3
Try it first, then show our solutionSolution
1. Arc length $= \frac{90}{360} \times 2 π r = \frac{1}{4} (2 π r)$ .
2. Ratio of arc to circumference $= 1 : 4$ . Option (B).
  1 mark
Answer: 1 : 4, option (B)
Where the marks are
- Uses $\frac{θ}{360}$ fraction to get 1 : 4 (1 mark)
Q6 · 1 mark · MCQAreas Related to Circles
The area of the sector of a circle of radius 12 cm is $60 π$ cm $^{2}$ . The central angle of this sector is :
1. (A) $12 0^{\circ}$
2. (B) $6^{\circ}$
3. (C) $7 5^{\circ}$
4. (D) $15 0^{\circ}$
Try it first, then show our solutionSolution
1. $\frac{θ}{360} \times π (12)^{2} = 60 π \Rightarrow \frac{144 θ}{360} = 60$ .
2. $θ = \frac{60 \times 360}{144} = 15 0^{\circ}$ . Option (D).
  1 mark
Answer: $15 0^{\circ}$ , option (D)
Where the marks are
- Uses sector area formula to get $15 0^{\circ}$ (1 mark)
Q7 · 1 mark · MCQStatistics
If the difference of mode and median of a data is 24, then the difference of its median and mean is :
1. (A)12
2. (B)24
3. (C)8
4. (D)36
Try it first, then show our solutionSolution
1. Empirical relation: Mode $= 3$ Median $- 2$ Mean, so Mode $-$ Median $= 2 (Median - Mean)$ .
2. $24 = 2 (Median - Mean) \Rightarrow$ Median $-$ Mean $= 12$ . Option (A).
  1 mark
Answer: 12, option (A)
Where the marks are
- Uses empirical relation to get 12 (1 mark)
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Q8 · 1 mark · MCQProbability
Two dice are tossed simultaneously. The probability of getting odd numbers on both the dice is :
1. (A) $\frac{6}{36}$
2. (B) $\frac{3}{36}$
3. (C) $\frac{12}{36}$
4. (D) $\frac{9}{36}$
Try it first, then show our solutionSolution
1. Total outcomes $= 36$ . Each die odd (1, 3, 5): $3 \times 3 = 9$ favourable outcomes.
2. $P = \frac{9}{36}$ . Option (D).
  1 mark
Answer: $\frac{9}{36}$ , option (D)
Where the marks are
- Counts 9 favourable out of 36 (1 mark)
Q9 · 1 mark · MCQSurface Areas and Volumes
The ratio of total surface area of a solid hemisphere to the square of its radius is :
1. (A) $2 π : 1$
2. (B) $4 π : 1$
3. (C) $3 π : 1$
4. (D) $1 : 4 π$
Try it first, then show our solutionSolution
1. TSA of solid hemisphere $= 2 π r^{2} + π r^{2} = 3 π r^{2}$ .
2. $3 π r^{2} : r^{2} = 3 π : 1$ . Option (C).
  1 mark
Answer: $3 π : 1$ , option (C)
Where the marks are
- Uses TSA $3 π r^{2}$ to get $3 π : 1$ (1 mark)
Q10 · 1 mark · MCQIntroduction to Trigonometry
If $sin θ = 1$ , then the value of $\frac{1}{2} sin (\frac{θ}{2})$ is :
1. (A) $\frac{1}{2 2}$
2. (B) $\frac{1}{2}$
3. (C) $\frac{1}{2}$
4. (D)0
Try it first, then show our solutionSolution
1. $sin θ = 1 \Rightarrow θ = 9 0^{\circ}$ , so $\frac{θ}{2} = 4 5^{\circ}$ .
2. $\frac{1}{2} sin 4 5^{\circ} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2 2}$ . Option (A).
  1 mark
Answer: $\frac{1}{2 2}$ , option (A)
Where the marks are
- Finds $θ = 9 0^{\circ}$ and evaluates to $\frac{1}{2 2}$ (1 mark)
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Q11 · 1 mark · MCQPair of Linear Equations in Two Variables
Two lines are given to be parallel. The equation of one of these lines is $5 x -3 y = 2$ . The equation of the second line can be :
1. (A) $- 15 x -9 y = 5$
2. (B) $15 x + 9 y = 5$
3. (C) $9 x -15 y = 6$
4. (D) $- 15 x + 9 y = 5$
Try it first, then show our solutionSolution
1. Parallel lines need $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
2. For $- 15 x + 9 y = 5$ : $\frac{5}{- 15} = \frac{- 3}{9} = - \frac{1}{3}$ and $\frac{2}{5} \neq = - \frac{1}{3}$ . Parallel. Option (D).
  1 mark
Answer: $- 15 x + 9 y = 5$ , option (D)
Where the marks are
- Checks coefficient ratios for parallel lines (1 mark)
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Q12 · 1 mark · MCQArithmetic Progressions
Three numbers in A.P. have the sum 30. What is its middle term ?
1. (A)4
2. (B)10
3. (C)16
4. (D)8
Try it first, then show our solutionSolution
1. Take the terms as $a - d, a, a + d$ ; sum $= 3 a = 30$ .
2. $a = 10$ . Option (B).
  1 mark
Answer: 10, option (B)
Where the marks are
- Uses $a - d, a, a + d$ to get middle term 10 (1 mark)
Q13 · 1 mark · MCQTriangles
In $△$ ABC, DE $∥$ BC (as shown in the figure). If AD = 4 cm, AB = 9 cm and AC = 13.5 cm, then the length of EC is :
1. (A)6 cm
2. (B)7.5 cm
3. (C)9 cm
4. (D)5.7 cm
Figure in the paper: Triangle ABC with vertex A at the top. D lies on side AB and E lies on side AC, with segment DE drawn parallel to the base BC. B and C are at the bottom corners. See it in the official paper.
Try it first, then show our solutionSolution
1. By BPT [DE ∥ BC]: $\frac{A D}{A B} = \frac{A E}{A C} \Rightarrow \frac{4}{9} = \frac{A E}{13.5} \Rightarrow A E = 6$ cm.
2. $E C = 13.5 - 6 = 7.5$ cm. Option (B).
  1 mark
Answer: 7.5 cm, option (B)
Where the marks are
- Applies BPT to get AE = 6 and EC = 7.5 cm (1 mark)
Q14 · 1 mark · MCQCompetency-basedSome Applications of Trigonometry
At some time of the day, the length of the shadow of a tower is equal to its height. Then, the Sun's altitude at that time is :
1. (A) $3 0^{\circ}$
2. (B) $4 5^{\circ}$
3. (C) $6 0^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. $tan θ = \frac{height}{shadow} = 1$ .
2. $θ = 4 5^{\circ}$ . Option (B).
  1 mark
Answer: $4 5^{\circ}$ , option (B)
Where the marks are
- Uses $tan θ = 1$ to get $4 5^{\circ}$ (1 mark)
Q15 · 1 mark · MCQCircles
In the given figure, AB and AC are tangents to the circle. If $∠ A B C = 4 2^{\circ}$ , then the measure of $∠ B A C$ is :
1. (A) $9 6^{\circ}$
2. (B) $4 2^{\circ}$
3. (C) $10 6^{\circ}$
4. (D) $8 6^{\circ}$
Figure in the paper: A circle with an external point A. Two tangents AB and AC touch the circle at B and C. Chord BC is drawn, and the angle between BC and the tangent at B (angle ABC) is marked as 42 degrees. The angle at A (BAC) is marked. See it in the official paper.
Try it first, then show our solutionSolution
1. AB = AC [tangents from an external point], so $∠ A C B = ∠ A B C = 4 2^{\circ}$ .
2. $∠ B A C = 18 0^{\circ} - 8 4^{\circ} = 9 6^{\circ}$ . Option (A).
  1 mark
Answer: $9 6^{\circ}$ , option (A)
Where the marks are
- Uses equal tangents and angle sum to get $9 6^{\circ}$ (1 mark)
Q16 · 1 mark · MCQCoordinate Geometry
The fourth vertex D of a parallelogram ABCD whose three vertices are A(– 2, 3), B(6, 7) and C(8, 3) is :
1. (A)(0, 1)
2. (B)(0, – 1)
3. (C)(– 1, 0)
4. (D)(1, 0)
Try it first, then show our solutionSolution
1. Diagonals bisect each other: midpoint of AC $= (3, 3)$ = midpoint of BD.
2. $\frac{6 + x}{2} = 3, \frac{7 + y}{2} = 3 \Rightarrow D (0, - 1)$ . Option (B).
  1 mark
Answer: (0, −1), option (B)
Where the marks are
- Uses equal midpoints of diagonals to find D (1 mark)
Q17 · 1 mark · MCQProbability
For an event E, if $P (E) + P (\overset{ˉ}{E}) = q$ , then the value of $q^{2} -4$ is :
1. (A) $- 3$
2. (B)3
3. (C)5
4. (D) $- 5$
Try it first, then show our solutionSolution
1. $P (E) + P (\overset{ˉ}{E}) = 1$ , so $q = 1$ .
2. $q^{2} - 4 = 1 - 4 = - 3$ . Option (A).
  1 mark
Answer: −3, option (A)
Where the marks are
- Uses $q = 1$ to get $- 3$ (1 mark)
Q18 · 1 mark · MCQCircles
In the given figure, QR is a common tangent to the two given circles touching externally at A. The tangent at A meets QR at P. If AP = 4.2 cm, then the length of QR is :
1. (A)4.2 cm
2. (B)2.1 cm
3. (C)8.4 cm
4. (D)6.3 cm
Figure in the paper: Two circles touching externally at point A. A horizontal common tangent line touches the left circle at Q and the right circle at R. The common tangent at A is drawn vertically and meets QR at P, with P between Q and R. See it in the official paper.
Try it first, then show our solutionSolution
1. Tangents from P: PQ = PA and PR = PA [tangents from an external point are equal], so PQ = PR = 4.2 cm.
2. QR = PQ + PR = 8.4 cm. Option (C).
  1 mark
Answer: 8.4 cm, option (C)
Where the marks are
- Uses equal tangents from P to get QR = 8.4 cm (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedCoordinate Geometry
Assertion (A): Mid-point of a line segment divides the line segment in the ratio $1 : 1$ .
Reason (R): The ratio in which the point $(-3, k)$ divides the line segment joining the points $(-5, 4)$ and $(-2, 3)$ is $1 : 2$ .
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A is true: the mid-point is equidistant from both ends, so ratio 1 : 1.
2. R: with ratio $k : 1$ , $\frac{- 2 k - 5}{k + 1} = - 3 \Rightarrow k = 2$ , so the ratio is 2 : 1, not 1 : 2. R is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows ratio is 2 : 1 so R false, A true (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedAreas Related to Circles
Assertion (A): If the circumference of a circle is 176 cm, then its radius is 28 cm.
Reason (R): Circumference $= 2 π \times$ radius of a circle.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. R is the standard formula, true.
2. $2 \times \frac{22}{7} \times r = 176 \Rightarrow r = 28$ cm, so A is true and follows from R. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Uses $C = 2 π r$ to verify $r = 28$ cm (1 mark)

Section B

Very short answer; internal choice in 22 and 23

Q21 · 2 marks · Very short answerCompetency-basedReal Numbers
Three bells toll at intervals of 9, 12 and 15 minutes respectively. If they start tolling together, after what time will they next toll together ?
Try it first, then show our solutionSolution
1. Prime factorise the intervals: $9 = 3^{2}$ , $12 = 2^{2} \times 3$ , $15 = 3 \times 5$ .
  1 mark
2. The bells next toll together after the LCM of the intervals: $LCM = 2^{2} \times 3^{2} \times 5 = 180$ . So they toll together again after 180 minutes, i.e. 3 hours.
  1 mark
Answer: After 180 minutes (3 hours)
Where the marks are
- Correct prime factorisation of 9, 12 and 15 (1 mark)
- LCM = 180, so the bells toll together after 180 min (1 mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerCompetency-basedAreas Related to Circles
The minute hand of a clock is 14 cm long. Find the area on the face of the clock described by the minute hand in 5 minutes.
Try it first, then show our solutionSolution
1. In 60 minutes the minute hand turns $36 0^{\circ}$ , so in 5 minutes it turns $θ = \frac{36 0 ^{\circ}}{60} \times 5 = 3 0^{\circ}$ .
  ½ mark
2. Area swept = area of sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$
  1 mark
3. $= \frac{154}{3} cm^{2} \approx 51.33 cm^{2}$
  ½ mark
Answer: $\frac{154}{3}$ cm² ≈ 51.33 cm²
Where the marks are
- Angle turned in 5 minutes = 30° (½ mark)
- Correct sector area formula with substitution (1 mark)
- Area = 154/3 cm² ≈ 51.33 cm² (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerAreas Related to Circles
Find the length of the arc of a circle which subtends an angle of $6 0^{\circ}$ at the centre of the circle of radius 42 cm.
Try it first, then show our solutionSolution
1. Length of arc $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 42$
  1½ marks
2. $= \frac{1}{6} \times 264 = 44$ cm
  ½ mark
Answer: 44 cm
Where the marks are
- Correct arc-length formula with substitution (1½ marks)
- Arc length = 44 cm (½ mark)
Open Q22(b) on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerIntroduction to Trigonometry
Evaluate : $\frac{5 c o s ^{2} 6 0 ^{\circ} + 4 s e c ^{2} 3 0 ^{\circ} - t a n ^{2} 4 5 ^{\circ}}{s i n ^{2} 3 0 ^{\circ} + s i n ^{2} 6 0 ^{\circ}}$
Try it first, then show our solutionSolution
1. Substitute $cos 6 0^{\circ} = \frac{1}{2}$ , $sec 3 0^{\circ} = \frac{2}{3}$ , $tan 4 5^{\circ} = 1$ , $sin 3 0^{\circ} = \frac{1}{2}$ , $sin 6 0^{\circ} = \frac{3}{2}$ :
  $\frac{5 ( \frac{1}{2} ) ^{2} + 4 ( \frac{2}{3} ) ^{2} - ( 1 ) ^{2}}{( \frac{1}{2} ) ^{2} + ( \frac{3}{2} ) ^{2}} = \frac{\frac{5}{4} + \frac{16}{3} - 1}{\frac{1}{4} + \frac{3}{4}}$
  1½ marks
2. $= \frac{\frac{15 + 64 - 12}{12}}{1} = \frac{67}{12}$
  ½ mark
Answer: $\frac{67}{12}$
Where the marks are
- Correct substitution of all trigonometric values (1½ marks)
- Simplifies to 67/12 (½ mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerIntroduction to Trigonometry
If $sin (A - B) = \frac{1}{2}$ , $cos (A + B) = \frac{1}{2}$ ; $0 < A + B \leq 9 0^{\circ}$ , $A > B$ ; find $∠ A$ 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): $2 A = 9 0^{\circ} \Rightarrow A = 4 5^{\circ}$ ; then $B = 6 0^{\circ} - 4 5^{\circ} = 1 5^{\circ}$ .
  1 mark
Answer: ∠A = 45°, ∠B = 15°
Where the marks are
- A − B = 30° (½ mark)
- A + B = 60° (½ mark)
- Solving to get A = 45°, B = 15° (1 mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerCircles
In the given figure, O is the centre of the circle. If $∠ A O B = 14 5^{\circ}$ , then find the value of $x$ .
Figure in the paper: A circle with centre O. Chord AB is drawn with the central angle AOB marked as 145 degrees. A point C lies on the circle on the arc opposite to the major side near the bottom, and the angle ACB is marked as x, with segments CA and CB drawn. See it in the official paper.
Try it first, then show our solutionSolution
1. Take a point P on the major arc AB and join PA and PB.
  ½ mark
2. $∠ A P B = \frac{1}{2} ∠ A O B = \frac{1}{2} \times 14 5^{\circ} = 72. 5^{\circ}$ [angle at the centre is double the angle at the remaining part of the circle]
  ½ mark
3. APBC is a cyclic quadrilateral, so $∠ A P B + ∠ A C B = 18 0^{\circ}$ [opposite angles of a cyclic quadrilateral are supplementary]
  ½ mark
4. $∠ A C B = 18 0^{\circ} - 72. 5^{\circ} = 107. 5^{\circ}$ , so $x = 107. 5^{\circ}$ .
  ½ mark
Answer: x = 107.5°
Where the marks are
- Takes a point P on the major arc and joins AP, BP (½ mark)
- ∠APB = 72.5° (½ mark)
- Uses ∠APB + ∠ACB = 180° (½ mark)
- x = 107.5° (½ mark)
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Q25 · 2 marks · Very short answerTriangles
In the given figure, $△$ AHK $\sim$ $△$ ABC. If AK = 8 cm, BC = 3.2 cm and HK = 6.4 cm, then find the length of AC.
Figure in the paper: A horizontal line through C, A and K (C on the left, K on the right). Point B lies vertically below C, with BC drawn vertically and a right angle at C. Point H lies vertically above K, with HK vertical and a right angle at K. The line from B passes through A and continues up to H. See it in the official paper.
Try it first, then show our solutionSolution
1. Since $△ A H K \sim △ A B C$ , corresponding sides are proportional: $\frac{H K}{B C} = \frac{A K}{A C} \Rightarrow \frac{6.4}{3.2} = \frac{8}{A C}$
  1 mark
2. $2 = \frac{8}{A C} \Rightarrow A C = 4$ cm
  1 mark
Answer: AC = 4 cm
Where the marks are
- Writes correct ratio of corresponding sides and substitutes (1 mark)
- AC = 4 cm (1 mark)
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Section C

Short answer; internal choice in 27 and 29

Q26 · 3 marks · Short answerIntroduction to Trigonometry
Prove that $\frac{s i n θ - c o s θ + 1}{s i n θ + c o s θ -1} = \frac{1}{s e c θ - t a n θ}$
Try it first, then show our solutionSolution
1. Dividing numerator and denominator of the LHS by $cos θ$ : LHS $= \frac{t a n θ - 1 + s e c θ}{t a n θ + 1 - s e c θ}$
  1 mark
2. Replace 1 in the denominator by $sec^{2} θ - tan^{2} θ$ [identity $sec^{2} θ - tan^{2} θ = 1$ ]: $= \frac{t a n θ - 1 + s e c θ}{( t a n θ - s e c θ ) + ( s e c ^{2} θ - t a n ^{2} θ )}$
  1 mark
3. Factorise the denominator: $(tan θ - sec θ) + (sec θ - tan θ) (sec θ + tan θ) = (sec θ - tan θ) (tan θ + sec θ - 1)$ , so LHS $= \frac{t a n θ + s e c θ - 1}{( s e c θ - t a n θ ) ( t a n θ + s e c θ - 1 )}$
  ½ mark
4. Cancelling the common factor: LHS $= \frac{1}{s e c θ - t a n θ} =$ RHS. Hence proved.
  ½ mark
Answer: LHS = 1/(secθ − tanθ) = RHS, hence proved
Where the marks are
- Divides by cos θ to get expression in tan θ and sec θ (1 mark)
- Replaces 1 by sec²θ − tan²θ (1 mark)
- Factorises the denominator (½ mark)
- Cancels to reach RHS (½ mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerProbability
Three coins are tossed simultaneously. What is the probability of getting
1. (i)
  at least one head ?
  1 mark
2. (ii)
  exactly two tails ?
  1 mark
3. (iii)
  at most one tail ?
  1 mark
Try it first, then show our solutionSolution
1. Outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT — total 8.
2. (i) At least one head: all except TTT, so 7 outcomes. $P = \frac{7}{8}$
  1 mark
3. (ii) Exactly two tails: HTT, THT, TTH — 3 outcomes. $P = \frac{3}{8}$
  1 mark
4. (iii) At most one tail: HHH, HHT, HTH, THH — 4 outcomes. $P = \frac{4}{8} = \frac{1}{2}$
  1 mark
Answer: (i) 7/8; (ii) 3/8; (iii) 1/2
Where the marks are
- P(at least one head) = 7/8 (1 mark)
- P(exactly two tails) = 3/8 (1 mark)
- P(at most one tail) = 4/8 = 1/2 (1 mark)
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OR
Q27(b) · 3 marks · Short answerProbability
A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a :
1. (i)
  2-digit number less than 40.
  1 mark
2. (ii)
  number divisible by 5 and greater than 50.
  1 mark
3. (iii)
  a perfect square number.
  1 mark
Try it first, then show our solutionSolution
1. Total number of outcomes = 90.
2. (i) 2-digit numbers less than 40 are 10, 11, ..., 39: 30 numbers. $P = \frac{30}{90} = \frac{1}{3}$
  1 mark
3. (ii) Numbers divisible by 5 and greater than 50: 55, 60, 65, 70, 75, 80, 85, 90 — 8 numbers. $P = \frac{8}{90} = \frac{4}{45}$
  1 mark
4. (iii) Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 — 9 numbers. $P = \frac{9}{90} = \frac{1}{10}$
  1 mark
Answer: (i) 1/3; (ii) 4/45; (iii) 1/10
Where the marks are
- P(2-digit number < 40) = 30/90 = 1/3 (1 mark)
- P(divisible by 5 and > 50) = 8/90 = 4/45 (1 mark)
- P(perfect square) = 9/90 = 1/10 (1 mark)
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Q28 · 3 marks · Short answerCompetency-basedPair of Linear Equations in Two Variables
Rehana went to a bank to withdraw ₹ 2,000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Rehana got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 did she receive.
Try it first, then show our solutionSolution
1. Let the number of ₹50 notes be $x$ and of ₹100 notes be $y$ . Total notes: $x + y = 25$ ...(i)
  1 mark
2. Total amount: $50 x + 100 y = 2000 \Rightarrow x + 2 y = 40$ ...(ii)
  1 mark
3. Subtracting (i) from (ii): $y = 15$ ; then $x = 25 - 15 = 10$ . So she received 10 notes of ₹50 and 15 notes of ₹100.
  1 mark
Answer: 10 notes of ₹50 and 15 notes of ₹100
Where the marks are
- Forms x + y = 25 (1 mark)
- Forms 50x + 100y = 2000 i.e. x + 2y = 40 (1 mark)
- Solves: x = 10, y = 15 (½ + ½ marks)
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Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerPolynomials
Find the zeroes of the polynomial $4 x^{2} + 4 x -3$ and verify the relationship between zeroes and coefficients of the polynomial.
Try it first, then show our solutionSolution
1. Split the middle term: $4 x^{2} + 4 x - 3 = 4 x^{2} + 6 x - 2 x - 3 = (2 x + 3) (2 x - 1)$
  1 mark
2. Zeroes: $2 x + 3 = 0 \Rightarrow x = - \frac{3}{2}$ ; $2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$ .
  1 mark
3. Sum of zeroes $= - \frac{3}{2} + \frac{1}{2} = - 1$ , and $\frac{- ( coefficient of x )}{coefficient of x ^{2}} = \frac{- 4}{4} = - 1$ . Verified.
  ½ mark
4. Product of zeroes $= - \frac{3}{2} \times \frac{1}{2} = - \frac{3}{4}$ , and $\frac{constant term}{coefficient of x ^{2}} = \frac{- 3}{4}$ . Verified.
  ½ mark
Answer: Zeroes are −3/2 and 1/2; sum = −1 = −b/a, product = −3/4 = c/a
Where the marks are
- Factorises as (2x + 3)(2x − 1) (1 mark)
- Zeroes −3/2 and 1/2 (1 mark)
- Verifies sum of zeroes = −b/a (½ mark)
- Verifies product of zeroes = c/a (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerPolynomials
If $α$ and $β$ are the zeroes of the polynomial $x^{2} + x -2$ , then find the value of $\frac{α}{β} + \frac{β}{α}$ .
Try it first, then show our solutionSolution
1. For $x^{2} + x - 2$ : $α + β = - \frac{1}{1} = - 1$ , $α β = \frac{- 2}{1} = - 2$ .
  1 mark
2. $\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β} = \frac{( α + β ) ^{2} - 2 α β}{α β}$
  1 mark
3. $= \frac{( - 1 ) ^{2} - 2 ( - 2 )}{- 2} = \frac{1 + 4}{- 2} = - \frac{5}{2}$
  1 mark
Answer: $- \frac{5}{2}$
Where the marks are
- α + β = −1 and αβ = −2 (1 mark)
- Expresses the sum as [(α+β)² − 2αβ]/αβ (1 mark)
- Value = −5/2 (1 mark)
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Q30 · 3 marks · Short answerReal Numbers
Prove that $\frac{2- 3}{5}$ is an irrational number, given that $3$ is an irrational number.
Try it first, then show our solutionSolution
1. Assume, to the contrary, that $\frac{2 - 3}{5}$ is rational. Then $\frac{2 - 3}{5} = \frac{p}{q}$ , where $p, q$ are integers and $q \neq = 0$ .
  ½ mark
2. Then $2 - 3 = \frac{5 p}{q} \Rightarrow 3 = 2 - \frac{5 p}{q} = \frac{2 q - 5 p}{q}$ .
  1 mark
3. Since $p, q$ are integers, $\frac{2 q - 5 p}{q}$ is rational, so the RHS is rational while the LHS $3$ is irrational (given).
  ½ mark
4. This is a contradiction, so our assumption is wrong.
  ½ mark
5. Hence $\frac{2 - 3}{5}$ is an irrational number.
  ½ mark
Answer: (2 − √3)/5 is irrational (proved by contradiction)
Where the marks are
- Assumes (2 − √3)/5 = p/q, q ≠ 0 (½ mark)
- Rearranges to √3 = (2q − 5p)/q (1 mark)
- States RHS rational but LHS irrational (½ mark)
- Concludes the assumption is wrong (½ mark)
- Concludes the number is irrational (½ mark)
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Q31 · 3 marks · Short answerCircles
Prove that the parallelogram circumscribing a circle is a rhombus.
Try it first, then show our solutionSolution
1. Figure: Draw parallelogram ABCD with all four sides touching a circle. The points of contact are P on AB, Q on BC, R on CD and S on DA.
  Given: ABCD is a parallelogram circumscribing the circle.
  To prove: ABCD is a rhombus.
  ½ mark
2. Tangents drawn from an external point to a circle are equal in length. So:
  $A P = A S, B P = B Q, C R = C Q, D R = D S$
  1 mark
3. Adding these four equations:
  $(A P + B P) + (C R + D R) = (A S + D S) + (B Q + C Q)$
  $\Rightarrow A B + C D = A D + B C$
  ½ mark
4. In a parallelogram opposite sides are equal: $A B = C D$ and $B C = A D$ .
  So $2 A B = 2 B C$ , which gives $A B = B C$ .
  ½ mark
5. Therefore $A B = B C = C D = D A$ . A parallelogram with all sides equal is a rhombus, so ABCD is a rhombus. Hence proved.
  ½ mark
Answer: AB + CD = AD + BC together with AB = CD and BC = AD gives AB = BC = CD = DA, so ABCD is a rhombus
Where the marks are
- Correct labelled figure (1/2 mark)
- Writes the four equal-tangent relations (1 mark)
- Adds them to get $A B + C D = A D + B C$ (1/2 mark)
- Uses the opposite sides of the parallelogram to get $A B = B C$ (1/2 mark)
- Concludes all sides are equal, so ABCD is a rhombus (1/2 mark)
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Section D

Long answer; internal choice in 33 and 34

Q32 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry
Two pillars of equal lengths stand on either side of a road which is 100 m wide, exactly opposite to each other. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are $6 0^{\circ}$ and $3 0^{\circ}$ . Find the length of each pillar and distance of the point on the road from the pillars. (Use $3 = 1 \cdot 732$ )
Try it first, then show our solutionSolution
1. Figure: pillars AB and CD, each of height $h$ m, stand at B and D with BD = 100 m. P is a point on BD with PD = $x$ m and BP = $(100 - x)$ m; $∠ C P D = 6 0^{\circ}$ , $∠ A P B = 3 0^{\circ}$ , and $∠ B = ∠ D = 9 0^{\circ}$ .
  1 mark
2. In right $△ C D P$ : $tan 6 0^{\circ} = \frac{C D}{P D} \Rightarrow 3 = \frac{h}{x}$
  1 mark
3. $\Rightarrow h = 3 x$ ...(i)
  ½ mark
4. In right $△ A B P$ : $tan 3 0^{\circ} = \frac{A B}{B P} \Rightarrow \frac{1}{3} = \frac{h}{100 - x}$
  1 mark
5. $\Rightarrow h = \frac{100 - x}{3}$ ...(ii)
  ½ mark
6. From (i) and (ii): $3 x = \frac{100 - x}{3} \Rightarrow 3 x = 100 - x \Rightarrow x = 25$
  ½ mark
7. $h = 253 = 25 \times 1.732 = 43.3$ m. So each pillar is 43.3 m high, and the point is 25 m from the pillar CD (elevation 60°) and 75 m from the pillar AB (elevation 30°).
  ½ mark
Answer: Height of each pillar = 43.3 m; the point is 25 m and 75 m from the pillars
Where the marks are
- Correct labelled figure (1 mark)
- tan 60° = h/x (1 mark)
- h = √3 x (½ mark)
- tan 30° = h/(100 − x) (1 mark)
- h = (100 − x)/√3 (½ mark)
- x = 25 (½ mark)
- h = 43.3 m and distances 25 m, 75 m (½ mark)
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Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerTriangles
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $△$ ABE $\sim$ $△$ CFB.
Try it first, then show our solutionSolution
1. Figure: parallelogram ABCD with AD produced beyond D to E; BE cuts CD at F. Triangles ABE and CFB are marked.
  2 marks
2. In $△ A B E$ and $△ C F B$ : $∠ E A B = ∠ B C F$ [opposite angles of parallelogram ABCD are equal, $∠ A = ∠ C$ ]
  1 mark
3. $∠ A E B = ∠ C B F$ [alternate angles, since AE ∥ BC (AD ∥ BC) and BE is a transversal]
  1 mark
4. Therefore $△ A B E \sim △ C F B$ [AA similarity criterion].
  1 mark
Answer: △ABE ∼ △CFB by AA similarity
Where the marks are
- Correct figure (2 marks)
- ∠EAB = ∠BCF, opposite angles of parallelogram (1 mark)
- ∠AEB = ∠CBF, alternate angles as AE ∥ BC (1 mark)
- Concludes similarity by AA (1 mark)
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OR
Q33(b) · 5 marks · Long answerTriangles
Sides AB, BC and the median AD of $△$ ABC are respectively proportional to sides PQ, QR and the median PM of another $△$ PQR. Prove that $△$ ABC $\sim$ $△$ PQR.
Try it first, then show our solutionSolution
1. Figure: $△ A B C$ with median AD (D mid-point of BC) and $△ P QR$ with median PM (M mid-point of QR). Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ .
  1 mark
2. Since $B C = 2 B D$ and $QR = 2 QM$ : $\frac{A B}{P Q} = \frac{2 B D}{2 QM} = \frac{A D}{P M} \Rightarrow \frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ ...(i)
  1 mark
3. From (i), $△ A B D \sim △ P QM$ [SSS similarity criterion].
  1 mark
4. Hence $∠ B = ∠ Q$ [corresponding angles of similar triangles] ...(ii)
  1 mark
5. In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ (given) and $∠ B = ∠ Q$ from (ii). Therefore $△ A B C \sim △ P QR$ [SAS similarity criterion].
  1 mark
Answer: △ABC ∼ △PQR (via △ABD ∼ △PQM by SSS, then SAS)
Where the marks are
- Correct figure with medians (1 mark)
- Uses BC = 2BD, QR = 2QM to get AB/PQ = BD/QM = AD/PM (1 mark)
- △ABD ∼ △PQM by SSS (1 mark)
- ∠B = ∠Q (1 mark)
- △ABC ∼ △PQR by SAS (1 mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerCompetency-basedQuadratic Equations
A train travels a distance of 90 km at a constant speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Try it first, then show our solutionSolution
1. Let the original speed be $x$ km/h; the increased speed is $(x + 15)$ km/h.
  ½ mark
2. Time = distance/speed, and 30 minutes = $\frac{1}{2}$ h: $\frac{90}{x} - \frac{90}{x + 15} = \frac{1}{2}$
  1½ marks
3. $\frac{90 ( x + 15 ) - 90 x}{x ( x + 15 )} = \frac{1}{2} \Rightarrow 2700 = x^{2} + 15 x \Rightarrow x^{2} + 15 x - 2700 = 0$
  1 mark
4. $x^{2} + 60 x - 45 x - 2700 = 0 \Rightarrow (x + 60) (x - 45) = 0$
  1 mark
5. $x = - 60$ or $x = 45$ . Speed cannot be negative, so $x = 45$ . The original speed of the train is 45 km/h.
  1 mark
Answer: 45 km/h
Where the marks are
- Takes original speed x and new speed x + 15 (½ mark)
- Forms 90/x − 90/(x+15) = 1/2 (1½ marks)
- Reduces to x² + 15x − 2700 = 0 (1 mark)
- Factorises (x + 60)(x − 45) = 0 (1 mark)
- Rejects −60, speed = 45 km/h (1 mark)
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OR
Q34(b) · 5 marks · Long answerQuadratic Equations
Find the value of 'c' for which the quadratic equation $(c + 1) x^{2} -6 (c + 1) x + 3 (c + 9) = 0$ ; $c \neq = -1$ has real and equal roots.
Try it first, then show our solutionSolution
1. Here $a = c + 1$ , $b = - 6 (c + 1)$ , $c_{0} = 3 (c + 9)$ . For real and equal roots, $D = b^{2} - 4 a c_{0} = 0$ : ${- 6 (c + 1)}^{2} - 4 (c + 1) \times 3 (c + 9) = 0$
  2 marks
2. $36 (c + 1)^{2} - 12 (c + 1) (c + 9) = 0 \Rightarrow 12 (c + 1) [3 (c + 1) - (c + 9)] = 0 \Rightarrow 12 (c + 1) (2 c - 6) = 0$
  2 marks
3. Since $c \neq = - 1$ , $c + 1 \neq = 0$ , so $2 c - 6 = 0 \Rightarrow c = 3$ .
  1 mark
Answer: c = 3
Where the marks are
- Sets discriminant {−6(c+1)}² − 4(c+1)·3(c+9) = 0 (2 marks)
- Simplifies to 12(c + 1)(2c − 6) = 0 (2 marks)
- Uses c ≠ −1 to get c = 3 (1 mark)
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Q35 · 5 marks · Long answerCompetency-basedStatistics
The following table shows the ages of the patients admitted in a hospital during a year :
Age (in years) 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65
Number of patients 6 11 21 23 14 5
Find the mode and mean of the data given above.
Try it first, then show our solutionSolution
1. Take class marks $x_{i} = \frac{lower limit + upper limit}{2}$ and form the table:
  Age $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  5–15 6 10 60
  15–25 11 20 220
  25–35 21 30 630
  35–45 23 40 920
  45–55 14 50 700
  55–65 5 60 300
  Total $Σ f_{i} = 80$ $Σ f_{i} x_{i} = 2830$
  1½ marks
2. Mean $\overset{x}{ˉ} = \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{2830}{80}$
  ½ mark
3. $\overset{x}{ˉ} = 35.375$ years
  ½ mark
4. The highest frequency is 23, so the modal class is 35–45. Here $l = 35$ , $f_{1} = 23$ , $f_{0} = 21$ , $f_{2} = 14$ and $h = 10$ .
  ½ mark
5. Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 35 + (\frac{23 - 21}{46 - 21 - 14}) \times 10 = 35 + \frac{20}{11}$
  1 mark
6. Mode $= 35 + 1.82 \approx 36.82$ years. So the mean age is 35.375 years and the modal age is about 36.82 years.
  1 mark
Answer: Mean = 35.375 years; Mode ≈ 36.82 years
Where the marks are
- Correct table of class marks and $f_{i} x_{i}$ with $Σ f_{i} = 80$ and $Σ f_{i} x_{i} = 2830$ (1½ marks)
- Uses the mean formula $Σ f_{i} x_{i} /Σ f_{i}$ (½ mark)
- Gets mean = 35.375 (½ mark)
- Identifies modal class 35–45 (½ mark)
- Substitutes correctly in the mode formula (1 mark)
- Gets mode ≈ 36.82 (1 mark)
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Age (in years)	5 – 15	15 – 25	25 – 35	35 – 45	45 – 55	55 – 65
Number of patients	6	11	21	23	14	5

Age	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
5–15	6	10	60
15–25	11	20	220
25–35	21	30	630
35–45	23	40	920
45–55	14	50	700
55–65	5	60	300
Total	$Σ f_{i} = 80$		$Σ f_{i} x_{i} = 2830$

Section E

Case study based; internal choice in each case study (sub-part (ii) in 36, 37, 38)

Q36 · 4 marks · Case-basedCompetency-basedCoordinate Geometry
Ryan, from a very young age, was fascinated by the twinkling of stars and the vastness of space. He always dreamt of becoming an astronaut one day. So he started to sketch his own rocket designs on the graph sheet. One such design is given below :
Based on the above, answer the following questions :
1. (i)
  Find the mid-point of the segment joining F and G.
  1 mark
2. (ii)
  (a) What is the distance between the points A and C ?
  2 marks
3. OR
  (ii)
  (b) Find the coordinates of the point which divides the line segment joining the points A and B in the ratio $1 : 3$ internally.
  2 marks
4. (iii)
  What are the coordinates of the point D ?
  1 mark
Figure in the paper: A rocket design drawn on a graph sheet with X and Y axes (x from -6 to 6, y from -5 to 5). Labelled vertices: A(3, 4), B(3, 2), C(-1, -2), D(-2, -5), E(-5, 0), F(-3, 0), G(1, 4). The rocket outline runs A to B on the right, B to C diagonally, C down to a notch near (-1, -4), up to D at the bottom, and then the left side with zigzag fins; segment EF and FG form the upper-left edge with G at (1, 4). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) From the graph, F(−3, 0) and G(1, 4). Mid-point $= (\frac{- 3 + 1}{2}, \frac{0 + 4}{2}) = (- 1, 2)$
  1 mark
2. (ii)(a) From the graph, A(3, 4) and C(−1, −2). $A C = (- 1 - 3)^{2} + (- 2 - 4)^{2} = 16 + 36$
  1 mark
3. (ii)(a) $A C = 52 = 213$ units
  1 mark
4. (ii)(b) From the graph, A(3, 4) and B(3, 2), with $m : n = 1 : 3$ . By the section formula, the point is $(\frac{1 \times 3 + 3 \times 3}{1 + 3}, \frac{1 \times 2 + 3 \times 4}{1 + 3})$
  1 mark
5. (ii)(b) $= (\frac{12}{4}, \frac{14}{4}) = (3, \frac{7}{2})$
  1 mark
6. (iii) Reading from the graph, D is at $(- 2, - 5)$ .
  1 mark
Answer: (i) (−1, 2); (ii)(a) $52 = 213$ units; (ii)(b) $(3, 7/2)$ ; (iii) D(−2, −5)
Where the marks are
- (i) Mid-point of FG = (−1, 2) (1 mark)
- (ii)(a) Correct substitution in the distance formula (1 mark)
- (ii)(a) $A C = 213$ (1 mark)
- (ii)(b) Correct substitution in the section formula (1 mark)
- (ii)(b) Point $(3, 7/2)$ (1 mark)
- (iii) D(−2, −5) (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions
Treasure Hunt is an exciting and adventurous game where participants follow a series of clues/numbers/maps to discover hidden treasures. Players engage in a thrilling quest, solving puzzles and riddles to unveil the location of the coveted prize.
While playing a treasure hunt game, some clues (numbers) are hidden in various spots collectively forming an A.P. If the number on the $n^{t h}$ spot is $20 + 4 n$ , then answer the following questions to help the players in spotting the clues :
1. (i)
  Which number is on first spot ?
  1 mark
2. (ii)
  (a) Which spot is numbered as 112 ?
  2 marks
3. OR
  (ii)
  (b) What is the sum of all the numbers on the first 10 spots ?
  2 marks
4. (iii)
  Which number is on the $(n -2)^{t h}$ spot ?
  1 mark
Try it first, then show our solutionSolution
1. (i) $a_{n} = 20 + 4 n$ , so $a_{1} = 20 + 4 \times 1 = 24$
  1 mark
2. (ii)(a) Put $a_{n} = 112$ : $20 + 4 n = 112$
  1 mark
3. (ii)(a) $4 n = 92 \Rightarrow n = 23$ , so the 23rd spot is numbered 112.
  1 mark
4. (ii)(b) $a_{2} = 28$ , so $d = a_{2} - a_{1} = 28 - 24 = 4$
  ½ mark
5. (ii)(b) $S_{10} = \frac{10}{2} [2 \times 24 + (10 - 1) \times 4] = 5 (48 + 36)$
  1 mark
6. (ii)(b) $S_{10} = 420$
  ½ mark
7. (iii) $a_{n - 2} = 20 + 4 (n - 2) = 4 n + 12$
  1 mark
Answer: (i) 24; (ii)(a) 23rd spot; (ii)(b) 420; (iii) $4 n + 12$
Where the marks are
- (i) First spot number 24 (1 mark)
- (ii)(a) Sets up $20 + 4 n = 112$ (1 mark)
- (ii)(a) Gets n = 23 (1 mark)
- (ii)(b) Finds d = 4 (½ mark)
- (ii)(b) Correct use of $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ (1 mark)
- (ii)(b) Gets $S_{10} = 420$ (½ mark)
- (iii) $(n - 2)$ th term $= 4 n + 12$ (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes
Tamper-proof tetra-packed milk guarantees both freshness and security. This milk ensures uncompromised quality, preserving the nutritional values within and making it a reliable choice for health-conscious individuals.
500 mL milk is packed in a cuboidal container of dimensions 15 cm $\times$ 8 cm $\times$ 5 cm. These milk packets are then packed in cuboidal cartons of dimensions 30 cm $\times$ 32 cm $\times$ 15 cm.
Based on the above given information, answer the following questions :
1. (i)
  Find the volume of the cuboidal carton.
  1 mark
2. (ii)
  (a) Find the total surface area of a milk packet.
  2 marks
3. OR
  (ii)
  (b) How many milk packets can be filled in a carton ?
  2 marks
4. (iii)
  How much milk can the cup (as shown in the figure) hold ?
  1 mark
Figure in the paper: Three images: a plain cuboidal carton (unlabelled), a tetra-pack milk packet labelled MILK, and a cylindrical cup. The cup is drawn with its height labelled 7 cm (height of the cup, with a handle on the right side) and its base labelled 5 cm. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Volume of carton $= l \times b \times h = 30 \times 32 \times 15$
  ½ mark
2. (i) $= 14400 cm^{3}$
  ½ mark
3. (ii)(a) TSA of the packet $= 2 (l b + bh + h l) = 2 (15 \times 8 + 8 \times 5 + 5 \times 15) = 2 (120 + 40 + 75)$
  1 mark
4. (ii)(a) $= 470 cm^{2}$
  1 mark
5. (ii)(b) Volume of one packet $= 15 \times 8 \times 5 = 600 cm^{3}$ , so the number of packets $= \frac{30 \times 32 \times 15}{15 \times 8 \times 5} = \frac{14400}{600}$
  1 mark
6. (ii)(b) $= 24$ packets. They fit exactly, as $2 \times 4 \times 3$ along the 30 cm, 32 cm and 15 cm edges.
  1 mark
7. (iii) The cup is a cylinder with radius 5 cm and height 7 cm. Volume $= π r^{2} h = \frac{22}{7} \times 5 \times 5 \times 7$
  ½ mark
8. (iii) $= 550 cm^{3} = 550$ mL
  ½ mark
Answer: (i) 14400 cm³; (ii)(a) 470 cm²; (ii)(b) 24 packets; (iii) 550 cm³ (550 mL)
Where the marks are
- (i) Uses $l \times b \times h$ (½ mark) and gets 14400 cm³ (½ mark)
- (ii)(a) Uses $2 (l b + bh + h l)$ with correct values (1 mark)
- (ii)(a) Gets 470 cm² (1 mark)
- (ii)(b) Divides the carton volume by the packet volume (1 mark)
- (ii)(b) Gets 24 packets (1 mark)
- (iii) Uses $π r^{2} h$ with r = 5, h = 7 (½ mark)
- (iii) Gets 550 cm³ or 550 mL (½ mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.