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Mathematics Standard – Class X (2025–26)

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Mathematics Standard – Class X (2025–26) > Question Paper

Question Paper

Maximum Marks: 80 | Time: 3 Hours

  1. This question paper contains 38 questions. All questions are compulsory.

  2. This question paper is divided into 5 Sections — A, B, C, D and E.

  3. Section A: Question numbers 1–18 are MCQs and questions 19 and 20 are Assertion–Reason based questions of 1 mark each.

  4. Section B: Question numbers 21–25 are Very Short Answer (VSA) type questions carrying 2 marks each.

  5. Section C: Question numbers 26–31 are Short Answer (SA) type questions carrying 3 marks each.

  6. Section D: Question numbers 32–35 are Long Answer (LA) type questions carrying 5 marks each.

  7. Section E: Question numbers 36–38 are Case Study–based questions carrying 4 marks each with sub-parts of 1, 1 and 2 marks respectively.

  8. There is no overall choice. However, an internal choice in 2 questions of Section B, 2 questions of Section C and 2 questions of Section D has been provided. Internal choice has been provided in all 2-mark questions of Section E.

  9. Draw neat and clean figures wherever required. Take π = 22/7 wherever required if not stated.

  10. Use of calculators is not allowed.

SECTION A: Multiple Choice Questions (MCQs)

Questions 1-18: 1 mark each

MARKING SCHEME (Total: 1 mark) Correct Answer: (C) 3 [1 mark] SOLUTION: LCM(a, b, c) = 2² × 3ˣ × 5 × 7 = 3780 140 × 3ˣ = 3780 3ˣ = 27 = 3³ ∴ x = 3 Answer:C MARKING SCHEME (Total: 1 mark) Correct Answer: (A) 2 [1 mark] SOLUTION: The shortest distance from any point to the y-axis is the perpendicular distance For a point (x, y), the shortest distance from y-axis = |x| As shortest distance from (2, 3) to y-axis is the x coordinate, i.e., 2. Answer: A MARKING SCHEME (Total: 1 mark) Correct Answer: (B) k ≠ 15/4 [1 mark] SOLUTION: For parallel lines: a₁/a₂ = b₁/b₂ 3/2 ≠ 2k/5, hence k ≠ 15/4 Answer: B MARKING SCHEME (Total: 1 mark) Correct Answer: (C) 6 cm [1 mark] SOLUTION: Property: AB + CD = AD + BC (opposite sides sum equal in circumscribed quadrilateral) AB + 4 = 3 + 7 AB = 6 cm Answer: C MARKING SCHEME (Total: 1 mark) Correct Answer: (D) 1/x [1 mark] SOLUTION: 1/(secθ + tanθ) = (secθ - tanθ)/[(secθ + tanθ)(secθ - tanθ)] = (secθ - tanθ)/1 = secθ - tanθ So secθ - tanθ = 1/x Answer: D MARKING SCHEME (Total: 1 mark) Correct Answer: (D) [1 mark] SOLUTION: (D) (x+2)(x+1) = x²+2x+3 x²+3x+2 = x²+2x+3 gives x-1=0 It's not a quadratic equation (it's linear). Answer: D MARKING SCHEME (Total: 1 mark) Correct Answer: (D) 8[π/6 − √3/4] cm² [1 mark] SOLUTION: Required Area = 8 × area of one segment (with r = 1cm and θ = 60°) = 8 × (60°/360° × π × 1² − √3/4 × 1²) = 8[π/6 − √3/4] cm² Answer: D MARKING SCHEME (Total: 1 mark) Correct Answer: (B) 31/36 [1 mark] SOLUTION: Probability of getting sum 8 is 5/36 Probability of not getting sum 8 is 31/36 Answer: B MARKING SCHEME (Total: 1 mark) Correct Answer: (B) 12° [1 mark] SOLUTION: sin 5x = √3/2 So, 5x = 60° And hence x = 12° Answer: B MARKING SCHEME (Total: 1 mark) Correct Answer: (C) 4 [1 mark] SOLUTION: Since HCF=81, the numbers can be 81x and 81y 81x + 81y = 1215 x + y = 15 which gives four coprime pairs as (1,14), (2,13), (4,11), (7,8) Answer: C MARKING SCHEME (Total: 1 mark) Correct Answer: (D) 5 cm [1 mark] SOLUTION: πr² = 51 V = (1/3) × πr² × h 85 = (1/3) × 51 × h h = 85/17 = 5 cm Answer: D MARKING SCHEME (Total: 1 mark) Correct Answer: (D) c and a have same signs [1 mark] SOLUTION: For equal roots to the corresponding equation: b² = 4ac Hence ac = b²/4 And hence ac > 0 ⇒ c and a must have same signs Answer: D MARKING SCHEME (Total: 1 mark) Correct Answer: (C) 231 [1 mark] SOLUTION: Area of sector = (1/2) × l × r = (1/2) × 22 × 21 = 231 cm² Answer: C MARKING SCHEME (Total: 1 mark) Correct Answer: (C) 18 cm [1 mark] SOLUTION: ΔABC ~ ΔDEF AB/DE = BC/EF = AC/DF = Perimeter of ΔABC / Perimeter of ΔDEF 6/9 = Perimeter of ΔABC / 27 Perimeter of ΔABC = 18 cm Answer: C MARKING SCHEME (Total: 1 mark) Correct Answer: (B) 9/4 [1 mark] SOLUTION: Probability of getting vowels in the word Mathematics is 4/11 So, 2/(2x+1) = 4/11 ⇒ x = 9/4 Answer: B MARKING SCHEME (Total: 1 mark) Correct Answer: (C) Parallelogram [1 mark] SOLUTION: By visualising the figure by plotting points in co-ordinate plane, it can be concluded it is a Parallelogram. Answer: C MARKING SCHEME (Total: 1 mark) Correct Answer: (A) median is increased by 2 [1 mark] SOLUTION: Adding constant to all values increases median by same constant New median = 20.5 + 2 = 22.5 (increases by 2) Answer: A MARKING SCHEME (Total: 1 mark) Correct Answer: (A) 40 cm [1 mark] SOLUTION: Since, tangent is perpendicular to the radius at the point of contact In ΔOPT, right angled at T OP² = OT² + TP² 41² = 9² + TP² TP² = 1681 - 81 = 1600 TP = 40 cm Answer: A

DIRECTIONS: In the question number 19, a statement of Assertion (A) is followed by a statement of Reason (R).

Choose the correct option:

(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

(B) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

(C) Assertion (A) is true but reason (R) is false.

(D) Assertion (A) is false but reason (R) is true.

MARKING SCHEME (Total: 1 mark) Correct Answer: (a) Both A and R are true, R is correct explanation of A [1 mark] SOLUTION: R is TRUE (ending with 0 needs factors 2 and 5) A is TRUE (5ⁿ has only 5s, no 2s, so cannot end with 0) R correctly explains A Answer: A

DIRECTIONS: In the question number 20, a statement of Assertion (A) is followed by a statement of Reason (R).

Choose the correct option:

(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

(B) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

(C) Assertion (A) is true but reason (R) is false.

(D) Assertion (A) is false but reason (R) is true.

MARKING SCHEME (Total: 1 mark) Correct Answer: (a) Both A and R are true, R is correct explanation of A [1 mark] SOLUTION: cosA + cos²A = 1 ---------(i) gives cos A = sin²A ------(ii) (using sin²A + cos²A = 1) Substituting value of cos A from (ii) in (i) sin²A + sin⁴A = 1 ∴ Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion Answer: A

SECTION B: Very Short Answer Questions (VSA)

Questions 21-25: 2 marks each

 MARKING SCHEME (Total: 2 marks) Question 21A: Question 21A METHOD 1: Using a₅₁ as the first term • n = 60, a = 8, d = 2 [½ mark] • t₅₁ = 8 + 50(2) = 108 [½ mark] • S₁₀ = (10/2)[2(108) + 9(2)] correctly evaluated [½ mark] • Final answer: Sum = 1170 [½ mark] Question 21A METHOD 2: Using difference of sums (S₆₀ - S₅₀) • n = 60, a = 8, d = 2 identified [½ mark] • S₆₀ = (60/2)[2(8) + 59(2)] = 4020 calculated correctly [½ mark] • S₅₀ = (50/2)[2(8) + 49(2)] = 2850 calculated correctly [½ mark] • S₁₀ = S₆₀ - S₅₀ = 1170 [½ mark] Question 21A METHOD 3: Using a₆₀ and working backwards • n = 60, a = 8, d = 2 identified [½ mark] • a₆₀ = 8 + 59(2) = 126 calculated correctly [½ mark] • S₁₀ = (10/2)[2(126) + 9(-2)] with correct setup [½ mark] • Final answer: Sum = 1170 [½ mark] Note: Student can use ANY ONE method. Award full marks if any single method is correctly completed. Question 21A SOLUTION: Given AP = 8, 10, 12, … First term a = 8, common difference d = 2 Total terms = 60 Question 21A METHOD 1: Find the 51st term (first of last 10 terms) a₅₁ = a + 50d = 8 + 50(2) = 8 + 100 = 108 Apply sum formula for last 10 terms S₁₀ = (n/2)[2a₅₁ + (n-1)d] where n = 10 (last 10 terms) Substitute values S₁₀ = (10/2)[2(108) + 9(2)] = 5[216 + 18] = 5 × 234 = 1170 Question 21A METHOD 2: Find S₆₀ and S₅₀ and subtract them S₆₀ = (60/2)[2(8) + 59(2)] = 30[16 + 118] = 30 × 134 = 4020 S₅₀ = (50/2)[2(8) + 49(2)] = 25[16 + 98] = 25 × 114 = 2850 S₁₀ = S₆₀ - S₅₀ = 4020 - 2850 = 1170 Question 21A METHOD 3: Find last term a₆₀ and work backwards with d=-2 a₆₀ = a + 59d = 8 + 59(2) = 8 + 118 = 126 For last 10 terms: first term = 126, d = -2, n = 10 S₁₀ = (10/2)[2(126) + 9(-2)] = 5[252 - 18] = 5 × 234 = 1170 Answer: Sum of last 10 terms = 1170 Note: User can do any one method out of the three mentioned above. --------------------------------------------- Question 21B: Question 21B Step-by-step marking: • 230 = 6 + (n-1)7 gives n = 33 [1 mark] • ∴ Middle Term = t₁₇ = 6 + (16)(7) = 118 [1 mark] Question 21B SOLUTION: Given AP = 6, 13, 20, …, 230 First term a = 6, common difference d = 7 Last term l = 230 Find number of terms aₙ = a + (n-1)d 230 = 6 + (n-1)7 224 = (n-1)7 n - 1 = 32 n = 33 Find middle term position For odd number of terms, middle term = (n+1)/2 th term Middle term position = (33+1)/2 = 17th term Calculate 17th term a₁₇ = a + 16d = 6 + 16(7) = 6 + 112 = 118 Answer: Middle term = 118 MARKING SCHEME (Total: 2 marks) Step-by-step marking: • A+B = 90° and A - B = 30° [1 mark] • A=60° and B =30° [1 mark] SOLUTION: Given: sin(A + B) = 1 and cos(A - B) = √3/2 Constraint: 0° < A, B < 90° Solve sin(A + B) = 1 sin(A + B) = 1 but sin(90°) = 1 A + B = 90° ... (1) Solve cos(A - B) = √3/2 cos(A - B) = √3/2 but cos(30°) = √3/2 A - B = 30° ... (2) Add equations (1) and (2) (A + B) + (A - B) = 90° + 30° 2A = 120° A = 60° Subtract equation (2) from (1) (A + B) - (A - B) = 90° - 30° 2B = 60° B = 30° Answer: A = 60°, B = 30° MARKING SCHEME (Total: 2 marks) Step-by-step marking: • △ABC ~ △DEF ⇒ AB/DE = BC/EF [½ mark] • AB/DE = 2BP/2EQ (AP and DQ are the medians) [½ mark] • In △ABP and △DEQ: AB/DE = BP/EQ and ∠B = ∠E ⇒ △ABP ~ △DEQ [½ mark] • Hence, AB/DE = AP/DQ [½ mark] SOLUTION: Given: ΔABC ~ ΔDEF AP and DQ are medians of ΔABC and ΔDEF respectively To Prove: AB/DE = AP/DQ Proof user has to draw two triangles ABC and DEF and mark two points P and Q as midpoints of BC and EF respectively. Also user has to join AP and DQ. Since ΔABC ~ ΔDEF ∠A = ∠D, ∠B = ∠E, ∠C = ∠F AB/DE = BC/EF = CA/FD ... (1) Since AP is a median of ΔABC BC=2BP Since DQ is a median of ΔDEF EF=2EQ From (1), BC/EF = 2BP/2EQ ... (2) In ΔABP and ΔDEQ AB/DE = BC/EF (from step 1) BP/EQ = 2BP/2EQ= BP/EQ (from step 2) ∠B = ∠E (corresponding angles) By SAS similarity ΔABP ~ ΔDEQ Therefore, AB/DE = BP/EQ = AP/DQ Hence proved: AB/DE = AP/DQ Consider if your uses a different method but solution is correct. MARKING SCHEME (Total: 2 marks) Question24A Question24A:Step-by-step marking: • Area of grass field that can be grazed = (θ₁/360°)×πr² + (θ₂/360°)×πr² + (θ₃/360°)×πr² • = (πr²/360°) (θ₁ + θ₂ + θ₃) = (πr²/360°) × 180° • = 1/2 × 22/7 × 14 × 14 = 308 m² [2 marks] SOLUTION: Given: Triangular field ABC, Rope length r = 14m. user has to draw a triangle ABC and mark three points A, B and C. user has to indicate an arc near angle A, B and C where they have to indicate the distance from point A or B or C touching the triangle sides must be equal to the rope length-14m. Area grazed = Area of 3 sectors with angles A, B, C Total Area = (A/360 + B/360 + C/360) × πr² = ((A+B+C)/360) × πr² = (180/360) × (22/7) × (14)² = (1/2) × 22 × 14 x 2= 308 m² Answer: Area grazed = 308 m² In the above solution user can use anything in place of A,B,C. So consider it and give full marks. ------------------------------ Question24B Question24B:Step-by-step marking: • Area of sector = (90°/360°) × πr² = 25π/4 cm² [½ mark] • Area of triangle = (1/2) × r² = 25/2 cm² [½ mark] • Area of minor segment = (25π/4 - 25/2) cm² [Full 1 mark if above two correct] • Area of major segment = 25π - (25π/4 - 25/2) = (75π/4 + 25/2) cm² [1 mark] Alternate marking distribution: • Finding area of minor segment correctly [1 mark] - Area of sector calculation [½ mark] - Subtracting area of triangle [½ mark] • Finding area of major segment correctly [1 mark] - Using: Area of circle - Area of minor segment SOLUTION: Given: Radius r = 5 cm, angle at center θ = 90° user has to draw a circle with radius mentioned as 5cm and mark a point O as center. draw a chord and join a chord to center O. Mark the angle at center as 90°. They have to shade the major segment. Find area of the major segment = Area of circle - Area of minor segment. Area of circle = πr² = (22/7) × 5² = 25π Area of circle = 25π cm² STEP 1: Find Area of minor segment Area of minor segment = Area of sector - Area of triangle Area of sector = (θ/360°) × πr² = (90°/360°) × π × 5² = (1/4) × 25π = 25π/4 cm² Triangle formed is right-angled (since θ = 90°) Area of triangle = (1/2) × r × r = (1/2) × 5 × 5 = 25/2 cm² Area of minor segment = Area of sector - Area of triangle = 25π/4 - 25/2 = (25π - 50)/4 cm² STEP 2: Find Area of major segment Area of major segment = Area of circle - Area of minor segment = 25π - (25π/4 - 25/2) = 25π - 25π/4 + 25/2 = (100π - 25π)/4 + 25/2 = 75π/4 + 25/2 = (75π + 50)/4 = 25(3π + 2)/4 cm² Final Answer: Area of major segment = 25(3π + 2)/4 cm² or (75π/4 + 25/2) cm² Note: Accept any equivalent form: • (75π + 50)/4 cm² • 25(3π + 2)/4 cm² • (75π/4 + 25/2) cm² MARKING SCHEME (Total: 2 marks) Step-by-step marking: Let r be the radius of the inscribed circle BD = BE = 10cm, CD = CF = 8cm, Let AF = AE = x [½ mark] ar(△ABC) =ar(△AOC) + ar(△BOC) + ar(△AOB)=(1/2)rAC+(1/2)rBC+(1/2)rAB [½ mark] 90 =(1/2)×4(𝑥+8+18+𝑥+10) x = 4.5 cm [½ mark] ∴ AB = 5+10 = 14.5 cm, AC = 4.5+8 = 12.5 cm [½ mark] Alternate marking distribution: • Using tangent property correctly and setting up x [½ mark] • Setting up area equation with all three triangles [½ mark] • Solving for x correctly [½ mark] • Finding AB and AC correctly [½ mark] SOLUTION: User must draw the diagram with a circle inside the triangle labeling A,B,C,D. Marking center O and points of contact E, F on AC and AB respectively. Join OA, OB, OC. Given: Circle with radius r = 4 cm inscribed in ΔABC BD = 10 cm, DC = 8 cm, Area of ΔABC = 90 cm² STEP 1: Use tangent property Tangents from external point are equal So, BD = BE = 10 cm DC = CF = 8 cm Let AE = AF = x cm Therefore AB = AF + FB = x + 10 BC = BD + DC = 10 + 8 = 18 AC = AE + EC = x + 8 STEP 2: Express area using sum of three triangles Area of ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC) Since OD ⊥ BC, OE ⊥ AC, OF ⊥ AB (radius ⊥ tangent at point of contact) and OD = OE = OF = r = 4 cm = (1/2) × AB × OF + (1/2) × BC × OD + (1/2) × AC × OE = (1/2) × (x+10) × 4 + (1/2) × 18 × 4 + (1/2) × (x+8) × 4 = (1/2) × 4 [(x+10) + 18 + (x+8)] = 2[x + 10 + 18 + x + 8] = 2[2x + 36] = 4x + 72 STEP 3: Equate to given area Area of ΔABC = 90 cm² 4x + 72 = 90 4x = 18 x = 4.5 cm STEP 4: Calculate AB and AC AB = x + 10 = 4.5 + 10 = 14.5 cm AC = x + 8 = 4.5 + 8 = 12.5 cm Answer: AB = 14.5 cm, AC = 12.5 cm

SECTION C: Short Answer Questions (SA)

Questions 26-31: 3 marks each

MARKING SCHEME (Total: 3 marks) Step-by-step marking: • In ΔAPO and ΔACO: AP=AC (Tangents), AO=AO (common), OP=OC (radii) ∴ ΔAPO ≅ ΔACO [1 mark] • ∠POQ=180° (PQ is diameter), ∠POA+∠COA+∠QOB+∠COB=180° 2∠COA+2∠COB=180° [1 mark] • ∴ ∠AOB = 90° [1 mark] SOLUTION: In ΔOPA and ΔOCA OP = OC (Radii of the circle) AP = AC (Tangents from external point A are equal) AO = AO (Common) By SSS congruency, ΔOPA ≅ ΔOCA Therefore, ∠POA = ∠AOC ... (1) Similarly, in ΔOCB and ΔOQB OC = OQ (Radii of the circle) BC = BQ (Tangents from external point B are equal) OB = OB (Common) By SSS congruency, ΔOCB ≅ ΔOQB Therefore, ∠COB = ∠BOQ ... (2) PQ is a diameter, hence a straight line and ∠POQ = 180° ∠POQ = ∠POA + ∠AOC + ∠COB + ∠BOQ ∠POA + ∠AOC + ∠COB + ∠BOQ = 180° Using equations (1) and (2): 2∠AOC + 2∠COB = 180° 2(∠AOC + ∠COB) = 180° ∠AOC + ∠COB = 90° From figure, ∠AOC + ∠COB = ∠AOB Therefore, ∠AOB = 90° Hence Proved: ∠AOB = 90° MARKING SCHEME (Total: 3 marks) Step-by-step marking: • HCF (36,60,84) =12 [1½ marks] • Required number of rooms = 36/12 + 60/12 + 84/12 [1 mark] • =15 [½ mark] SOLUTION: Given: English teachers = 36 Hindi teachers = 60 Science teachers = 84 Condition: Same number of teachers per room, all from same subject We need to find maximum number of teachers per room This is the HCF of 36, 60, and 84 Find HCF using prime factorization 36 = 2² × 3² 60 = 2² × 3 × 5 84 = 2² × 3 × 7 The highest number that divides all given numbers without leaving a remainder. HCF(36, 60, 84) = 2² × 3 = 12 Calculate number of rooms for each subject. Rooms for English = 36/12 = 3 Rooms for Hindi = 60/12 = 5 Rooms for Science = 84/12 = 7 Total minimum rooms = 3 + 5 + 7 = 15 Answer: Minimum number of rooms required = 15 MARKING SCHEME (Total: 3 marks) Step-by-step marking: • Factorization: (2x - 1)(x - √2) [1 mark] • Zeroes are 1/2 and √2 [1 mark] • Verification of sum and product -Sum of zeroes = -b/a and verification [½ mark] -Product of zeroes = c/a and verification [½ mark] SOLUTION: Given: f(x) = 2x² - (1 + 2√2)x + √2 Factorize the polynomial 2x² - (1 + 2√2)x + √2 = 2x² - x - 2√2x + √2 = x(2x - 1) - √2(2x - 1) = (2x - 1)(x - √2) Find zeroes f(x) = 0 (2x - 1)(x - √2) = 0 Either 2x - 1 = 0 or x - √2 = 0 x = 1/2 or x = √2 Zeroes are α = 1/2 and β = √2 Verify relationship (Sum of zeroes) Sum of zeroes = α + β = 1/2 + √2 = (1 + 2√2)/2 From polynomial: a=2, b=-(1+2√2), c=√2, Sum = -b/a = (1 + 2√2)/2 Verify relationship (Product of zeroes) Product of zeroes = α × β = 1/2 × √2 = √2/2 From polynomial: Product = c/a = √2/2 Hence, the relationship between zeroes and coefficients is verified. MARKING SCHEME (Total: 3 marks) Question 29A: Question 29A:Step-by-step marking: • (sinθ + cosθ)² = 3 gives 1 + 2sinθcosθ = 3 [1 mark] • ⇒ sinθcosθ = 1 [1 mark] • ∴ tanθ + cotθ = 1/(sinθcosθ) = 1 [1 mark] (here user may give additional steps in between steps mentioned in the marking scheme) Question 29A:SOLUTION: Given: sinθ + cosθ = √3 To Prove: tanθ + cotθ = 1 Proof Square both sides of given equation (sinθ + cosθ)² = (√3)² sin²θ + 2sinθcosθ + cos²θ = 3 Use sin²θ + cos²θ = 1 1 + 2sinθcosθ = 3 2sinθcosθ = 2 sinθcosθ = 1 ... (1) Express tanθ + cotθ interms of sin and cos tanθ + cotθ = sinθ/cosθ + cosθ/sinθ = (sin²θ + cos²θ)/(sinθcosθ) Substitute known values = 1/(sinθcosθ) = 1/1 (from equation 1) = 1 Hence proved: tanθ + cotθ = 1 ----------------- Question 29B: Question 29B:Step-by-step marking: METHOD 1: • Dividing by sinA and simplification [1 mark] • Using identity cosec²A - cot²A = 1 [1 mark] • Final simplification to cosecA + cotA [1 mark] METHOD 2: Multiplying by (cosA + sinA + 1) • Multiplying numerator and denominator by (cosA + sinA + 1) and using a² - b² identity for denominator [1 mark] • Simplifying denominator: (cosA + sinA)² - 1 = 2sinAcosA using sin²A + cos²A = 1 [1 mark] • Simplifying numerator and final result to get cosecA + cotA [1 mark] Question 29B:SOLUTION: To Prove: (cosA - sinA + 1)/(cosA + sinA - 1) = cosecA + cotA Proof Take LHS and divide numerator and denominator by sinA LHS = (cosA - sinA + 1)/(cosA + sinA - 1) = [(cosA - sinA + 1)/sinA] / [(cosA + sinA - 1)/sinA] Simplify = (cosA/sinA - 1 + 1/sinA) / (cosA/sinA + 1 - 1/sinA) = (cotA - 1 + cosecA) / (cotA + 1 - cosecA) = (cotA + cosecA - 1) / (cotA - cosecA + 1) Use identity cosec²A - cot²A = 1 cosec²A - cot²A = 1 (cosecA - cotA)(cosecA + cotA) = 1 cosecA - cotA = 1/(cosecA + cotA) ... (i) Multiply and divide by (cosecA + cotA) LHS = (cotA + cosecA - 1) / (cotA - cosecA + 1) = (cotA + cosecA - 1) × (cosecA + cotA) / [1 × (cosecA + cotA)] Simplify numerator Numerator = (cotA + cosecA - 1)(cosecA + cotA) Let x = cotA + cosecA = (x - 1)x = x² - x From equation (i) cotA - cosecA + 1 = 1/(cosecA + cotA) Denominator = cosecA + cotA Substitute values = (x - 1)/x = (x² - x)/x = x - 1 Hence proved: (cosA - sinA + 1)/(cosA + sinA - 1) = cosecA + cotA Question 29B: Alternative approach = (cosA - sinA + 1)/(cosA + sinA - 1) Multiply num. and den. by (cosA + sinA + 1) = [(cosA - sinA + 1)(cosA + sinA + 1)] / [(cosA + sinA - 1)(cosA + sinA + 1)] Denominator = (cosA + sinA)² - 1 (since a²-b²=(a-b)(a+b)) = cos²A + sin²A + 2cosAsinA - 1 (sin²θ + cos²θ = 1) = 1 + 2cosAsinA - 1 = 2cosAsinA Numerator = (cosA + 1)² - sin²A = cos²A + 2cosA + 1 - sin²A = 2cosA + cos²A + cos²A = 2cosA(1 + cosA) = 2cosA(1 + cosA) / (2sinAcosA) = (1 + cosA)/sinA = 1/sinA + cosA/sinA = cosecA + cotA = RHS Hence proved. MARKING SCHEME (Total: 3 marks) Step-by-step marking: • P(Vidhi drives the car) = 3/8 as favourable outcomes are HHT,THH,HHH [1 mark] • P(Unnati drives the car) = 4/8 as favourable outcomes are HHT,HTT,THH,THT [1 mark] • As 4/8 > 3/8, Unnati has greater probability [1 mark] SOLUTION: Total outcomes when flipping a coin 3 times = 2³ = 8 Sample Space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} For Vidhi (Two heads in a row) Favorable outcomes = {HHH, HHT, THH} Number of favorable outcomes = 3 Probability P(Vidhi) = 3/8 For Unnati (Head immediately followed by Tail) Favorable outcomes = {THH, HTH, HTT, THT} Number of favorable outcomes = 4 Probability P(Unnati) = 4/8 = 1/2 Comparison Since 4/8 > 3/8, P(Unnati) > P(Vidhi) Answer: Unnati has the greater probability to drive the car. MARKING SCHEME (Total: 3 marks) Question 31A: Question 31A Step-by-step marking: • Let income be 3x and 4x, expenditure be 5y and 7y [1 mark] • Forming equations: 3x - 5y = 15000, 4x - 7y = 15000 and solving [1 mark] - Any valid method (substitution/elimination/cross-multiplication) - Award ½ mark for correct equation formation - Award ½ mark for correct solving process to get x = 30000 • Their incomes = ₹90,000 and ₹1,20,000 respectively [1 mark] Question 31A SOLUTION: Given: Income ratio = 3:4 Expenditure ratio = 5:7 Savings of each = ₹15,000 Express incomes Let Aryan's income = 3x Let Babban's income = 4x Express expenditures Let Aryan's expenditure = 5y Let Babban's expenditure = 7y Use savings formula (Income - Expenditure = Savings) For Aryan: 3x - 5y = 15,000 ... (1) For Babban: 4x - 7y = 15,000 ... (2) Solve equations Multiply equation (1) by 7: 21x - 35y = 105,000 ... (3) Multiply equation (2) by 5: 20x - 35y = 75,000 ... (4) Subtract (4) from (3) 21x - 20x = 105,000 - 75,000 x = 30,000 Note: Here user can use any method to solve the equations. Find incomes Aryan's income = 3x = 3 × 30,000 = ₹90,000 Babban's income = 4x = 4 × 30,000 = ₹1,20,000 Answer: Aryan's income = ₹90,000, Babban's income = ₹1,20,000 ----------------- Question 31B: Question 31B Step-by-step marking: • Correct graph with both lines plotted accurately [2 marks] - First line 2x + y = 6 with at least 3 points plotted correctly [1 mark] - Second line 2x - y = 2 with at least 3 points plotted correctly [1 mark] - Lines should be straight and properly labeled - Award full 2 marks if graph shows correct intersection point • Hence, the solution is x = 2, y = 2 [½ mark] • Area = 2 sq. units [½ mark] Question 31B Detailed marking breakdown: • Graph plotting [2 marks total] - Table of values for first equation (at least 3 points) [½ mark] - Plotting first line accurately [½ mark] - Table of values for second equation (at least 3 points) [½ mark] - Plotting second line accurately [½ mark] - If graph shows correct intersection clearly, award full 2 marks • Solution from graph [½ mark] - Reading intersection point correctly as (2, 2) - Stating x = 2, y = 2 • Area calculation [½ mark] - Identifying triangle vertices correctly - Correct area formula application - Final answer: 2 sq. units Question 31B SOLUTION: Given equations: 2x + y = 6 ... (1) 2x - y - 2 = 0 or 2x - y = 2 ... (2) STEP 1: Create table of values for equation (1): 2x + y = 6 Rewrite as: y = 6 - 2x x=0,y=6 x=1,y=4 x=3,y=0 Points-(0,6) , (1,4) , (3,0) STEP 2: Create table of values for equation (2): 2x - y = 2 Rewrite as: y = 2x - 2 x=0,y=-2 x=1,y=0 x=2,y=2 Points-(0,-2) , (1,0) , (2,2) STEP 3: Plot the graph User should have: • Drawn x-axis and y-axis with proper scale • Plotted points for first equation: (0,6), (1,4), (3,0) • Drawn a straight line through these points and labeled as "2x + y = 6" • Plotted points for second equation: (0,-2), (1,0), (2,2) • Drawn a straight line through these points and labeled as "2x - y = 2" • Marked the point of intersection clearly STEP 4: Find the solution from graph From the graph, the two lines intersect at point A(2, 2) Hence, the solution is x = 2, y = 2 STEP 5: Find the area of triangle formed by two lines and x-axis Triangle vertices are: A(2, 2), B(3, 0), C(1, 0) Method 1: Using base and height Base BC = distance between B(3,0) and C(1,0) = 3 - 1 = 2 units Height = perpendicular distance from A to x-axis = y-coordinate of A = 2 units Area = ½ × base × height = ½ × 2 × 2 = 2 sq. units

SECTION D: Long Answer Questions (LA)

Questions 32-35: 5 marks each

MARKING SCHEME (Total: 5 marks) Step-by-step marking: • Let the original speed of train be x km/hr Distance = 63 km, time(t₁) = 63/x hrs [½ mark] Faster speed = (x + 6) km/hr, time(t₂) = 72/(x+6) hrs [½ mark] Now t₁ + t₂ = 3 hrs So 63/x + 72/(x+6) = 3 [1 mark] 63(x+6) + 72x = 3x(x+6) 135x + 378 = 3x² + 18x 3x² - 117x - 378 = 0 x² - 39x - 126 = 0 [1 mark] x² - 42x + 3x - 126 = 0 gives (x + 3)(x - 42) = 0 [1 mark] As x can't be negative, so x = 42 km/hr The original speed of train = 42 km/hr [1 mark] SOLUTION: Given: Total journey time = 3 hours Part 1: Distance = 63 km, Speed = Original speed Part 2: Distance = 72 km, Speed = Original speed + 6 km/hr STEP 1: Define variables Let the original average speed of the train be x km/hr. Then, the new speed for the second part is (x + 6) km/hr. STEP 2: Express time taken for each part Time = Distance / Speed Time taken for first part (t₁) = 63/x hours Time taken for second part (t₂) = 72/(x + 6) hours STEP 3: Form the equation Total time = t₁ + t₂ = 3 hours 63/x + 72/(x + 6) = 3 STEP 4: Solve the equation Divide the entire equation by 3 to simplify: 21/x + 24/(x + 6) = 1 Take LCM: [21(x + 6) + 24x] / [x(x + 6)] = 1 21x + 126 + 24x = x(x + 6) 45x + 126 = x² + 6x Rearrange to form a quadratic equation: x² + 6x - 45x - 126 = 0 x² - 39x - 126 = 0 STEP 5: Factorize the quadratic equation Find two numbers whose product is -126 and sum is -39. Factors of 126: 1×126, 2×63, 3×42... Numbers are -42 and +3. x² - 42x + 3x - 126 = 0 x(x - 42) + 3(x - 42) = 0 (x - 42)(x + 3) = 0 STEP 6: Find roots x - 42 = 0 ⇒ x = 42 x + 3 = 0 ⇒ x = -3 STEP 7: Final Conclusion Since speed cannot be negative, we reject x = -3. Therefore, x = 42 km/hr. Answer: The original average speed of the train is 42 km/hr. MARKING SCHEME (Total: 5 marks) Step-by-step marking: • Correct given figure and construction [2 marks] • Correct Proof [2 marks] • Application: PL/LQ = PM/MR,5.7/9.5 = PM/5.5 solving gives PM = 3.3 cm [1 mark] SOLUTION: Statement: 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. Proof Given: In ΔABC, D is a point on AB and E is a point on AC such that DE ∥ BC. To Prove: AD/DB = AE/EC Construction: User should draw a triangle ABC and draw a line DE parallel to BC intersecting AB at D and AC at E. User should draw perpendiculars EN ⊥ AB and DM ⊥ AC. User should join BE and CD to form triangles ADE and DEC. Consider triangles ADE and BDE on the same altitude EN. ar(ΔADE) = (1/2) × AD × EN ar(ΔBDE) = (1/2) × DB × EN Taking ratio: ar(ΔADE)/ar(ΔBDE) = AD/DB ...(1) Consider triangles ADE and DEC on the same altitude DM. ar(ΔADE) = (1/2) × AE × DM ar(ΔDEC) = (1/2) × EC × DM Taking ratio: ar(ΔADE)/ar(ΔDEC) = AE/EC ...(2) Apply property of triangles between parallel lines. Since ΔBDE and ΔDEC are on the same base DE and between the same parallels DE ∥ BC: ar(ΔBDE) = ar(ΔDEC) ...(3) (Triangles on same base and between same parallels have equal areas) From equations (1), (2), and (3) Since ar(ΔBDE) = ar(ΔDEC), the denominators in equations (1) and (2) are equal. Therefore, AD/DB = AE/EC Hence proved. APPLICATION: Given: In ΔPQR, LM || QR PL = 5.7 cm, PQ = 15.2 cm, MR = 5.5 cm To Find: PM Find LQ = PQ - PL = 15.2 - 5.7 = 9.5 cm Using Basic Proportionality Theorem: PL/LQ = PM/MR Substituting values: 5.7/9.5 = PM/5.5 PM = (5.7/9.5) × 5.5 Ratio 5.7/9.5 = 57/95 = 3/5 = 0.6 PM = 0.6 × 5.5 PM = 3.3 cm Answer: PM = 3.3 cm MARKING SCHEME (Total: 5 marks) Question 34A: Question 34A Step-by-step marking: • Slant height of the cone L = √(R² + H²) = √(12² + 6²) = 3√20 cm [½ mark] • Curved Surface area of cone = πRL = π × 12 × 3√20 = (36√20)π cm² [1 mark] • Area of base circle of cone (= area of outer circle - area of inner circle + top circular area of cylinder) = 144π cm² [1 mark] • Curved Surface area of cylinder = 2πrh = 2π × 4 × 3 = 24π cm² [1 mark] • Surface area of the remaining solid = Curved surface of cone + area of base circle of cone + curved surface area of cylinder = (36√20)π + 144π + 24π = (168 + 36√20)π cm² [1½ marks] - Award 1 mark for correct formula/approach - Award ½ mark for correct final answer Question 34A Detailed marking breakdown: • Step 1: Finding slant height [½ mark] - Use formula: L = √(R² + H²) - Correct substitution and calculation - L = 3√20 cm or 6√5 cm • Step 2: Curved surface area of cone [1 mark] - Formula: CSA = πRL - Correct substitution: π × 12 × 3√20 - Answer: (36√20)π cm² or 72π√5 cm² • Step 3: Area of base circle [1 mark] - This is the circular ring + top of cylinder - Area of outer circle: πR² = π(12)² = 144π - Award full mark for identifying this correctly • Step 4: Curved surface area of cylinder [1 mark] - Formula: CSA = 2πrh - Correct substitution: 2π × 4 × 3 - Answer: 24π cm² • Step 5: Total surface area [1½ marks] - Correctly add all three components [1 mark] - Final simplification: (168 + 36√20)π cm² [½ mark] Question 34A SOLUTION: Given: Solid right circular cone • Height H = 6 cm • Base radius R = 12 cm Cylindrical cavity hollowed out • Height h = 3 cm • Radius r = 4 cm • Bases of cone and cylinder are concentric circles User should draw a cone and a cylinder of given dimensions such that: • A solid right circular cone, labeled height = 6cm and radius of base = 12cm • A right circular cylinder inside the cone, labeled height = 3cm and radius = 4cm • Bases of cone and cylinder form concentric circles • Shade the cylinder inside the cone to show it's hollowed out Find slant height of cone Formula: L = √(R² + H²) Where: R = radius of cone = 12 cm H = height of cone = 6 cm L = √(12² + 6²) = √(144 + 36) = √180 = √(36 × 5) = 6√5 cm = 3√20 cm [Both forms are acceptable] Slant height L = 3√20 cm = 6√5 cm ✓ Find Curved Surface Area of Cone Formula: CSA of cone = πRL Where: R = 12 cm L = 3√20 cm CSA of cone = π × R × L = π × 12 × 3√20 = 36π√20 cm² = (36√20)π cm² = 72π√5 cm² [Alternative form] Curved Surface Area of cone = (36√20)π cm² ✓ CSA of cone = π × R × L = π × 12 × 3√20 = 36π√20 cm² = (36√20)π cm² = 72π√5 cm² [Alternative form] Curved Surface Area of cone = (36√20)π cm² ✓ Area of base circle of cone This is the circular ring + top of cylinder Area of outer circle: πR² = π(12)² = 144π Area of base circle of cone = 144π cm² ✓ Area of base circle of cone = 144π cm² ✓ The base of the remaining solid consists of: • The circular ring (area between outer circle and inner circle) • The top circular area of the cylinder Total area = (Area of outer circle - Area of inner circle) + Area of top of cylinder = [πR² - πr²] + πr² = πR² = π(12)² = 144π cm² Question 34A Alternative explanation: Since the cylinder is removed, we have: • Outer circle of radius 12 cm (cone base) • Inner circle of radius 4 cm is removed BUT • Top of cylinder of radius 4 cm is added back • Net result: Full circle of radius 12 cm = πR² = 144π cm² Area of base circle = 144π cm² STEP 4: Find Curved Surface Area of Cylinder Formula: CSA of cylinder = 2πrh Where: r = radius of cylinder = 4 cm h = height of cylinder = 3 cm CSA of cylinder = 2πrh = 2 × π × 4 × 3 = 24π cm² Curved Surface Area of cylinder = 24π cm² STEP 5: Find Total Surface Area of Remaining Solid The surface area of the remaining solid consists of: 1. Curved surface of cone 2. Area of base circle of cone 3. Curved surface area of cylindrical cavity Total Surface Area = CSA of cone + Base area + CSA of cylinder TSA = (36√20)π + 144π + 24π TSA = (36√20 + 144 + 24)π TSA = (36√20 + 168)π cm² TSA = (168 + 36√20)π cm² Alternative form: TSA = (72√5 + 168)π cm² [Using 36√20 = 72√5] FINAL ANSWER Surface area of the remaining solid = (168 + 36√20)π cm² OR = (168 + 72√5)π cm² Both forms are acceptable. ---------------------------- Question 34B: Question 34B Step-by-step marking: • Volume of cone = (1/3)πr²h=(1/3) × π × (3)² × 12 = 36π cm³ [2 marks] • Volume of ice-cream in cone = (5/6) × 36π = 30π cm³ [Included above] • Volume of hemisphere = (2/3)πr³ = (2/3) × π × (3)³ = 18π cm³ [1½ marks] • Total = 48π ≈ 150.86 cm³ [1½ marks] Question 34B SOLUTION: Given: Cone - radius r = 3 cm, height h = 12 cm Empty portion at bottom = (1/6)th of cone volume Hemisphere on top with radius = 3 cm User should draw a cone with a hemisphere on top such that: Draw a solid right circular cone with vertex pointing downward, labelled height = 12 cm and base radius = 3 cm. On top of the cone's circular base, draw a hemisphere of radius = 3 cm sitting perfectly on the circular opening. Mark a horizontal line near the bottom (vertex) of the cone to show the empty unfilled portion, label this region as "Empty (1/6)th volume". Shade the ice-cream filled portion (upper part of cone and entire hemisphere) to distinguish it from the empty lower portion. The bases of cone and hemisphere are concentric circles (same center). Find the volume of the complete cone. Volume of cone = (1/3)πr²h Vcone = (1/3) × π × (3)² × 12 Vcone = (1/3) × π × 9 × 12 Vcone = (1/3) × 108π Vcone = 36π cm³ Find the volume of the empty (unfilled) part. Empty volume = (1/6) × Volume of cone Vempty = (1/6) × 36π Vempty = 6π cm³ Find the volume of ice-cream filled in the cone part. Vfilled_cone = Total volume - Empty volume Vfilled_cone = 36π - 6π Vfilled_cone = 30π cm³ Find the volume of the hemisphere on top. Radius of hemisphere = radius of cone = 3 cm Volume of hemisphere = (2/3)πr³ = (2/3) × π × (3)³ = (2/3) × π × 27 = (2 × 27π)/3 = 54π/3 = 18π cm³ Find the total volume of ice-cream. Total ice-cream volume = Volume in cone + Volume of hemisphere V_ice-cream = V_filled_cone + V_hemisphere V_ice-cream = 30π + 18π V_ice-cream = 48π cm³ Converting to decimal form V_ice-cream = 48 × (22/7) = 1056/7 ≈ 150.86 cm³ OR using π ≈ 3.14 V_ice-cream = 48 × 3.14 ≈ 150.72 cm³ Final Answer: Volume of ice-cream = 48π cm³ or approximately 150.86 cm³ MARKING SCHEME (Total: 5 marks) Question 35A: Question 35A Step-by-step marking: • Mode of the frequency distribution = 55 Modal class is 45-60, Lower limit is 45, Class Interval (h) = 15 [½ mark] • Now, Mode = l + ((f₁-f₀)/(2f₁-f₀-f₂)) × h 55 = 45 + ((15-x)/(30-x-10)) × 15 Solving: x = 5 [1 mark] • Creating frequency distribution table with class marks and fᵢxᵢ [1 mark] • Completing the table with all calculations [1½ marks] - Calculate xᵢ for each class [½ mark] - Calculate fᵢxᵢ for each class [½ mark] - Find Σfᵢ and Σfᵢxᵢ [½ mark] • Mean = x̄ = Σfᵢxᵢ/Σfᵢ = 2872.5/59 = 48.68 [1 mark] Question 35A Detailed marking breakdown: • Step 1: Identifying modal class [½ mark] - Mode = 55 lies in class 45-60 - l = 45, h = 15, f₁ = 15, f₀ = x, f₂ = 10 • Step 2: Using mode formula and solving for x [1 mark] - Correct formula application [½ mark] - Correct solving to get x = 5 [½ mark] • Step 3: Table preparation [1 mark] - Correct class intervals and frequencies - Setting up columns for xᵢ and fᵢxᵢ • Step 4: Table calculations [1½ marks] - All class marks (xᵢ) calculated correctly [½ mark] - All fᵢxᵢ values calculated correctly [½ mark] - Correct totals: Σfᵢ = 59, Σfᵢxᵢ = 2872.5 [½ mark] • Step 5: Mean calculation [1 mark] - Correct formula application - Final answer: 48.68 (or approximately 48.7) Question 35A SOLUTION: Given: Mode of the distribution = 55 Class intervals: 0-15, 15-30, 30-45, 45-60, 60-75, 75-90 Frequencies: 10, 7, x, 15, 10, 12 STEP 1: Identify the modal class Given: Mode = 55 Since Mode = 55, it must lie in the class interval containing 55. 55 lies between 45 and 60. Therefore, Modal class = 45-60 STEP 2: Identify parameters for mode formula For modal class 45-60: • l = Lower limit of modal class = 45 • h = Class width = 60 - 45 = 15 • f₁ = Frequency of modal class = 15 • f₀ = Frequency of class preceding modal class = x (class 30-45) • f₂ = Frequency of class succeeding modal class = 10 (class 60-75) STEP 3: Apply mode formula Formula: Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h Substitute the values: 55 = 45 + [(15 - x)/(2(15) - x - 10)] × 15 55 = 45 + [(15 - x)/(30 - x - 10)] × 15 55 = 45 + [(15 - x)/(20 - x)] × 15 STEP 4: Solve for x 55 - 45 = [(15 - x)/(20 - x)] × 15 10 = [(15 - x)/(20 - x)] × 15 10/15 = (15 - x)/(20 - x) 2/3 = (15 - x)/(20 - x) Cross multiply: 2(20 - x) = 3(15 - x) 40 - 2x = 45 - 3x 40 - 45 = -3x + 2x -5 = -x x = 5 STEP 5: Verify the modal class condition For 45-60 to be the modal class, its frequency (15) must be the highest. Frequencies: 10, 7, 5, 15, 10, 12 Highest frequency = 15 ✓ Therefore, x = 5 is correct. Answer: x = 5 Create frequency distribution table with x = 5 User should create a table with columns: Class Interval (CI), Frequency (fᵢ), Class Mark (xᵢ), fᵢxᵢ Calculate class marks (xᵢ) Formula: xᵢ = (Lower limit + Upper limit)/2 For class 0-15: x₁ = (0 + 15)/2 = 15/2 = 7.5 For class 15-30: x₂ = (15 + 30)/2 = 45/2 = 22.5 For class 30-45: x₃ = (30 + 45)/2 = 75/2 = 37.5 For class 45-60: x₄ = (45 + 60)/2 = 105/2 = 52.5 For class 60-75: x₅ = (60 + 75)/2 = 135/2 = 67.5 For class 75-90: x₆ = (75 + 90)/2 = 165/2 = 82.5 Complete the frequency table Class Interval (CI), Frequency (fᵢ) ,Class Mark (xᵢ) ,fᵢxᵢ as header.where user has to find fᵢxᵢ by multiplying fᵢ and xᵢ For class Interval-0-15, Frequency is 10, Class Mark is 7.5, fᵢxᵢ is 75.0 For class Interval-15-30, Frequency is 7, Class Mark is 22.5, fᵢxᵢ is 157.5 For class Interval-30-45, Frequency is 5, Class Mark is 37.5, fᵢxᵢ is 187.5 For class Interval-45-60, Frequency is 15, Class Mark is 52.5, fᵢxᵢ is 787.5 For class Interval-60-75, Frequency is 10, Class Mark is 67.5, fᵢxᵢ is 675.0 For class Interval-75-90, Frequency is 12, Class Mark is 82.5, fᵢxᵢ is 990.0 Total Frequency is 59, Total fᵢxᵢ is 2872.5 Calculate totals Σfᵢ = 10 + 7 + 5 + 15 + 10 + 12 = 59 Σfᵢxᵢ = 75 + 157.5 + 187.5 + 787.5 + 675 + 990 = 2872.5 Calculate mean Formula: Mean (x̄) = Σfᵢxᵢ / Σfᵢ x̄ = 2872.5 / 59 x̄ = 48.686... x̄ ≈ 48.68 or 48.7 Answer: Mean = 48.68 (approximately) VERIFICATION Check if mode is correctly calculated with x = 5: Modal class: 45-60 l = 45, f₁ = 15, f₀ = 5, f₂ = 10, h = 15 Mode = 45 + [(15-5)/(2(15)-5-10)] × 15 Mode = 45 + [10/(30-15)] × 15 Mode = 45 + [10/15] × 15 Mode = 45 + 10 Mode = 55 The mode is indeed 55, confirming x = 5 is correct. ------------------------------------- Question 35B: Question 35B Step-by-step marking: • Converting "less than" cumulative frequency to simple frequency and creating frequency distribution table [1 mark] • Finding median class (145-150) and applying median formula Median = l + ((n/2 - cf)/f) × h [1 mark] • Median = 145 + ((25.5 - 11)/18) × 5 = 149.03 cm [1 mark] • Using empirical relationship: 3 Median = Mode + 2 Mean 3 × 149.03 = 148.05 + 2 Mean [1 mark] • Mean = 149.52 cm [1 mark] Question 35B Detailed marking breakdown: • Step 1: Frequency conversion [1 mark] - Convert cumulative to simple frequency correctly [½ mark] - Create proper table with class intervals [½ mark] • Step 2: Median class identification and formula [1 mark] - Find n/2 = 25.5, identify median class as 145-150 [½ mark] - Set up formula with correct values [½ mark] • Step 3: Median calculation [1 mark] - Correct substitution in formula [½ mark] - Final answer: Median = 149.03 cm [½ mark] • Step 4: Empirical formula setup [1 mark] - Correct formula: 3 Median = Mode + 2 Mean [½ mark] - Correct substitution with given mode [½ mark] • Step 5: Mean calculation [1 mark] - Correct solving for Mean [½ mark] - Final answer: Mean = 149.52 cm [½ mark] Question 35B SOLUTION: Given: "Less than" cumulative frequency distribution of heights of 51 girls Given: Mode = 148.05 cm FINDING MEDIAN HEIGHT Convert cumulative frequency to simple frequency Given "less than" type cumulative frequency. Convert to simple frequency by subtraction For class 135-140: Frequency = 4 - 0 = 4 For class 140-145: Frequency = 11 - 4 = 7 For class 145-150: Frequency = 29 - 11 = 18 For class 150-155: Frequency = 40 - 29 = 11 For class 155-160: Frequency = 46 - 40 = 6 For class 160-165: Frequency = 51 - 46 = 5 CREATE FREQUENCY DISTRIBUTION TABLE where header is Class Interval (CI), Frequency (fᵢ), Cumulative Frequency (cf) For class Interval-135-140, Frequency is 4, Cumulative Frequency is 4 For class Interval-140-145, Frequency is 7, Cumulative Frequency is 11 For class Interval-145-150, Frequency is 18, Cumulative Frequency is 29 For class Interval-150-155, Frequency is 11, Cumulative Frequency is 40 For class Interval-155-160, Frequency is 6, Cumulative Frequency is 46 For class Interval-160-165, Frequency is 5, Cumulative Frequency is 51 Total Frequency is 51, Total Cumulative Frequency is 51 FINDING MEDIAN CLASS Total number of observations, n = 51 n/2 = 51/2 = 25.5 We need to find the class whose cumulative frequency is just greater than or equal to 25.5. From the table: • Class 135-140: cf = 4 (less than 25.5) • Class 140-145: cf = 11 (less than 25.5) • Class 145-150: cf = 29 (greater than 25.5) Therefore, Median class = 145-150 For median class 145-150: • l = Lower limit of median class = 145 • n = Total frequency = 51 • cf = Cumulative frequency of class preceding median class = 11 • f = Frequency of median class = 18 • h = Class width = 150 - 145 = 5 APPLYING MEDIAN FORMULA Formula: Median = l + [(n/2 - cf)/f] × h Median = 145 + [(51/2 - 11)/18] × 5 Median = 145 + [(25.5 - 11)/18] × 5 Median = 145 + [14.5/18] × 5 Median = 145 + (14.5 × 5)/18 Median = 145 + 72.5/18 Median = 145 + 4.0277... Median = 149.0277... Median ≈ 149.03 cm Answer: Median height of girls = 149.03 cm RECALLING THE EMPirical RELATIONSHIP The empirical relationship between Mean, Median, and Mode is: 3 Median = Mode + 2 Mean OR Mode = 3 Median - 2 Mean OR Mean = (3 Median - Mode)/2 Given: • Mode = 148.05 cm • Median = 149.03 cm (calculated above) To find: Mean Using: 3 Median = Mode + 2 Mean Substitute the values: 3 × 149.03 = 148.05 + 2 × Mean 447.09 = 148.05 + 2 × Mean SOLVING FOR MEAN 2 × Mean = 447.09 - 148.05 2 × Mean = 299.04 Mean = 299.04/2 Mean = 149.52 cm Answer: Mean height of girls = 149.52 cm

SECTION E: Case Study Questions

Questions 36-38: 4 marks each (1+1+2)

MARKING SCHEME (Total: 4 marks = 1+1+2) Step-by-step marking: (i) Common difference of first progression = 3 Common difference of second progression = -3 Sum of common difference = 0 [1 mark] (ii) t₃₄ = 187 + (34-1)(-3) So, t₃₄ = 88 [1 mark] (iii)(A) Sum = (10/2)[2(-5) + (10-1)(3)] = 85 [1 mark + 1 mark] OR (iii)(B) -5 + (n-1)3 = 187 + (n-1)(-3) n = 33 [1 mark + 1 mark] SOLUTION: Given: Aryan's sequence: -5, -2, 1, 4, ... Roshan's sequence: 187, 184, 181, ... (i) FINDING SUM OF COMMON DIFFERENCES For Aryan's progression: First term a₁ = -5 Second term a₂ = -2 Common difference d₁ = a₂ - a₁ = -2 - (-5) = -2 + 5 = 3 For Roshan's progression: First term a₁ = 187 Second term a₂ = 184 Common difference d₂ = a₂ - a₁ = 184 - 187 = -3 Sum of common differences = d₁ + d₂ = 3 + (-3) = 0 Answer: Sum of common differences = 0 (ii) FINDING 34th TERM OF ROSHAN'S PROGRESSION For Roshan's progression: First term a = 187 Common difference d = -3 n = 34 Formula: tₙ = a + (n-1)d t₃₄ = 187 + (34-1)(-3) = 187 + (33)(-3) = 187 - 99 = 88 Answer: 34th term = 88 (iii)(A) SUM OF FIRST 10 TERMS OF ARYAN'S PROGRESSION For Aryan's progression: First term a = -5 Common difference d = 3 n = 10 Formula: Sₙ = (n/2)[2a + (n-1)d] S₁₀ = (10/2)[2(-5) + (10-1)(3)] = 5[-10 + 9(3)] = 5[-10 + 27] = 5[17] = 85 Answer: Sum of first 10 terms = 85 OR (iii)(B) WHICH TERM HAS THE SAME VALUE IN BOTH PROGRESSIONS For Aryan's progression: tₙ = a + (n-1)d = -5 + (n-1)(3) For Roshan's progression: tₙ = a + (n-1)d = 187 + (n-1)(-3) Set them equal: -5 + (n-1)(3) = 187 + (n-1)(-3) -5 + 3n - 3 = 187 - 3n + 3 3n - 8 = 190 - 3n 3n + 3n = 190 + 8 6n = 198 n = 33 Verification: Aryan's 33rd term = -5 + (33-1)(3) = -5 + 96 = 91 Roshan's 33rd term = 187 + (33-1)(-3) = 187 - 96 = 91 Answer: 33rd term is the same in both progressions MARKING SCHEME (Total: 4 marks = 1 + ½ + ½ + 2) Step-by-step marking: (i) PR = √[(8-2)² + (3-5)²] = 2√10 [1 mark] (ii) Co-ordinates of Q (4,4). The mid-point of PR is (5,4) ∴ Q is not the mid-point of PR [½ mark + ½ mark] (iii)(A) Let the point be (x,0) So, √[(2-x)² + 25] = √[(4-x)² + 16] Hence x = 3/4. Therefore the point is (3/4, 0). [1 mark + 1 mark] OR (iii)(B) The coordinates of S will be ((2×4+3×2)/(2+3), (2×4+3×5)/(2+3)) = (14/5, 23/5) [1 mark + 1 mark] SOLUTION: Given: Coordinates from figure: P(2, 5), Q(4, 4), R(8, 3) (i) FIND DISTANCE PR Using distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²] PR = √[(8 - 2)² + (3 - 5)²] = √[6² + (-2)²] = √[36 + 4] = √40 = √(4 × 10) = 2√10 units Answer: PR = 2√10 units (ii) CHECK IF Q IS MIDPOINT OF PR Co-ordinates of Q are (4, 4) Midpoint of PR using formula: M = [(x₁ + x₂)/2, (y₁ + y₂)/2] Midpoint of PR = [(2 + 8)/2, (5 + 3)/2] = [10/2, 8/2] = (5, 4) Since (5, 4) ≠ (4, 4), Q is NOT the midpoint of PR. Answer: No, Q is not the midpoint of PR. The midpoint is (5, 4). (iii)(A) POINT ON X-AXIS EQUIDISTANT FROM P AND Q Let the point be A(x, 0) on x-axis Distance AP = Distance AQ √[(x - 2)² + (0 - 5)²] = √[(x - 4)² + (0 - 4)²] Squaring both sides: (x - 2)² + 25 = (x - 4)² + 16 x² - 4x + 4 + 25 = x² - 8x + 16 + 16 x² - 4x + 29 = x² - 8x + 32 -4x + 29 = -8x + 32 -4x + 8x = 32 - 29 4x = 3 x = 3/4 Answer: Point on x-axis = (3/4, 0) or (0.75, 0) OR (iii)(B) COORDINATES OF S DIVIDING PQ IN RATIO 2:3 Given: P(2, 5), Q(4, 4), ratio m:n = 2:3 Using section formula: S = [(m×x₂ + n×x₁)/(m+n), (m×y₂ + n×y₁)/(m+n)] x-coordinate of S = (2×4 + 3×2)/(2+3) = (8 + 6)/5 = 14/5 y-coordinate of S = (2×4 + 3×5)/(2+3) = (8 + 15)/5 = 23/5 Coordinates of S = (14/5, 23/5) or (2.8, 4.6) Answer: S = (14/5, 23/5) MARKING SCHEME (Total: 4 marks = ½ + ½ + ½ + ½ + 2) Step-by-step marking: (i) Distance from India gate = 41m, Height of monument = 42m, Shreya's height = 1m So, tan θ = 41/41 = 1 Angle of elevation = θ = 45° [½ mark + ½ mark] (ii) Angle of elevation = 60°, Perpendicular = 41m Let the distance from the India Gate be x m Hence tan 60° = 41/x ⇒ x = 41/√3 ∴ Shreya is standing at a distance of (41√3)/3 m [½ mark + ½ mark] (iii)(A) Distance from the India Gate = 41 m Let the distance moved back be x m Then, tan30° = 41/(41+x) x = (41√3 - 41) m = 41(√3-1) m ∴ The distance moved back = 41(√3-1) m [1 mark + 1 mark] OR (iii)(B) Let the angle of elevation be θ Now, tan θ = 41/(41√3) = 1/√3 This gives θ = 30° [1 mark + 1 mark] Note: The marking scheme image shows θ = 60° which appears to be an error. The correct answer is θ = 30°. SOLUTION: Given: Height of India Gate = 42m Shreya's height = 1m Effective height for angle calculation = 42 - 1 = 41m (i) User should have drawn the figure: Draw a right-angled triangle one rectangle below triangle where bases are same-to represent shreya's height and effective height of India Gate: • Vertical side represents the effective height from Shreya's eye to top of India Gate = 42m - 1m = 41m • Horizontal side (base) represents the distance from India Gate = 41m • The angle at Shreya's position (angle of elevation θ) is to be foundOF ELEVATION WHEN DISTANCE = 41mDistance from India Gate = 41m Height of monument = 42m Shreya's height = 1m Effective height (perpendicular) = 42 - 1 = 41m Base (distance) = 41m tan θ = Perpendicular/Base = 41/41 = 1 Since tan θ = 1, θ = 45° Answer: Angle of elevation = 45° (ii)DIAGRAM DESCRIPTION: Draw a right-angled triangle one rectangle below triangle where bases are same-to represent shreya's height and effective height of India Gate: • Vertical side = 41m (effective height) • Horizontal side = x (to be found) • Angle of elevation = 60° DISTANCE WHEN ANGLE OF ELEVATION = 60° Angle of elevation = 60° Perpendicular = 41m Let the distance from the India Gate be x m tan 60° = 41/x √3 = 41/x x = 41/√3 Rationalizing: x = (41/√3) × (√3/√3) = (41√3)/3 m Answer: Distance = (41√3)/3 m or 41/√3 m (iii)(A)DIAGRAM DESCRIPTION: Draw a right-angled triangle with two hypotenuses,one rectangle below triangle where bases are same-to represent shreya's height and effective height of India Gate. • Vertical side = 41m (angle 45°) • Horizontal side (distance moved back)= x (from angle 45° to 30°) • Angle of elevation = 45° and 30° DISTANCE MOVED BACK WHEN ANGLE CHANGES FROM 45° TO 30° Original position: Distance = 41m, Angle = 45° New position: Let distance = (41 + x) m, Angle = 30° where x is the distance moved back tan 30° = 41/(41 + x) 1/√3 = 41/(41 + x) 41 + x = 41√3 x = 41√3 - 41 = 41(√3 - 1) m Answer: Distance moved back = 41(√3 - 1) m OR (iii)(B)DIAGRAM DESCRIPTION: Draw a right-angled triangle ans one rectangle below triangle where bases are same-to represent shreya's height and effective height of India Gate. • Vertical side = 41m (angle 45°) • Horizontal side = 41√3 m (from angle 45° to 30°) • Angle of elevation = θ ANGLE OF ELEVATION WHEN DISTANCE = 41√3 m Distance from India Gate = 41√3 m Perpendicular = 41m Let angle of elevation = θ tan θ = Perpendicular/Base = 41/(41√3) = 1/√3 Since tan θ = 1/√3, θ = 30° Answer: Angle of elevation = 30° NOTE: The marking scheme image shows θ = 60° for part (iii)(B), but this appears to be an error. When tan θ = 1/√3, the angle is 30°, not 60°. (tan 60° = √3, not 1/√3)