All questions
- Q.3 (B)(4) How many solid cylinders of radius 10 cm and height 6 cm can be made by melting a solid sphere of radius 30 cm ? Answer: 60 cylinders Volume of sphere = 4/3π r³ = 4/3π(30)³ = 36000π cm³. Volume of one cylinder = π r² h = π(10)²(6) = 600π cm³. Number of... 1
- Q.4 (1) In the following figure DE || BC, then :[Figure: Δ ABC with D on side AB and E on side AC, DE parallel to BC.](i) If DE = 4 cm, BC = 8 cm, A(Δ ADE) = 25 cm2, find A(Δ ABC).(ii) If DE : BC = 3 : 5, then find A(Δ ADE) : A(□ DBCE). Answer: (i) A(Δ ABC) = 100 cm²; (ii) A(Δ ADE) : A(□ DBCE) = 9 : 16 Since DE || BC, Δ ADE ~ Δ ABC (AA test: ∠A common, ∠ADE ≅ ∠ABC as corresp... 1
- Q.4 (2) Δ ABC ~ Δ PQR. In Δ ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of Δ ABC and Δ PQR are in the ratio 2 : 3, construct Δ ABC and Δ PQR. Answer: PQ = 5.4 cm, QR = 6 cm, PR = 6.3 cm; construct both triangles from their sides. AB/PQ = BC/QR = AC/PR = 2/3 . PQ = 3/2 × 3.6 = 5.4 c... 1
- Q.4 (3) The radii of the circular ends of a frustum of a cone are 14 cm and 8 cm. If the height of the frustum is 8 cm, find : (π = 3.14)(i) Curved surface area of frustum.(ii) Total surface area of the frustum.(iii) Volume of the frustum. Answer: (i) 690.8 cm²; (ii) 1507.2 cm²; (iii) 3114.88 cm³ r₁ = 14, r₂ = 8, h = 8. Slant height l = √(h² + (r₁ - r₂)²) = √(64 + 36) = 10 cm.... 1
- Q.5 (1) ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm, then find the values of BD and BX. Answer: BD = 15 cm, BX = 9.6 cm In right Δ BAD (∠A = 90º): BD² = AB² + AD² = 144 + 81 = 225, so BD = 15 cm. ∠AXD is an angle inscribed in a... 1
- Q.5 (2) Taking θ = 30º to verify the following Trigonometric identities :(i) \(\sin^2\theta + \cos^2\theta = 1\)(ii) \(1 + \tan^2\theta = \sec^2\theta\)(iii) \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\). Answer: All three identities hold for θ = 30º. sin 30º = 1/2 , cos 30º = sqrt3/2 , tan 30º = 1/sqrt3 , sec 30º = 2/sqrt3 , cot 30º = sqrt3 ,... 1
- Q.1 (A)(iv) Which number cannot represent a probability ? Answer: B. 15/10 Answer: (B) 15/10 Probability always lies between 0 and 1. 15/10 = 1.5 > 1 , so it cannot be a probability. 2/3 , 15% = 0.1... 1
- Q.1 (B)(i) If 2x + y = 7 and x + 2y = 11, then find the value of x + y. Answer: x + y = 6 Adding the two equations: 3x + 3y = 18 . Dividing by 3, x + y = 6 . 1
- Q.1 (B)(ii) Find the first term of the given sequence : \(t_n = 3n - 4\). Answer: t₁ = -1 Put n = 1 : t₁ = 3(1) - 4 = -1 . 1
- Q.1 (B)(iii) How many alpha numerals are there in the format of GSTIN ? Answer: 15 GSTIN (Goods and Services Tax Identification Number) is a 15-character alphanumeric code: 2 digits for the state code, 10 charact... 1
- Q.1 (B)(iv) Two coins are tossed simultaneously. Write the sample space S. Answer: S = HH, HT, TH, TT Each coin shows H or T, so there are 2 × 2 = 4 outcomes; n(S) = 4. 1
- Q.2 (A)(i) Complete the following table to draw the graph of x + 2y = 4.Activity :x–2____y____1(x, y)________ Answer: x = –2 gives y = 3; y = 1 gives x = 2; points (–2, 3) and (2, 1). x + 2y = 4. When x = –2: –2 + 2y = 4, so 2y = 6, y = 3 → (–2, 3).... 1
- Q.2 (A)(ii) Complete the following activity to form a quadratic equation.Activity :I am a quadratic equation. → My standard form is ____ . → My roots are 5 and 12. → Sum of my roots ____ . → Product of my roots ____ . → My quadratic equation is ____… Answer: x² - 17x + 60 = 0 Standard form: ax² + bx + c = 0 , a ≠ 0 . Sum of roots = 5 + 12 = 17 . Product of roots = 5 × 12 = 60 . Equation:... 1
- Q.2 (A)(iii) Pushpmala has invested ₹ 24,000 and purchased share of FV ₹ 20 at a premium of ₹ 4. Complete the following activity to find the number of shares she purchased.Activity :FV = ₹ 20Premium = ₹ 4MV = FV + ____ = 20 + ____ = ₹ 24Number of sha… Answer: 1000 shares MV = FV + premium = 20 + 4 = ₹ 24. Number of shares = Total investment ÷ MV = 24,000 ÷ 24 = 1000. 1
- Q.2 (B)(i) Solve the following simultaneous equations : x + y = 3; 3x – 2y = 4 Answer: x = 2, y = 1 Multiply x + y = 3 by 2: 2x + 2y = 6. Add to 3x – 2y = 4: 5x = 10, so x = 2. Then y = 3 – 2 = 1. Check: 3(2) – 2(1) = 4... 1
- Q.2 (B)(ii) Solve the following quadratic equation by factorisation method : \(m^2 + 14m + 13 = 0\) Answer: m = –1 or m = –13 m² + 13m + m + 13 = 0 m(m + 13) + 1(m + 13) = 0 (m + 13)(m + 1) = 0 So m + 13 = 0 or m + 1 = 0 , giving m = –13 or... 1
- Q.2 (B)(iii) Find the 19th term of the following A.P. : 7, 13, 19, 25, ... Answer: t₁₉ = 115 a = 7, d = 13 – 7 = 6. t_n = a + (n - 1)d , so t₁₉ = 7 + 18 × 6 = 7 + 108 = 115 . 1
- Q.2 (B)(iv) A share is sold for the market value of ₹ 2,000. Brokerage is paid at the rate of 0.5%. What is the amount received after the sale ? Answer: ₹ 1,990 Brokerage = 0.5% of 2000 = 0.5/100 × 2000 = ₹ 10. On a sale, brokerage is deducted: amount received = 2000 – 10 = ₹ 1,990. 1
- Q.2 (B)(v) The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by the students for their studies.Class Time (In hours)Class Marks (xi)No. of Students (fi)fixi0–218082–4314424–6… Answer: Mean time = 5 hours Σ f_i = 8 + 14 + 18 + 10 + 10 = 60 Σ f_i x_i = 8 + 42 + 90 + 70 + 90 = 300 x = Σ f_i x_i/Σ f_i = 300/60 = 5 hour... 1
- Q.3 (A)(i) Shri Maniklal has purchased 300 shares of F.V. ₹ 100, for M.V. ₹ 120. Company has paid dividend at 7%. Complete the following activity to find the rate of return on his investment.Activity :F.V. = ₹ 100, Number of shares = 300, Market valu… Answer: Rate of return = 55/6 % ≈ 5.83% (a) Sum invested = 120 × 300 = ₹ 36,000. (b) Dividend per share = 100 × 7/100 = ₹ 7; total dividend... 1
- Q.3 (A)(ii) A two digit number is to be formed from the digits 2, 3, 5 without repetition of the digits. Complete the following activity to find the probability that the number so formed is an odd number.Activity :Let S be the sample space.S = {23, 2… Answer: P(A) = 2/3 S = 23, 25, 32, 35, 52, 53, n(S) = 6. Odd numbers end in 3 or 5: A = 23, 25, 35, 53, n(A) = 4. P(A) = n(A)/n(S) = 4/6 = 2... 1
- Q.3 (B)(i) Solve the following simultaneous equations by Cramer’s rule : 4x + 3y = 18; 3x – 2y = 5 Answer: x = 3, y = 2 D = beginvmatrix 4 & 3 \\ 3 & -2 endvmatrix = -8 - 9 = -17 D_x = beginvmatrix 18 & 3 \\ 5 & -2 endvmatrix = -36 - 15 =... 1
- Q.3 (B)(ii) Solve the following quadratic equation by using formula method : \(x^2 - 2x - 3 = 0\) Answer: x = 3 or x = –1 a = 1, b = –2, c = –3. b² - 4ac = 4 + 12 = 16 . x = -b ± √(b² - 4ac)/2a = 2 ± 4/2 x = 6/2 = 3 or x = -2/2 = -1 . 1
- Q.3 (B)(iii) A committee of two members is to be formed from three boys and two girls. Find the probability of the following events :Event A : At least one girl must be a member of the committee.Event B : Committee must be of one boy and one girl. Answer: P(A) = 7/10, P(B) = 3/5 Let the boys be B₁, B₂, B₃ and the girls G₁, G₂. S = B₁B₂, B₁B₃, B₂B₃, B₁G₁, B₁G₂, B₂G₁, B₂G₂, B₃G₁, B₃G₂, G... 1