Mathematical Modelling — Telangana (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Mathematical Modelling" — important questions with detailed answers, download PDF for board exam preparation.

Mathematical Modelling — Telangana (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Mathematical Modelling" — important questions with detailed answers, download PDF for board exam preparation.

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TL;DR: Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Mathematical Modelling" — important questions with detailed answers, download PD…

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Mathematical Modelling — Textbook Questions Answered (8)

Q1: A farmer has 100 meters of fencing to enclose a rectangular garden. If one side is x meters, express the area as a function of x and find dimensions for maximum area.

Step 1: Let length = x, width = w Step 2: Perimeter = 2(x + w) = 100 Step 3: x + w = 50, so w = 50 - x Step 4: Area A(x) = x × w = x(50 - x) = 50x - x² Step 5: For maximum area, dA/dx = 50 - 2x = 0 Step 6: x = 25 meters Step 7: w = 50 - 25 = 25 meters Step 8: Maximum area = 25 × 25 = 625 m² Final Answer: A(x) = 50x - x²; Maximum area when length = width = 25 m (square); Area = 625 m²

Q2: A ball is thrown upward from ground with initial velocity 40 m/s. Height h(t) = 40t - 5t² where t is time in seconds. Find maximum height and time to reach it.

Step 1: h(t) = 40t - 5t² Step 2: For maximum height, dh/dt = 40 - 10t = 0 Step 3: t = 4 seconds Step 4: Maximum height h(4) = 40(4) - 5(4)² = 160 - 80 = 80 meters Step 5: Ball returns to ground when h(t) = 0 Step 6: 40t - 5t² = 0, t(40 - 5t) = 0 Step 7: t = 0 or t = 8 seconds Final Answer: Maximum height = 80 m; Time to reach = 4 seconds; Returns to ground at t = 8 seconds

Q3: The cost of manufacturing x units is C(x) = 50 + 20x. Revenue from selling x units is R(x) = 40x. Find break-even point and profit function.

Step 1: Cost function C(x) = 50 + 20x Step 2: Revenue function R(x) = 40x Step 3: Profit P(x) = R(x) - C(x) = 40x - (50 + 20x) = 20x - 50 Step 4: Break-even when Profit = 0 Step 5: 20x - 50 = 0 Step 6: x = 2.5 units Step 7: At break-even: Revenue = Cost = 40(2.5) = 100 Final Answer: Profit function P(x) = 20x - 50; Break-even at 2.5 units; Break-even cost/revenue = 100

Q4: Population of a city grows exponentially. If initial population is 100,000 and growth rate is 5% per year, find population after 10 years.

Step 1: Initial population P₀ = 100,000 Step 2: Growth rate r = 5% = 0.05 per year Step 3: Formula P(t) = P₀(1 + r)ᵗ Step 4: After t = 10 years: P(10) = 100,000 × (1.05)¹⁰ Step 5: (1.05)¹⁰ ≈ 1.6289 Step 6: P(10) = 100,000 × 1.6289 = 162,889 Final Answer: Population after 10 years ≈ 162,889

Q5: A rectangular box has a square base of side x cm. If volume is fixed at 500 cm³, express surface area as function of x.

Step 1: Square base with side x, let height = h Step 2: Volume V = x²h = 500 Step 3: h = 500/x² Step 4: Surface area SA = 2x² + 4xh (2 bases + 4 sides) Step 5: SA(x) = 2x² + 4x(500/x²) = 2x² + 2000/x Step 6: To minimize: dSA/dx = 4x - 2000/x² = 0 Step 7: 4x = 2000/x², 4x³ = 2000, x³ = 500 Step 8: x = ∛500 ≈ 7.94 cm Final Answer: SA(x) = 2x² + 2000/x; Minimum surface area at x ≈ 7.94 cm

Q6: A ladder 10 m long leans against a wall. If bottom slides away at 2 m/s, find rate at which top slides down when bottom is 6 m from wall.

Step 1: Ladder length L = 10 m (constant) Step 2: Let x = distance of bottom from wall, y = height on wall Step 3: x² + y² = 100 (Pythagorean) Step 4: Given: dx/dt = 2 m/s (bottom sliding away) Step 5: When x = 6: y² = 100 - 36 = 64, so y = 8 m Step 6: Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0 Step 7: 2(6)(2) + 2(8)(dy/dt) = 0 Step 8: 24 + 16(dy/dt) = 0 Step 9: dy/dt = -24/16 = -1.5 m/s Final Answer: Top slides down at 1.5 m/s (negative indicates downward motion)

Q7: A water tank fills at rate 30 L/min and drains at rate 5 L/min. Tank capacity is 10,000 L and starts empty. How long to fill?

Step 1: Fill rate = 30 L/min, drain rate = 5 L/min Step 2: Net fill rate = 30 - 5 = 25 L/min Step 3: Tank capacity = 10,000 L Step 4: Time to fill = Capacity / Net rate = 10,000 / 25 Step 5: Time = 400 minutes Step 6: Time = 400/60 hours = 6 hours 40 minutes Final Answer: Tank fills in 400 minutes or 6 hours 40 minutes

Q8: An investment of 5,000 rupees grows at compound interest rate of 8% per annum. How much after 3 years?

Step 1: Principal P = 5,000 rupees Step 2: Rate r = 8% = 0.08 per annum Step 3: Time t = 3 years Step 4: Formula A = P(1 + r)ᵗ Step 5: A = 5,000 × (1.08)³ Step 6: (1.08)³ = 1.259712 Step 7: A = 5,000 × 1.259712 = 6,298.56 rupees Step 8: Interest earned = 6,298.56 - 5,000 = 1,298.56 rupees Final Answer: Amount after 3 years = 6,298.56 rupees; Interest earned = 1,298.56 rupees

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