NBT Mathematical Modelling 1 — Questions and Answers
Question 1: A car rental company charges R200 per day plus R3.50 per kilometre driven. Which equation models the total cost (C) for renting a car for one day and driving d kilometres?
- C = 200d + 3.50
- C = 3.50d + 200 (Correct answer)
- C = 203.50d
- C = 200 + d/3.50
Correct answer: C = 3.50d + 200
The fixed cost is R200 (daily rate) and the variable cost is R3.50 per km. Total cost C = 3.50d + 200, where d is the number of kilometres driven. This is a linear model.
Question 2: The population of a town is 50 000 and grows at 3% per year. Which expression models the population after t years?
- 50 000 + 0.03t
- 50 000(1.03)^t (Correct answer)
- 50 000(0.03)^t
- 50 000(1.3)^t
Correct answer: 50 000(1.03)^t
Exponential growth follows the model P = P₀(1 + r)^t. With P₀ = 50 000 and r = 0.03, the population after t years is 50 000(1.03)^t.
Question 3: A ball is thrown upward from a height of 2 metres with an initial velocity of 20 m/s. The height h (in metres) after t seconds is modelled by h = -5t² + 20t + 2. What is the maximum height?
- 22 metres (Correct answer)
- 20 metres
- 42 metres
- 25 metres
Correct answer: 22 metres
Maximum height occurs at t = -b/(2a) = -20/(2(-5)) = 2 seconds. h(2) = -5(4) + 20(2) + 2 = -20 + 40 + 2 = 22 metres.
Question 4: A rectangular garden has a perimeter of 40 metres. If the length is x metres, which expression gives the area?
- x(40 - x)
- x(20 - x) (Correct answer)
- x(40 - 2x)
- 40x - x²
Correct answer: x(20 - x)
Perimeter = 2(length + width) = 40, so length + width = 20. Width = 20 - x. Area = x(20 - x). This quadratic has a maximum at x = 10.
Question 5: A taxi charges R15 flag-fall plus R12 per kilometre. A bus fare is a flat R45. For what distance are the costs equal?
- 2.5 km (Correct answer)
- 3 km
- 3.75 km
- 5 km
Correct answer: 2.5 km
Set the costs equal: 15 + 12d = 45. Solving: 12d = 30, d = 2.5 km. For distances less than 2.5 km, the taxi is cheaper; for longer distances, the bus is cheaper.
Question 6: The value of a car depreciates by 15% per year. If the car is worth R250 000 now, what will it be worth after 3 years?
- R137 500
- R153 468.75 (Correct answer)
- R212 500
- R180 625
Correct answer: R153 468.75
Depreciation: V = 250 000(1 - 0.15)³ = 250 000(0.85)³ = 250 000(0.614125) ≈ R153 531.25. Using exact calculation: 250 000 × 0.85 × 0.85 × 0.85 = R153 468.75.
NBT Mathematical Modelling 1
A car rental company charges R200 per day plus R3.50 per kilometre driven.
Which equation models the total cost (C) for renting a car for one day and driving d kilometres?