1. **Problem statement:** a. Define what 1 unit of water consumed means. b. Calculate the electricity charge for 45 minutes at 5A. c. Find the water charge for 75 units. d. Compare two trips based on distance, waiting time, and fare rules. 2. **Definitions and formulas:** - 1 unit of water consumed = 1000 litres. - Electricity charge formula: $$\text{Charge} = \text{Service charge} + (\text{Energy charge per hour} \times \text{time in hours})$$ - Water charge = Minimum charge + Extra consumption charge if consumption exceeds minimum. - Fare calculation: $$\text{Fare} = \text{Minimum fare} + \left(\frac{\text{distance in meters}}{200}\right) \times \text{fare per 200m} + \left(\frac{\text{waiting time in minutes}}{2}\right) \times \text{waiting charge}$$ 3. **Step-by-step solutions:** a. 1 unit of water consumed means 1000 litres of water. b. Electricity charge for 45 minutes (0.75 hours) at 5A: Given service charge for 31-50 units = 50.00 Energy charge per unit = 8.00 Since 5A current is given but no voltage or power, assume 5 units consumed in 1 hour for calculation (or interpret 5A as 5 units for simplicity). For 0.75 hours, units consumed = $5 \times 0.75 = 3.75$ units Energy charge = $3.75 \times 8.00 = 30.00$ Total charge = Service charge + Energy charge = $50.00 + 30.00 = 80.00$ c. Water charge for 75 units: Minimum consumption = 56,000 litres = 56 units (since 1 unit = 1000 litres) Consumption = 75 units > 56 units Extra units = $75 - 56 = 19$ units Minimum charge = 3,900 Extra charge per unit = 71 Extra charge = $19 \times 71 = 1,349$ Total water charge = Minimum charge + Extra charge = $3,900 + 1,349 = 5,249$ d. Compare two trips: Trip 1: Distance = 9.8 km = 9,800 m, Waiting time = 5 minutes Trip 2: Distance = 10 km = 10,000 m, Waiting time = 0 minutes Minimum fare = 50 Fare per 200 m = 10 Waiting charge per 2 minutes = 10 Calculate fare for Trip 1: Distance fare = $\frac{9,800}{200} \times 10 = 49 \times 10 = 490$ Waiting charge = $\frac{5}{2} \times 10 = 2.5 \times 10 = 25$ Total fare Trip 1 = $50 + 490 + 25 = 565$ Calculate fare for Trip 2: Distance fare = $\frac{10,000}{200} \times 10 = 50 \times 10 = 500$ Waiting charge = 0 Total fare Trip 2 = $50 + 500 + 0 = 550$ **Comparison:** Trip 1 costs 565, Trip 2 costs 550, so Trip 2 is cheaper by 15. 4. **Final answers:** a. 1 unit water consumed = 1000 litres. b. Electricity charge for 45 minutes = 80.00 c. Water charge for 75 units = 5,249 d. Trip 1 fare = 565, Trip 2 fare = 550; Trip 2 is cheaper by 15.