Village Water Tank: Capacity and Paint Cost
A village water committee is installing a new tank. The tank is a cylinder of radius 1.5 m and height 2.8 m, with a hemispherical roof on top. The flat circular base sits on a concrete slab and is not painted. The external curved surface - the cylinder side and the hemispherical roof - must be painted with two coats. One litre of paint covers 6 m² per coat, and paint is sold only in 5-litre tins at M85 per tin. The committee needs the tank to hold at least 25,000 litres of water. A supplier claims the tank will hold 30,000 litres. The committee must check the capacity, calculate the paint needed including a 10% allowance for wastage, and decide how many tins to buy. They also want to know the cost per 1,000 litres of capacity to compare with a rival tank priced at M9,500.
1. Calculate the capacity of the tank in litres. Check the supplier's claim that it will hold 30,000 litres and the committee's minimum of 25,000 litres.
Read the worked answer
Cylinder volume = πr²h = π × 1.5² × 2.8 = π × 2.25 × 2.8 = 6.3π = 19.792 m³. Hemisphere volume = (2/3)πr³ = (2/3)π × 1.5³ = (2/3)π × 3.375 = 2.25π = 7.069 m³. Total = 8.55π = 26.861 m³. 1 m³ = 1000 L, so capacity ≈ 26,861 L. This is above 25,000 L, so it meets the committee's minimum, but it is below 30,000 L, so the supplier's claim is false.
2. Calculate the external curved area to be painted for two coats, including 10% wastage. How many 5-litre tins are needed, and what is the paint cost?
Read the worked answer
Curved cylinder area = 2πrh = 2π × 1.5 × 2.8 = 8.4π = 26.389 m². Hemispherical roof area = 2πr² = 2π × 2.25 = 4.5π = 14.137 m². One coat = 40.526 m². Two coats = 81.052 m². Add 10% wastage: 81.052 × 1.10 = 89.157 m². Paint needed = 89.157 ÷ 6 = 14.86 L. Tins are 5 L, so 3 tins give 15 L, which is enough. Cost = 3 × 85 = M255. The base is not painted, as stated.
3. If the tank is filled to 90% of capacity, how many full 20-litre buckets can be filled? Comment on whether the committee should use this as a practical distribution plan.
Read the worked answer
90% of 26,861 L = 0.90 × 26,861 = 24,174.9 L. Number of full 20 L buckets = 24,174.9 ÷ 20 = 1208.745, so 1,208 full buckets, with about 0.745 × 20 = 14.9 L left over. The plan is practical in principle, but it leaves a small remainder and assumes no spillage or draw-off losses. If the committee wants every bucket full, it should plan for 1,208 buckets and treat the remainder as reserve or cleaning water, not as a full bucket.