Capacity of the hostel water tank
A rural school buys a cylindrical water tank for the hostel kitchen. The supplier's label reads: external diameter 1.80 m; internal height 2.40 m; wall and base thickness 40 mm.
The head teacher asks the clerk to work out how many 200-litre drums the tank can fill completely, and how long the tank takes to fill from a borehole pump delivering 25 litres per minute. The clerk computes π × (0.90)² × 2.40 = 6.11 m³, writes 'capacity 6 110 litres, 30 drums, filling time 4 hours 4 minutes', and submits the figures for the school budget.
The municipality bills water at N$14.50 per kilolitre (1 kL = 1 000 L), and the tank must stand on an existing rectangular concrete plinth of 2.10 m by 1.90 m. The supplier requires at least 100 mm clearance between the outer edge of the tank and the edge of the plinth on every side.
1. Recalculate the capacity of the tank in litres, identifying which dimensions are internal, and quantify the clerk's error and the cost of filling the tank.
Read the worked answer
The 1.80 m is an external diameter, so 0.90 m is an outside radius. For capacity we need internal dimensions: internal radius = 0.90 − 0.04 = 0.86 m. The 2.40 m is already stated as internal height, so no adjustment is needed there.
V = πr²h = π × 0.86² × 2.40 = π × 0.7396 × 2.40 = 5.5765 m³ ≈ 5 577 litres (using π = 3.14 gives 5 574 L).
The clerk used the external radius, which is the volume of the steel plus the water, and overstated the capacity by 6.108 − 5.577 = 0.531 m³ = 531 L, about 10%.
Cost of one full tank = 5.5765 kL × N$14.50 = N$80.86; the clerk's figure would have given N$88.57, an error of N$7.71 per filling.
Judgement: the decision is which radius is internal. A written reason ('the label gives an external diameter, so subtract the wall thickness twice from the diameter') earns the method mark even if the arithmetic slips.
2. How many 200-litre drums can be filled completely, and how long does the pump take to fill the tank? Give the time in hours and minutes.
Read the worked answer
Drums: 5 576.5 ÷ 200 = 27.88, so 27 drums can be filled completely. The 28th drum would be only 0.88 full (about 176 L), so 28 drums are needed if all the water is to be carried away.
Time: 5 576.5 L ÷ 25 L/min = 223.06 minutes = 3 h 43 min (223 min), not 4 h 4 min as the clerk reported.
Judgement: 'how many drums can be filled completely' (27) and 'how many drums must be brought' (28) are different questions, and rounding must match the wording. The clerk's time also came from the wrong volume: 6 108 ÷ 25 = 244 min = 4 h 4 min, so the time error is a direct consequence of the capacity error, not a separate mistake. When one wrong number feeds the next, marks are still available for correct method applied to your own value.
3. Decide whether the existing plinth satisfies the supplier's clearance requirement, showing the calculation that supports your decision.
Read the worked answer
The tank's footprint is a circle of outside diameter 1.80 m, so the plinth must extend at least 0.10 m beyond it on every side. Required plinth size in each direction = 1.80 + 2(0.10) = 2.00 m.
Existing plinth is 2.10 m × 1.90 m.
- 2.10 m direction: (2.10 − 1.80) ÷ 2 = 0.15 m clearance each side ✔ (≥ 0.10 m).
- 1.90 m direction: (1.90 − 1.80) ÷ 2 = 0.05 m clearance each side ✘ (less than 0.10 m).
So the plinth is not acceptable: it is 0.10 m too short in one direction and must be at least 2.00 m.
Judgement: rotating the tank does not help, because the footprint is circular - 1.90 m is too small whatever the orientation. The clearance rule exists for support and for access to the base, so the practical decision is to enlarge the plinth rather than to interpret 100 mm loosely as 'about 50 mm on each side'.