Repainting the Club Room: Area, Coverage and Leftovers
A youth club has permission to repaint the main room of its building. The room is rectangular, 7.2 m long and 5.4 m wide, with a ceiling height of 2.7 m. There are two identical windows, each 1.2 m wide and 1.5 m high, and one door 0.9 m wide and 2.1 m high. Only the four walls are to be painted, and the committee has decided on two coats everywhere. The chosen paint covers 12 m² per litre per coat and is sold in 5 litre tins at €38.50 or 2.5 litre tins at €22.00. The club also wants new skirting board around the full perimeter of the floor except across the doorway, bought in 2.4 m lengths at €14.20 each. A volunteer has estimated 'about 70 square metres and four big tins' and wants the order placed before the weekend. The wall behind the radiators is scuffed and the committee would like a third coat on that one wall if paint is left over.
1. Calculate the net wall area to be painted, the litres required for two coats, and the cheapest way to buy that paint.
Read the worked answer
Step 1 – Gross wall area. Perimeter = 2(7.2 + 5.4) = 25.2 m. Gross area = 25.2 × 2.7 = 68.04 m².
Step 2 – Deduct openings (they are not painted). Windows: 2 × (1.2 × 1.5) = 2 × 1.8 = 3.6 m². Door: 0.9 × 2.1 = 1.89 m². Total deductions = 5.49 m².
Step 3 – Net area. 68.04 − 5.49 = 62.55 m².
Step 4 – Two coats. 62.55 × 2 = 125.1 m² of painted surface.
Step 5 – Litres. 125.1 ÷ 12 = 10.425 L. Paint cannot be bought in fractions, so round up.
Step 6 – Compare purchase options.
- Three 5 L tins = 15 L, €115.50.
- Two 5 L + one 2.5 L = 12.5 L, 77.00 + 22.00 = €99.50.
- Five 2.5 L tins = 12.5 L, €110.00.
- The volunteer's four 5 L tins = 20 L, €154.00 - nearly €55 wasted.
Best buy: two 5 L tins and one 2.5 L tin, €99.50, giving 12.5 L against the 10.425 L required.
2. Calculate the skirting board required and the number of lengths to buy, and comment on the offcut.
Read the worked answer
Step 1 – Run required. Skirting follows the floor perimeter minus the doorway: 25.2 − 0.9 = 24.3 m.
Step 2 – Lengths. 24.3 ÷ 2.4 = 10.125. Part-lengths cannot be bought, so round up to 11 lengths.
Step 3 – Cost. 11 × €14.20 = €156.20.
Step 4 – Offcut. 11 × 2.4 = 26.4 m supplied; 26.4 − 24.3 = 2.1 m left over. That is almost a full length, so 10 lengths (24.0 m) would be 0.3 m short - an expensive mistake to correct after the shop closes.
Step 5 – Practical judgement. The 2.1 m offcut is not really waste: 24.3 m of continuous run cannot be cut from whole lengths without joints, and corners and the door frame need mitred cuts that consume material. A sensible order is 11 lengths, keeping the offcut as a spare for a miscut or for a damaged section later. Note also that the skirting is measured in linear metres while the paint was measured in square metres - mixing these units is the most common error in this kind of task.
3. The committee wants a third coat on the back wall, which is 7.2 m long and 2.7 m high, using only leftover paint. Decide whether this is possible.
Read the worked answer
Step 1 – Check paint left over. Purchased 12.5 L; used for two coats 10.425 L. Leftover = 12.5 − 10.425 = 2.075 L.
Step 2 – Area of the back wall. 7.2 × 2.7 = 19.44 m². This wall has no windows or door, so nothing is deducted.
Step 3 – Paint needed for one coat. 19.44 ÷ 12 = 1.62 L.
Step 4 – Compare. 1.62 L needed against 2.075 L available → yes, with about 0.455 L to spare.
Step 5 – Caution. The coverage figure of 12 m² per litre is a manufacturer's estimate for a smooth, sealed surface. Bare or porous plaster, or a strong colour change, can reduce coverage by 10–20%. At 20% worse coverage the wall would need 1.62 × 1.2 ≈ 1.94 L, which still fits within 2.075 L - but only just. The third coat should therefore be applied last and watched, rather than assumed. This is the point of buying 12.5 L rather than exactly 10.425 L: the surplus is doing real work rather than sitting in a shed.