Painting the hall: cheaper tins, dearer paint
A school hall is 12 m long, 8 m wide and 3.5 m high. Only the four walls are to be painted; the ceiling and floor are not. There are two doors, each 1 m by 2 m, and six windows, each 1.5 m by 1.2 m. Doors and windows are not painted. Two coats are needed. Brand A costs P285 for a 5 L drum or P68 for a 1 L tin, and covers 9 m² per litre. Brand B costs P310 for a 5 L drum or P72 for a 1 L tin, and covers 11 m² per litre. The bursar says Brand A must be cheaper because its drum and tin are both cheaper. There is also an unopened 4 L drum of Brand A left over from last term, which can be used if the school buys Brand A again.
1. Find the area to be painted and the number of litres of each brand needed for two coats.
Read the worked answer
Wall area = distance round the room × height = 2(12 + 8) × 3.5 = 40 × 3.5 = 140 m².
Openings to subtract:
- Doors: 2 × (1 × 2) = 4 m²
- Windows: 6 × (1.5 × 1.2) = 6 × 1.8 = 10.8 m²
- Total openings = 14.8 m²
Net wall = 140 − 14.8 = 125.2 m². Two coats = 2 × 125.2 = 250.4 m² of paint film.
- Brand A: 250.4 ÷ 9 = 27.82 L → must buy 28 L (paint cannot be bought in fractions of a litre and running short leaves a patchy coat).
- Brand B: 250.4 ÷ 11 = 22.76 L → must buy 23 L.
Two common traps: forgetting the second coat (which halves the litres needed and under-buys by a factor of two), and forgetting to deduct the doors and windows (which over-buys by about 15 m², or 1.7 L of A).
2. Find the cheapest way to buy each brand from scratch. Is the bursar right?
Read the worked answer
Drum unit prices: A = 285 ÷ 5 = P57 per litre (tin P68/L); B = 310 ÷ 5 = P62 per litre (tin P72/L). Drums are always cheaper per litre, so use as many drums as possible without buying more litres than needed, and top up with tins.
Brand A - need 28 L:
- 5 drums (25 L) + 3 tins (3 L) = 1425 + 204 = P1 629
- 4 drums (20 L) + 8 tins (8 L) = 1140 + 544 = P1 684
- 6 drums (30 L) = P1 710
Cheapest A = P1 629.
Brand B - need 23 L:
- 4 drums (20 L) + 3 tins (3 L) = 1240 + 216 = P1 456
- 5 drums (25 L) = P1 550
Cheapest B = P1 456. B is cheaper by 1629 − 1456 = P173.
The bursar is wrong: he compared the price of a container instead of the price of covering a square metre. B's drum costs P25 more, but 1 L of B covers 11 m² instead of 9 m² (22% further), so 23 L do the job instead of 28 L. The valid comparison is cost per square metre: A = 1629 ÷ 250.4 = P6.51/m²; B = 1456 ÷ 250.4 = P5.82/m².
3. Now include the unopened 4 L drum of Brand A already in the storeroom. Does the answer change, and what assumptions does your answer depend on?
Read the worked answer
Yes - the decision flips back to Brand A.
With 4 L of A in stock, only 28 − 4 = 24 L of A must be bought: 4 drums (20 L) + 4 tins (4 L) = 1140 + 272 = P1 412. That beats Brand B's P1 456 by P44.
Reasoning point: the P already spent on the 4 L is a sunk cost - it is the same whether or not the school buys A - so the fair comparison is cash still to be paid: P1 412 against P1 456. Ignoring the stock would overstate A's cost by the full value of 4 L (4 × 57 = P228 at drum rate).
Assumptions to check:
- The drum is genuinely unopened and usable - paint that has skinned over or frozen is worthless, and then B wins again.
- The 4 L must be the same product and colour; you cannot assume Brand B behaves like Brand A on the same wall, so the leftover cannot be used to cut B's requirement.
- Mixing is only safe within one brand; if the second coat were a different brand, coverage and finish could differ.
- The school may want a little spare paint for touch-ups, which slightly favours whichever option leaves more in the drum.