Flooring two classrooms: counting tiles along walls versus using area
A school committee is flooring two classrooms with square tiles 40 cm by 40 cm. Room A measures 8 m by 6 m. Room B measures 8.2 m by 4.6 m. Both rooms are rectangles with straight walls and no pillars. Tiles are sold only in boxes of 12, and one box costs MK 30 000. Room A also needs skirting board along every wall except a 1 m doorway; the skirting costs MK 1 800 per metre. The tiler's rule is to count the tiles needed along each wall, round each count up to a whole tile, and multiply the two counts; off-cuts are treated as waste and are not reused. The chairperson says: 'There is no need for that. Just take the floor area in square metres, divide by 0.16 to get the number of tiles, round up and buy that many.'
1. Find the number of tiles and the cost of tiles for Room A. Should the committee buy an extra box as a precaution?
Read the worked answer
Convert the tile side: 40 cm = 0.4 m.
Along the 8 m wall: 8 ÷ 0.4 = 20 tiles exactly. Along the 6 m wall: 6 ÷ 0.4 = 15 tiles exactly.
Tiles = 20 × 15 = 300. Boxes = 300 ÷ 12 = 25 exactly, no remainder.
Cost = 25 × MK 30 000 = MK 750 000.
Check by area: 8 × 6 = 48 m²; 48 ÷ 0.16 = 300 tiles - the same answer, because 0.4 × 0.4 = 0.16 m².
Extra box? No. Both room dimensions are exact multiples of 0.4 m, so no tile needs cutting, and a 26th box (MK 30 000, about 4% of the tile budget) would be pure waste. Only order extra if the tiler expects breakages, and then agree in writing that unopened boxes can be returned.
2. Work out the cost of the skirting board for Room A and the total cost of finishing Room A.
Read the worked answer
Perimeter of Room A = 2 × (8 + 6) = 28 m.
The doorway is 1 m wide and needs no skirting, so skirting length = 28 − 1 = 27 m.
Cost = 27 × MK 1 800 = MK 48 600 (27 × 18 = 486, then add two zeros).
Total for Room A = MK 750 000 + MK 48 600 = MK 798 600.
Watch out: skirting runs round the edge of the floor, so it uses the perimeter, not the area of 48; and forgetting the doorway gives 28 × 1 800 = MK 50 400, which overstates the cost by MK 1 800.
3. Apply both methods to Room B: the tiler's rule and the chairperson's area method. Which gives the right number of boxes, and what does this show about shortcuts?
Read the worked answer
Tiler's rule:
- 8.2 ÷ 0.4 = 20.5 → round up to 21 tiles along the length.
- 4.6 ÷ 0.4 = 11.5 → round up to 12 tiles along the width.
- Tiles = 21 × 12 = 252; boxes = 252 ÷ 12 = 21 exactly; cost = 21 × MK 30 000 = MK 630 000.
Chairperson's method:
- Area = 8.2 × 4.6 = 37.72 m²; 37.72 ÷ 0.16 = 235.75 → 236 tiles; 236 ÷ 12 = 19.67 → 20 boxes = 240 tiles.
- That is 252 − 240 = 12 tiles short, i.e. one whole box short - the floor cannot be finished.
Why: the area method assumes off-cuts can be pieced together. Under the tiler's rule each row needs its own cut tile, so the fractional parts (0.5 of a tile in each direction) each cost a full tile instead of being averaged away.
Judgement: the shortcut happens to work for Room A only because 8 m and 6 m are exact multiples of the tile size - a coincidence that makes a wrong rule look reliable. Count tiles along each wall, round up each count, then multiply.