The Hall Floor Tile Contract
A school in Mysuru is retiling a rectangular multipurpose hall measuring 12.96 m by 8.64 m. The contractor will lay square tiles edge to edge, with no cutting allowed. Tiles are available only in side lengths that are whole numbers of centimetres, and the school has capped the tile side at 80 cm because bigger tiles are difficult to store and handle. The contractor suggests 90 cm tiles because they would cover the floor faster, but the PE teacher points out that a 90 cm tile cannot fit the width exactly. The school wants the largest permissible square tile that divides both dimensions exactly, then the number of tiles to order with 5% extra for breakage, rounded up to a whole tile, and finally the number of boxes, since tiles are packed six to a box. A second supplier offers 54 cm tiles at ₹95 each, while the 72 cm tiles cost ₹180 each. The management wants a material-cost comparison before deciding, although labour will be higher with smaller tiles.
1. Determine the largest tile side the school may use, and show why the contractor's 90 cm suggestion is not admissible.
Read the worked answer
Convert to centimetres: 12.96 m = 1296 cm and 8.64 m = 864 cm. For no cutting, the tile side s must divide both dimensions, so s must divide gcd(1296, 864).
1296 = 2^4 × 3^4 and 864 = 2^5 × 3^3, so gcd = 2^4 × 3^3 = 432 cm.
A 90 cm tile fails because 1296 ÷ 90 = 14.4 and 864 ÷ 90 = 9.6, so both edges would need cutting.
The permissible sides are divisors of 432. The largest divisor not exceeding 80 cm is 72 cm, since 72 divides both (1296 ÷ 72 = 18 and 864 ÷ 72 = 12), while 80 does not divide 432 and there is no larger divisor below 80. So the tile side is 72 cm.
2. Using the 72 cm tile, find the exact number of tiles, the number to order with 5% extra rounded up, and the number of boxes required.
Read the worked answer
Exact tiles = 18 × 12 = 216.
Extra 5% = 0.05 × 216 = 10.8, so required = 216 + 10.8 = 226.8 tiles. Since tiles are whole, round up to 227 tiles.
Boxes of 6: 227 ÷ 6 = 37.83..., so 38 boxes are needed, giving 228 tiles. Order 38 boxes.
3. Compare the material cost of 72 cm tiles with 54 cm tiles, and state whether the cheaper material is automatically the cheaper contract.
Read the worked answer
54 cm divides both dimensions: 1296 ÷ 54 = 24 and 864 ÷ 54 = 16, so exact tiles = 384.
With 5%: 384 × 1.05 = 403.2, so 404 tiles. Boxes: 404 ÷ 6 = 67.33..., so 68 boxes = 408 tiles. Cost = 408 × 95 = ₹38,760.
For 72 cm: 38 boxes = 228 tiles, cost = 228 × 180 = ₹41,040.
Material saving with 54 cm = ₹41,040 − ₹38,760 = ₹2,280. However, the 54 cm option needs 384 tile-setting operations instead of 216 and more grout/labour. The lower tile cost does not automatically make it the cheaper contract; the material comparison favours 54 cm, but total cost may not.