Materials: 75 m³ brick 1:3, 115 m² PCC floor

Assumptions: dry volume = 1.54 × wet volume for concrete and 1.27 × wet volume for mortar; 1 m³ of cement = 28.8 bags of 50 kg; no wastage is added. Brick size is not given, so a standard brick of 230 × 110 × 55 mm with 10 mm joints is assumed.

(i) 75 m³ brickwork in 1:3 mortar

• Brick with mortar $= 0.240 \times 0.120 \times 0.065 = 1872.0\times 10^{-6}$ m³; bricks per m³ $= 1/1872.0\times10^6 = 534.19$
• Bricks for 75.00 m³ $= 534.19 \times 75.00 = 40,064$ nos (say 40,065)
• Volume of bricks only $= 0.230\times0.110\times0.055 = 1391.5\times10^{-6}$ m³ each; wet mortar per m³ $= 1 - 534.19\times1391.5\times10^{-6} = 0.2567$ m³
• Wet mortar for 75.00 m³ $= 19.251$ m³; dry $= 1.27\times19.251 = 24.449$ m³
• Cement $= 1/4\times24.449 = 6.112$ m³ $= 176.0$ bags
• Sand $= 3/4\times24.449 = 18.336$ m³

(ii) 115 m² × 75 mm PCC (1:2:4)

• Wet volume $= 8.625$ m³; dry volume $= 1.54 \times 8.625 = 13.283$ m³ (sum of ratio $= 7$)
• Cement $= 1/7 \times 13.283 = 1.898$ m³ $= 1.898 \times 28.8 = 54.6$ bags (50 kg)
• Sand $= 2/7 \times 13.283 = 3.795$ m³
• Coarse aggregate $= 4/7 \times 13.283 = 7.590$ m³

Answer: (i) bricks 40,065; cement 176.0 bags; sand 18.34 m³. (ii) cement 54.6 bags; sand 3.80 m³; aggregate 7.59 m³.