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Materials: 115 m² PCC floor, 110 m² plaster

Short answer

(i) 115 m² of 75 mm PCC (1:3:6) needs cement 38.3 bags, sand 3.98 m³, aggregate 7.97 m³. (ii) 110 m² of 12.5 mm plaster (1:4) needs cement 10.1 bags, sand 1.40 m³.

Calculate the quantities of materials required for the following items of work: (i) 115 m² of 75 mm thick PCC (1:3:6) in floor. (ii) 110 m² of 12.5 mm thick cement sand plaster (1:4) in wall.

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.

(i) 115 m² of 75 mm thick PCC (1:3:6)

  • Wet volume =115×0.075=8.625= 115 \times 0.075 = 8.625 m³; dry volume =1.54×8.625=13.283= 1.54 \times 8.625 = 13.283 m³ (sum of ratio =10= 10)
  • Cement =1/10×13.283=1.328= 1/10 \times 13.283 = 1.328 m³ =1.328×28.8=38.3= 1.328 \times 28.8 = 38.3 bags (50 kg)
  • Sand =3/10×13.283=3.985= 3/10 \times 13.283 = 3.985 m³
  • Coarse aggregate =6/10×13.283=7.970= 6/10 \times 13.283 = 7.970 m³

(ii) 110 m² of 12.5 mm plaster (1:4)

  • Wet mortar =110.0×0.0125=1.3750= 110.0 \times 0.0125 = 1.3750 m³; dry volume =1.27×1.3750=1.7463= 1.27 \times 1.3750 = 1.7463 m³
  • Cement =1/5×1.7463=0.3493= 1/5 \times 1.7463 = 0.3493 m³ =10.06= 10.06 bags
  • Sand =4/5×1.7463=1.3970= 4/5 \times 1.7463 = 1.3970 m³

Answer: (i) cement 38.3 bags, sand 3.98 m³, aggregate 7.97 m³. (ii) cement 10.1 bags, sand 1.40 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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