---
title: "Materials: 115 m² PCC floor, 110 m² plaster"
subject: "Estimating and Costing"
chapter: 4
key: "semester_7/estimating_and_costing:materials-pcc-floor-plaster-i"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/materials-pcc-floor-plaster-i
site: BCT Engineering
---

# Materials: 115 m² PCC floor, 110 m² plaster

Estimating and Costing · Chapter 4: Analysis of Rates

- Asked in: 2076 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/materials-pcc-floor-plaster-i
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-4

## Question

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.

## 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.

## Answer

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

#### (ii) 110 m² of 12.5 mm plaster (1:4)

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which dry-volume factors are used?
   A. 1.27 for concrete and 1.54 for mortar
   B. 1.54 for both
   C. 1.30 for both
   D. 1.54 for concrete and 1.27 for mortar
   Answer: D. Concrete dry volume is 1.54 times wet, mortar 1.27 times wet. (Key point: Dry volume factor)

2. For 115 m^2 x 75 mm PCC 1:3:6, what is the cement?
   A. 1/6 of 8.625 = 1.44 m^3
   B. 1/9 of 13.283 = 1.476 m^3
   C. 1/10 of dry 13.283 m^3 = 1.328 m^3 = 38.3 bags
   D. 1/10 of 8.625 = 0.86 m^3
   Answer: C. The ratio sum is 10 and dry volume 1.54 x 8.625 = 13.283 m^3. (Key point: PCC 1:3:6 cement)

3. For 110 m^2 of 12.5 mm plaster 1:4, how is cement found?
   A. 1.375 x 1/4 = 0.344 m^3
   B. Dry mortar 1.7463 m^3 x 1/5 = 0.3493 m^3 = 10.06 bags
   C. 1.7463 x 1/4 = 0.4366 m^3
   D. 1.7463 x 4/5 = 1.397 m^3
   Answer: B. The ratio sum is 5, so cement is 1/5 of dry mortar 1.27 x 1.375. (Key point: Plaster 1:4 cement)

## Close variants set in other exams

- [Materials: 105 m³ PCC 1:4:8, 725 m² plaster](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Amaterials-pcc-plaster-r) — Estimating and Costing, chapter 4, asked 2×

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