---
title: "Materials: 75 m³ brick 1:3, 115 m² PCC floor"
subject: "Estimating and Costing"
chapter: 4
key: "semester_7/estimating_and_costing:materials-brick-75m3-pcc-floor"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/materials-brick-75m3-pcc-floor
site: BCT Engineering
---

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

Estimating and Costing · Chapter 4: Analysis of Rates

- Asked in: 2074 Asoj, 4 marks
- Video: [Brick Quantity calculation for house | How to calculate sand & Cement for Brick work By CivilGuruji](https://www.youtube.com/watch?v=bFY5ZaOrY7U) — The Institute of Construction Managment 
- 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-brick-75m3-pcc-floor
- 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) 75 m³ of brick work in (1:3) cement mortar (ii) 115 m² of 75 mm thick PCC (1:2:4) in floor.

## Short answer

For 75 m³ brickwork in 1:3 mortar: bricks 40,065 nos, cement 176.0 bags, sand 18.34 m³. For 115 m² 75 mm thick PCC floor in 1:2:4: wet volume 8.625 m³, dry 13.283 m³; cement 54.6 bags, sand 3.80 m³, aggregate 7.59 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. 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.**

## Check yourself

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

1. Brick size is not given. What is assumed?
   A. Nominal 9 x 4.5 x 3 in
   B. 240x115x60
   C. Standard 230x110x55 mm with 10 mm joint
   D. None, can't solve
   Answer: C. Standard brick assumed. (Key point: Assumed brick)

2. Bricks for 75 m^3?
   A. About 5,342
   B. About 40,065
   C. About 37,500
   D. About 49,907
   Answer: B. 534.19 x 75. (Key point: Bricks)

3. Cement for 75 m^3 at 1:3?
   A. Dry 24.449 m^3; 1/4 = 6.112 m^3 = 176 bags
   B. 1/3 of 19.251
   C. 3/4 of 24.449
   D. 1/4 of 19.251 = 4.8
   Answer: A. Ratio sum 4. (Key point: Cement 1:3)

4. Cement for 115 m^2 x 75 mm PCC 1:2:4?
   A. 1/4 x 13.283
   B. 1/7 x 8.625
   C. 1/6 x 13.283
   D. 1/7 x 13.283 = 1.898 m^3 = 54.6 bags
   Answer: D. Sum 7. (Key point: PCC)

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