---
title: "Materials for 100 m brick wall 1:6"
subject: "Estimating and Costing"
chapter: 4
key: "semester_7/estimating_and_costing:materials-brick-wall-100m"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/materials-brick-wall-100m
site: BCT Engineering
---

# Materials for 100 m brick wall 1:6

Estimating and Costing · Chapter 4: Analysis of Rates

- Asked in: 2076 Asoj, 6 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-wall-100m
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-4

## Question

Calculate the quantities of materials required for a 100 m long, 23 cm thick and 1.20 m high wall in (1:6) cement mortar. (Assume size of brick is 235 × 110 × 57 mm and thickness of mortar 10 mm.)

## Short answer

Wall volume 27.60 m³ yields 14,012 bricks (nominal size 235×110×57 mm with 10 mm joint). Mortar dry volume 8.832 m³ for 1:6 mix: cement 36.3 bags, sand 7.57 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.

#### Volume of wall

$V = 100 \times 0.23 \times 1.20 = 27.60$ m³ (23 cm = two bricks of 110 mm plus one 10 mm joint).

#### Quantities (brick 235 × 110 × 57 mm, joint 10 mm, mortar 1:6)

**Brickwork 1:6**
- Brick with mortar $= 0.245 \times 0.120 \times 0.067 = 1969.8\times 10^{-6}$ m³; bricks per m³ $= 1/1969.8\times10^6 = 507.67$
- Bricks for 27.60 m³ $= 507.67 \times 27.60 = 14,012$ nos (say 14,012)
- Volume of bricks only $= 0.235\times0.110\times0.057 = 1473.4\times10^{-6}$ m³ each; wet mortar per m³ $= 1 - 507.67\times1473.4\times10^{-6} = 0.2520$ m³
- Wet mortar for 27.60 m³ $= 6.955$ m³; dry $= 1.27\times6.955 = 8.832$ m³
- Cement $= 1/7\times8.832 = 1.262$ m³ $= 36.3$ bags
- Sand $= 6/7\times8.832 = 7.571$ m³

**Answer: bricks = 14,012 nos; cement = 36.3 bags; sand = 7.57 m³.**

## Check yourself

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

1. How is the volume of the 23 cm wall found?
   A. 100 x 0.23 x 1.20 = 27.60 m^3
   B. 100 x 0.11 x 1.20
   C. 100 x 1.20 = 120 m^3
   D. 100 x 0.245 x 1.20
   Answer: A. A 23 cm wall = two 110 mm bricks and a 10 mm joint. (Key point: Wall volume)

2. How are bricks per m^3 found?
   A. 1 divided by brick size without mortar = 678.7
   B. 1 divided by 0.235 x 0.11 x 0.057 only, then minus 1.27
   C. Divide by 1.54
   D. 1 divided by volume of brick with mortar (0.245 x 0.120 x 0.067) = 507.67
   Answer: D. Each brick occupies the brick plus joint volume, so bricks per m^3 = 507.67. (Key point: Bricks per m^3)

3. How is wet mortar per m^3 of brickwork found?
   A. 1.27 x brick volume
   B. 1 minus brick-with-mortar volume
   C. 1 minus the volume of bricks only, giving 0.2520 m^3
   D. Always 30% of wall
   Answer: C. 507.67 bricks of 1473.4 x 10^-6 m^3 leave 0.252 m^3 for mortar. (Key point: Wet mortar per m^3)

4. Wet mortar 6.955 m^3 gives which cement?
   A. 6.955/7 = 0.99 m^3
   B. Dry 8.832 m^3, cement 1/7 = 1.262 m^3 = 36.3 bags
   C. 8.832/6 = 1.472 m^3
   D. 8.832 x 6/7 = 7.57 m^3
   Answer: B. Dry = 1.27 x 6.955; cement is 1 of 7 parts. (Key point: Cement for 1:6)

## Related in Estimating and Costing

- [Materials for brick wall 10 m (1:5)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Amaterials-brick-wall-10m-1-5) — Estimating and Costing, chapter 4, asked 2×
- [Materials for brick wall 20 m (1:5)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Amaterials-brick-wall-20m) — Estimating and Costing, chapter 4, asked 2×
- [Materials for brick wall, two windows 1.5 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Amaterials-brick-wall-windows-1-5) — Estimating and Costing, chapter 4
- [Materials: 10 m³ brick 1:4, 100 m³ PCC 1:3:6](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Amaterials-brick-pcc-v) — Estimating and Costing, chapter 4, asked 2×
- [Brick (1:6) materials; PCC (1:2:4) rate](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Amaterials-brick-1-6-and-pcc-rate) — Estimating and Costing, chapter 4
- [Materials for 10 m³ brick masonry 1:3](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Amaterials-brick-10m3-1-3-old-size) — Estimating and Costing, chapter 4

## Related in other subjects

- [Brick and mortar for 23 cm wall (165 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-23cm-165kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
- [Brick and mortar: 20 cm wall, 3.4 m](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-20cm-100kn-3-4m) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
- [Brick and mortar for 230 mm wall (100 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-230mm-100kn-5m) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
- [Brick and mortar for 250 mm wall (350 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-250mm-350kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×