---
title: "Rate analysis: brick masonry 1:5, superstructure"
subject: "Estimating and Costing"
chapter: 4
key: "semester_7/estimating_and_costing:rate-brick-1-5-superstructure"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/rate-brick-1-5-superstructure
site: BCT Engineering
---

# Rate analysis: brick masonry 1:5, superstructure

Estimating and Costing · Chapter 4: Analysis of Rates

- Asked in: 2068 Baisakh (old course), 4 marks
- Video: [Estimation of Brickwork | Rate analysis](https://www.youtube.com/watch?v=GgzGTfH45dw) — Civil Site visit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/rate-brick-1-5-superstructure
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-4

## Question

Prepare an analysis of rate of brick masonry in (1:5) cement mortar in super structure. Assume size of brick 240 × 130 × 65 mm and thickness of mortar joint is 12 mm.

## Short answer

Brick 240x130x65 mm, 12 mm mortar yields 363 bricks/m³. Per m³: bricks 362.9, cement 1.609 bags, sand 0.2794 m³. Direct cost NPR 14,089.10; total with 15% overhead and 10% profit = NPR 17,611.38 per m³.

## Answer

Rates and labour norms below are illustrative (DUDBC-style norms; use the current District Rate and norms of the project area for a real estimate): cement NPR 1,000/bag, sand NPR 3,500/m³, 20 mm aggregate NPR 4,500/m³, skilled labour NPR 1,500/day, unskilled labour NPR 1,100/day, water lump sum, contractor's overhead 15% and profit 10% on direct cost (VAT excluded). Brick is taken at NPR 18 each.

#### Quantities per m³ of brickwork

- Brick with mortar $= 0.252 \times 0.142 \times 0.077 = 2,755\times10^{-6}$ m³; bricks per m³ $= 362.9$
- Brick volume $= 0.24 \times 0.13 \times 0.065 = 2,028\times10^{-6}$ m³; wet mortar $= 1 - 362.9\times2,028\times10^{-6} = 0.2640$ m³
- Dry mortar $= 1.27 \times 0.2640 = 0.3353$ m³; cement $= 1/6 \times 0.3353 = 0.0559$ m³ $= 1.609$ bags; sand $= 5/6 \times 0.3353 = 0.2794$ m³
- Labour (superstructure, assumed norm): mason 1.6 days, labourer 2.2 days per m³.

#### Analysis

| Item | Qty | Unit | Rate (NPR) | Amount (NPR) |
|---|---:|---|---:|---:|
| Bricks 240 × 130 × 65 mm | 362.9 | no | 18.00 | 6,532.20 |
| Cement | 1.609 | bag | 1,000.00 | 1,609.00 |
| Sand | 0.2794 | m³ | 3,500.00 | 977.90 |
| Skilled labour (mason) | 1.6 | day | 1,500.00 | 2,400.00 |
| Unskilled labour | 2.2 | day | 1,100.00 | 2,420.00 |
| Water and scaffolding (lump) | 1 | LS | 150.00 | 150.00 |
| **Direct cost** | | | | **14,089.10** |
| Contractor's overhead @ 15% | | | | 2,113.37 |
| Contractor's profit @ 10% | | | | 1,408.91 |
| **Total (rate per m³)** | | | | **17,611.38** |

**Rate of 1:5 brickwork in superstructure = NPR 17,611.38 per m³.**

## Check yourself

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

1. How is the number of bricks per m3 of brickwork found in this analysis?
   A. 1 divided by the volume of the nominal brick 0.24 x 0.13 x 0.065 m
   B. Volume of the wall divided by the volume of mortar
   C. Number of bricks in one course multiplied by 10
   D. 1 divided by the volume of one brick including its mortar joint (0.252 x 0.142 x 0.077 m)
   Answer: D. With 12 mm joints each brick occupies 0.252 x 0.142 x 0.077 m, giving 362.9 bricks per m3. (Key point: Bricks per m3)

2. How is the wet mortar needed per m3 of brickwork obtained?
   A. 1 m3 minus the volume of the bricks alone (362.9 x 0.002028 m3), giving 0.2640 m3
   B. Equal to the volume of bricks, 0.736 m3
   C. 1 m3 plus the volume of the bricks
   D. One-tenth of the brickwork volume taken as a thumb rule
   Answer: A. Whatever space in 1 m3 is not occupied by the brick bodies is filled by mortar. (Key point: Wet mortar volume)

3. For 1:5 mortar, how are cement and sand quantities found from the wet mortar?
   A. Take 1/5 as cement and 4/5 as sand of the wet volume
   B. Multiply wet mortar by 1.27 for dry volume, then take 1/6 as cement and 5/6 as sand
   C. Take 1/5 as cement and 5/5 as sand of the dry volume
   D. Use the wet mortar volume directly with 1/6 and 5/6
   Answer: B. Dry volume = 1.27 x 0.2640 = 0.3353 m3; sum of ratio is 1+5 = 6, so cement 1/6 and sand 5/6. (Key point: Dry mortar and cement)

4. In the final total, how are overhead (15%) and profit (10%) applied?
   A. Profit is taken on direct cost plus overhead
   B. They are taken only on the material cost
   C. Both are taken on the direct cost (14,089.10) and added to it, giving 17,611.38 per m3
   D. They are deducted from the direct cost
   Answer: C. Overhead and profit are each a percentage of direct cost: 14,089.10 + 2,113.37 + 1,408.91 = 17,611.38. (Key point: Overhead and profit)

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