Analysis of Rates

Basis

Rate analysis for supplying and fixing one W.C. commode (pan) with a low-level cistern, complete with flush pipe, fixing and testing. Quantities follow the DUDBC-style norm for sanitary fixtures; the rates are illustrative (replace by the District Rate): cement Rs 900/bag, sand Rs 2,500/m³, brick Rs 15 each, skilled labour Rs 1,200/day, unskilled Rs 900/day.

Analysis (per set / number)

Answer: Rate of W.C. pan with low-level cistern = Rs 14,691.25 per set (about Rs 14,690), excluding 13% VAT.

Steps in work: the pan is bedded on a 1:4 cement mortar bed over brick filling, the trap is connected to the soil pipe, the cistern is fixed on the wall with brackets at low level, the flush pipe is connected, joints are sealed, and the unit is tested for flushing and leaks.

Rate analysis

Rate analysis is the determination of the cost of executing one unit of an item of work (per m³, m², m, kg or no.), by adding the cost of the materials, labour and equipment needed for that unit, with the overhead and profit.

Purposes

1. To find the unit rate of each item for the abstract of cost and the BOQ.
2. To check the reasonableness of tendered rates and of rates for extra and substitute items.
3. To fix the rates for variations and day-work in a contract.
4. To compare the cost of alternative materials and methods.
5. To calculate the requirement of materials and labour for each unit of work.
6. To prepare the District Rate and department norms.

Factors affecting rate analysis

1. Location and lead: distance from the source of materials to the site (lead) and the lift; remote hill districts have a high transport cost.
2. Availability and price of materials: local or imported materials, royalty, taxes and market fluctuation.
3. Labour: wage rates, availability of skilled labour and productivity (output per day).
4. Specification and quality: mix proportion, grade, thickness and finish required.
5. Quantity of work: a large quantity lowers the unit rate through economy of scale.
6. Equipment and method of construction: manual or mechanical, the cost of machines, fuel and operators.
7. Climate, season and site conditions: rain, altitude, access, difficult soil or water, working space and height of work (lifting).
8. Time of completion: a short time needs more resources and overtime.
9. Wastage and norms: the standard wastage allowance and the consumption norms of the department.
10. Contractor's overhead and profit, taxes (VAT) and insurance, and the mode of payment and contract conditions.

Significance of rate analysis in civil projects

1. Gives the unit rate of every item, which with the quantities gives the cost of the project.
2. Gives the consumption of materials, labour and equipment, for procurement and scheduling.
3. Provides a sound basis for tender evaluation, and for judging abnormally low or high bids.
4. Helps to price extra items and variations fairly.
5. Allows comparison of alternatives (materials, methods, mix proportions).
6. Helps in cost control during construction, and standardizes the rates in the government departments.

Requirements for rate analysis

1. Drawings and specifications of the item (to know the dimensions, quality and method).
2. Material rates: local market rates and the District Rate of the materials, with the transport (lead) and loading/unloading charges.
3. Labour rates: wages of skilled and unskilled labour from the District Rate, and the labour productivity.
4. Norms: the standard consumption of materials and labour per unit of work (DUDBC norms for building and sanitary work, Department of Roads norms for road work).
5. Equipment data: hire or running cost of machines, with their output.
6. Wastage allowances and conversion factors, such as dry volume of mortar and compaction factors.
7. Overhead, profit and tax: contractor's overhead and profit (15% in the DUDBC norms), VAT 13% and any royalty or insurance.
8. Site data: lead, lift, location and conditions of the work.

Basis

Rate per m³ of finished sal wood door and window frames (planed, joined, fixed and treated). The norm is: 1 m³ finished frame needs about 1.10 m³ of sawn sal wood (wastage in planing and cutting), carpenter 12 md and helper 6 md. Rates are illustrative: sal wood Rs 140,000/m³ (about Rs 4,000 per cft), skilled labour Rs 1,200/day and unskilled Rs 900/day.

Analysis (per m³)

Answer: Rate of sal wood frames = Rs 203,895.00 per m³ (excluding VAT).

Check of the unit price: Rs 5,773.66 per cft.

Basis

12.5 mm thick plaster in cement mortar 1:5 on brick wall, worked for 10 m² and divided by 10. Rates are illustrative.

Mortar quantity

• Wet volume per m² = 0.0125 m³; add 20% for filling uneven wall surface and wastage: 0.0150 m³.
• Dry mortar = 0.0150 x 1.30 = 0.01950 m³ per m².
• For 10 m²: dry mortar = 0.1950 m³; cement = 0.1950/6 = 0.03250 m³ = 46.8 kg = 0.936 bags; sand = 0.1625 m³.

Analysis (per 10 m²)

Answer: Rate of 12.5 mm 1:5 plaster = 4,333.95 / 10 = Rs 433.39 per m² (excluding VAT).

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).

Quantities per 10 m²

• Wet mortar $= 10 \times 0.012 = 0.120$ m³; dry $= 1.27 \times 0.120 = 0.1524$ m³
• Cement $= 1/4 \times 0.1524 = 0.0381$ m³ $= 1.097$ bags; sand $= 3/4 \times 0.1524 = 0.1143$ m³
• Labour for ceiling (overhead work, about 25% more than wall plaster): mason 1.2 days, labourer 1.3 days; plus staging.

Analysis

Rate = NPR 6,283.82 per 10 m² (NPR 628.38 per m²).

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).

Quantities per 100 m²

• Wet mortar $= 100 \times 0.02 = 2.00$ m³; dry $= 1.27 \times 2.00 = 2.540$ m³
• Cement $= 1/5 \times 2.540 = 0.5080$ m³ $= 14.63$ bags; sand $= 4/5 \times 2.540 = 2.032$ m³
• Labour (assumed norm for a 20 mm two-coat plaster): mason 10 days, labourer 12 days per 100 m².

Analysis

Rate = NPR 64,677.50 per 100 m² (NPR 646.77 per m²).

Requirements of rate analysis

1. Drawings and specifications of the item (to know the dimensions, quality and method).
2. Material rates: local market rates and the District Rate of the materials, with the transport (lead) and loading/unloading charges.
3. Labour rates: wages of skilled and unskilled labour from the District Rate, and the labour productivity.
4. Norms: the standard consumption of materials and labour per unit of work (DUDBC norms for building and sanitary work, Department of Roads norms for road work).
5. Equipment data: hire or running cost of machines, with their output.
6. Wastage allowances and conversion factors, such as dry volume of mortar and compaction factors.
7. Overhead, profit and tax: contractor's overhead and profit (15% in the DUDBC norms), VAT 13% and any royalty or insurance.
8. Site data: lead, lift, location and conditions of the work.

Factors affecting rate analysis

1. Location and lead: distance from the source of materials to the site (lead) and the lift; remote hill districts have a high transport cost.
2. Availability and price of materials: local or imported materials, royalty, taxes and market fluctuation.
3. Labour: wage rates, availability of skilled labour and productivity (output per day).
4. Specification and quality: mix proportion, grade, thickness and finish required.
5. Quantity of work: a large quantity lowers the unit rate through economy of scale.
6. Equipment and method of construction: manual or mechanical, the cost of machines, fuel and operators.
7. Climate, season and site conditions: rain, altitude, access, difficult soil or water, working space and height of work (lifting).
8. Time of completion: a short time needs more resources and overtime.
9. Wastage and norms: the standard wastage allowance and the consumption norms of the department.
10. Contractor's overhead and profit, taxes (VAT) and insurance, and the mode of payment and contract conditions.

Basis

Rate per m³ of 1st class brickwork in 1:6 cement-sand mortar in the upper floor, standard brick 230 x 110 x 55 mm with 10 mm joints (nominal size 240 x 120 x 65 mm). All rates are illustrative.

Materials per m³

• Bricks (net) = $\dfrac{1}{0.24 \times 0.12 \times 0.065} = 534.19$ nos.; with 5% breakage = 561 nos.
• Volume of bricks without mortar = 534.19 x 0.23 x 0.11 x 0.055 = 0.7433 m³, so wet mortar = 1 - 0.7433 = 0.2567 m³.
• Dry mortar = 0.2567 x 1.30 = 0.3337 m³ (30% for filling voids and shrinkage).
• Cement = 0.3337/7 = 0.0477 m³ x 1,440 kg/m³ = 68.6 kg = 1.37 bags.
• Sand = 0.3337 x 6/7 = 0.286 m³.

Analysis (per m³)

Answer: Rate of 1:6 brickwork in the upper floor = Rs 18,357.45 per m³ (excluding VAT). The extra labour and scaffolding for the upper floor is the difference from the ground-floor rate.

Basis

Rate for providing, laying and consolidating 40 mm thick premix asphalt carpeting, worked out for 100 m² and then divided to give the rate per m². Quantities follow the usual DUDBC / DoR (Department of Roads) practice; rates are illustrative.

Assumptions:

• Density of compacted mix = 2,300 kg/m³, so mass per m² = 0.04 x 2,300 = 92 kg.
• Bitumen content 5.5% of the mix = 5.06 kg/m² (plus 0.25 kg/m² for the tack coat on the base); aggregate = 86.94 kg/m², and bulk density of loose aggregate 1,600 kg/m³ gives 0.0543 m³/m².

Analysis (per 100 m²)

Answer: Rate = 117,662.25 / 100 = Rs 1,176.62 per m² (excluding VAT).

Net volume of masonry

• Gross volume = 20 x 0.23 x 3 = 13.800 m³
• Openings = 2 x 1.2 x 1.2 x 0.23 = 0.6624 m³
• Net volume = 13.800 - 0.6624 = 13.1376 m³

Bricks

Using the standard brick 230 x 110 x 55 mm with 10 mm mortar joints, the nominal size is 240 x 120 x 65 mm.

$$\text{Bricks per m}^3 = \frac{1}{0.240 \times 0.120 \times 0.065} = 534.19$$

Number of bricks = 534.19 x 13.1376 = 7018 nos. (about 7369 nos. with 5% wastage).

Mortar (1:5)

• Volume of bricks without mortar = 7018 x 0.23 x 0.11 x 0.055 = 9.7655 m³
• Wet mortar = 13.1376 - 9.7655 = 3.3721 m³
• Dry mortar = 3.3721 x 1.30 = 4.3838 m³
• Cement = 4.3838/6 = 0.7306 m³ = 1052.1 kg = 21.04 bags (50 kg)
• Sand = 4.3838 x 5/6 = 3.653 m³

Answer: Bricks = 7018 nos.; cement = 21.04 bags (0.731 m³); sand = 3.653 m³.

Net volume of masonry

• Gross volume = 10 x 0.23 x 3 = 6.900 m³
• Openings = 2 x 1.2 x 1.2 x 0.23 = 0.6624 m³
• Net volume = 6.900 - 0.6624 = 6.2376 m³

Bricks

Using the standard brick 230 x 110 x 55 mm with 10 mm mortar joints, the nominal size is 240 x 120 x 65 mm.

$$\text{Bricks per m}^3 = \frac{1}{0.240 \times 0.120 \times 0.065} = 534.19$$

Number of bricks = 534.19 x 6.2376 = 3332 nos. (about 3499 nos. with 5% wastage).

Mortar (1:5)

• Volume of bricks without mortar = 3332 x 0.23 x 0.11 x 0.055 = 4.6365 m³
• Wet mortar = 6.2376 - 4.6365 = 1.6011 m³
• Dry mortar = 1.6011 x 1.30 = 2.0814 m³
• Cement = 2.0814/6 = 0.3469 m³ = 499.5 kg = 9.99 bags (50 kg)
• Sand = 2.0814 x 5/6 = 1.734 m³

Answer: Bricks = 3332 nos.; cement = 9.99 bags (0.347 m³); sand = 1.734 m³.

Basis

Rate for providing, laying and consolidating 10 cm thick compacted gravel in the sub-grade, worked out for 100 m² and divided for 1 m². Assumptions (illustrative, in the style of Department of Roads norms):

• Compacted volume = 100 x 0.10 = 10.0 m³; loose volume = compacted x 1.30 (compaction factor) = 13.0 m³.
• Water for compaction = 12% of compacted volume = 1.20 m³.
• Labour 0.4 unskilled md per m³ and 0.05 skilled md per m³ of compacted gravel; roller 0.05 hr per m³.

Analysis (per 100 m²)

Answer: Rate = 52,957.50 / 100 = Rs 529.58 per m² (excluding VAT).

Basis

Rate for providing, laying and consolidating 30 cm thick compacted gravel in the sub-grade, per 100 m², laid and compacted in two layers of 15 cm (a single layer thicker than about 15 to 20 cm cannot be compacted properly). Assumptions (illustrative):

• Compacted volume = 100 x 0.30 = 30.0 m³; loose volume = compacted x 1.30 = 39.0 m³.
• Water for compaction = 12% of compacted volume = 3.60 m³.
• Labour 0.4 unskilled md and 0.05 skilled md per m³ of compacted gravel; roller 0.05 hr per m³.

Analysis (per 100 m²)

Answer: Rate = 158,872.50 / 100 = Rs 1,588.72 per m² (excluding VAT).

Check: the 30 cm rate is three times the 10 cm rate, Rs 1,588.72 per m², as every quantity scales with thickness.

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)

(i) PCC 1:3:6, 115 m² × 0.075 m

• Wet volume $= 8.625$ 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)

(ii) Plaster 1:4, 12.5 mm, 110 m²

• 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.

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) 105 m³ PCC (1:4:8) in foundation

(i) 105 m³ PCC 1:4:8

• Wet volume $= 105.000$ m³; dry volume $= 1.54 \times 105.000 = 161.700$ m³ (sum of ratio $= 13$)
• Cement $= 1/13 \times 161.700 = 12.438$ m³ $= 12.438 \times 28.8 = 358.2$ bags (50 kg)
• Sand $= 4/13 \times 161.700 = 49.754$ m³
• Coarse aggregate $= 8/13 \times 161.700 = 99.508$ m³

(ii) 725 m² of 20 mm plaster (1:4)

(ii) 725 m² plaster 20 mm, 1:4

• Wet mortar $= 725.0 \times 0.0200 = 14.5000$ m³; dry volume $= 1.27 \times 14.5000 = 18.4150$ m³
• Cement $= 1/5 \times 18.4150 = 3.6830$ m³ $= 106.07$ bags
• Sand $= 4/5 \times 18.4150 = 14.7320$ m³

Answer: (i) cement 358.2 bags, sand 49.75 m³, aggregate 99.51 m³. (ii) cement 106.1 bags, sand 14.73 m³.

Assumptions: dry volume 1.54 × wet for concrete and 1.27 × wet for mortar; 1 m³ cement = 28.8 bags. The brick size is not given, so 230 × 110 × 55 mm with 10 mm joints is taken.

(i) 10 m³ brickwork in 1:6 mortar

(i) 10 m³ brickwork, mortar 1:6

• 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 10.00 m³ $= 534.19 \times 10.00 = 5,342$ nos (say 5,342)
• 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 10.00 m³ $= 2.567$ m³; dry $= 1.27\times2.567 = 3.260$ m³
• Cement $= 1/7\times3.260 = 0.466$ m³ $= 13.4$ bags
• Sand $= 6/7\times3.260 = 2.794$ m³

(ii) 10 m³ PCC (1:3:6)

(ii) 10 m³ PCC (1:3:6)

• Wet volume $= 10.000$ m³; dry volume $= 1.54 \times 10.000 = 15.400$ m³ (sum of ratio $= 10$)
• Cement $= 1/10 \times 15.400 = 1.540$ m³ $= 1.540 \times 28.8 = 44.4$ bags (50 kg)
• Sand $= 3/10 \times 15.400 = 4.620$ m³
• Coarse aggregate $= 6/10 \times 15.400 = 9.240$ m³

Answer: (i) bricks 5,342, cement 13.4 bags, sand 2.79 m³. (ii) cement 44.4 bags, sand 4.62 m³, aggregate 9.24 m³.

Assumptions: dry volume 1.54 × wet for concrete and 1.27 × wet for mortar; 1 m³ cement = 28.8 bags. Brick size not given: 230 × 110 × 55 mm with 10 mm joints.

(i) 10 m³ brickwork in 1:4 mortar

(i) 10 m³ brickwork, 1:4

• 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 10.00 m³ $= 534.19 \times 10.00 = 5,342$ nos (say 5,342)
• 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 10.00 m³ $= 2.567$ m³; dry $= 1.27\times2.567 = 3.260$ m³
• Cement $= 1/5\times3.260 = 0.652$ m³ $= 18.8$ bags
• Sand $= 4/5\times3.260 = 2.608$ m³

(ii) 100 m³ PCC (1:3:6)

(ii) 100 m³ PCC (1:3:6) in foundation

• Wet volume $= 100.000$ m³; dry volume $= 1.54 \times 100.000 = 154.000$ m³ (sum of ratio $= 10$)
• Cement $= 1/10 \times 154.000 = 15.400$ m³ $= 15.400 \times 28.8 = 443.5$ bags (50 kg)
• Sand $= 3/10 \times 154.000 = 46.200$ m³
• Coarse aggregate $= 6/10 \times 154.000 = 92.400$ m³

Answer: (i) bricks 5,342, cement 18.8 bags, sand 2.61 m³. (ii) cement 443.5 bags, sand 46.20 m³, aggregate 92.40 m³.

Factors affecting rate analysis

1. Location and lead: distance from the source of materials to the site (lead) and the lift; remote hill districts have a high transport cost.
2. Availability and price of materials: local or imported materials, royalty, taxes and market fluctuation.
3. Labour: wage rates, availability of skilled labour and productivity (output per day).
4. Specification and quality: mix proportion, grade, thickness and finish required.
5. Quantity of work: a large quantity lowers the unit rate through economy of scale.
6. Equipment and method of construction: manual or mechanical, the cost of machines, fuel and operators.
7. Climate, season and site conditions: rain, altitude, access, difficult soil or water, working space and height of work (lifting).
8. Time of completion: a short time needs more resources and overtime.
9. Wastage and norms: the standard wastage allowance and the consumption norms of the department.
10. Contractor's overhead and profit, taxes (VAT) and insurance, and the mode of payment and contract conditions.

Necessity of rate analysis in civil engineering

• The cost of a work cannot be found without the unit rates, and these are not constant; they depend on place, time and specification, so each item must be analysed.
• It is required to prepare estimates, BOQs and tender documents, to evaluate bids, and to value variations.
• It shows the quantity of materials and labour needed per unit, so planning and procurement are accurate.
• It allows comparison of alternative designs and methods, and helps keep the cost under control.

The unit rate of an item of work depends on:

1. Location and lead: the distance of material sources from the site, and the lift; transport can double the cost in remote hills.
2. Availability and price of materials: local or imported, royalty, taxes and market fluctuations.
3. Wage rates and labour productivity: availability of skilled labour and the output per day.
4. Specification and quality: mix proportions, grade, thickness and finish.
5. Quantity of work: a large quantity lowers the unit rate.
6. Equipment and method: manual or mechanical work, the cost of machines and fuel.
7. Climate, season and site conditions: rain, altitude, access, soil and working space, height of work.
8. Time of completion: a short time needs more resources.
9. Norms and wastage allowances adopted.
10. Overhead, profit, taxes and the terms of the contract.

Rate analysis is the process of finding the cost of one unit of an item of work from the materials, labour and equipment needed, plus overhead and profit. The following factors affect the rate.

1. Location and lead: the distance of material sources from the site, and the lift; transport can double the cost in remote hills.
2. Availability and price of materials: local or imported, royalty, taxes and market fluctuations.
3. Wage rates and labour productivity: availability of skilled labour and the output per day.
4. Specification and quality: mix proportions, grade, thickness and finish.
5. Quantity of work: a large quantity lowers the unit rate.
6. Equipment and method: manual or mechanical work, the cost of machines and fuel.
7. Climate, season and site conditions: rain, altitude, access, soil and working space, height of work.
8. Time of completion: a short time needs more resources.
9. Norms and wastage allowances adopted.
10. Overhead, profit, taxes and the terms of the contract.

For example, the rate of brickwork in a remote hill district is much higher than in the Terai, because bricks, cement and sand must be carried a long way and skilled masons are scarce.

The purposes of rate analysis are:

1. To find the unit rate of each item for use in the abstract of cost and the BOQ of an estimate.
2. To fix the rates of extra and substitute items and of variations in a contract.
3. To check the reasonableness of the rates quoted in tenders and to judge unbalanced or abnormally low bids.
4. To determine the quantity of materials, labour and equipment needed per unit of work.
5. To compare alternative materials, mixes and methods of construction.
6. To prepare standard rates such as the District Rate and departmental norms.
7. To control the cost of the project and settle claims.

Importance of rate analysis

• It gives the unit rate of every item of work, without which an estimate or a bill cannot be prepared.
• It gives the materials, labour and machinery per unit, so procurement and planning can be done.
• It is the basis for evaluating tenders, valuing variations and settling extra-item rates.
• It allows comparing alternative designs and methods and keeps the cost under control.

How rates are taken

1. The quantity of materials, labour and equipment per unit of work is taken from the departmental norms (DUDBC norms for buildings, Department of Roads norms for road work).
2. The rates of the materials and labour are taken from the District Rate of the concerned district (or from a market survey if the material is not listed).
3. Transport is added for the lead from the source to the site.
4. Cost per unit = norm x rate for each resource; the sum is increased by 15% contractor's overhead and profit.
5. VAT (13%) is added in the estimate.

Rate analysis

Rate analysis is the process of finding the cost of executing one unit of an item of work (for example per m³ of brickwork or per m² of plaster), by adding up the cost of the materials, labour, equipment, and the contractor's overhead and profit needed for that unit.

Importance

1. It gives the unit rates used in the abstract of cost and the BOQ, hence the project cost.
2. It determines the quantity of materials, labour and machines needed, for procurement and planning.
3. It is the basis for tender evaluation, and for fixing the rates of extra and substituted items.
4. It makes comparison of different materials and construction methods possible.
5. It helps control cost and avoid disputes on payment.
6. It gives a standard basis for government offices, such as the District Rate.

For rate analysis of building construction the following data are required:

1. Drawings and specification of the item: dimensions, thickness, mix proportion, grade and quality of materials.
2. Norms (standard consumption): quantity of each material, number of skilled and unskilled labour days, and equipment hours per unit of work (DUDBC norms).
3. Rates of materials at the site: the District Rate of cement, sand, aggregate, bricks, steel, timber and so on.
4. Rates of labour: wages of skilled (mason, carpenter, plumber) and unskilled workers.
5. Lead and lift from the source to the site and the transport cost per unit and per km.
6. Equipment hire and running costs (mixer, vibrator, pump), with output.
7. Wastage allowances and conversion factors (for example dry volume of mortar and concrete).
8. Overhead and profit (usually 15%), taxes (VAT 13%), royalty and insurance.
9. Conditions of the site, such as height of work, the season and the working space.

Departmental norms used in Nepal

Each norm is used with the District Rate of the district.

Data given in the norms

• Item of work with its unit and description (mix, thickness, and specification).
• Materials per unit, with the wastage allowance.
• Skilled and unskilled labour in man-days per unit.
• Equipment and tools, with hours of use and output.
• Conversion factors and the standard output per day (task or out-turn).
• Percentage allowances: overhead and profit (15%), and for transport.

Norms

Government agencies in Nepal (DUDBC, Department of Roads, DWSSM, DoLI, local governments) use published norms for rate analysis. A norm tells, for one unit of an item of work (such as 1 m³ of 1:4 brickwork), the quantity of each material (with wastage), the days of skilled and unskilled labour, and the equipment hours needed. These norms are standardized so that all offices analyse an item in the same way.

Rates

• Every district has a District Rate (published yearly by the District Rate Fixation Committee) giving the prices of materials at the district centre or at the source, and the daily wages of the labour, with the transport rates.
• If a material is not listed, its rate is found by a market survey.
• Transport cost for the lead from the source or district centre to the site is added to the material rate (by road, mule, or porter, as the case may be).

Rate calculation

1. Cost of each resource = norm quantity x rate (including transport).
2. Sum of all costs is the direct cost.
3. Contractor's overhead and profit, 15% of the direct cost, is added to get the unit rate.
4. VAT of 13% is added in the estimate, and contingency separately.

These rates are then used in the engineer's estimate under the Public Procurement Act.

Lead is the horizontal distance over which a material (or excavated earth) is carried, from its source or place of loading to the point of use or disposal. It is measured in metres or kilometres, and for materials it decides the transport cost.

Lift is the vertical height through which the material is raised or lowered, for example from the excavated trench to the ground, or from the ground to the upper floor of the building. It is measured in metres.

Both add to the cost of the item, as extra labour or transport, over and above the initial lead and lift that are already included in the norm. Example: bricks carried 5 km from the kiln to the site (lead) and lifted to a 6 m high floor (lift).

i) Overhead charge

Overhead charges are the indirect expenses of the contractor that cannot be allotted to a particular item. They are of two types:

• Establishment (general) overheads: head office rent, staff salary, insurance, licence fees, interest on capital and so on.
• Job (site) overheads: site office, supervisors, watchmen, temporary sheds, water and electricity, and tools. In the Nepal rate analysis the contractor's overhead and profit together are allowed as 15% of the direct cost.

ii) Task or out-turn of work

The task (out-turn) of work is the quantity of work that a worker or a gang of standard composition can do in one working day (8 hours), for example the area of plaster one mason does per day, or the volume of brickwork laid. Task work gives the labour required per unit of work (labour days = 1 / out-turn) and is the basis of labour in the norms. It depends on the skill, the conditions and the method of work.

Mortar and brick quantities for 10 m³

Brick with mortar: $240 \times 120 \times 65$ mm (10 mm joints).

• Number of bricks per m³ = $\dfrac{1}{0.24 \times 0.12 \times 0.065} = 534.19$; for 10 m³ = 5341.88 nos. (net). Add 5% breakage: 5609 nos.
• Volume of bricks alone = 5341.88 x 0.23 x 0.11 x 0.055 = 7.433 m³.
• Wet mortar = 10 - 7.433 = 2.567 m³.
• Dry mortar = 2.567 x 1.30 = 3.337 m³ (1.30 allows for voids in sand and loss).
• For 1:4: cement = 3.337/5 = 0.6674 m³ x 1,440 = 961.0 kg = 19.22 bags; sand = 3.337 x 4/5 = 2.669 m³.

Analysis for 10 m³ (rates illustrative; labour norms assumed in DUDBC style)

Answer: Cost of 10 m³ of 1:4 brickwork in superstructure = Rs 169,174.20, i.e. Rs 16,917.42 per m³ (excluding VAT).

Basis

Rate for supplying and fixing one W.C. pan with a high-level cistern (cistern fixed near the ceiling at about 2 m above the floor, with a long flush pipe), complete with testing. Rates are illustrative (cement Rs 900/bag, sand Rs 2,500/m³, brick Rs 15, skilled Rs 1,200/day, unskilled Rs 900/day); norms follow DUDBC-style quantities.

Analysis (per number)

Answer: Rate of W.C. with high-level cistern = Rs 15,082.25 per number (excluding VAT). The plumber's time is a little more than for the low-level type because of the longer flush pipe and the fixing at height.

Basis

M20 concrete has the nominal mix 1:1.5:3 (cement : sand : 20 mm aggregate), as per IS 456 Table 9 for nominal mixes. Mixing is by hand (manual). Dry volume of materials = 1.54 x wet volume. "PCC for RCC work" here means the concrete only; formwork and reinforcement are paid separately. Rates are illustrative.

Quantities per m³

Sum of ratio = 1 + 1.5 + 3 = 5.5; dry volume = 1.54 m³.

• Cement = 1.54 x 1/5.5 = 0.2800 m³ = 403.2 kg = 8.06 bags.
• Sand = 1.54 x 1.5/5.5 = 0.4200 m³.
• Aggregate = 1.54 x 3/5.5 = 0.8400 m³.

Rate analysis (per m³)

Cost of the column

Volume = 0.3 x 0.3 x 4 = 0.36 m³.

Cost = 0.36 x 21,900.60 = Rs 7,884.22.

Answer: Rate of M20 concrete = Rs 21,900.60 per m³; cost of the 0.3 m x 0.3 m x 4 m column concrete = Rs 7,884.22 (excluding formwork, steel and VAT).

Net volume of masonry

• Gross volume = 10 x 0.23 x 3 = 6.900 m³
• Openings = 2 x 1.5 x 1.5 x 0.23 = 1.0350 m³
• Net volume = 6.900 - 1.0350 = 5.8650 m³

Bricks

Using the average brick 230 x 115 x 55 mm with 12 mm mortar joints, the nominal size is 242 x 127 x 67 mm.

$$\text{Bricks per m}^3 = \frac{1}{0.242 \times 0.127 \times 0.067} = 485.63$$

Number of bricks = 485.63 x 5.8650 = 2848 nos. (about 2991 nos. with 5% wastage).

Mortar (1:5)

• Volume of bricks without mortar = 2848 x 0.23 x 0.115 x 0.055 = 4.1435 m³
• Wet mortar = 5.8650 - 4.1435 = 1.7215 m³
• Dry mortar = 1.7215 x 1.30 = 2.2380 m³
• Cement = 2.2380/6 = 0.3730 m³ = 537.1 kg = 10.74 bags (50 kg)
• Sand = 2.2380 x 5/6 = 1.865 m³

Answer: Bricks = 2848 nos.; cement = 10.74 bags (0.373 m³); sand = 1.865 m³.

Assumption

The mix of the PCC is not given; the usual nominal mix 1:2:4 (M15) is assumed. Dry volume = 1.54 m³ per m³ of concrete.

Cost of cement at site

Carriage = 0.05 Rs/kg/km x 50 kg x 100 km = Rs 250 per bag. Cement at site = 600 + 250 = Rs 850 per bag.

Quantities per m³ of PCC (1:2:4, sum = 7)

• Cement = 1.54/7 = 0.2200 m³ x 1,440/50 = 6.34 bags
• Sand = 1.54 x 2/7 = 0.440 m³
• Aggregate = 1.54 x 4/7 = 0.880 m³

Cost of 1 m³ PCC

Answer: Cost of 1 m³ of PCC (1:2:4) at the site = Rs 12,629.00 (direct cost); with 15% overhead and profit the rate = Rs 14,523.35.

Assumptions

For rubble stone masonry in cement mortar: stone required = 1.2 m³ per m³ of masonry (about 20% for voids is filled with mortar, and wastage), wet mortar = 30% of the masonry volume, and dry volume of mortar = 1.30 x wet volume.

Calculation for 5 m³

• Stone = 1.2 x 5 = 6.00 m³
• Wet mortar = 0.30 x 5 = 1.50 m³
• Dry mortar = 1.50 x 1.30 = 1.950 m³
• Mortar ratio 1:6 (sum 7):
  • Cement = 1.950/7 = 0.2786 m³ x 1,440 kg/m³ = 401.1 kg = 8.02 bags (50 kg)
  • Sand = 1.950 x 6/7 = 1.671 m³

Answer: Stone = 6.0 m³; cement = 8.0 bags (0.279 m³); sand = 1.67 m³.

Basis

Raft = 10 x 10 x 1 = 100 m³ of M20 concrete (nominal mix 1:1.5:3, IS 456 Table 9). Dry volume = 1.54 x wet volume. Rates are illustrative; mixing is manual (a concrete mixer would be preferred for this volume and would reduce labour).

Quantities per m³

• Cement = 1.54/5.5 = 0.2800 m³ = 8.06 bags
• Sand = 1.54 x 1.5/5.5 = 0.4200 m³
• Aggregate = 1.54 x 3/5.5 = 0.8400 m³

Rate analysis (per m³)

Quantities and cost for the raft (100 m³)

• Cement = 806.4 bags; sand = 42.0 m³; aggregate = 84.0 m³
• Skilled labour = 100 md; unskilled = 700 md
• Cost = 100 x 21,900.60 = Rs 2,190,060.00

Answer: Rate of M20 PCC = Rs 21,900.60 per m³; cost of the 100 m³ raft concrete = Rs 2,190,060.00 (excluding formwork, reinforcement and VAT).

Rate analysis is the work of finding the cost of one unit of an item from the quantities of materials, labour and plant needed and their current rates. 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).

Rate analysis: 12.5 mm plaster 1:3, wall, per 100 m²

• Wet mortar $= 100 \times 0.0125 = 1.2500$ m³; dry mortar $= 1.27 \times 1.2500 = 1.5875$ m³
• Cement $= 1/4 \times 1.5875 = 0.3969$ m³ $= 11.43$ bags; sand $= 3/4 \times 1.5875 = 1.1906$ m³
• Labour (illustrative norm): mason 8 days, helper 10 days per 100 m², including mixing, curing and cleaning.

Rate of 12.5 mm plaster (1:3), wall = NPR 50,123.13 per 100 m² (NPR 501.23 per m²).

Modification for ceiling plaster

The material quantity stays the same for the same thickness (in practice ceiling plaster is often only 6–10 mm, which reduces the mortar proportionally). The changes are:

• Labour is increased by about 20–25% because the mason works overhead, output falls and wastage from dropping mortar is higher.
• A scaffold or staging cost is added (and cost for lifting material to the floor level, if any).

Modified rate (labour +25%, staging NPR 3,000):

Rate of ceiling plaster = NPR 59,185.63 per 100 m².

Why lead and lift matter

• Lead is the distance from the source (quarry, market, store) to the work site; transport cost rises with lead (it is paid per m³ or tonne per km as per the district rate transport norms), so the same material has a different rate at different sites.
• Lift is the vertical height through which material is raised above ground level, e.g. to upper floors; it needs extra labour or hoisting, so cost rises with floor height.
• Without lead and lift the rate of the same item cannot be fixed fairly; they are added to the basic rate of materials.

The rate is for supplying, fixing and testing one wash basin of 800 × 400 mm with fittings, per number. All prices are illustrative market rates (use the local rate list and brand specified); labour is at NPR 1,500 (skilled) and NPR 1,100 (unskilled) per day; overhead 15% and profit 10%.

Scope of work

Wash basin with a pair of brackets or pedestal, two CP pillar taps, waste coupling, bottle trap, flexible connectors, supply connection, waste connection to the drain, fixing to the wall with rawl plugs and screws, jointing with white cement, testing for leakage.

Analysis of rate per number

Rate = NPR 19,812.50 per number (direct cost NPR 15,850.00).

Quantities and rates change with the quality of the fittings; the labour part (plumber 1 day, mason 0.25 day, helper 0.75 day) follows normal DUDBC-type norms for sanitary fixing.

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.

Volumes

• Gross volume of roof slab $= 0.10 \times 20 \times 25 = 50.00$ m³
• Steel (0.8%) $= 0.008 \times 50.00 = 0.400$ m³, which is deducted because steel occupies this space.
• Net volume of concrete $= 50.00 - 0.400 = 49.600$ m³

Materials for concrete 1:1.5:3

Concrete 1:1.5:3

• Wet volume $= 49.600$ m³; dry volume $= 1.54 \times 49.600 = 76.384$ m³ (sum of ratio $= 5.5$)
• Cement $= 1/5.5 \times 76.384 = 13.888$ m³ $= 13.888 \times 28.8 = 400.0$ bags (50 kg)
• Sand $= 1.5/5.5 \times 76.384 = 20.832$ m³
• Coarse aggregate $= 3/5.5 \times 76.384 = 41.664$ m³

Steel

Mass of steel $= 0.400 \times 7850 = 3,140$ kg $= 3.140$ tonne.

Answer: cement = 400.0 bags (13.89 m³); sand = 20.83 m³; coarse aggregate = 41.66 m³; steel = 3,140 kg.

The rate covers supplying, cutting, bending, placing and tying the steel bars as per drawing and bar-bending schedule, per metric tonne (MT). Rates are assumed: steel (Fe 500D) NPR 100/kg, binding wire NPR 130/kg, skilled labour NPR 1,500/day, unskilled NPR 1,100/day, overhead 15%, profit 10%. The labour norms are read as 12 skilled and 12 unskilled man-days per MT, as given.

Quantities per MT

• Steel $= 1000$ kg plus 3% for cutting waste and laps $= 1030$ kg
• Binding wire $= 8$ kg per MT (DUDBC-type norm)
• Skilled labour $= 12$ days; unskilled labour $= 12$ days

Analysis

Rate = NPR 169,050.00 per MT (about NPR 169.05 per kg).

If the contract pays only for the steel in the drawing (as per bar-bending schedule), the wastage allowance is not paid separately but is included in the rate as above.

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.

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).

Quantities per m² of 40 mm PCC (1:2:4)

• Wet volume $= 1 \times 1 \times 0.04 = 0.04$ m³; dry volume $= 1.54 \times 0.04 = 0.0616$ m³
• Cement $= 1/7 \times 0.0616 = 0.00880$ m³ $= 0.253$ bags
• Sand $= 2/7 \times 0.0616 = 0.0176$ m³; aggregate $= 4/7 \times 0.0616 = 0.0352$ m³
• Labour for mixing, laying, compacting, levelling and curing: mason 0.08 day, labourer 0.20 day per m² (assumed norm).

Analysis of rate per m²

Rate = NPR 1,022.50 per m².

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.

The "normal" (nominal) brick size of 9" × 4½" × 3" is taken as the size including mortar, with a joint of 10 mm; the actual brick is then 10 mm smaller in each direction.

Bricks

• Nominal size $= 0.2286 \times 0.1143 \times 0.0762 = 1,991\times10^{-6}$ m³
• Bricks per m³ $= 1/(0.2286\times0.1143\times0.0762) = 502.25$
• Bricks for 10 m³ $= 5,023$ nos

Mortar

• Actual brick size $= 0.2186 \times 0.1043 \times 0.0662 = 1,509\times10^{-6}$ m³
• Volume of bricks in 1 m³ $= 502.25 \times 1,509\times10^{-6} = 0.7581$ m³; so wet mortar $= 0.2419$ m³ per m³, i.e. 2.419 m³ for 10 m³
• Dry mortar $= 1.27 \times 2.419 = 3.072$ m³
• Cement $= 1/4 \times 3.072 = 0.768$ m³ $= 22.1$ bags
• Sand $= 3/4 \times 3.072 = 2.304$ m³

Answer: bricks = 5,023 nos; cement = 22.1 bags; sand = 2.30 m³.

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). Aggregate of 12 mm size is taken at NPR 4,800/m³.

Quantities per 100 m²

• Wet volume $= 100 \times 0.025 = 2.5$ m³; dry volume $= 1.54 \times 2.5 = 3.850$ m³
• Cement $= 1/7 \times 3.850 = 0.550$ m³ $= 15.84$ bags
• Sand $= 1.100$ m³; aggregate $= 2.200$ m³
• Labour (assumed): mason 8 days, labourer 14 days per 100 m² for laying, levelling, finishing and curing.

Analysis

Rate = NPR 72,687.50 per 100 m² (NPR 726.88 per m²).

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.

12 mm plaster 1:6 per 10 m²

• Wet mortar $= 10.0 \times 0.0120 = 0.1200$ m³; dry volume $= 1.27 \times 0.1200 = 0.1524$ m³
• Cement $= 1/7 \times 0.1524 = 0.0218$ m³ $= 0.63$ bags
• Sand $= 6/7 \times 0.1524 = 0.1306$ m³

Answer per 10 m²: cement = 0.63 bags (0.0218 m³); sand = 0.131 m³.

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).

Quantities per m³ of PCC (1:3:4)

• Dry volume $= 1.54$ m³; sum of ratio $= 8$
• Cement $= 1/8 \times 1.54 = 0.1925$ m³ $= 5.544$ bags
• Sand $= 3/8 \times 1.54 = 0.5775$ m³
• Aggregate $= 4/8 \times 1.54 = 0.7700$ m³
• Labour: 1.0 skilled and 6.0 unskilled days per m³ for mixing, placing, compacting and curing (assumed norm).

Analysis

Rate of PCC (1:3:4) = NPR 23,975.32 per m³.

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. No volume of brickwork is given, so the quantities are worked per 1 m³ and per 10 m³ (multiply for the actual volume).

Part 1: Quantities of materials for brickwork (brick 230 × 110 × 55 mm, joint 10 mm, mortar 1:6)

Per 1 m³ of brickwork

• 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 1.00 m³ $= 534.19 \times 1.00 = 534$ nos (say 535)
• 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 1.00 m³ $= 0.257$ m³; dry $= 1.27\times0.257 = 0.326$ m³
• Cement $= 1/7\times0.326 = 0.047$ m³ $= 1.3$ bags
• Sand $= 6/7\times0.326 = 0.279$ m³

Per 10 m³ of brickwork

• 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 10.00 m³ $= 534.19 \times 10.00 = 5,342$ nos (say 5,342)
• 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 10.00 m³ $= 2.567$ m³; dry $= 1.27\times2.567 = 3.260$ m³
• Cement $= 1/7\times3.260 = 0.466$ m³ $= 13.4$ bags
• Sand $= 6/7\times3.260 = 2.794$ m³

Answer: per m³ of brickwork about 535 bricks, 1.34 bags cement and 0.279 m³ sand; per 10 m³: 5,342 bricks, 13.4 bags cement and 2.79 m³ sand.

Part 2: Rate analysis of PCC (1:2:4) per m³

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).

• Dry volume $= 1.54$ m³; cement $= 1/7\times1.54 = 0.2200$ m³ $= 6.336$ bags; sand $= 2/7 \times 1.54 = 0.4400$ m³; aggregate $= 4/7\times1.54 = 0.8800$ m³
• Labour: 1.0 skilled and 6.0 unskilled days per m³.

Rate of PCC (1:2:4) = NPR 24,982.50 per m³.

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.

Asphalt concrete (AC) wearing coat is a hot-mixed, hot-laid mix of graded aggregate and bitumen. Illustrative rates: aggregate NPR 2,800/tonne, bitumen NPR 95,000/tonne, plant and paver costs as shown, labour NPR 1,500 and NPR 1,100 per day, overhead 15%, profit 10%.

Quantities per 10 m²

• Compacted volume $= 10 \times 0.04 = 0.40$ m³; density of compacted mix $= 2.3$ t/m³; mass of mix $= 0.920$ t
• Bitumen content 5.5% of mix $= 0.0506$ t; aggregate $= 0.869$ t
• Tack coat 0.25 kg/m² $= 2.5$ kg per 10 m²

Analysis

Rate = NPR 13,974.63 per 10 m² (NPR 1,397.46 per m²).

Premix carpet is a 20–25 mm layer of pre-coated stone chips mixed with bitumen, laid hot and rolled, followed by a seal coat. Illustrative rates: stone chips NPR 4,500/m³, bitumen NPR 95/kg, labour NPR 1,500 and NPR 1,100 per day, overhead 15%, profit 10%.

Quantities per m² (IRC-type values, assumed)

• Loose chips for premix $= 0.025 \times 1.25 = 0.0312$ m³ (1.25 for compaction)
• Bitumen in premix $= 1.75$ kg; tack coat $= 0.25$ kg; seal coat binder $= 0.73$ kg and seal chips $= 0.009$ m³
• Labour and roller as shown

Analysis

Rate = NPR 865.32 per m².

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. The slab is taken as one rectangle of the outside dimensions given and the volume of steel is not deducted.

• Volume of slab $= 4.20 \times 3.00 \times 0.12 = 1.512$ m³

Concrete 1:1½:3

• Wet volume $= 1.512$ m³; dry volume $= 1.54 \times 1.512 = 2.328$ m³ (sum of ratio $= 5.5$)
• Cement $= 1/5.5 \times 2.328 = 0.423$ m³ $= 0.423 \times 28.8 = 12.2$ bags (50 kg)
• Sand $= 1.5/5.5 \times 2.328 = 0.635$ m³
• Coarse aggregate $= 3/5.5 \times 2.328 = 1.270$ m³

Answer: cement = 12.2 bags (0.423 m³); sand = 0.635 m³; coarse aggregate = 1.270 m³.

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 240 × 115 × 60 mm, mortar joint 12 mm.

115 m³ brickwork, mortar 1:3

• Brick with mortar $= 0.252 \times 0.127 \times 0.072 = 2304.3\times 10^{-6}$ m³; bricks per m³ $= 1/2304.3\times10^6 = 433.97$
• Bricks for 115.00 m³ $= 433.97 \times 115.00 = 49,907$ nos (say 49,907)
• Volume of bricks only $= 0.240\times0.115\times0.060 = 1656.0\times10^{-6}$ m³ each; wet mortar per m³ $= 1 - 433.97\times1656.0\times10^{-6} = 0.2813$ m³
• Wet mortar for 115.00 m³ $= 32.354$ m³; dry $= 1.27\times32.354 = 41.090$ m³
• Cement $= 1/4\times41.090 = 10.272$ m³ $= 295.8$ bags
• Sand $= 3/4\times41.090 = 30.817$ m³

Answer: bricks = 49,907 nos; cement = 295.8 bags; sand = 30.82 m³.

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. The volume of brickwork is not stated, so quantities are found per 1 m³ and per 10 m³.

Per 1 m³ brickwork

• 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 1.00 m³ $= 534.19 \times 1.00 = 534$ nos (say 535)
• 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 1.00 m³ $= 0.257$ m³; dry $= 1.27\times0.257 = 0.326$ m³
• Cement $= 1/5\times0.326 = 0.065$ m³ $= 1.9$ bags
• Sand $= 4/5\times0.326 = 0.261$ m³

Per 10 m³ brickwork

• 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 10.00 m³ $= 534.19 \times 10.00 = 5,342$ nos (say 5,342)
• 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 10.00 m³ $= 2.567$ m³; dry $= 1.27\times2.567 = 3.260$ m³
• Cement $= 1/5\times3.260 = 0.652$ m³ $= 18.8$ bags
• Sand $= 4/5\times3.260 = 2.608$ m³

Answer: per m³ about 535 bricks, 1.88 bags cement, 0.261 m³ sand; per 10 m³: 5,342 bricks, 18.8 bags cement, 2.61 m³ sand.

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

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