2. Introduction
What is Rate Analysis?
★ Rate Analysis is the determination of rate per unit of a particular item of work from
the cost of materials, cost of labour and other miscellenious expenses requires its
completion is known as Analysis of Rate / Rate Analysis.
★ Reasonable profit usually 10 % to 15 % for the contractor is also included to Rate
Analysis.
★ Rate of materials are usually taken as the rates delivered at the site of work and
include the initial cost ( cost at origin) and also add transportation, freight charges if
any and any other taxes etc.,
3. ★ If the materials are to be carried from a distance place more than 8 km, then the
transportation cost is also included.
★ The rates of materials and labour vary from place to place and therefore, the
rates of different items of works also vary from place to place.
Purpose of Rate Analysis:
- Main purpose of rate analysis are the following:
a) To determine the current rate per unit of an item at the locality.
b) To examine the viability of rates offered by contractors.
c) To calculate the quantity of materials and labour strength required for project
planning.
d) To fix up labour contract rates.
4. The rates of particular items of work depends on the following :-
1) Specifications of works and materials, quality of material proportion of mortar,
methods of constructional operations etc.,
2) Quantity of materials and their rates, number of different types of labours and
their rates.
3) Location of the site of work and its distances from the sources of water
(construction of roads and irrigation)
4) Traffics and miscellenious and overhead expenses of contractor.
Overhead Cost:
Overhead cost includes general office expenses, rents, taxes, supervision costs and
other costs. Which are indirect expenses and not productive expenses on the job.
5. The miscellenious expenses are overheads may be under the following heads:
1) General Overheads -
a) Office Staff
b) Stationary, Printing etc.,
c) Travelling expenses
d) Telephone costs
e) Rents and taxes
2) Job Overheads -
a) Salary of engineers, supervisors etc.,
b) Handling of materials
c) Repairs, carriage and appreciation of tools and plants
d) Amenities of labour
e) Workman insurance, compensation
f) Interest on instruments
6. ★ The contractor may be allowed a net profit of 6 - 8 % and miscellenious overhead
expenses may come to about 5 - 10 %.
★ For overall expenses and contractor profit 15% of the actual cost may be
reasonable amount but it is usually practiced to add 10% for all these under the
head profit.
★ The analysis of rate is usually worked out for the unit of payment of the
particular item of work under two heads.
(i) Material (ii) Labour
★ The cost of materials are taken as delivery at site inclusive of the transportation,
local charges and other taxes.
★ For tool and plants and miscellenious pity expenses which cannot be accounted
in details. Lump Sum provision is made.
7. Sl.
No
Particular of Item Quantity Per Day
1 Brickwork in lime or cement mortar in foundation and
plinth
1.25 M³ Per Mason
2 Brickwork in lime or cement mortar in super structure 1 M³ Per Mason
3 Brickwork in cement or lime mortar in arches 0.55 M³ Per Mason
4 Brickwork in cement or lime mortar in jack arches 0.55 M³ Per Mason
5 Half brick wall in partition 5 M² Per Mason
6 Course rubble stone in lime or cement mortar including
dressing
0.80 M³ Per Mason
7 Random Rubble stone in lime or cement mortar 1 M³ Per Mason
8 Stone arch work 0.4 M³ Per Mason
8. Sl.
No
Particular of Item Quantity Per Day
9 Lime concrete in foundation or floor 8.50 M³ Per Mason
10 Lime concrete in roof terracing 6.0 M³ Per Mason
11 Cement concrete M15 (1:2:4) 5.0 M³ Per Mason
12 Reinforced Brickwork 1.0 M³ Per Mason
13 R.C.C. Work 3.0 M³ Per Mason
14 12 mm thick plastering with cement or lime mortar 8.0 M² Per Mason
15 Pointing with cement or lime mortar 10.0 M² Per Mason
16 White washing or colour washing - 3 coats 70.0 M² Per white washer
17 White washing or colour washing - 1 coat 200.0M² Per white washer
9. Sl.
No
Particular of Item Quantity Per Day
18 Painting of doors or windows - 1 coat 25.0 M² Per Painter
19 Painting large surface - 1 coat 35.0 M² Per Painter
20 Distempering - 1 coat 35.0 M² Per Painter
21 2.5 cm C.C. Floor 7.50 M² Per Painter
22 Brick ballast 40 mm gauge 0.75 M³ Per Labourer
23 Breaking of brick ballast 25 mm gauge 0.55 M³ Per Labourer
24 Breaking of stone ballast 40 mm (1 ½ ) gauge 0.40 M³ Per Labourer
25 Breaking of stone ballast 25 mm (1″) gauge 0.25 M³ Per Labourer
26 Flagstone dressing 1.50 M² Per Stone Cutter
10. Sl.
No
Particular of Item Quantity Per Day
27 Earthwork excavation in ordinary soils 3.00 M³ Per Beldar/Mazdoor
28 Earthwork excavation in hard soils 2.00 M³ Per Beldar/Mazdoor
29 Excavation in rock 1.00 M³ Per Beldar/Mazdoor
30 Sand filling in plinth 4.00 M³ Per Beldar/Mazdoor
31 No. of Bricks laid by a mason in brickwork upto a
height of 3 M
600 Bricks Per Mason
32
Amount of work done by a Mazdoor per day.
(i) Mix
(ii) Deliver Brick
(iii) Deliver Mortar
3 M³
4000 No’s
5.50 M³
Mortar per Mazdoor
Per Mazdoor
Per Mazdoor
11. Labour Required for Different Works:
I. Earthwork per 28.30 Cu.m : -
1) Excavation in foundation, trenches etc., in ordinary soil including disposal upto 30 m
and lift of 1.5 m - 5 beldar, 4 mazdoor can do 28.30 cu .m of work per day.
2) Refilling excavated earth in foundation, plinth etc., including consolidation in 15 cm
layer - 3 beldars, 2 mazdoor and 1/2 Bisti can do 28.30 cu .m per day.
3) Disposal of surplus earth within a lead of 30 m that requires - 1 Mazdoor can do 2.83
cu . m per day.
II. Cement Concrete Work per 2.83 Cu . m : -
1) Laying of cement concrete requires - 2 beldars, 3 Mazdoors, 3/4 Bisti and 1/4 Mason
can do a 2.83 cu . m of work per day.
12. III. R. C. C. Work : -
1) Laying of Reinforced concrete requires - 3 Beldars, 3 Mazdoors, 4/3 Bisti and 1/2
Mason can do a 2.83 cu . m per day.
2) Centering and shuttering for flat surface requires - 4 Beldars and 4 Carpenters can do
9.60 Sq .m per day.
3) Reinforcement work for R.C.C requires - 1 Black smith , 1 Beldar can bend and place in
position one Quintal of steel.
IV. Stone Work for 2.83 cu . m : -
1) Random Rubble Masonry for foundation requires - 3 Masons, 3 Beldars, 2 Mazdoors and
1/4 Bisti can do 2.83 cu . m of work per day.
V. Brick Work for 2.83 cu . m : -
1) First Class Brickwork in 1:4 cement mortar in super structure, partition works, junctions
of roof, parapet wall etc., requires - 2 ¼ Masons, 4 ½ Mazdoors, and 1/2 Bisti can do
2.83 cu .m of work per day.
13. VI. Wood Work : -
1) For the frames of doors and windows - 2 Carpenters and 1 Beldar can work 0.18 cu . m
of wood equivalent to 4 door frames 7.50 X 10.00 cm of 1.20 m X 2.10 m per day.
2) For panel, glazed, shutters requires - 15 Carpenters and 4 Beldars can make and fix 4
shutters 40 mm thick of size 2 M X 1.15 M per day. Quantity of wood for shutter is
0.075 m³.
VII. Iron Work : -
1) Fixing 40 mm X 3 mm X 38 cm flat iron fasteners requires - 1 Blacksmith, 1 Mason, 1
Beldar can fix 36 fasteners per day.
2) Fixing 16 mm dia mild steel rod requires - 1 Blacksmith, 2 Carpenters and 3 Beldars
can fix 16.50 m per day.
14. VIII. Flooring : -
1) 4 cm thick cement concrete flooring of 40 Sq . m requires - 5 Masons, 4 Beldars, 3
Mazdoors and 1 Bisti per day. For mixing, laying and finishing.
IX. Finishing : -
1) plastering with any mortar 12 mm thick requires - 3 Masons, 3 Mazdoors and 1
Bhisti can plaster 40 sq. m per day.
2) White Washing / Colour Washing ( 3 coats) requires - 1 White washer and 1
Mazdoor can do 60 sq. m per day.
3) Painting 2 coats on wood (or) steel requires - 3 Painters and 2 Mazdoors can paint
10 sq. m per day.
15. Rate of Material: ( Consider as per the local market)
Cement - ₹. 350 / bag
Steel - ₹. 45,000 / tonne
20 mm Coarse Aggregate - ₹. 5,200 / unit
40 mm Coarse Aggregate - ₹. 4,500 / unit
Bricks - ₹ 4,500 / 1000 No’s
Sand - ₹. 3,500 / unit
Labour - 1st Class Mason - ₹ 700 / day
Labour - 2nd Class Mason - ₹. 500 / day
Man Mazdoor - ₹. 350 / day
Women Mazdoor - ₹. 250 / day
16. Lead Statement:
- The distance between the source of availability of material and construction site is
known as “Lead” and is expected in Km. the cost of conveyance of material depends on
lead.
- This statement will give the total cost of materials per unit item. It includes initial cost,
conveyance loading, unloading, stacking charges etc.,
- The rate shown in the lead statement are for metalled road and include loading and
stacking charges.
- The environmental lead on the metalled roads are arrived by multiplying by a factor
a) For metal tracks - Lead X 1.0
b) For cartz tracks - Lead X 1.1
c) For sandy tracks - Lead X 1.4
17. 1 Unit = 100 Cu feet = 2.83 Cu m
Unit weight of Steel = 7850 Kg / m³
Unit weight of Cement = 1440 Kg / m³
Unit weight of Fine Aggregate = 1450 to 1600 Kg/m³
Unit weight of Coarse Aggregate = 1450 to 1500 Kg/m³
1 Cu m of cement is how many bags?
1 Bag of Cement = 50 Kg
Density of Cement = 1440 Kg/m³
1 Bag of cement Volume is---
1 m³ bag has 1440 kg of cement
For 50 kg cement bag volume is 0.034 m
1/0.034 = 28.8 bags - Say 30 Bags.
18. Note: For 1 m³ wet concrete = 1.52 m³ dry volume of concrete approximately
Ex : M15 Grade 10 Cu m wet concrete material (1:2:4) but entire 52% to 54 %
of dry material will be added to it.
= 10.00 + 10.00 X (52 / 100)
= 10.00 + 5.20
= 15.20
1 Cu m of wet concrete in dry material is = 15.2/10 = 1.52
1 Cu m of wet concrete = 1.52 unit dry material.
M5 = 1:5:10 ( Cement : Sand : Aggregate)
M7.5 = 1:4:8 ( Cement : Sand : Aggregate)
M10 = 1:3:6 ( Cement : Sand : Aggregate)
M15 = 1:2:4 ( Cement : Sand : Aggregate)
M20 = 1:1.5:3 ( Cement : Sand : Aggregate)
M25 = 1:1:2 ( Cement : Sand : Aggregate)
19. Consider for 10 m³ M5 (1:5:10)
15.20/16 = 0.95 cu m
Cement = 0.95x1 = 0.95 = 0.95 X 1440 = 1368 kg = 1368/50 = 27.36 bags
Sand = 0.95 X 5 = 4.75 Cu m
Coarse Aggregate = 0.95 x 10 =9.50 Cu m
Cost of Cement = 27.36 X 350/bag = ₹. 9576.00
Cost of F.A / Sand = 4.75 X 1236.75/Cu m = ₹. 5874.56
Cost of C . A = 9.50 X 1590.11/ cu m = ₹. 15106.04
20. M7.5 ( 1:4:8) Calculate for the 10 cu m
15.20/(1+4+8) = 15.20/13 = 1.17 Cu m
Cement = 1.17x1 = 1.17 cu m = 1.17x 1440 = 1684.8 /50 = 33.70 Bags
Sand = 1.17 x 4 = 4.68 Cu m
C.A = 1.17 x 8 = 9.36 cu m
Cost of the Materials:
Cement = 33.70 x 350 = ₹. 11795.00
Sand/ F.A = 4.68 x 1236.75 = ₹. 5787.99
C.A = 9.36 x 1590.11 = ₹. 14883.43
21. M10 (1:3:6) Consider 10 cu m
15.20 / 10 = 1.52 cu m
Quantity of the Materials:
Cement = 1.52 Cu m = 1.52 x 1440 = 2188.80 Kg’s = 2188.8/50 = 43.78 Baga
Sand = 1.52 x 3 = 4.56 cu m
C.A = 1.52 x 6 = 9.12 cu m
Cost of Material:
Cement = 43.78 x 350 = ₹. 15323.00
Sand = 4.56 x 1236.75 = ₹. 5639.58
C.A = 9.12 x 1590.11 = ₹. 14501.80
22. M15 ( 1:2:4) Consider for 10 Cu m.
15.20/7 = 2.17 cu m
Quantity of Material:
Cement = 2.17 cu m = 2.17 x 1440 =3124.80 kg = 3124.8/50 = 62.50 Bags
Sand= 2.17 x 2 = 4.34 cu m
C.A = 2.17 x 4 = 8.68 cum
Cost of Materials:
Cement = 62.50 x 350 = ₹. 21875.00
Sand = 4.34 x 1236.75 = ₹. 5367.49
C.A = 8.68 x 1590.11 = ₹. 13802.15
23. M 20 (1:1.5:3) - Consider 10 Cu m
15.2/1+1.5+3) = 15./ 5.5 = 2.76 cu m
Quantity of the materials:
Cement = 2.76x1 = 2.76 cu m = 2.76 x 1440 = 3974.40kg = 79.50 bags
Sand = 2.76 x 1.5 = 4.14 cu m
C.A = 2.76 x 3 = 8.28 cu m
Cost of the Materials:
Cement = 79.50 bags x 350/bag = 27825.00/ 10 cu m = 2782.50/ cu m
Sand = 4.14 x 1236.75 = 5120.14 / 10 cu m = 512.00 / cu m
C.A = 8.28 x 1590.11 = 13166.00/ 10 cu m = 1316.60/ cu m
24. M25 ( 1:1:2) Consider 10 Cu m
15.20/4 = 3.80 cu m
Quantity of the Materials:
Cement = 3.80 cu m = 3.80 x 1440 = 5472 kgs = 109.44 Bags
Sand/ F.A = 3.80 x 1 = 3.80 cu m
C.A = 3.80 x 2 = 7.60 cu m
Cost of the Materials:
Cement = 109.44 x 350/bag = 38304.00/10 cu m = 3830.40/ 1 cu m
Sand/ F.A = 3.80 x 1236.75 = 4699.65/10 cu m = 470.00/ 1 cu m
C.A = 7.60 x 1590.11 = 12084.83/ 10 cu m = 1208.48 / 1 cu m
25. Example - 1.
Calculate the cost of the concrete of M5 grade, take 10 cu m and find the 1 cu m cost?
1st class mason - 0.5 No’s
2nd Class Mason - 1.5 No’s
Man Mazdoor - 12 No’s
Women Mazdoor - 18 No’s and
Curing purpose - 4 No’s
Assume water charges 1 ½ % of total cost and Contractor profit is 15% of the total cost.
Solution:
For M5 Grade of Concrete mix proportion is 1:5:10
15.2/(1+5+10) = 15.2/16 = 0.95 m³
26. Sl.
No
Particulars Quantity Rate in ₹. Per
unit
Amount in ₹.
1 Materials
Cement
Fine Aggregate/Sand
Coarse Aggregate
0.95 x 1 = 0.95 m³
0.95 x 1440 = 1368 kg’s
1368/50 = 27.36 bags
0.95 x 5 = 4.75 m³
0.95 x 10 = 9.50 m³
350/-
1236.75
1590.11
Bag
M³
m³
9576.00
5874.56
15106.04
2 Labour
1st Class Mason
2nd Class Mason
Man Mazdoor
Women Mazdoor
Curing purpose
0.5
1.5
12
18
4
700
500
350
250
250
Day
Day
Day
Day
Day
350.00
750.00
4200.00
4500.00
1000.00
TOTAL COST ₹. 41356.60
27. Add water charges (1 ½ %) of total cost.
= ₹. 41,356.60 X (1.5/100) = 620.35 Say ₹, 620/-
= ₹. 620.00
Add Contractor Profit (15%) of total cost.
= ₹. 41,356.60 X (15/100) = 6203.49 Say ₹. 6203.00
= ₹. 6203.00
Total Cost = ₹. 41,356.60 + 620.00 + 6203.00 = ₹. 48,179.60
Sundries = ₹. 820.40
Total Cost for 10 m³ = ₹. 49,000=00
Cost of 1 m³ = ₹. 49,000/10 = ₹. 4,900.00
28. Example - 2:
Calculate the cost of the concrete of M15 Grade, take 10 m³ and find the 1 m³cost?. 1st Class
mason - 0.5 No’s, 2nd class mason - 1 No, Man mazdoor - 12 No’s, women mazdoor - 18 No’s
and curing purpose - 4 no’s.
Solution:
M15 - 1:2:4
15.2 / 7 = 2.17 m³
Sl.
No
Particulars Quantity Rate in ₹. Per unit Amount in .Rs
1 Materials
Cement
Sand
Coarse Aggregate
2.17*1440=
3124.80kgs=62.50 bags
2.17*2 = 4.34m³
2.17*4 = 8.68m³
350
1236.75
1590.11
Bag
M³
m³
21875.00
5367.50
13802.15
29. Sl.
No
Particulars Quantity Rate in ₹. Per unit Amount in .Rs
2 Labour
1st class Mason
2nd class mason
Man mazdoor
Women mazdoor
0.5
1.0
12.0
18+4=22.0
700
500
350
250
Day
Day
Day
day
350.00
500.00
4200.00
5500.00
Total Cost ₹. 51,594.65
Water Charges - (1 ½ %) of total cost: = 51594.65 x (1.5/100) = 773.91 say Rs.774.00
Add contractor profit (15%) of total cost = 51594.65 x (15/100) = Rs.7739.00
Total Cost = 51,594.65 + 774.00 + 7739.00 = Rs. 60,107.65
Sundries = Rs.892.35 = Rs. 892.35
Cost of 10 m³ cement concrete of M15 grade = Rs. 61,000.00
Cost of 1 m³ cement concrete of M15 Grade = 61,000/10 = Rs. 6,100.00 / m³
30. Example - 3:
Calculate the cost of the concrete of M25 grade and find the one cu m cost?. 1st class
mason - 1.5 No’s, 2nd class mason - 2 No’s, Man mazdoor - 12 No’s, Women mazdoor
- 18+4, water charges 1 ½ % of total cost and consider the contractor profit is 12 %
of total cost.
Solution:
Grade of Concrete = M25. Mix Proportion: 1:1:2
1.52 / 4 = 0.38 m³
Sl.
No
Particulars Quantity Rate in ₹. Per unit Amount in Rs.
1 Materials
Cement
Sand
Coarse Aggregate
0.38*1440 =547.20
kg’s = 10.90 bags
0.38*1 = 0.38 m³
0.38*2 = 0.76 m³
350
1236.75
1590.11
Bag
M³
m³
3815.00
470.00
1208.48
31. Sl.
No
Particulars Quantity Rate Per unit Amount in Rs.
2 Labour
1st Class mason
2nd Class Mason
Man mazdoor
Women mazdoor
1.5
2.0
12
22
700/-
500/-
350/-
250/-
Day
Day
Day
Day
1050.00
1000.00
4200.00
5500.00
Total Cost ₹. 17259.00
Add Water Charges 1 ½ % of Total cost = 17259 x (1.5/100) = 258.88 Say ₹. 259.00
Add Contractor Profit 12% of total cost = 17259 x (12/100) = ₹. 2071.00
Cost of 1 m³ cement concrete of M25 Grade = 17259 + 259 + 2071= ₹. 19589.00
Sundries = ₹. 411.00 = ₹. 20,000.00
32. BRICKWORK
Brickwork with Standard Bricks ( Calculation of material required for brickwork)
1. Take a wall of 1 ½ brick thickness i.e. 30 cm, nominal thickness of 20 m length and 5 m
height.
2. Nominal volume = Length X width X Height
= 20.00 X 0.30 X 5.00
= 30.00 Cu .m
3. Normal mortar joint will be less than 1 cm, taking 1 cm mortar joint, the actual
thickness of the wall be 29 cm.
Actual Volume = Length X Width X Height
= 20.00 X 0.29 X 5.00
= 29.00 Cu . m
4. Number of Standard bricks ( 20 X 10 X 10 Cm) = 29.00 / (0.2x0.1x0.1)
= 14500 No’s
33. 5. This 14500 no. of bricks is for 29 M³
Therefore Number of bricks per M³ ( Nominal Volume)
= 14500 / 30
= 483.30 - Say 484 No’s
6. Consider 5% of breakage, wastage etc.,. This may be taken as 500 no.of bricks.
Therefore 1 M³ requires 500 no. of bricks.
Mortar Calculations:
Mortar requirement = Actual Volume of brickwork - Net volume of bricks.
= 29.00 - (0.19 x 0.09 x 0.09) x 14500
= 6.68 M³
For frog filling, for use of cut bricks, for bonding, for uniform joints, wastages etc., 15% extra
mortar may be required.
Therefore the Volume of Mortar = 6.68 + (6.685 X 15/100)
= 7.688 M³ ( Wet Volume)
34. For dry volume increase the wet volume by 25%
= 1.25 X 7.68 M³
= 9.61 M³
For 30 Cu .m of brick work, dry volume of mortar = 9.61 M³
For 1 Cu .m of brick work = 0.32 M³
For 10 Cu .m of brick work, dry volume of mortar = 0.32 / 10
= 3.20 M³
For 10 Cu .m of brick work, No.of bricks = 5000 No’s
Dry volume of mortar = 3.20 M³
Brick Masonry :
Brick size with mortar = (20 x 10 x10) Cm³
Brick size without mortar = (19 x 9 x 9) Cm³
35. Length = 20 m, Thickness = 1 ½ brick wall Height = 3 m.
Nominal volume of Brick wall = 20 x 0.3 x 3 = 18 M³
Actual volume of 1 brick = 0.19 x 0.09 x 0.09 = 1.539 x 10⁻ ³
Example : - 1 Brick volume = 20 x 10 x 10 cm³
= 2000 cm³ 0.002 M³
For a volume of 18 m³, No.of bricks = 18 / 0.002 = 9000 bricks
1 M³ = 9000 / 18 = 500 Bricks.
36. Example : -
Total volume of 500 bricks = 500 x 0.20 x 0.10 x 0.10 = 1 M³
Modular brick volume for 500 bricks = 500 x 0.19 x 0.09 x 0.09 = 0.769 M³
Volume of mortar = 1 - 0.769 = 0.231 M³
Cement = (0.231 x 1)/7 = 0.033 M³
Mortar = (0.231 /7) x 6 = 0.198 M³
Density of Cement = Weight / Volume = 1440 Kg / m³
1 M³ = 1440 kg
1 Bag = 50 kg
1 Bag = 50 / 1440 = 0.034 M³
37. Cement Mortar :- 1:6
Wall = 10.00 X 0.23 X 3.00 m = 6.90 m³
For 1 m³ = 1/Brick volume= 1/ (0.1 x 0.1 x 0.2) = 1/0.002 = 500 No’s
Bricks for 1 cu m = 500 No’s
For 6.90 Cu m = 6.90 X 500 = 3450 No’s (or) 6.9/0.002 = 3450 No’s
For 10 Cu m = 10/0.002 = 5000 No’s
For 10 cu m of Brick work with CM (1:6)
70% of Brick volume = 7 cu m
30% of Cement Mortar = 3 cu m
38. The quantity of material for cement mortar - (1:6)
Cement = 3 / (1+6) = 3/7 = 0.42 m³
Sand/F.A= 0.42 X 6 = 2.52 m³
Cement Mortar for 10 Cu m of Brick Work
Mortar No of Bricks Cement Fine Aggregate/Sand
1:2 5000 1.0 m³ 2.0 m³
1:3 5000 0.75 m³ 2.25 m³
1:4 5000 0.60 m³ 2.40 m³
1:5 5000 0.50 m³ 2.50 m³
1:6 5000 0.43 m³ 2.58 m³
39. Labour Required for 10 Cu m Brick Work:
1st Class Mason - 0.50 No’s
2nd Class Mason - 1 No.
Man Mazdoor - 12 No’s
Women Mazdoor - 12 No’s
Watering / Curing - 2 No’s
40. Example - 4:
Calculate the quantity of cement, number of bricks, quantity of sand for C.M
(1:6) for 1 ½ brick thick, take the length of wall is 20 m and height of wall is 5 m.
Solution:
Volume of wall = Length X Breadth X Height
20.00 X 0.30 X 5.00 = 30 Cu m
Number of bricks = 30 /0.002 or 30 X 500 = 15,000 No’s
Cement Mortar = 1:6
Quantity of cement mortar ( 30%) - 9 m³
Quantity of Cement = 9 / (1+6) = 9/7 = 1.29 m³= 1.29 x 1440
= 1857.6 kg = 37.15 bags X 350/- = Rs.13003.00
Quantity of F.A. = 1.29 X 6 = 7.74 cu m = 1236.75 = Rs. 9572.00
Bricks = 15000 = 4500/1000 = Rs. 67500.00
41. For 10 Cu m For 30 Cu m
1st ClassMason - 0.50 No’s 1 st Class Mason - 1.50 No’s
2nd Class mason - 1 No 2nd Class Mason - 3.00 No’s
Man Mazdoor - 12 No’s Man Mazdoor - 36 No’s
Women Mazdoor - 12 No’s Women Mazdoor - 36 No’s
Curing - 2 No’s Curing - 6 No’s
1St Class Mason - 1.50 X 700 = Rs.1050.00
2nd Class Mason - 3.00 X 500 = Rs.1500.00
Man Mazdoor - 36 X 350 = Rs. 12600.00
Women Mazdoor - 36 X 250 = Rs. 9000.00
Curing - 6 X 250 = Rs. 1500.00
Total Cost = Cost of the Material + Cost of the Labour = Rs. 1,15,725.00
42. Add water charges @ 1 ½ % = 115725x1.5/100 = Rs.1736.00
Add Contractor Profit @ 15% = 115725 X 15/100 = Rs.17359.00
TOTAL COST IN Rs. 1,15,725 + 1,736 + 17,359 = Rs. 1,34,820.00
Sundries = Rs. 180.00
Scaffolding = Rs. 5,000.00
Total Cost in Rs. 1,40,000.00
Example - 5:
Calculate the quantity of cement, number of bricks, quantity of sand for C.M (1:5)
for 1 ½ brick thick, take the length of wall is 15 m and height of wall is 3.5 m.
43. Solution:
Given data - Length of Wall = 15.00 m
Height of Wall = 3.50 m
Thickness of Wall = 1 ½ brick wall = 0.30 m
Cement Mortar = 1:5
Volume of Wall= 15.0 X 0.30 X 3.50 = 15.75 M³
Quantity of cement Mortar (30%) = 4.73 M³
Quantity of Cement = 4.73 /(1+5) = 4.73 / 6 = 0.79 m³
= 0.79 / 0.034 = 23 Bags.
Quantity of Sand / F.A = 0.79 X 5 = 3.95 m³
Number of bricks = 15.75 / 0.002 = 7875 No’s
Cost of Cement = 23 bags X 350/- = ₹. 8050.00
Cost of Sand/F.A = 3.95 X 1236.75 = ₹. 4885.00
Cost of Brick = 7875 X 4.50 = ₹. 35438.00
Total Cost of the Material - Rs. 48373.00
44. Labour for 15.75 Cu . m
1st Class Mason - 0.788 No’s X Rs. 700 = Rs. 551.60
2nd Class Mason - 1.575 No’s X Rs. 500 = Rs. 787.50
Man Mazdoor - 18.90 No’s X Rs. 350 = Rs. 6615.00
Women Mazdoor - 18.90 No’s X Rs. 250 = Rs. 4725.00
Curing purpose - 3.15 No’s X Rs. 250 = Rs. 787.50
Total Labour Cost = Rs. 13466.60
TOTAL COST = Total cost of the Material + Total cost of the Labour
= Rs. 48,373.00 + 13,466.60
= Rs. 61,839.60
Add Water charges (1 ½ %) = Rs. 61,839.60 X 1.5/100= Rs. 928.00
Add Contractor Profit (10%) = Rs. 61,839.60 X 10/100 = Rs. 6184.00
Total Cost = Rs. 61,839.6 + 928.0 + 6184.0 = Rs. 68,951.60
Sundries = Rs. 48.40
Scaffolding = Rs. 1,000.00
TOTAL COST = Rs. 70,000.00
45. Example - 6:
Estimate the 1st class brick work of a room size 3.50 m X 4.0 m and height is
3 m thickness of wall is 1 ½ brick wall. Mortar proportion is 1:4.
Solution:
Given data - Room dimensions - 3.5 X 4 m
Thickness of wall - 0.30 m
Height of wall - 3 m
Total Volume of brick work = 2(0.15 + 4.00 + 0.15) + 2(0.15 + 3.5 + 0.15)
= 8.60 + 7.60 = 16.20 m
Volume = 16.20 X 0.30 X 3.00 = 14.58 Cu .m
Volume of 1 m³ of bricks = 500 No’s
Volume of 14.58 m³ bricks 500 X 14.58 = 7290 No’s
46. Mortar proportion is 1:4
Dry Volume of mortar - 1 m³ = 0.3 m³
For 14.58 m³ = 0.3 X 14.58 = 4.374 m³
Volume of cement= 4.374 / (1+4) = 4.374 / 5 = 0.875 m³
No of bags = 0.875/0.034 = 25 Bags
Volume of F.A / Sand = 0.875 x 4 = 3.5 m³
Cost of Bricks = 7290 x 4.5/- = Rs. 32,805.00
Cost of Cement = 25 X 350/- = Rs. 8750.00
Cost of F.A/Sand = 3.5 X 1236.75= Rs. 4328.60
Total cost of the Material = Rs. 45,883.60
Labour for 14.58 cu .m
1st class Mason - 0.73 No’s Man Mazdoor - 17.50 No’s
2nd Class Mason - 1.46 No’s Women Mazdoor - 17.50 No’s
Curing - 3 No’s
47. 1st Class Mason - 0.73 X 700 = Rs. 511.00
2nd Class Mason - 1.46 X 500 = Rs. 730.00
Man Mazdoor - 17.5 X 350 = Rs. 6,125.00
Women Mazdoor - 17.50 X 250 = Rs. 4,375.00
Curing - 3 X 250 = Rs. 750.00
Total Labour Cost = Rs. 12,491.00
TOTAL COST = Cost of the Material + Cost of the Labour
= Rs. 45,883.60 + 12,491.00 = Rs.58374.60
Add water charges (1 ½ %) = 58,374.60 X 1.5/100 = Rs.875.60
Add contractor charges (10%) = 58,374.60 X 10/100 = Rs. 5837.46
TOTAL COST = 58374.60+875.60+5837.46 = Rs. 65,087.66
Sundries = Rs. 412.34
Scaffolding = Rs. 4500.00
TOTAL COST OF THE 1ST CLASS BRICKWORK = Rs. 70,000.00
48. Random Rubble Masonry
For 10 Cu . m of Random Rubble Stone masonry. Add 25% of dry volume of material
10.00 + [10.00 X (25/100)] = 12.50 Cu . m
Take 42% of Cement Mortar = 4.20 Cu m
Cement , sand mix Proportion is - 1 : 6
Volume of Cement = 4.20 /(1+6) = 4.20/7 = 0.60 Cu . m
Volume of Sand/F.A = 0.60 X 6 = 3.60 Cu . m
49. Example : 7
Calculate the cost of the R.R.S masonry for 1 Cu . m. The ratio of mortar is 1:6. The
cost of boulders for 1 Cu . m is ₹.800/-, Labour 1st class mason - 0.5, 2nd class mason - 12 No’s,
man mazdoor - 10 No’s, women mazdoor - 10 No’s. For R.R.S masonry volume 10 Cu .m is taken
as 12.5 cu . m. For 1 Cu . m = 1.25 Cu . m.
MORTAR CEMENT SAND
1: 3 ( Cement : Sand) 1.0 cu . m 3.0 Cu . m
1: 4 ( Cement : Sand) 0.85 Cu . m 3.40 Cu . m
1: 5 ( Cement : Sand) 0.70 Cu . m 3.5 Cu . m
1 : 6 ( Cement : Sand) 0.60 Cu . m 3.60 Cu . m
1 : 2 ( Lime : Sand) 1.60 Cu . m of Lime 3.20 Cu . m
1 : 3 ( Lime : Sand) 1.20 Cu . m of Lime 3.60 Cu . m
50. Solution:
Given data:- CM (1:6),
Volume of cement mortar for 10 M³ - 42% = 4.20 M³
Volume of Cement = 4.20 / (1+6) = 4.20/7 = 0.60 m³
0.60 / 0.034 = 17.64 bags Say 18 Bags
Volume of Sand = 0.60 X 6 = 3.60 m³
Volume of Boulders = 12.50 m³
Sl.No Particulars Quantity Rate Per Amount in
₹.
1 Materials
a. Cement
b. Sand
c. Boulders
18 bags
3.60 cu .m
12.50 cu .m
350.00
1236.75
800.00
Bag
Cu m
Cu m
6300.00
4452.30
10,000.00
51. Sl.
No
Particulars Quantit
y
Rate Per Amount in
₹.
2 Labour
a. 1st class mason
b. 2nd class mason
c. Man mazdoor
d. Women mazdoor
0.50 No’s
12 No’s
10 No’s
10 No’s
700.00
500.00
350.00
250.00
Day
Day
Day
Day
350.00
6000.00
3500.00
2500.00
33,102.30
3 Add water charges - (1 ½ %) 496.50
4 Add contractor charges 10% 3310.23
Sundries 90.97
TOTAL COST FOR 10 M³ 37,000.00
For 1 M³ Cost of R.R.S Masonry 37000/10 ₹. 3,700.00
52. U. C. R. S. MASONRY
U. C. R. S. MASONRY FOR 10 Cu. M
Take Cement Mortar - 1:6
U.C.R.Stone for 10 Cu m - Add 25% dry volume of material
= [10.00 + (10.00+25/100)]
= 12.50 Cu m
Volume of Stone can take = 12.50 Cu.m for 10 Cu . m
Volume of cement mortar take - 40 % = 4.00 Cu . m for 10 Cu .m
4.00 Cu. m Cement Mortar
Volume of cement = 4.00 / (1+6) = 4/7 = 0.57 Cu . m
Volume of sand = 0.57 X 6 = 3.42 Cu . m
53. MORTAR CEMENT SAND
1: 3 ( Cement : Sand) 1.00 Cu . m 3.00 Cu . m
1: 4 ( Cement : Sand) 0.80 Cu . m 3.20 Cu . m
1: 5 ( Cement : Sand) 0.67 Cu . m 3.35 Cu . m
1 : 6 ( Cement : Sand) 0.60 Cu . m 3.60 Cu . m
Example : 8
Calculate the cost of the U.C.R.S masonry for 1 Cu . m. The ratio of mortar is 1:4. The
cost of boulders for 1 Cu . m is ₹.1200/-, Labour 1st class mason - 0.5, 2nd class mason - 12 No’s,
man mazdoor - 10 No’s, women mazdoor - 10 No’s. For U.C.R.S masonry volume 10 Cu .m is
taken as 12.5 cu . m.
54. Solution:
Given data:- CM (1:4),
Volume of cement mortar for 10 M³ - 40% = 4.00 M³
Volume of Cement = 4.00 / (1+4) = 4.00/5 = 0.80 m³
0.80 / 0.034 = 23.53 bags Say 24 Bags
Volume of Sand = 0.80 X 4 = 3.20 m³
Volume of Stone = 12.50 m³
Sl.No Particulars Quantity Rate Per Amount in
₹.
1 Materials
a. Cement
b. Sand
c. Boulders
24 bags
3.2 cu .m
12.50 cu .m
350.00
1236.75
1200.00
Bag
Cu m
Cu m
8400.00
3957.60
15,000.00
55. Sl.
No
Particulars Quantity Rate Per Amount in
₹.
2 Labour
a. 1st class mason
b. 2nd class mason
c. Man mazdoor
d. Women mazdoor
0.50 No’s
12 No’s
10 No’s
10 No’s
700.00
500.00
350.00
250.00
Day
Day
Day
Day
350.00
6000.00
3500.00
2500.00
39,707.60
3 Add water charges - (1 ½ %) 595.60
4 Add contractor charges 10% 3970.76
Sundries & L.S 726.04
TOTAL COST FOR 10 M³ 45,000.00
For 1 M³ Cost of R.R.S Masonry ₹. 45,000/10 ₹. 4,500.00
56. PLASTERING
Calculation of quantity mortar and materials Area X thickness gives the quantity of
mortar for uniform thickness, for filling of the joints and to make up non uniform
surface of wall. This may be increased by 30% which will give wet mixed mortar to
get the total dry volume of materials it may be further increased by 25%.
Material for 12 mm thick plastering in wall for 100 sq . m
Wet mixed mortar for uniform layer = 12 mm
= 0.012 x 100
= 1.20 Cu . m
Adding 30% to fill up the joints, uneven surface etc.,
Now the wet volume of mortar = {1.20 + [1.20 x (30/100)]}
= 1.56 Cu . m
57. 1.56 Cu . m of wet volume is converted into dry volume by increasing 25%
Now dry volume of mortar = 1.56 + [1.56 x (25/100)]
= 1.95 Cu . m Say 2.00 Cu . m
Therefore Dry volume of mortar is 2 Cu . m ( For 100 Sq . m at 12 mm thickness)
Material for 20 mm thick plastering
As the thickness of the plaster is more, 20% of mortar is taken to fill up the joints,
unevenness etc.,
The quantity of wet mortar = 0.02 X 100 = 2.00 Cu . m
Now the wet volume of mortar = 2.00 + [2.00 x (20/100)] = 2.40 Cu . m
Volume of dry mortar = 2.40 + [ 2.40 x (25/100)] = 3.00 Cu . m
60. Requirement of Workmanship for Plastering for 100 Sq.m :
Labour Quantity
1st Class Mason 0.35 No’s
2nd Class Mason 10 No’s
Man Mazdoor 15 No’s
Women Mazdoor / Curing 0.75 No’s
61. Example - 9 :
Estimate the plastering work thickness of 12 mm, area 100 sq.m for 1:6
Solution :
volume = Wet mixed mortar for uniform layer = 12 mm
= 0.012 x 100
= 1.20 Cu . m
Thickness = 12 mm; Area= 100 sq.m
1.20 + 1.20 (30/100)= 1.56 Cu.m ( Wet Volume)
1.56 + 1.56 (25/100)= 1.95 Cu.m ( Dry Volume)
Dry Volume = 1.95 Cu.m ≃ 2 Cu.m
Cement = 2.00/(1+6) = 0.285 Cu.m X 1 = 0.285 Cu.m
= 0.285 / 0.034 = 8.38 ≃ 8 Bags
Sand = 0.285 X 6 = 1.71 Cu.m
62. Sl.No Particulars Quantity Rate Per Amount in ₹.
1
Materials
a. Cement
b. Sand
8 bags
1.71 cu .m
350.00
1236.75
Bag
Cu m
8400.00
3957.60
2
Labour
a. 1st Class Mason
b. 2nd Class Mason
c. Man Mazdoor
d. Curing
0.35 No
10 No’s
15 No’s
0.75 No
700.00
500.00
350.00
250.00
Day
Day
Day
Day
245.00
5000.00
5250.00
187.50
₹. 23,040.10
3 Add Water Charges (1 ½ %) ₹. 345.60
4 Add Contractor Profit (10%) ₹. 2304.01
Sundries & L.S Charges ₹. 310.29
TOTAL COST FOR 100 Sq. M ₹. 26,000.00
63. For Pointing in Brick work for total dry volume of material is taken
as 0.60 Cu.m for 100 Sq.m
Example - 10 :
Estimate the pointing work quantity 1:2 for 1200 Sq.m ?
Solution :
0.60 Cu.m - 100 Sq.m
? - 1200 Sq.m
1200 Sq.m = (0.60 X 1200) / 100 = 7.20 Cu.m
Cement = 7.20 / (1+2) = 2.40 Cu.m X 1 = 2.40 Cu.m
= 2.40 / 0.034 = 70.58 ≃ 71 Bags
Sand = 2.40 X 2 = 4.80 Cu.m
64. Sl.No Particulars Quantity Rate Per Amount in ₹.
1
Materials
a. Cement
b. Sand
71 Bags
4.80 Cu.m
350.00
1,236.75
Bag
Cu.m
24,850.00
5,936.40
2
Labour
a. 1st Class Mason
b. 2nd Class Mason
c. Man Mazdoor
d. Curing
0.35 No
10 No’s
10No’s
0.50 No
700.00
500.00
350.00
250.00
Day
Day
Day
Day
245.00
5,000.00
3,500.00
125.00
₹.39,656.40
3 Add Water Charges (1 ½ %) ₹. 594.85
4 Add Contractor Profit (10%) ₹. 3.965.64
Sundries & L.S Charges ₹. 283.11
TOTAL COST FOR 100 Sq. M ₹.44,500.00
65. Cement Concrete Floor
- The quantity of cement concrete may be calculated by multiplying area of floor
by the thickness.
- The quantity of each material may be found same as the plain concrete of
cement (PCC) .
1. For 2.5 cm cement concrete floor for 100 M² of area the quantity of cement
concrete is 2.5 M³. Now add 10% extra for unevenness of base concrete.
2.5 + 2.5 X (10/100) = 2.75 M³
For neat cement surface finishing additionally 0.20 M³ (6 Bags) are required
2. For 100 M³ cement concrete, the total dry volume of material is approximately
increased by 50%.
2.75 + 2.75 X (50/100) = 4.125 M³
3. For 2 cm thick cement concrete floor for 100 M² area, the dry volume is 4.125 M³.
66. 4. For 4 cm thick cement concrete floor for 100 M² area, the dry volume is 4 M³
Add 10%= 4 + 4 x (10/100) = 4.40 M³ ( Wet Volume)
Add 50% Dry volume
= 4.40 + 4.40 X (50/100) = 6.60 M³
Example - 11 :
Estimate the 2.5 cm cement concrete floor of 1:2:4 for 1 Sq.m ?
Solution :
100 Sq . m = 2.50 M³ of quantity of dry volume is = 4.125 M³
Cement = 4.125 /(1+2+4) = 0.59 M³ X 1 = 0.59 M³
1 Bag = 0.034 M³
0.59 M³ = ?
No.of bags = 0.59 / 0.034 = 17.35 ≃ 17 Bags
Sand = 0.59 X 2 = 1.18 M³
Coarse Aggregates = 0.59 X 4 = 2.36 M³
67. Note : [ For neat finishing of floor, the thickness of neat cement layer is 1.50mm.
∴ The cement paste requirement for 100 M²
= 0.0015 X 100 = 0.15 M³ .
Dry volume of cement is increased by 25% (For 100M² only)
= 0.15 + 0.15 X (25/100)
= 0.187 M³ ≃ 0.20 M³ ]
Surface finishing for 100 Sq . m = 0.20 Cu . m
68. Sl.No Particulars Quantity Rate Per Amount in ₹.
1
Materials
a. Cement
b. Sand
c. Coarse Aggregate
17 Bags
1.18 Cu .m
2.36 Cu.m
350.00
1,236.75
1590.11
Bag
Cu.m
Cu.m
5950.00
1459.36
3752.65
2
Labour
a. 1st Class Mason
b. 2nd Class Mason
c. Man Mazdoor
d. Curing
0.75 No
10 No’s
10 No’s
2 No’s
700.00
500.00
350.00
250.00
Day
Day
Day
Day
525.00
5,000.00
3,500.00
500.00
₹. 20,687.01
3 Add Water Charges (1 ½ %) ₹. 310.30
4 Add Contractor Profit (10%) ₹. 2068.70
Sundries & L.S Charges ₹. 933.99
TOTAL COST FOR 100 Sq. M ₹. 24,000.00
Cost for 1 Sq.M = ₹. 24000 / 100 = ₹. 240.00 ₹. 240.00
69. Example - 12 :
Estimate the White Washing of 1 coat for 1 Sq.m area, Taking 100 Sq.m as the area
10 kg of white lime unslaked ?
Solution :
Sl.No Particulars Quantity Rate Per Amount in ₹.
1
Materials
White Lime unslaked
10 Kg’s 950.00
9.50
Quintal
Kg
95.00
2 Glue Powder Lumpsum 30.00 L.S 30.00
3
Labour
a. White Washer
b. Men Mazdoor
0.70 (or) 2/3 No’s
0.70 (or) 2/3 No’s
500.00
350.00
Day
Day
350.00
245.00
4 Sundries & Lump Sum 80.00
Total Cost in Rs. 800.00
70. Add Water charges (1 ½ %) = Rs. 800 X )1.5/100) = Rs. 12.00
Add Contractor Charges (10%) = Rs. 800 X (10/100) = Rs. 80.00
Total Cost for 100 Sq.m = Rs. 800.00 + 12.00 + 80.00 = Rs. 892.00
Cost for 1 Sq.m of White Washing of 1 coat = Rs. 892.00 / 100 = Rs. 8.92