Tiling Calculation for Home Renovation
VerifiedAdded on 2020/05/28
|13
|2079
|254
AI Summary
This assignment involves calculating the total tiling area needed for a house renovation project, encompassing the family room, meals room, kitchen, and foyer/stair area. It requires determining the area of each space, adding a 10% wastage factor, and then calculating the number of boxes of floor and wall tiles required based on the given coverage per box.
Contribute Materials
Your contribution can guide someone’s learning journey. Share your
documents today.
ESTIMATED COSTS FOR BUILDING PROJECTS 1
Name
Course
Professor
Date
Name
Course
Professor
Date
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
ESTIMATED COSTS FOR BUILDING PROJECTS 2
Estimated Costs for Building and Construction Projects
Question 1: Bored excavation for 300mm diameter piers
Total volume is calculated by considering bulking factor of soil (Peterson, 2014).
Ite
m
Particulars Unit Quantity Rate $ C
1 Bored excavation volume for 53 Ø300mm
and 2.5m deep piers
Cross sectional area of pier = πr2 = π x 0.152 =
0.07m2
Volume of pier = cross sectional area x depth
= 0.07 x 2.5 = 0.18m3
Volume of 53 piers = 0.18m3 x 53 = 9.54m3
Assume bulking factor of 1.4 (clay soil), total
volume of soil material to be excavated for
boring of piers = 1.4 x 9.54 = 13.6m3
m3 14
Summary:
A total of 53 53 Ø300mm and 2.5m deep pier bores shall be excavated. For all the 53 piers, a
total volume of about 14m3 of soil material will be excavated for the piers to be bored.
Estimated Costs for Building and Construction Projects
Question 1: Bored excavation for 300mm diameter piers
Total volume is calculated by considering bulking factor of soil (Peterson, 2014).
Ite
m
Particulars Unit Quantity Rate $ C
1 Bored excavation volume for 53 Ø300mm
and 2.5m deep piers
Cross sectional area of pier = πr2 = π x 0.152 =
0.07m2
Volume of pier = cross sectional area x depth
= 0.07 x 2.5 = 0.18m3
Volume of 53 piers = 0.18m3 x 53 = 9.54m3
Assume bulking factor of 1.4 (clay soil), total
volume of soil material to be excavated for
boring of piers = 1.4 x 9.54 = 13.6m3
m3 14
Summary:
A total of 53 53 Ø300mm and 2.5m deep pier bores shall be excavated. For all the 53 piers, a
total volume of about 14m3 of soil material will be excavated for the piers to be bored.
ESTIMATED COSTS FOR BUILDING PROJECTS 3
Question 2: Bulk excavation for general site preparation (DelPico, 2012)
Ite
m
Particulars Unit Quantity Rate $ C
2 Bulk excavation m3 150
Overall bulk area of excavation (plus working
space) = (12.83 + 4.08 + 2.00) x (11.87 +
2.00)
= 18.91 x 13.87 = 262.28
Estimated depths at corners: NE – 0.69m,
NW - 0.55m, SE – 0.50 and SW – 0.45)
Average depth = (0.690 + 0.54 + 0.45 +
0.40)/4 = 0.55m
Volume of bulk excavation = area x depth
= 262.28m2 x 0.55m = 144.25
Assume that only one side is battered =
18.91m x 1.0m wide x 0.55m = 10.40m3
Since battering is done at 45 degrees, battered
volume is = 0.5 x 10.40 = 5.20
Total volume of bulk excavation = 144.25 +
5.20 = 149.45
Summary:
A total of 150m3 of shall be excavated from the site during general preparation of the site. This
bulk volume includes working space, which extends outside the building towards the boundary
by 1m. The longest side of the building shall also be battered at an angle of 45 degrees and a
width of 1m from the building. The total bulk volume of 150m3 to be excavated includes the
actual site of the house, working space and battered volume. The subcontractor has to use this
value to determine the most suitable equipment to use for bulk excavation and if the excavated
material will have to be cart away.
Question 2: Bulk excavation for general site preparation (DelPico, 2012)
Ite
m
Particulars Unit Quantity Rate $ C
2 Bulk excavation m3 150
Overall bulk area of excavation (plus working
space) = (12.83 + 4.08 + 2.00) x (11.87 +
2.00)
= 18.91 x 13.87 = 262.28
Estimated depths at corners: NE – 0.69m,
NW - 0.55m, SE – 0.50 and SW – 0.45)
Average depth = (0.690 + 0.54 + 0.45 +
0.40)/4 = 0.55m
Volume of bulk excavation = area x depth
= 262.28m2 x 0.55m = 144.25
Assume that only one side is battered =
18.91m x 1.0m wide x 0.55m = 10.40m3
Since battering is done at 45 degrees, battered
volume is = 0.5 x 10.40 = 5.20
Total volume of bulk excavation = 144.25 +
5.20 = 149.45
Summary:
A total of 150m3 of shall be excavated from the site during general preparation of the site. This
bulk volume includes working space, which extends outside the building towards the boundary
by 1m. The longest side of the building shall also be battered at an angle of 45 degrees and a
width of 1m from the building. The total bulk volume of 150m3 to be excavated includes the
actual site of the house, working space and battered volume. The subcontractor has to use this
value to determine the most suitable equipment to use for bulk excavation and if the excavated
material will have to be cart away.
ESTIMATED COSTS FOR BUILDING PROJECTS 4
Question 3: N25 concrete for the ground slab (beams, patio slab and pad footings)
Thickness of slab = 150mm = 0.15m
Thickness of beam = 450mm = 0.35m
Width of beam = 100mm = 0.1m
Ite
m
Particulars Unit Quantity Rate $ C
3 N25 Concrete for ground slab m3
Slab on ground = length x width x slab
thickness
= 12.83m x 11.87m x 0.15m = 22.84
m3 23
Patio slab = length x width x thickness
= 6.35m x 1.67m x 0.15m = 1.59
m3 2
Family room = length x width x thickness
= 5.51m x 4.08m x 0.15m = 3.37
m3 4
Beams on slab = total length of beams x
width of beam x depth of beam
= 82.82m x 0.1m x 0.45m = 3.73
m3 4
Pad Footings
Area = 10(1.2 x 0.9) + (0.35 x 2.4)
= 10.80 + 0.84 = 11.64m2
Volume = area x depth
= 11.64 x 0.35 = 4.074
m3 5
Total m3 38
Summary:
A total volume of 38m3 of N25 concrete shall be required for the whole ground slab, including
beams, patio slab and pad footings but excluding piers. Depending on the mix design
Question 3: N25 concrete for the ground slab (beams, patio slab and pad footings)
Thickness of slab = 150mm = 0.15m
Thickness of beam = 450mm = 0.35m
Width of beam = 100mm = 0.1m
Ite
m
Particulars Unit Quantity Rate $ C
3 N25 Concrete for ground slab m3
Slab on ground = length x width x slab
thickness
= 12.83m x 11.87m x 0.15m = 22.84
m3 23
Patio slab = length x width x thickness
= 6.35m x 1.67m x 0.15m = 1.59
m3 2
Family room = length x width x thickness
= 5.51m x 4.08m x 0.15m = 3.37
m3 4
Beams on slab = total length of beams x
width of beam x depth of beam
= 82.82m x 0.1m x 0.45m = 3.73
m3 4
Pad Footings
Area = 10(1.2 x 0.9) + (0.35 x 2.4)
= 10.80 + 0.84 = 11.64m2
Volume = area x depth
= 11.64 x 0.35 = 4.074
m3 5
Total m3 38
Summary:
A total volume of 38m3 of N25 concrete shall be required for the whole ground slab, including
beams, patio slab and pad footings but excluding piers. Depending on the mix design
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
ESTIMATED COSTS FOR BUILDING PROJECTS 5
recommended by the engineer, it noes becomes easier to determine the volume of concrete
ingredients (concrete, coarse and fine aggregates and sand) needed for the ground slab.
Question 4: SL72 reinforcing mesh for the ground slabs
Ite
m
Particulars Unit Quantity Rate $ C
4 Number of SL72 Reinforcing Mesh Sheets
4 (a) Area of ground slab to be reinforced with
mesh (including 50mm cover)
= length x width = 11.98m x 11.02m
= 132.02m2
Area of family room to be reinforced with
mesh
= length x width = 4.66m x 3.68m = 17.15m2
Total area of slab to be covered = 132.02 +
17.15 = 149.17m2
m2 150
4 (b) Number of sheets = total area of slab/
effective area of SL72
Total area of slab to be reinforced with SL72
mesh = 150m2
Effective area of SL72 sheet = 5.8m x 2.2m =
12.76m2
Number of SL reinforcing mesh sheets =
150/12.76 = 11.76
No 12
Summary:
The total area of slab that requires to be covered by the SL72 reinforcing mesh is 150m2. Based
on this area (and putting into consideration the required overlap length and cover at the edges),
the number of 6.0m x 2.4m with 400mm lap SL72 reinforcing sheets required to complete the
slab is 12.
recommended by the engineer, it noes becomes easier to determine the volume of concrete
ingredients (concrete, coarse and fine aggregates and sand) needed for the ground slab.
Question 4: SL72 reinforcing mesh for the ground slabs
Ite
m
Particulars Unit Quantity Rate $ C
4 Number of SL72 Reinforcing Mesh Sheets
4 (a) Area of ground slab to be reinforced with
mesh (including 50mm cover)
= length x width = 11.98m x 11.02m
= 132.02m2
Area of family room to be reinforced with
mesh
= length x width = 4.66m x 3.68m = 17.15m2
Total area of slab to be covered = 132.02 +
17.15 = 149.17m2
m2 150
4 (b) Number of sheets = total area of slab/
effective area of SL72
Total area of slab to be reinforced with SL72
mesh = 150m2
Effective area of SL72 sheet = 5.8m x 2.2m =
12.76m2
Number of SL reinforcing mesh sheets =
150/12.76 = 11.76
No 12
Summary:
The total area of slab that requires to be covered by the SL72 reinforcing mesh is 150m2. Based
on this area (and putting into consideration the required overlap length and cover at the edges),
the number of 6.0m x 2.4m with 400mm lap SL72 reinforcing sheets required to complete the
slab is 12.
ESTIMATED COSTS FOR BUILDING PROJECTS 6
Question 5: Lengths of SL12TM for ground slab construction
Ite
m
Particulars Unit Quantity Rate $ C
5 Length and number of SL12TM Trench Mesh
5
(a)
Length of trench mesh (including 200mm lap)
= length of one trench x number of trenches
= 13.23m x 3 = 39.69m
= length of one trench x number of trenches
= 12.27m x 3 = 36.81m
= length of one trench x number of trenches
= 4.48m x 2 = 8.96m
= length of one trench x number of trenches
= 5.91m x 1 = 5.91m
Total length of trench mesh = 39.69 + 36.81 +
8.96 + 5.91 = 91.37
m 92
5 (b) Number of SL12TM trench mesh
16 sheets
= Length of trenches/length of trench sheets
= 92/6 = 15.33
No 16
Summary:
The total length of SL12TM trench mesh needed for ground lab (including overlap of 200mm at
each end) is 92m. Since each trench measure comes in 6m lengths, the number of lengths of
SL12TM trench mesh needed is 16. Therefore the contractor will have to supply 16 sheets of 6m
long SL12TM trench mesh.
Question 5: Lengths of SL12TM for ground slab construction
Ite
m
Particulars Unit Quantity Rate $ C
5 Length and number of SL12TM Trench Mesh
5
(a)
Length of trench mesh (including 200mm lap)
= length of one trench x number of trenches
= 13.23m x 3 = 39.69m
= length of one trench x number of trenches
= 12.27m x 3 = 36.81m
= length of one trench x number of trenches
= 4.48m x 2 = 8.96m
= length of one trench x number of trenches
= 5.91m x 1 = 5.91m
Total length of trench mesh = 39.69 + 36.81 +
8.96 + 5.91 = 91.37
m 92
5 (b) Number of SL12TM trench mesh
16 sheets
= Length of trenches/length of trench sheets
= 92/6 = 15.33
No 16
Summary:
The total length of SL12TM trench mesh needed for ground lab (including overlap of 200mm at
each end) is 92m. Since each trench measure comes in 6m lengths, the number of lengths of
SL12TM trench mesh needed is 16. Therefore the contractor will have to supply 16 sheets of 6m
long SL12TM trench mesh.
ESTIMATED COSTS FOR BUILDING PROJECTS 7
Question 6: Structural steel
Ite
m
Particulars Unit Quantity Rate $ C
6 Structural steel tonnes 1.78
B1
Weight = 25.7kg/m x 12.83m = 329.73kg
B2
Weight = 32kg/m x 12.83m = 410.56kg
B3
Weight = 14kg/m x 11.87m = 166.18kg
B4
Weight = 14kg/m x 11.87m = 166.18kg
B6
Weight = 14kg/m x 11.87m = 166.18kg
B5
Weight = 31.6kg/m x 12.83m = 405.43kg
L1
Weight = 14.6kg/m x 4.08m = 59.57kg
L2
Weight = 14.6kg/m x 5.51m = 80.45kg
Total weight = 329.73 + 410.56 + 166.18 +
166.18 + 166.18 + 405.43 + 59.57 + 80.45 =
1784.28kg
Summary:
Question 6: Structural steel
Ite
m
Particulars Unit Quantity Rate $ C
6 Structural steel tonnes 1.78
B1
Weight = 25.7kg/m x 12.83m = 329.73kg
B2
Weight = 32kg/m x 12.83m = 410.56kg
B3
Weight = 14kg/m x 11.87m = 166.18kg
B4
Weight = 14kg/m x 11.87m = 166.18kg
B6
Weight = 14kg/m x 11.87m = 166.18kg
B5
Weight = 31.6kg/m x 12.83m = 405.43kg
L1
Weight = 14.6kg/m x 4.08m = 59.57kg
L2
Weight = 14.6kg/m x 5.51m = 80.45kg
Total weight = 329.73 + 410.56 + 166.18 +
166.18 + 166.18 + 405.43 + 59.57 + 80.45 =
1784.28kg
Summary:
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
ESTIMATED COSTS FOR BUILDING PROJECTS 8
The total weight of structural steel needed for the project is 1.78 tonnes.
Question 7: Plasterboard
Ite
m
Particulars Unit Quantity Rate $ C
7 Plasterboard
7
(a)
Area of plasterboard
Assume that the height of plasterboard wall is
2.4m
Area of wall = length of wall x height of wall
A1 = 11.93m x 2.4m = 28.63m2
A2 = 11.93m x 2.4m = 28.63m2
A3 = 11.93m x 2.4m = 28.63m2
A4 = 11.93m x 2.4m = 28.63m2
A5 = 10.97m x 2.4m = 26.33m2
A6 = 10.97m x 2.4m = 26.33m2
A7 = 10.97m x 2.4m = 26.33m2
A8 = 10.97m x 2.4m = 26.33m2
A9 = 3.63m x 2.4m = 8.71m2
A10 = 3.63m x 2.4m = 8.71m2
A11 = 4.61m x 2.4m = 11.06m2
A12 = 4.61m x 2.4m = 11.06m2
Total area of walls = 4(28.63) + 4(26.33) +
2(8.71) + 2(11.06) = 259.38m2
Deduct area of doors and windows
Total area of doors = 5(2m x 0.9m) = 9.00m2
Area of windows = 6(1.2m x 0.8m) = 5.76m2
m2 245
The total weight of structural steel needed for the project is 1.78 tonnes.
Question 7: Plasterboard
Ite
m
Particulars Unit Quantity Rate $ C
7 Plasterboard
7
(a)
Area of plasterboard
Assume that the height of plasterboard wall is
2.4m
Area of wall = length of wall x height of wall
A1 = 11.93m x 2.4m = 28.63m2
A2 = 11.93m x 2.4m = 28.63m2
A3 = 11.93m x 2.4m = 28.63m2
A4 = 11.93m x 2.4m = 28.63m2
A5 = 10.97m x 2.4m = 26.33m2
A6 = 10.97m x 2.4m = 26.33m2
A7 = 10.97m x 2.4m = 26.33m2
A8 = 10.97m x 2.4m = 26.33m2
A9 = 3.63m x 2.4m = 8.71m2
A10 = 3.63m x 2.4m = 8.71m2
A11 = 4.61m x 2.4m = 11.06m2
A12 = 4.61m x 2.4m = 11.06m2
Total area of walls = 4(28.63) + 4(26.33) +
2(8.71) + 2(11.06) = 259.38m2
Deduct area of doors and windows
Total area of doors = 5(2m x 0.9m) = 9.00m2
Area of windows = 6(1.2m x 0.8m) = 5.76m2
m2 245
ESTIMATED COSTS FOR BUILDING PROJECTS 9
Total area of doors and windows = 9.00 +
5.76 = 14.76m2
Net measure = 259.38 – 14.76 = 244.62m2
7 (b) Number of plasterboard sheets
The selected size of plasterboard sheets is
1200mm wide and 2400mm long
Area of one plasterboard sheet = 1.2m x 2.4m
= 2.88m2
Number of sheets = net measure of wall/area
of one sheet
= 245m2/2.88m2 =85.07
No 86
Summary:
The total area of wall that will be covered with plasterboard is 245m2 and a total of 86
plasterboard sheets measuring 1200mm wide and 2400mm long shall be required for the project.
Total area of doors and windows = 9.00 +
5.76 = 14.76m2
Net measure = 259.38 – 14.76 = 244.62m2
7 (b) Number of plasterboard sheets
The selected size of plasterboard sheets is
1200mm wide and 2400mm long
Area of one plasterboard sheet = 1.2m x 2.4m
= 2.88m2
Number of sheets = net measure of wall/area
of one sheet
= 245m2/2.88m2 =85.07
No 86
Summary:
The total area of wall that will be covered with plasterboard is 245m2 and a total of 86
plasterboard sheets measuring 1200mm wide and 2400mm long shall be required for the project.
ESTIMATED COSTS FOR BUILDING PROJECTS 10
Question 8: 66x18mm finished DAR primed pine skirting
Ite
m
Particulars Unit Quantity Rate $ C
8 66x18mm finished DAR primed pine skirting
8 (a) Length of skirting
L1 = 11.93m
L2 = 11.93m
L3 = 11.93m
L4 = 11.93m
L5 = 10.97m
L6 = 10.97m
L7 = 10.97m
L8 = 10.97m
L9 = 3.63m
L10 = 3.63m
L11 = 4.61m
L12 = 4.61m
Total length of skirting = 4(11.93) + 4(10.97)
+ 2(3.63) + 2(4.61) = 108.08m
Plus 5% wastage
Total length of skirting (including wastage) =
1.05 x 108.08 = 113.48
m 114
8 (b) Number of skirting lengths
The selected length of 66x18mm finished
DAR primed pine skirting is 2.7m
No 48
Question 8: 66x18mm finished DAR primed pine skirting
Ite
m
Particulars Unit Quantity Rate $ C
8 66x18mm finished DAR primed pine skirting
8 (a) Length of skirting
L1 = 11.93m
L2 = 11.93m
L3 = 11.93m
L4 = 11.93m
L5 = 10.97m
L6 = 10.97m
L7 = 10.97m
L8 = 10.97m
L9 = 3.63m
L10 = 3.63m
L11 = 4.61m
L12 = 4.61m
Total length of skirting = 4(11.93) + 4(10.97)
+ 2(3.63) + 2(4.61) = 108.08m
Plus 5% wastage
Total length of skirting (including wastage) =
1.05 x 108.08 = 113.48
m 114
8 (b) Number of skirting lengths
The selected length of 66x18mm finished
DAR primed pine skirting is 2.7m
No 48
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
ESTIMATED COSTS FOR BUILDING PROJECTS 11
Number of skirting lengths = length of
skirting/ selected length of skirting
L1: 11.93m/2.7m = 4.42 = 5 sheets
L2: 11.93m/2.7m = 4.42 = 5 sheets
L3: 11.93m/2.7m = 4.42 = 5 sheets
L4: 11.93m/2.7m = 4.42 = 5 sheets
L5: 10.97m/2.7m = 4.06 = 5 sheets
L6: 10.97m/2.7m = 4.06 = 5 sheets
L7: 10.97m/2.7m = 4.06 = 5 sheets
L8: 10.97m/2.7m = 4.06 = 5 sheets
L9: 3.63m/2.7m = 1.34 = 2 sheets
L10: 3.63m/2.7m = 1.34 = 2 sheets
L11: 4.61m/2.7m = 1.71 = 2 sheets
L12: 4.61m/2.7m = 1.71 = 2 sheets
Total sheets = 8(5) + 4(2) = 48
Summary:
The total length of skirting is 114m and a total of 48 lengths of 66x18mm finished DAR primed
pine skirting measuring 2.7m long shall be required for the project.
Number of skirting lengths = length of
skirting/ selected length of skirting
L1: 11.93m/2.7m = 4.42 = 5 sheets
L2: 11.93m/2.7m = 4.42 = 5 sheets
L3: 11.93m/2.7m = 4.42 = 5 sheets
L4: 11.93m/2.7m = 4.42 = 5 sheets
L5: 10.97m/2.7m = 4.06 = 5 sheets
L6: 10.97m/2.7m = 4.06 = 5 sheets
L7: 10.97m/2.7m = 4.06 = 5 sheets
L8: 10.97m/2.7m = 4.06 = 5 sheets
L9: 3.63m/2.7m = 1.34 = 2 sheets
L10: 3.63m/2.7m = 1.34 = 2 sheets
L11: 4.61m/2.7m = 1.71 = 2 sheets
L12: 4.61m/2.7m = 1.71 = 2 sheets
Total sheets = 8(5) + 4(2) = 48
Summary:
The total length of skirting is 114m and a total of 48 lengths of 66x18mm finished DAR primed
pine skirting measuring 2.7m long shall be required for the project.
ESTIMATED COSTS FOR BUILDING PROJECTS 12
Question 9: Ceramic floor tiles
Ite
m
Particulars Unit Quantity Rate $ C
9 Ceramic floor tiles
9 (a) Amount of ceramic floor tiles
Family room = 4.08m x 5.51m = 22.48m2
Plus 10% wastage = 1.10 x 22.48m2 =
24.73m2 ≈ 25m2
Meals room = 2.43m x 4.68m = 11.37m2
Plus 10% wastage = 1.10 x 11.37m2 =
12.51m2 ≈ 13m2
Kitchen
Floor tiles: 4.59m x 6.11m = 28.04m2
Plus 10% wastage = 1.10 x 28.04m2 =
30.85m2
Wall tiles (under cupboards) = 2(4.59m x
1.8m) + 2(6.11m x 1.8m) = 38.52m2
Plus 10% wastage = 1.10 x 38.52m2 =
42.37m2
Total kitchen = 30.85 + 42.37 = 73.22m2 ≈
74m2
Foyer/stair area = 1.71m x 1.09m = 1.86m2
Plus 10% wastage = 1.10 x 1.86m2 = 2.05m2
≈ 3m2
Total = 25 + 13 + 74 + 3 = 115
m2 115
Question 9: Ceramic floor tiles
Ite
m
Particulars Unit Quantity Rate $ C
9 Ceramic floor tiles
9 (a) Amount of ceramic floor tiles
Family room = 4.08m x 5.51m = 22.48m2
Plus 10% wastage = 1.10 x 22.48m2 =
24.73m2 ≈ 25m2
Meals room = 2.43m x 4.68m = 11.37m2
Plus 10% wastage = 1.10 x 11.37m2 =
12.51m2 ≈ 13m2
Kitchen
Floor tiles: 4.59m x 6.11m = 28.04m2
Plus 10% wastage = 1.10 x 28.04m2 =
30.85m2
Wall tiles (under cupboards) = 2(4.59m x
1.8m) + 2(6.11m x 1.8m) = 38.52m2
Plus 10% wastage = 1.10 x 38.52m2 =
42.37m2
Total kitchen = 30.85 + 42.37 = 73.22m2 ≈
74m2
Foyer/stair area = 1.71m x 1.09m = 1.86m2
Plus 10% wastage = 1.10 x 1.86m2 = 2.05m2
≈ 3m2
Total = 25 + 13 + 74 + 3 = 115
m2 115
ESTIMATED COSTS FOR BUILDING PROJECTS 13
9 (b)
(i)
Number of boxes of floor tiles
Number of boxes of floor tiles = area of floor
tiles/1.2m2 = (25 + 13 + 31 + 3)/1.2 =
72m2/1.2m2 = 60
No 60
9 (b)
(ii)
Number of boxes of wall tiles
Number of wall tiles = area of wall tiles/1.4m2
= 43m2/1.4m2 = 30.71
No 31
Summary:
A total area of 115m2 (including family room, meals room, kitchen and foyer/stair area) shall be
tiled. 60 boxes of floor tiles and 31 boxes of wall tiles shall be required for the project.
References
DelPico, W., 2012. Estimating Building Costs for the Residential and Light Commercial
Construction Professional. New York: John Wiley & Sons.
Peterson, S., 2014. Estimating in Building Construction. Chicago: Pearson Education.
9 (b)
(i)
Number of boxes of floor tiles
Number of boxes of floor tiles = area of floor
tiles/1.2m2 = (25 + 13 + 31 + 3)/1.2 =
72m2/1.2m2 = 60
No 60
9 (b)
(ii)
Number of boxes of wall tiles
Number of wall tiles = area of wall tiles/1.4m2
= 43m2/1.4m2 = 30.71
No 31
Summary:
A total area of 115m2 (including family room, meals room, kitchen and foyer/stair area) shall be
tiled. 60 boxes of floor tiles and 31 boxes of wall tiles shall be required for the project.
References
DelPico, W., 2012. Estimating Building Costs for the Residential and Light Commercial
Construction Professional. New York: John Wiley & Sons.
Peterson, S., 2014. Estimating in Building Construction. Chicago: Pearson Education.
1 out of 13
Related Documents
Your All-in-One AI-Powered Toolkit for Academic Success.
+13062052269
info@desklib.com
Available 24*7 on WhatsApp / Email
Unlock your academic potential
© 2024 | Zucol Services PVT LTD | All rights reserved.