1.You are short in the number of piers, you need to ensure no pier is no more than 1800 from centre to centre. Revise this and check the measurements. The piers on the dining living areas and family room are too far apart. Do not use linear or lineal metres this is for measuring a different type of dimension.
2. You have not followed the general methodology of how to calculate bulk excavations. The volume is short. The excavation would most likely be done in overall dimensions of (4.08 + 12.83) x 11.87 plus the working space as this will create a useable flat area at the back door. It will be difficult for the owner if the area in the SE corner of the yard was left there.
3.Industry practice is to write the quantity against the last line of the description. This description should also list what is included, supply? Place? You have crunched the numbers into a small space. It is ok to spread the dimensions out, to make it easier to read and it should be on P10 paper format.
I have no idea which area or beam you are measuring. All the quantities need to be added together as this is one item of work and all areas will be poured together. So a description and a list of all dimensions including description of where the measurements come from and then all put into one quantity.
Answer:
Bored excavation for 300mm diameter piers
Total volume is calculated by considering bulking factor of soil (Peterson, 2014).
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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1
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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
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m3
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14
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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.
Question 2: Bulk excavation for general site preparation (DelPico, 2012)
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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2
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Bulk excavation
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m3
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150
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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
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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 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
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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3
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N25 Concrete for ground slab
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m3
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Slab on ground = length x width x slab thickness
= 12.83m x 11.87m x 0.15m = 22.84
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m3
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23
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Patio slab = length x width x thickness
= 6.35m x 1.67m x 0.15m = 1.59
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m3
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2
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Family room = length x width x thickness
= 5.51m x 4.08m x 0.15m = 3.37
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m3
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4
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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
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m3
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4
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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
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m3
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5
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Total
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m3
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38
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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 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
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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4
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Number of SL72 Reinforcing Mesh Sheets
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4 (a)
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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
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m2
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150
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4 (b)
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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
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No
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12
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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.
Question 5: Lengths of SL12TM for ground slab construction
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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5
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Length and number of SL12TM Trench Mesh
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5 (a)
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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
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m
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92
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5 (b)
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Number of SL12TM trench mesh
16 sheets
= Length of trenches/length of trench sheets
= 92/6 = 15.33
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No
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16
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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 6: Structural steel
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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6
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Structural steel
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tonnes
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1.78
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B1
Weight = 25.7kg/m x 12.83m = 329.73kg
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B2
Weight = 32kg/m x 12.83m = 410.56kg
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B3
Weight = 14kg/m x 11.87m = 166.18kg
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B4
Weight = 14kg/m x 11.87m = 166.18kg
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B6
Weight = 14kg/m x 11.87m = 166.18kg
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B5
Weight = 31.6kg/m x 12.83m = 405.43kg
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L1
Weight = 14.6kg/m x 4.08m = 59.57kg
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L2
Weight = 14.6kg/m x 5.51m = 80.45kg
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Total weight = 329.73 + 410.56 + 166.18 + 166.18 + 166.18 + 405.43 + 59.57 + 80.45 = 1784.28kg
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Summary:
The total weight of structural steel needed for the project is 1.78 tonnes.
Question 7: Plasterboard
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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7
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Plasterboard
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7 (a)
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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
Total area of doors and windows = 9.00 + 5.76 = 14.76m2
Net measure = 259.38 – 14.76 = 244.62m2
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m2
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245
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7 (b)
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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
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No
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86
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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.
Question 8: 66x18mm finished DAR primed pine skirting
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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8
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66x18mm finished DAR primed pine skirting
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8 (a)
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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
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m
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114
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8 (b)
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Number of skirting lengths
The selected length of 66x18mm finished DAR primed pine skirting is 2.7m
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
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No
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48
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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.
Question 9: Ceramic floor tiles
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Item
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Particulars
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Unit
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Quantity
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Rate
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$ C
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9
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Ceramic floor tiles
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9 (a)
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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
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m2
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115
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9 (b) (i)
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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
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No
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60
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9 (b) (ii)
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Number of boxes of wall tiles
Number of wall tiles = area of wall tiles/1.4m2 = 43m2/1.4m2 = 30.71
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No
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31
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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.
