INDENT OF MATERIAL

Item Item Head Quantity Calculation Material Mat Unit

No Executed Qty.

1 Murum filling 65 cum Basic quantity + additional quantity due to

compaction factor 20%
Therefore, (1 cum + 0.2 cum) x 65cum Murum 78 cum

2 Rubble soling 85 cum Basic quantity + additional quantity due to

Dressing wastage factor 10%
Therefore, (1 cum + 0.1 cum) x 85 cum Rubble 94 cum

3 All Concrete items

a) PCC 1:3:6 mix 50 cum

a) Cement 1 cum = 30 bags
b) Concrete Add 52% in basic quantity as wet to dry ratio
1 cum of Wet mix = 1.52 cum of Dry mix
c) Working out
ingradients Therefore, 1.52 cum/ (Sum of Proportion)
Therefore, 1.52 cum/ (1+3+6)
Therefore Value of 1 = 1.52/10 = 0.152 cum
Value 1 is Cement = 1 x 0.152 x 30 x 50 Cement 228 Bags
Value 3 is Sand = 1 x 0.152 x 3 x 50 Sand 23 cum
Value 6 is Metal = 1 x 0.152 x 6 x 50 Metal 46 cum

b) RCC 1:1.5:3 mix 75 cum

a) Cement 1 cum = 30 bags
b) Concrete Add 52% in basic quantity as wet to dry ratio
1 cum of Wet mix = 1.52 cum of Dry mix
c) Working out
ingradients Therefore, 1.52 cum/ (Sum of Proportion)
Therefore, 1.52 cum/ (1+1.5+3)
Therefore Value of 1 = 1.52/5.5 = 0.276 cum
Value 1 is Cement = 1 x 0.276 x 30 x 75 Cement 621 Bags
Value 1.5 is Sand = 1 x 0.276 x 1.5 x 75 Sand 31 cum
Value 3 is Metal = 1 x 0.276 x 3 x 75 Metal 62 cum

4 Uncoursed rubble
masonary in cm 1:5 50 cum
 a) Rubble Basic quantity + additional quantity due to
Dressing wastage factor 40%
Therefore, (1 cum + 0.4 cum) x 50 cum Rubble 70 cum

b) Mortor Factor Dry mortor = 0.40 cum/ 1 cum

c) Working out
ingradients Therefore, 0.40 cum/ (Sum of Proportion)
Therefore, 0.40 cum/ (1+5)
Therefore Value of 1 = 0.40/6 = 0.067 cum
Value 1 is Cement = 1 x 0.067 x 30 x 50 Cement 101 Bags
Value 5 is Sand = 1 x 0.067 x 5 x 50 Sand 17 cum

5 230mm thick Brick

masonary in cm 1:5 100 Sq.M
(23 cu.m)
a) Bricks 500 nos per cum x 23 cum Bricks 11500 Nos
b) Mortor Factor Dry mortor = 0.25 cum/ 1 cum
c) Working out
ingradients Therefore, 0.25 cum/ (Sum of Proportion)
Therefore, 0.25 cum/ (1+5)
Therefore Value of 1 = 0.25/6 = 0.042 cum
Value 1 is Cement = 1 x 0.042 x 30 x 23 Cement 29 Bags
Value 5 is Sand = 1 x 0.042 x 5 x 23 Sand 5 cum

6 115mm thick Brick

masonary in cm 1:4 80 Sq.M
a) Bricks 60 nos per Sqm x 80 Sq.M Bricks 4800 Nos
b) Mortor Factor Dry mortor = 0.022 cum/ 1 Sqm
c) Working out
ingradients Therefore, 0.022 cum/ (Sum of Proportion)
Therefore, 0.022 cum/ (1+4)
Therefore Value of 1 = 0.022/5 = 0.0044 cum
Value 1 is Cement = 1 x 0.0044 x 30 x 80 Cement 11 Bags
Value 4 is Sand = 1 x 0.0044 x 4 x 80 Sand 1 cum

7 Neeru plaster for

ceiling 8 to 10mm
thick in cm 1:3 120 Sq.M
a) Mortor Factor Dry mortor = 0.012 cum/ 1 Sqm
b) Working out
ingradients Therefore, 0.012 cum/ (Sum of Proportion)
Therefore, 0.012 cum/ (1+3)
Therefore Value of 1 = 0.022/4 = 0.0055 cum
Value 1 is Cement = 1 x 0.0055 x 30 x 120 Cement 20 Bags
 Value 3 is Sand = 1 x 0.0055 x 3 x 1.20 x120 Sand 2 cum
(Factor 1.2 multiplied for screening wastage)
c) Neeru 0.1 bag per Sqm x 120 sq.m Sanala 12 Bags

8 Neeru plaster for

wall 12 to 15mm
thick in cm 1:4 300 Sq.M
a) Mortor Factor Dry mortor = 0.018 cum/ 1 Sqm
b) Working out
ingradients Therefore, 0.018 cum/ (Sum of Proportion)
Therefore, 0.018 cum/ (1+4)
Therefore Value of 1 = 0.018/5 = 0.0036 cum
Value 1 is Cement = 1 x 0.0036 x 30 x 300 Cement 32 Bags
Value 4 is Sand = 1 x 0.0036 x 4 x 1.20 x300 Sand 5 cum
(Factor 1.2 multiplied for screening wastage)
c) Neeru 0.1 bag per Sqm x 300 sq.m Sanala 30 Bags

9 Sandfaced double
coat plaster to wall
18 to 20mm thick
in cm 1:4 280 Sq.M
a) Mortor Factor Dry mortor = 0.024 cum/ 1 Sqm
b) Working out
ingradients Therefore, 0.024 cum/ (Sum of Proportion)
Therefore, 0.024 cum/ (1+4)
Therefore Value of 1 = 0.024/5 = 0.0048 cum
Value 1 is Cement = 1 x 0.0048 x 30 x 280 Cement 40 Bags
Value 4 is Sand = 1 x 0.0048 x 4 x 1.20 x280 Sand 6 cum
(Factor 1.2 multiplied for screening wastage)

10 Kota stone flooring

on 50mm thick sub
base in cm 1:5 185 Sq.m

a) Kota stone Add 20% on basic quantity for dressing wastage

Therefore, 1 Sqm + 0.2 Sqm
Therefore, 1.2 Sq.m x 185 Kotah stone 222 Sq.m

b) Mortor Dry mortor = 0.06 cum/ 1 Sqm

Therefore, 0.06 cum/ (Sum of Proportion)
Therefore, 0.06 cum/ (1+5)
Therefore Value of 1 = 0.06/6 = 0.01 cum
Value 1 is Cement = 1 x 0.01 x 30 x 185 Cement 56 Bag
Value 5 is Sand = 1 x 0.01 x 5 x 185 Sand 9 cum
 c) Extra Cement 0.1 bag per 1 Sqm for Cement slurry backing
Therefore, 0.1 bag x 185 sq.m Cement 19 Bag

11 Vitrified flooring
on 35mm thick sub
base in cm 1:5 145 Sq.m
a) Vitrified tile 1 Sq.m x 145 Sq.m Tile 145 Sq.m
b) Mortor Dry mortor = 0.042 cum/ 1 Sqm
Therefore, 0.042 cum/ (Sum of Proportion)
Therefore, 0.042 cum/ (1+5)
Therefore Value of 1 = 0.042/6 = 0.007 cum
Value 1 is Cement = 1 x 0.007 x 30 x 145 Cement 30 Bag
Value 5 is Sand = 1 x 0.007 x 5 x 145 Sand 5 cum

c) Extra Cement 0.1 bag per 1 Sqm for Cement slurry backing
Therefore, 0.1 bag x 145 sq.m Cement 15 Bag

12 Vitrified wall tiles

on 15mm thick
backing mortor in
cm 1:2 70 Sq.m
a) Vitrified tile 1 Sq.m x 70 Sq.m Tile 70 Sq.m
b) Mortor Dry mortor = 0.018 cum/ 1 Sqm
Therefore, 0.018 cum/ (Sum of Proportion)
Therefore, 0.018 cum/ (1+2)
Therefore Value of 1 = 0.018/3 = 0.006 cum
Value 1 is Cement = 1 x 0.006 x 30 x 70 Cement 13 Bag
Value 2 is Sand = 1 x 0.006 x 2 x 1.2 x 70 Sand 1 cum
(Factor 1.2 multiplied for screening wastage to
make sand fine)

13 100mm thick Brick

bat waterproof
treatment in cm 1:4 230 Sq.m
a) Bricks 35 nos per Sqm x 230 Sq.m Bricks 8050 Nos
b) Mortor Dry mortor = 0.053 cum/ 1 Sqm
Therefore, 0.053 cum/ (Sum of Proportion)
Therefore, 0.053 cum/ (1+4)
Therefore Value of 1 = 0.053/5 = 0.011 cum
Value 1 is Cement = 1 x 0.011 x 30 x 230 Cement 76 Bag
Value 4 is Sand = 1 x 0.011 x 4 x 230 Sand 10 cum
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