Answer / rajarshi basu

Let Area to be plastered=100 m^2
& Thickness of plaster= 12mm=(12/1000)m= 0.012m
Volume of mortar required= 100 x 0.012= 1.20 m^3
Add 20% extra mortar for filling joint,depressions and
wastage.

Volume of the mortar= [0.20 X 1.20]+ 1.20=1.44 m^3
[Wet Volume]

Increase further by One-third of the volume to get Dry
volume
Quantity becomes= [(1/3) X 1.44]+ 1.44=1.92 m^3

Summation of proportion=1+5=6
For 12mm plastering in 1:5 cement:sand mortar mix,

1)Cement required= 1.92/6= 0.32 cu. m
No. of bags of cement= 0.320/0.0347=9.22 bags
>>9 bags(say)
Kgs of cement required=9 x 50= 450 kg (approx)

2)Sand required= [1.92 x (5/6)]= 1.60 cu. m


[Weight of 1 bag of cement= 50 kg
Cement should be always measured by weight to avoid
variation.

One 50 kg bag of cement weighs 1440 kg/cu. m
Volume of 1 bag of cement= 50/1440=0.0347 cu. m ]

Therefore to convert Cu. m into No of bags of 50 kg
cement,just divide the volume of cement by 0.0347

Answer / sandeep tv

let Area to be plastered=100 m^2
& Thickness of plaster= 12mm=(12/1000)m= 0.012m
Volume of mortar required= 100 x 0.012= 1.20 m^3
Add 20% extra mortar for filling joint,depressions and
wastage.

Volume of the mortar= [0.20 X 1.20]+ 1.20=1.44 m^3
[Wet Volume]

Increase further by One-third of the volume to get Dry
volume
Quantity becomes= [(1/3) X 1.44]+ 1.44=1.92 m^3

Summation of proportion=1+5=6
For 12mm plastering in 1:5 cement:sand mortar mix,

1)Cement required= 1.92/6= 0.32 cu. m
No. of bags of cement= 0.320/0.0347=9.22 bags
>>9 bags(say)
Kgs of cement required=9 x 50= 450 kg (approx)

2)Sand required= [1.92 x (5/6)]= 1.60 cu. m


[Weight of 1 bag of cement= 50 kg
Cement should be always measured by weight to avoid
variation.

One 50 kg bag of cement weighs 1440 kg/cu. m
Volume of 1 bag of cement= 50/1440=0.0347 cu. m ]

Therefore to convert Cu. m into No of bags of 50 kg
cement,just divide the volume of cement by 0.0347

Answer / ashwin

cement:sand= 1:5
for 1cum= 1/6
=0.166
so cement for 1 cum=0.166cum
but volume of 1 cement bag is=.033cum
bag required per cum is=.166/.033
=5 bags
sand required is=0.166*5
=0.833 cum

Answer / munir

above ansewr is corect

Answer / yuy

anser plz

Answer / jegan

1:5 PLASTERING CEMENT REQUIRED IS 288 KGS

Answer / yogesh khandelwal

Let Area to be plastered=100 m^2
& Thickness of plaster= 12mm=(12/1000)m= 0.012m
Volume of mortar required= 100 x 0.012= 1.20 m^3
Add 20% extra mortar for filling joint,depressions and
wastage.

Volume of the mortar= [0.20 X 1.20]+ 1.20=1.44 m^3
[Wet Volume]

Increase further by One-third of the volume to get Dry
volume
Quantity becomes= [(1/3) X 1.44]+ 1.44=1.92 m^3

Summation of proportion=1+5=6
For 12mm plastering in 1:5 cement:sand mortar mix,

1)Cement required= 1.92/6= 0.32 cu. m
No. of bags of cement= 0.320/0.0347=9.22 bags
>>9 bags(say)
Kgs of cement required=9 x 50= 450 kg (approx)

2)Sand required= [1.92 x (5/6)]= 1.60 cu. m


[Weight of 1 bag of cement= 50 kg
Cement should be always measured by weight to avoid
variation.

One 50 kg bag of cement weighs 1440 kg/cu. m
Volume of 1 bag of cement= 50/1440=0.0347 cu. m ]

Therefore to convert Cu. m into No of bags of 50 kg
cement,just divide the volume of cement by 0.0347

Answer / hidayathulla

why should we have to take 20% of wet mortar & 1/3rd of Dry mortar......is there eny code for that...or plz tell me where it coms from......plz clear me ....

Answer / adz

1:5 proportion for (1)cement @ 40kg): (5)sand @ 33 kg(S.G
@2.6) gives you approximate volume of 0.142 cubic meter.

if you want exact volume, provide the specific gravity of
your materials and include air content and water for your
volume...

Answer / abrar

c:m ratio for plastering (1:5) for 1 cum

sand 1cum = 1.25cum .... constant in all ratios

cement 1cum = 1.25 x 1440
------------ = 360 kgs = 7.2 bags
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