Answer Posted / rustam sahu

Suppose we have a room size of 10 ft. x 10 ft. where height of room is 11 feets and width of plastering is 1/2 inch.
1/2 inch = 1/2 divided by 12 = 0.04167 feet
now, let us find out the wet volume for plastering just for one side of wall..
i.e., (length x width x height ) =
= 10 x 0.04167 x 11
= 4.583 c.ft.
for dry volume, segregation value 1.52
Dry volume = wet vol. x seg. value
= 4.583 x 1.52
= 6.97 c.ft.
also, plastering ratio (1:6) sum of ratio = 1+6= 7
per quantity = dry vol. divided by sum of ratio
i.e., = 6.97 / 7 = 0.9957
note: 1 bag cement carries 1.25 c.ft
for cement per bag = 0.9957 /1.25 = 0.7965 bag (for 1 wall)
since a room have 4 walls. so,
0.7965 x 4 = 3.1862 bag
add 10% extra for irregularities at joints or corners
i.e., 3.186 + 0.3186 = 3.5 bags
for sand = per quantity x per ratios x no. of walls
= 0.9957 x 6 x 4
= 23.8968 c.ft

Answer = a room size (10 x 10) ft. having 11 feets height and 1/2 inch width of plastering will required 3.50 bag of cement and 23.89 c.ft. of sand

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