The number of bricks,each measuring 25,cm tim | vedclass

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(C) Volume of the wall $= 6\,m 	imes 5\,m 	imes 0.5\,m = 15\,m^3$.Since the mortar occupies $5\%$ of the volume,the volume occupied by bricks $= 95\

Vedclass · Accepted Answer

$6080$

Vedclass · Answer

(C) Volume of the wall $= 6\,m 	imes 5\,m 	imes 0.5\,m = 15\,m^3$.Since the mortar occupies $5\%$ of the volume,the volume occupied by bricks $= 95\%$ of $15\,m^3 = 0.95 	imes 15\,m^3 = 14.25\,m^3$.Convert the volume of the wall into cubic centimeters: $14.25\,m^3 = 14.25 	imes (100\,cm)^3 = 14.25 	imes 1,000,000\,cm^3 = 14,250,000\,cm^3$.Volume of $1$ brick $= 25\,cm 	imes 12.5\,cm 	imes 7.5\,cm = 2343.75\,cm^3$.Number of bricks $= \frac{	ext{Volume of bricks}}{	ext{Volume of } 1\, 	ext{brick}} = \frac{14,250,000}{2343.75} = 6080$.

The number of bricks,each measuring $25\,cm \times 12.5\,cm \times 7.5\,cm$,required to construct a wall $6\,m$ long,$5\,m$ high,and $0.5\,m$ thick,if the mortar occupies $5\%$ of the volume of the wall,is:

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