A wall 24 , m long,0.4 , m thick and 6 , m hi | vedclass

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(D) Given that,a wall is constructed with the help of bricks and mortar.Number of bricks $= \frac{	ext{Volume of wall} - (\frac{1}{10} 	ext{th volum

Vedclass · Accepted Answer

$12960$

Vedclass · Answer

(D) Given that,a wall is constructed with the help of bricks and mortar.Number of bricks $= \frac{	ext{Volume of wall} - (\frac{1}{10} 	ext{th volume of wall})}{	ext{Volume of a brick}}$ .........$(i)$Length of wall $(l) = 24 \, m$,Thickness $(b) = 0.4 \, m$,Height $(h) = 6 \, m$.Volume of wall $= l 	imes b 	imes h = 24 	imes 0.4 	imes 6 = 57.6 \, m^3$.Volume occupied by mortar $= \frac{1}{10} 	imes 57.6 = 5.76 \, m^3$.Volume occupied by bricks $= 57.6 - 5.76 = 51.84 \, m^3$.Dimensions of a brick $= 25 \, cm 	imes 16 \, cm 	imes 10 \, cm = 0.25 \, m 	imes 0.16 \, m 	imes 0.10 \, m$.Volume of one brick $= 0.25 	imes 0.16 	imes 0.10 = 0.004 \, m^3$.Number of bricks $= \frac{51.84}{0.004} = \frac{51840}{4} = 12960$.Hence,the required number of bricks used in constructing the wall is $12960$.

Part $I$	Part $II$
$1.$ Volume of a cuboid	$a.$ $\frac{1}{3} \pi r^{2} h$
$2.$ Volume of a cone	$b.$ $\pi r^{2} h$
$3.$ Volume of a cylinder	$c.$ $\frac{4}{3} \pi r^{3}$
$4.$ Volume of a sphere	$d.$ $lbh$

$A$ wall $24 \, m$ long,$0.4 \, m$ thick and $6 \, m$ high is constructed with bricks each of dimensions $25 \, cm \times 16 \, cm \times 10 \, cm$. If the mortar occupies $\frac{1}{10}$th of the volume of the wall,then find the number of bricks used in constructing the wall.

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