 # A mason constructs a wall of dimensions

Question:

A mason constructs a wall of dimensions $270 \mathrm{~cm} \times 300 \mathrm{~cm} \times 350 \mathrm{~cm}$ with the bricks each of size $22.5 \mathrm{~cm} \times$ $11.25 \mathrm{~cm} \times 8.75 \mathrm{~cm}$ and it is assumed that $\frac{1}{8}$ space is covered by the mortar. Then, the

number of bricks used to construct the wall is

(a) 11100

(b) 11200

(c) 11000

(d) 11300

Solution:

(b) Volume of the wall $=270 \times 300 \times 350=28350000 \mathrm{~cm}^{3}$

$[\because$ volume of cuboid $=$ length $\times$ breadth $\times$ height $]$

Since, $\frac{1}{8}$ space of wall is covered by mortar.

So, remaining space of wall = Volume of wall - Volume of mortar

$=28350000-3543750=24806250 \mathrm{~cm}^{3}$

Now, volume of one brick $=22.5 \times 1125 \times 8.75=2214.844 \mathrm{~cm}^{3}$

[ $\because$ volume of cuboid $=$ length $\times$ breadth $\times$ height]

$\therefore$ Required number of bricks $=\frac{24806250}{2214.844}=11200$ (approx)

Hence, the number of bricks used to construct the wall is 11200 .