A cistern, internally measuring $ 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} $, has $ 129600 \mathrm{~cm}^{3} $ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being $ 22.5 \mathrm{~cm} \times 7.5 \mathrm{~cm} \times 6.5 \mathrm{~cm} $ ?
Given:
A cistern, internally measuring \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \), has \( 129600 \mathrm{~cm}^{3} \) of water in it.
Porous bricks are placed in the water until the cistern is full to the brim.
Each brick absorbs one-seventeenth of its own volume of water.
Dimensions of each is \( 22.5 \mathrm{~cm} \times 7.5 \mathrm{~cm} \times 6.5 \mathrm{~cm} \).
To do:
We have to find the number of bricks that can be put in without overflowing the water.
Solution:
The dimensions of the cistern are \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \)
This implies,
Volume of the cistern $= 1980000\ cm^3$
Volume of water in the cistern $=129600 \mathrm{~cm}^{3}$
Volume of the bricks to be filled in cistern $= 1980000 - 129600\ cm^3$
$=1850400\ cm^3$
Dimensions of each is \( 22.5 \mathrm{~cm} \times 7.5 \mathrm{~cm} \times 6.5 \mathrm{~cm} \).
Let the number of bricks placed be $n$.
Therefore,
Volume of $n$ bricks $= n \times 22.5 \times 7.5 \times 6.5$
Each brick absorbs one-seventeenth of its own volume.
The volume of water absorbed by $n$ bricks $=\frac{n}{17} \times 22.5 \times 7.5 \times 6.5$
Therefore,
The volume of $n$ bricks $=$ Volume of water absorbed by $n$ bricks $+$ Volume to be filled in cistern
$n \times 22.5 \times 7.5 \times 6.5=1850400+\frac{n}{17} \times 22.5 \times 7.5 \times 6.5$
$\frac{17n-n}{17} \times 22.5 \times 7.5 \times 6.5=1850400$
$\frac{16n}{17} \times 22.5 \times 7.5 \times 6.5=1850400$
$n = 1792.41$
Hence, the number of bricks which can be put in the cistern without overflowing the water is $1792$.
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