# 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 $1792$.