Question

A cistern, internally measuring $150cm\times 120cm\times 110cm$ has $129600{cm}^3$ 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 the water overflowing, each brick being $22.5cm\times 7.5cm\times 6,5cm$?

Answer 1

We have
volume of cistern $=150\times 120\times 110{cm}^3=1980000{cm}^3$

Volume of water in cister $=129600{cm}^3$

Volume of one brick $=22.5\times 7.5\times 6.5{cm}^3=1096.875{cm}^3$

volume of water absorbed by one brick $=\frac{1}{17}\times 1096.875{cm}^3$

Let $n$ be the total number of bricks which can be put in the cistern without water overflowing.

then volume of water absorbed by $n$ bricks $=n\times \frac{1}{17}\times 1096.875{cm}^3$

Volume of water left in the cistern $=(129600-\frac{n}{17}\times 1096.875){cm}^3$

Since the cistern is filled upto the brim

volume of water left in the cistern + volume of bricks = volume of the cistern

$\Rightarrow 129600-\frac{n}{17}\times 1096.875+n\times 1096.875=1980000$

$\Rightarrow 1096.875\times (n-\frac{n}{17})=1850400$

$\Rightarrow n=\frac{1850400}{17550}=1792.41\cong 1792$