Question

A cistern, internally measuring $150$cm $\times 120$cm $\times 110$cm, has $129600c{m}^3$ 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 $22.5$cm$\times 7.5$cm$\times 6.5$cm?

Answer 1

Volume of cistern = 150 x 120 x 110
= 1980000 cm${}^3$
Volume of water in cistern
= 1980000 - 129600
= 1850400 cm${}^3$
volume of one brick = 22.5 x 7.5 x 6.5
= 1096.875 cm${}^3$
Let 'N' no of bricks can be placed into the cistern
Hence volume of 'N' no of bricks
= N x 1096.875 cm${}^3$
One bricks absorb water $\frac{1}{17}$ of his volume hence water absorb by 1 brick $=\frac{1096.875}{17}c{m}^3$
Hence total N no. of brick absorb water
$=\frac{1096.875}{17}\times Nc{m}^3$
Volume to be filled in tank + water absorbs by N no. of bricks = volume of N No of bricks.
$1850400+\frac{1096.875N}{17}=N\times 1097.875$
$1850400=N\times 1096.875-\frac{1096.875N}{17}$
$=N\times 1096.875[1-\frac{1}{17}]$
$1850400=\frac{N\times 1096.875\times 16}{17}$
$N=\frac{1850400\times 17}{16\times 1096.875}$
$=\frac{115650\times 17}{10968.875}$
$=1792.4$
$\therefore$ No. of bricks can be put in cistern without over flowing water.