# A cistern, internally measuring $150cm \times 120 \times 110cm$, 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 $22.5cm \times 7.5cm \times 6.5cm$.

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Hint: The bricks are porous. So, they will absorb some water. The empty space left in the cistern will be covered by the volume of the brick.

Dimensions of the cistern is given as $150cm \times 120 \times 110cm$.

Volume of the cistern $ = 150 \times 120 \times 110 = 1980000c{m^3}$

Volume of water filled in the cistern $ = 129600c{m^3}$

Empty space left in the cistern $ = 1980000 - 129600 = 1850400c{m^3}$

Dimensions of brick are given as $22.5cm \times 7.5cm \times 6.5cm$.

Volume of each brick $ = 22.5 \times 7.5 \times 6.5 = 1096.875c{m^3}$

Let there be $x$ number of bricks that can be placed in the cistern just to avoid overflow. Then,

Total volume of all the bricks $ = 1096.875x$

It is also given that each brick absorbs one-seventeenth of its own volume of water. Then we hane:

Total volume of water absorbed by all the bricks $ = \dfrac{1}{{17}} \times 1096.875x$

This will increase the empty spaces in the cistern. So we have:

Total empty space left in the cistern$ = 1850400 + \dfrac{1}{{17}} \times 1096.875x$

This empty space will be capitalized by the space taken by the bricks themselves. So, we have:

$

\Rightarrow 1850400 + \dfrac{1}{{17}} \times 1096.875x = 1096.875x, \\

\Rightarrow 1096.875x - \dfrac{1}{{17}} \times 1096.875x = 1850400, \\

\Rightarrow \dfrac{{16}}{{17}} \times 1096.875x = 1850400, \\

\Rightarrow x = \dfrac{{17 \times 115650}}{{1096.875}}, \\

\Rightarrow x = 1792.41 \\

$

Therefore, a maximum of 1792 bricks can be placed in the cistern without overflowing the water.

Note: Above, we are getting the value of $x$as 1792.41. The number of bricks can only be integer.

If we take 1793 as the total number of bricks, this will result in some amount of water overflow because 1793 is greater than 1792.41. That’s why we can take 1792 as the maximum number of bricks.

Dimensions of the cistern is given as $150cm \times 120 \times 110cm$.

Volume of the cistern $ = 150 \times 120 \times 110 = 1980000c{m^3}$

Volume of water filled in the cistern $ = 129600c{m^3}$

Empty space left in the cistern $ = 1980000 - 129600 = 1850400c{m^3}$

Dimensions of brick are given as $22.5cm \times 7.5cm \times 6.5cm$.

Volume of each brick $ = 22.5 \times 7.5 \times 6.5 = 1096.875c{m^3}$

Let there be $x$ number of bricks that can be placed in the cistern just to avoid overflow. Then,

Total volume of all the bricks $ = 1096.875x$

It is also given that each brick absorbs one-seventeenth of its own volume of water. Then we hane:

Total volume of water absorbed by all the bricks $ = \dfrac{1}{{17}} \times 1096.875x$

This will increase the empty spaces in the cistern. So we have:

Total empty space left in the cistern$ = 1850400 + \dfrac{1}{{17}} \times 1096.875x$

This empty space will be capitalized by the space taken by the bricks themselves. So, we have:

$

\Rightarrow 1850400 + \dfrac{1}{{17}} \times 1096.875x = 1096.875x, \\

\Rightarrow 1096.875x - \dfrac{1}{{17}} \times 1096.875x = 1850400, \\

\Rightarrow \dfrac{{16}}{{17}} \times 1096.875x = 1850400, \\

\Rightarrow x = \dfrac{{17 \times 115650}}{{1096.875}}, \\

\Rightarrow x = 1792.41 \\

$

Therefore, a maximum of 1792 bricks can be placed in the cistern without overflowing the water.

Note: Above, we are getting the value of $x$as 1792.41. The number of bricks can only be integer.

If we take 1793 as the total number of bricks, this will result in some amount of water overflow because 1793 is greater than 1792.41. That’s why we can take 1792 as the maximum number of bricks.

Last updated date: 27th Sep 2023

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