- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# 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 $1792$.

- Related Articles
- The paint in a certain container is sufficient to paint an area equal to \( 9.375 \mathrm{~m}^{2} \). How many bricks of dimensions \( 22.5 \mathrm{~cm} \times 10 \mathrm{~cm} \times 7.5 \mathrm{~cm} \) can be painted out of this container?
- Find the volume of the cuboid having the following dimensions.\( 12 \mathrm{cm} \times 5 \mathrm{cm} \times 8 \mathrm{cm} \).
- 16 glass spheres each of radius \( 2 \mathrm{~cm} \) are packed into a cuboidal box of internal dimensions \( 16 \mathrm{~cm} \times 8 \mathrm{~cm} \times 8 \mathrm{~cm} \) and then the box is filled with water. Find the volume of water filled in the box.
- Find the length of the longest rod that can be put in the room of dimensions \( 10 \mathrm{~cm} \times 6 \mathrm{~cm} \) \( \times 4 \mathrm{~cm} . \quad \)
- How many spherical lead shots each of diameter \( 4.2 \mathrm{~cm} \) can be obtained from a solid rectangular lead piece with dimensions \( 66 \mathrm{~cm} \times 42 \mathrm{~cm} \times 21 \mathrm{~cm} \).
- How many coins \( 1.75 \mathrm{~cm} \) in diameter and \( 2 \mathrm{~mm} \) thick must be melted to form a cuboid \( 11 \mathrm{~cm} \times 10 \mathrm{~cm} \times 7 \mathrm{~cm} ? \)
- A matchbox measures \( 4 \mathrm{~cm} \times 2.5 \mathrm{~cm} \times 1.5 \mathrm{~cm} \). What will be the volume of a packet containing 12 such boxes?
- Find the area of a quadrilateral \( \mathrm{ABCD} \) in which \( \mathrm{AB}=3 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}, \mathrm{CD}=4 \mathrm{~cm} \), \( \mathrm{DA}=5 \mathrm{~cm} \) and \( \mathrm{AC}=5 \mathrm{~cm} \).
- A solid consisting of a right circular cone of height \( 120 \mathrm{~cm} \) and radius \( 60 \mathrm{~cm} \) standing on a hemisphere of radius \( 60 \mathrm{~cm} \) is placed upright in a right circular cylinder full of water such that it touches the botioms. Find the volume of water left in the cylinder, if the radius of the cylinder is \( 60 \mathrm{~cm} \) and its height is \( 180 \mathrm{~cm} \).
- A canal is \( 300 \mathrm{~cm} \) wide and \( 120 \mathrm{~cm} \) deep. The water in the canal is flowing with a speed of \( 20 \mathrm{~km} / \mathrm{h} \). How much area will it irrigate in 20 minutes if \( 8 \mathrm{~cm} \) of standing water is desired?
- Find the perimeter of a triangle with sides measuring \( 10 \mathrm{~cm}, 14 \mathrm{~cm} \) and \( 15 \mathrm{~cm} \).
- The circumference of the base of a cylindrical vessel is \( 132 \mathrm{~cm} \) and its height is \( 25 \mathrm{~cm} \). How many litres of water can it hold? \( \left(1000 \mathrm{~cm}^{3}=1 l\right) \).
- Construct a triangle with sides \( 5 \mathrm{~cm}, 5.5 \mathrm{~cm} \) and \( 6.5 \mathrm{~cm} \). Now, construct another triangle whose sides are \( 3 / 5 \) times the corresponding sides of the given triangle.
- \( \triangle \mathrm{ABC} \sim \triangle \mathrm{ZYX} . \) If \( \mathrm{AB}=3 \mathrm{~cm}, \quad \mathrm{BC}=5 \mathrm{~cm} \), \( \mathrm{CA}=6 \mathrm{~cm} \) and \( \mathrm{XY}=6 \mathrm{~cm} \), find the perimeter of \( \Delta \mathrm{XYZ} \).
- Metal spheres, each of radius \( 2 \mathrm{~cm} \), are packed into a rectangular box of internal dimension \( 16 \mathrm{~cm} \times 8 \mathrm{~cm} \times 8 \mathrm{~cm} \) when 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. [Use \( \pi=669 / 213] \)