Question

The length of the classroom is two times its height and its breadth is $1\dfrac{1}{2}$ times its height. The cost of whitewashing the walls at the rate of $Rs.1.60\,per{m^2}$ is $Rs.179.20$. Find the cost of tiling the floor at the rate of $Rs.6.75\,per{m^2}$

Answer 1

The whitewashing is done on the four walls. The Area of the four walls is given by $2\left( {l + b} \right)h$ . The tiling the floor we will find the area of the floor, Area is $l \times b$. The product of the area of the walls and the rate of whitewashing gives the total cost.

Complete step-by-step answer:
Let the height of the classroom be $x$
Given, the length of the classroom is two times its height
$\Rightarrow l = 2x$
And, the breadth is $1\dfrac{1}{2}$ times its height
$\Rightarrow b = 1\dfrac{1}{2}h = \dfrac{3}{2}h$
To find the area of 4 walls of the room,
The whitewashing is done on the four walls, the rate is $Rs.1.60\,per{m^2}$
The total cost of whitewashing is $Rs.179.20$
The product of the area of the walls and the rate of whitewashing gives the total cost.
$\Rightarrow Area \times 1.60 = 179.20$
$\Rightarrow Area = \dfrac{{179.20}}{{1.60}} = 112{m^2}$
Area of the four walls is given by $2\left( {l + b} \right)h$
$\Rightarrow 2(l + b)h = 112 \\\ \Rightarrow (l + b) \times h = 56 \\\$
Since, $h = x$, $l = 2x$and b=$\dfrac{3}{2}x$
$\Rightarrow (2x + \dfrac{3}{2}x) \times x = 56 \\\ \Rightarrow \dfrac{7}{2}{x^2} = 56 \\\ \Rightarrow {x^2} = 16 \\\ \Rightarrow x = \pm 4 \\\$
Dimension is not acceptable in the negative.
Therefore, $x = 4$
So height=4m
Length=8m
Breadth=6m
For tiling the floor we will find the area of the floor, Area is $l \times b$
$\Rightarrow Area = 8 \times 6 = 48 {m^2}$
The rate given of tiling the floor is $Rs.6.75\,per{m^2}$
The total cost of tiling the floor $= 48 \times 6.75 = Rs.324$
The required answer is Rs.324

Note: The room is in cuboidal shape. The lateral surface area of the cuboid is given by $2\left( {l + b} \right)h$ which is the area of the four walls. The area of the floor, Area is $l \times b$. The total area of the cuboidal room is the sum of four walls and the floor i.e., $2(l \times b) \times h + l \times b$