Question

A classroom is $7m$ long, $6m$ broad and $4m$ high. The doors and windows occupy an area of $7{{m}^{2}}$. Find the cost of whitewashing the walls and roof at the rate of $Rs.15$ per ${{m}^{2}}$.

Answer 1

Hint: We will find the total surface area of cuboid or we can use the total surface area of the classroom and subtract the surface area of windows and doors which are $7{{m}^{2}}$. Also subtract the surface area of the floor from the total surface; finally we will multiply the net surface area to get the cost of total whitewashing as $Rs.15$.

Complete step-by-step answer:
It is given in the question that the length, breadth and height of a classroom is $7m$, $6m$ and $4m$ long. Also, the doors and windows occupy an area of $7{{m}^{2}}$. The cost of whitewashing is $Rs.15$ per ${{m}^{2}}$. We have to find the total cost of whitewashing for a classroom.
Assume the shape of the classroom as cuboid whose length is $7m$, breadth is $6m$, and height is $4m$.

We know that total surface area of a cuboid is given by formula,
$cuboid=2(length\times breadth+breadth\times height+height\times length)$.

Total surface area of classroom is

$\begin{align}

& =2(7m\times 6m+6m\times 4m+4m\times 7m) \\

& =2(42{{m}^{2}}+24{{m}^{2}}+28{{m}^{2}}) \\

& =2(94{{m}^{2}}) \\

& =188{{m}^{2}} \\

\end{align}$

But we have to subtract the surface area of doors, windows and floor as we don’t whitewash them.

So, the surface area of the floor is given by the multiplication of length and breadth as it is in the shape of a rectangle and the area of the rectangle is $length\times breadth$. So, the surface area of floor is

$=7m\times 6m$

$=42{{m}^{2}}$

Also, it is given in the question that the surface area of windows and doors is $7{{m}^{2}}$
So, total area in which we are not whitewashing is

$=$ Area of floor$+$area of doors and windows

\[\begin{align}

& =42{{m}^{2}}+7{{m}^{2}} \\

& =49{{m}^{2}} \\

\end{align}\]

And the area in which we will do whitewash,

$=$ (Total surface area of classroom)$-$(Area in which we are not doing whitewash)

\[\begin{align}

& =188{{m}^{2}}-49{{m}^{2}} \\

& =139{{m}^{2}} \\

\end{align}\]

Thus, we have to whitewash only in the $139{{m}^{2}}$ area.

Cost per ${{m}^{2}}$ is $Rs.15$

Using unitary method, we get

Cost of $139{{m}^{2}}$ whitewashing $=$ cost of $1{{m}^{2}}$ whitewashing $\times 139{{m}^{2}}$

$=15\times 139$

$=Rs.2085$

Therefore, the total cost for whitewashing a classroom is $Rs.2085$.

Note: You can find the surface area of whitewashing by finding the curved surface area of the classroom$+$surface area of roof$-$ surface area of windows and doors. We know that curved surface area
$\begin{align}
& =2\times \left( l\times b+b\times h \right)+l\times b-7{{m}^{2}} \\
& =2\left( 7\times 4+6\times 4 \right)+7\times 6-7{{m}^{2}} \\
& =2\left( 28+24 \right)+42-7{{m}^{2}} \\
& =2\left( 52 \right)+42-7{{m}^{2}} \\
& =(104+35) \\
& =139{{m}^{2}} \\
\end{align}$

Now, we can calculate the cost of whitewashing by multiplying $139{{m}^{2}}$ with $Rs.15$. We get
$\begin{align}
& =139\times 15 \\
& =Rs.2085 \\
\end{align}$