A rectangular water tank is open at the top. Its capacity is $24{{m}^{3}}$. Its length and breadth are $4m$ and $3m$ respectively. Ignoring the thickness of the material used for the building the tank, the total cost of painting the inner and outer surfaces of the tank at Rs. $10$ per ${{m}^{2}}$ ,is
A. Rs. $400$
B. Rs. $500$
C. Rs. $600$
D. Rs. $800$

Hint: In the problem we have the length, breadth and volume of the cuboid. From the relation between length, breadth, height and volume of the cuboid we will calculate the height of the cuboid. They have mentioned that they are painting the tank which is opened at top, so we need to calculate the surface area of the cuboid which is opened at top. It will give the only outer surface are, but they have mentioned they are painting the inner part also and the thickness is ignored, so we need to multiply the surface area of the tank which is opened at top with $2$. Now we have the total surface area to be painted, from this we can calculate the cost of painting from the given data.

Complete step by step answer:
Given that,
A rectangular water tank is open at the top. Its capacity is $24{{m}^{3}}$. Its length and breadth are $4m$ and $3m$ respectively. Let the water tank is shown below

We know that the volume of the cuboid is the product of the height, width and length, then
$\begin{align}
  & 4m\times 3m\times h=24{{m}^{3}} \\
 & \Rightarrow h=2m \\
\end{align}$
Now the top of the tank is opened, then the total surface area can be calculated as
$\begin{align}
  & S=2lh+2hb+lb \\
 & \Rightarrow S=2\left( 4 \right)\left( 2 \right)+2\left( 2 \right)\left( 3 \right)+\left( 4 \right)\left( 3 \right) \\
 & \Rightarrow S=16+12+12 \\
 & \Rightarrow S=40 \\
\end{align}$
The above value is only the outer surface area, but they have mentioned that they are painting the inner surface also, so we need to calculate the inner surface area also. For this they have mentioned that the thickness of the wall is ignored, so the inner surface area is equal to the outer surface area. Hence the total surface area to be painted is given by
$\begin{align}
  & A=2S \\
 & \Rightarrow A=2\times 40 \\
 & \Rightarrow A=80{{m}^{2}} \\
\end{align}$
Given that the cost for painting $1{{m}^{2}}$ is Rs. $10$, now the total cost for painting $80{{m}^{2}}$ is Rs.$80\times 10=800$.

So, the correct answer is “Option D”.

Note: In this problem they have mentioned that the tank is open at top, so we neglected that surface while calculating the surface area. If they have mentioned that they are not going to paint the bottom surface, then we need to neglect the bottom surface from the surface area of the tank.