Question

A cellar has dimensions $ 4 $ m by $ 3 $ m by $ 2.4 $ m high. Find the cost of painting the walls and ceiling of the cellar if $ 1 $ liter of paint costs \[Rs. 24.90\] and each liter covers $ 15 $ square meters.

Answer 1

Hint: From the given dimensions we can evaluate that the cellar is cuboidal in shape. Now, we have to find the total surface area of this cuboid and remove the area of the floor. This gives the total area that needs to be painted.

Complete step-by-step answer:
Given to us are dimensions of a cellar as $ 4 $ m by $ 3 $ m by $ 2.4 $ m high. This cellar is a cuboid so in order to find the area that needs to be painted, let us first calculate its total surface area.
The formula for total surface of a cuboid is given as $ T.S.A = 2\left( {lb + bh + lh} \right) $
Length of the given cellar is $ 4 $ meters.
Breadth of the cellar is $ 3 $ meters.
Height of the cellar is $ 2.4 $ meters.
By substituting these values in the formula, we get
 $ T.S.A = 2\times[4 \times 3 + 3 \times 2.4 + 4 \times 2.4] = 57.6 $ square meters. But it is given that only the ceiling and the walls are being painted so the area of the floor must be removed from this area.
The area of the floor would be an area of a rectangle with dimensions $ 4 $ m by $ 3 $ m. Area of a rectangle is $ l \times b = 4 \times 3 = 12 $ square meters.
So, the total area that needs to be painted is $ 57.6 - 12 = 45.6 $ square meters.
Given that one liter of paint covers $ 15 $ square meters. The amount of paint required for $ 45.6 $ square meters would be $ \dfrac{{45.6}}{{15}} = 3.04 $ liters.
One liter of paint costs \[Rs. 24.90\] so $ 3.04 $ liters of paint would cost $ 24.90 \times 3.04 = 27.89 $
Hence, the cost of paint is \[Rs. 75.69\]
So, the correct answer is “\[Rs. 75.69\]”.

Note: Important formulas to remember:
The total surface area of a cuboid is $ 2\left( {lb + bh + lh} \right) $ where l, b, h are length, breadth and height respectively.
The area of a rectangle is $ l \times b $ where l and b are length and breadth.
Since the floor has to be excluded and the floor is of rectangular shape as the base of cuboid is of rectangular shape and hence we remove the area of the floor.