Question

Preeti is painting the walls and ceiling of a hall whose dimensions are $ 18m \times 15m \times 5m $ . From each can of paint $ 120{m^2} $ of area is painted. How many cans of paint does she need to paint the hall?

Answer 1

Hint: The cuboid can be defined as the box-like three dimensional shaped which can be either solid or hollow and has six rectangular faces. The faces of cuboids are opposite to each other. Here, we will use the formula to find the surface area as the surface area of the cuboid $ = 2(lb + bh + hl) $ and then find the number of cans required to paint the area of the cuboid.

Complete step-by-step answer:
Let the given data be,
Length, $ l = 18m $
Height, $ h = 15m $
Breadth, $ b = 5m $
Area of the walls and ceiling subtracting the area of the floor $ = 2lh + 2bh + lb $
Place the given values in the above equation –
Area $ = 2(18)(5) + 2(15)(5) + (18)(15) $
Simplify the above expression finding the product of the terms.
Area $ = 180 + 150 + 270 $
Simplify the above expression finding the sum of the terms
Area $ = 600m{}^2 $
Given that one can paint $ 120{m^2} $ ,
Number of cans required to paint $ 600{m^2} = ? $
Number of cans $ = \dfrac{{600}}{{120}} $
Common factors from the numerator and the denominator cancels each other.
Number of cans $ = 5 $
Hence, Preeti needs $ 5 $ cans of paint to paint the hall.
So, the correct answer is “5”.

Note: Before solving the numerical, always check the system of units applied. System of units for the quantities should be the same. Always remember the conversational relation between the system of units and apply it accordingly. Be good in multiples and division and always remember that the common factors from the numerator and the denominator cancels each other.