Given that, Length (l) = 15 m,
breadth (b) = 10 m,
height (h) = 7 m
Area of the hall to be painted = Area of the wall + Area of the ceiling
= 2h (l + b) + lb
= [2(7) (15 + 10) + 15 ×10] m2
= [14(25) + 150] m2
= 500 m2
It is given that 100 m2 area can be painted from each can.
Number of cans required to paint an area of 500 m2
\(= \frac{500}{100} = 5\)
Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.