Question

Arun is painting the walls and ceiling of a cuboidal hall with length,breadth and height of
15m, 10m and 7m respectively. From each can of paint 100m² of area is painted, how
many cans of paint will he need to paint the room?​

Answer 1

Answer:

5 cans are required for painting

Answer 2

GIVEN :-

• Length → 15m
• Breadth → 10m
• Height → 7m
• From each can , Arun can paint 100m².

TO FIND :-

• Number of cans required to paint the walls and ceiling.

TO KNOW :-

★ Total surface area of Cuboid = 2(lh+bh+lb)

★ Area of base of Cuboid = l × b

SOLUTION :-

We will find area of the room which is to be painted. The floor of the room is not painted . So, we will subtract the Base area from Total surface area.

Area(painted) = 2(lh+bh+lb) - lb

We have ,

• l = 15m
• b = 10m
• h = 7m

Putting values,

→ Area = 2[(15)(7) + (10)(7) + (15)(10)] - (15)(10)

→ Area = 2[105 + 70 + 150] - 150

→ Area = 2[325] - 150

→ Area = 650 - 150

→ Area = 500m²

Hence , area of 500m² is to be painted by Arun.

_________________

Now , we know that with 1 Can , he can paint 100m². So , we have to find for 500m².

♦ 100m² → 1can

♦ 500m² → (1/100) × 500 = 5cans

Hence , 5cans are required to paint the 4 walls and the ceiling.