Question

A room is 16 m long, 9 m wide and 3 m high. It has two doors, each of dimensions (2
m x 2.5 m) and three windows, each of dimensions (1.6 m x 75 cm). Find the cost of
distempering the walls of the room from inside at the rate of Rs. 8 per sq. metre​

Answer 1

Answer:- Rs. 2963.2

Given the dimension of the room as:-

• Length, L = 16 m
• Breadth, B = 9 m
• Height, H = 3 m

The room has two doors whose dimension is given as :-

• length, l = 2 m
• breadth, b = 2.5

Furthermore, It has three windows as well of which the measurements can be given as:-

• Length, l₁ = 1.6 m
• Breadth, b₁ = 75 cm or 0.75 m

We have to find the cost of distempering the wall at the rate of Rs. 8 per sq. m.

The room is in the shape of a cuboid which means its area of the walls can be given as:-

⇒ A = (2×L×H) + (2×L×B)

⇒ A = 2L (H + B)

⇒ A = 2×16 ( 9 + 3 )

⇒ A = 32 × 12

⇒ A = 384 m²

This area also includes the area of the all the doors and windows, which we won't distemper, so let's subtract the areas of unnecessary things:-

Because the doors and the windows are in the shape of a rectangle hence the area of each of them can be given by the formula:

• Area = Length × Breadth

Now, Required area is given by:-

⇒ Area of the walls - (2 × Area of door) - (3 × Area of window)

⇒ 384 - (2 × 2 × 2.5) - (3 × 1.6 × 0.75)

⇒ 384 - 10 - 3.6

⇒ 384 - 13.6

⇒ 370.4 m²

So, the required area to be distempered is 370.4 m² and the cost of distempering this much area is given by:-

⇒ Area × Rate

⇒ 370.4 × 8

⇒ Rs. 2963.2

Note:- The question has asked for only the walls and not the roof.

Answer 2

Given :-

A room is 16 m long, 9 m wide and 3 m high. It has two doors, each of dimensions (2 m x 2.5 m) and three windows, each of dimensions (1.6 m x 75 cm).

To Find :-

The cost of  distempering the walls of the room from inside at the rate of Rs. 8 per sq. meter

​Solution :-

At first area of wall

LSA of cuboid = Area of wall

2[l + b]h = Area of wall

2[16 + 9] 3

2[25] × 3

50 × 3

150 m

Now

We know that

1 m = 100 cm

75 cm = 75/100 = 0.75 m

Area to be distempering =  150 - (2 × 2 × 2.5) - (3 × 1.6 × 0.75)

= 150 - (4 × 2.5) - (4.8 × 0.75)

= 150 - 10 - 3.6

= 150 - 13.6

= 136.4 m

Now

Cost = 136.4 × 8

Cost = 1091.2