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

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Total surface area of the room=2(lh+lb+hb)m²

=2[9×6.5+9×8+8×6.5]m²

=365m²

Area of the door =l×b=2×1.5=3m²

Area of the windows=l×b=1.5×1=1.5m²

Surface area of walls=TSA of room −Area of door−Area of windows

=365−3−1.5=360.5m²

Cost of whitewashing per square meter=Rs. 6.40

∴ Cost of whitewashing the walls=Rs. 6.40×360.5

=Rs. 2307.2.