Question

A room is 6m long, 5m broad and 4m high. It has two doors, each measuring 1m by 1m. Find the cost of distempering the walls at ₹25/m2

Answer 1

Answer:

Cost of distempring @ of 25m^2 = Rs 2150

Explanation:

>> We are given with dimensions of the room, which are, 6m long, 5m broad, and 4m high.

>> 2 Doors are there of 1m by 1m.

>> Target is to find the cost of distempering walls at Rs 25/m^2

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We have to basically find out the area of 4 walls excluding the area of windows as no paint is to be applied there.

Our room is a cuboid, so we know that the lateral surface area of a cuboid is = $\sf 2h(l+w)$

So, let's put the known values into this formula and equate them.

$\bf \implies Lateral \ surface \ area = 2h(l+w)$

$\implies \sf Surface \ area\ of \ walls \ of \ the room = 2(4)[6m+5m]$

$\implies \sf Surface \ area \ of \ walls \ of \ the \ room = 8m[11m]$

$\implies \sf Surface \ area \ of \ walls \ of \ the \ room = 88m^2$

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Now, let's find the area of a window measuring 1m by 1m. Don't forget that there are 2 such windows!

So, the area of those windows, = 2 × [1m ×1m]

Area of windows = 2m²

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So the area we need to apply distemper is = 88m² - 2m² = 86m²

Now, what's left? We have to find the cost of destempering them.

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Given that distempering is done @ of Rs 25/m^2 so,

let's apply direct proportion:

Cost of distempring @ of 25m^2 = 25 × 86

=> Cost of distempering @ of 25m^2 = Rs 2150

Answer 2

Answer:

₹2050

Step-by-step explanation:

Length of room- 6m= L

Breadth of room- 5m= B

Height of room- 4m= H

Dimensions of the door= 3m*1m

Therefore area of the door= 3m^2

Dimensions of the windows= 1.5m*1m

Area of the windows= 2(1.5*1) [as there are two windows]

=3m^2

Area of 4 walls= 2(l+b)*h

Area of 4 walls= Lateral surface area (L.S.A)= 2(6+5)*4

L.S.A= 2*11*4

L.S.A= 88m^2

Area of 4 walls to be plastered= (L.S.A)-(Area of door+Area of two windows)

=88-(3+3)

=82m^2

Rate of plastering= Rs. 25 per square metre

Rate of plastering walls= 82×25

= Rs. 2050