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The cost of plastering the four walls of a room is \[{\rm{Rs}}.25\]. The cost of plastering a room twice in length breadth and height will be
A. Rs. 50
B. Rs. 75
C. Rs. 100
D. Rs.200
Answer
442.2k+ views
Hint: Here, we will find the area which is plastered in the previous room and the cost of plastering that area is given as \[{\rm{Rs}}.25\]. We will find the area of four walls of the new room and compare it with the area and cost of the previous one. This will give us the required cost of plastering the new room.
Formula Used:
We will use the formula Area of four walls of a room \[ = 2h\left( {l + b} \right)\] square units.
Complete step-by-step answer:
Let the length, breadth and height of the room be \[l,b\] and \[h\] respectively.
Now, formula for area of four walls of a room \[ = 2lh + 2bh = 2h\left( {l + b} \right){{\rm{m}}^2}\]
According to the question, the cost of plastering the four walls of a room is \[{\rm{Rs}}.25\].
Hence, the cost of plastering the area \[2h\left( {l + b} \right){{\rm{m}}^2}\] is \[{\rm{Rs}}.25\]……………………………….\[\left( 1 \right)\]
Now, if the measurements of the room are such that it is twice in length, breadth and height than the previous room, then, the length, breadth and height of this room will be \[2l,2b,2h\]respectively.
Therefore, area of four walls of this room \[ = 2 \times 2h\left( {2l + 2b} \right)\]
\[ \Rightarrow \] Area of four walls of this room \[ = 2 \times 4h\left( {l + b} \right)\]
Multiplying the terms, we get
\[ \Rightarrow \] Area of four walls of this room \[ = 8h\left( {l + b} \right){{\rm{m}}^2}\]
Now, this can also be written as:
\[ \Rightarrow \] Area of four walls of this room \[ = 4 \times 2h\left( {l + b} \right){{\rm{m}}^2}\]
From equation \[\left( 1 \right)\], we know that, the cost of plastering the area \[2h\left( {l + b} \right){\rm{ }}{{\rm{m}}^2}\] is \[{\rm{Rs}}.25\].
Therefore, the cost of plastering the area \[4 \times 2h\left( {l + b} \right){{\rm{m}}^2}\] \[ = 4 \times 25 = {\rm{Rs}}.100\]
Hence, the cost of plastering a room twice in length, breadth and height will be \[{\rm{Rs}}.100\].
Hence, option C is the correct answer.
Note:
Here the shape of the room is not given but we will assume it to be a cuboid because the dimensions of the room are given as length, breadth and height. A cuboid is a three dimensional shape whose sides are rectangular. In real life, rooms are usually in the shape of cuboid but if in case all the 4 walls and the top and bottom of a room have the same length, breadth and height, this would mean that it is cubical in shape as a cube has all the sides equal and square shaped. Some other examples of cuboidal figures which we see in our day to day life are matchboxes, shoeboxes, books, etc. All of these have a length, breadth and height.
Formula Used:
We will use the formula Area of four walls of a room \[ = 2h\left( {l + b} \right)\] square units.
Complete step-by-step answer:
Let the length, breadth and height of the room be \[l,b\] and \[h\] respectively.
Now, formula for area of four walls of a room \[ = 2lh + 2bh = 2h\left( {l + b} \right){{\rm{m}}^2}\]
According to the question, the cost of plastering the four walls of a room is \[{\rm{Rs}}.25\].
Hence, the cost of plastering the area \[2h\left( {l + b} \right){{\rm{m}}^2}\] is \[{\rm{Rs}}.25\]……………………………….\[\left( 1 \right)\]
Now, if the measurements of the room are such that it is twice in length, breadth and height than the previous room, then, the length, breadth and height of this room will be \[2l,2b,2h\]respectively.
Therefore, area of four walls of this room \[ = 2 \times 2h\left( {2l + 2b} \right)\]
\[ \Rightarrow \] Area of four walls of this room \[ = 2 \times 4h\left( {l + b} \right)\]
Multiplying the terms, we get
\[ \Rightarrow \] Area of four walls of this room \[ = 8h\left( {l + b} \right){{\rm{m}}^2}\]
Now, this can also be written as:
\[ \Rightarrow \] Area of four walls of this room \[ = 4 \times 2h\left( {l + b} \right){{\rm{m}}^2}\]
From equation \[\left( 1 \right)\], we know that, the cost of plastering the area \[2h\left( {l + b} \right){\rm{ }}{{\rm{m}}^2}\] is \[{\rm{Rs}}.25\].
Therefore, the cost of plastering the area \[4 \times 2h\left( {l + b} \right){{\rm{m}}^2}\] \[ = 4 \times 25 = {\rm{Rs}}.100\]
Hence, the cost of plastering a room twice in length, breadth and height will be \[{\rm{Rs}}.100\].
Hence, option C is the correct answer.
Note:
Here the shape of the room is not given but we will assume it to be a cuboid because the dimensions of the room are given as length, breadth and height. A cuboid is a three dimensional shape whose sides are rectangular. In real life, rooms are usually in the shape of cuboid but if in case all the 4 walls and the top and bottom of a room have the same length, breadth and height, this would mean that it is cubical in shape as a cube has all the sides equal and square shaped. Some other examples of cuboidal figures which we see in our day to day life are matchboxes, shoeboxes, books, etc. All of these have a length, breadth and height.
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