Answer
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Hint: We are given the total cost of papering four walls and the rate of papering one square metre using which we get ${\text{Total cost = Area needed to be papered} \times 3}$. Since the shape of the room is cuboid we need to find the lateral surface area which is given by $LSA = 2h\left( {l + b} \right)sq.units$ where h is the height, l is the length and b is the breadth of the cuboid. Since we are given that the length and breadth are in a ratio 4 : 1 we get that l = 4 b and using this in the main equation we get the value of l and b.
Complete step by step solution:
We are given that the total cost of papering the four walls is Rs.3160
And the rate of papering one square metre is Rs.3
Hence the total cost is given by
$ \Rightarrow {\text{Total cost = Area needed to be papered} \times 3}$
We know that the shape of the room is cuboid
Since only the four walls are needed papering its enough if we find the lateral surface area
Hence the lateral surface area of a cuboid is given by
$ \Rightarrow LSA = 2h\left( {l + b} \right)sq.units$ where h is the height , l is the length and b is the breadth of the cuboid
We are given that the height of the room is 5 m
$
\Rightarrow LSA = 2\times 5\left( {l + b} \right)sq.units \\
\Rightarrow LSA = 10\left( {l + b} \right)sq.units \\
$
Now we are given that the length and breadth of the room are in a ratio 4 : 1
This can be written as
$ \Rightarrow \dfrac{l}{b} = \dfrac{4}{1}$
Cross multiplying we get
$ \Rightarrow l = 4b$
Using this the LSA formula we get
$
\Rightarrow LSA = 10\left( {4b + b} \right)sq.units \\
\Rightarrow LSA = 10\left( {5b} \right)sq.units \\
\Rightarrow LSA = 50b{m^2} \\
$
Let's use this in equation (1)
$
\Rightarrow 3150 = 50b\times 3 \\
\Rightarrow 3150 = 150b \\
\Rightarrow \dfrac{{3150}}{{150}} = b \\
\Rightarrow b = 21m \\
$
Since l = 4b
We get
$ \Rightarrow l = 4\times 21 = 84m$
Therefore the length and breadth of the room is 84 m and 21 m
Therefore none of the options are correct.
Note :
A cuboid is a 3D shape. Cuboids have six faces, which form a convex polyhedron. Broadly, the faces of the cuboid can be any quadrilateral. More narrowly, rectangular cuboids are made from 6 rectangles, which are placed at right angles.
Complete step by step solution:
We are given that the total cost of papering the four walls is Rs.3160
And the rate of papering one square metre is Rs.3
Hence the total cost is given by
$ \Rightarrow {\text{Total cost = Area needed to be papered} \times 3}$
We know that the shape of the room is cuboid
Since only the four walls are needed papering its enough if we find the lateral surface area
Hence the lateral surface area of a cuboid is given by
$ \Rightarrow LSA = 2h\left( {l + b} \right)sq.units$ where h is the height , l is the length and b is the breadth of the cuboid
We are given that the height of the room is 5 m
$
\Rightarrow LSA = 2\times 5\left( {l + b} \right)sq.units \\
\Rightarrow LSA = 10\left( {l + b} \right)sq.units \\
$
Now we are given that the length and breadth of the room are in a ratio 4 : 1
This can be written as
$ \Rightarrow \dfrac{l}{b} = \dfrac{4}{1}$
Cross multiplying we get
$ \Rightarrow l = 4b$
Using this the LSA formula we get
$
\Rightarrow LSA = 10\left( {4b + b} \right)sq.units \\
\Rightarrow LSA = 10\left( {5b} \right)sq.units \\
\Rightarrow LSA = 50b{m^2} \\
$
Let's use this in equation (1)
$
\Rightarrow 3150 = 50b\times 3 \\
\Rightarrow 3150 = 150b \\
\Rightarrow \dfrac{{3150}}{{150}} = b \\
\Rightarrow b = 21m \\
$
Since l = 4b
We get
$ \Rightarrow l = 4\times 21 = 84m$
Therefore the length and breadth of the room is 84 m and 21 m
Therefore none of the options are correct.
Note :
A cuboid is a 3D shape. Cuboids have six faces, which form a convex polyhedron. Broadly, the faces of the cuboid can be any quadrilateral. More narrowly, rectangular cuboids are made from 6 rectangles, which are placed at right angles.
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