Answer
Verified
395.7k+ views
Hint: The room is in the form of a cuboid and we have to cover the walls of the room. The room has 4 walls so find the surface area of 4 walls which is equal to $2h\left( l+b \right)$. In this formula, l, b and h are the length, breadth and height of the room. After that we have given the breadth of the paper but we require length of the paper which we will calculate by dividing area by breadth then we have given the cost for 1 m so multiply the length in metres by 0.80 Rs to get the cost of the paper.
Complete step by step answer:
We have given a room with length, breadth and height as 7.5 m, 5.5 m and 5 m respectively. The room is given in the form of a cuboid.
In the below figure, we have shown the room with length, breadth and height as 7.5 m, 5.5 m and 5 m respectively.
Now, we have to cover the walls of this room with paper. The walls that need to be covered are (ACDB, EGHF, AEFB and CGHD). We are going to find the surface area of these four walls.
Let us assume that length, breadth and height of the room is l, b and h respectively then surface area of the four walls that we have mentioned above is:
$2h\left( l+b \right)$
Substituting the value of l, b and h as 7.5 m, 5.5 m and 5 m respectively in the above formula we get,
$\begin{align}
& 2\left( 5 \right)\left( 7.5+5.5 \right) \\
& =10\left( 13.0 \right) \\
& =130{{m}^{2}} \\
\end{align}$
From the above calculations, we have found the surface area of four walls as $130{{m}^{2}}$.
Now, we have given the paper of 40 cm broad or (breadth) and we require the length of the paper. The paper is in the form of rectangle so we know that,
Area of rectangle $=\left( length \right)\left( breadth \right)$…………. Eq. (1)
The total area of four walls we know so equating that area with the multiplication of length and breadth we get,
$130=\left( length \right)\left( 40cm \right)$………. Eq. (2)
Now, area is given in ${{m}^{2}}$ and breadth of the paper is given in cm so we have to convert breadth of the paper into m by dividing 40 by 100 we get,
$\begin{align}
& \dfrac{40}{100}m \\
& =0.40m \\
\end{align}$
Substituting this value in eq. (1) in place of 40 cm we get,
$130=\left( length \right)\left( 0.40m \right)$
Dividing 0.40 on both the sides of the above equation we get,
$\begin{align}
& \dfrac{130}{0.4}=length \\
& \Rightarrow 325m=length \\
\end{align}$
The cost of 1 m paper is given as 80 paise.
Converting 80 paise into rupees by dividing 80 by 100 we get,
$\begin{align}
& \dfrac{80}{100}Rs \\
& 0.80Rs \\
\end{align}$
Now, the cost of 1 m paper is Rs 0.80 so the cost of 325 m is calculated by multiplying 325 by 0.80.
Multiplying 325 by 0.80 we get,
$\begin{align}
& 325\left( 0.8 \right) \\
& =Rs 260 \\
\end{align}$
Hence, the cost of covering the four walls by paper is Rs 260.
Note: The blunder that could happen in this problem is in writing the curved surface. Generally, in the rush of completing the exam students tend to write the total surface area of the cuboid which we are shown below:
$2\left( lb+bh+hl \right)$
But the question is saying, we just have to cover the walls of the room and exclude the ceiling and floor of the room so be careful while writing the surface areas of the walls of the room which is $2h\left( l+b \right)$.
Complete step by step answer:
We have given a room with length, breadth and height as 7.5 m, 5.5 m and 5 m respectively. The room is given in the form of a cuboid.
In the below figure, we have shown the room with length, breadth and height as 7.5 m, 5.5 m and 5 m respectively.
Now, we have to cover the walls of this room with paper. The walls that need to be covered are (ACDB, EGHF, AEFB and CGHD). We are going to find the surface area of these four walls.
Let us assume that length, breadth and height of the room is l, b and h respectively then surface area of the four walls that we have mentioned above is:
$2h\left( l+b \right)$
Substituting the value of l, b and h as 7.5 m, 5.5 m and 5 m respectively in the above formula we get,
$\begin{align}
& 2\left( 5 \right)\left( 7.5+5.5 \right) \\
& =10\left( 13.0 \right) \\
& =130{{m}^{2}} \\
\end{align}$
From the above calculations, we have found the surface area of four walls as $130{{m}^{2}}$.
Now, we have given the paper of 40 cm broad or (breadth) and we require the length of the paper. The paper is in the form of rectangle so we know that,
Area of rectangle $=\left( length \right)\left( breadth \right)$…………. Eq. (1)
The total area of four walls we know so equating that area with the multiplication of length and breadth we get,
$130=\left( length \right)\left( 40cm \right)$………. Eq. (2)
Now, area is given in ${{m}^{2}}$ and breadth of the paper is given in cm so we have to convert breadth of the paper into m by dividing 40 by 100 we get,
$\begin{align}
& \dfrac{40}{100}m \\
& =0.40m \\
\end{align}$
Substituting this value in eq. (1) in place of 40 cm we get,
$130=\left( length \right)\left( 0.40m \right)$
Dividing 0.40 on both the sides of the above equation we get,
$\begin{align}
& \dfrac{130}{0.4}=length \\
& \Rightarrow 325m=length \\
\end{align}$
The cost of 1 m paper is given as 80 paise.
Converting 80 paise into rupees by dividing 80 by 100 we get,
$\begin{align}
& \dfrac{80}{100}Rs \\
& 0.80Rs \\
\end{align}$
Now, the cost of 1 m paper is Rs 0.80 so the cost of 325 m is calculated by multiplying 325 by 0.80.
Multiplying 325 by 0.80 we get,
$\begin{align}
& 325\left( 0.8 \right) \\
& =Rs 260 \\
\end{align}$
Hence, the cost of covering the four walls by paper is Rs 260.
Note: The blunder that could happen in this problem is in writing the curved surface. Generally, in the rush of completing the exam students tend to write the total surface area of the cuboid which we are shown below:
$2\left( lb+bh+hl \right)$
But the question is saying, we just have to cover the walls of the room and exclude the ceiling and floor of the room so be careful while writing the surface areas of the walls of the room which is $2h\left( l+b \right)$.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE