Question

The length, breadth, and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will
(a) remain the same
(b) decrease by 30%
(c) decrease by 15%
(d) decrease by 18.75%

Answer 1

Hint: We are given that the length, breadth, and height of the room are in the ratio 3:2:1. So, consider them as 3x, 2x and x respectively. Then find the area of the four walls of the room by using 2h(l +b). Then by changing l, b, and h according to the given information, again find the new area by using the same formula. Then, find the % change in the area by using \[\left[ \dfrac{\left( \text{New Area} \right)-\left( \text{Old Area} \right)}{\left( \text{Old Area} \right)}\times 100 \right]\].

Complete step-by-step answer:
We are given the length, breadth, and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then we have to find the change in the total area of the four walls of the room. First of all, we know that the rooms are in the shape of a cuboid. We also know that the total area of the four faces that is the lateral surface area of the cuboid = 2h(l + b)…..(i)
where h = height of the cuboid, l = length of the cuboid, and b = breadth of the cuboid.
So, also we get the area of the four walls of the room = 2h(l + b)….(ii)
We are given that the length, breadth, and height of a room are in the ratio 3:2:1. So, let us consider the length, breadth, and height of the room as 3x, 2x and x respectively. Substituting the value of l = 3x, b = 2x and h = x in equation (ii), we get,
The total area of the four walls of the room,
\[\begin{align}
  & {{A}_{1}}=2x\left( 3x+2x \right) \\
 & {{A}_{1}}=2x\left( 5x \right) \\
 & {{A}_{1}}=10{{x}^{2}}....\left( iii \right) \\
\end{align}\]
Now, we are given that the breadth and height of the room is halved while the length is doubled, so we get,
The new length of the room (\[{{l}_{1}}\]) = 2.(3x) = 6x
The new breadth of the room (\[{{b}_{1}}\]) = \[\dfrac{2x}{2}=x\]
The new height of the room (\[{{h}_{1}}\]) = \[\dfrac{x}{2}\]
So, we get the new total area of the four walls of the room,
\[{{A}_{2}}=2{{h}_{1}}\left( {{l}_{1}}+{{b}_{1}} \right)\]
By substituting the value of \[{{h}_{1}},{{l}_{1}}\] and \[{{b}_{1}}\], we get,
\[\begin{align}
  & {{A}_{2}}=2\left( \dfrac{x}{2} \right)\left[ 6x+x \right] \\
 & {{A}_{2}}=x.\left( 7x \right) \\
 & {{A}_{2}}=7{{x}^{2}}.....\left( iv \right) \\
\end{align}\]
Now, we get the change in the area of the four walls of the room (C) = New area – Old area
\[C={{A}_{2}}-{{A}_{1}}\]
By substituting the values of \[{{A}_{1}}\] and \[{{A}_{2}}\] from equation (iii) and (iv), we get, change in the area of four walls of the room,
\[\begin{align}
  & C=7{{x}^{2}}-10{{x}^{2}} \\
 & C=-3{{x}^{2}}.....\left( v \right) \\
\end{align}\]
Now, the percentage change in the area of the four walls of the room,
\[P=\left( \dfrac{\text{Change in area}}{\text{Old area}}\times 100 \right)\]%
\[P=\left( \dfrac{C}{{{A}_{1}}}\times 100 \right)\]%
By substituting the values of C and \[{{A}_{1}}\] from equation (v) and (iii) respectively, we get, the percentage change in the area of the four walls of the room,
\[P=\left( \dfrac{-3{{x}^{2}}}{10{{x}^{2}}}\times 100 \right)\]%
By canceling the like terms, we get,
\[P=\left[ \dfrac{-3}{10}\times 100 \right]\]%
P = – 30 %
So, – 30 % signifies that there is a decrease in the area of the four walls of the room by 30 % with respect to the original area.
Hence, option (b) is the right answer.

Note: Students must note that whenever there is a negative sign with the change of any entity that means that the value of the new entity is less than the value of the old entity or there is a decrease of value in comparison to the old entity. Also, students are advised to first read properly which parameter is decreased or increased by which factor and then only solve because students often interchange the values for different factors for parameters and get the wrong results. So, this must be taken care of.