Answer
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Hint: Firstly we find the total area of the walls by using the given conditions in the problem. Then similarly find the total area of the floor using the other conditions. Then using the length of the room the width and breadth can be found easily.
Complete step by step solution: Let, the height of room be, h meters
And the width of the room be, w meters and the length be l meters.
Now, we have, our,
⇒ Area of walls \[ = 2\left( {lh + wh} \right)\;\]\[{m^2}\]
\[ = 2h\left( {l + w} \right)\]\[{m^2}\]
So, now, the cost of papering of the walls at the rate of Rs 0.95 per \[{m^2}\] \[ = Rs.2h\left( {l + w} \right)\left( {0.95} \right)\]
According to the question, \[2h\left( {l + w} \right)\left( {0.95} \right) = 212.80\]
\[ \Rightarrow h = \dfrac{{212.80}}{{2(l + w) \times 0.95}} = \dfrac{{112}}{{2(l + w)}}\]
Now, again, Area of floor = \[\;lw\;\]
So, the total cost, at the rate of 0.50 Rs per \[{m^2}\] is, \[0.50\;lw = 97.50\;\]
\[
\Rightarrow lw = 195 \\
\Rightarrow w = 195/15 = 13m \\
\] as the value of \[l = 15m\]
So, now, \[h = \dfrac{{112}}{{2(l + w)}}\]
\[h = \dfrac{{112}}{{2(15 + 13)}}\]\[ \Rightarrow h = \dfrac{{112}}{{2 \times 28}}\]\[ = 4m\].
Hence the height of the wall is 4m.
Hence, the correct option is (b).
Note: The area of walls in the room are\[ = 2\left( {lh + wh} \right)\;\]\[{m^2}\]. And the area of the floor is\[\;lw\;\]\[{m^2}\]. Though there might be a case where the walls are not only 4 but more than that. We need to think differently in that case.
Complete step by step solution: Let, the height of room be, h meters
And the width of the room be, w meters and the length be l meters.
Now, we have, our,
⇒ Area of walls \[ = 2\left( {lh + wh} \right)\;\]\[{m^2}\]
\[ = 2h\left( {l + w} \right)\]\[{m^2}\]
So, now, the cost of papering of the walls at the rate of Rs 0.95 per \[{m^2}\] \[ = Rs.2h\left( {l + w} \right)\left( {0.95} \right)\]
According to the question, \[2h\left( {l + w} \right)\left( {0.95} \right) = 212.80\]
\[ \Rightarrow h = \dfrac{{212.80}}{{2(l + w) \times 0.95}} = \dfrac{{112}}{{2(l + w)}}\]
Now, again, Area of floor = \[\;lw\;\]
So, the total cost, at the rate of 0.50 Rs per \[{m^2}\] is, \[0.50\;lw = 97.50\;\]
\[
\Rightarrow lw = 195 \\
\Rightarrow w = 195/15 = 13m \\
\] as the value of \[l = 15m\]
So, now, \[h = \dfrac{{112}}{{2(l + w)}}\]
\[h = \dfrac{{112}}{{2(15 + 13)}}\]\[ \Rightarrow h = \dfrac{{112}}{{2 \times 28}}\]\[ = 4m\].
Hence the height of the wall is 4m.
Hence, the correct option is (b).
Note: The area of walls in the room are\[ = 2\left( {lh + wh} \right)\;\]\[{m^2}\]. And the area of the floor is\[\;lw\;\]\[{m^2}\]. Though there might be a case where the walls are not only 4 but more than that. We need to think differently in that case.
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