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Question
The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3 m × 1.5 m and 10 windows each of size 1.5 m × 1 m. If the cost of white-washing the walls of the hall at the rate of Rs 1.20 per m2 is Rs 2385.60, fidn the breadth of the hall.
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Solution
\[\text { Suppose that the breadth of the hall is b m . } \]
\[\text { Lenght of the hall = 80 m }\]
\[\text { Height of the hall = 8 m }\]
\[\text { Total surface area of 4 walls including doors and windows }= 2 \times \text { (length} \times\text { height + breadth } \times \text { height) }\]
\[ = 2 \times (80 \times 8 + b \times 8)\]
\[ = 2 \times (640 + 8b)\]
\[ = 1280 + 16b m^2 \]
\[\text { The walls have 10 doors each of dimensions 3 m } \times 1 . 5 m . \]
\[\text { i . e . , area of a door = 3 } \times 1 . 5 = 4 . 5 m^2 \]
\[ \therefore \text { Area of 10 doors = 10 } \times 4 . 5 = 45 m^2 \]
\[\text { Also, there are 10 windows each of dimensions 1 . 5 m } \times 1 m . \]
\[\text { i . e . , area of one window = 1 . 5 }\times 1 = 1 . 5 m^2 \]
\[ \therefore \text { Area of 10 windows = 10 } \times 1 . 5 = 15 m^2 \]
\[\text { Thus, total area to be whitwashed = (total area of 4 walls) - (areas of 10 doors + areas of 10 windows) }\]
\[ = (1280 + 16b) - (45 + 15)\]
\[ = 1280 + 16b - 60\]
\[ = 1220 + 16b m^2 \]
\[\text { It is given that the cost of whitewashing 1 } m^2 of area = Rs 1 . 20\]
\[ \therefore\text { Total cost of whitewashing the walls }= (1220 + 16b) \times 1 . 20\]
\[ = 1220 \times 1 . 20 + 16b \times 1 . 20\]
\[ = 1464 + 19 . 2b\]
\[\text { Since the total cost of whitewashing the walls is Rs 2385 . 60, we have: } \]
\[1464 + 19 . 2b = 2385 . 60\]
\[ \Rightarrow 19 . 2b = 2385 . 60 - 1464\]
\[ \Rightarrow 19 . 2b = 921 . 60\]
\[ \Rightarrow b = \frac{921 . 60}{19 . 2} = 48 m\]
\[ \therefore \text { The breadth of the central hall is 48 m } .\]
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