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.
Suppose that the breadth of the hall is $b \mathrm{~m}$.
Lenght of the hall $=80 \mathrm{~m}$
Height of the hall $=8 \mathrm{~m}$
Total surface area of 4 walls including doors and windows $=2 \times($ length $\times$ height $+$ breadth $\times$ height $)$
$=2 \times(80 \times 8+\mathrm{b} \times 8)$
$=2 \times(640+8 \mathrm{~b})$
$=1280+16 \mathrm{~b} \mathrm{~m}^{2}$
The walls have 10 doors each of dimensions $3 \mathrm{~m} \times 1.5 \mathrm{~m} .$
i. e., area of a door $=3 \times 1.5=4.5 \mathrm{~m}^{2}$
$\therefore$ Area of 10 doors $=10 \times 4.5=45 \mathrm{~m}^{2}$
Also, there are 10 windows each of dimensions $1.5 \mathrm{~m} \times 1 \mathrm{~m} .$
i. e., area of one window $=1.5 \times 1=1.5 \mathrm{~m}^{2}$
$\therefore$ Area of 10 windows $=10 \times 1.5=15 \mathrm{~m}^{2}$
Thus, total area to be whitwashed $=($ total area of 4 walls $)-($ areas of 10 doors $+$ areas of 10 windows $)$
$=(1280+16 b)-(45+15)$
$=1280+16 b-60$
$=1220+16 b \mathrm{~m}^{2}$
It is given that the cost of whitewashing $1 \mathrm{~m}^{2}$ of area $=\mathrm{Rs} 1.20$
$\therefore$ Total cost of whitewashing the walls $=(1220+16 b) \times 1.20$
$=1220 \times 1.20+16 b \times 1.20$
$=1464+19.2 b$
Since the total cost of whitewashing the walls is Rs $2385.60$, we have:
$1464+19.2 b=2385.60$
$\Rightarrow 19.2 b=2385.60-1464$
$\Rightarrow 19.2 b=921.60$
$\Rightarrow b=\frac{921.60}{19.2}=48 \mathrm{~m}$
$\therefore$ The breadth of the central hall is $48 \mathrm{~m}$.