The central hall of a school is 80 m long and 8 m high.

Solution:

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 $48 \mathrm{~m}$.