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.

Answer 1

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