Question

(xi) When a bucket is filled $\frac{3}{8}$ of its capacity its weight is 10 kg. If it is filled $\frac{3}{5}$ of its capacity its weight is 14.5 kg. How much will its weight be when it is filled to the full capacity?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{weight}\mathrm{of}\text{the}\mathrm{vacant}\mathrm{bucket}=x\mathrm{kg} \\ \mathrm{Weight}\mathrm{of}\mathrm{material}\mathrm{which}\mathrm{is}\mathrm{filled}\mathrm{to}\mathrm{the}\mathrm{full}\mathrm{capacity}\mathrm{of}\text{the}\mathrm{bucket}=y\mathrm{kg} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ x+\frac{3}{8}y=10 \\ \Rightarrow x=10-\frac{3}{8}y...\left(1\right) \\ x+\frac{3}{5}y=14.5...\left(2\right) \\ \mathrm{Substituting}\mathrm{eq}\left(1\right)\mathrm{in}\mathrm{eq}\left(2\right),\mathrm{we}\mathrm{get}: \\ 10-\frac{3}{8}y+\frac{3}{5}y=14.5 \\ \Rightarrow \frac{3}{5}y-\frac{3}{8}y=4.5 \\ \Rightarrow \frac{24y-15y}{40}=4.5 \\ \Rightarrow \frac{9y}{40}=4.5 \\ \Rightarrow y=\frac{4.5\times 40}{9} \\ \Rightarrow y=20\mathrm{kg} \\ \mathrm{Substituting}y=20\mathrm{in}\mathrm{eq}\left(1\right), \\ x=10-\frac{3}{8}\times 20 \\ \Rightarrow x=2.5\mathrm{kg} \\ \mathrm{Therefore},\text{the}\mathrm{weight}\mathrm{of}\mathrm{bucket}\mathrm{when}\mathrm{it}\mathrm{is}\mathrm{filled}\mathrm{to}\text{its}\mathrm{full}\mathrm{capacity}=20\mathrm{kg}+2.5\mathrm{kg}=22.5\mathrm{kg}. \end{gathered}$$