Question

The cost of a washing machine depreciates by rs 720 during the second year and rs648 during the third year.Calculate
1) the rate of depreciation per annum
2)the original cost of the machine
3)the value of the machine at the end of third year.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{the}\mathrm{washing}\mathrm{machine}=\mathrm{Rs}\mathrm{P} \\ \mathrm{Let}r\%\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{depriciation}\mathrm{on}\mathrm{per}\mathrm{annum} \\ \mathrm{then},\mathrm{cost}\mathrm{of}\mathrm{washing}\mathrm{machine}\mathrm{after}{1}^{\mathrm{st}}\mathrm{year}=\mathrm{Rs}\mathrm{P}\left(1-\frac{r}{100}\right) \\ \mathrm{and}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{washing}\mathrm{machine}\mathrm{at}\mathrm{the}\mathrm{end}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{year}=\mathrm{Rs}\mathrm{P}{\left(1-\frac{r}{100}\right)}^2 \\ \therefore \mathrm{P}\left(1-\frac{r}{100}\right)-P{\left(1-\frac{r}{100}\right)}^2=720 \\ \Rightarrow \mathrm{P}\left(1-\frac{r}{100}\right)\left[1-\left(1-\frac{r}{100}\right)\right]=720 \\ \Rightarrow \mathrm{P}\left(1-\frac{r}{100}\right)\left(1-1+\frac{r}{100}\right)=720 \\ \Rightarrow \mathrm{P}\left(1-\frac{r}{100}\right)\left(\frac{r}{100}\right)=720...........\left(1\right) \\ \mathrm{The}\mathrm{cost}\mathrm{of}\mathrm{washing}\mathrm{machine}\mathrm{at}\mathrm{the}\mathrm{end}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{year}=\mathrm{P}{\left(1-\frac{r}{100}\right)}^3 \\ \therefore \mathrm{P}{\left(1-\frac{r}{100}\right)}^2-P{\left(1-\frac{r}{100}\right)}^3=648 \\ \Rightarrow \mathrm{P}{\left(1-\frac{r}{100}\right)}^2\left[1-\left(1-\frac{r}{100}\right)\right]=648 \\ \Rightarrow \mathrm{P}{\left(1-\frac{r}{100}\right)}^2\left(1-1+\frac{r}{100}\right)=648 \\ \Rightarrow \mathrm{P}{\left(1-\frac{r}{100}\right)}^2\left(\frac{r}{100}\right)=648...........\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Dividing}\left(2\right)\mathrm{by}\left(1\right),\mathrm{we}\mathrm{get}, \\ \frac{\mathrm{P}{\left(1-{\displaystyle \frac{r}{100}}\right)}^2\left({\displaystyle \frac{r}{100}}\right)}{\mathrm{P}\left(1-{\displaystyle \frac{r}{100}}\right)\left({\displaystyle \frac{r}{100}}\right)}=\frac{648}{720} \\ \Rightarrow 1-\frac{r}{100}=\frac{9}{10} \\ \Rightarrow 1-\frac{9}{10}=\frac{r}{100} \\ \Rightarrow \frac{1}{10}=\frac{r}{100} \\ \Rightarrow 10r=100 \\ \Rightarrow r=10 \end{gathered}$$

$$\begin{gathered} \underline{\left(\mathbf{1}\right)\mathbf{.}} \\ \mathrm{Rate}\mathrm{of}\mathrm{depriciation}\mathrm{on}\mathrm{per}\mathrm{annum}=10\% \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathit{}r\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}, \\ \mathrm{P}\left(1-\frac{10}{100}\right)\left(\frac{10}{100}\right)=720 \\ \Rightarrow \mathrm{P}\left(\frac{90}{100}\times \frac{10}{100}\right)=720 \\ \Rightarrow \mathrm{P}\times \frac{9}{100}=720 \\ \Rightarrow \mathrm{P}=720\times \frac{100}{9} \\ \Rightarrow \mathrm{P}=80\times 100=\mathrm{Rs}8000 \\ \underline{\left(\mathbf{2}\right)\mathbf{.}} \\ \mathrm{The}\mathrm{original}\mathrm{cost}\mathrm{of}\mathrm{the}\mathrm{machine}=\mathrm{Rs}8000 \\ \underline{\left(\mathbf{3}\right)\mathbf{.}} \\ \mathrm{The}\mathrm{cost}\mathrm{of}\mathrm{machine}\mathrm{at}\mathrm{the}\mathrm{end}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{year}=\mathrm{P}{\left(1-\frac{r}{100}\right)}^3 \\ =8000{\left(1-\frac{10}{100}\right)}^3 \\ =8000{\left(\frac{90}{100}\right)}^3=8000\times \frac{9}{10}\times \frac{9}{10}\times \frac{9}{10} \\ =8\times 729=\mathrm{Rs}5832 \end{gathered}$$