Rahul purchased a Maruti van for Rs. 1, 96,000 and the rate of depreciation is 14(²/₇) % per annum. Find the value of the van after two years.

1Rs. 1,40,000
2Rs.1,44,000
3Rs. 1,50,000
4Rs. 1,60,000

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Value of maruti Van, \\ V_{0} = R s . 196000 \\ Rate of depreciation, \\ r = 14 (\frac{2}{7}) % = \frac{100}{7} %; \\ Time, t = 2 years \\ Let V_{1} is the value after depreciation . \\ V_{1} = V_{0} \times {[1 - (\frac{r}{100})]}^{t} \\ V_{1} = 196000 \times {[1 - (\frac{(\frac{100}{7})}{100})]}^{2} \\ V_{1} = 196000 \times {(\frac{6}{7})}^{2} \\ V_{1} = \frac{(196000 \times 36)}{49} \\ V_{1} = R s . 144000 \end{aligned}

$\eqalign{ & {\text{Value}}\,{\text{of}}\,{\text{maruti}}\,{\text{Van}},\, \cr & {V_0} = Rs.\,196000 \cr & {\text{Rate}}\,{\text{of}}\,{\text{depreciation}},\, \cr & r = 14\left( {\frac{2}{7}} \right)\% = \frac{{100}}{7}\% ; \cr & {\text{Time}},\,t = 2\,{\text{years}} \cr & {\text{Let}}\,{V_1}\,{\text{is}}\,{\text{the}}\,{\text{value}}\,{\text{after}}\,{\text{depreciation}}. \cr & {V_1} = {V_0} \times {\left[ {1 - \left( {\frac{r}{{100}}} \right)} \right]^t} \cr & {V_1} = 196000 \times {\left[ {1 - \left( {\frac{{\left( {\frac{{100}}{7}} \right)}}{{100}}} \right)} \right]^2} \cr & {V_1} = 196000 \times {\left( {\frac{6}{7}} \right)^2} \cr & {V_1} = \frac{{\left( {196000 \times 36} \right)}}{{49}} \cr & {V_1} = Rs.\,144000 \cr}$