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Question

# The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 291600, for how much was it purchased?

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Solution

## $\mathrm{Let}\mathrm{the}\mathrm{initial}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine},P\mathrm{be}\mathrm{Rs}x.\phantom{\rule{0ex}{0ex}}\mathrm{Rate}\mathrm{of}\mathrm{depreciation},\mathit{}R=10%\phantom{\rule{0ex}{0ex}}\mathrm{Time},n=3\mathrm{years}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{present}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}\mathrm{is}\mathrm{Rs}291600.\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{initial}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}\mathrm{Value}=P×{\left(1-\frac{R}{100}\right)}^{n}\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}.\mathit{}x×{\left(1-\frac{10}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}.x×{\left(\frac{100-10}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}.x×{\left(\frac{90}{100}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}.x×{\left(\frac{9}{10}\right)}^{3}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Present}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}=\mathrm{Rs}291600\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{Rs}291600=\mathrm{Rs}\mathit{}x×\left(\frac{9}{10}\right)×\left(\frac{9}{10}\right)×\left(\frac{9}{10}\right)\phantom{\rule{0ex}{0ex}}⇒x\mathit{}=\mathrm{Rs}\frac{291600×10×10×10}{9×9×9}\phantom{\rule{0ex}{0ex}}⇒x=\mathrm{Rs}\frac{291600000}{729}\phantom{\rule{0ex}{0ex}}⇒x\mathit{}=\mathrm{Rs}400000\phantom{\rule{0ex}{0ex}}\therefore \mathrm{The}\mathrm{initial}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{machine}\mathrm{is}\mathrm{Rs}400000.$

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