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# A piece of equipment costs$Rs6,00,000$ in a certain factory. If it depreciates in value$15%$ the first year, $13.5%$ the next year, $12%$the third year, and so on. What will be its value at the end of$10$ years, all percentages applying to the original costs $?$

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## Finding the equipment cost:The cost of equipment $=Rs6,00,000$Step 1 - Finding the type of series The value depreciates as$15%,13.5%,12%,--------$Percentage of depreciation in the first year $=15%$Percentage of depreciation in the second year $=13.5%$Percentage difference $=$ $15%-13.5%$ $=-1.5%$Percentage of depreciation in the third year $=12%-13.5%$ $=-1.5%$Since, the percentage of depreciation is constant $=-1.5%$Therefore, the series is in arithmetic progression.Step 2- Value of equipment after $10$years Since the rate of depreciation is in arithmetic progression So, the first term $\left(a\right)=15%$ Common difference $\left(d\right)=-1.5%$ Number of terms $\left(n\right)=10$Total percentage depreciation in $10$ years $s=\frac{n}{2}\left[2a+\left(n-1\right)×d\right]$$s=\frac{10}{2}\left[2×15+\left(10-1\right)×\left(-1.5\right)\right]$$s=5\left[30+9×\left(-1.5\right)\right]$$s=5\left[30-13.5\right]$$s=5×16.5$$s=82.5$Therefore, percentage depreciation in $10$ years $=82.5%$Value of depreciation $=$$\left(\frac{100-\mathrm{depreciation}%}{100}\right)×$ cost of equipment$=\left(\frac{100-82.5}{100}\right)×6,00,000\phantom{\rule{0ex}{0ex}}=\frac{17.5}{100}×6,00,000\phantom{\rule{0ex}{0ex}}=17.5×6000\phantom{\rule{0ex}{0ex}}=Rs105000$Hence, the value of the equipment after $10$ years is $Rs105000$.  Suggest Corrections  0      Similar questions  Related Videos   Formula for Sum of N Terms of an AP
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