A machinery worth rs10, 500 depreciated …
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Sia ? 2 years, 6 months ago
P = Rs. 10500
R = 5% per annum
h = 1 year
{tex}\therefore A = P{\left( {1 - \frac{R}{{100}}} \right)^n}{/tex}
{tex} = 10500{\left( {1 - \frac{5}{{100}}} \right)^1}{/tex}
{tex} = 10500\left( {1 - \frac{1}{{20}}} \right){/tex}
{tex} = 10500 \times \frac{{19}}{{20}}{/tex}
= 9975
Hence, value after 1 years is Rs. 9975
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