A firm must replace a machine in 5 years for ₹1,00,000. It puts an equal sum into a sinking fund at the end of every year, earning 10% a year. To the nearest rupee, what is each deposit? (Take $(1.1)^{5} = 1.61051$.)

Explanation

1. 01

   The deposits $R$ must grow to $A$: $R \times \frac{(1 + i)^{n} - 1}{i} = A$, so $R = \frac{A\,i}{(1 + i)^{n} - 1}$.

2. 02

   $R = \frac{1{,}00{,}000 \times 0.1}{0.61051} \approx 16{,}380$ rupees.