Commercial Mathematics - Banking

A person deposits a certain sum of money each month in a recurring deposit account of a bank. If the rate of interest is $8\%$ per annum and the person gets ₹ $8088$ from the bank after $3$ years, find the amount of money deposited each month.


Let money deposited per month $= P = $ ₹ $x$

Rate of interest $= r = 8 \%$ per annum

Number of months $= n = 36$ $\;\;\;$ [$3$ years $= 36$ months]

Interest $= I = P \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{r}{100}$

i.e. $\;$ $I = $ ₹ $x \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{8}{100} = $ ₹ $4.44 \; x$

Money deposited in $36$ months $= $ ₹ $36 \; x$

$\therefore \;$ Maturity value (M.V) $= $ ₹ $36 \; x + $ ₹ $4.44 \; x = $ ₹ $40.44 \; x$

Given: $\;$ M.V $= $ ₹ $8088$

$\implies$ $40.44 \; x = 8088$ $\implies$ $x = \dfrac{8088}{40.44} = 200$

$\therefore \;$ Money deposited per month $= $ ₹ $200$