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 value of the monthly installment.

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

Time for which money is deposited $= n = 3$ years $= 36$ months

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

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

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

i.e. $\;$ $I =$ ₹ $\left[\dfrac{111 x}{25}\right] = $ ₹ $4.44 x$

Total money deposited in $36$ months $= $ ₹ $36 x$

Maturity value (MV) $=$ Total money deposited $+ $ Interest

i.e. $\;$ $MV = $ ₹ $\left(36 x + 4.44 x\right) = $ ₹ $40.44 x$

Given: $\;$ $MV =$ ₹ $8088$

$\implies$ $40.44x = 8088$ $\implies$ $x = 200$

i.e. $\;$ money deposited per month $= $ ₹ $200$

Commercial Mathematics - Banking

A person has a recurring deposit account of ₹ $400$ per month at $10 \%$ per annum. If the person gets ₹ $260$ as interest at the time of maturity, find the total time for which the account was held.

Let the account be held for $n$ months.

Money deposited per month $= P =$ ₹ $400$

$\therefore \;$ Total money deposited $= $ ₹ $\left(400 \times n\right)$

Rate of interest $= r = 10\%$

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

i.e. $\;$ $I = $ ₹ $\left[400 \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{10}{100}\right]$

i.e. $\;$ $I = $ ₹ $\left[\dfrac{5 n \left(n + 1\right)}{3}\right]$

But, interest earned $= $ ₹ $260$

$\therefore \;$ $\dfrac{5n \left(n + 1\right)}{3} = 260$

i.e. $\;$ $5n^2 + 5n = 780$

i.e. $\;$ $n^2 + n = 156$

i.e. $\;$ $n^2 + n - 156 = 0$

i.e. $\;$ $n^2 + 13 n - 12n - 156 = 0$

i.e. $\;$ $n \left(n + 13\right) - 12 \left(n + 13\right) = 0$

i.e. $\;$ $\left(n + 13\right) \left(n - 12\right) = 0$

i.e. $\;$ $n + 13 = 0$ $\;$ or $\;$ $n - 12 = 0$

i.e. $\;$ $n = -13$ $\;$ or $\;$ $n = 12$

Since time cannot be negative, the time for which the account was held $= n = 12$ months $= 1$ year

Commercial Mathematics - Banking

A person has a recurring deposit account in a bank for $2$ years at $9\%$ per annum. If the person gets ₹ $7837.50$ at the time of maturity, find the monthly installment.

Let money deposited per month $= P =$ ₹ $100$

Number of months $= n = 24$ $\;\;\;$ [$2$ years]

Rate of interest $= r = 9 \%$

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

i.e. $\;$ $I =$ ₹ $\left(100 \times \dfrac{24 \times 25}{12} \times \dfrac{9}{100}\right) = $ ₹ $450$

Money deposited in $24$ months $= 24 \times $ ₹ $100 = $ ₹ $2400$

Maturity value (M.V) $= $ ₹ $\left(2400 + 450\right) = $ ₹ $2850$

When M.V is ₹ $2850$, monthly installment $= $ ₹ $100$

When M.V is ₹ $7837.50$, monthly installment $= $ ₹ $\dfrac{100 \times 7837.50}{2850} = $ ₹ $275$

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 $= $ ₹ $200$

Commercial Mathematics - Banking

If ₹ $\; 15,000$ is earned as interest on a monthly deposit of ₹ $\; 5,000$ for $2$ years. Find the rate of interest on the recurring deposit.

Money deposited each month $= P =$ ₹ $5,000$

Time for which money deposited $= n = 2 $ years $= 24 $ months

Let, rate of interest $= r \%$

Interest received $= $ ₹ $15,000$

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

i.e. $\;$ $15000 = 5000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100}$

i.e. $r = \dfrac{15000 \times 2 \times 12 \times 100}{5000 \times 24 \times 25} = 12 \%$

$\therefore \;$ Rate of interest on the recurring deposit $= 12 \%$

Commercial Mathematics - Banking

A person deposited ₹ $ 400$ at the beginning of every month in a recurring deposit account and received ₹ $ 16,398$ at the end of $3$ years. Find the rate of interest given by the bank.

Money deposited each month $= P =$ ₹ $ 400$

Time for which money deposited $= n = 3 $ years $= 36 $ months

Let, rate of interest $= r \%$

Amount received on maturity $= $ ₹ $ 16,398$

$\begin{aligned} \text{Interest} & = P \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{r}{100} \\\\ & = 400 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{r}{100} \\\\ & = 222 \; r \end{aligned}$

i.e. $\;$ Interest $= $ ₹ $ 222 \; r$

Total money deposited $= 400 \times 36 = $ ₹ $ 14,400$

$\text{Amount received on maturity} = \text{Money deposited} + \text{Interest}$

i.e. $\;$ $16398 = 14400 + 222 \; r$

i.e. $\;$ $222 \; r = 1998$

i.e. $\;$ $r = 9$

$\therefore \;$ Rate of interest given by the bank $= 9 \%$