Question

A recurring deposit account of $₹1200$ per month has a maturity value of $₹12440$. If the rate of interest is 8% and the interest is calculated at the end of every month, find the time (in months) of this Recurring Deposit Account.

Answer 1

Given

The principal amount $P=₹1200$

Maturity Value $=₹12440$

Rate$=8\%$

We have to determine the interest is calculated at the end of every month and also the time (in months) of this Recurring Deposit Account.

Let us assume the number of months $=n$

hence the total amount deposited be $=1200n$

From the simple interest formula, we know that

Simple interest formula is given as:

$SI=\frac{PTR}{100}$

Where SI = simple interest

P = principal

R = interest rate (in percentage)

T = time duration (in years)

Simple interest for the recurring time period $=P\times R\times \left(\frac{n-1}{100\times 12}\right)$

On substituting the known values we get

$$\begin{gathered} =1200\times \left[\frac{n\left(n+1\right)\times 8}{2\times 12\times 100}\right] \\ =4n\left(n+1\right) \end{gathered}$$

Maturity value $=1200n+4n\left(n+1\right)=12440$

$$\begin{gathered} 12440=1200n+4n^2+4n \\ 4n^2+1204n-12440=0 \\ {n}^2+301n-3110=0 \end{gathered}$$

On factorizing we get
$$\begin{gathered} n^2+311n-10n-3110=0 \\ n\left(n+311\right)-10\left(n+311\right)=0 \\ \left(n+311\right)\left(n-10\right)=0 \end{gathered}$$

$n+311=0$ or $n-10=0$

$n=-311$ or $n=10$

months cannot be negative

Therefore, the number of months $n=10$

Hence, the number of months $n=10$