Question

A recurring deposit account of Rs. 1,200 per month has maturity value of Rs. 12,440. 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

Hint: When we borrow money from bank in the form of loan which is known as principal (P) amount, for a particular time period ‘t’ at an interest rate 0f R% then the extra amount of money that is paid to bank during payback is known as Simple interest (S.I).
Simple interest \[ = \dfrac{{P \times R \times t}}{{100}}\]
And the total amount paid to the bank \[ = P + S.I\]
When any depositor deposits a fixed amount of money for a specific time period in a regular interval of time then this type of deposit is known as recurring deposit.
And the amount that person gets together with the interest is known as Maturity Value.
For example, if depositor deposits money every month then the money gets added monthly at a particular rate,
i.e.
Suppose n is the no. of month,
Then S.I for n months\[ = \dfrac{{P \times R \times n}}{{100 \times 12}}\]
S.I for n-1 months\[ = \dfrac{{P \times R \times (n - 1)}}{{100 \times 12}}\]
.
.
.
S.I for 1 month\[ = \dfrac{{P \times R \times 1}}{{100 \times 12}}\]
Total S.I \[ = \dfrac{{P \times R}}{{100 \times 12}}(1 + 2 + ... + (n - 1) + n) = \dfrac{{P \times R \times n(n + 1)}}{{100 \times 12 \times 2}}\]

Complete step-by-step answer:
Given P \[ = R.s1,200\]
Let us suppose no. of months \[ = n\]
Then total amount deposited \[ = 1200n\]
Maturity Value \[ = R.s12,440\]
Rate \[ = 8\% \]
And we know that S.I for recurring time period \[ = \dfrac{{P \times R \times (n - 1)}}{{100 \times 12}}\]
By putting the values we get,
S.I\[ = \dfrac{{1200 \times n(n + 1) \times 8}}{{2 \times 12 \times 100}}\]
\[ \Rightarrow S.I = 4n(n + 1)\]
Hence, maturity value \[ = 1200n + 4n(n + 1) = Rs.12,440\]
\[\begin{gathered}
   \Rightarrow 12440 = 1200n + 4{n^2} + 4n \\
   \Rightarrow 4{n^2} + 1204n - 12440 = 0 \\
\end{gathered} \]
By factorisation we get,
\[ \Rightarrow (n + 311)(n - 10) = 0\]
But months can’t be negative hence \[n = 10\]
That is time for this recurring deposit account\[ = 10months\]

Note: Always read the question properly don’t confuse Simple Interest with Compound Interest,
\[C.I.{\text{ }} = {\text{ }}P{\left( {1 + R / 100} \right)^t}\; - {\text{ }}P\]
And if the amount is compounded annually then,
\[A{\text{ }} = {\text{ }}P{\left( {1 + R/100} \right)^t}\]