Question

Priyanka has a recurring deposit account of Rs.\[1,000\] per month at \[10\% \] per annum. If she gets Rs. \[5,550\] as interest at the time of maturity. Find the total time for which the account was held.

Answer 1

Hint: In recurring deposit, an amount is deposited in the account at continuous intervals of time, and a rate of interest is given through which the interest is evaluated on the deposited amount. In this question, simple interest is evaluated on the amount deposited.
The formulae for simple interest is given by
\[I = \dfrac{{P \times r \times t}}{{100}}\]
Where,
 $I$- interest received on the principal amount
$P$- Principal amount (Deposited amount)
$r$ - annual rate of interest
$t$ - duration in years

Complete step by step answer:
From the information given in the question, we have the following information,
Amount deposited monthly for recurring deposit (\[P\]) = Rs.\[1,000\]
Rate of interest per annum (\[r\]) = \[10\% \]
Interest received at the time of maturity (\[S.I.\]) = Rs. \[5,550\]
We are required to find the time (\[n\]) in months for which this recurring deposit account was held.
Since in a recurring deposit, Rs.\[1,000\] amount is deposited monthly, so monthly interest (given annual rate of interest \[10\% \] ) is calculated as shown below:
Rate of interest for one month, is given by $\dfrac{r}{{12}} = \dfrac{{10}}{{12}}$%
For first month,
${1^{st}} = P \times \dfrac{r}{{100}} \times \dfrac{1}{{12}}$
$ = 1000 \times \dfrac{{10}}{{12}} \times \dfrac{1}{{100}}$
For second month,
${2^{nd}} = P \times \dfrac{r}{{100}} \times \dfrac{2}{{12}}$
$ = 1000 \times \dfrac{{10}}{{12}} \times \dfrac{2}{{100}}$
Similarly, for ${n^{th}}$month, ${n^{th}} = P \times \dfrac{r}{{12}} \times \dfrac{n}{{100}}$
So, the total interest received at the end of ${n^{th}}$ month is
$\dfrac{{P \times r \times 1}}{{12 \times 100}} + \dfrac{{P \times r \times 2}}{{12 \times 100}} + \dfrac{{P \times r \times 3}}{{12 \times 100}} + ... + \dfrac{{P \times r \times n}}{{12 \times 100}}$
$\dfrac{{1000 \times 10 \times 1}}{{12 \times 100}} + \dfrac{{1000 \times 10 \times 2}}{{12 \times 100}} + \dfrac{{1000 \times 10 \times 3}}{{12 \times 100}} + ... + \dfrac{{1000 \times 10 \times n}}{{12 \times 100}}$
\[\left( {\dfrac{{1000 \times 10}}{{12 \times 100}}} \right) \times (1 + 2 + 3 + ... + n)\]
Using the formulae for the sum of first n natural numbers,
\[\left( {\dfrac{{1000 \times 10}}{{12 \times 100}}} \right) \times \left( {\dfrac{{n(n + 1)}}{2}} \right)\]
We get :
\[\left( {\dfrac{{25}}{6}} \right) \times n(n + 1)\]
Since, it is given that total interest received at the end of n months is Rs. 5550, so the above term which represent the total interest received will be equal to 5550,
\[5,550 = \dfrac{{25}}{6} \times n(n + 1)\]
Simplify the above equation,
\[5,550 \times \dfrac{6}{{25}} = n(n + 1)\]
\[1332 = n(n + 1)\]
Convert the above equation into the standard form of quadratic equation,
 \[{n^2} + n - 1332 = 0\]
Solve for \[n\] using factorization method,
Split the term \[1n\] into $37n$ and $ - 36n$ such that the product of $37$ and $ - 36$ is $ - 1332$.
\[{n^2} + 37n - 36n - 1332 = 0\]
\[n(n + 37) - 36(n + 37) = 0\]
\[(n + 37)(n - 36) = 0\]
\[(n + 37) = 0\] or \[(n - 36) = 0\]
\[n = - 37\] or \[n = 36\]
Since the value of \[n\] cannot be negative, so \[n = - 37\] is discarded, so \[n = 36\].
Hence, the recurring deposit was held for \[36\] months or $3$ years.

Note:
To solve any quadratic equation of the form \[a{x^2} + bx + c = 0\] using the factorization method, we have to make the quadratic equation of the form $(x - \alpha )(x - \beta ) = 0$where $\alpha $and $\beta $ are the roots of the given equation.