Question

Mrs. Goswami deposits RS. 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the maturity value.
(a) RS. 40040
(b) RS. 40444
(c) RS. 40440
(d) RS. 40044

Answer 1

Hint: We solve this problem first by finding the interest of the recurring account.
The formula for interest is given as
\[I=P\times \dfrac{n\left( n+1 \right)}{2}\times \dfrac{R}{12\times 100}\]
Where, \[P\] is the amount deposited per month, \[n\] is the number of months and \[R\] is rate of interest per month.
Then we calculate the maturity value. The formula for matured value is given as
\[M.V=nP+I\]

Complete step by step answer:
We are given that the amount deposited per month is RS. 1000
Let us assume that the amount deposited per month as
\[\Rightarrow P=1000\]
We are given that the time period is 3 years.
We know that there are 12 months in a year.
Let us assume that the number of months of the period as \[n\] then we get
\[\begin{align}
  & \Rightarrow n=12\times 3 \\
 & \Rightarrow n=36 \\
\end{align}\]
We are also given that the rate of interest per annum as 8%
Let us assume that the rate of interest as
\[\Rightarrow R=8%\]
Let us assume that the interest for 3 years as \[I\]
We know that the formula for interest is given as
\[I=P\times \dfrac{n\left( n+1 \right)}{2}\times \dfrac{R}{12\times 100}\]
Where, \[P\] is the amount deposited per month, \[n\] is the number of months and \[R\] is rate of interest per month.
By using the above formula we get the interest as
\[\begin{align}
  & \Rightarrow I=1000\times \dfrac{36\times 37}{2}\times \dfrac{8}{12\times 100} \\
 & \Rightarrow I=\dfrac{13320}{3} \\
 & \Rightarrow I=4440 \\
\end{align}\]
Now, let us assume that the matured value as \[M\]
We know that the formula for matured value is given as
\[M.V=nP+I\]
By using the above formula we get the maturity value as
\[\begin{align}
  & \Rightarrow M=36\times 1000+4440 \\
 & \Rightarrow M=36000+4440 \\
 & \Rightarrow M=40440 \\
\end{align}\]
Therefore we can conclude that the matured value is RS. 40440

So, the correct answer is “Option c”.

Note: Students may make mistakes in taking the formula of interest in a recurring account.
We have the formula for interest is given as
\[I=P\times \dfrac{n\left( n+1 \right)}{2}\times \dfrac{R}{12\times 100}\]
Where, \[P\] is the amount deposited per month, \[n\] is the number of months and \[R\] is rate of interest per month.
Here we are given the rate of interest as per annum. But we are taking the amount deposited as per month and the period as the number of months.
So, we get 12 in the denominator because one year has 12 months.
Bu students may do mistake and take the formula as
\[I=P\times \dfrac{n\left( n+1 \right)}{2}\times \dfrac{R}{100}\]
This gives the wrong answer because all the terms are not in the same parameter. The values of \[P\] and \[n\] are in months but the value of \[R\] is in terms of year.