Question

Puneet has a recurring deposit account in the Bank of Baroda and deposits Rs.140 per month for 4 years. If he gets Rs.8092 on maturity, find the rate of interest given by the bank.

Answer 1

Hint: In this question, we are given monthly installments, time period, and total amount after maturity. We need to find the rate of interest per annum. For this, we will first find simple interest in terms of r and then calculate the total amount after maturity. Since we are already given a total amount, equating them will give us the value of r. Here, r is the rate of interest per annum. Formula for finding simple interest is given by: $SI=P\times \dfrac{n\left( n+1 \right)}{2\times 12}\times \dfrac{r}{100}$ where, P is the monthly installment, n is the number of installment (number of months) and r is the rate of interest per annum. $\text{Maturity value}=\left( P\times n \right)+SI$.

Complete step-by-step solution
Here, we are given an installment per month as Rs.140. Therefore, we can say P = Rs.140.
We are given a time period of 4 years. Since the installment was made monthly, therefore we need to find the number of months.
1 year has 12 months. So, 4 years will have ... months. Therefore, 4 years have 48 months.
The number of months is 48. Hence, the number of installments is 48. So we can say n = 48.
Let us suppose “r” to be the rate of interest per annum.
We know, formula for calculating simple interest is given by: $SI=P\times \dfrac{n\left( n+1 \right)}{2\times 12}\times \dfrac{r}{100}$ where, SI is the simple interest after 4 years, P is the monthly installment, n is the number of installments and r is the rate of interest per annum. Hence, we get,
\[\begin{align}
  & \Rightarrow SI=140\times \dfrac{48\left( 48+1 \right)}{2\times 12}\times \dfrac{r}{100} \\
 & \Rightarrow SI=140\times \dfrac{2352}{24}\times \dfrac{r}{100} \\
 & \Rightarrow SI=137.20r \\
\end{align}\]
Hence simple interest is Rs.137.20r where, r is the rate of interest.
Now, let us find maturity value.
As we know, maturity value is given by $\text{Maturity value}=\left( P\times n \right)+SI$ so we get:
\[\begin{align}
  & \text{Maturity value}=Rs.\left( 140\times 48 \right)+Rs.137.20r \\
 & \Rightarrow Rs.\left( 6720+137.20r \right) \\
\end{align}\]
Now we are already given maturity value as Rs.8092. So both these values will be equal. Hence,
\[\begin{align}
  & \Rightarrow Rs.\left( 6720+137.20r \right)=Rs.8092 \\
 & \Rightarrow 6720+137.20r=8092 \\
 & \Rightarrow 137.20r=8092-6720 \\
 & \Rightarrow 137.20r=1372 \\
\end{align}\]
Dividing both sides by 137.20 we get:
\[\begin{align}
  & \Rightarrow r=\dfrac{1372}{137.20} \\
 & \Rightarrow r=10\% \\
\end{align}\]
Hence rate of interest per annum is 10%.

Note: Students should know that here n should be converted in months only since installments are made monthly. Students should know that here, we don't need to divide the final answer by 100 to get a percentage. Here, r is already in percentage because we already divided r by 100 in the formula. Take care while solving decimal numbers.