Question

Anuradha invests INR 500 every month in a recurring deposit scheme for 3 years. The interest earned is INR 2220. Find the rate of interest and the maturity values of the investment.

Answer 1

Hint: To solve the given question, we will first assume that the rate of interest is r %. Then we will calculate the simple interest for one month, two months and so on up to 3 years, i.e. 36 months. Then we will add all these interests to get the total simple interest. For calculating the simple interest, we will use the formula \[SI=\dfrac{P\times R\times T}{100}.\] After doing this, we will get the value of R. Then we will make use of the fact that the maturity level of the investment will be the sum of the total simple interest and the total principal for 3 years. From here, we will get the maturity value of the investment.

Complete step-by-step answer:
To start with, we will assume that the rate of the interest of the given scheme is r %. Now, it is given in the question that money is invested every month. So, we will find the simple interest for each month up to 3 years, i.e. 36 months. The formula of calculating SI is given as \[SI=\dfrac{P\times R\times T}{100}.\]
In our case, P = INR 500, R = r. Thus, the simple interest for the first month will be
\[S{{I}_{1}}=\dfrac{500\times r\times 1}{12\times 100}\]
Similarly,
\[S{{I}_{2}}=\dfrac{500\times r\times 2}{12\times 100}\]
\[S{{I}_{3}}=\dfrac{500\times r\times 3}{12\times 100}\]
\[S{{I}_{36}}=\dfrac{500\times r\times 36}{12\times 100}\]
Now, the sum of all these interests will be equal to INR 2220. Thus, we can say that,
\[\text{Total Interest}=S{{I}_{1}}+S{{I}_{2}}+S{{I}_{3}}+.....+S{{I}_{36}}\]
\[\Rightarrow \text{Total Interest}=\left( \dfrac{500\times r\times 1}{12\times 100} \right)+\left( \dfrac{500\times r\times 2}{12\times 100} \right)+\left( \dfrac{500\times r\times 3}{12\times 100} \right)+.....+\left( \dfrac{500\times r\times 36}{12\times 100} \right)\]
\[\Rightarrow \text{Total Interest}=\dfrac{500\times r}{12\times 100}\left( 1+2+3+....+36 \right)\]
\[\Rightarrow \text{Total Interest}=\dfrac{5r}{12\times 100}\left( 1+2+3+....+36 \right)\]
Now, we know that the sum of the first n terms of AP given as 1, 2, 3, ….n is \[\dfrac{n\left( n+1 \right)}{2}.\] In our case, n = 36. Thus, we will get,
\[\Rightarrow \text{Total Interest}=\dfrac{5r}{12}\times \dfrac{\left( 36\times 37 \right)}{2}\]
\[\Rightarrow \text{Total Interest}=\dfrac{5r\times 3\times 37}{2}\]
Now, it is given in the question that the sum of all these interests equal to INR 2220. Thus, we will get,
\[2220=\dfrac{5r\times 3\times 37}{2}\]
\[\Rightarrow r=\dfrac{2220\times 2}{5\times 3\times 37}\]
\[\Rightarrow r=\dfrac{2\times 111\times 10\times 2}{5\times 111}\]
Therefore, r = 8 %.
Thus, the rate of interest in the investment is 8 %. Now, we have to find the maturity level of the given investment. This will be equal to the sum of the principal for 3 years (i.e. 36 months) and the total simple interest. The principal for one month is 500, so the principal for 36 months will be \[500\times 36.\] Thus, we have,
Maturity Level = Principal for 36 months + Total Simple Interest
\[\Rightarrow \text{Maturity Level}=\text{INR}\left( 500\times 36 \right)+\text{INR }2220\]
\[\Rightarrow \text{Maturity Level}=\text{INR }18000+\text{INR }2220\]
\[\Rightarrow \text{Maturity Level}=\text{INR }20220\]
Thus, the maturity level of this investment is INR 20220.

Note: If we do not remember the formula of the sum of the first n terms of 1, 2, 3,…., then we can also find the sum of (1 + 2 + 3 + …. + 36) by using the concept of AP. We can see that the given series is an AP that has the total terms equal to 36. We know that if an AP has the first term \[{{a}_{1}}\] and last term \[{{a}_{n}}\] then the sum of the n terms will be
\[{{S}_{n}}=\dfrac{n}{2}\left[ {{a}_{1}}+{{a}_{n}} \right]\]
In our case, \[{{a}_{1}}=1,{{a}_{n}}=36.\] Thus,
\[{{S}_{n}}=\dfrac{36}{2}\left[ 1+36 \right]\]
\[\Rightarrow {{S}_{n}}=18\times 37\]
\[\Rightarrow {{S}_{n}}=666\]