Question

The maturity value of a R.D. Account is Rs. 16,176. If the monthly installment is Rs. 400 and the rate of interest is 8 %. Find the time (period) of this R.D. Account.

Answer 1

Hint: To solve the given question, we will first assume that the total number of months for the RD account to mature is n months. Then, we will calculate the simple interest for one month, two months and so on up to n months, and then we will add all these simple interests to get the total simple interest. For calculating the simple interest, we will use the formula \[SI=\dfrac{P\times R\times T}{100}\] where T is the time period in years. Then we will make use of the fact that the total maturity level will be equal to the sum of the total simple interest and the total principal for n months. From here, we will calculate the value of n.

Complete step by step solution:
To start with, we will assume that the total number of months required to reach the maturity level is n. Now, we will find the simple interest for each month up to n months. The formula for calculating the simple interest is given below.
\[SI=\dfrac{P\times R\times T\left( \text{in years} \right)}{100}\]
where P is the principal value, R is the rate of interest and T is the time in years. But we have to calculate the interest monthly. So, the modified formula becomes,
\[SI=\dfrac{P\times R\times T\left( \text{in months} \right)}{12\times 100}\]
In our case, P = Rs. 400 and R = 8 %. Thus, the simple interest for the first month will become
\[S{{I}_{1}}=\dfrac{400\times 8\times 1}{12\times 100}\]
Similarly,
\[S{{I}_{2}}=\dfrac{400\times 8\times 2}{12\times 100}\]
\[S{{I}_{3}}=\dfrac{400\times 8\times 3}{12\times 100}\]
\[S{{I}_{n}}=\dfrac{400\times 8\times n}{12\times 100}\]
Now, the sum of all these interests will be the total interest. Thus, the total interest becomes
\[\text{Total Interest}=\dfrac{400\times 8\times 1}{12\times 100}+\dfrac{400\times 8\times 2}{12\times 100}+\dfrac{400\times 8\times 3}{12\times 100}+.......+\dfrac{400\times 8\times n}{12\times 100}\]
\[\Rightarrow \text{Total Interest}=\dfrac{400\times 8}{12\times 100}\left( 1+2+3+.....+n \right)\]
Now, we know that according to the given formula, the sum of 1 + 2 + 3 + …… + i is
\[1+2+3+......+i=\dfrac{i\left( i+1 \right)}{2}\]
Now, the value of i = n in our case, thus,
\[\Rightarrow \text{Total Interest}=\dfrac{400\times 8}{12\times 100}\times \dfrac{n\left( n+1 \right)}{2}\]
\[\Rightarrow \text{Total Interest}=\dfrac{4}{3}n\left( n+1 \right)......\left( i \right)\]
Now, the total principal for n months will be obtained by multiplying the principal of one month by n. Thus, we will get,
\[\text{Total principal for n months}=400\times n.....\left( ii \right)\]
Now, we will make use of the fact that the total maturity level will be equal to the sum of the total simple interest and the total principal for n months. Thus, we will get,
Total Maturity Level = Total interest + Total Principal for n months
Now, we will put the value of total interest and total principal for n months from (i) and (ii) to the above equation.
\[\Rightarrow \text{Total maturity level}=\dfrac{4}{3}n\left( n+1 \right)+400n\]
\[\Rightarrow \text{Total maturity level}=\dfrac{4{{n}^{2}}+4n}{3}+400n\]
\[\Rightarrow \text{Total maturity level}=\dfrac{4{{n}^{2}}+4n+1200n}{3}\]
\[\Rightarrow \text{Total maturity level}=\dfrac{4{{n}^{2}}+1204n}{3}\]
Now, it is given in the question that the maturity level of the given RD account is Rs. 16176. Thus, we will get,
\[\Rightarrow 16176=\dfrac{4{{n}^{2}}+1204n}{3}\]
\[\Rightarrow 48528=4{{n}^{2}}+1204n\]
\[\Rightarrow 4{{n}^{2}}+1204n-48528=0\]
On dividing the whole equation by 4, we will get,
\[\Rightarrow {{n}^{2}}+301n-12132=0\]
Now, we have got a quadratic equation in which we will solve with the help of the quadratic formula. If the quadratic equation is of the type \[a{{x}^{2}}+bx+c=0\] then its roots will be
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Thus, we have,
\[n=\dfrac{-301\pm \sqrt{{{\left( 301 \right)}^{2}}-4\left( 1 \right)\left( -12132 \right)}}{2\left( 1 \right)}\]
\[\Rightarrow n=\dfrac{-301\pm \sqrt{90601+48528}}{2}\]
\[\Rightarrow n=\dfrac{-301\pm \sqrt{139129}}{2}\]
\[\Rightarrow n=\dfrac{-301\pm 373}{2}\]
\[\Rightarrow n=\dfrac{-301+373}{2}\text{ or }n=\dfrac{-301-373}{2}\]
\[\Rightarrow n=36\text{ or }n=-337\]
Now, we know that n is the time period, so it will be positive. Thus, the value of n is 36.
Hence, the time period of this account will be 36 months.

Note: We can solve the quadratic equation with the help of completing the square method. In this method, we will try to form the square terms on both sides. Thus,
\[{{n}^{2}}+301n-12132=0\]
\[\Rightarrow {{n}^{2}}+2\times n\times \left( \dfrac{301}{2} \right)+\dfrac{90601}{4}-\dfrac{90601}{4}-12132=0\]
\[\Rightarrow {{\left( n+\dfrac{301}{2} \right)}^{2}}=\dfrac{90601+48528}{4}\]
\[\Rightarrow {{\left( n+\dfrac{301}{2} \right)}^{2}}=\dfrac{139129}{4}\]\[\Rightarrow {{\left( n+\dfrac{301}{2} \right)}^{2}}={{\left( \dfrac{373}{2} \right)}^{2}}\]
On taking the square root on both the sides, we will get,
\[\Rightarrow \left( n+\dfrac{301}{2} \right)=\pm \left( \dfrac{373}{2} \right)\]
\[\Rightarrow \left( n+\dfrac{301}{2} \right)=+\left( \dfrac{373}{2} \right)\text{or }\left( n+\dfrac{301}{2} \right)=-\left( \dfrac{373}{2} \right)\]
\[\Rightarrow n=36\text{ or }n=-337\]
Thus, the value of n is 36.