More Examples of Compound Interest

To Find Principal
We are given n, A, and r
We have to find P by the formula
{\text{A}} = {\text{P}}{\left( {1 + {\text{r}}} \right)^n} \Rightarrow {\text{P}} = \frac{{\text{A}}}{{{{\left( {1 + {\text{r}}} \right)}^n}}}

Example 01:
Find principal, when compound interest for 4 years at 6% is 525.

Solution:
Let Principal = P, n = 4, r = 6%, C.I = 525
We know that

\begin{gathered} {\text{P}} = \frac{{{\text{P}} + 525}}{{{{\left( {1 + r} \right)}^n}}} \\ \Rightarrow {\text{P}} = \frac{{525 + {\text{P}}}}{{{{\left( {1 + .06} \right)}^4}}} \\ \Rightarrow {\text{P}} = \frac{{525 + {\text{P}}}}{{{{\left( {1.06} \right)}^4}}} \\ \Rightarrow {\text{P}}{\left( {1.06} \right)^4} = 525 + {\text{P}} \\ \Rightarrow \left[ {{\text{P}}{{\left( {1.06} \right)}^4} - 1} \right] = 525 \\ \Rightarrow {\text{P}} = \frac{{525}}{{{{\left( {1.06} \right)}^4} - 1}}\,\,\,\,\, = 2019.23 \\ \end{gathered}


To Find Rate
When P, n, A are given

We have to find r by the using same formula
{\text{A}} = {\text{P}}{\left( {1 + {\text{r}}} \right)^n}

Example 02:
Find the rate of interest compounded per annum for 3 years so that 4000 becomes 4635.50

Solution:
Let A = 4635.50, P = 4000, n = 3, r = ?
Using {\text{A}} = {\text{P}}{\left( {1 + {\text{r}}} \right)^n}

\begin{gathered} {\text{4635}}{\text{.50}} = 4000{\left( {1 + {\text{r}}} \right)^3} \\ \Rightarrow \frac{{{\text{4635}}{\text{.50}}}}{{4000}} = {\left( {1 + {\text{r}}} \right)^3} \\ \Rightarrow 1.158 = {\left( {1 + {\text{r}}} \right)^3} \\ \Rightarrow {\left( {1.158} \right)^{\frac{1}{3}}} = 1 + {\text{r}} \\ \Rightarrow 1.05 = 1 + {\text{r}} \\ \Rightarrow \frac{{\text{r}}}{{100}} = 0.05\,\, \Rightarrow \,{\text{r}} = 5\% \\ \end{gathered}