# More Examples of Compound Interest

To find the 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 the principal when the 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 the rate
When P, n, A are given

We have to find r by using the 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 Rs.4000 becomes 4635.50

Solution:
Let A = 4635.50, P = 4000, $n = 3$, r = ?

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