Compound Interest is calculated on the initial payment and also on the interest of previous periods.
Example: Suppose you give \$100 to a bank which pays you 10% compound interest at the end of every year. After one year you will have \$100 + 10% = \$110, and after two years you will have \$110 + 10% = \$121.
Problem
Suppose that a savings account is compounded yearly with a principal of $1700. After 2 years years, the amount increased to $1910. What was the per annum interest rate?
Result:
Interest rate per anum was 6%.
Explanation
To find interest rate we use formula:
$$ A = P \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}} $$ |
A = total amount P = principal or amount of money deposited, r = annual interest rate n = number of times compounded per year t = time in years |
In this example we have
$$ A = $1910 ~,~ P = $1700 , t = 2 ~ \text{years} ~~ \text{and} ~ n = 1$$After plugging the given information we have
$$ \begin{aligned} 1910 &= 1700 \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 1 \cdot 2 }} \\ 1910 &= 1700 \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 2 }} \\ \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 2 }} &= \frac{ 1910 }{ 1700 }\\ \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 2 }} &= 1.12353 ~~~ \text{ Take the natural logarithm of each side} \\ ln \left( 1 + \frac{ r }{ 1 }\right) ^{\Large{ 2 }} &= ln(1.12353) \\ 2 \cdot ln \left( 1 + \frac{ r }{ 1 }\right) &= ln(1.12353) \\ ln \left( 1 + \frac{ r }{ 1 }\right) &= \frac{ln(1.12353)}{ 2} \\ ln \left( 1 + \frac{ r }{ 1 }\right) &= 0.05824 \\ 1 + \frac{ r }{ 1 } &= e^{ 0.05824 } \\ 1 + \frac{ r }{ 1 } &= 1.0599693578315 \\ \frac{ r }{ 1 } &= 0.059969357831492 \\ r &= 0.059969357831492 \approx 6\% \end{aligned}$$Share this result with others by using the link below.
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