Compound interest is calculated on both the initial payment and the interest earned in previous periods.
problem
Suppose that a savings account is compounded monthly with a principal of $1350. After 8 months years, the amount increased to $1424. What was the per annum interest rate?
solution
Interest rate per anum was 8%.
explanation
STEP 1: Convert 8 months into years.
$$ \text{ 8 months } = \frac{ 8 }{12} \text{ years} = 0.6667 \text{ years}$$STEP 2: To find interest rate we use formula:
$$ A = P \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}} $$ |
A = total amount P = principal (amount of money deposited) r = annual interest rate n = number of times compounded per year t = time in years |
In this example we have
$$ A = $1424 ~,~ P = $1350 , t = 0.6667 ~ \text{years} ~~ \text{and} ~ n = 12$$After plugging the given information we have
$$ \begin{aligned} 1424 &= 1350 \left( 1 + \frac{ r }{ 12 } \right)^{\Large{ 12 \cdot 0.6667 }} \\ 1424 &= 1350 \left( 1 + \frac{ r }{ 12 } \right)^{\Large{ 8.0004 }} \\ \left( 1 + \frac{ r }{ 12 } \right)^{\Large{ 8.0004 }} &= \frac{ 1424 }{ 1350 }\\ \left( 1 + \frac{ r }{ 12 } \right)^{\Large{ 8.0004 }} &= 1.0548 ~~~ \text{ Take the natural logarithm of each side} \\ ln \left( 1 + \frac{ r }{ 12 }\right) ^{\Large{ 8.0004 }} &= ln(1.0548) \\ 8.0004 \cdot ln \left( 1 + \frac{ r }{ 12 }\right) &= ln(1.0548) \\ ln \left( 1 + \frac{ r }{ 12 }\right) &= \frac{ln(1.0548)}{ 8.0004} \\ ln \left( 1 + \frac{ r }{ 12 }\right) &= 0.0067 \\ 1 + \frac{ r }{ 12 } &= e^{ 0.0067 } \\ 1 + \frac{ r }{ 12 } &= 1.0067 \\ \frac{ r }{ 12 } &= 0.0067 \\ r &= 0.0804 \approx 8\% \end{aligned}$$Please tell me how can I make this better.