Question

Suppose that a savings account is compounded yearly with a principal of $80000. After 3 years years, the amount increased to $106480. What was the per annum interest rate?

Answer 1

Step-by-step explanation:

Given that

principal = 8000$

amount = ,106480$

time = 3 years

according to questions:-

$a = p(1 + \frac{r}{100} ) {}^{t} \\ \\ 106480 = 8000(1 + \frac{r}{100} ) {}^{3} \\ \\ \frac{106480}{8000} = (1 + \frac{r}{100} ) {}^{3} \\ \\ \frac{1331}{1000} = (1 + \frac{r}{100} ) {}^{3} \\ \\ \sqrt{ \frac{1331}{1000} } = 1 + \frac{r}{100} \\ \\ \frac{11}{10} = 1 + \frac{r}{100} \\ \\ \frac{11}{10} - 1 = \frac{r}{100} \\ \\ \frac{11 - 10}{10} = \frac{r}{100} \\ \\ \frac{1}{10} = \frac{r}{100} \\ \\ 100 = 10r \\ \\ r = \frac{100}{10} \\ \\ r = 10 \\ \\ therefore \: the \: rate \: is \: 10\%$

Answer 2

Step-by-step explanation:

Given that

principal = 8000$

amount = ,106480$

time = 3 years

according to questions:-

a = p(1 + \frac{r}{100} ) {}^{t} \\ \\ 106480 = 8000(1 + \frac{r}{100} ) {}^{3} \\ \\ \frac{106480}{8000} = (1 + \frac{r}{100} ) {}^{3} \\ \\ \frac{1331}{1000} = (1 + \frac{r}{100} ) {}^{3} \\ \\ \sqrt{ \frac{1331}{1000} } = 1 + \frac{r}{100} \\ \\ \frac{11}{10} = 1 + \frac{r}{100} \\ \\ \frac{11}{10} - 1 = \frac{r}{100} \\ \\ \frac{11 - 10}{10} = \frac{r}{100} \\ \\ \frac{1}{10} = \frac{r}{100} \\ \\ 100 = 10r \\ \\ r = \frac{100}{10} \\ \\ r = 10 \\ \\ therefore \: the \: rate \: is \: 10\%