Question

f $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula

I = 1,000 ( (1 + r/100)^n - 1 ),

where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent?

(1) The deposit earns a total of $210 in interest in the first two years.
(2) ( 1 + r/100 )^2 > 1.5

Answer 1

Answer:

(1) The annual interest rate is 10%, that is greater than 8%

(2) The annual interest is bigger than 22.47% and that is greater than 8%

Step-by-step explanation:

We have the equation:

$I=1,000((1+\frac{r}{100} )^{n} -1)$

Where I is the earns in interest, n is the number of years and r is the annual interest rate.

For the first case, we can replace I by $210 and n by 2 as:

$210=1,000((1+\frac{r}{100} )^{2} -1)$

Solving for r, we get:

$\frac{210}{1,000} = (1+\frac{r}{100})^{2} -1$

$0.21 +1 = (1+\frac{r}{100})^{2}$

$1.21 = (1+\frac{r}{100})^{2}$

$\sqrt{1.21} = 1 + \frac{r}{100}$

$\sqrt{1.21} - 1 = \frac{r}{100}$

$(\sqrt{1.21} - 1)100=r$

$10=r$

So, for the first case, the interest rate paid by the bank is 10% and it is greater than 8%

For the second case, we need to take the equation and solve for r as:

$(1+\frac{r}{100} )^{2} >1.5$

$1+\frac{r}{100} > \sqrt{1.5}$

$\frac{r}{100} > \sqrt{1.5}-1$

$r > (\sqrt{1.5}-1)*100$

$r > 22.47$

So, for the second case the rate need to be bigger than 22.47% and that is bigger than 8%