Question

9. A bank pays interest of 11% on $6000 in a deposit account

Answer 1

Answer:

Approximately 6.642 years

Step-by-step explanation:

The given parameters are;

The amount in the deposit account = $6,000

The time in which the money will have trebled at 11% compound interest, is given as follows;

$A = P\times \left (1 + \dfrac{r}{n} \right )^{n\cdot t}$

Where;

A = The amount at the end of the period

P = The amount in the deposit = $6,000.00

r = The rate of interest = 11%

t =  The periods that elapsed

n = The number of times the interest is applied per period of time, t = 1

For the money to have trebled, the amount generated at the end of the period will be 200% the amount deposited

Therefore, we have;

Amount, A, at the end of period = 200/100× $6,000 = $12,000

Substituting the values into the formula for the formula, we have;

$\ 12,000 = \ 6,000\times \left (1 + \dfrac{0.11}{1} \right )^{1\times t}$

Which gives;

$\ 12,000 = \ 6,000\times \left (1 + {0.11} \right )^{t}$

$\left (1 .11} \right )^{t} = \dfrac{ \ 12,000}{\ 6,000} = 2$

t = ㏒(2)/(㏒(1.11)) ≈ 6.642 years which is approximately 6 years, 7 months and 24 days