Question

A person invests 1000 dollars in a bank. The bank pays 7% interest compounded quarterly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 3400 dollars?

Answer 1

Using compound interest, it is found that the person must leave money in the bank for 17.6 years.

What is compound interest?

The amount of money earned, in compound interest, after t years, is given by:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

In which:

• A(t) is the amount of money after t years.

• P is the principal(the initial sum of money).

• r is the interest rate(as a decimal value).

• n is the number of times that interest is compounded per year.

• t is the time in years for which the money is invested or borrowed.

In this problem, we want to find t when:

$A(t) = 3400, P = 1000, r = 0.07, n = 4$.

Hence:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$3400 = 1000\left(1 + \frac{0.07}{4}\right)^{4t}$

$(1.0175)^{4t} = 3.4$

$\log{(1.0175)^{4t}} = \log{3.4}$

$4t\log{1.0175} = \log{3.4}$

$t = \frac{\log{3.4}}{4\log{1.0175}}$

$t = 17.6$

The person must leave money in the bank for 17.6 years.

More can be learned about compound interest at brainly.com/question/25781328