Question

A person invests 8000 dollars in a bank. The bank pays 6.75% interest

compounded quarterly. To the nearest tenth of a year, how long must
the person leave the money in the bank until it reaches 10500 dollars?
r
A = P(1 + -)nt
1+
n

Answer 1

To find how long it takes for the investment to reach $10,500, we can use the compound interest formula A = P(1 + r/n)^(nt), where A is the final amount, P is the principal (initial investment), r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years. Plugging in the given values, we can solve for t and find that it takes approximately 3.23 years for the investment to reach $10,500.

Step-by-step explanation:

To find how long it takes for the investment to reach $10,500, we can use the compound interest formula: A = P(1 + r/n)^(nt), where A is the final amount, P is the principal (initial investment), r is the annual interest rate (as a decimal), n is the number of times the interest is compounded per year, and t is the number of years.

In this case, we have A = $10,500, P = $8,000, r = 6.75% (or 0.0675 as a decimal), and n = 4 (quarterly compounding).

Plugging in these values, we have 10500 = 8000(1 + 0.0675/4)^(4t). We can solve for t by taking logarithms:

log((1 + 0.0675/4)^(4t)) = log(10500/8000)

(4t)log(1 + 0.0675/4) = log(10500/8000)

4t = log(10500/8000) / log(1 + 0.0675/4)

t = (log(10500/8000) / log(1 + 0.0675/4)) / 4

Using a calculator, we find that t ≈ 3.23 years.