Question

A person invests 4000 dollars in a bank. The bank pays 5.75% interest compounded quarterly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 5900 dollars?

Answer 1

Given:

A person invests 4000 dollars in a bank.

so, the initial balance = P = 4000

The interest rate = r = 5.75% = 0.0575

Compounded quarterly, n = 4

We will find the time (t) to reach 5900

We will use the following formula:

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

Substitute with the given values then solve for (t)

$\begin{gathered} 5900=4000(1+\frac{0.0575}{4})^{4t} \\ \frac{5900}{4000}=1.014375^{4t} \\ \end{gathered}$

Taking the natural logarithm for both sides:

$\begin{gathered} \ln \frac{5900}{4000}=4t\cdot\ln 1.014375 \\ \\ t=\frac{\ln \frac{5900}{4000}}{4\ln 1.014375}\approx6.8077 \end{gathered}$

Rounding to the nearest tenth of a year

So, the answer will be t = 6.8 years