Question

Question 14>If 3000 dollars is invested in an account for 10 years. Find the value of the investment at the end of10 years if the interest is:(a) 6.2% compounded monthly: $(b) 6.2% compounded daily (ignore leap years): ¢

Answer 1

Given:

Principal amount, P = $3000

Time, t = 10 years

Let's solve for the following:

• (a). Find the value of the investment if the interest is 6.2% compounded monthly.

Apply the compound interest formula:

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

Where:

• P is the principal = $3000

,

• r is the interest rate = 6.2% = 0.062

,

• n is the number of times compounded = monthly = 12 months a year

,

• t is the time in years = 10 years

Plug in these values and solve for the value, A:

$\begin{gathered} A=3000(1+\frac{0.062}{12})^{12*10} \\ \\ A=3000(1+0.00516667)^{120} \\ \\ A=3000(1.00516667)^{120} \\ \\ A=3000(1.855963241) \\ \\ A=5567.89 \end{gathered}$

Therefore, if it is compounded monthly, the value at the end of 10 years is $5567.89

• (b). 6.2% compounded daily

In this case, the number of times compounded will be:

n = 365 times a year.

Thus, we have:

$\begin{gathered} A=3000(1+\frac{0.062}{365})^{365\times10} \\ \\ A=3000(1.000169863)^{3650} \\ \\ A=3000(1.858830169) \\ \\ A=5576.49 \end{gathered}$

If it is compounded daily, the value at the end of 10 years will be $5576.49

ANSWER:

(a). $5567.89

(b).