Question

Ella is going to invest $11,000 and leave it in an account for 16 years. Assuming the interest is compounded monthly, what interest rate, to the nearest hundredth of a percent, would be required in order for Ella to end up with $14,000?

Answer 1

Using the formula for compounded interest, it is found that an interest rate of 1.56% would be required.

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The compound interest formula is given by:

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

• 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.

• Invest $11,000, thus $P = 11000$
• 16 years, thus $t = 16$
• End up with $14,000, thus $A(t) = 14000$
• Compounded monthly, thus $n = 12$.

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

$14000 = 11000(1 + \frac{r}{12})^{12(16)}$

$(1 + \frac{r}{12})^{192} = \frac{14}{11}$

$\sqrt[192]{(1 + \frac{r}{12})^{192}} = \sqrt[192]{\frac{14}{11}}$

$1 + \frac{r}{12} = (\frac{14}{11})^{\frac{1}{192}}$

$1 + \frac{r}{12} = 1.0013$

$\frac{r}{12} = 0.0013$

$r = 0.0013(12)$

$r = 0.0156$

An interest rate of 1.56% would be required.

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