Question

Oliver is going to invest $2,500 and leave it in an account for 18 years. Assuming the interest is compounded quarterly, what interest rate, to the nearest tenth of a percent, would be required in order for Oliver to end up with $4,600?

Answer 1

The rate of interest is 3.4% (to the nearest tenth of a percent)

With compound interest, the old amount is added to the interest gotten from the old amount to get the new amount. This process is repeated at regular intervals over a period of time.

The compounded amount formula is given by

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

where

$A=\text{the amount gotten after t years}\\P=\text{the original investment}\\r=\text{the rate, as a decimal fraction}\\n=\text{the number of times the interest is compounded, per year}\\t=\text{the overall duration of the deposit, in years}$

From the question,

• The compounding is done quarterly, $n=4$

• The duration is 18 years, $t=18$

• The original investment is $2,500, $P=2500$

• The amount after 18 years is $4,600, $A=4600$

we want to use the given information to calculate the interest rate Oliver will need in order to end up with $4,600.

First rearrange the formula so that the rate becomes the subject.

$A=P \left(1+\dfrac{r}{n}\right)^{nt}\\\\\dfrac{A}{P}=\left(1+\dfrac{r}{n}\right)^{nt}\\\\\sqrt[4t]{\dfrac{A}{P}}=1+\dfrac{r}{n}\\\\\sqrt[4t]{\dfrac{A}{P}}-1=\dfrac{r}{n}\\\\n\left(\sqrt[4t]{\dfrac{A}{P}}-1\right)=r$

substituting the values in the problem into the rearranged formula, we can calculate the decimal value of $r$

$r=n\left(\sqrt[4t]{\dfrac{A}{P}}-1\right)\\\\=4\left(\sqrt[4(18)]{\dfrac{4600}{2500}}-1\right)\\\\\approx0.034$

The rate of interest, to the nearest tenth of a percent, is

$0.034\times 100\%=3.4\%$

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