Question

Alma invests $300 in an account that compounds interest annually. After 2 years, the balance of the account is $329.49. To the nearest tenth of a percent, what is the rate of interest on the account?

Answer 1

Invested amount (P) = $300.

Time in years (t) = 2 years.

Balance after 2 years (A) = $329.49.

Let us assume rate of interest = r % compounds annually.

We know, formula for compound interest

$A=P(1+r)^t$

Plugging values in formula, we get

$329.49=300(1+r)^2$

$\mathrm{Divide\:both\:sides\:by\:}300$

$\frac{300\left(1+r\right)^2}{300}=\frac{329.49}{300}$

$\left(1+r\right)^2=1.0983$

Taking square root on both sides, we get

$1+r=\sqrt{1.0983}$

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$1+r-1=\sqrt{1.0983}-1$

$r=\sqrt{1.0983}-1$

$r=1.048-1$

r=0.048.

Converting it into percentage by multiplying by 100.

r=0.048 × 100

r  = 4.8 %

Therefore, the rate of interest on the account is 4.8% compounds annually.