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
[tex]A=P(1+r)^t[/tex]
Plugging values in formula, we get
[tex]329.49=300(1+r)^2[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}300[/tex]
[tex]\frac{300\left(1+r\right)^2}{300}=\frac{329.49}{300}[/tex]
[tex]\left(1+r\right)^2=1.0983[/tex]
Taking square root on both sides, we get
[tex]1+r=\sqrt{1.0983}[/tex]
[tex]\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}[/tex]
[tex]1+r-1=\sqrt{1.0983}-1[/tex]
[tex]r=\sqrt{1.0983}-1[/tex]
[tex]r=1.048-1[/tex]
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.
