Explanation
We are asked to find the interest rate that will yield $1500 for a sum of $1000 invested quarterly
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]In our case, we have
[tex]\begin{gathered} P=\text{ principal=1000} \\ A=final\text{ amount=}1500 \\ n=number\text{ of times compounded yearly=}4 \\ t=1 \end{gathered}[/tex]Thus, we will have
[tex]\begin{gathered} 1500=1000(1+\frac{r}{4})^{4(1)} \\ \\ \frac{1500}{1000}=(1+0.25r)^4 \\ \\ 1.5=(1+0.25r)^4 \end{gathered}[/tex]Solving for r
we will have
[tex]r=0.42672,\:r=-8.4267[/tex]Since the value increased, the rate will be positive
Therefore
[tex]\begin{gathered} The\text{ rate will be 0.42672} \\ 42.67\text{ \%} \end{gathered}[/tex]To the nearest per cent, we will have the annual rate as 42.7 %
