SAT Mathematics Practice Test 2

If ${\mathit{I}}_{\mathit{A}}$ is the intensity measured by Observer A from a distance of ${\mathit{r}}_{\mathit{A}}$ and ${\mathit{I}}_{\mathit{B}}$ is the intensity measured by Observer B from a distance of ${\mathit{r}}_{\mathit{B}}$,then ${\mathit{I}}_{\mathit{A}}=16{\mathit{I}}_{\mathit{B}}$ Using the formula $\mathit{I}=\frac{\mathit{P}}{4{\mathit{\pi }}^2}$,he intensity measured by Observer A is ${\mathit{I}}_{\mathit{A}}=\frac{\mathit{P}}{4\mathit{\pi }{\mathit{r}}_{\mathit{A}}^2}$,which can also be written in terms of ${\mathit{I}}_{\mathit{B}}$ as ${\mathit{I}}_{\mathit{A}}=16{\mathit{I}}_{\mathit{B}}=16\left(\frac{\mathit{P}}{4\mathit{\pi }{\mathit{r}}_{\mathit{B}}^2}\right)$.Setting the right-hand sides of these two equations equal to each other gives $\frac{\mathit{P}}{4\mathit{\pi }{\mathit{r}}_{\mathit{A}}^2}=16\left(\frac{\mathit{P}}{4\mathit{\pi }{\mathit{r}}_{\mathit{B}}^2}\right)$ which relates the distance of Observer A from the radio antenna to the distance ofObserver B from the radio antenna. Canceling the common factor $\frac{\mathit{p}}{4\mathit{\pi }}$ and rearranging the equation gives ${\mathit{r}}_{\mathit{B}}^2=16{\mathit{r}}_{\mathit{A}}^2$ Taking the square root of each side of ${\mathit{r}}_{\mathit{B}}^2=16{\mathit{r}}_{\mathit{A}}^2$ gives ${\mathit{r}}_{\mathit{B}}=4{\mathit{r}}_{\mathit{A}}$ and then dividing each side by 4 yields ${\mathit{r}}_{\mathit{A}}=\frac{1}{4}{\mathit{r}}_{\mathit{B}}$.Therefore, the distance of Observer A from the radio antenna is $\frac{1}{4}$ the distance of Observer B from the radio antenna.
Choices B, C, and D are incorrect and may result from errors in deriving or using the formula $\frac{\mathit{P}}{4\mathit{\pi }{\mathit{r}}_{\mathit{A}}^2}=\left(16\right)\left(\frac{\mathit{P}}{4\mathit{\pi }{\mathit{r}}_{\mathit{B}}^2}\right)$