The intensity of a sound detected by a receiver is inversely proportional to the square of the distance of the receiver from the source of the sound. When the distance is $r$ metres, the intensity of the sound detected is $q$ decibels. When the distance is reduced by 40%,

1. Find the new distance after a 40% reduction.

A reduction of $40\%$ means the new distance is $60\%$ of the original distance.

New distance $=60\% \times r=0.6r=\dfrac{3r}{5}$.

2. Apply inverse square proportionality to find the new intensity.

Since intensity $I$ is inversely proportional to the square of the distance,

$I \propto \dfrac{1}{d^2}.$

When $d=r$, $I=q$.

When $d=\dfrac{3r}{5}$,

$\frac{I_{\text{new}}}{q}=\frac{r^2}{\left(\frac{3r}{5}\right)^2}=\frac{r^2}{\frac{9r^2}{25}}=\frac{25}{9}.$

So

$I_{\text{new}}=\frac{25}{9}q.$

1. Find the percentage difference between the new intensity and the original intensity.

From part (a), the new intensity is $\dfrac{25q}{9}$.

Percentage difference from the original intensity $q$ is
$\frac{\frac{25q}{9}-q}{q}\times 100\%$
$=\frac{\frac{16q}{9}}{q}\times 100\%$
$=\frac{16}{9}\times 100\%$
$=177.8\% \text{ correct to 1 decimal place}.$