Sound in decibels is measured by comparing the sound intensity, $I$, to a benchmark sound $I_{0}$ with intensity $10^{-16}$ watts/ $\mathrm{cm}^{2}$. Then,
\[
\text { Noise level in decibels }=10 \log \left(\frac{I}{I_{0}}\right)
\]
Sound $A$ measures 32 decibels and sound $B$ is 4 times as loud as sound $A$. What is the decibel rating of sound $B$ to the nearest integer?
The sound rating of sound $B$ is

Step 1 ：We are given that sound A measures 32 decibels and sound B is 4 times as loud as sound A. The formula for noise level in decibels is given as \(10 \log \left(\frac{I}{I_{0}}\right)\), where \(I\) is the sound intensity and \(I_{0}\) is the benchmark sound intensity.

Step 2 ：Since sound B is 4 times as loud as sound A, we can say that the intensity of sound B is 4 times the intensity of sound A.

Step 3 ：We can substitute this into the formula to find the decibel rating of sound B.

Step 4 ：Let's denote the intensity of sound A as \(I_A\), and the intensity of sound B as \(I_B\). Then, \(I_B = 4I_A\).

Step 5 ：Substituting \(I_B\) into the formula, we get \(10 \log \left(\frac{4I_A}{I_{0}}\right)\).

Step 6 ：Solving this equation, we find that the decibel rating of sound B is approximately 38.

Step 7 ：Final Answer: The sound rating of sound B is \(\boxed{38}\) decibels.