Two tuning forks A and B produce 10 beats per second when sounded together. On slightly loading fork  A with a little wax, it was observed that 15 beats are heard per second. If the frequency of fork B is 480 Hz, what is the frequency of fork A before it was loaded?

Let ${\mathrm{v}}_1$ be the frequency of A and ${\mathrm{v}}_2$ be that of fork B. There are two possibilities $\left(\mathrm{a}\right) {\mathrm{v}}_1>{\mathrm{v}}_2$   and   $\left(\mathrm{b}\right) {\mathrm{v}}_1<{\mathrm{v}}_2$.

$$\begin{gathered} \left(\mathrm{a}\right) {\mathrm{v}}_1>{\mathrm{v}}_2:{\mathrm{v}}_2=480 \mathrm{Hz}, {\mathrm{v}}_{\mathrm{b}}=10 \mathrm{Hz} \\ \mathrm{Now}, {\mathrm{v}}_{\mathrm{b}}={\mathrm{v}}_1-{\mathrm{v}}_2 \\ {\mathrm{v}}_1={\mathrm{v}}_2+{\mathrm{v}}_{\mathrm{b}}=480+10=490 \mathrm{Hz} \end{gathered}$$

On loading fork A its frequency decreases. In other words, ${\mathrm{v}}_1$ decreases.

${\mathrm{v}}_1 - {\mathrm{v}}_2 = {\mathrm{v}}_{\mathrm{b}}$ should also decrease, But  ${\mathrm{v}}_{\mathrm{b}}$ increase to 15 beats/s. Hence  ${\mathrm{v}}_1$ is not greater than ${\mathrm{v}}_2$.

$$\begin{gathered} \left(\mathrm{b}\right) {\mathrm{v}}_1 < {\mathrm{v}}_2 : \mathrm{In} \mathrm{this} \mathrm{case} {\mathrm{v}}_2 - {\mathrm{v}}_1 = {\mathrm{v}}_{\mathrm{b}} \\ {\mathrm{v}}_1= {\mathrm{v}}_2 - {\mathrm{v}}_{\mathrm{b}} =480 - 10 = 470 \mathrm{Hz} \end{gathered}$$