Explanation

(c) The apparent frequency heard by the man is given by
$v_{app} = v\left( \frac{v \pm v_{o}}{v \pm v_{s}} \right)$
On our case,$v_{o} = 0$
$\therefore v_{app} = v\left( \frac{v}{v \pm v_{s}} \right)$
Given, $v = 300ms^{- 1},\ v_{s} = 4ms^{- 1},\ v = 240\ Hz$
In first case, train is approaching the man, so frequency heard
$v_{1} = v\left( \frac{v}{v \pm v_{s}} \right) = 240\left( \frac{300}{300 - 4} \right)$
$= \frac{240 \times 300}{296} = 243.24\ Hz$
In second case, train is going away from the man, so frequency heard
$v_{2} = v\left( \frac{v}{v + v_{s}} \right) = 240\ \left( \frac{300}{300 + 4} \right)$
$= \frac{240 \times 30}{304} = 236.84Hz$
Hence, number of beats heard by the man per second
$= v_{1} - v_{2}$
$= 243.24 - 236.84 = 6.4 \approx 6$