A source emitting a pure tone of frequency 720 Hz moves with speed 20 m s^-1 along a straight line towards a large flat wall that is perpendicular to…

Direct sound: the source recedes from the observer, so $f_d=f_0\dfrac{v}{v+v_s}=720\cdot\dfrac{340}{360}=680\ \mathrm{Hz}$. Option A is correct. Sound travelling towards the wall is emitted by an approaching source: wavelength $\lambda=\dfrac{v-v_s}{f_0}=\dfrac{320}{720}=0.444\ \mathrm{m}$, so the frequency received by the wall is $f_w=v/\lambda=765\ \mathrm{Hz}$. D is correct. A stationary wall re-emits this same frequency and the observer is stationary, so the reflected sound is heard at $765\ \mathrm{Hz}$ (B correct). Beat frequency $=765-680=85\ \mathrm{Hz}$, not $80\ \mathrm{Hz}$, so C is wrong. The key idea is that the wall, being stationary, re-radiates exactly the frequency it receives, so the only Doppler shifts are those caused by the moving source in each of the two paths.