Steven W. answered 08/19/16
Tutor
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Physics Ph.D., college instructor (calc- and algebra-based)
I know it does not matter at this point, but the previous solution will give you an incorrect answer. That is because, while it accurately includes the Doppler shift of the frequency heard by the observer directly from the source, it neglects to include that the frequency the observer hears echoing off the wall will be Doppler-shifted the other way, as the source is moving away from the wall. In this case, the frequency heard echoing off the wall is:
Fwall = (340/(v+vs))f = (340/(340+5))f = (340/345)f
So the beat frequency expression should actually be:
fbeat = (F - Fwall) = ((340/335)f - (340/345)f) = 5 Hz
Solving this expression for f will give one of the answers listed exactly.

Hilton T.
08/19/16