Speed!

Let $t =$ Time taken , $s =$ Distance travelled and $a =$ Overall average speed

At $20mh^{-1}$ it takes the man (let's call him Steve) $t$ time in hours to get to his location which is $s$ miles away

At $30mh^{-1}$ it takes Steve $\frac {2t}{3}$ hours to go $s$ miles

We don't know $s$ or $t$ but that doesn't matter.

The equation for average speed is

$a = \frac {\bigtriangleup s}{\bigtriangleup t}$

Okay so maybe we do need to know $\bigtriangleup s$

$\bigtriangleup s = 20(t) + 30(\frac {2t}{3}) = 40t$

And $\bigtriangleup t$

$\bigtriangleup t = t + \frac {2t}{3} = \frac {5t}{3}$

So

$a = \frac {40t}{\frac {5t}{3}} = 24$

So the answer is $24$

for average speed use 2.v1.v2\v1+v2

=2.30.20\50=24

if speeds are x and y distance/hr. while coming and going

then avg. speed = 2xy/(x + y)