So you have defined two different functions. You have
$\displaystyle s:[0,20] \to [0,1]$
and
$\displaystyle r:[0,10] \to [0,1]$
For these functions to be compatable they must agree on where you define the top and the bottom.
If you defince the top to 1 and the bottom to be 0 then you must have these values
$\displaystyle s(0)=0 \quad s(20)=1$
and
$\displaystyle r(0)=1 \quad r(10)=0$
So you get the two functions
$\displaystyle s(t)=\frac{t}{20} \quad r(t)=1-\frac{t}{10}$
Now if you define their difference it will only exist on the intersection of their domains!
$\displaystyle f:[0,10] \to [-1,.5] \quad f(t) =s(t)-r(t)=\frac{3t}{20}-1$
But notice that
$\displaystyle f(0) =-1 \text{ and } f(10)=.5$
Since f is continous the intermediate value theorem says there exists a $\displaystyle c \in [0,10]$ such that $\displaystyle f(c)=0$
So we get that
$\displaystyle f(c)=s(c)-r(c) =0 \iff 0=s(c)-r(c) \iff s(c)=r(c)$
So the hiker was at the same place at the same time.
Note: Since the function is linear would could solve for the value of t explicitly, but the power of the IVT is we don't have to be able to find it.