Gently down the stream

Let us call the speed of the current $v$ .

Note the following: From our adventurous mathematician's dropping of his stick and pursuing the hat until the rescue of the hat the distance between the hat and the stick does not change, since they are both being moved by the current.

The man's speed relative to the stick after the pursuit is $\frac{3}{2}v+v=\frac{5}{2}v$ for $10$ minutes, because the stick and man are approaching one another. Moreover, this is equal to the the distance which he ran $\frac{4}{2}v \times x$ minutes.

Solve using proportions:

$\frac{5}{2}v\times 10=\frac{4}{2}v \times x$

$x=12.5$

He ran twice as fast as he walked, so he is $12.5+25=37.5$ minutes late.

Let us begin from the moment the mathematician throws the stick.

Let the speed of the current be $v$ . After throwing the stick, he turns back and goes after his hat at a speed of $3v$ . Let the time taken to catch up with his hat be $t_1$ . Given that distance = speed $\times$ time, we can say that the mathematician ran a distance of $3vt_1$ . During that period of time, the stick traveled a distance of $vt_1$ . The situation is illustrated in the diagram below:

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ <------- $(vt_1)$ --------

<------------------------------- $(3vt_1)$ -----------------------------

After grabbing his hat, the mathematician turned back to his original direction and walked at a speed of $1.5v$ . $10$ minutes later, he met his stick. During this time period, the mathematician walked a distance of $10 \times 1.5v = 15v$ . At the same time, the stick continued to move an additional distance of $10 \times v = 10v$ . The movement can be shown below:

$\quad\quad\quad\quad\quad\quad\quad\quad$ <---- $(10v)$ ----<------- $(vt_1)$ --------

----------- $(15v)$ --------->

<------------------------------- $(3vt_1)$ -----------------------------

From the diagram above, we know that:

$3vt_1 = 10v + 15v + vt_1$

$3t_1 = 25 + t_1 \implies 2t_1 = 25 \implies t_1=12.5$ .

Now, we want to find the number of minutes he was late. That would be the amount of time he spent running back to his hat and walking back to the point where he threw his stick. We know that he walked back in the initial direction at a speed of $1.5v$ . Now let $t_2$ be the time taken to walk back to the point where he threw the stick. The distance covered is the same, therefore:

$3vt_1 = 1.5vt_2$

$3(12.5) = 1.5t_2 \implies t_2 = 25$ .

Therefore, the total amount of minutes he was late is $t_1 + t_2 = 12.5 + 25 = \boxed{37.5}$