The foolish guy

Let,

time spend for the full journey be t,

distance between the 2 boatyards be d,

the uniform swimming speed of the man be s.

$\boxed{time=\frac{distance}{speed}}$ ,

$t=\frac{d}{4ms^{-1}}+\frac{d}{s-4ms^{-1}}$ ------------------- $\boxed{1}$

$t-15s=\frac{d}{10ms^{-1}}+\frac{d}{s-4ms^{-1}}$ ------------------ $\boxed{2}$

$\boxed{1}-\boxed{2}$

$15s=\frac{d}{4ms^{-1}}-\frac{d}{10ms^{-1}}$

$15s=\frac{10d-4d}{40ms^{-1}}$

$15s=\frac{6d}{40ms^{-1}}$

$6d=600m$

$d=100m$

So the answer in meters will be -------------------------------100

We know that: $distance=velocity \times time$

We don't know the time exactly but we know the speed. Imagine that t is time needed to cross the river with boat without any force by the man.

$d=4.t=4t$ and also $d=(4+6).(t-1/4minutes)=10t-2.5minutes$

$d=d$

$4t=10t-2.5minutes$

$6t=2.5minutes$

$6t=2.5x60second$

$t=25second$

input t to the distance will give us $d=4.25=100m$