recall your d = rt, distance = rate * time.
the lake is only 3 miles wide, and these guys are really fast.
now, when Jim and John meet, Jim has covered say "d" miles, and John has covered by then, the slack from 3 and d, namely "3 - d", the lake is only 3 miles wide.
By the time they meet, Jim has been paddling "t" hours, and John has been paddling "t" hours as well, because both started at the same time, though John with the faster rate, covered more ground.
[tex]\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Jim&d&4&t\\
John&3-d&5&t
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=4t\\
3-d=5t\\
------\\
3-\boxed{4t}=5t
\end{cases}
\\\\\\
3=9t\implies \cfrac{3}{9}=t\implies \stackrel{hours}{\cfrac{1}{3}}=t\implies \stackrel{minutes}{20}[/tex]
