Let \(v_b\) be the speed of the boat in still water and \(v_w\) be the speed of water.

Travelling downstream: \(V = v_b + v_w \)

\(V_{down} = 8 + 2\)

\(V_{down} = 10\)

Travelling upstream: \(V_{up}= v_b - v_w \)

\(V_{up} = 8 - 2\)

\(V_{up} = 6\)

\(V = \frac{distance}{time}\)

Let \(t\) be the total travel time. Thus,
\(t = t_{up} + t_{down}\)
\(8 = t_{up} + t_{down}\).

The distance travelled in upstream is same distance travelled downstream, \(d\).

\(V_{down} = 10 = \frac{d}{t_{down}}\)

\( d = 10 t_{down}\)

\(V_{up} = 6 = \frac{d}{t_{up}}\)

\( d = 6 t_{up}\)

\(6 t_{up} = 10 t_{down}\)

\(t_{down} = \frac{3}{5} t_{up}\) and \(8 = t_{up} + t_{down}\).

\(8 = t_{up} +\frac{3}{5} t_{up} \).

\(8 = \frac{8}{5} t_{up} \).

\(\frac{40}{8} = t_{up} \).

\(t_{up} = 5 ~hrs\). Answer