\[ \textbf{Reasoning:} \] Let us examine the height at the end of each climbing phase (i.e., just before the slip). $$ \text{ After the 1st climb (end of first hour): height \(=30\) ft.} $$ $$ \text{ After the 1st slip (immediately after): height \(=30-20=10\) ft.} $$ $$ \text{ After the 2nd climb: height \(=10+30=40\) ft.} $$ $$ \text{ After the 2nd slip: height \(=40-20=20\) ft.} $$ \end So the heights just {at the end of the \(k\)-th climb} form the sequence \[ 30,\;40,\;50,\;\dots \] which can be written as \[ H_k = 30 + (k-1)\cdot 10 = 20 + 10k\quad\text{(after the \(k\)-th climb, before slipping).} \] We need the smallest \(k\) with \(H_k \ge 120\): \[ 20+10k \ge 120 \;\Longrightarrow\; 10k \ge 100 \;\Longrightarrow\; k \ge 10. \] Thus the monkey reaches \(120\) ft exactly at the end of the \(10\)-th climb. \[ \text{Start time } = 8{:}00\ \text{a.m.} \quad\Rightarrow\quad 8{:}00 + 10\ \text{hours} = 6{:}00\ \text{p.m.} \] \[ \boxed{\text{The monkey first touches the flag at }6{:}00\ \text{p.m.}} \]