\nLet the distance to the destination be D.
\nLet A’s speed be $V_A$ and B’s speed be $V_B$.
\nWe are given that B travelled at $\\frac{5}{7}$ of A’s speed, so $V_B = \\frac{5}{7} V_A$.
\nThe time taken by A to reach the destination is $T_A = \\frac{D}{V_A}$.
\nThe time taken by B to reach the destination is $T_B = \\frac{D}{V_B}$.
\nSubstitute $V_B = \\frac{5}{7} V_A$ into the expression for $T_B$:
\n$T_B = \\frac{D}{(5\/7)V_A} = \\frac{7D}{5V_A} = \\frac{7}{5} \\left(\\frac{D}{V_A}\\right) = \\frac{7}{5} T_A$.
\nWe are given that B reached the destination 1 hour 20 minutes after A.
\n1 hour 20 minutes $= 1 + \\frac{20}{60}$ hours $= 1 + \\frac{1}{3}$ hours $= \\frac{4}{3}$ hours.
\nThe difference in time is $T_B – T_A = \\frac{4}{3}$ hours.
\nSubstitute $T_B = \\frac{7}{5} T_A$:
\n$\\frac{7}{5} T_A – T_A = \\frac{4}{3}$.
\n$(\\frac{7}{5} – 1) T_A = \\frac{4}{3}$.
\n$(\\frac{7-5}{5}) T_A = \\frac{4}{3}$.
\n$\\frac{2}{5} T_A = \\frac{4}{3}$.
\nNow solve for $T_A$:
\n$T_A = \\frac{4}{3} \\times \\frac{5}{2} = \\frac{20}{6} = \\frac{10}{3}$ hours.
\nWe need to find $T_B$:
\n$T_B = \\frac{7}{5} T_A = \\frac{7}{5} \\times \\frac{10}{3} = \\frac{7 \\times 2}{3} = \\frac{14}{3}$ hours.
\nConvert $\\frac{14}{3}$ hours into hours and minutes.
\n$\\frac{14}{3} = 4 \\frac{2}{3}$ hours.
\n$\\frac{2}{3}$ hours $= \\frac{2}{3} \\times 60$ minutes $= 40$ minutes.
\nSo, $T_B = 4$ hours 40 minutes.
\n<\/section>\n