Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minutes late. His usual time (in hours) to reach the destination is:
Step 1 — Use the inverse ratio of speed and time:
For a fixed distance, time is inversely proportional to speed. At 60% speed (\(\tfrac{3}{5}S\)), the new time is \(\tfrac{5}{3}T\).
Step 2 — Express the delay in hours:
\(1\text{ hr }40\text{ min}=\tfrac{5}{3}\) hours.
Step 3 — Form and solve the equation:
\[\tfrac{5}{3}T-T=\tfrac{5}{3}\;\Rightarrow\;\tfrac{2}{3}T=\tfrac{5}{3}\;\Rightarrow\;T=\tfrac{5}{2}\]
Therefore the usual time is \(\tfrac{5}{2}=\mathbf{2\tfrac{1}{2}}\) hours.
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