Walking $\frac{6}{7}$ th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is

### Concept & Inverse Proportionality For a fixed distance, time and speed are inversely related. If speed becomes $\frac{a}{b}$ of its usual value, the time taken becomes $\frac{b}{a}$ of its usual value. ### Step-by-Step Solution * Let the usual time taken be $T$. * The new speed is $\frac{6}{7}$ of the usual speed. * Therefore, the new time taken is $\frac{7}{6}$ of the usual time, which is $\frac{7}{6}T$. * The difference between the new time and the usual time is the amount he is late. * Late time = $\frac{7}{6}T - T = \frac{1}{6}T$. * We are given that he is 12 minutes late, so $\frac{1}{6}T = 12$ minutes. * $T = 12 \times 6 = 72$ minutes. * Convert minutes to hours: 72 minutes = 1 hour 12 minutes. ### Exam Strategy & Shortcut If new speed is $\frac{x}{y}$ of original speed (where $x < y$), then usual time $= \frac{x}{y - x} \times$ (late time). Here, $x=6, y=7$. Usual time $= \frac{6}{7 - 6} \times 12 = 6 \times 12 = 72$ minutes, which is 1 hr 12 min. ### Common Pitfall Failing to convert the final answer from minutes (72) back into the hour-minute format required by the options. ### Final Answer Therefore, the correct answer is **1 hr 12 min**.