MAT Exam 2022 Paper

Concept:When the same distance is covered at two different speeds, the total difference in time taken is the sum of the late and early times. Use $\text{Distance} = \text{Speed} \times \text{Time}$.Explanation:Let the distance to school be $d$ km and the scheduled time be $T$ hours.6 minutes $= \frac{6}{60} = \frac{1}{10}$ hours.First day, speed $= \frac{5}{2}$ km/h.Time taken $= \frac{d}{5/2} = \frac{2d}{5}$ hours.He is late by 6 minutes, so $\frac{2d}{5} = T + \frac{1}{10}$.Next day, increased speed $= \frac{5}{2} + 1 = \frac{7}{2}$ km/h.Time taken $= \frac{d}{7/2} = \frac{2d}{7}$ hours.He reaches 6 minutes early, so $\frac{2d}{7} = T - \frac{1}{10}$.Subtract the second equation from the first:$\frac{2d}{5} - \frac{2d}{7} = \frac{1}{10} + \frac{1}{10}$$\frac{14d - 10d}{35} = \frac{1}{5}$$\frac{4d}{35} = \frac{1}{5}$$d = \frac{35}{20} = \frac{7}{4}$ km.Answer:The school is $\frac{7}{4}$ km from his house, which is Option B.