MAT Exam 2021 Paper

Concept:Time difference is written using distance and speed, then solved for distance.Explanation:Let distance be $d$ km and normal speed be $s$ km/h.When speed increases by $4$ km/h, time reduces by $30$ min $= 0.5$ h.So,$$\frac{d}{s} - \frac{d}{s+4} = \frac{30}{60}$$This gives,$$8d = s^2 + 4s \quad \text{...(1)}$$When speed decreases by $2$ km/h, time increases by $20$ min $= \frac{1}{3}$ h.So,$$\frac{d}{s-2} - \frac{d}{s} = \frac{20}{60}$$This gives,$$6d = s^2 - 2s \quad \text{...(2)}$$From equations (1) and (2),$$\frac{s^2 + 4s}{8} = \frac{s^2 - 2s}{6}$$$$6s^2 + 24s = 8s^2 - 16s$$$$2s^2 - 40s = 0$$$$2s(s - 20) = 0$$Since speed cannot be zero,$$s = 20 \text{ km/h}$$Using equation (2),$$d = \frac{20^2 - 2 \times 20}{6}$$$$d = \frac{400 - 40}{6} = \frac{360}{6} = 60 \text{ km}$$Answer:Distance covered is $60$ km.Correct option: B.