MAT Exam 2025 Paper

Concept:When rowing downstream, the current helps the boat, so effective speed is $x + c$.When rowing upstream, the current resists the boat, so effective speed is $x - c$.Here, $x$ is the speed in still water and $c$ is the current speed.Explanation:Upstream time is given as $24$ minutes.Convert this into hours: $24$ minutes $= \frac{24}{60} = 0.4$ hours.Let the downstream time be $t$ hours.The upstream time is one-fifth more than the downstream time.Therefore: $0.4 = t + \frac{1}{5}t = \frac{6}{5}t$.Solving for $t$: $t = 0.4 \times \frac{5}{6} = \frac{1}{3}$ hours.Downstream speed $= \frac{10 \text{ km}}{\frac{1}{3} \text{ h}} = 30$ km/h.Upstream speed $= \frac{10 \text{ km}}{0.4 \text{ h}} = 25$ km/h.Now form the equations:Downstream: $x + c = 30$.Upstream: $x - c = 25$.Add the two equations: $(x + c) + (x - c) = 30 + 25$.This gives $2x = 55$.So, $x = \frac{55}{2} = 27.5$ km/h.Hence, the speed of rowing in still water is $27.5$ km/h.Answer:Option A: $27.5$ kmph.