Multiple choice

Speeds of Boat A and B in still water are in the ratio of 3:2 Rate of current is 10 Km/hr. Both Boats started from Point P to point Q downstream at the same time. After Boat B reaching Point Q, in return journey, it is powered by engine due to which the speed of the boat in still water is increased by 70%, while retuned Boat A returned to Point Q as usual. Both the boats returned back to point P at the same time. Then what is the speed of Boat A?

  1. 20 Km/hr

  2. 30 Km/hr

  3. 40 Km/hr

  4. 50 Km/hr

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let Boat A's still-water speed be 3x and Boat B's be 2x. Equating their total travel times, including Boat B's increased return speed, gives x = 10, so Boat A's speed is 3x = 30 km/hr.

AI explanation

Let the speeds of boat A and B in still water be 3x and 2x respectively; since the distance is constant, the times to travel downstream to point Q are proportional to their downstream speeds, so boat A takes time D divided by 3x plus 10 and boat B takes D divided by 2x plus 10. For the return upstream trip, boat B's new speed is 1.7 times 2x, making its upstream speed 3.4x minus 10, while boat A's upstream speed is 3x minus 10, and since their total travel times are equal, D divided by 3x plus 10 equals D divided by 2x plus 10 plus D divided by 3.4x minus 10. Dividing by D and solving this equation yields x equals 10, so the speed of boat A is 3x, resulting in 30 km/hr.

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