The speed of a motorboat in still water and that of the current of water is in a ratio of 27:5. The boat goes along the current from point A to point B in 3 hours 40 minutes. How much time will it take to come back from B to A?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5 hours 20 minutes
Let the speeds of the boat in still water and of the current be 27k and 5k. Then
\(\text{Downstream speed} = 27k + 5k = 32k\)
\(\text{Upstream speed} = 27k - 5k = 22k\)
Going along the current takes 3 hours 40 minutes, that is \(\frac{11}{3}\) hours, so
\( \displaystyle AB = 32k \times \frac{11}{3} \)
Coming back is against the current, at speed 22k:
\( \displaystyle \text{Time} = \frac{32k \times 11}{3 \times 22k} = \frac{16}{3} \text{ hours} \)
\( \displaystyle \frac{16}{3} \text{ hours} = 5 \text{ hours } 20 \text{ minutes} \)