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?

The boat's speed in still water and the current speed are in the ratio 27:5.

Let the boat speed in still water = $27x$ and current speed = $5x$

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When going downstream (A to B), the water helps the boat:

Speed = $27x + 5x$

Speed = $32x$

When going upstream (B to A), the water opposes the boat:

Speed = $27x - 5x$

Speed = $22x$

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The distance from A to B equals the distance from B to A.

Time for downstream journey = 3 hours 40 minutes

Converting to hours: 3 hours 40 minutes = $3 + \frac{40}{60}$ = $3 + \frac{2}{3}$ = $\frac{11}{3}$ hours

Distance = Speed × Time

Distance = $32x \times \frac{11}{3}$

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For the upstream journey, using the same distance:

$22x \times \text{Time} = 32x \times \frac{11}{3}$

$\text{Time} = \frac{32x \times \frac{11}{3}}{22x}$

$\text{Time} = \frac{32 \times 11}{22 \times 3}$

$\text{Time} = \frac{352}{66}$

$\text{Time} = \frac{16}{3}$ hours

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Converting back to hours and minutes:

$\frac{16}{3}$ hours = $5\frac{1}{3}$ hours

$\frac{1}{3}$ hour = 20 minutes

Time = 5 hours 20 minutes