A boat covers a roundtrip journey between two points A and B in a river in T hours. If its speed in still water becomes 2 times, it would take $\frac{80}{161}$ hours for the same journey. Find the ratio of its speed in still water to the speed of the river.

11 : 1

Let the speed of boat in still water = x km/hr

Speed of stream = y km/hr

Let the distance between them be d 

Downstream = (x + y) km/hr

Upstream = (x – y) km/hr

According to the question

= d/(x + y) + d/(x – y) = T

= d(1/x + y) + (1/x – y) = T

= d[(x – y) + (x + y)/(x² – y²)] = T

= d[(2x/(x² – y²) = T

= d/T = x² – y²/2x .....(1)

Again,

= d/(2x + y) + d/(2x – y) = 80T/161

= d(1/2x + y + 1/2x – y) = 80T/161

= d(2x – y+ 2x + y)/4x² – y² = 80T/161

= d(4x/4x² – y²) = 80T/161

= d/T = 80(4x² – y²)/4x × 161 ......(2)

Now, substituting both the equation, we get

d/T = x² – y²/2x =  80(4x² – y²)/4x × 161

= x² – y² = 40(4x² – y²)/161

= (161x² – 161y²) = (160x² – 40y²)

= (161x² – 160x²) = (161y² – 40y²)

= x² = 121y²

= x/y = 11/1

∴ The required ratio is 11 : 1