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.
Answer & explanation
Correct answer: option 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