Given:
A boat covers a round trip journey between two points A and B in a river in T hours
Formula used:
Downstream = (u + v) km/hr
Upstream = (u – v) km/hr
Calculation:
Let the speed of boat in still water be xkm/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) = 161T
⇒ d(x - y + x + y)/ x2 - y2 = 161T
⇒d(2x)/161(x2 - y2) = T ......... (1)
Again,
⇒ d/(2x + y) + d/(2x – y) = 80T
⇒d(2x - y + 2x + y)/(2x)2 - (y)2 = 80T
⇒d(4x) / 80(4x2 - y2) = T ..........(2)
Now equalling both the equations,
⇒d(2x)/161(x2 - y2) = d(4x) / 80(4x2 - y2)
⇒1/161(x2 - y2) = 2/ 80(4x2 - y2)
⇒1/(161x2 - 161y2) = 1/(160x2 - 40y2)
⇒160x2 - 40y2 = 161x2 - 161y2
⇒x2 = 121y2
⇒x/y = 11/1
∴ The required ratio is 11 : 1
Alternate Method
Let the speed of boat in still water be xkm/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
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Hint
In this type of question, always try to make two equations equal to the same variable so that we can equalize the two equations.