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Question 10: A motorboat takes the passengers from Rishikesh to Haridwar and back. Both the cities, Rishikesh and Haridwar are located on the banks of River Ganga. During Kumbh Mela ,to earn more money, the owner of the motorboat decided to have more trips from Rishikesh to Haridwar and back, so he increased the speed of the motorboat in still water, by 50%. By increasing the speed, he was able to cut down the travel time from Rishikesh to Haridwar and back, by 60%. What is the ratio of the speed of motorboating still water to that of the speed of river Ganga?
Let the speed of the boat be ‘b', the speed of the stream be ‘s', and the distance travelled be ‘d'.
Then the speed of the boat going upstream will be b-s and going downstream will be b+s.
Time taken for a round trip: T = \\frac{d}{b+s}) + \\frac{d}{b-s})
Now when the owner increases the speed of the boat by 50%:
The new speed of boat will be 1.5b
The new speed of boat going upstream will be 1.5b-s and while going downstream it will be 1.5b+s.
Then the time taken for a round trip: \\frac{2}{5T})= \\frac{d}{1.5b+s}) + \\frac{d}{1.5b-s})
Dividing the two equations we get:
\\frac{2b}{b^2-s^2}) ˙ \\frac{2.25b^2-s^2}{3b}) = \\frac{5}{2})
\\frac{2.25b^2-s^2}{b^2-s^2}) = \\frac{5}{2}) * \\frac{3}{2})
4(2.25b2-s2) = 15(b2-s2)
9b2 – 4s2 = 15b2 – 15s2
11s2 = 6b2
So, \\frac{b}{s}) = √\\frac{11}{6})
The question is "What is the ratio of the speed of motorboating still water to that of the speed of river Ganga?"
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