Question:
A boat covers a round trip in a river in a certain time.If its speed in still water is doubled and the speed of the stream tripled it would take the same time for the round trip journey.Find the ratio of the speed of the boat in still water and the speed of the stream.
Solution:
A boat covers a round trip in a river in a certain time.
Let the speed of boat in still water= x
Let speed of the stream = y
Let d be the one way distance.
So, total time taken by boat in round trip = time taken in downstream + time taken in upstream
total time for round trip =\({d\over x+y}+{d\over x-y}\) .....1
If its speed in still water is doubled ie 2x and the speed of the stream tripled ie 3y.
Total time for round trip=\( {d\over 2x+3y}+{d\over 2x-3y}\) ....2
As time consumed same in the round trip journey.
From eqn 1 and 2
\({d \over x+y}+{d \over x-y}={d \over 2x+3y}+{d \over 2x-3y}\)
Cancel out d on both side.
\(\implies {1\over x+y}+{1 \over x-y}={1\over 2x+3y}+{1 \over 2x-3y}\)
\(\implies {{(x-y)+(x+y)}\over{(x+y)(x-y)}}={{(2x-3y)+(2x+3y)}\over {(2x+3y)(2x-3y)}}\)
\(\implies {{2x}\over{{x^2}-{y^2}}}={{4x}\over {{4x^2}-{9y^2}}}\)
\(\implies {{2x}\times {({4x^2}-{9y^2})}}={{4x}\times {({x^2}-{y^2})}}\)
\(\implies { {8x^3}-{18xy^2}}={{4x^3}-{4xy^2}}\)
\(\implies { {8x^3}-{4x^3}}={-{4xy^2}+{18xy^2}}\)
\(\implies { 4x^3}={14xy^2}\)
\(\implies {x^3\over xy^2}={14 \over 4}\)
\(\implies {x^2\over y^2}={14 \over 4}\)
\(\implies {x^2\over y^2}={14 \over 4}\)
\(\implies {x\over y}={\sqrt {14}\over 2}\) on taking square root on both sides
Ratio of the speed of the boat in still water and the speed of the stream
\(\implies {x : y}={\sqrt {14} : 2}\)