Q.A man swimming down stream overcomes a float at a point M.After travelling distance D,he turned back and passed the float at a distance of D/2 from the point M,then the ratio of speed of swimmer with respect to still water to the speed of river is:
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The float is swimming along with the stream with speed say, r
While the man swimming downstream has a speed of (s+r), where, s is the speed of his swimming in still water, and r the speed of the river stream.
After reaching point M, he travels a distance D, then turns back and passes the float at a distance of D/2 from the point M.
Therefore, swimming downstream, he travels distance D with speed (s+r), then swims upstream till distance D/2 with a speed of (s-r). By that time the float has travelled distance D/2 with a speed of r.
Total time taken by the man in his swim to and fro =D/(s+r) + D/2/(s-r)
And by the float = D/2/r
`D/(s+r) + D/(2(s-r)) = D/2r`
`rArr 1/(s+r) + 1/(2(s-r)) = 1/2r`
`rArr (2s-2r+s+r)/(2(s+r)(s-r)) = 1/2r`
`rArr 2r(3s-r)= 2 (s+r)(s-r)`
`rArr 3sr-r^2 = s^2-r^2`
`rArr s/r =3`
So, the ratio of speed of the man in still water to the speed of the river stream was 3.
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