A man can row against the current 3/4 of a ki | vedclass

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(D) Let the speed of the boat be $x$ $km/h$ and the speed of the stream be $y$ $km/h$.Speed against the current (upstream) $= x - y = \frac{	ext{Dist

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$5:1$

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(D) Let the speed of the boat be $x$ $km/h$ and the speed of the stream be $y$ $km/h$.Speed against the current (upstream) $= x - y = \frac{	ext{Distance}}{	ext{Time}} = \frac{3/4 	ext{ km}}{15/60 	ext{ h}} = \frac{3}{4} 	imes 4 = 3 	ext{ km/h}$.So,$x - y = 3$ ... $(i)$Speed with the current (downstream) $= x + y = \frac{	ext{Distance}}{	ext{Time}} = \frac{3/4 	ext{ km}}{10/60 	ext{ h}} = \frac{3}{4} 	imes 6 = 4.5 	ext{ km/h} = \frac{9}{2} 	ext{ km/h}$.So,$x + y = 4.5$ ... $(ii)$Adding equations $(i)$ and $(ii)$:$(x - y) + (x + y) = 3 + 4.5$$2x = 7.5 \Rightarrow x = 3.75 	ext{ km/h} = \frac{15}{4} 	ext{ km/h}$.Substituting $x$ in equation $(i)$:$3.75 - y = 3 \Rightarrow y = 0.75 	ext{ km/h} = \frac{3}{4} 	ext{ km/h}$.The ratio of the speed of the man to the speed of the current is $\frac{x}{y} = \frac{15/4}{3/4} = \frac{15}{3} = \frac{5}{1}$.Thus,the ratio is $5:1$.

$A$ man can row against the current $3/4$ of a kilometre in $15$ $min$ and returns the same distance in $10$ $min.$ The ratio of his speed to that of the current is:

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