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Question

(C) Let the total distance be $D$.Distance covered in mud $= \frac{D}{6}$.Distance covered on tar road $= \frac{D}{2}$.Distance covered by boat (water

Vedclass · Accepted Answer

$10:15:18$

Vedclass · Answer

(C) Let the total distance be $D$.Distance covered in mud $= \frac{D}{6}$.Distance covered on tar road $= \frac{D}{2}$.Distance covered by boat (water) $= D - (\frac{D}{6} + \frac{D}{2}) = D - (\frac{D+3D}{6}) = D - \frac{4D}{6} = D - \frac{2D}{3} = \frac{D}{3}$.Given the ratio of speeds in mud,water,and tar road is $3:4:5$. Let the speeds be $3v, 4v, 5v$ respectively.Time taken for mud $= \frac{D/6}{3v} = \frac{D}{18v}$.Time taken for water $= \frac{D/3}{4v} = \frac{D}{12v}$.Time taken for tar road $= \frac{D/2}{5v} = \frac{D}{10v}$.Ratio of times $= \frac{D}{18v} : \frac{D}{12v} : \frac{D}{10v} = \frac{1}{18} : \frac{1}{12} : \frac{1}{10}$.To simplify,multiply by the $LCM$ of $18, 12, 10$,which is $180$.Ratio $= (\frac{1}{18} 	imes 180) : (\frac{1}{12} 	imes 180) : (\frac{1}{10} 	imes 180) = 10 : 15 : 18$.

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