The ratio of the speed of a motorboat to that | vedclass

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(C) Let the speed of the motorboat be $v_b = 36x$ and the speed of the current be $v_c = 5x$.The speed of the motorboat along the current (downstream)

Vedclass · Accepted Answer

$6 \, h \, 50 \, min$

Vedclass · Answer

(C) Let the speed of the motorboat be $v_b = 36x$ and the speed of the current be $v_c = 5x$.The speed of the motorboat along the current (downstream) is $v_d = v_b + v_c = 36x + 5x = 41x$.The speed of the motorboat against the current (upstream) is $v_u = v_b - v_c = 36x - 5x = 31x$.Let the distance be $d$. The time taken to travel downstream is $t_d = 5 \, h \, 10 \, min = 5 + \frac{10}{60} \, h = 5 + \frac{1}{6} \, h = \frac{31}{6} \, h$.Since $d = v_d 	imes t_d$,we have $d = 41x 	imes \frac{31}{6}$.The time taken to return (upstream) is $t_u = \frac{d}{v_u} = \frac{41x 	imes \frac{31}{6}}{31x} = \frac{41}{6} \, h$.Converting $\frac{41}{6} \, h$ into hours and minutes: $\frac{41}{6} \, h = 6 \frac{5}{6} \, h = 6 \, h + (\frac{5}{6} 	imes 60) \, min = 6 \, h \, 50 \, min$.

The ratio of the speed of a motorboat to that of the water current is $36 : 5$. The motorboat travels along with the current in $5 \, h \, 10 \, min$. Find the time taken by the motorboat to return.

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