A boat covers 24 , km upstream and 36 , km do | vedclass

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(C) Let the velocity of the boat in still water be $u \, km/h$ and the velocity of the current be $x \, km/h$.The speed upstream is $(u - x) \, km/h$

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$2$

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(C) Let the velocity of the boat in still water be $u \, km/h$ and the velocity of the current be $x \, km/h$.The speed upstream is $(u - x) \, km/h$ and the speed downstream is $(u + x) \, km/h$.According to the problem:$\frac{24}{u-x} + \frac{36}{u+x} = 6$ ... $(1)$$\frac{36}{u-x} + \frac{24}{u+x} = 6.5$ ... $(2)$Multiply equation $(1)$ by $1.5$:$1.5 	imes \left( \frac{24}{u-x} + \frac{36}{u+x} ight) = 1.5 	imes 6$$\frac{36}{u-x} + \frac{54}{u+x} = 9$ ... $(3)$Subtract equation $(2)$ from $(3)$:$\left( \frac{36}{u-x} + \frac{54}{u+x} ight) - \left( \frac{36}{u-x} + \frac{24}{u+x} ight) = 9 - 6.5$$\frac{30}{u+x} = 2.5$$u + x = \frac{30}{2.5} = 12 \, km/h$Substitute $(u + x) = 12$ into equation $(1)$:$\frac{24}{u-x} + \frac{36}{12} = 6$$\frac{24}{u-x} + 3 = 6$$\frac{24}{u-x} = 3$$u - x = \frac{24}{3} = 8 \, km/h$Now,we have:$u + x = 12$$u - x = 8$Subtracting the two equations: $(u + x) - (u - x) = 12 - 8$$2x = 4$$x = 2 \, km/h$Thus,the velocity of the current is $2 \, km/h$.

$A$ boat covers $24 \, km$ upstream and $36 \, km$ downstream in $6 \, hours,$ while it covers $36 \, km$ upstream and $24 \, km$ downstream in $6 \frac{1}{2} \, hours.$ Then the velocity of the current is..........$km/h$

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