A motor boat whose speed is 18, km/h in still | vedclass

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(C) Let the speed of the stream be $x\, km/h$.Therefore,the speed of the boat upstream is $(18-x)\, km/h$ and the speed of the boat downstream is $(18

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$6$

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(C) Let the speed of the stream be $x\, km/h$.Therefore,the speed of the boat upstream is $(18-x)\, km/h$ and the speed of the boat downstream is $(18+x)\, km/h$.The time taken to go upstream is $\frac{	ext{distance}}{	ext{speed}} = \frac{24}{18-x}\, 	ext{hours}$.Similarly,the time taken to go downstream is $\frac{24}{18+x}\, 	ext{hours}$.According to the problem,the difference in time is $1\, 	ext{hour}$:$\frac{24}{18-x} - \frac{24}{18+x} = 1$Multiplying by $(18-x)(18+x)$,we get:$24(18+x) - 24(18-x) = (18-x)(18+x)$$432 + 24x - 432 + 24x = 324 - x^2$$48x = 324 - x^2$$x^2 + 48x - 324 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-48 \pm \sqrt{48^2 - 4(1)(-324)}}{2(1)}$$x = \frac{-48 \pm \sqrt{2304 + 1296}}{2} = \frac{-48 \pm \sqrt{3600}}{2}$$x = \frac{-48 \pm 60}{2}$Since speed cannot be negative,we take $x = \frac{-48 + 60}{2} = \frac{12}{2} = 6\, km/h$.

$A$ motor boat whose speed is $18\, km/h$ in still water takes $1\, hour$ more to go $24\, km$ upstream than to return downstream to the same spot. Find the speed of the stream. (in $ km/h$)

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