A man rows upstream 12 text{ km} and downstre | vedclass

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Question

(D) Let the man's rowing speed in still water be $x 	ext{ km/h}$ and the speed of the current be $y 	ext{ km/h}$.Speed upstream $= (x - y) 	ext{ km

Vedclass · Accepted Answer

$1 \frac{3}{5}$

Vedclass · Answer

(D) Let the man's rowing speed in still water be $x 	ext{ km/h}$ and the speed of the current be $y 	ext{ km/h}$.Speed upstream $= (x - y) 	ext{ km/h}$.Speed downstream $= (x + y) 	ext{ km/h}$.Given that the time taken for both trips is $5 	ext{ hours}$,we have:$x - y = \frac{12}{5} = 2.4 	ext{ km/h}$ (Equation $1$)$x + y = \frac{28}{5} = 5.6 	ext{ km/h}$ (Equation $2$)Subtracting Equation $1$ from Equation $2$:$(x + y) - (x - y) = 5.6 - 2.4$$2y = 3.2$$y = 1.6 	ext{ km/h}$.Converting $1.6$ to a fraction: $1.6 = \frac{16}{10} = \frac{8}{5} = 1 \frac{3}{5} 	ext{ km/h}$.

$A$ man rows upstream $12 \text{ km}$ and downstream $28 \text{ km}$,taking $5 \text{ hours}$ each time. The velocity of the water current is ......... $\text{km/h}$.

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