A man can row 30 text{ km} upstream and 44 te | vedclass

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Question

(A) Let the upstream speed be $x 	ext{ km/h}$ and the downstream speed be $y 	ext{ km/h}$.According to the problem,we have the equations:$\frac{30}{

Vedclass · Accepted Answer

$3 	ext{ km/h}, 8 	ext{ km/h}$

Vedclass · Answer

(A) Let the upstream speed be $x 	ext{ km/h}$ and the downstream speed be $y 	ext{ km/h}$.According to the problem,we have the equations:$\frac{30}{x} + \frac{44}{y} = 10$  --- $(1)$$\frac{40}{x} + \frac{55}{y} = 13$  --- $(2)$Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. Then the equations become:$30u + 44v = 10$  --- $(3)$$40u + 55v = 13$  --- $(4)$Multiply equation $(3)$ by $4$ and equation $(4)$ by $3$:$120u + 176v = 40$$120u + 165v = 39$Subtracting the two equations: $11v = 1 \implies v = \frac{1}{11}$.Substituting $v = \frac{1}{11}$ into equation $(3)$: $30u + 44(\frac{1}{11}) = 10 \implies 30u + 4 = 10 \implies 30u = 6 \implies u = \frac{1}{5}$.Thus,$x = 5 	ext{ km/h}$ and $y = 11 	ext{ km/h}$.Speed of the man in still water $= \frac{y + x}{2} = \frac{11 + 5}{2} = 8 	ext{ km/h}$.Rate of the current $= \frac{y - x}{2} = \frac{11 - 5}{2} = 3 	ext{ km/h}$.

$A$ man can row $30 \text{ km}$ upstream and $44 \text{ km}$ downstream in $10 \text{ hours}$. Also,he can row $40 \text{ km}$ upstream and $55 \text{ km}$ downstream in $13 \text{ hours}$. Find the rate of the current and the speed of the man in still water.

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