Question

sailor goes 8 km downstream and return in 1 hour . determine the speed of the sail in still water and speed of the current​

Answer 1

Answer:

$x = 10km \: per \: h \\ y = \: \: 2km \: per \: h$

Step-by-step explanation:

$let \: us \: take \: \\ speed \: of \: boat = x \: km \: per \: hour \\ speed \: of \: current = y \: km \: per \: hour$

remaining i will attach

hope it helps!!

Answer 2

Step-by-step explanation:

$We \: know, \\ \text{Speed}=\frac{\text {Distance}}{\text {Time}}$

$\begin{lgathered}\begin{array}{l}{(x+y)=\frac{8}{\frac{40}{60}} \frac{k m}{h r}} \\ {(x+y)=\frac{8 \times 60}{40} \frac{k m}{h r}}\end{array}\end{lgathered} </em></p><p><em>[tex]\begin{lgathered}\begin{array}{l}{(x+y)=\frac{8}{\frac{40}{60}} \frac{k m}{h r}} \\ {(x+y)=\frac{8 \times 60}{40} \frac{k m}{h r}}\end{array}\end{lgathered} \:$

$</em></p><p><em>[tex]\begin{lgathered}\begin{array}{l}{x+y=12 \quad \rightarrow(i)} \\ {(x-y)=\frac{8}{1} \frac{k m}{h r}} \\ {x-y=8 \quad \rightarrow(i i)}\end{array}\end{lgathered} </em></p><p><em>[tex]\begin{lgathered}\begin{array}{l}{x+y=12 \quad \rightarrow(i)} \\ {(x-y)=\frac{8}{1} \frac{k m}{h r}} \\ {x-y=8 \quad \rightarrow(i i)}\end{array}\end{lgathered} \:$

$</em></p><p></p><p><em>[tex](i) \rightarrow x+y=12 \\ </em></p><p></p><p><em>[tex](i) \rightarrow x+y=12 \\ (ii) \rightarrow x-y= 8$

On solving equation (i) and equation (ii), we get

On solving equation (i) and equation (ii), we get2 x = 20

$x=\frac{20}{2} \:$

x=10

Substitute x=10 in (i), we get,

Substitute x=10 in (i), we get,10+y=12

Substitute x=10 in (i), we get,10+y=12y=12-10

Substitute x=10 in (i), we get,10+y=12y=12-10y=2

Substitute x=10 in (i), we get,10+y=12y=12-10y=2Hence, the