Question

A boat sails downstream from point $A$ to point $B$, which is $10$ km away from $A$ and then returns to $A$. If the actual speed of the boat (in still water) is $3$ km/h, the trip from $A$ to $B$ takes $2$ h $30$ min less than that from $B$ to $A$. What must the actual speed of the boat be for the trip from $A$ to $B$ to take $2$ hours?

Answer 1

speed $=\frac{distance}{time}⟹time=\frac{distance}{speed}$

let speed of flow$=\text{x}$

speed in ascent$=$speed of boat in still water $+$ speed of water flow

speed in decent$=$speed of boat in still water$-$ speed of water flow

let time of decent$=\text{y}$

$⟹\text{time of ascent}=\text{y}-2.5$

time of ascent$=\frac{10}{3+\text{x}}$

time of decent$=\frac{10}{3-\text{x}}$

$⟹\frac{10}{3+\text{x}}=\frac{10}{3-\text{x}}-2.5$

$10(3-\text{x})-10(3+\text{x})=-2.5(3+\text{x})(3-\text{x})$

$⟹{\text{x}}^2+8\text{x}-9=0$

$\text{x}=1(\because \text{x}\text{is positive})$

given time of ascent $=2$

$2=\frac{10}{\text{actual speed of boat}+1}$

$⟹$ actual speed of boat required is $\text{4km/hr}$