Question

The speed of a motorboat in still water is 25 km/hr. In a river, it goes 60 km downstream

and comes back the same distance upstream in 5 hours. Find the speed of the current of the river. (Speed of the current of the river is less than the speed of the motorboat in still water.)​

Answer 1

Answer:

I hope that you are able to understand.

Answer 2

Solution :

$\underline{\bf{Given\::}}$

The speed of a motorboat in still water is 25 km/hrs, it goes 60 km downstream & comes back the same distance upstream in 5 hours .

$\underline{\bf{Explanation\::}}$

Let the speed of current of the river be r km/hrs.

The speed of the motorboat in still water = 25 km/hrs.

The speed of motorboat for upstream = (25 - r)

The speed of motorboat for downstream = (25 + r)

The distance covered by motorboat = 60 km

As we know that formula of the time;

${\boxed{\bf{Time=\frac{Distance}{Speed} }}}$

A/q

$\longrightarrow\sf{\dfrac{60}{25-r} +\dfrac{60}{25+r} =5}\\\\\\\longrightarrow\sf{\dfrac{60(25+r) + 60(25-r)}{(25)^{2} - (r)^{2}} =5}\\\\\\\longrightarrow\sf{\dfrac{60(25\cancel{+r} + 25 \cancel{- r} )}{(25)^{2} -(r)^{2} } =5}\\\\\\\longrightarrow\sf{\dfrac{60\times 50}{625-(r)^{2}} =5}\\\\\\\longrightarrow\sf{3000=5(625 -r^{2})}\\\\\longrightarrow\sf{3000 = 3125 - 5r^{2}}\\\\\longrightarrow\sf{-5r^{2} =3000-3125}\\\\\longrightarrow\sf{-5r^{2} =-125}$

$\longrightarrow\sf{r^{2} = \cancel{-125/-5}}\\\\\longrightarrow\sf{r^{2} = 25}\\\\\longrightarrow\sf{r=\pm\sqrt{25} }\\\\\longrightarrow\sf{r=\pm5}\\\\\longrightarrow\bf{ r = 5 \:\:Or\:\:r\neq -5}$

As we know that negative value is not acceptable.

Thus;

The speed of the current of the river will be 5 km/hrs .