Question

In a stream running at 3 km/h. a motorboat goes 12 km upstream and back to the starting point in 60 min.Find the speed of the motorboat in still water.
a) 2(2+√17) b) 2(4+√15) c) 3(4+√17) d) 3(2+√17)

Answer 1

Solution :-

Given :-

Speed of stream = 3km/h

Let :-

Speed of motorboat in still water = x km/h

Upstream :-

Distance travelled = 12 km

Speed = ( x - 3 ) km/h

Time taken = Distance/speed

                   $= \sf \dfrac{12}{x-3}$

Downstream :-

Distance travelled = 12 km

Speed = ( x + 3 ) km/h

Time taken = Distance/speed

                   $= \sf \dfrac{12}{x+3}$

According to the question :-

$\longrightarrow \sf \dfrac{12}{x-3} +\dfrac{12}{x+3} = \dfrac{60}{60} \\\\\longrightarrow 12 \left( \dfrac{1}{x-3} +\dfrac{1}{x+3} \right) = 1 \\\\\longrightarrow 12 \left( \dfrac{x+3+x-3}{(x-3)(x+3)} \right) = 1 \\\\\longrightarrow 12 \left( \dfrac{2x}{x^2-9} \right) = 1 \\\\\longrightarrow 12 \times 2x = x^2 - 9 \\\\\longrightarrow x^2-9 = 24 x \\\\\longrightarrow x^2-24x-9 = 0$

We know :-

$\boxed{\bf x = \dfrac{-b \pm \sqrt{b^2-4ac} }{ 2a}}$

$\longrightarrow \sf x = \dfrac{-(-24) \pm \sqrt{(-24)^2- ( 4 \times -9 \times 1)} }{2 \times 1} \\\\\longrightarrow x = \dfrac{24 \pm \sqrt{576 -(-36)} }{2} \\\\\longrightarrow x = \dfrac{24 \pm \sqrt{576+ 36 } }{2} \\\\\longrightarrow x = \dfrac{24 \pm \sqrt{612} }{2} \\\\\longrightarrow x = \dfrac{24 \pm \sqrt{153 \times 4} }{2} \\\\\longrightarrow x = \dfrac{24\pm 2 \sqrt{153} }{2} \\\\\longrightarrow x = 12 \pm \sqrt{153} \\\\\longrightarrow x = 12 \pm \sqrt{17 \times 9 }$

$\longrightarrow \sf x = 12 \pm 3\sqrt{17} \\\\\longrightarrow \bf x = 3(4 \pm \sqrt{17} )$

According to the given options,

$\sf x = 3( 4 + \sqrt{17} )$

∴ Option C is correct