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 motor boat in still water. (in km/h)

Speed of stream = 3 km/hr

Let the speed of motor boat be v.

Relative speed in upstream = v - 3 km/hr
Relative speed in upstream = v + 3 km/hr

Distance = 12 km

Time = 60 min = 1 hr

Time = Distance/speed 

$$\frac{12}{v - 3} + \frac{12}{v + 3} = 1$$

12(v + 3) + 12(v - 3) = (v + 3)(v - 3)

24v = $$v^2$$ - 3^2

$$v^2 -24v - 9 = 0$$

From Dharacharya Formula,

x = $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

x =  $$\frac{24 \pm \sqrt{24^2 - 4(1)(-9)}}{2}$$

x = $$\frac{24 \pm \sqrt{576 + 36}}{2}$$

x = $$12 \pm \sqrt{153}$$

x = $$12 \pm \sqrt{17 \times 9}$$

x =  $$3(4 \pm \sqrt{17})$$

According to option,

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