A car moving in the morning fog passes a man walking at 4km/h. in the same direction. The man can see the car for 3 minutes and visibility is upto a distance of 130m. The speed of the car is :

Let the speed of car = $$s$$ km/hr and speed of man = 4 km/hr

Since, they both are moving in same direction, thus relative speed = $$(s-4)$$ km/hr

Visibility of the car = 130 m for 3 minutes

=> speed = $$\frac{130}{3}$$ m/min

= $$(\frac{130}{3} \times \frac{60}{1000})$$km/hr = $$\frac{13}{5}$$ km/hr

According to ques, => $$(s-4)=\frac{13}{5}$$

=> $$s=\frac{13}{5}+4$$

=> $$s=\frac{33}{5}=6\frac{3}{5}$$ km/hr

=> Ans - (B)