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)
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