In a journey of three unequal laps, a car covers a distance of 200 km in 4 h in the first lap, while another 162 km at the speed of 15 m/s in the second lap. It covered the remaining distance of the final lap in 4 h such that the average speed of the car for entire journey was 50 km/h. What was the speed of the car in the third lap of the journey?

47 km/h

Distance covered in first lap = 200 km, Time taken = 4 h

Distance covered in second lap = 162 km, speed = 15 m/s

= 15 × \(\frac{18}{5}\)

= 54 km/h

Time taken = \(\frac{162}{54}\) = 3h

Let the distance covered in the 3rd lap = S km

Time taken = 4h

Average Speed = \(\frac{200 + 162 + S}{4 + 3 + 4}\)

According to question ,

  \(\frac{200 + 162 + S}{4 + 3 + 4}\) = 50km/h

 362 + x = 50 × 11

 x = 550 - 362 = 188 km

So, the speed in 3rd lap =  \(\frac{188}{ 4}\)

= 47km/h

Therefore, the speed of the car in the third lap of the journey is 47 km/h