In a journey of three unequal laps a car covers a distance of 200 km in 4 h in the first lap while a...........

Quantitative Aptitude (CPO) Time & distance

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?

This questions was previously asked in

SSC CPO 10th Nov 2022 Shift 2

47 km/h

52 km/h

42 km/h

45 km/h

- The total journey consists of three laps with an average speed of 50 km/h.

- First Lap: Distance = 200 km, Time = 4 hours, Speed = 50 km/h.

- Second Lap:

- Distance = 162 km.

- Speed = 15 m/s = 54 km/h.

- Time = Distance/Speed = 162 km / 54 km/h = 3 hours.

- Total distance of the journey = Distance of all laps = \(200 + 162 + d_3\).

- Total time = Time of all laps = \(4 + 3 + 4 = 11\) hours.

- Total distance in the entire journey = Average Speed × Total Time = 50 km/h × 11 h = 550 km.

- Hence, Distance of third lap (\(d_3\)) = Total Distance - (Distance of first + second lap) = 550 - (200 + 162) = 188 km.

- Speed in third lap = \(d_3\) / Time in third lap = 188 km / 4 h = 47 km/h.

- The speed in the third lap was 47 km/h.

Correct Answer: Option 1, 47 km/h

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