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A man covers equal distances at 10 km/h, 15 km/h, and 30 km/h. Find average speed.
16.36┬аkm/h
15┬аkm/h
18┬аkm/h
20┬аkm/h
16.36┬аkm/h
Assume the total distance for each leg to be the LCM of speeds (10,15,30), which is 30┬аkm. Then total distance = 3├Ч30=90┬аkm, and total time = 3+2+1=6┬аhours, giving average speed = 90/6=15┬аkm/h? Wait, let's recalculate carefully: LCM of 10, 15, 30 is 30. Distance for each = 30 km. Time taken = 30/10 + 30/15 + 30/3 = 3 + 2 + 1 = 6 hours. Total distance = 3 * 30 = 90 km. Average speed = 90 / 6 = 15 km/h.
Equal distances covered at speeds of 10┬аkm/h, 15┬аkm/h, and 30┬аkm/h.
Average┬аSpeed=v1тАЛdтАЛ+v2тАЛdтАЛ+v3тАЛdтАЛ3dтАЛ=v1тАЛ1тАЛ+v2тАЛ1тАЛ+v3тАЛ1тАЛ3тАЛ
Assume the total distance for each leg to be the LCM of speeds (10,15,30), which is 30┬аkm. Then total distance = 3├Ч30=90┬аkm, and total time = 3+2+1=6┬аhours, giving average speed = 90/6=15┬аkm/h? Wait, let's recalculate carefully: LCM of 10, 15, 30 is 30. Distance for each = 30 km. Time taken = 30/10 + 30/15 + 30/3 = 3 + 2 + 1 = 6 hours. Total distance = 3 * 30 = 90 km. Average speed = 90 / 6 = 15 km/h.
Students incorrectly take the arithmetic mean of the speeds: (10+15+30)/3=55/3=18.33, which leads to a wrong answer.
Identify Given Data and Formula
Since the distances are equal, the harmonic mean formula for average speed is applied: Average┬аSpeed=v1тАЛ1тАЛ+v2тАЛ1тАЛ+v3тАЛ1тАЛ3тАЛ
Average┬аSpeed=101тАЛ+151тАЛ+301тАЛ3тАЛ
Calculate Total Time for Equal Distances
Assume a convenient distance for each segment, equal to the LCM of 10,15, and 30, which is 30┬аkm. The times taken for the three legs are 30/10=3┬аhrs, 30/15=2┬аhrs, and 30/30=1┬аhr.
Total┬аTime=3+2+1=6┬аhours
Compute Average Speed
Total distance covered for the three equal segments is 3├Ч30=90┬аkm. Divide total distance by total time to get the average speed.
Average┬аSpeed=690тАЛ=15┬аkm/h
B is correct because the calculated average speed using the harmonic mean for equal distances is 15┬аkm/h.
Whenever distances are equal, the average speed can always be efficiently solved using the harmonic mean or by assuming a distance equal to the LCM of the speeds.
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