The following table shows the average speed (in km per hour) of five different trains A-E during six days of a week from Monday through Saturday. Some data is missing in the table ( indicated as '-') that you are expected to calculate, if required. Based on the data in the table, answer the question that follows: Train - wise speed (in km/hour) on different DaysTrain Average Speed (in km/hr) of Trains Monday Tuesday Wednesday Thursday Friday Saturday A 72 80 - 64 54 - B 88 - 80 84 72 90 C 54 70 72 - 64 60 D - 120 95 - 90 110 E 72 80 84 75 - -
Train D runs 150 km more on Monday than that on Thursday and times taken by the train on Monday and Thursday are 5 hour and 4 hours, respectively. If the average speed of this train on Monday is 15 km/hour more than that on Thursday, then the mean of the distance travelled by the train on Monday and Thursday is
375 km
Let train D's Thursday speed = s km/hour. Then Monday speed = s + 15 (given 15 more).
Distance = speed × time. Monday time = 5 h, Thursday time = 4 h.
Monday distance = 5(s + 15); Thursday distance = 4s.
Monday runs 150 km more than Thursday: 5(s + 15) − 4s = 150.
Expand: 5s + 75 − 4s = 150 ⇒ s + 75 = 150 ⇒ s = 75.
So Thursday speed = 75, Monday speed = 90.
Monday distance = 5 × 90 = 450 km; Thursday distance = 4 × 75 = 300 km.
Mean distance = (450 + 300) / 2 = 750 / 2 = 375 km.
Hence 375 km is correct. The other figures come from arithmetic slips in solving for s.
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