There are A station and B station and another station C is between A and B.
Train X will leave A station at 23:10 and will arrive at intermediate C station at 2:10 on the next day.
After stopping for 20 minutes, train X will leave C station at 2:30 and it will arrive at B station at 6:10.
The speed of this train is 2.675 km per hour faster than the speed which is calculated the distance between A and B is divided by 7 hours which is actual time taken from A to B.
Find the distance of A station and B station.
Train X will leave A station at 23:10 and will arrive at intermediate C station at 2:10 on the next day.
After stopping for 20 minutes, train X will leave C station at 2:30 and it will arrive at B station at 6:10.
The speed of this train is 2.675 km per hour faster than the speed which is calculated the distance between A and B is divided by 7 hours which is actual time taken from A to B.
Find the distance of A station and B station.
Answer
7 hour - 20 minutes = 20/3 hours.
The ratio of the time is 20/3 : 7 = 20/3 : 21/3 = 20 : 21.
The ratio of the speed is an inverse ratio and is 21 : 20.
Since 21 - 20 = 1 of the difference of this ratio is equivalent to 2.675 km per hour which is a difference of actual speed, the speed of this train is 2.675 km per hour × 21.
Since the running time is 20/3 hours, the distance between A and B is 2.675 × 21 × 20/3 = 374.5km.
374.5 km
Solution
The actual time the train ran on the way from A station to B station is calculated by stoppage-time 20 minutes is subtracted from 7 hours.
7 hour - 20 minutes = 20/3 hours.
The ratio of the time is 20/3 : 7 = 20/3 : 21/3 = 20 : 21.
The ratio of the speed is an inverse ratio and is 21 : 20.
Since 21 - 20 = 1 of the difference of this ratio is equivalent to 2.675 km per hour which is a difference of actual speed, the speed of this train is 2.675 km per hour × 21.
Since the running time is 20/3 hours, the distance between A and B is 2.675 × 21 × 20/3 = 374.5km.
