Direction: Study the given chart carefully and answer the following questions. | |||||||||
Train A | |||||||||
Station | Arrival time | Departure time | Distance from origin (in km) | Number of passengers boarding at each station | Fare (in Rs.) | ||||
Ahmedabad | Starting | 5.00 pm | - | 400 | - | ||||
Vadodara | 630 pm | 635 pm | 100 | 100 | 50 | ||||
Bharuch | 8:50 pm | 9:00 pm | 250 | 90 | 120 | ||||
Mumbai | 400am | 4:10 am | 800 | 300 | 400 | ||||
Pune | 730 am | 7:45 am | 1050 | 150 | 500 | ||||
Solapur | 1020 am | Terminates | 1280 | - | 620 | ||||
Trial B | |||||||||
Station | Arrival time | Departure time | Distance from origin | Number of passengers boarding at each station | Fare (in Rs.) | ||||
Solapur | Starting | 6:00 pm | - | 300 | - | ||||
Pune | 7:40 pm | 7:45 pm | 230 | 150 | 120 | ||||
Mumbai | 9:30 pm | 9:35 pm | 480 | 270 | 220 | ||||
Bharuch | 5:40 am | 5:55 am | 1030 | 50 | 500 | ||||
Vadodara | 9:00 am | 9:10 am | 1180 | 100 | 570 | ||||
Ahmedabad | 12:00 noon | Terminates | 1280 | - | 620 |
A) 7:45 am
B) 9:45 am
C) 8:45 am
D) 10:45 am
E) 11:45 am
Correct Answer: C
Solution :
If the average speed of Train A increases by 10%, then its new speed \[=73.84\times \frac{110}{100}=814.22\,kmph\] Time taken by Train A during the journey \[=\frac{1280}{81.22}=15.75\]hours =15 hours 45 minutes The time when the train will reach its destination = 5 pm +15 hours 45 minutes = 8:45 amYou need to login to perform this action.
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