| 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) 2.73kmph
B) 1.97kmph
C) 3.6kmph
D) 2.62kmph
E) 3.9kmph
Correct Answer: A
Solution :
Speed of Train A \[=\frac{1280}{10:20\,am-5:00\,pm}\] \[=\frac{1280}{17\,hours\,20\,\min utes}=\frac{1280\times 3}{52}=73.84\,kmph\] Speed of Train B \[=\frac{1280}{12:00\,noon-6:00\,pm\,}\] \[=\frac{1280}{18\,hours}=71.11\,kmph\] \[\therefore \]Difference between the speed of Train A and Train \[B=73.84-71.11=2.73kmph\]
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