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A train after travelling 100 kms from P meets with an accident and then proceeds
at the terminus Q 90 minutes late. Had the accident occurred 60 kms further on, it would have reached 15 minutes sooner. Find the original speed of the train and the distance PQ.at 3 th of its original speed and arrives 4
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- 65 kmph; 480 km
- 75 kmph; 450 km
- 80 kmph; 460 km
- 85 kmph; 460 km
Correct Option: C
Let P be the starting point, Q the terminus, M and N the places where accidents occur.
At (3/4)th of the original speed, the train will take 4/3 of its usual time to cover the same distance i.e., (1/3)rd more than the usual time.
(1/3)rd of the usual time to travel a distance of 60 kms between MN = 15 min.
∴ Usual time to travel 60 kms
| = 15 × 3 = 45 min. = | hr. | |
| 4 |
∴ Usual speed of the train per hour
| = 60 × | = 80 km per hr. | |
| 3 |
Usual time taken to travel MQ = 90 × 3
| = 270 min. or | hrs. | |
| 2 |
| = 80 × | = 360 km. | |
| 2 |
Therefore, the total distance PQ
= PM + MQ
= 100 + 360 = 460 kms.
