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Given below is information about 2 trains X and Y.
Train X: It starts from station A and after travelling 60 km, it reduces its speed by 25%. Hence, it reaches station B at 10 AM.
Train Y: It starts from station A at same time when train X starts and after travelling 90 km, it reduces its speed by 25%. Hence it reaches station B at 9 : 45 AM.
Had both travelled at their original speeds, then, they reach B at 8 : 30 AM.
If speed of train Z is 200% more than speed of train X. Then, find the time taken by it to cover distance AB.
3.1
4
2
1.9
None of these
- Train X travels at an original speed, reduces by 25% after 60 km, and reaches station B at 10 AM.
- Train Y follows similar behavior but changes speed after 90 km and reaches 15 minutes earlier than Train X at 9:45 AM.
- If both trains maintained their original speed, they would arrive at 8:30 AM.
- Train Z is 200% faster than Train X.
- If Train X's speed is 100 units, Train Z's speed is 300 units (because 200% more than 100 = 300).
- Train Z covers the AB distance in one-third the time at Train Z's higher speed.
- Comparing Train X's adjusted time, get Train Z's time by considering the "faster" aspect.
- Calculate using original arrival (8:30 AM): 90 minutes faster till 10 AM, third of the 90-minute time = 30 minutes.
- Option 3 (2 hours) is correct.
By: Parvesh Mehta ProfileResourcesReport error
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