A train starting from a station X was to arrive at a station Y at 6:06 PM. It could travel at 62.5% of its usual
speed and reach Y at 7 PM. At what time did it start from X?
Answers
Step-by-step explanation:
Let it start from station X at the speed of 100.
Time of reaching at Y=6:06
Now when, Speed is 62.5% i. e. 62.5
Time of reaching at Y=7:00
Let depart time be t
Then for 7:00 it will be t+54
Equating 2 situations:
100×t=62.5×(t+54)
It gives t=90
So it will start 90 mins earlier than 6:06 PM
Hence Answer=4:36PM
Given :- A train starting from a station X was to arrive at a station Y at 6:06 PM. It could travel at 62.5% of its usual
speed and reach Y at 7 PM. At what time did it start from X ?
Solution :-
Let us assume that, the actual speed of the train be 100x km/hr and the actual time taken be y hours. Then,
→ Distance covered = Speed * Time = (100xy)km ..
now, given that, if train travels at 62.5% of its usual
speed and reach Y at 7 PM.
So,
→ New speed = 62.5% of 100x = 62.5x km/h .
→ increased time = 7PM - 6:06 = 54 minutes = (54/60) = (9/10) hours.
→ New total time = (y + 9/10) hours.
Then,
→ Distance covered = New Speed * New Time = (62.5x) * {y + (9/10)} km .
since distance is same in both case from X to Y stations.
→ 100xy = 62.5x * {y + (9/10)}
→ 100y = 62.5(10y + 9)/10
→ 1000y = 625y + 562.5
→ 1000y - 625y = 562.5
→ 375y = 562.5
dividing both sides by 375 ,
→ y = 1.5 hours.
Therefore,
→ Train reached at Y with usual speed = 6:06 PM
→ Time taken by train = 1.5 hour = 1 hour , 30 minutes.
So,
→ Train start from X at = 6 : 06 - 1 : 30 = 5 : 66 - 1 : 30 = 4 : 36 PM. (Ans.)
Hence, The Train starts from station X at 4 : 36 PM..
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