A train travels a distance of 300km at a constant speed. if the speed of the train is increased by 5km, the journey would have taken 2 hours less. find the original speed of the train
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I thought over this quite a lot. Very interesting. I am of the opinion that the bee will travel an infinite distance. This might be some kind of a paradox.
Think of it like this, each time the bee starts from just above one train(A) and starts flying towards the other one(B), it meets B at a point 2/3rd from its original position (as its speed is twice that of the train). That is, if the original distance between A and B at the start of any randomly chosen iteration is x, the bee meets B at a point 2x/3 from initial position of A and x/3 from initial position of B.
Thus each time the bee turns around, the distance that it sees between it and the train in front is 1/3rd of that in the previous turn (as the two trains each wipe off 1/3rd each i.e, 2/3rd in all of the distance). But the bee will cover 2/3 rd of THIS distance (as its speed is twice that of the train facing it).
So it effectively boils down to this-
1)Suppose in the current iteration the trains are separated by x. The bee covers 2x/3 in this iteration.
2)In the next iteration the bee sees a distance of x/3 between itself and the train, and covers 2/3 of this i.e, 2/3 * x/3 = 2x/9
3)This time the bee sees a distance of x/3/3 i.e x/9 in the beginning and covers x/9 * 2/3 before turning back i.e, 2x/27.
Notice the pattern? The distances covered by the bee in successive iterations:
2x/3 -> 2x/9 -> 2x/27 -> so on.......... (infinite times)
This is a never-ending chain!!
AND SO, FINALLY : THE BEE WILL TRAVEL AN INFINITE DISTANCE!
I say it again, nice question, my grey cells got a nice little workout. I think this should be the right solution!!
Think of it like this, each time the bee starts from just above one train(A) and starts flying towards the other one(B), it meets B at a point 2/3rd from its original position (as its speed is twice that of the train). That is, if the original distance between A and B at the start of any randomly chosen iteration is x, the bee meets B at a point 2x/3 from initial position of A and x/3 from initial position of B.
Thus each time the bee turns around, the distance that it sees between it and the train in front is 1/3rd of that in the previous turn (as the two trains each wipe off 1/3rd each i.e, 2/3rd in all of the distance). But the bee will cover 2/3 rd of THIS distance (as its speed is twice that of the train facing it).
So it effectively boils down to this-
1)Suppose in the current iteration the trains are separated by x. The bee covers 2x/3 in this iteration.
2)In the next iteration the bee sees a distance of x/3 between itself and the train, and covers 2/3 of this i.e, 2/3 * x/3 = 2x/9
3)This time the bee sees a distance of x/3/3 i.e x/9 in the beginning and covers x/9 * 2/3 before turning back i.e, 2x/27.
Notice the pattern? The distances covered by the bee in successive iterations:
2x/3 -> 2x/9 -> 2x/27 -> so on.......... (infinite times)
This is a never-ending chain!!
AND SO, FINALLY : THE BEE WILL TRAVEL AN INFINITE DISTANCE!
I say it again, nice question, my grey cells got a nice little workout. I think this should be the right solution!!
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Answer:
Let the original speed of the train be y km / hr .
A.T.Q.
300 ( y + 5 ) - 300 y = 2 y ( y + 5 )
y² + 5 y - 750 = 0
y² + 30 y - 25 y - 750 = 0
( y + 30 ( y -25 ) = 0
y = - 30 or y = 25
Since , speed of train can't be negative .
Therefore , the original speed of the train is 25 km / hr .
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