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| Powersdale To Quicksville (Posted on 2007-03-20) |
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Train X moves at an uniform speed from Powersdale to Quicksville; two stations 240 kilometers apart. Train Y starts from Powersdale precisely one hour after Train X departed (also from Powersdale) and, after two hours, comes to a point that Train X had passed 45 minutes previously.
Train Y now increases its speed by 5 kilometers per hour and it overtakes Train X as soon it reaches Quicksville.
Determine the original speed of both the trains.
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Submitted by K Sengupta
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Rating: 2.0000 (2 votes)
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Solution:
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Let the speed of the Train X be x kilometers per hour while that of Train Y be y kilometers per hour.
By conditions of the problem, the distance described by the Train X in 9/4 hours is the same as the distance traversed by the Train Y in 2 hours.
Hence, 2y = (9/4)*x, giving 8y = 9x……..(i)
Also, (240 – 3x)/x = (240-2y)/(y+5)
Or, (240 – 3x)/x = (960 – 9x)/(9x/2 + 20)
Or, 3*x^2 – 40x – 3200 = 0 (upon simplification)
Or, (x-40)(3x+80) = 0
Or, x = 40; ignoring the negative value which is inadmissible.
So, y = 40*(9/8) = 45
Consequently, the speed of Train X is 40 kilometers per hour while the speed of Train Y is 45 kilometers per hour.
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