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Question
A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?
Solution
Let the original average speed be x km/hr
Since time = distance/speed`
Therefore
`63/x + 72/(x + 6) = 3`
`63(x+6) + 72x = 3x(x + 6)`
`21(x+ 6) + 24x = x(x+6)`
`x^2+ 6x= 21x + 126 + 24x`
`x^2 + 6x = 45x + 126`
`x^2 - 39x - 126 = 0`
`x^2 - 42x + 3x - 126 = 0`
(x - 42)(x + 3) = 0
`=> x = 42 or x = -3`
Since speed can not be negative
Therefore the original speed is 42 km/hr
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