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Question
A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed, If it takes 3 hours to complete total journey, what is its original average speed?
Solution
Let the original average speed of the train be x km/hr.
Time taken to cover `63 \text { km } = 63/xx` hours
Time taken to cover 72 km when the speed is increased by `6 (km) /(hr) = 72/(x + 6)` hours
According to given information, we have
`63/ xx + 72/(x + 6) = 3`
⇒`21/x + 24/(x +6) = 1`
⇒ `(21x + 126 + 24x)/(x^2 + 6x) = 1`
⇒ `45x + 126 = x^2 + 6x`
⇒`x^2 - 39x - 126 = 0`
⇒ `x^2 - 42x + 3x - 126 = 0`
⇒` x(x - 42) + 3(x - 42) = 0`
Since the speed cannot be negative, x ≠-3
⇒ `(x - 42)(x + 3) = 0`
⇒ `x-42 =0 or x + 3 = 0`
⇒ x = 42
Thus, the original average speed of the train is 42 km/hr.
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