A train travels at a certain average speed for a distance 630 kilometre and then travels a distance of
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Given that distance = 63 km.
Let original speed of train = x km/hr.
time = distance / time = 63/x hrs.
And it travels a distance of 72 km at a average speed of 6 km/hr more than the original speed.
distance = 72 km ; speed = (x + 6) km/hr .
time = 72/(x+6) hrs.
If it takes 3 hours to complete the whole journey
63/x + 72/(x + 6) = 3 hrs
⇒ 63(x + 6) + 72x = 3x(x + 6)
⇒ 21(x + 6) + 24x = x(x+6)
⇒ 45x + 21×6 = x2 + 6x
⇒ x2 - 39x - 126 = 0
⇒ x2 - 39x - 126 = 0
⇒ (x - 42)(x + 3) = 0
∴ x = 42 km/hr
∴ the original average speed = 42 km/hr
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