A train travels at a certain average speed for distance 54 kilometres and then Travels a distance of 63 km at an average speed of 6 km per hour more than the first speed if it takes 3 hours to complete the journey what is its first speed
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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 kkm 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
And it travels a distance of 72 kkm 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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