a train Travels at a certain speed of a distance of 63 kilometre and then Travels a distance of 72 km and the average speed of 6 km per hour and then his original speed is 3 hours to complete total then he was the original and average speed
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Let the original speed of train is x km/h
time taken to cover 63 km with speed x km/h, T₁ = distance/time = 63/x hours
Again, question said , speed of train now (x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h , T₂= 72/(x + 6) km/h
A/C to question ,
T₁ + T₂ = 3 hours
⇒63/x + 72/(x + 6) = 3
⇒ 21/x + 24/(x + 6) = 1
⇒ (21x + 126 + 24x) = x(x + 6)
⇒45x + 126 = x² + 6x
⇒ x² - 39x - 126 = 0
⇒x² - 42x + 3x - 126 = 0
⇒ x(x - 42) + 3(x - 42) = 0
⇒(x + 3)(x - 42) = 0
∴x = 42 and -3 , but x ≠ -3 ∵ speed doesn't negative
Hence, original speed of train = 42 km/h
time taken to cover 63 km with speed x km/h, T₁ = distance/time = 63/x hours
Again, question said , speed of train now (x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h , T₂= 72/(x + 6) km/h
A/C to question ,
T₁ + T₂ = 3 hours
⇒63/x + 72/(x + 6) = 3
⇒ 21/x + 24/(x + 6) = 1
⇒ (21x + 126 + 24x) = x(x + 6)
⇒45x + 126 = x² + 6x
⇒ x² - 39x - 126 = 0
⇒x² - 42x + 3x - 126 = 0
⇒ x(x - 42) + 3(x - 42) = 0
⇒(x + 3)(x - 42) = 0
∴x = 42 and -3 , but x ≠ -3 ∵ speed doesn't negative
Hence, original speed of train = 42 km/h
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