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 per hour more than its original speed if it takes three of the complete total journey what is its original average speed

Answer 1

$\huge\red {<strong>Heya</strong>\:<strong>Mate</strong>}$......

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

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HOPE THAT THIS WILL HELP YOU. ......