Question

a train travels at a certain speed of a distance of 63 km and then travel a distance of 72 km at an average speed 6km/hr more than its original speed it takes 3 hour to complete the journey what is the original speed

Answer 1

Let x and (x + 6kmph) represent the speeds traveling 63km and 72km respectively
Question States***total time of trip 3hr. t = D/r
63/x + 72/(x+6) = 3 |Multiplying each term by x(x+6) so as all denominators= 1
63(x+6) + 72x = 3x(x+6)
63x + 6*63 + 72x = 3x^2 + 18x
3x^2 -117x - 378 = 0 3(x^2 - 39x - 126) = 0
factoring
(x+3)(x-42) = 0
(x+3) = 0 |tossing out negative solution
(x-42) = 0
x = 42kmph, speed of train while traveling 63kmph, the original average speed
CHECKING our Answer ***
63km/42kmph + 72km/48kmph = 1.5hr + 1.5hr = 3hr

Answer 2

Step-by-step explanation:

here is the answer u can refer this