Question

A train travels at a certain speed of a distance of 63 km and then a distance of 72 km at an average speed. 6 km/hr more than its original speed it takes 3 hrs less to complete total journey, what is the original average speed.

Answer 1

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

Hope This Helps :)

Answer 2

Solutions :-

Given

Distance travels on certain speed = 63 km
Let Original speed of train be x
Distance travels at speed of 6 km/hr = 72 km
Speed = (x + 6) km per hour
Time = 72/(x+6) hrs
Total journey completed in 3 hrs.

we know that

Distance = Speed × time
Speed = distance/time
Time = distance/speed


we have to Find the value of x :-

A/q

=> 63/x + 72/(x + 6) = 3
=> 63(x + 6) + 72x = 3x(x+ 6)
=> 21(x + 6) + 24x = x(x+6)
=> 45x + 21×6 = x2 + 6x
=> x^2 - 39x - 126 = 0
=> x^2 - 39x - 126 = 0
=> x^2-42x+3x -126=0
=> x(x-42)+3(x-42)=0
=> (x - 42)(x + 3) = 0
=> x = 42 or x = -3

Hence

Original average speed is 42 km / hr ( distance positive )