Question

a train travels at a certain average speed for a distance 63 km and then Travels a distance of 72 km at an average speed of 60 km / hour more than the original speed if it takes 3 hour to complete total journey what is its 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

Answer 2

Solutions :-

We know,
Distance = Speed × time
Speed = distance/time
Time = distance/speed

We have,
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

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


Answer : Original average speed = 42 km per hr


Note :- Distance is always in positive