Question

A train travel at a certain average speed for a distance of 63 km and then travel at a distance of 72 km ata an average speed kf 6km/hr more then it's original speed.If it takes 3hours to complete total journey, what is the original average speed

Answer 1

let the original speed by x
time= distance/speed
t1= 63/x
t2= 72/(x+6)

so (63/x) + 72/(x+6)=3
solving this, x= 42 kmph
the origibnal speed was 42kmph, the later was 48

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 )