Question

A train travel at a certain speed pf a distance of 63km and then travel a distance of 72 at an average speed 6km/h man then its original speed it takes 3 hours to comelete total journey what is the original average

Answer 1

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

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 )