Question

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

Answer 1

Here is your solution

Given that  :-

Distance = 63 km. 

Let,

original speed of train = x km/hr. 

Time = distance / time =  63/x hrs.

Also

it travels a distance of 72 km at a average speed of 6 km/hr more than the original speed.

We have

Distance = 72 km

speed = (x + 6) km/hr

Time = 72/(x+6) hrs

it is given that it takes 3 hours to complete the full journey .

A/q

=>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

=>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,

The original average speed = 42 km/hr

(Distance always take positive )

Hope it helps you

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