A train travels at a certain average speed for a distance of 63 km and then Travels at a distance of 72 km at an average speed of 6 km per hour more than its original speed if it takes 3 hours to complete total journey what is the original average speed
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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
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