A train travel at a certain speed of a distance of 63 km and than travel a distance of 72 km/hr more than original speed ,it takes 3 hr to complete journey. what is original speed
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Hi
Just do factorisation for the equation and u will get the answer....
Hope this helps u... All the best for ur exam....
Just do factorisation for the equation and u will get the answer....
Hope this helps u... All the best for ur exam....
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amrit263:
thank you
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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 )
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 )
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