Q. 335.0( 3 Votes )

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 kmph more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Answer :

Let the speed of the train and the time taken to cover the same be ‘X’ km/hr and ‘Y’ hrs respectively.



As per the question,


XY = 90 –––––– (i)


(X + 15)(Y – 0.5) = 90 ––––– (ii)


LHS = RHS


Equating the LHS of both equations, and simplyifying it further


XY = XY – 0.5X + 15Y – 7.5


0.5X – 15Y + 7.5 = 0 ––––––– (iii)


Multiplying the above equation by 10,


5X – 150Y + 75 = 0


Simplyfying it further,


X – 30Y + 15 = 0


Putting the value of Y, in equation (iii)



X2 + 15X – 2700 = 0


On applying Sreedhracharya formula







X = – 60 or X = 45


the original speed of train is 45 km/hr as speed can’t be negative.


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