# 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.

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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Get To Know About Quadratic Formula42 mins  Quiz | Knowing the Nature of Roots44 mins  Take a Dip Into Quadratic graphs32 mins  Foundation | Practice Important Questions for Foundation54 mins  Nature of Roots of Quadratic EquationsFREE Class  Getting Familiar with Nature of Roots of Quadratic Equations51 mins  Quadratic Equation: Previous Year NTSE Questions32 mins  Champ Quiz | Quadratic Equation33 mins  Balance the Chemical Equations49 mins  Champ Quiz | Quadratic Equation48 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 