Question

. A train covers a distance of 90 km at a uniform speed. Had the speed Been 15 km/hr more, it would have taken half an hour less the journey. Final original speed of train.

Answer 1

Let the original speed of train be x
Time taken by train in x km/h for 90 km distance= $\frac{90}{x}$
Time taken by train in (X+ 15 ) km/h for 90 km distance=$\frac{90}{x+15}$
ATQ,$\frac{90}{x+15} = \frac{90}{x} -1$
      =$\frac{90}{x+15} = \frac{90-x}{x}$
      =>(x+15)(90-x)             = 90x
      =>90x -x² + 1350 - 15 x= 90x
      => x²+15x - 1350 =0
         By quadratic formula 
          a= 1
          b= 15
          c= 1350
          D= b²-4ad
            = 225+ 1350x 4
            = 225+ 5400
            = 5625
           
Two roots are $\frac{-b+√D}{2a} and \frac{-b+√D}{2a}$
                    $\frac{-15+75}{2} and \frac{-15 - 75}{2}$
                    30 and -40 
x= 30 km/h
so the speed of train is 30 km/h
the second value -40 rejected because speed cant be in negative
_______________________________________________________thanks_