A superfast train runs having the speed 15 kms / hr more than that of an express train. Leaving same station the superfast train reached at a station of 180 kms distance 1 hour before than the express train. Determine the speed of the superfast train in km/ hr.

Explanation

Let the speed of the superfast train was x km/hr.
`:.` the speed of the express train was (x-5)km/hr.
`:.` to travel 180 km, time taken by the superfast train is `(180)/(x)` hr and by express train is `(180)/(x-15)` hr As per questions, `(180)/(x-15)-(180)/(x)=1`
or, `180((1)/(x-15)-(1)/(x))=1or,(x-x+15)/((x-15)x)=(1)/(180)`
or, `(15)/(x^(2)-15x)=(1)/(180)`
or, `x^(2)-15x=2700`
or, `x^(2)-15x-2700=0...............(1)`
Comparing (1) with `ax^(2)+bx+c=0,(ane0)` we get, `a=1,b=-15 andc=-2700`
`:.` by Sreedhar Acharya's formula we get, `x=(-(-15)pmsqrt((15)^(2)-4xx1xx(-2700)))/(2xx1)`
or, `x=(15pmsqrt(225+10800))/(2)`
or, `x=(15pmsqrt(11025))/(2)or,x=(15pm105)/(2)`
`:.x=(15+105)/(2)` (taking +sign) and `:.x=(15-105)/(2)` (taking -sign)
`=(120)/(2)andx=(-90)/(2)`
=60 and x=-45
But the speed of a train connot be negative, `:.x=60`.
Hence the speed of the superfast train was 60 km/hr.