Let the usual (original) speed of train x=km/hr
and the increased speed =(x+5)km/hr
Usual distance covered =300 km
`because` Time taken in original speed `=(300)/(x)`
and time taken in increased speed `=(300)/(x+5)`
But the difference in time in both cases = 2hrs
`because(300)/(x)=(300)/(x+5)=2implies(300(x+5-x))/(x(x+5))=2`
`impliesx^(2)+5x=750impliesx^(2)+5x-750=0`
`implies(x+30)(x-25)=0`
`impliesx=-30orx=25`
`becausex=25km//hr` (rejecting x=30, because speed cannot be neagtive)
`because` The original speed =25 km/hr