25mph.
using the equation distance = speed x time, you can make 2 equations. let x be the original speed of the train, and t be the original time it takes. the first equation is the original trip, the 2nd is your new speed and shortened time.
[tex]150=xt \\ 150=(x+5)(t-1)[/tex]
this is because you are adding 5mph, and taking off 1 hour of time, to still cover the 150 miles. you can solve the first equation for t, and substitute into the second, and then solve for x.
[tex]t= \frac{150}{x} \\ 150=(x+5)(t-1)=xt+5t-x-5 \\ 150=150+5( \frac{150}{x})-x-5 \\ 5= \frac{750}{x} -x \\ 5+x= \frac{750}{x} \\ x^2+5x=750 \\ x^2+5x-750=0[/tex]
and using the quadratic formula you can find x=25.
