Maximum speed at which car can take turn on banked road is given as
[tex]v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}[/tex]
now we know that
v = 85 km/h = 23.61 m/s
R = 78 m
[tex]\mu = 0.30[/tex]
now we will have
[tex]23.61 = \sqrt{(78)(9.8)(\frac{0.30 + tan\theta}{1 - 0.30tan\theta}}[/tex]
now solving above equation we will have
[tex]0.73(1 - 0.30 tan\theta) = 0.30 + tan\theta[/tex]
[tex]0.43 = 1.22tan\theta[/tex]
[tex]tan\theta = 0.35[/tex]
Now for the minimum speed
[tex]v_{min} = \sqrt{Rg(\frac{tan\theta - \mu}{1 + \mu tan\theta})}[/tex]
[tex]v_{min} = \sqrt{78(9.8)(\frac{0.35 - 0.30}{1+(0.35)(0.30)})}[/tex]
[tex]v_{min} = 5.88 m/s = 21.2 km/h[/tex]
so speed range will be 21.2 km/h to 85 km/h
