Determine velocity if ratio of force acting when car goes down the valley and up the hill is known

Question

I have problem from schoolbook. My solution differes from the solution in the book. Is there an error in the book or am I wrong?

Problem: A car drives up the hill and then down into a cavity. Both the cavity and the hill have the same radius R=40 m. It is known that force with which the car is acting on the supporting surface at the peak of the hill is 2 times smaller than the force with which the that car is acting on the supporting surface at the bottom of the cavity.

My answer is 11.6 m/s, while the book provides answer 20 m/s.

My solution:

1. Picture one. Cavity. Ox is pointing from car to the center of cavity. $F_g$ is pointed in the direction opposite to the center of the cavity (therefore it is with sign -1) $F_c$ is pointed in the direction to the center of the cavity (therefore it is with sign +1). $F_r$ is pointed in the direction of the center of the cavity (therefore it is with sign +1).

By the Newton's 2'nd law: $\frac{mv^2}{R}=-mg+F_{r1}$

2. Picture two. Hill. Ox is pointing from the car to the center of the hill. $F_g$ is pointed in the direction of the center of the hill(therefore it is with sign +1), $F_c$ is pointed in the direction of hill's center(therefore it is with sign +1). $F_r$ is pointed in the direction opposite to the center of the cavity (therefore it is with sign -1).

By the Newton's 2'nd law: $\frac{mv^2}{R}=mg-F_{r2}$

3. We know that $F_{r2}=0.5F_{r1}$

4. So we have system of equations

$$ \left\{ \begin{array}{c} \frac{mv^2}{R}=-mg+F_{r1} \\ \frac{mv^2}{R}= mg-F_{r2} \\ F_{r2}=0.5 F_{r1} \end{array} \right. $$ my solution is $v=\sqrt{\frac{R g}{3}} $