Rates, Distance and Time Problem 45 2 vehicles left the town $A$ and $C$ at the same time to ride to the other town, crossing each other at town $B$ and both traveling at different steady speeds. The vehicle from town $A$ finished the trip from town $B$ to town $C$ in 45 minutes at a constant speed of 64 kilometers per hour. The vehicle from town $C$ finished the trip from town $B$ to town $A$ in 20 minutes. I know from this info that the 2 vehicles meet at town $B$, so the time it takes to go from town $B$ to town $A$ for the vehicle from town $A$ is equal to the time it takes to go from town $C$ to town $B$ for the vehicle from town $C$. But I do not know how to use this information to solve the problem. Please help me with figuring out what the speed of the vehicle from town $C$ would be and how you would calculate this. Thanks :)

Let $X_{BC}$ be the distance between towns B and C. Then: $$X_{BC}=64km/h(\frac {45min}{60min/h})=48km$$

So, it takes C 20 minutes to get from B to A. Let $V_c$ be the speed of C, and $t=20min=1/3h$. Then:

$$X_{AB}=V_c t=\frac {V_c}{3}h$$

Knowing that the time that takes A to get to B ($t_{AB}$) is equal the time that takes C to get to B ($t_{BC}$). Hence, $t_{AB}=t_{BC}$:

$$\frac {X_{AB}}{V_a}=\frac {X_{BC}}{V_c}$$ $$\frac {\frac {V_c}{3}h}{64km/h}=\frac {48km}{V_c}$$ $$\frac {V_c^2}{3}=3072km^2/h^2$$ $$V_c^2=9216km^2/h^2$$ $$V_c=96km/h$$