24) The answer is (D). Indicating with $v_{1}, v_{2}$, and $v_{3}$ the speeds of the three cyclists, it is evident that $v_{1}<v_{2}<v_{3}$. Let $t_{3}$ be the time, expressed in hours, required for the third cyclist to catch up with the first, then $v_{3} t_{3}=v_{1}\left(t_{3}+1\right)$ and thus $t_{3}=\frac{v_{1}}{v_{3}-v_{1}}$. Similarly, the time $t_{2}$, expressed in hours, required for the second cyclist to catch up with the first will be $t_{2}=\frac{v_{1}}{v_{2}-v_{1}}$. Since $t_{3}+2=t_{2}$, we have $\frac{v_{1}}{v_{3}-v_{1}}+2=\frac{v_{1}}{v_{2}-v_{1}}$, from which, setting $x=\frac{v_{1}}{v_{3}}$, dividing numerators and denominators by $v_{3}$ and considering that $\frac{v_{2}}{v_{3}}=\frac{2}{3}$, we obtain $\frac{x}{1-x}+2=\frac{x}{\frac{2}{3}-x}$. Eliminating the denominators yields the equation $6 x^{2}-11 x+4=0$ which has solutions $\frac{4}{3}$ and $\frac{1}{2}$, of which only the latter is acceptable, given the inequalities on the speeds.
24) Three cyclists set off from Cesenatico along the road to Cortona. The first one leaves an hour before the other two, who start together. Each cyclist maintains a constant speed. After a certain time, the third cyclist catches up with the first, and two hours later, the second also catches up with him. The ratio of the second cyclist's speed to the third cyclist's speed is $\frac{2}{3}$. What is the ratio of the first cyclist's speed to that of the third?
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{2}{5}$
(D) $\frac{1}{2}$
(E) $\frac{2}{3}$.