Problem source

A tortoise and a hare have a race to the vegetable patch, a distance $X$ kilometres from the starting post, and back.
The tortoise sets off immediately, at a steady $v$ kilometers per hour. The hare goes to sleep for half an hour and then sets off at a steady speed $V$ kilometres per hour. The hare overtakes the tortoise half a kilometre from the starting post, and continues on to the vegetable patch, where she has another half an hour's sleep before setting  off for the return journey at her previous pace. One and quarter kilometres from the vegetable patch, she passes the tortoise, still plodding gallantly and steadily towards the vegetable  patch. Show that 
\[
V= \frac{10}{4X-9}
\]
and find $v$ in terms of $X$. 
Find $X$ if the hare  arrives back at the starting post one and a half  hours after the start of the race.

Solution source

If $T_1$ is the time they meet after $\frac12$km from the starting points and $T_2$ is the time they meet a second time, then

\begin{align*}
&& \frac12 &= vT_1 \\
&&&= V(T_1-\tfrac12) \\
&& X - \frac54 &= vT_2 \\
&& X + \frac54 &= V(T_2 - 1) \\
&& \frac{T_2}{T_1} &= \frac{4X-5}{2} \\
&& X + \frac54 + V &= VT_2 \\
&& \frac12 + \frac12 V &= VT_1 \\
\Rightarrow && \frac{T_2}{T_1} &= \frac{4X + 5 + 4V}{2(1+V)} \\
\Rightarrow && \frac{4X-5}{2}&=\frac{4X + 5 + 4V}{2(1+V)}  \\
\Rightarrow && V(4X-9) &= 10 \\
\Rightarrow && V &= \frac{10}{4X-9} \\
\\
&& T_1 &= \frac{1}{2V} + \frac12 \\
&&&= \frac{4X+1}{20} \\
&& v &= \frac{1}{2T_1} \\
&&&= \frac{10}{4X+1} 
\end{align*}

\begin{align*}
&& 2X &= \frac12 V \\
\Rightarrow && 2X(4X-9) &= 5 \\
\Rightarrow && 0 &= 8X^2-18X-5 \\
&&&= (4X+1)(2X-5) \\
\Rightarrow && X &= -\frac14, \frac52
\end{align*}

Since $X$ is positive, we must have $X = \frac52$km