Three cars leave $A$ for $B$ in equal time intervals. They reach $B$ simultaneously and then leave for point $C$ which is $120$ km from $B$. The first car arrives there an hour after the second car, and the third car, having reached $C$, immediately reverses the direction and $40$ km from $C$ meets the first car. Find the speed of the first car.
Let the first car be $C_1$, second car be $C_2$ and the third car be $C_3$.
Let $C_1$ cover distance $AB$ in time $t_1$ hours.
Let $C_2$ start after $C_1$ at point $A$ after $t$ hours
and $C_3$ start after $C_2$ at point $A$ after another $t$ hours.
Since cars $C_1, \; C_2, \; C_3$ reach point $B$ simultaneously
$\implies$ car $C_2$ covers distance $AB$ in $\left(t_1 - t\right)$ hours
and car $C_3$ covers distance $AB$ in $\left(t_1 - 2t\right)$ hours
$\therefore \;$ Speed of first car $C_1$ is $= s_1 = \dfrac{AB}{t_1}$ kmph
Speed of second car $C_2$ is $= s_2 = \dfrac{AB}{t_1 - t}$ kmph
Speed of third car $C_3$ is $s_3 = \dfrac{AB}{t_1 - 2t}$ kmph
Given: $\;$ Distance $BC = 120$ km
Time taken by $C_1$ (with speed $s_1$) to cover distance $BC$
$= T_1 = \dfrac{BC}{s_1} = \dfrac{120}{AB / t_1} = \dfrac{120 \times t_1}{AB}$ hours
Time taken by $C_2$ (with speed $s_2$) to cover distance $BC$
$= T_2 = \dfrac{BC}{s_2} = \dfrac{120}{AB / \left(t_1 - t\right)} = \dfrac{120 \times \left(t_1 - t\right)}{AB}$ hours
Given: $\;$ $T_1 = \left(T_2 + 1\right)$ hours
i.e. $\;$ $\dfrac{120 \times t_1}{AB} = \dfrac{120 \times \left(t_1 - t\right)}{AB} + 1$
i.e. $\;$ $120 t_1 = 120 t_1 - 120 t + AB$
i.e. $\;$ $AB = 120 t$ $\implies$ $t = \dfrac{AB}{120}$ $\;\;\; \cdots \; (1)$
Let cars $C_1$ and $C_3$ meet at a point $P$.
Then, distance $CP = 40$ km and distance $BP = 80$ km
Since $C_1$ and $C_3$ meet at point $P$
$\implies$ Time taken by $C_1$ to cover distance $BP$
$\hspace{2cm}$ $=$ Time taken by $C_3$ to cover distance $BC + CP$
i.e. $\;$ $\dfrac{BP}{s_1} = \dfrac{BC + CP}{s_3}$
Substituting the values of $BP$, $BC$, $CP$, $s_1$ and $s_3$ gives
$\dfrac{80}{AB / t_1} = \dfrac{120 + 40}{AB / \left(t_1 - 2t\right)}$
i.e. $\;$ $\dfrac{80 t_1}{AB} = \dfrac{160 \left(t_1 - 2t\right)}{AB}$
i.e. $\;$ $t_1 = 2\left(t_1 - 2t\right)$
i.e. $\;$ $t_1 = 2 t_1 - 4t$
i.e. $\;$ $t_1 = 4t = 4 \times \dfrac{AB}{120}$ $\;\;\;$ [by equation $(1)$]
i.e. $\;$ $t_1 = \dfrac{AB}{30}$ hours
$\therefore \;$ Speed of first car $C_1$ $= s_1 = \dfrac{AB}{t_1} = \dfrac{AB}{AB / 30} = 30$ kmph