Two cyclists left the same point simultaneously and traveled in the same direction. The speed of the first was $15$ km/h and that of the second was $12$ km/h. Half an hour later, another cyclist left the same point and traveled in the same direction. Some time later, he overtook the second cyclist and another hour and a half later he overtook the first cyclist. Find the speed of the third cyclist.
Let cyclists $C_1$ and $C_2$ leave point $A$ simultaneously and travel in the same direction.
After $\dfrac{1}{2}$ hr, let cyclist $C_3$ start from point $A$.
Speed of $C_1 = u_1 = 15$ kmph
and speed of $C_2 = u_2 = 12$ kmph
Let $C_1$ and $C_2$ start from point $A$ at time $t = 0$
In time $= \dfrac{1}{2}$ hr, distance covered by $C_1 = 15 \times \dfrac{1}{2} = 7.5$ km
In time $= \dfrac{1}{2}$ hr, distance covered by $C_2 = 12 \times \dfrac{1}{2} = 6$ km
Let speed of third cyclist $C_3 = u_3$ kmph
Let $C_3$ overtake $C_2$ after a time $T$ hr
Distance covered by $C_2$ in time $T$ $= \left(12 \times T\right)$ km
$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,
distance covered by $C_2$ in time $T$ $= \left(12T + 6\right)$ km $\;\;\; \cdots \; (1)$
Distance covered by $C_3$ in time $T$ $= u_3 \times T$ km $\;\;\; \cdots \; (2)$
$\because \;$ $C_3$ overtakes $C_2$, we have from equations $(1)$ and $(2)$
$12 T + 6 = u_3 T$ $\implies$ $T = \dfrac{6}{u_3 - 12}$ $\;\;\; \cdots \; (3)$
Now, $C_3$ overtakes $C_1$ after a time $\left(T + 1.5\right)$ hr
Distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr is $= 15 \left(T + 1.5\right) = \left(15 T + 22.5\right)$ km
$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,
distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr
$= \left(15T + 22.5 + 7.5\right) = 15 T + 30$ km $\;\;\; \cdots \; (4)$
Distance covered by $C_3$ in time $\left(T + 1.5\right)$ hr
$= u_3 \times \left(T + 1.5\right) = u_3 T + 1.5 u_3$ km $\;\;\; \cdots \; (5)$
$\because \;$ $C_3$ overtakes $C_1$, we have from equations $(4)$ and $(5)$
$15 T + 30 = u_3 T + 1.5 u_3$
i.e. $\;$ $T = \dfrac{1.5 u_3 - 30}{15 - u_3}$ $\;\;\; \cdots \; (6)$
$\therefore \;$ We have from equations $(3)$ and $(6)$,
$\dfrac{6}{u_3 - 12} = \dfrac{1.5 u_3 - 30}{15 - u_3}$
i.e. $\;$ $1.5 u_3^2 - 42 u_3 + 270 = 0$ $\;\;\; \cdots \; (7)$
Solving quadratic equation $(7)$ gives $\;$ $u_3 = 10$ $\;$ or $\;$ $u_3 = 18$
Speed of $C_3$ has to be greater than the speeds of $C_1$ and $C_2$ for $C_3$ to overtake them.
$\implies$ $u_3 = 10$ is not an acceptable solution.
$\therefore \;$ Speed of cyclist $C_3$ $= u_3 = 18$ kmph