A pedestrian and a cyclist start simultaneously towards each other from towns $A$ and $B$ which are $40$ km apart and meet two hours after the start. Then they resumed their trips and the cyclist arrives at $A$ $7$ h $30$ min earlier than the pedestrian arrives at $B$. Find the speeds of the pedestrian and the cyclist.
The pedestrain starts from point $A$ and the cyclist from point $B$.
Distance between $A$ and $B$ $= 40$ km (given)
Let the pedestrian and the cyclist meet at point $P$.
Let the speed of the pedestrian be $= S_p$ kmph
and the speed of the cyclist be $= S_c$ kmph
In $2$ hours,
distance covered by the pedestrian $= d_{AP} = S_p \times 2$ km
distance covered by the cyclist $= d_{BP} = S_c \times 2$ km
Since the pedestrian and the cyclist meet after $2$ hours
$\implies$ $2 \; S_p + 2 \; S_c = 40$
i.e. $\;$ $S_p + S_c = 20$
i.e. $\;$ $S_p = 20 - S_c$ $\;\;\; \cdots \; (1)$
Remaining distance covered by the pedestrian $= d_{PB} = 2 \; S_c$ km
Since the speed of the pedestrian is $S_p$
$\therefore \;$ Time taken by the pedestrian from $P$ to $B$ is $= t_{PB} = \dfrac{d_{PB}}{S_p}$
i.e. $\;$ $t_{PB} = \dfrac{2 \; S_c}{S_p}$ $\;\;\; \cdots \; (2)$
Remaining distance covered by the cyclist $= d_{PA} = 2 \; S_p$ km
Since the speed of the cyclist $= S_c$
$\therefore \;$ Time taken by the cyclist from $P$ to $A$ $= t_{PA} = \dfrac{d_{PA}}{S_c}$
i.e. $\;$ $t_{PA} = \dfrac{2 \; S_p}{S_c}$ $\;\;\; \cdots \; (3)$
Given: $\;$ The cyclist arrives at point $A$ $7.5$ hours before the pedestrian arrives at point $B$.
i.e. $\;$ $t_{PA} = t_{PB} - 7.5$
i.e. $\;$ $\dfrac{2 \; S_p}{S_c} = \dfrac{2 \; S_c}{S_p} - 7.5$
i.e. $\;$ $\dfrac{2 \left(20 - S_c\right)}{S_c} = \dfrac{2 \; S_c}{20 - S_c} - 7.5$ $\;$ [in view of equation $(1)$]
i.e. $\;$ $\dfrac{2 \; S_c}{20 - S_c} - \dfrac{2 \left(20 - S_c\right)}{S_c} = 7.5$
i.e. $\;$ $2 \; S_c^2 - 2 \left(400 - 40 \; S_c + S_c^2\right) = 7.5 \left(20 \; S_c - S_c^2\right)$
i.e. $\;$ $-800 + 80 \; S_c = 150 \; S_c - 7.5 \; S_c^2$
i.e. $\;$ $7.5 \; S_c^2 - 70 \; S_c - 800 = 0$ $\;\;\; \cdots \; (4)$
Solving the quadratic equation $(4)$ gives
$S_c = 16$ $\;$ or $\;$ $S_c = \dfrac{-20}{3}$
Since the speed of the cyclist cannot be negative
$\implies$ Speed of cyclist $= S_c = 16$ kmph
Substituting the value of $S_c$ in equation $(1)$ gives
$S_p = 20 - 16 = 4$
i.e. $\;$ Speed of pedestrian $= S_p = 4$ kmph