Two cyclists started simultaneously towards each other from points $A$ and $B$ which are $28$ km apart. An hour later they met and kept pedalling with the same speed without stopping. The first cyclist arrived at $B$ $35$ minutes earlier than the second arrived at $A$. Find the speed of each cyclist.
Let cyclist $1$ start from point $A$ towards point $B$
and cyclist $2$ start from $B$ towards $A$.
Let the two cyclists meet at point $P$ after time $= t = 1$ hr
Let cyclist $1$ cover a distance $x$ km in $1$ hour
i.e. $d_{AP} = x$ km
Then cyclist $2$ covers a distance $\left(28 - x\right)$ km in $1$ hour
i.e. $\;$ $d_{BP} = \left(28 - x\right)$ km
$\therefore \;$ Speed of cyclist $1$ $= s_1 = \dfrac{d_{AP}}{t} = \dfrac{x}{1} = x$ kmph
and speed of cyclist $2$ $= s_2 = \dfrac{d_{BP}}{t} = \dfrac{28 - x}{1} = \left(28 - x\right)$ kmph
To reach point $B$,
cyclist $1$ covers a distance $= d_{PB} = \left(28 - x\right)$ km with a speed of $x$ kmph
$\therefore \;$ Time taken by cyclist $1$ to reach point $B$ $= t_1 = \left(\dfrac{28 - x}{x}\right)$ hr
To reach point $A$,
cyclist $2$ covers a distance $= d_{PA} = x$ km with speed $\left(28 - x\right)$ kmph
$\therefore \;$ Time taken by cyclist $2$ to reach point $A$ $= t_2 = \left(\dfrac{x}{28 - x}\right)$ hr
Given: $\;$ Cyclist $1$ arrives at point $B$ $35$ minutes $\left(= \dfrac{35}{60} = \dfrac{7}{12} \text{ hr}\right)$ before cyclist $2$ arrives at point $A$.
i.e. $\;$ $t_1 = t_2 - \dfrac{7}{12}$
i.e. $\;$ $\dfrac{28 - x}{x} = \dfrac{x}{28 - x} - \dfrac{7}{12}$
i.e. $\;$ $\dfrac{x}{28 - x} - \dfrac{28 - x}{x} = \dfrac{7}{12}$
i.e. $\;$ $12 \left(x^2 - 784 + 56x - x^2\right) = 7x \left(28 - x\right)$
i.e. $\;$ $672x - 9408 = 196x - 7x^2$
i.e. $\;$ $7 x^2 + 476 x - 9408 = 0$
i.e. $\;$ $x^2 + 68 x - 1344 = 0$
i.e. $\;$ $\left(x + 84\right) \left(x - 16\right) = 0$
i.e. $\;$ $x = -84$ $\;$ or $\;$ $x = 16$
Since distance covered cannot be negative,
therefore, $x = 16$ is the only acceptable solution.
$\therefore \;$ Speed of cyclist $1$ $= s_1 = x = 16$ kmph
and speed of cyclist $2$ $= s_2 = 28 - x = 28 - 16 = 12$ kmph