Two people started simultaneously towards each other from points $A$ and $B$ which are $50$ km apart. They met $5$ hours later. After their meeting, the first person who traveled from $A$ to $B$, decreased his speed by $1$ kmph and the other person who traveled from $B$ to $A$ increased her speed by $1$ kmph. The first person is known to arrive at $B$ two hours earlier than the second person arrived at $A$. Find the initial speed of the first person.
Let the two people meet at point $P$ after $5$ hours.
Let distance $AP = x$ km
Then, distance $PB = \left(50 - x\right)$ km
$\therefore \;$ Speed of person traveling from $A$ $= \left(\dfrac{x}{5}\right)$ kmph
and speed of person traveling from $B$ $= \left(\dfrac{50 - x}{5}\right)$ kmph
New speed of person (who started from $A$) traveling from $P$ towards $B$ [distance $PB$] is
$S_{PB} = \left(\dfrac{x}{5} - 1\right) = \left(\dfrac{x - 5}{5}\right)$ kmph
New speed of person (who started from $B$) traveling from $P$ towards $A$ [distance $PA$] is
$S_{PA} = \left(\dfrac{50 - x}{5} + 1\right) = \left(\dfrac{55 - x}{5}\right)$ kmph
$\therefore \;$ Time taken by person traveling from $A$ to $B$ to arrive at point $B$ is
$= T_{AB} = \dfrac{PB}{S_{PB}} = \dfrac{50 - x}{\left(x - 5\right) / 5} = \left(\dfrac{250 - 5x}{x - 5}\right)$ hr
and time taken by person traveling from $B$ to $A$ to arrive at point $A$ is
$= T_{BA} = \dfrac{PA}{S_{PA}} = \dfrac{x}{\left(55 - x\right) / 5} = \left(\dfrac{5x}{55 - x}\right)$ hr
Given: $\;$ $T_{AB} = \left(T_{BA} - 2\right)$ hr
i.e. $\;$ $\dfrac{250 - 5x}{x - 5} = \dfrac{5x}{55 - x} - 2$
i.e. $\;$ $\dfrac{250 - 5x}{x - 5} = \dfrac{5x - 110 + 2x}{55 - x}$
i.e. $\;$ $\dfrac{250 - 5x}{x - 5} = \dfrac{7x - 110}{55 - x}$
i.e. $\;$ $13750 - 275 x - 250 x + 5x^2 = 7x^2 - 110 x - 35x + 550$
i.e. $\;$ $2x^2 + 380 x - 13200 = 0$
i.e. $\;$ $x^2 + 190 x - 6600 = 0$
i.e. $\;$ $\left(x + 220\right) \left(x - 30\right) = 0$
i.e. $\;$ $x = - 220$ $\;$ or $\;$ $x = 30$
$\because \;$ Distance covered cannot be negative $\implies$ $x = 30$ is the only acceptable solution.
$\therefore \;$ Initial speed of the person who traveled from $A$ to $B$ (i.e. the first person) is
$= \dfrac{x}{5}$ kmph $= \dfrac{30}{5} = 6$ kmph