Two buses $$X,Y$$ started from two different places $$P_1$$ and $$P_2$$ respectively, moving towards each other. The ratio of the speed of bus $$X$$ to that of bus $$Y$$ was $$5:4$$. After they meet, the speed of bus $$X$$ was reduced by $$20\%$$ and the speed of $$Y$$ was increased by $$20\%$$. When bus $$X$$ arrived at $$P_2$$, bus $$Y$$ was still $$10\text{ km}$$ away from $$P_1$$. The distance between the places $$P_1$$ and $$P_2$$, in kilometers, is

Let the original speeds be $$5v$$ and $$4v$$. The distances covered before meeting are therefore in the ratio $$5:4$$, say $$5k$$ and $$4k$$, so the total distance is $$9k$$. After meeting, the speeds become $$4v$$ and $$4.8v$$. Bus $$X$$ takes $$k$$ units of time to cover the remaining $$4k$$, during which bus $$Y$$ covers $$4.8k$$. Hence the remaining distance of bus $$Y$$ from $$P_1$$ is $$5k-4.8k=0.2k=10$$, giving $$k=50$$. Therefore the distance between $$P_1$$ and $$P_2$$ is $$9k=450\text{ km}$$.