Two persons $A$ and $B$ start moving from point $P$ towards $Q$ which are $1400$ $Km$ apart. Speed of $A$ is $50 Km/hr$ and that of $B$ is $20 Km/hr$. How far is $A$ from $Q$ when he meets $B$ for the $22nd$ time?
How can we solve this? I tried but I was not able to.
1st time they will met at 1000 km from P and 2nd time as per calculation 933.33 kms from Q when P will return back. But how can we simplify this for checking this for 22nd time? I looked on the web but i could not find a satisfactory and lucid explanation.
Please show me how to solve this with a little explanation.
Thanks in advance !
The only way that this problem makes sense, is if it is assumed that $A$ always immediately turns around from point $Q$ and go back towards person $B$.
Further, the only way that they meet $22$ times is if it is also assumed that each time that they meet, person $A$ immediately turns around and starts heading back to point $Q$.
In the following analysis, kilometers and hours will be assumed for the appropriate time/distance/speed units of measurement.
Suppose that $A,B$ are either at the start, or are meeting in the middle, between points $P$ and $Q$, where they are a distance $d$ from $Q$.
Then $A$ will take $t_1 = (d/50)$ to hit point Q.
During time $t_1,$ $B$ will travel $20 \times (d/50) = (2d/5).$
Then, $A$ and $D$ will be $(3d/5)$ apart, and approaching each other at a combined speed of $(70)$.
So the time back will be $t_2 = (3d/5) \times (1/70) = (3d/350).$
During time $t_2$, $B$ will travel $(60d/350).$
This means that during the combined time of
$(t_1 + t_2) = (d/50) + (3d/350) = (10d/350) = (d/35),$
$B$ will have moved forward $(20d/35) = (4/7)d$.
This means that if meeting $n$ is at a distance of $d$ from point $Q$, then meeting $(n+1)$ is at a distance of $(3/7)d$ from point $Q$
At meeting $0$, $A$ is $1400$ from point $Q$.
At meeting $1$, $A$ is $1400 \times 3/7$ from point $Q$.
At meeting $2$, $A$ is $1400 \times (3/7)^2$ from point $Q$.
At meeting $22$, $A$ is $1400 \times (3/7)^{(22)}$ from point $Q$.