how many returning ants will meet an ant which is going from $A\to B$? A stream of ants go and return via same path from point $A\to B\to A$ with constant velocity, FROM ANY GIVEN POINT TWO ANTS PASS BY PER SECOND IN ONE WAY.It takes one minute to go $A\to B$ for any ant. how many returning ants will meet an ant which is going from $A\to B$?
Honestly I am not understanding the problem, mainly I do not understand the situation I have written in block letters. could any one explain me what is happening? may be i am not understanidng this simple problem as I am not good in english language.
 A: The capitalized section seems to mean that if you’re standing at any point $P$ beside the path, in every second two ants travelling from $A$ to $B$ will pass by you. In other words, the ants are half a second apart: if one ant is passing you right now, the next one will pass you half a second from now. Let $d$ be the spacing between one ant and the next in centimetres, say; each ant travels a distance $d$ in half a second, or $2d$ cm/s. 
Now consider an ant $P$ leaving $A$. On its way to $B$ this ant will pass a number of ants travelling from $B$ to $A$; we want to know how many. There’s no real harm in supposing that an ant arrives at $A$ just as $P$ leaves; call this ant $Q_0$. Let $Q_1$ be the first ant that $P$ passes, $Q_2$ the second, and so on. Start a clock at time $t_0=0$ when $P$ leaves and $Q_0$ arrives at $A$, and let $t_k$ be the time in seconds when $P$ meets $Q_k$. 
$P$ and $Q_1$ meet at time $t_1$. At time $t_0$ the ant $P$ is moving at $2d$ cm/s in one direction, and $Q_1$ is moving at $2d$ cm/s in the opposite direction towards him, so they are approaching each other at a speed of $4d$ cm/s. $Q_1$ is $d$ cm behind $Q_0$ on the path from $B$ to $A$, so at time $t_0=0$ he’s $d$ cm away from $P$. $P$ and $Q_1$ are approaching each other at a relative speed of $4d$ cm/s, so they will meet in $\frac14$ s, at time $t_1=t_0+\frac14=\frac14$. 


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*Use the same ideas to find a simple formula for $t_k$ in terms of $k$, a formula that tells you exactly how long after $P$ leaves $A$ he will meet $Q_k$.  

*Combine that with the known length of time needed for $P$ to travel from $A$ to $B$ to find out how many ants he will meet.

A: If an ant do not walk and stand at a point, it will meet 60*2 = 120 ants. when it will walk with same speed in opposite direction it will meet twice of ants. so answer is 240. Another way to look at the problem is that we can think it as a problem of relative velocity.
A: The moment an ant $M$ starts travelling from $A$ to $B$, there are a number of ants on the return path. From the time $M$ starts travelling until she arrives at $B$, more ants will leave. $M$ must encounter both sets of these ants.
Since it takes a minute, and there are two ants returning per second, there must be 120 ants on the return path when $M$ starts. 
Since it takes $M$ a minute to cover the path, another 120 ants will start the return journey.
Hence $M$ encounters 240 ants.
A: When one ant is moving from A to B, it meets (60x2)=120 ants(from A to B)
At the same time ants are also returning from B to A. So, one ant which is moving from A to B, can meet the ants which are moving from A to B as well as B to A. So, it meets other ants (60x2)=120 ants (from B to A). 
Total number of ants it can meet during its journey from A to B= (120+120)=240 ants. (ans)
