I am trying quantitative aptitude questions of a qualifying exam and I couldn't solve this question.
A stream of ants go from point A to point B and return to A along the same path. All the ants move at a constant speed and from any given point 2 ants pass per second one way. It takes 1 minute for an ant to go from A to B. Ho many returning ants will an ant meet in its journey from A to B?
I marked A as that ant will meet 120 ants as at a point only 2 ants can pass per second.
Consider the situation at the moment the ant is about to start moving from A to B.
The ant will meet all the ants that are already returning from B to A. It will also meet all the ants that are ahead of him and moving to B, because they will all reach B before him and turn around and meet him while he is not yet at B.
To answer the question we therefore have to work out
the total number of ants that are currently between A and B.
Consider the first ant he meets.
This ant left A two minutes ago since he went to B and back. Our ant will therefore meet all the ants that left point A in the last two minutes. At 2 ants per second, that is 240 ants.
120 Ants would be correct if the other line of ants was stationary. Because the ants in the other line are also moving, they pass each other on the half-tick as well as the full-tick (each "tick" here being half a second, as 2 ants pass each point in a second)
This has the effect of doubling the number of ants encountered.
Stating the question in another way: pick any point between A and B and you will see two ants pass that point each second in each direction. It doesn't say two ants per second pass each other. I think that's where your confusion is coming from.
I'm taking this as it is a constant stream of ants and we're looking at an ant starting while the stream has been going on for at least 2 minutes.
If two ants pass any point each second, then there must be 120 ants on the path from B to A heading towards A.
If we divide the path into 60 segments since it takes 60 seconds to get from A to B,
each segment will have two ants in it. In once second they will pass the point
dividing segments and move to the next segment.
Since it is a constant stream of ants, there will be more coming while our ant makes
his journey. How many? The same number that reach point B and start back towards
A while our ant is making his journey from A to B. That's the ants that are on the path from A to B when our ant starts, 2 per second over 60
seconds, or another 120.
There are 120 ants in motion from B to A and another 120 in motion from A to B that
will turn around at B and meet us on their return journey, so the answer is