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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?

a: 120
b: 60
c: 240
d: 180

I marked A as that ant will meet 120 ants as at a point only 2 ants can pass per second.

But answer is

C

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    $\begingroup$ Please make the effort to type in the question instead of just posting a picture. This ensures that the question will be found when someone uses the search function. Making the title descriptive is helpful too. $\endgroup$
    – Jaap Scherphuis
    Commented Jun 22, 2020 at 13:36
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    $\begingroup$ In the last couple of days you've posted no fewer than seven of these questions. Maybe dial it back a bit? $\endgroup$
    – Gareth McCaughan ♦
    Commented Jun 22, 2020 at 15:05
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    $\begingroup$ @user795826 It's a typo; he meant HNQ, which means "hot network questions". See the sidebar on the right. $\endgroup$
    – Joseph Sible-Reinstate Monica
    Commented Jun 23, 2020 at 5:29
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    $\begingroup$ The other bit of context needed to understsand Andrew's comment is that unfortunately Puzzling questions that get to HNQ are not the best ones. The HNQ algorithm prefers questions with lots of answers, and on Puzzling that's usually a sign either of a question that's too broad and vague, or of a question that for some reason attracts lazy answerers who haven't bothered to read what's already there. IQ-test-type questions often produce both of those effects. $\endgroup$
    – Gareth McCaughan ♦
    Commented Jun 23, 2020 at 6:48
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    $\begingroup$ If someone posts a really great puzzle here and it attracts a really great answer, it might get HNQed, but usually not. (Because the formula has a term in it like "number of answers times question score", and really great puzzles don't usually produce a lot of answers, and to be HNQ-worthy with only one or two answers the question's score needs to be very high.) $\endgroup$
    – Gareth McCaughan ♦
    Commented Jun 23, 2020 at 6:50

5 Answers 5

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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.

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    $\begingroup$ This is a more intuitive answer than mine. $\endgroup$
    – Damila
    Commented Jun 22, 2020 at 13:40
  • $\begingroup$ I think this expressly means that what what we understand by meeting. If you think that Newton means Bing between point ine and point b then pause the answer would be 240 but effect means passing by then the answer should actually 120. $\endgroup$
    – Pandey_Ji
    Commented Jun 22, 2020 at 15:00
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    $\begingroup$ The one point is that this assumes that our ant is somewhere in the middle of the stream, starting after at least a minute has passed. For example, if he is first, he will not pass any on his way from A to B. $\endgroup$
    – Damila
    Commented Jun 22, 2020 at 20:16
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    $\begingroup$ But, tell me, would you be confused if 239, would have been an option as well? Because, for 240, we are counting the first ant which our ant has crossed entering while entering the line A, and hence it may not be counted as for moving from A to B. $\endgroup$
    – Pandey_Ji
    Commented Jun 23, 2020 at 19:04
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    $\begingroup$ @Pandey_Ji I agree. The first ant it meets after leaving A will itself have left A just under 2 minutes ago. The returning ant that left A exactly 2 minutes ago will be also be at A at the start. A good argument could certainly be made that 239 is a better answer. If ants not only turned around at B but also at A, then you would have 240 ants in total, and our ant would meet the 239 other ants exactly once on his trip from A to B. $\endgroup$
    – Jaap Scherphuis
    Commented Jun 23, 2020 at 19:43
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The answer is

$C$

The reason is

Our ant's relative velocity to the returning ants is 4 ants per second

The explanation:

-If our ant was not moving, then 2 ant-lengths per second would pass him, and 120 ants per minute would pass him.

-But our ant is moving from A to B at a velocity of 2 ant-lengths per second.

-Returning ants are moving from B to A at a velocity of 2 ant-lengths per second.

-Therefore, our ant is moving relative to the returning ants at 4 ant-lengths per second (2+2).

-60 seconds per minute * 4 ant-lengths per second = 240 ant-lengths per minute.

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While the reasoning below has already been given, here is a more visual example.

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.

A Red ant moves up, while Green, Blue and Yellow ants move down

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  • $\begingroup$ That looks like ants from SimAnt $\endgroup$
    – justhalf
    Commented Jun 24, 2020 at 8:46
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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.

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  • $\begingroup$ Yes you are right!! $\endgroup$
    – user795826
    Commented Jun 23, 2020 at 11:59
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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.

But also

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.

Therefore

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 240 ants.

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    $\begingroup$ What do you mean on the field? I would take it to be the ants going from $A$ to $B$ plus the ants going from $B$ to $A$ at any instant. In that case there are $240$ at any point in time. Our ant meets the $240$ who started before him. The $241st$ was off the field before our ant started, The $240th$ leaves just as our ant arrives. $\endgroup$
    – Ross Millikan
    Commented Jun 23, 2020 at 1:58
  • $\begingroup$ I meant on the return path $\endgroup$
    – Jason Goemaat
    Commented Jun 23, 2020 at 16:28

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