Passengers, to the train!

Passengers, to the train!

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A train leaves the station with half of the seats occupied. In the first stop as many people went up as they went down. In the second stop, half of the people who got off at the first stop went up and twice as many people went up at the first stop. As many people as the sum of the people who got off at the first two stops went up on the last stop and as many people went down as the sum of the people who got on the first two stops.

How many seats are free now?


If you look, we see that as many people go up as they go down, so the occupation will be the same, so if initially half of the seats were occupied, now, half of the seats will be free.

We can see it mathematically as follows:

We define the following variables:
Y = people on the train initially
X = people who got on the first stop
Z = people left on the train

We can propose the following equation from the statement data:
Z = Y + X - X + X / 2 - 2 * X + (X + 2 * X) - (X + X / 2)

We can see that all terms with X cancel each other out
Z = Y + X / 2 - 2X + X + 2X - X - X / 2 = Y

So we have Z = Y from where we deduce that the train carries the same number of passengers as at
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