One of Sam Loyd's puzzles reads:

When they started off on the great annual picnic wagon in town was pressed into service.

Half way to the picnic ground ten wagons broke down, so it was necessary for each of the remaining wagons to carry one more person.

When they started home it was discovered that fifteen more wagons were out of commission, so on the return trip there were three persons more in each wagon than when they started out in the morning.

Now, who can tell me how many people attended the great annual picnic?

One solution is given here: http://www.mathsisfun.com/puzzles/great-picnic-puzzle-solution.html

I come to a different answer when I attempt to solve this algebraically. Can anyone tell me where I went wrong?

Let x be the number of wagons in the morning, n be the number of people per wagon in the morning, and k to be the total number of people who attended the picnic. Then

$nx = k$.

10 wagons break down and each of the remaining wagons yields an additional person. So we must have

$(x-10)(n+1) = k$.

Another words:

$xn + x - 10n - 10 = k $.

$xn + x - 10n - 10 = xn $

So that $x = 10(n+1)$.

We also know that an additional 15 wagons break down and an additional 3 folk hop on each remaining wagon, so that

$(x-25)(n+4) = k$.


$xn + 4x - 25n - 100 = k = xn$.


$4(10(n+1)) - 25n - 100 = 0$.

This simplifies to

n = 4.

Therefore x = 10(n+1) = 50 and k = xn = 200. So the number of folk that attend the picnic is 200, contrary to what the solution stated.


The question stated that there were three more people than when the wagon started out in the morning, not three more than in the last breakdown.

  • $\begingroup$ Ah excellent. In that case we have (x-25)(n+3)=k=xn, which simplifies down to n=9, x = 100, and k = 900. $\endgroup$ – user81362 Jun 7 '13 at 23:45

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