A group of farmers need to harvest 2 farms, one of which is twice the size of the other.

They worked half of the first day on the bigger farm. Then they split into two equal groups to work on harvesting both farms.

By end of the first day, the bigger farm was completely harvested. The smaller farm had a some work left to be done, but a single farmer was able to finish harvesting it by the end of the second day.

How many farmers were in the group?

  • $\begingroup$ @RobWatts, I appreciate your edit :) $\endgroup$
    – Rafe
    Jan 13, 2015 at 19:35
  • 5
    $\begingroup$ This question appears to be off-topic because it belongs on Math.SE. $\endgroup$
    – xnor
    Jan 13, 2015 at 21:49

2 Answers 2


Let's say that there are $x$ farmers. Since they spent half of the day working on the larger farm, we know that the larger farm requires at least $\frac{x}{2}$ man-days for harvest. Then, for the second half of the first day, $\frac{x}{2}$ farmers worked on each farm. Since the larger farm was then finished, we know it requires $\frac{x}{2}+\frac{\frac{x}{2}}{2}=\frac{3x}{4}$ man-days.

The smaller farm received $\frac{x}{4}$ man-days of work on the first day, and since it is half the size of the larger farm should require $\frac{3x}{8}$ total man-days of work to finish. This means $\frac{3x}{8}-\frac{x}{4}=\frac{x}{8}$ man-days of work are left, and since that was done by one farmer in one day, $\frac{x}{8}=1$ so there are 8 farmers in the group.


Whoops, read that as 3 days total. 8 farmers, same farmer-days needed
We know the large field takes $3X/4$ farmer-days to complete, and the smaller field takes $X/4 + 1$ farmer-days to complete. Since it is half the size, we have the equality $X/2 + 2 = 3X/4$ which resolves to $X = 8$.
There were 4 farmers. The big field took 6 farmer-days, the second field took 3 farmer-days (yeah, I made the unit up).


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