Ancient mathematical task from Lion Tolstoy

[Zorro]

Ancient mathematical task from the Russian writer Lion Nikolayevich Tolstoy, which he asked schoolchildren 12-13 years old Can you solve this task?

The workers group needed to mow two meadows. The first meadow is twice the second meadow. Half of the day the group mowed a big meadow. Then the group was divided in half. One half of the group remained on a large meadow and completed its work in the evening; Another half of the group went to work on a small meadow and in the evening they had an unprocessed small part of the meadow. The next morning, one worker went to finish the work, and in one day he coped with this mission. Question: How many people were in the group?

 

[Minimalist]

Good one, took me less than 10 minutes...

 

[Zorro]

Oh, the previous semester my university professor gave us that task and said, that those students, who will solve it in 1:30 hours will be exempt from writing a semester essay. I was doing my best to solve it, because his essays are always so complicated, but failed. Only one girl from the whole group managed to solve it, and the rest needed to write that essay. But thanks to the educational resources online, especially to Edubirdie I managed to write it on time. But still, sometimes I think about that day, and I would like to find the solution.

The solution of the problem. In this case, in addition to the main unknown - the number of workers, which we will denote by x - it is convenient to introduce an auxiliary one, namely, the size of the plot mowed by one worker in 1 day; let's denote it by y.
Although the problem does not require its definition, it will make it easier for us to find the main unknown. Express in terms of x and y the area of a large meadow.
This meadow was mowed for half a day by x workers; they mowed - x × 1/2 × y = xy/4. In the second half of the day, only half of the group mowed it, that is, x / 2 workers; they mowed x/2 × 1/2 × y = xy/4.
Since the whole meadow was mown by evening, its area is equal to xy/2 + xy/4 = 3xy/4. Let us now express the area of the smaller meadow in terms of x and y. It was mowed by x/2 workers for half a day and the area was mowed by x/2 × 1/2 × y = xy/4.
Let us add an uncultivated area just equal to y (the area cultivated by one worker in 1 working day), and we get the area of the smaller meadow: xy/4 + y = (xy + 4y)/4.
It remains to translate into the language of algebra the phrase: "the first meadow is twice as large as the second" - and the equation is composed: 3xy/4 : (xy + 4y)/4 = 2, or 3xy/ (xy + 4y) = 2. Reduce the fraction on the left side of the equation by y; the auxiliary unknown is thus eliminated, and the equation takes the form 3x/(x + 4) = 2, or 3x = 2x + 8. In the end we get x = 8. There were 8 workers in the group.

You can also say this:

Let "s" be the unit of area mowed by one worker in one day.
Let "x" be the number of workers equal to half the group.
Then, 2x - is a whole group.
Then, xs is the area mowed by half of the group for the whole day.
xs / 2 - the area that is mowed by half of the group in half a day.
First half of the day: a whole gang mowed 2 * (xs / 2) = xs of the area from a large meadow.
Second half of the day: half of the artel mowed xs/2 area out of a large meadow.
The other half mowed the same area from a small meadow: xs/2.
Second day: one worker mowed from a small meadow "s" square in one day.
Total area of a large meadow: S1 = xs + xs/2
Total area of a small meadow: S2 = xs/2 + s
Because a large meadow is twice as large as a small one, S1 = 2S2,
xs + xs/2 = 2 * (xs/2 + s)
xs + xs/2 = 2*xs/2 + 2s
xs - 2s = 2*xs/2 - xs/2
xs - 2s = xs/2
2(xs - 2s) = xs
2xs - 4s = xs
2xs - xs = 4s
xs = 4s
x = 4 (workers) - this is half of the group.
x*2 = 8
Consequently, there were 8 workers in the group.

 

[FosterWest]

It`s interesting but very difficult for me)