The Farmer's Oxen
(
MISCELLANEOUS PUZZLES)
A child may propose a problem that a sage cannot answer. A farmer propounded the following question: "That ten-acre meadow of mine will feed twelve bullocks for sixteen weeks or eighteen bullocks for eight weeks. How many bullocks could I feed on a forty-acre field for six weeks, the grass growing regularly all the time?"
It will be seen that the sting lies in the tail. That steady growth of the grass is such a reasonable point to be considered, and yet to some readers it will cause considerable perplexity. The grass is, of course, assumed to be of equal length and uniform thickness in every case when the cattle begin to eat. The difficulty is not so great as it appears, if you properly attack the question.
Answer:
Sir Isaac Newton has shown us, in his Universal Arithmetic, that we may divide the bullocks in each case in two parts—one part to eat the increase, and the other the accumulated grass. The first will vary directly as the size of the field, and will not depend on the time; the second part will also vary directly as the size of the field, and in addition inversely with the time. We find from the farmer's statements that 6 bullocks keep down the growth in a 10-acre field, and 6 bullocks eat the grass on 10 acres in 16 weeks. Therefore, if 6 bullocks keep down the growth on 10 acres, 24 will keep down the growth on 40 acres.
Again, we find that if 6 bullocks eat the accumulated grass on 10 acres in 16 weeks, then
12 |
eat the |
grass on |
10 |
acres in |
8 |
weeks, |
48 |
" |
" |
40 |
" |
8 |
" |
192 |
" |
" |
40 |
" |
2 |
" |
64 |
" |
" |
40 |
" |
6 |
" |
Add the two results together (24 + 64), and we find that 88 oxen may be fed on a 40-acre meadow for 6 weeks, the grass growing regularly all the time.
