More sheep, fewer days

A flock of one hundred and twenty sheep lives on one field. Each sheep eats two kilograms of grass a day, so the flock eats two hundred and forty kilograms a day.

A flock of $120$ sheep, $2\mathrm{kg}$ of grass per sheep per day: the flock eats $120\times 2=240\mathrm{kg}$ a day.

sheep $s$, days $d$: $s=120,2s=240\mathrm{kg}$ a day.

The field holds one thousand six hundred and eighty kilograms of grass. Dividing, the grass lasts seven days.

The field has $1680\mathrm{kg}$ of grass, and $1680÷240=7$ days.

$d=1680÷240=7$.

Now twenty more sheep arrive. That is only a sixth more sheep. But one hundred and forty sheep eat two hundred and eighty kilograms a day, and the grass lasts only six days. Twenty extra sheep cost a whole day.

$20$ more sheep arrive: $140\times 2=280\mathrm{kg}$ a day, and $1680÷280=6$ days. Twenty extra sheep cost a whole day.

$s=140\Rightarrow d=1680÷280=6$.

The rule: when the supply is fixed, the number of days is inversely proportional to the number of sheep. More mouths, fewer days; the product of sheep and days stays the same, at eight hundred and forty.

The rule: the supply is fixed, so days are inversely proportional to sheep. $120\times 7=140\times 6=840$.

$d\propto \frac{1}{s}$ and $sd=120\times 7=140\times 6=840$.