$44$ cows can graze a field in $9$ days. How many less / more cows will graze the same field in $12$ days?

This problem follows inverse variation rules, governed by the formula:
$$M_1\times D_1=M_2\times D_2$$
Where:$M_1=44$ cows$D_1=9$ days$D_2=12$ days$M_2=$ number of cows needed for $12$ daysSubstitute the parameters:
$$44\times 9=M_2\times 12$$
$$M_2=\frac{44\times 9}{12}$$
Simplifying by dividing by $4$ and $3$:
$$M_2=\frac{11\times 9}{3}=11\times 3=33\text{ cows}$$
Initially, there were $44$ cows. Now, $33$ cows are required.
$$\text{Difference}=44-33=11\text{ cows less}$$