We assume that each scenario we begin with the same amount of grass, say g₀, and that the grass grows at the same rate, say r_g inches/day, and that each cow eats at the same rate, say r_c inches/day/cow. From this we may establish the following equations

\[ \begin{eqnarray*} g_0 + 24r_g &=& 24\cdot 70r_e, \\ g_0 + 60r_g &=& 60\cdot 30r_e, \\ g_0 + 96r_g &=& 96cr_e \end{eqnarray*} \]

where c is the number of cows being sought. Subtracting the first equation from the second (after a little algebra) yields r_e = 3/10 r_g. Subtracting the second equation from the third gives 36r_g = r_e(96c - 60·30). Using the substitution r_e = 3/10 r_g in this last equation we see that r_g will cancel out and after a little bit of simplification, we get c = 20.