10 cows can graze in a field for 15 days and 20 cows can graze in the same field for 10 days. For how many days can 30 cows graze in the field?

### Concept & Formula This is a classic uniform growth problem (Newton's Grazing Problem). The total consumption equals the initial quantity plus the growth during that period. $$\\text{Total Consumption} = \\text{Initial Amount} + (\\text{Growth Rate} \\times \\text{Time})$$ ### Step-by-Step Solution * Let the initial grass in the field be $g$ units, and the grass growth rate per day be $r$ units. Let each cow consume 1 unit of grass per day. * Case 1: 10 cows for 15 days consume $10 \\times 15 = 150$ units. $$g + 15r = 150 \\quad \\text{--- (Equation 1)}$$ * Case 2: 20 cows for 10 days consume $20 \\times 10 = 200$ units. $$g + 10r = 200 \\quad \\text{--- (Equation 2)}$$ * Subtract Equation 2 from Equation 1: $$5r = -50 \\implies r = -10$$ *(Note: A negative growth rate means the grass is depleting over time naturally, or it acts as a constant reduction factor).* * Substitute $r = -10$ into Equation 2: $$g + 10(-10) = 200 \\implies g - 100 = 200 \\implies g = 300\\text{ units}$$ * Case 3: Let 30 cows graze for $d$ days. $$g + d \\cdot r = 30 \\cdot d$$ $$300 + d(-10) = 30d$$ $$300 = 40d \\implies d = \\frac{300}{40} = 7.5 = 7\\frac{1}{2}\\text{ days}$$ ### Exam Strategy & Shortcut Notice the arithmetic progression in cow counts: 10, 20, 30. Total consumption for 10 cows = 150. Total consumption for 20 cows = 200. The change per 10 additional cows is $+50$ units over a reduction of 5 days. For 30 cows, the total work required follows a highly predictable system of linear equations that you can rapidly solve by computing the differences in total cow-days. ### Common Pitfall Treating this as a simple inverse variation problem ($M_1 D_1 = M_2 D_2$) without accounting for the dynamically changing grass volume will lead to incorrect answers like 5 days. ### Final Answer Therefore, the correct answer is 7\\frac{1}{2} days.