Explanation:

This problem relies on the principle that when a fruit or vegetable loses water, the amount of its dry matter (non-water content) remains constant. **Initial State:**\nTotal weight of cucumbers = $100 \text{ kg}$.\nWater content = $99\%$.\nDry matter content = $100\% - 99\% = 1\%$. \nCalculate the actual weight of dry matter:\nWeight of dry matter = $1\%$ of $100 \text{ kg} = \frac{1}{100} \times 100 \text{ kg} = 1 \text{ kg}$. **Final State (After 30 days):**\nAfter losing water, the dry matter content remains $1 \text{ kg}$.\nNew water content = $98\%$.\nNew dry matter content = $100\% - 98\% = 2\%$. \nLet the new total weight of cucumbers be $W_{\text{new}}$.\nWe know that $2\%$ of $W_{\text{new}}$ is the dry matter, which is $1 \text{ kg}$.\nSo, $2\%$ of $W_{\text{new}} = 1 \text{ kg}$. $\frac{2}{100} \times W_{\text{new}} = 1$ $W_{\text{new}} = \frac{1 \times 100}{2}$ $W_{\text{new}} = \frac{100}{2} = 50 \text{ kg}$. \nLet's evaluate the options:\nOption (1) $99 \text{ kg}$: If the new weight is $99 \text{ kg}$, the dry matter ($1 \text{ kg}$) would be $\frac{1}{99} \times 100\% \approx 1.01\%$, not $2\%$.\nOption (2) $50 \text{ kg}$: This matches our calculated new total weight. If the new weight is $50 \text{ kg}$, the dry matter ($1 \text{ kg}$) would be $\frac{1}{50} \times 100\% = 2\%$. This is correct.\nOption (3) $75 \text{ kg}$: If the new weight is $75 \text{ kg}$, the dry matter ($1 \text{ kg}$) would be $\frac{1}{75} \times 100\% \approx 1.33\%$, not $2\%$.\nOption (4) $2 \text{ kg}$: If the new weight is $2 \text{ kg}$, the dry matter ($1 \text{ kg}$) would be $\frac{1}{2} \times 100\% = 50\%$, not $2\%$. \nTherefore, the total weight of cucumbers now is $50 \text{ kg}$. The correct option is (2) 50 kg.