A large watermelon weighs 20 kg with 96% of its weight being water. It is allowed to stand in the sun and some of the water evaporates so that only 95% of its weight is water. Its reduced weight will be

### Concept & Logic The **Constant Pulp Concept** applies perfectly here. While the water evaporates in the sun, the solid mass (pulp) of the watermelon remains identical. $$ \text{Initial Solid Mass} = \text{Final Solid Mass} $$ ### Step-by-Step Solution * **Given:** * Initial weight = $20$ kg * Initial water = $96\%$ * Final water = $95\%$ * **Calculation:** * Initial pulp percentage = $100\% - 96\% = 4\%$. * Weight of initial pulp = $4\%$ of $20 = 0.04 \times 20 = 0.8$ kg. * After evaporation, water is $95\%$, so the final pulp percentage = $100\% - 95\% = 5\%$. * Let the reduced total weight of the watermelon be $x$. * The new pulp weight must equal the old pulp weight: $5\%$ of $x = 0.8$ kg. * $\frac{5}{100} \times x = 0.8$ * $x = 0.8 \times \frac{100}{5}$ * $x = 0.8 \times 20 = 16$ kg. ### Exam Strategy & Shortcut **Inverse Ratio Method:** Initial Pulp percentage = $4\%$, Final Pulp percentage = $5\%$. Pulp Percentage Ratio = $4 : 5$. Since the pulp weight is completely constant, the Total Weight Ratio is the inverse = $5 : 4$. We are given the initial weight ($5$ units) = $20$ kg. So, $1$ unit = $4$ kg. The final weight ($4$ units) = $4 \times 4 = 16$ kg. ### Common Pitfall A frequent error is assuming a $1\%$ drop in water percentage means a $1\%$ drop in total weight (e.g., subtracting $0.2$ kg to get $19.8$ kg). Percentages are always relative to the total mass at that specific time, which is constantly shrinking. Always equate absolute constant values. ### Final Answer **Therefore, the correct answer is 16 kg.**