When water is changed into ice, its volume increases by 9%. If ice changes into water, the percentage decrease in volume is

### Concept & Logic This problem revolves around the concept of changing base values in percentages. When a value increases by a certain percentage to reach a new amount, returning to the original value requires calculating the decrease relative to that *new, larger base*. If a quantity increases by $R\%$, the percentage decrease required to return to the original quantity is: $$Decrease \% = \frac{R}{100 + R} \times 100$$ ### Step-by-Step Solution * **Establish Initial Values:** Let the initial volume of water be $100$ units. * **Calculate Increased Volume:** When it turns to ice, the volume increases by $9\%$. Volume of ice = $100 + 9 = 109$ units. * **Calculate the Reversal:** When ice changes back to water, the volume must decrease from $109$ units back to $100$ units. Absolute decrease in volume = $109 - 100 = 9$ units. * **Calculate Percentage Decrease:** The decrease is calculated over the *volume of the ice*. Percentage decrease = $(\frac{\text{Absolute Decrease}}{\text{Base Volume}}) \times 100$ Percentage decrease = $(\frac{9}{109}) \times 100 = \frac{900}{109}\%$. * **Convert to Mixed Fraction:** Divide $900$ by $109$: $109 \times 8 = 872$. Remainder = $900 - 872 = 28$. Result = $8 \frac{28}{109}\%$. ### Exam Strategy & Shortcut Directly apply the base reversal formula: $\frac{R}{100+R} \times 100$. Here, $R = 9$. Answer = $\frac{9}{100+9} \times 100 = \frac{900}{109}\%$. A quick estimation: $\frac{900}{110} \approx 8.18\%$. The answer must be slightly higher than $8.18\%$, leading you directly to $8 \frac{28}{109}\%$ without full division. ### Common Pitfall The most frequent error is assuming that a $9\%$ increase in one direction means a $9\%$ decrease in the reverse direction. Percentages are always relative to their starting base, and the base changes from water (100) to ice (109). ### Final Answer **Therefore, the correct answer is $8 \frac{28}{109}\%$.**