Directions: Study the table carefully to answer the questions that follow: Number of Boys and Girls (in Hundreds) in Six Different Years in 5 Different Schools | Years | A (Boys) | A (Girls) | B (Boys) | B (Girls) | C (Boys) | C (Girls) | D (Boys) | D (Girls) | E (Boys) | E (Girls) | |---|---|---|---|---|---|---|---|---|---|---| | 2005 | 3.3 | 3.6 | 5.2 | 3.1 | 5.5 | 4.5 | 2.4 | 1.4 | 6.5 | 6.6 | | 2006 | 6.6 | 4.2 | 4.9 | 2.2 | 6.9 | 3.3 | 4.4 | 2.3 | 5.5 | 3.6 | | 2007 | 9.3 | 6.9 | 4.7 | 4.2 | 5.8 | 4.9 | 6.4 | 3.3 | 2.7 | 2.4 | | 2008 | 5.4 | 9.6 | 6.3 | 5.4 | 6.6 | 5.2 | 5.3 | 5.4 | 5.4 | 5.7 | | 2009 | 8.4 | 12.9 | 7.5 | 5.9 | 8.7 | 6.6 | 12.1 | 5.2 | 6.8 | 6.5 | | 2010 | 12.3 | 14.4 | 9.8 | 4.4 | 11.7 | 4.2 | 12.2 | 9.4 | 10.8 | 12.7 | What is the approximate percentage decrease in the number of boys in School D in the year 2008 as compared to that in the previous year ?

### Concept & Percentage Decrease To find the percentage decrease between two values, we use the percentage change formula: $$ \text{Percentage Decrease} = \frac{\text{Initial Value} - \text{Final Value}}{\text{Initial Value}} \times 100 $$ ### Step-by-Step Solution * Number of boys in School D in the previous year (2007) = $6.4$ (in hundreds). * Number of boys in School D in the year 2008 = $5.3$ (in hundreds). * Decrease in number of boys = $6.4 - 5.3 = 1.1$ * Percentage Decrease = $\frac{1.1}{6.4} \times 100$ * Percentage Decrease = $\frac{11}{64} \times 100 \approx 17.1875\%$ * The approximate percentage decrease is $17\%$. ### Exam Strategy & Shortcut For quick estimation, observe that $1.1$ out of $6.4$ is roughly $1$ out of $6$, which corresponds to $16.66\%$. The closest given option is $17\%$. ### Common Pitfall A common mistake is using the final year's value as the denominator. Always use the base year (in this case, the previous year, 2007) for calculating percentage change. ### Final Answer Therefore, the correct answer is **17%**.