In the year 2010, 5000 students were admitted in a college. It is found that the number of students admitted is constantly increasing by 24 percent per year. How many students will be admitted in the college in the year 2012?

### Concept & Formula This problem applies the formula for compound interest to population or quantity growth over time. When a quantity increases by a constant percentage annually, the final amount is calculated as: $$ A = P \left(1 + \frac{R}{100}\right)^T $$ Where $P$ is the initial population/quantity, $R$ is the annual rate of increase, and $T$ is the number of years. ### Step-by-Step Solution 1. **Given:** * Initial number of students ($P$) = $5000$ * Rate of increase ($R$) = $24\%$ * Time ($T$) = $2012 - 2010 = 2$ years 2. **Apply the formula:** * $$ \text{Students in 2012} = 5000 \times \left( 1 + \frac{24}{100} \right)^2 $$ * $$ = 5000 \times \left( \frac{124}{100} \right)^2 $$ 3. **Simplify and calculate:** * Simplify the fraction $\frac{124}{100}$ to $\frac{31}{25}$. * $$ = 5000 \times \left( \frac{31}{25} \right) \times \left( \frac{31}{25} \right) $$ * $$ = 5000 \times \frac{961}{625} $$ * Divide $5000$ by $625$, which is exactly $8$. * $$ = 8 \times 961 $$ * $$ = 7688 $$ ### Exam Strategy & Shortcut Use the successive percentage increase shortcut formula for 2 years: $X + Y + \frac{X \times Y}{100}$. Net increase $\%$ = $24 + 24 + \frac{24 \times 24}{100} = 48 + 5.76 = 53.76\%$. Now, simply find $153.76\%$ of $5000$. $153.76 \times 50 = \frac{15376}{2} = 7688$. This method avoids squaring large two-digit numbers and relies purely on mental addition and halving. ### Common Pitfall The most frequent error is treating this as simple interest and just adding $24\%$ twice ($48\%$ total). $48\%$ of $5000$ is $2400$, leading to a total of $7400$, which is deliberately placed as Option (a) to trap students. Growth on top of growth must always be compounded. ### Final Answer **Therefore, the correct answer is $7688$.**