In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was

The correct answer is 55.

Let's represent the initial number of boys in the class as $B$ and the initial number of girls as $G$.

According to the problem, the number of boys is more than 10. We can write this as:

$B>10$

Next, we are told that 40% of the girls left the class. This means that 100% - 40% = 60% of the girls remained. The number of remaining girls is:

$0.6G$

Similarly, 60% of the boys left the class, which means 100% - 60% = 40% of the boys remained. The number of remaining boys is:

$0.4B$

We are given that the remaining number of girls was 8 more than the remaining number of boys. We can set up the following equation:

$0.6G=0.4B+8$

To make the equation easier to work with, let's multiply the entire equation by 10 to remove the decimals:

$6G=4B+80$

We can simplify this further by dividing the entire equation by 2:

$3G=2B+40$

Since the number of people must be a whole number, $B$ and $G$ must be integers. Furthermore, because people left the class in exact percentages, the number of people who left must also be whole numbers.

40% of girls left, which is equivalent to the fraction 2/5 of the girls. For this to be a whole number, $G$ must be a multiple of 5.

60% of boys left, which is equivalent to the fraction 3/5 of the boys. For this to be a whole number, $B$ must also be a multiple of 5.

Let's express $G$ and $B$ in terms of variables $x$ and $y$, where $x$ and $y$ are positive integers:
$G=5x$
$B=5y$

We already know that $B>10$. Substituting $B=5y$ into this inequality:

$5y>10$

$y>2$

Since $y$ is an integer greater than 2, its minimum possible value is 3.

Now, let's substitute $G=5x$ and $B=5y$ back into our simplified equation:

$3\left(5x\right)=2\left(5y\right)+40$

$15x=10y+40$

Divide the entire equation by 5 to simplify:

$3x=2y+8$

We want to find the minimum possible number of total students initially, which means we need to find the lowest valid combination of $x$ and $y$. We will test values for $y$ starting from its minimum possible value, 3, until we find an $x$ that is a whole number.

Case 1: Let $y=3$

$3x=2\left(3\right)+8$
$3x=14$
$x=\frac{14}{3}$

Since $x$ must be an integer, this is not a valid solution.

Case 2: Let $y=4$

$3x=2\left(4\right)+8$
$3x=16$
$x=\frac{16}{3}$

Again, $x$ is not an integer, so we reject this value.

Case 3: Let $y=5$

$3x=2\left(5\right)+8$
$3x=18$
$x=6$

Here, $x$ is a valid integer. This gives us our minimum valid combination.

Now, let's find the original number of boys and girls using these values for $x$ and $y$:

Number of boys: $B=5y=5\left(5\right)=25$

Number of girls: $G=5x=5\left(6\right)=30$

The problem asks for the minimum possible number of students initially in the class, which is the sum of the boys and girls:

$\text{Total Students}=B+G=25+30=55$

Therefore, the minimum possible initial number of students is 55.