School Party

Let $N_{G}$ and $N_{B}$ be the number of girls and boys at the beginning of the party, and $N_{T} = N_{G} + N_{B}$ be the number of total people.

The problem statement says that at the beginning of the party $N_{G} = 0.6 N_{T}$ thus $N_{B} = 0.4 N_{T}$ , which follows the following expression: $N_{G} = \frac{3}{2}N_{B}$ , Eq. (1).

In a second moment, boys and girls left the party, and that can be written as $N_{G} - 8 = 2(N_{B} - 12)$ , Eq. (2).

Substituting Eq. (1) in Eq. (2), it follows that $\frac{3}{2}N_{B} - 8 = 2(N_{B} - 12)$ , which leads to $N_{B} = 32$ . Using this result in Eq. (1), it follows that $N_{G} = 48$ , so $N_{T} = 80$ .

Let initial number of people be x

Number of girls = $\frac{60 \times x}{100}$ = $\frac{3x}{5}$

Number of boys = $\frac{40 \times x}{100}$ = $\frac{2x}{5}$

Now 12 boys and 8 girls went away

New number of girls = $\frac{3 \times x}{5}$ - 8 = $\frac{3x - 40}{5}$

New number of boys = $\frac{2 \times x}{5}$ - 12 = $\frac{2x - 60}{5}$

Now , Number of girls = 2 x Number of boys

$\Rightarrow \frac{3x - 40}{5}$ = $\frac{2(2x - 60)}{5}$

$\Rightarrow$ 3x - 40 = 2(2x -60)

$\Rightarrow$ 3x - 40 = 4x - 120

$\Rightarrow$ 4x - 3x = 120 - 40

$\Rightarrow$ x = 80

Therefore , the number of people initially present in the party is 8

Let x be the number of people initially at the party, B be the number of boys at the party, and G be the number of girls at the party.

From the problem,

x = B + G 0.6x = G x = 0.6x + B B = 0.4x

2(B-12) = G - 8

Since B = 0.4x and G = 0.6x, we can substitute and solve.

2(0.4x - 12) = 0.6x - 8 0.8x - 24 = 0.6x - 8 0.2x = 16

x = 80