In a class, 60% students are boys and 40% are girls. By admitting 16 boys and 8 girls

Question

In a class, 60% students are boys and 40% are girls. By admitting 16 boys and 8 girls the ratio of boys to girls becomes 8:5. What must be the number of boys and number of girls originally in the class?

Detailed Solution

Step 1: Understand the Initial Ratio

First, let's determine the ratio of boys to girls based on the given percentages.

Given:

• Boys = 60%
• Girls = 40%

The ratio of Boys to Girls is:

$$ \frac{60}{40} = \frac{3}{2} $$

So, we can assume the original number of boys is \( 3x \) and the original number of girls is \( 2x \), where \( x \) is a common multiplier.

Step 2: Apply the Changes

According to the problem, new students are admitted:

• Number of boys admitted = 16
• Number of girls admitted = 8

The new quantities become:

$$ \text{New Boys} = 3x + 16 $$ $$ \text{New Girls} = 2x + 8 $$

Step 3: Formulate the Equation

The problem states that after these admissions, the new ratio of boys to girls becomes 8:5.

We can write this as an equation:

$$ \frac{3x + 16}{2x + 8} = \frac{8}{5} $$

Step 4: Solve for x

Cross-multiply to solve the equation:

$$ 5(3x + 16) = 8(2x + 8) $$

Expand the brackets:

$$ 15x + 80 = 16x + 64 $$

Rearrange the terms to isolate \( x \):

$$ 80 - 64 = 16x - 15x $$ $$ 16 = x $$

Step 5: Calculate Original Numbers

Now that we know \( x = 16 \), we can find the original number of students.

Original Number of Boys (\( 3x \)):

$$ 3 \times 16 = 48 $$

Original Number of Girls (\( 2x \)):

$$ 2 \times 16 = 32 $$

Answer: Originally, there were 48 Boys and 32 Girls.