The ratio of students in class A to class B is $5:3$. The number of boys in class B is $40\%$ of the total students in class B. The number of girls in class A is $11\frac{1}{9}\%$ more than the number of girls in class B. The average number of girls in both classes is $19$.

What is the percentage of girls in class B with respect to the number of boys in class A?

Question

The ratio of students in class A to class B is $5:3$. The number of boys in class B is $40\%$ of the total students in class B. The number of girls in class A is $11\frac{1}{9}\%$ more than the number of girls in class B. The average number of girls in both classes is $19$.

What is the percentage of girls in class B with respect to the number of boys in class A?

Accepted Answer

**Solution:**

1. Let the total number of students in class A be $5x$ and in class B be $3x$.

2. Number of boys in class B = $40\%$ of $3x = 0.4 	imes 3x = 1.2x$.

3. Therefore, number of girls in class B = total students in class B - boys in class B = $3x - 1.2x = 1.8x$.

4. Girls in class A are $11\frac{1}{9}\% = \frac{100}{9}\%$ more than girls in class B.

So, girls in class A = girls in class B $+ \frac{100}{9}\%$ of girls in class B = $1.8x + \frac{100}{9} 	imes \frac{1.8x}{100} = 1.8x + 0.2x = 2x$.

5. Average number of girls in both classes = $19$.

So, $\frac{	ext{girls in A} + 	ext{girls in B}}{2} = 19$.

Substitute values:

$\frac{2x + 1.8x}{2} = 19 \implies \frac{3.8x}{2} = 19 \implies 3.8x = 38 \implies x = 10$.

6. Now, boys in class A = total students in A - girls in A = $5x - 2x = 3x = 3 	imes 10 = 30$.

7. Girls in class B = $1.8x = 1.8 	imes 10 = 18$.

8. Required percentage = $\frac{	ext{girls in B}}{	ext{boys in A}} 	imes 100 = \frac{18}{30} 	imes 100 = 60\%$.

**Answer:** $60\%$

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