In a 10th class, there are four sections: A, B, C, and D. The overall ratio of boys to girls is $8:9$. In section B, the ratio of boys to girls is $1:3$. The number of girls in sections A and B are equal. In section C, the number of boys is 10 less than the number of girls. The number of girls in section D is 5 less than the number of boys in section B. The total number of girls is 225, out of which 55 are in section C. The ratio of boys in sections A and D is $5:8$.
What is the percentage difference of the number of girls in section D compared to the number of boys in section B? Is it more or less, and by what percent?

Question

Accepted Answer

**Solution:**

1. Total girls = 225.
2. Girls in section C = 55.
3. Girls in sections A and B are equal. Let girls in A = girls in B = x.

Therefore, girls in D = 225 - (55 + x + x) = 225 - 55 - 2x = 170 - 2x.

4. Boys to girls ratio overall is 8:9.

Total girls = 225, so total boys = $ $\frac{8}{9} \times 225 = 200$ $.

5. In section B, boys to girls ratio is 1:3.

Girls in B = x, so boys in B = $ $\frac{1}{3} x $$.

6. Number of girls in D is 5 less than boys in B:

Girls in D = boys in B - 5 = $ $\frac{x}{3} - 5$ $.

7. From step 3, girls in D = 170 - 2x.

Equate:

$$ $170 - 2x = \frac{x}{3} - 5$ $$

Multiply both sides by 3:

$$ 510 - 6x = x - 15 $$

$$ 510 + 15 = x + 6x $$

$$ 525 = 7x $$

$$ x = 75 $$

8. Girls in B = 75.

Boys in B = $ $\frac{1}{3} \times 75 = 25 $$.

9. Girls in D = boys in B - 5 = 25 - 5 = 20.

10. Percentage difference of girls in D compared to boys in B:

$$
  $  \text{Percentage difference} = \frac{\text{Girls in D} - \text{Boys in B}}{\text{Boys in B}} \times 100 = \frac{20 - 25}{25} \times 100 = \frac{-5}{25} \times 100 = -20\% $
    $$

Negative sign indicates girls in D are 20\% less than boys in B.

**Answer:** 20\%

Answer

Answer

Answer

Answer