In a school’s annual sports meet, 1/4 of the boys and 3/8 of the girls participated. What fractional part of the total student population (boys + girls) took part in the event?

Introduction / Context:
In problems that ask for a fraction or percentage of a combined population, you must know either the actual counts or at least the ratio between the groups. Here, different fractions of boys and girls participated. Without knowing how many boys there are relative to girls, the overall (combined) fraction cannot be uniquely determined.

Given Data / Assumptions:

• Participating boys = 1/4 of all boys.
• Participating girls = 3/8 of all girls.
• Total students = boys + girls.
• No ratio or count of boys to girls is given.

Concept / Approach:
Overall participation fraction = (participating boys + participating girls) / (boys + girls). This simplifies to ( (1/4)*B + (3/8)*G ) / (B + G ). The value depends on the ratio B:G. Without B:G, there is no single numeric answer.

Step-by-Step Solution:
Let B = number of boys, G = number of girls.Participating = (1/4)B + (3/8)G.Total = B + G.Overall fraction = [ (1/4)B + (3/8)G ] / (B + G ).This expression varies with B:G; e.g., if B = G, fraction = (1/4 + 3/8) / 2 = (5/8) / 2 = 5/16 = 31.25%.If B = 2G, fraction = [ (1/4)*(2G) + (3/8)G ] / (3G) = [ (1/2)G + (3/8)G ] / (3G) = (7/8) / 3 = 7/24 ≈ 29.17%.

Verification / Alternative check:
Try different B:G ratios (1:1, 2:1, 1:2). Each gives a different combined percentage, confirming non-uniqueness.

Why Other Options Are Wrong:
32%: Only true for specific B:G, not universally.20%: Also depends on B:G; not guaranteed.36%: Same issue—no unique ratio provided.24%: Arbitrary without B:G information.

Common Pitfalls:
Assuming B = G without being told; treating weighted fractions as simple averages; ignoring the dependence on the population mix.

Final Answer:
data inadequate