Weighted-average ambiguity check (boys vs. girls): In a class of 30 students, the average weight of boys is 20 kg and the average weight of girls is 25 kg. What fraction of the class are boys?

Introduction / Context:
This question probes understanding of weighted averages. Knowing subgroup averages (boys vs. girls) is not enough to determine subgroup sizes unless the overall average (or equivalent information) is also provided.

Given Data / Assumptions:

• Total students = 30.
• Average weight of boys = 20 kg.
• Average weight of girls = 25 kg.
• Overall average of the entire class is not given.

Concept / Approach:
Let the number of boys be b and girls be g with b + g = 30. The overall average A must satisfy 20*b + 25*g = A*(b + g) = 30A. Without A (or an equivalent constraint), infinitely many (b, g) pairs can satisfy different overall averages, making the fraction of boys indeterminate.

Step-by-Step Solution:
20b + 25g = 30A.But g = 30 − b, so 20b + 25(30 − b) = 30A.This reduces to an equation containing A and b. Without A, b cannot be uniquely determined.

Verification / Alternative check:
Pick two different overall averages between 20 and 25 (say 22 and 24). Each will lead to different b values that still respect the subgroup averages and total count, demonstrating non-uniqueness of the answer.

Why Other Options Are Wrong:

• 4/5, 5/6, and 3/4 assert specific fractions without any overall average or other binding information to fix b uniquely.

Common Pitfalls:

• Assuming the overall average is the arithmetic mean of 20 and 25; it is not unless subgroup sizes are equal.
• Forgetting that weighted averages depend on counts as well as subgroup means.

Final Answer:
Data insufficient