Three numbers are related as follows: the first number is exactly twice the second number, and at the same time the first number is half of the third number. If the arithmetic average of all three numbers is 56, determine the smallest number among the three.

Introduction / Context:
This problem tests proportional reasoning with simultaneous relationships between three unknowns and the use of an average. By translating each sentence into equations, we can solve systematically for each number and then identify the smallest value.

Given Data / Assumptions:

• The first number equals twice the second: let first = a, second = b, third = c, so a = 2b.
• The first is half of the third: a = c / 2, so c = 2a.
• The average is 56, so (a + b + c) / 3 = 56 → a + b + c = 168.
• All numbers are real; context suggests positive values.

Concept / Approach:
Use substitution to reduce the number of variables. Express a and c in terms of b, then substitute into the sum a + b + c = 168. Solve for b, recover a and c, and compare to find the minimum. This approach ensures no step is skipped and that relationships are consistent.

Step-by-Step Solution:

Verification / Alternative check:
Average check: (48 + 24 + 96) / 3 = 168 / 3 = 56, matching the given average. Relationship checks: a = 2b → 48 = 2 * 24 (true) and a = c / 2 → 48 = 96 / 2 (true).

Why Other Options Are Wrong:
36, 40, and 48 do not arise from the solved system. 30 would yield sums and relationships that do not satisfy both conditions simultaneously.

Common Pitfalls:
Confusing which variable is double or half of which, or averaging incorrectly (e.g., dividing by 2 instead of 3). Another frequent error is forgetting that c depends on a, which itself depends on b.

Final Answer:
24