What is the exact average of the six numbers: n, 35, 39, 42, p, and w? I. n is six more than w. II. w is four less than p.

Introduction / Context:
We must determine the exact arithmetic mean of six values given only linear relations among three variables (n, p, w). The question is sufficient only if a unique numeric average (not an expression in a variable) results.

Given Data / Assumptions:

• I: n = w + 6.
• II: w = p − 4 (equivalently, p = w + 4).
• Constants: 35, 39, 42.

Concept / Approach:
Combine relations and attempt to express the average purely as a number. If a free variable remains, the data are not sufficient.

Step-by-Step Solution:

Verification / Alternative check:
Pick two different w values (e.g., 0 and 2) to obtain distinct averages (21 and 22), both consistent with I and II—confirming non-uniqueness.

Why Other Options Are Wrong:

• A/B/C/E: Neither statement alone nor together yields a fixed numeric mean; a parameter remains free.

Common Pitfalls:
Assuming hidden constraints (e.g., integers only still leaves infinitely many values); treating an expression in w as a “determined” average.

Final Answer:
Both statements are NOT sufficient.