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.
Correct Answer: If both the statements are NOT sufficient.
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:
1) Average A = (n + 35 + 39 + 42 + p + w)/6.2) Substitute I and II: n = w + 6 and p = w + 4 ⇒ numerator = (w + 6) + 35 + 39 + 42 + (w + 4) + w = 3w + (6 + 35 + 39 + 42 + 4) = 3w + 126.3) Hence A = (3w + 126)/6 = (w/2) + 21.4) The mean depends on w; without a numeric value for w, the average is not uniquely determined.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.