Difficulty: Medium
Correct Answer: W
Explanation:
Introduction / Context:This odd-letter question is based on a simple classification property. A common approach in letter-based aptitude questions is to classify letters by vowel/consonant, even/odd alphabet position, or shape. Here, the cleanest dominant rule is based on alphabet position parity (even/odd).
Given Data / Assumptions:
Concept / Approach:Compute the alphabet position of each letter and check whether most of them share the same parity (even or odd). The letter that differs is the odd one out.
Step-by-Step Solution:
B is the 2nd letter of the alphabet, so position(B)=2 (even). N is the 14th letter, so position(N)=14 (even). P is the 16th letter, so position(P)=16 (even). A is the 1st letter, position(A)=1 (odd) (added distractor). W is the 23rd letter, so position(W)=23 (odd).Verification / Alternative check:Among the original four alternatives B, N, P, W: B(2), N(14), and P(16) are all even-position letters, while W(23) is odd-position. Therefore W breaks the shared even-position pattern and is the odd one out in the intended set.
Why Other Options Are Wrong:
B: even-position letter, matches the pattern. N: even-position letter, matches the pattern. P: even-position letter, matches the pattern. A: odd-position (added distractor); however the intended odd among the provided set is W because it alone breaks the dominant even-position rule.Common Pitfalls:Students often try vowel/consonant first, but all four original letters are consonants, so that does not help. Another pitfall is forgetting the letter positions (for example, mixing up N as 13 instead of 14). When vowel/consonant fails, position parity (even/odd) is a strong next check.
Final Answer:W
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