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This question is based on the following letter-clusters. BOC TWE JVN HEQ In each of the letter-clusters, each vowel is changed to the letter immediately following it in the English alphabetical order and each consonant is changed to the letter immediately preceding it in the English alphabetical order. In how many letter-clusters thus formed, will no vowel appear?
One
Four
Three
Two
Two
Identify vowels first, as their transformation (V+1) is more likely to result in another vowel (e.g., E becomes F, I becomes J, O becomes P, U becomes V, A becomes B). Only vowels A, E, I, O, U can generate new vowels.
Identify vowels first, as their transformation (V+1) is more likely to result in another vowel (e.g., E becomes F, I becomes J, O becomes P, U becomes V, A becomes B). Only vowels A, E, I, O, U can generate new vowels.
Mapping Rules
Rule 1: Vowel (A, E, I, O, U) тЖТ Next letter (+1). Rule 2: Consonant тЖТ Previous letter (-1).
Transformation Analysis
BOC: B(C-1)=A, O(V+1)=P, C(C-1)=B. Result: APB (Contains vowel A). TWE: T(C-1)=S, W(C-1)=V, E(V+1)=F. Result: SVF (No vowels). JVN: J(C-1)=I, V(C-1)=U, N(C-1)=M. Result: IUM (Contains vowels I, U). HEQ: H(C-1)=G, E(V+1)=F, Q(C-1)=P. Result: GFP (No vowels).
Final Count
The transformed clusters are APB, SVF, IUM, and GFP. SVF and GFP contain no vowels. Total count is 2.
A: One is incorrect as SVF and GFP both satisfy the condition. B: Four is incorrect because APB and IUM clearly contain vowels. C: Three is incorrect because only two clusters satisfy the condition.
D is correct because applying the transformation rules results in SVF and GFP, which are the only two clusters containing no vowels among the transformed sets APB, SVF, IUM, and GFP.
When checking for vowels, always note that the transformation V+1 can only produce a vowel if the original vowel was 'A' (A+1=B, not a vowel), 'E' (E+1=F), 'I' (I+1=J), 'O' (O+1=P), or 'U' (U+1=V). Wait, actually E, I, O, U plus 1 never results in a vowel; only the consonant transformation needs careful checking.
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