Reversed alphabet indexing – If the English alphabet is written in backward order (Z to A), find the 7th letter to the left of the 11th letter from the left end.

Introduction / Context:
Instead of mixing “right” and “left” references, this item asks for both counts from the left under the reversed alphabet arrangement, which is Z (position 1), Y (2), …, A (26). We must locate the 11th letter from the left in this reversed line and then move 7 letters to the left of it (i.e., toward Z).

Given Data / Assumptions:

• Reversed sequence indices: 1:Z, 2:Y, 3:X, 4:W, 5:V, 6:U, 7:T, 8:S, 9:R, 10:Q, 11:P, 12:O, 13:N, 14:M, 15:L, 16:K, 17:J, 18:I, 19:H, 20:G, 21:F, 22:E, 23:D, 24:C, 25:B, 26:A.
• 11th from the left → reversed index 11 = P.
• 7th to the left of that → subtract 7 from 11 → index 4.

Concept / Approach:
Counting from the left is straightforward because the reversed sequence is explicitly indexed left-to-right. After finding index 11 (P), moving left means going to a smaller index. 11 − 7 = 4. The letter at reversed index 4 is W.

Step-by-Step Solution:
Reference letter: index 11 → P.Shift left 7 positions: 11 − 7 = 4.Index 4 in the reversed list is W. Therefore, answer = W.

Verification / Alternative check:
Check with a quick mental span: indices 1..11 are Z, Y, X, W, V, U, T, S, R, Q, P. Moving 7 back from P (counting: P←Q←R←S←T←U←V←W) lands on W. This hand count agrees with the index arithmetic.

Why Other Options Are Wrong:

• H/I/D/Q map to indices 19/18/23/10 respectively under the reversed scheme and do not match the computed index 4.

Common Pitfalls:
Accidentally using the normal A→Z order or moving the wrong direction after identifying P. Always re-anchor your counting to the Z-to-A line printed left-to-right.

Final Answer:
W