How many people are standing between A and D (Note: All are standing in a straight line facing north)? Statement I: K stands second from the left end of the line. Only four people stand between K and T. Y is an immediate neighbour of T. A stands second to the right of Y. As many people stand between K and D as between A and D. Statement II: A stands second from the right end of the line. Z stands third from the left end of the line. D stands exactly in the centre of the line. As many people stand between A and T as between D and Z.

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We must decide sufficiency of the two statements for finding the number of people between A and D.

Statement I — Not sufficient (example of ambiguity):
- From I: K is 2nd from left ⇒ K at position 2. There are exactly 4 people between K and T ⇒ T is at position 7 (so total n ≥ 7). Y is an immediate neighbour of T, and A is 2nd to the right of Y.
- Construct two valid arrangements consistent with I that give different numbers between A and D.
  - Example 1 (total n = 8): positions 1..8, K=2, T=7, take Y=6 ⇒ A is 2nd right of Y ⇒ A=8. The condition “as many between K and D as between A and D” gives |2−d|−1 = |8−d|−1 ⇒ |2−d| = |8−d| ⇒ d = 5. Then people between A and D = |8−5|−1 = 2.
  - Example 2 (total n = 10): positions 1..10, K=2, T=7, take Y=8 ⇒ A=10. Then |2−d| = |10−d| ⇒ d = 6. People between A and D = |10−6|−1 = 3.
- Since statement I allows at least two consistent arrangements with different answers (2 and 3), I alone is not sufficient.

Statement II — Sufficient:
- Let total number of people be $n$, which must be odd because $D$ stands exactly in the centre. Put $n=2k+1$ so centre $D$ is at position $k+1$.
- Given: $Z$ is 3rd from left ⇒ $Z$ at position 3. $A$ is 2nd from right ⇒ $A$ at position $n-1=2k$. "As many people stand between A and T as between D and Z."
- Number between $D$ and $Z$ is $(k+1)-3-1 = k-3$ (must be ≥ 0). Number between $A$ and $T$ equals this, so
  $|2k - t| - 1 = k - 3 \implies |2k - t| = k - 2.$
  Hence $t = 2k \pm (k-2)$. The two possibilities are
  $t_1 = k+2$ and $t_2 = 3k-2$.
- Feasibility requires $1 \le t \le n = 2k+1$. For $t_2$ we need $3k-2 \le 2k+1 \implies k \le 3$. Also $Z$ at 3 must be strictly left of centre: $3 < k+1 \implies k > 2$. Combining gives $k = 3$ uniquely.
- Thus $n = 2k+1 = 7$. With $n=7$: $D$ at position 4, $Z$ at 3, $A$ at position 6, so number between $A$ and $D$ is $|6-4|-1 = 1$.
- Because $k$ is forced to be $3$, statement II determines the arrangement enough to compute the unique number between A and D (which is $1$).

Conclusion: Statement I alone is not sufficient, while statement II alone is sufficient. FINAL_ANSWER: e

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