Six boys A, B, C, D, E and F were born on the same day but in different years…

Concept

In a ranking/linear-order puzzle, a statement is considered the correct answer only if it is necessarily implied by the given clues—that is, it must be true in every arrangement that satisfies all the constraints. A statement that is true in only one valid arrangement but false in another is not a definite conclusion and therefore cannot be the answer.

Setup

There are six distinct birth years from 1950 to 1955: 1950, 1951, 1952, 1953, 1954, and 1955.

Here, "oldest" means the earliest birth year (the smallest year).

Deduction

2. F is the oldest, so F = 1950, the earliest birth year.

4. F is 2 years older than D, so D = 1950 + 2 = 1952.

6. A was born in either 1952 or 1953. Since 1952 is already assigned to D, A = 1953.

8. B is older than D (1952) and also older than E, so B must have been born before 1952. The only unassigned year before 1952 is 1951, since 1950 is already assigned to F. Therefore, B = 1951.

10. A (1953) is older than C, so C must have been born after 1953, i.e., 1954 or 1955. The remaining two years, 1954 and 1955, are assigned to C and E.

Why the Answer Is Definite, but C Is Not

Steps 1–4 assign F, D, A, and B to unique birth years. Therefore, B = 1951 is true in every valid arrangement, making it a necessary conclusion.

However, C is not uniquely determined. Both of the following assignments satisfy all the clues:

• C = 1954, E = 1955

• C = 1955, E = 1954

In both cases, A = 1953 is older than C, so all the given conditions are satisfied.

Therefore, the statement "C was born in 1954" is true only in some valid arrangements, not in all of them. Hence, it is not a definite conclusion.

Conclusion

B was born in 1951 is the only statement that is true in every valid arrangement, making it the correct answer.