Teacher distributes ‘N’ chocolates to ‘3x’ students. Find the value of x. A.…

Concept

In a Data Sufficiency problem the goal is not to compute a value but to decide which combination of the given statements pins the unknown to exactly ONE value. A combination is sufficient only if every value consistent with it forces a single answer; if two or more values survive, that combination is insufficient. Here the unknown is x, where the number of students is 3x.

Translate each statement

• A: giving 6 each leaves 24, so N = 6(3x) + 24 = 18x + 24. One equation linking N and x — not enough alone.

• B: with x students (one-third of 3x) nothing is left, so N is exactly divisible by x. A divisibility condition — not enough alone.

• C: 30 < 3x < 40 gives x in {11, 12, 13}; and 200 < N < 300. A range — not enough alone.

Test every pair

1. A and C: N = 18x + 24 with 30 < 3x < 40 and 200 < N < 300 allows x = 11 (N = 222), x = 12 (N = 240) and x = 13 (N = 258). Three survivors, so x is not fixed.

2. B and C: x in {11, 12, 13} with N divisible by x and 200 < N < 300 leaves many N for each of the three x values; x is not fixed.

3. A and B: N = 18x + 24 and x divides N means x divides 24, so x in {1, 2, 3, 4, 6, 8, 12, 24}. Many survivors, so x is not fixed.

Conclusion

No pair of statements forces a single x. The unknown is pinned to one value only when all three are used at once: x must divide 24 (from A and B) and lie in {11, 12, 13} (from C), and 12 is the only number meeting both, giving N = 18(12) + 24 = 240 and 3x = 36 students. Since every offered choice relies on a pair (or 'any two'), and no pair is sufficient, the correct response is that the given statements (in the offered combinations) cannot answer the question.

Cross-check

Verify x = 12 against all three: A gives N = 240 (6 × 36 + 24 = 240, true); B needs 240 divisible by 12 (240 / 12 = 20, true); C needs 30 < 36 < 40 and 200 < 240 < 300 (both true). All three are needed simultaneously — confirming no pair alone is enough.