If n is divided by 7, the remainder is 4. If 3n +1 is divided by 7 the remainder is:

Work with the remainder directly. Since n leaves remainder 4 on division by 7, write n = 7k + 4. Then 3n + 1 = 21k + 12 + 1 = 21k + 13 = 7(3k + 1) + 6. The remainder is 6, so the answer is D.

A quicker check: 3 × 4 = 12, which leaves remainder 5 on division by 7. Adding 1 gives 6 again.

Why the others fail: working backwards, if 3n + 1 leaves remainder r, then n must leave remainder 5(r − 1) mod 7, since 5 is the inverse of 3 modulo 7. Remainder 0 forces n ≡ 2, remainder 3 forces n ≡ 3, and remainder 5 forces n ≡ 6. None of these matches n ≡ 4.

The core concept is modular arithmetic: remainders obey addition and multiplication, so you can reduce at each step. This sits under numerical and quantitative reasoning in the aptitude component of the exam.

Trap: Candidates who stop at 3 × 4 = 12, get remainder 5, and pick C forget the +1 in the question.