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Load this one up as a remainder/Diophantine setup dressed as DS, and the classic trap is picking C before actually checking if the statements are insufficient alone.

1. Group A divided into subgroups of 13, none left over, so total in A = 13m.
2. Group B divided into subgroups of 11, with 8 left over, so total in B = 11n + 8. That's what the question wants.
3. Statement 1: m = n. That just says A and B have the same number of subgroups, not what that number is. m could be 1, 4, 100, whatever, B changes every time. Insufficient.
4. Statement 2: 13m = 11n + 8. One equation, two unknowns. Try m=4, n=4: 52=52, works. Try m=15, n=17: 195=195, also works. Two different values for B. Insufficient alone.
5. Combine both: m=n and 13m=11n+8 becomes 13m=11m+8, so m=4, n=4, B=11(4)+8=52. One clean number.

Answer is C.

I got burned by almost this exact structure on a practice test, saw two equations and figured "sufficient, has to be C" without testing whether statement 2 alone pins down a unique value. It doesn't, a single linear equation in two integer unknowns almost always has multiple solutions unless something else bounds it.

Takeaway: whenever a DS statement gives you one equation with two unknowns, go hunt for a second solution before calling it insufficient, don't just trust your gut.
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