UGC NET Questions (Paper – 1)

The numeral 121 in base b expands as 1×b^2 + 2×b + 1. Setting this equal to 16 gives b^2 + 2b + 1 = 16. Simplifying, we get b^2 + 2b - 15 = 0. Solving this quadratic yields b = 3 or b = -5, and only the positive base 3 is valid, so b = 3.

Option A:
Option A, base 2, would make 121 equal to 1×2^2 + 2×2 + 1 = 9, not 16.

Option B:
Option B, base 4, gives 1×4^2 + 2×4 + 1 = 25, not 16.

Option C:
Option C, base 3, yields 1×3^2 + 2×3 + 1 = 16, matching the condition.

Option D:
Option D, base 5, yields 36, far from 16.