Decimal of binary 1111 – UGC NET Question

For binary 1111, we sum the place values corresponding to each bit that is 1. From right to left, the bits represent 1, 2, 4 and 8. Since all four bits are 1, the sum is 8 + 4 + 2 + 1. This equals 15, so the decimal equivalent is 15.

Option A:
Option A suggests 13, which would come from the pattern 1101₂ (8 + 4 + 1). In 1111, all four bits are set, so the sum is larger than 13. Therefore 13 cannot be correct.

Option B:
Option B gives 14, which corresponds to 1110₂ (8 + 4 + 2). The given number has a 1 also in the 1's place, making its value 15, not 14. Thus this option is too small by one.

Option C:
Option C suggests 12, which would arise from 1100₂ (8 + 4). The pattern 1111 includes additional contributions from the 2's and 1's places, so 12 is clearly incorrect.

Option D:
Option D is correct because 1111₂ = 8 + 4 + 2 + 1 = 15. All place values are used, giving the highest possible value for a 4-bit binary number. This matches the required decimal equivalent.