UGC NET Questions (Paper – 1)

In 8-bit one’s complement, a leading 1 indicates a negative number. To find its magnitude, invert all bits and read the result as an unsigned binary number. Inverting 10011001₂ gives 01100110₂, which equals 64 + 32 + 4 + 2 = 102. Therefore, 10011001₂ represents −102 in one’s complement.

Option A:
Option A is correct because one’s complement encodes −X as the bitwise inversion of +X. Since 10011001₂ inverts to 01100110₂ (= 102), the represented value is −102.

Option B:
Option B (−104) would require the inverted magnitude to be 104 (01101000₂). But the inversion here is 01100110₂ (=102), so it cannot be −104.

Option C:
Option C (−127) is the most negative magnitude possible in 8-bit one’s complement and corresponds to inverting 01111111₂. The given pattern does not invert to 127, so this is incorrect.

Option D:
Option D (−128) is not representable in 8-bit one’s complement (the range is −127 to +127, with both +0 and −0). Hence −128 cannot be the answer.

The 1's complement of a binary number is obtained by inverting each bit, changing 0 to 1 and 1 to 0. For 010101, flipping each bit produces 101010. This new pattern is the 1's complement and reflects bitwise inversion of the original number.

Option A:
Option A repeats the original pattern without any inversion. Since 1's complement requires changing each bit, leaving the number unchanged cannot be correct.

Option B:
Option B, 010110, differs from the original in only one bit, which is insufficient for a full complement. 1's complement must invert all bits, not just some.

Option C:
Option C is correct because it flips 0→1 and 1→0 at every position, giving 101010 from 010101. This is the precise definition of 1's complement and is used in some earlier signed number representations.

Option D:
Option D, 101101, changes several bits but not in a simple inversion pattern. Some positions do not reflect the opposite of the original bits, so it is not a valid 1's complement.