UGC NET Questions (Paper – 1)

Signed magnitude representation uses one bit, typically the most significant bit, to indicate the sign of the number. The remaining bits represent the magnitude and are the same for both positive and negative versions. Thus, negative numbers differ from their positive counterparts only in the sign bit. The magnitude bits remain unchanged.

Option A:
Option A, magnitude, is incorrect because the magnitude bits for +5 and -5 are identical in signed magnitude representation. Only the sign indicator changes.

Option B:
Option B, parity, refers to an error-checking mechanism where an extra bit indicates whether the number of ones is even or odd. It is not the primary difference between positive and negative values in this scheme.

Option C:
Option C is correct because the sign bit explicitly encodes whether the number is positive or negative while leaving the magnitude intact. This is why the representation is called signed magnitude.

Option D:
Option D, overflow, describes a condition when a calculation exceeds the representable range. It is typically recorded in a separate status flag rather than being the distinguishing bit between positive and negative values.

For an 8-bit two's complement number starting with 1, we know it is negative. To find its decimal value, we invert the bits and add 1. The inversion of 11001011₂ is 00110100₂, and adding 1 gives 00110101₂, which is 53 in decimal. Therefore, the original 11001011₂ represents −53.

Option A:
Option A correctly performs the two-step process of inversion and addition to decode the negative value. It recognises that the pattern encodes a negative number and finds its magnitude as 53, then assigns the negative sign. Thus, −53 is the correct interpretation.

Option B:
Option B, −75, would require the magnitude bits after inversion and increment to equal 75, which they do not. This choice indicates a miscalculation in either the inversion step or the final addition.

Option C:
Option C, −77, similarly suggests a different magnitude than the one actually encoded by the bit pattern. Since decoding yields 53, this option does not match the numeric content of the 8-bit sequence.

Option D:
Option D, −69, also conflicts with the calculated magnitude. It represents another negative value that is not supported by the structure of 11001011₂. Hence, it is incorrect.