What is Z = X + Y in IEEE 754 format in GATE CS 2026 Set-1 Q42?

Z = 35E80000. X = 35C00000 decodes to 1.1 × 2^-20. Y = 34A00000 decodes to 1.01 × 2^-22 = 0.0101 × 2^-20. Adding mantissas: 1.10000 + 0.01010 = 1.11010 × 2^-20. Encoding back: exponent = -20+127 = 107 = 01101011, mantissa = 11010000000000000000000. Full hex = 35E80000.

How do you decode IEEE 754 single precision hex values in GATE CS 2026 Set-1 Q42?

A 32-bit IEEE 754 single precision number has 1 sign bit, 8 exponent bits (biased by 127), and 23 mantissa bits with an implicit leading 1. For X=35C00000: binary = 0 01101011 10000000000000000000000, exponent = 107-127 = -20, value = 1.1 × 2^-20. For Y=34A00000: binary = 0 01101001 01000000000000000000000, exponent = 105-127 = -22, value = 1.01 × 2^-22.

What is exponent alignment in IEEE 754 addition in GATE CS 2026 Set-1 Q42?

Before adding two floating point numbers, their exponents must be equal. The number with the smaller exponent has its mantissa shifted right until both exponents match. Here Y has exponent -22 and X has exponent -20, so Y's mantissa is shifted right by 2 positions: 1.01 × 2^-22 becomes 0.0101 × 2^-20. Then the mantissas are added directly.

IEEE 754 Single Precision Floating Point Addition Z=X+Y | GATE CS 2026 Set-1 Q42

Computer Sciences > GATE 2026 SET-1 > Floating Point Representation

Consider the real valued variables X, Y and Z represented using the IEEE 754 single-precision floating-point format. The binary representations of X and Y in hexadecimal notation are as follows:
X: 35C00000 Y: 34A00000

Let Z = X + Y.
Which one of the following is the binary representation of Z, in hexadecimal notation?

35C80000

35CC0000

35E80000

35EC0000

Correct : c

Step 1 — Decode X = 35C00000:
In binary: 0 01101011 10000000000000000000000
Exponent = 107 − 127 = −20. Mantissa = 1.1₂ = 1.5
X = 1.5 × 2⁻²⁰ = 1.10000 × 2⁻²⁰
Step 2 — Decode Y = 34A00000:
In binary: 0 01101001 01000000000000000000000
Exponent = 105 − 127 = −22. Mantissa = 1.01₂ = 1.25
Y = 1.25 × 2⁻²² = 1.010 × 2⁻²²
Step 3 — Align exponents:
Shift Y right by 2 to match X''s exponent −20:
Y = 0.01010 × 2⁻²⁰
Step 4 — Add mantissas:
1.10000 × 2⁻²⁰
0.01010 × 2⁻²⁰
= 1.11010 × 2⁻²⁰
Step 5 — Encode Z:
Sign = 0, Exponent = −20 + 127 = 107 = 01101011₂
Mantissa field = 1101 0000 0000 0000 0000 000
Full bits: 0 01101011 11010000000000000000000
= 0011 0101 1110 1000 0000 0000 0000 0000
= 35E80000
Correct answer: C — 35E80000.

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