ITPEC FE Subject A October 2025 Question 1
In the single precision format of the “IEEE Standard for Binary Floating-Point Arithmetic” (IEEE 754), a 32-bit floating point number is represented as below:
If 0 < E < 255, then the value of the number is (−1)S × 2 (E−127) × (1+F)
where F represents the fractional part of the number, stored as a 23-bit binary sequence. This F is the sum of its weighted binary components: F=b1×2−1+b2×2−2+⋯+b23×2−23 , where b1, b2, …, b23 are the binary digits of F.
According to the IEEE-754 standard, which of the following is the decimal equivalent of the 32-bit floating point number given below?
01000001011000000000000000000000
Explanation
14.0 — split the 32 bits into sign, exponent and fraction, then apply the given formula.
Bits: 0 | 10000010 | 11000000000000000000000
- •S = 0, so the number is positive
- •E =
10000010= 128 + 2 = 130, so the true exponent is 130 − 127 = 3 - •F =
11followed by zeros, so F = 2−1 + 2−2 = 0.5 + 0.25 = 0.75 - •Value: (−1)0 × 23 × (1 + 0.75) = 8 × 1.75 = 14
Answer (d)
Why not others:
- •(c) 6.0 — drops the implicit leading 1 and multiplies by F alone: 23 × 0.75 = 6
- •(b) 0.4375 — equals 1.75 × 2−2: the right significand with a negative exponent, which would need E < 127, but E is 130
- •(a) 0.1875 — equals 0.75 × 2−2: drops the implicit 1 and also uses a negative exponent
Key rule: In a normalised IEEE 754 single the significand is always 1 + F and the true exponent is E − 127; forgetting the hidden 1 or the bias is the usual trap.
Basic Theory · AI-written, may contain errors
This question comes from an official ITPEC past paper. ITPEC Practice is an independent study tool and is not affiliated with ITPEC. See the official FE past-paper collection or Report an issue.