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CUET PG 2023 Computer PYQ — Given the following binary number in 32-bit (single precision) IE… | Mathem Solvex | Mathem Solvex

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CUET PG
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Q. 1675

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CUET PG 2023 — Computer PYQ

CUET PG | Computer | 2023

Given the following binary number in 32-bit (single precision) IEEE-754 format:
00111110011011010000000000000000
The decimal value closest to this floating-point number is

Choose the correct answer:

A.
1.45×10¹
B.
1.45×10⁻¹
C.
2.27×10⁻¹
(Correct Answer)
D.
2.27×10¹

Correct Answer:

2.27×10⁻¹

Explanation

$E = 0111110 0_{2} = 12 4_{10}$

$M = 1101101000000000000000 0_{2}$

$(- 1)^{s} \times 1. M \times 2^{E - 127}$

$= (- 1)^{0} \times 1.110110 1_{2} \times 2^{124 - 127}$

$= 1.110110 1_{2} \times 2^{- 3}$

$= (1 + 2^{- 1} + 2^{- 2} + 2^{- 4} + 2^{- 5} + 2^{- 7}) \times 2^{- 3}$

$= 0.231 \approx 2.31 \times 1 0^{- 1} \approx 2.27 \times 1 0^{- 1}$

Explanation

$E = 0111110 0_{2} = 12 4_{10}$

$M = 1101101000000000000000 0_{2}$

$(- 1)^{s} \times 1. M \times 2^{E - 127}$

$= (- 1)^{0} \times 1.110110 1_{2} \times 2^{124 - 127}$

$= 1.110110 1_{2} \times 2^{- 3}$

$= (1 + 2^{- 1} + 2^{- 2} + 2^{- 4} + 2^{- 5} + 2^{- 7}) \times 2^{- 3}$

$= 0.231 \approx 2.31 \times 1 0^{- 1} \approx 2.27 \times 1 0^{- 1}$

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