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FC0-U71 Tech Concepts and Terminology Practice Question

A technician is converting the decimal number 255 to binary. Which binary value is correct?

Answer choices

Why each option matters

Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.

Correct answer & explanation

✓

11111111

255 in binary is 11111111 (eight 1s).

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

  • ✗

    10000000

    Why it's wrong here

    10000000 equals 128, not 255, since only the most significant bit is set. It is tempting because it is the eight-bit value with the highest single bit, which people confuse with the maximum. The stem needs every bit set, giving 11111111.

  • ✗

    11111110

    Why it's wrong here

    11111110 equals 254, since the least significant bit is zero; 255 needs all eight bits set to one. This value is the correct answer when converting 254, a common off-by-one trap, but the stem specifies 255.

  • ✗

    1111111

    Why it's wrong here

    Seven bits cannot represent 255; the maximum with seven bits is 127. The stem requires eight bits, where 255 is 11111111. It is tempting because 1111111 looks like 'all ones', which people associate with the maximum value, but the bit count is short by one.

  • ✓

    11111111

    Why this is correct

    255 equals 2⁸ − 1, so every bit in an 8-bit octet is set. Converting by repeated division by 2 yields remainders of 1 at each step, giving eight ones: 11111111. This satisfies the stem's requirement for the binary representation of the maximum 8-bit decimal value.

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Written by Johnson Ajibi, MSc IT Security

Senior Network & Security Engineer · founder of Courseiva

This FC0-U71 practice question is part of Courseiva's free CompTIA certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the FC0-U71 exam.