FC0-U71 Tech Concepts and Terminology Practice Question
A technician needs to convert a decimal IP address octet value of 192 to binary. Which of the following is the correct binary representation?
⚠ Common exam trap
FC0-U71 often tests binary conversion with distractors that are valid-looking octets (170, 240, 129), catching candidates who guess by pattern rather than summing powers of two.
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
✓
11000000
To convert decimal 192 to binary, decompose it into powers of two: 128 + 64 = 192, so bits 7 and 6 (values 128 and 64) are set to 1 and the remaining six bits are 0. That yields 11000000. This is a standard octet value recognizable as the first octet of a Class C private range (192.168.x.x).
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
10101010
Why it's wrong here
192 decimal equals 11000000 in binary; 10101010 is 170. The option tempts candidates who recall the alternating-bit pattern used to illustrate subnet masks, but positional weighting of 128 and 64 is what produces 192, so this representation fails the conversion.
- ✓
11000000
Why this is correct
192 equals 128 + 64, so bits 7 and 6 are set while the remaining six bits stay 0, giving 11000000. This satisfies the stem's requirement to convert the decimal octet 192 into its exact eight-bit binary representation.
- ✗
11110000
Why it's wrong here
11110000 equals 240, not 192; the correct representation is 11000000. It is tempting because it sets the two high-order bits, which is the pattern for values in the 192–223 range, but 192 requires only those two bits set.
- ✗
10000001
Why it's wrong here
10000001 equals 129, not 192; the correct representation is 11000000. It is tempting because it sets the high-order bit and a low bit, resembling a class-based address pattern, but 192 requires the top two bits set with all others clear.
About these practice questions
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JA
Written and reviewed by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
Last reviewed September 2026 · checked against the official CompTIA exam blueprint
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.