100+ Free Ruankao Network Administrator (Junior) Practice Questions

Prepare for the Computer Technology and Software Professional Qualification (Level) Examination — Junior Network Administrator (计算机技术与软件专业技术资格（水平）考试 — 初级网络管理员) exam with instant access — no signup required.

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Ruankao Information System Management Engineer — Intermediate (软考中级信息系统管理工程师, China)

Practice Questions100 questions

2026 Statistics

Key Facts: Ruankao Network Administrator (Junior) Exam

45/75

Passing score on each of two subjects in the same sitting

National 软考 60% standard cited in 深人考发〔2026〕7号

240 minutes

Combined CBT time

软考 junior two-subject sitting rules

24–27 Oct 2026

Lower-half-year 2026 sitting window

计考办〔2026〕1号 / 深人考发〔2026〕7号

RMB 73/subject

Guangdong/Shenzhen fee example

粤发改价格函〔2024〕1073 via 深人考发〔2026〕7号

软考 Junior Network Administrator is a Chinese CBT qualification: knowledge MCQs plus an operations paper in one 240-minute sitting, 45/75 on each subject, no score rollover. The 2026 sitting is 24–27 October 2026. This 100-question English MCQ bank adapts Ethernet, TCP/IP, services, and small-LAN judgment; it is not an official paper.

Sample Ruankao Network Administrator (Junior) Practice Questions

Try these sample questions to test your Ruankao Network Administrator (Junior) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1In 软考网络管理员 computer-system teaching, which description matches the classic Von Neumann architecture?

A.Instructions and data use completely separate, non-shared memories with no common bus

B.A single main memory holds both instructions and data, fetched over a shared bus

C.The ALU stores the entire operating system in ROM and never reads RAM

D.I/O devices execute machine instructions instead of the CPU

Explanation: Von Neumann architecture stores programs and data in the same main memory and uses a shared bus, so instruction fetch and data access compete for that path (the Von Neumann bottleneck). Harvard architecture, by contrast, uses separate instruction and data memories. The ALU does not replace RAM as the OS store, and I/O devices do not execute the instruction stream instead of the CPU.

2Why do 网络管理员 textbooks place a small cache between the CPU and main memory?

A.To hold recently used instructions and data so average access time is lower than DRAM alone

B.To replace the ALU for floating-point arithmetic

C.To assign public IPv4 addresses to hosts

D.To convert analog voice to Ethernet frames

Explanation: Cache is a small, fast SRAM hierarchy (L1/L2/L3) that exploits locality: recently or nearby referenced blocks are reused, so the CPU waits on DRAM less often. Cache does not perform ALU math, does not allocate IP addresses, and is not a voice codec or Ethernet MAC.

3A 网络管理员 is asked to protect a file server against a single disk failure with the simplest redundant RAID. Which level mirrors entire disks?

A.RAID 0 striping with no redundancy

B.RAID 4 with a single dedicated parity disk only, without mirroring

C.RAID 1 mirroring (disk mirroring)

D.JBOD spanning with no redundancy

Explanation: RAID 1 writes the same data to two (or more) disks so one disk can fail without data loss. RAID 0 stripes for speed with zero redundancy. RAID 5/6 use distributed parity rather than full mirrors. JBOD concatenates disks with no RAID protection.

4What is the 8-bit binary representation of decimal 255, a value 网络管理员 candidates must convert when reading subnet masks?

A.01111111

B.10000000

C.11111110

D.11111111

Explanation: 2^8 − 1 = 255, so eight 1-bits are 11111111, which is also 0xFF. That pattern appears in 255.255.255.0 and similar masks. 01111111 is 127, 10000000 is 128, and 11111110 is 254.

5A data-communication item uses even parity on the 7-bit word 1101010. What parity bit is appended so the total number of 1-bits is even?

A.1, because the word already has an odd number of 1-bits

B.0, because the word already has an even number of 1-bits

C.1, because even parity always appends 1

D.The CRC-32 remainder 0x04C11DB7

Explanation: 1101010 contains four 1-bits (positions with 1: four of them), which is even. Even parity therefore appends 0 so the 8-bit frame still has an even count of 1s. Parity is a 1-bit check, not a CRC polynomial remainder. Even parity does not always append 1.

6Nyquist sampling of a 4 kHz analog voice channel (standard PCM teaching in 软考网络管理员) requires a sampling frequency of at least:

A.2 kHz

B.4 kHz

C.8 kHz

D.64 kHz

Explanation: The sampling theorem requires fs > 2fmax. For a 4 kHz voice band, the Nyquist rate is 8 kHz, which is why G.711 PCM uses 8000 samples/s. 2 kHz and 4 kHz undersample; 64 kHz is a bit-rate figure for 8-bit PCM at 8 kHz (8 × 8000 = 64 kbit/s), not the sampling frequency.

7Shannon’s noisy-channel capacity used in 网络管理员 data-communication items is which formula (C in bit/s, B in Hz, S/N linear power ratio)?

A.C = B log2(1 + S/N)

B.C = 2B log2(M) with no noise term

C.C = B × (S/N) with no logarithm

D.C = 2B only, independent of SNR and signalling levels

Explanation: The Shannon–Hartley theorem gives C = B log2(1 + S/N) as the AWGN capacity bound. Nyquist’s noiseless formula C = 2B log2(M) uses discrete signalling levels M and does not include SNR. Capacity is not B times SNR without a log, and it is not simply 2B.

8Compared with a Hamming code that can correct single-bit errors, what is the usual role of a CRC on an Ethernet frame?

A.CRC always corrects every bit error in the payload

B.CRC encrypts the payload with AES

C.CRC assigns the IPv4 address of the sender

D.CRC detects (but does not by itself correct) many bit errors via a polynomial remainder, as in the Ethernet FCS

Explanation: Cyclic redundancy checks detect accidental bit errors by comparing a polynomial remainder (Ethernet’s 32-bit FCS). They do not encrypt, do not assign IP addresses, and typical CRC usage is detection plus discard/retransmit, not Hamming-style single-bit correction in the NIC.

9A switched Ethernet link that sends and receives at the same time is operating in which data-communication mode?

A.Simplex

B.Full duplex

C.Half duplex with CSMA/CD collisions required

D.Broadcast radio only

Explanation: Full duplex allows simultaneous transmit and receive, which is how modern switched Ethernet typically runs, so CSMA/CD is not needed on that link. Simplex is one direction only. Half duplex shares the medium and uses CSMA/CD on classic shared Ethernet. Ordinary copper Ethernet is not 'broadcast radio only'.

10Why does the Internet (and 网络管理员 TCP/IP teaching) use packet switching rather than dedicating a circuit for every conversation?

A.Every packet is guaranteed a reserved 64 kbit/s circuit like PCM voice

B.Packet switching forbids sharing any router output link

C.Packets share links statistically, improving utilization when traffic is bursty

D.Packet switching eliminates the need for IP addresses

Explanation: Packet switching multiplexes independent datagrams on shared links. Bursty data uses capacity when other flows are idle, which is more efficient than a dedicated circuit per session. IP still needs addresses. Circuits (classic PSTN) reserve capacity whether used or not.

About the Ruankao Network Administrator (Junior) Exam

The Ruankao Junior Network Administrator exam (软考初级网络管理员) is a national computer-technology qualification under MIIT and MOHRSS. Official 考试说明 (https://www.ruankao.org.cn/article/content/bkzn/01_28.html) requires computer-system and data-communication basics, TCP/IP, Ethernet/LAN, interconnect devices, OS install/admin, DNS/WWW/MAIL/FTP/proxy, Web-site operations, cabling, and security. The 2026 sitting is in the lower-half window 24–27 October 2026 per 计考办〔2026〕1号 / 深人考发〔2026〕7号. Delivery is computer-based in Chinese: two subjects in one sitting, combined 240 minutes. Both papers must reach 45/75; scores do not roll. There is no education or experience restriction. Fees are set provincially. This OpenExamPrep bank is an English-language MCQ study adaptation, not an official translation or a substitute for the written operations paper.

Assessment

Computer-based combined sitting of 240 minutes. 基础知识: maximum 120 minutes and minimum 90 minutes. Remaining time is 网络系统的管理与维护. Both subjects must reach 45/75 in the same sitting; scores do not roll.

Time Limit

240 minutes combined

Passing Score

45/75 each subject

Exam Fee

Provincially set (e.g. Guangdong/Shenzhen RMB 73/subject per 粤发改价格函〔2024〕1073 and 深人考发〔2026〕7号) (MIIT & MOHRSS / ruankao.org.cn)

Ruankao Network Administrator (Junior) Exam Content Outline

12%

computer-systems-and-data-communication

Computer-system and data-communication basics from the official 考试说明

18%

architecture-tcpip-and-internet

Network architecture, TCP/IP protocol basics, and Internet applications

22%

ethernet-lan-and-interconnect

LAN architecture, Ethernet performance/cabling/simple management, interconnect devices, and media — including subnetting arithmetic

28%

os-services-and-web

Mainstream OS install/admin, DNS/WWW/MAIL/FTP/proxy, Web-site build/maintain, and webpage-making familiarity

20%

cabling-security-law-and-new-tech

Structured cabling basics, network-security issues and countermeasures, related law/informatization, and 了解-level new technology

How to Pass the Ruankao Network Administrator (Junior) Exam

What You Need to Know

Passing score: 45/75 each subject
Assessment: Computer-based combined sitting of 240 minutes. 基础知识: maximum 120 minutes and minimum 90 minutes. Remaining time is 网络系统的管理与维护. Both subjects must reach 45/75 in the same sitting; scores do not roll.
Time limit: 240 minutes combined
Exam fee: Provincially set (e.g. Guangdong/Shenzhen RMB 73/subject per 粤发改价格函〔2024〕1073 and 深人考发〔2026〕7号)

Keys to Passing

Work through all 100 available questions
Review every answer and explanation
Track weak areas and revisit them
Use our AI tutor for tough concepts

Ruankao Network Administrator (Junior) Study Tips from Top Performers

1Practice IPv4 subnetting: network ID, broadcast, usable hosts = 2^(host bits) − 2.

2Map OSI layers to TCP/IP and to devices (hub L1, switch L2, router L3).

3Know default ports: DNS 53, HTTP 80, HTTPS 443, FTP 21, SMTP 25, POP3 110.

4Distinguish hub vs switch vs router vs proxy when a case asks what to install.

5On security items, prefer the least-privilege, patched, and segmented answer over 'buy a new firewall only'.

Frequently Asked Questions

What is the Ruankao Junior Network Administrator qualification?

It is the Junior Network Administrator subject (初级网络管理员) in the 软考 scheme. Official 考试说明: https://www.ruankao.org.cn/article/content/bkzn/01_28.html.

How is the official exam structured and scored?

Two subjects in one 240-minute CBT sitting. 计算机与网络基础知识 is a choice paper (max 120 / min 90 minutes). Remaining time is 网络系统的管理与维护. Each subject is 75 points; both must reach 45 in the same sitting.

Who may sit, and what does it cost?

No education or experience restriction under 国人部发〔2003〕39号. Fees are provincial; Shenzhen/Guangdong charge RMB 73 per subject (深人考发〔2026〕7号).

When is the 2026 sitting?

Network Administrator is in the lower-half window 24–27 October 2026 (计考办〔2026〕1号 / 深人考发〔2026〕7号). Register at https://www.ruankao.org.cn/.

Why are these practice questions in English?

The official exam is in Chinese. This bank is an English-language MCQ study adaptation of syllabus knowledge and operations judgment, not an official translation or a substitute for the written operations paper.