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S6 ICT: Networking and Data Transmission – Practice Paper 2 (Intermediate, 10 Questions)
S6 ICT: Networking and Data Transmission – Practice Paper 2 (Intermediate, 10 Questions)
Paper two works through subnets and protocols: slash 24 and slash 26 counts, the block holding 200.10.5.130/26, standard ports, OSI layers and round trip budgets.
Quick answer The subnet of 200.10.5.130/26 is 128 with hosts 129 to 190 and broadcast 191; a slash 26 holds 64 addresses; HTTPS runs on 443 with encryption; routers work at layer three and switches at layer two; 200 Mbps times 40 ms is 8 Mb.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 18 minutes |
| Difficulty mix | Foundation 2 · Intermediate 4 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 ICT | Networking and Data Transmission: Bandwidth, Subnets and Latency |
| Level | Intermediate |
02 Questions (10 items)
- FoundationA slash 24 network holds 256 addresses. How many are usable, and why?
- FoundationHow many addresses does a slash 26 subnet contain, and how many hosts can it hold?
- IntermediateWhich subnet holds 200.10.5.130/26, and what host range can be assigned inside it?
- IntermediateWhich standard ports do HTTP, HTTPS, DNS and SMTP use, and which one is encrypted?
- IntermediateIn the seven layer OSI model, which layers do routers and switches mainly work at?
- IntermediateA 1 500 byte IP packet crosses a link whose MTU is 500 bytes. How many fragments are needed?
- ChallengeA 200 Mbps link has a 40 ms round trip. What is the bandwidth delay product?
- ChallengeA slash 24 network is split into eight equal subnets. What is the new mask and the usable hosts per subnet?
- ChallengeThe TCP handshake costs one round trip and ten requests cost one each, with a 60 ms round trip. What is the total?
- Error redoError retry: a student gave 200.10.5.192 as the broadcast address of 200.10.5.130/26. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 254 addresses | The network address ending in zero and the broadcast address ending in 255 cannot be assigned, so two are lost. |
| 2 | 64 addresses and 62 hosts | Six host bits give 64 addresses, and removing the two reserved ones leaves 62. |
| 3 | The subnet is 200.10.5.128 with a usable range of 129 to 190 | Blocks are 64 wide, so 130 sits in 128 to 191, and dropping both ends gives 129 to 190. |
| 4 | HTTP 80, HTTPS 443, DNS 53 and SMTP 25, with HTTPS encrypted | HTTPS runs over TLS on port 443, while the other three are plain by default and expose credentials. |
| 5 | Routers at the network layer and switches at the data link layer | Routers choose paths by IP address while switches forward frames by MAC address, which also sets broadcast domains. |
| 6 | Three fragments | 1 500 over 500 is three, each at most 500 bytes, though headers leave less room for data. |
| 7 | 8 Mb, which is 1 MB | 200 Mbps times 0.04 seconds is 8 Mb, so a TCP window of 1 MB is needed to fill the pipe. |
| 8 | Slash 27 with 32 addresses and 30 usable hosts each | Borrowing three bits gives eight subnets and a 27 bit mask, leaving five bits or 32 addresses less two. |
| 9 | 660 ms | Eleven round trips at 60 ms each give 660 ms, and keeping the connection open saves repeat handshakes. |
| 10 | 200.10.5.191 | The block starts at 128 and runs 64 addresses, so the last is 128 plus 63 or 191, while 192 opens the next block. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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