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S6 ICT: Networking and Data Transmission – Practice Paper 1 (Foundation, 10 Questions)
S6 ICT: Networking and Data Transmission – Practice Paper 1 (Foundation, 10 Questions)
Paper one pins the basic units: bits against bytes, IPv4 and MAC formats, star topology faults, the count behind 255.255.255.0, and file size over bandwidth.
Quick answer A 1 GB file needs 80 seconds at 100 Mbps and 250 MB needs 20; a 255.255.255.0 network has 256 addresses and 254 hosts; 500 computers need two slash 24 subnets; 10 spare Mbps carry twenty video calls; 100 Mbps is only 12.5 MB/s.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 15 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 ICT | Networking and Data Transmission: Bandwidth, Subnets and Latency |
| Level | Foundation |
02 Questions (10 items)
- FoundationA 1 GB file crosses a 100 Mbps link. What is the minimum time, taking 1 GB as 1 000 MB and ignoring overhead?
- FoundationHow many bits does an IPv4 address use, and how is it grouped?
- FoundationA MAC address is 48 bits long. How many hexadecimal digits does it normally use?
- FoundationIn a star topology one cable comes loose. Can the other computers keep communicating, and why?
- IntermediateA network with the mask 255.255.255.0 has how many addresses, and how many usable hosts?
- IntermediateHow long does a 250 MB file take over a 100 Mbps link at best?
- IntermediateRound trip time is 20 ms and five requests each need one round trip. What is the total delay?
- ChallengeA firm has 500 computers and each slash 24 subnet holds at most 254. What is the smallest subnet count?
- ChallengeA 50 Mbps leased line carries 40 Mbps at peak. How many 500 kbps video calls fit in the remainder?
- Error redoError retry: a student wrote that 100 Mbps means 100 MB every second. What is the correct rate in MB/s?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 80 seconds | One gigabyte is 8 000 megabits and dividing by 100 Mbps gives 80 seconds, though real times are longer. |
| 2 | 32 bits in four groups of eight | Thirty two bits allow 2 to the power 32, which is 4 294 967 296 addresses. |
| 3 | Twelve digits | Each hexadecimal digit carries four bits, and 48 divided by four is twelve, usually grouped in six pairs. |
| 4 | Yes, each of them still reaches the switch on its own cable | Every host has a dedicated line in a star, so one break isolates only that host while the switch keeps forwarding. |
| 5 | 256 addresses and 254 usable hosts | Eight host bits give 256 addresses, and removing the network and broadcast addresses leaves 254. |
| 6 | 20 seconds | 250 MB is 2 000 Mb, and dividing by 100 Mbps gives 20 seconds. |
| 7 | 100 ms | Five round trips at 20 ms each come to 100 ms, which is delay rather than bandwidth. |
| 8 | Two subnets | 500 over 254 rounds up to two, with 254 hosts in the first and the remaining 246 in the second. |
| 9 | Twenty calls | The spare 10 Mbps is 10 000 kbps, and dividing by 500 kbps gives twenty calls. |
| 10 | 12.5 MB/s | Bits and bytes differ by a factor of eight, so 100 over 8 is 12.5, with a lower case b for bit and an upper case B for byte. |
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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