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S4 ICT Practice Paper: Computer Systems, Data Representation and Networking (12 questions, 100 marks, 75 minutes)
S4 ICT: Computer Systems, Data Representation and Networking
S4 ICT practice paper: binary and hex conversion, bitmap and audio file sizes, transfer time, IPv4 subnetting and network security. 12 questions, 100 marks, with answers.
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Paper details Level: S4 | Subject: ICT | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Short Calculations (25 marks)
Answer all questions and show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Convert the binary number 10110110₂ to denary and show your full working. | 10110110 = 1×128 + 0x64 + 1×32 + 1×16 + 0x8 + 1×4 + 1×2 + 0x1 = 182. Marks: correct place values 4; final answer 182, 2. |
| 2 | Convert the hexadecimal number 2F₁₆ to denary and give its 8-bit binary form. | 2F = 2×16 + 15 = 47; as 8 bits it is 00101111. Marks: 47, 3; 00101111, 3. |
| 3 | A 1024 x 768 pixel 24-bit colour bitmap. Find its uncompressed size in MiB (1 MiB = 1,048,576 bytes). | Total bits = 1024 x 768 x 24 = 18,874,368 bits; bytes = 18,874,368 / 8 = 2,359,296 B; MiB = 2,359,296 / 1,048,576 = 2.25 MiB. Marks: bit count 3; conversion 2; answer 2. |
| 4 | A 60-second stereo recording at 44.1 kHz with 16-bit sampling. Find its uncompressed size in MB (1 MB = 1,000,000 bytes). | Size = 44,100 x 2 x 2 x 60 = 10,584,000 bytes = 10.584 MB. (16 bits = 2 bytes; stereo = 2 channels.) Marks: method 3; answer 3. |
Section B: Structured Questions (30 marks)
State the formula or steps you use.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A 512 GB USB drive (1 GB = 1,000,000,000 bytes) holds photos of 3.2 MB each (1 MB = 1,000,000 bytes). What is the maximum number of photos it can hold (whole number)? | Photos = 512,000,000,000 / 3,200,000 = 160,000. Note 3.2 MB = 3,200,000 bytes, not MiB. Marks: unit conversion 4; answer 4. |
| 6 | A 4,800 kB file is compressed to 720 kB. Find the compression ratio (a : 1, two decimal places) and the percentage saved (one decimal place). | Ratio = 4,800 / 720 = 6.67 : 1; saved = (4,800 – 720) / 4,800 x 100% = 85.0%. Marks: ratio 4; percentage 3. |
| 7 | A 240 MB file is sent over an 80 Mbps link. Find the shortest time needed in seconds. | Size = 240 x 8 = 1,920 Mb; time = 1,920 / 80 = 24 seconds. Marks: MB to Mb 4; answer 4. |
| 8 | For the network 192.168.10.0/26: (a) how many usable hosts does each subnet have? (b) how many /26 subnets can a /24 network be split into? | (a) Host bits = 32 – 26 = 6, so 2^6 = 64 addresses and 64 – 2 = 62 usable hosts. (b) 2^(26-24) = 4 subnets. Marks: (a) 4; (b) 3. |
Section C: Long Questions (45 marks)
Answer in full sentences or a table.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A tutorial centre keeps one flat order table: (OrderID, StudentName, ParentPhone, Subject, FeePerLesson, Lessons). (a) State two update anomalies. (b) Split the table into tables in third normal form (3NF) and mark each primary key and foreign key. | (a) Anomalies: the same parent phone repeats on many orders, so one change must be made in many rows (update anomaly); deleting the last order erases the student and parent record (delete anomaly); a new student with no order cannot be stored (insert anomaly). Any two. (b) 3NF: Student(StudentID PK, StudentName, ParentPhone); Subject(SubjectID PK, SubjectName, FeePerLesson); Order(OrderID PK, StudentID FK, SubjectID FK, Lessons). Marks: anomalies 4; tables 6; keys 2. |
| 10 | Trace the algorithm below, write the final output and state how many times the loop runs: total <- 0 FOR i <- 1 TO 8 total <- total + i * i NEXT i OUTPUT total |
Output = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204; the loop runs 8 times. Marks: running total 8; output 2; count 2. |
| 11 | Sort 7, 3, 9, 2, 8, 5 into ascending order with bubble sort. (a) Write the list after the first pass. (b) How many comparisons are needed in the worst case? | (a) Pass 1: swap 7>3 -> 3,7,9,2,8,5; 7<9 keep; swap 9>2 -> 3,7,2,9,8,5; swap 9>8 -> 3,7,2,8,9,5; swap 9>5 -> 3,7,2,8,5,9, so 9 is now in place. (b) Worst case = n(n-1)/2 = 6×5/2 = 15 comparisons over 5 passes. Marks: (a) 6; (b) 4. |
| 12 | (a) Explain why a website should store user passwords as hashes rather than encrypted values. (b) State the purpose of a salt. (c) Give two features that identify a phishing email. | (a) A hash is one-way, so a stolen database cannot be turned back into passwords; encryption is reversible, so a leaked key exposes every password. (b) A salt is a unique random value hashed with each password, so identical passwords give different hashes and rainbow tables fail. (c) Any two: sender domain differs from the official one; link points to a fake URL; urgent or threatening wording; asks you to type a password or code; suspicious attachment. Marks: (a) 4; (b) 4; (c) 3. |
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