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DSE Information and Communication Technology Practice Paper: Computer Systems, Data Representation and Networks (12 Questions, 100 Marks, 75 Minutes)
DSE ICT: Computer Systems, Data and Networks
DSE ICT practice paper: 12 original questions, 100 marks, 75 minutes on binary numbers, file sizes, data transmission, subnets, logic circuits and databases.
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Paper details Level: S6 | Subject: ICT | Time: 75 minutes | Marks: 100 | Questions: 12
Section A: Computer systems and data representation (Questions 1-4, 24 marks)
Answer all questions. Show your working in calculations and give units in answers.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) Convert the binary number 10110101 to denary. (b) Write the same value in hexadecimal. (c) Explain why hexadecimal is commonly used for memory addresses. (6 marks) | (a) 10110101 = 128 + 32 + 16 + 4 + 1 = 181. (b) 181 = 11 x 16 + 5, so it is B5 in hexadecimal. (c) Each hexadecimal digit maps exactly to 4 bits, so one byte is always two digits; this is shorter than binary and far less prone to reading errors. Marking: 2 marks conversion, 2 marks hexadecimal, 2 marks explanation. |
| 2 | An uncompressed bitmap is 1024 x 768 pixels with a colour depth of 24 bits. Find its file size in MiB (1 MiB = 1 048 576 bytes). (6 marks) | Pixels = 1024 x 768 = 786 432; each pixel uses 24 / 8 = 3 bytes, so the size is 786 432 x 3 = 2 359 296 bytes = 2 359 296 / 1 048 576 = 2.25 MiB. Marking: 2 marks pixel count, 2 marks bytes per pixel, 2 marks conversion. |
| 3 | A stereo recording is sampled at 44.1 kHz with 16-bit samples for 3 minutes. Find the uncompressed size in MiB, correct to 3 significant figures. (6 marks) | Sample rate 44 100 per second, 2 bytes per sample, 2 channels, 180 seconds: 44 100 x 2 x 2 x 180 = 31 752 000 bytes = 30.277… MiB, about 30.3 MiB to 3 significant figures. Marking: 2 marks sample rate, 2 marks bits and channels, 2 marks time conversion. |
| 4 | A plain-text file holds 1 200 Chinese characters stored in UTF-8 (3 bytes each) plus a 3-byte BOM. (a) Find the file size. (b) What problem arises if the same text is stored in ASCII? (6 marks) | (a) 1 200 x 3 + 3 = 3 603 bytes. (b) ASCII defines only 128 characters and cannot represent Chinese, so the text would be stored as question marks or mojibake and the data would be lost. Marking: 3 marks calculation, 3 marks explanation. |
Section B: Data communication, networks and spreadsheets (Questions 5-8, 32 marks)
Give answers to the required number of significant figures and name the protocol or tool used.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A 12 MB file (1 MB = 1 000 000 bytes) is sent over an 8 Mbps link. (a) Find the ideal transmission time. (b) If it actually takes 15 seconds, find the effective bandwidth in Mbps. (8 marks) | (a) 12 MB = 12 000 000 bytes = 96 000 000 bits; dividing by 8 000 000 bps gives 12 seconds. (b) 96 000 000 / 15 = 6 400 000 bps = 6.4 Mbps. Marking: 4 marks (a), 4 marks (b). |
| 6 | For the network 192.168.10.0/26: (a) find the total number of addresses and the number of usable hosts; (b) state the broadcast address; (c) explain why only 62 hosts are usable rather than 64. (8 marks) | (a) /26 means 26 network bits, leaving 32 – 26 = 6 host bits, so there are 2 to the power 6 = 64 addresses and 64 – 2 = 62 usable hosts. (b) The broadcast address is 192.168.10.63. (c) The first address .0 names the network itself and the last address .63 is reserved for broadcast, so neither can be given to a host. Marking: 3, 2, 3. |
| 7 | (a) The data 1011001 uses even parity. Find the parity bit. (b) State one kind of error that parity checking cannot detect, with an example. (8 marks) | (a) 1011001 contains four 1s, which is already even, so the parity bit is 0. (b) Parity detects only an odd number of bit errors; if two bits flip, for example 1011001 becomes 1001001, the count of 1s is still even and the error passes undetected. Marking: 3 marks, 5 marks with example. |
| 8 | Cells A1 to A4 of a spreadsheet hold 85, 72, 91 and 68. (a) Find AVERAGE(A1:A4). (b) Find the result of COUNTIF(A1:A4,”≥80″). (c) Explain the difference between absolute and relative references. (8 marks) | (a) (85 + 72 + 91 + 68) / 4 = 316 / 4 = 79. (b) Only 85 and 91 are at least 80, so the result is 2. (c) A relative reference shifts when the formula is copied, while an absolute reference such as $A$1 stays fixed. Marking: 2, 2, 4. |
Section C: Logic circuits, compression and databases (Questions 9-12, 44 marks)
Show derivations or truth tables in the long questions, and use standard SQL syntax in database questions.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A logic circuit produces the output (A AND B) OR NOT C. (a) Draw the full eight-row truth table. (b) State how many input combinations give an output of 1. (c) Explain how this circuit differs from an XOR gate. (11 marks) | (a) The eight rows run from 000 to 111; A AND B is 1 only for 110 and 111, and NOT C is 1 whenever C is 0, so the combined output is 1 for 000, 010, 100, 110 and 111. (b) Five combinations. (c) An XOR gate outputs 0 when both inputs are equal, whereas this circuit outputs 1 for A = B = 1 with C = 0, so the truth tables differ. Marking: 5 truth table, 3 count, 3 comparison. |
| 10 | An uncompressed image is 2 359 296 bytes and 294 912 bytes after compression. (a) Find the compression ratio. (b) Find the percentage saved. (c) Distinguish lossy from lossless compression and give one example of each. (11 marks) | (a) 2 359 296 / 294 912 = 8, a ratio of 8:1. (b) (2 359 296 – 294 912) / 2 359 296 x 100% = 87.5%. (c) Lossless compression restores every original bit, for example PNG or ZIP; lossy compression discards detail for a smaller file, for example JPEG or MP3. Marking: 3, 3, 5. |
| 11 | (a) Use run-length encoding (RLE) to compress the string AABBBCCCCD. (b) Find the percentage reduction in characters. (c) State the kind of data on which RLE performs worst. (11 marks) | (a) AABBBCCCCD becomes 2A3B4C1D. (b) The text shrinks from 10 characters to 8, a reduction of (10 – 8) / 10 x 100% = 20%. (c) RLE is worst on data with few runs, such as ABABAB or random bytes, because every character then carries a count and the encoded file is larger than the original. Marking: 4, 3, 4. |
| 12 | A database has a Student table with 5 rows and an Enrolment table with 8 rows, linked by StudentID. (a) Identify the primary and foreign keys. (b) Write the SQL to list the names of students taking ICT. (c) Explain the problems caused by storing both topics in one table. (11 marks) | (a) The primary key of Student is StudentID; the primary key of Enrolment is the pair (StudentID, Subject), in which StudentID is the foreign key pointing back to Student. (b) SELECT Name FROM Student, Enrolment WHERE Student.StudentID = Enrolment.StudentID AND Subject = “ICT”. (c) One table holding both topics repeats the name of every student who takes several subjects, so an update may miss a row and deleting the last enrolment can destroy the only copy of the student record. Marking: 4, 4, 3. |
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