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Question 1 Report
(a) (i) Define computer virus.
(i) List three types of computer network security breach.
(b) State five sources of computer network security breach.
Answer Details
None
Question 2 Report
(a) (i) What is a network cable?
(ii) State three types of cable used in networking
(b) Write the acronym VSAT in full.
(C) (i) Distinguish between an e-mail address and a website address.
(ii) List examples of each of: (1) e-mail address (2) website address
(a)(i) What is a network cable?
A network cable is a physical transmission medium (a wire or fibre) used to connect two or more computers or network devices together so that data signals can pass between them.
(a)(ii) Three types of cable used in networking
(b) VSAT in full
VSAT means Very Small Aperture Terminal.
(c)(i) Difference between an e-mail address and a website address
An e-mail address is the unique identifier of an electronic mailbox to which messages are sent; it always contains the @ symbol in the form username@domain. A website address (URL) is the unique location of a web page or site on the Internet that you type into a browser to view the site; it has no @ symbol and usually begins with a protocol such as http:// or www.
(c)(ii) Examples of each
Answer Details
(a)(i) What is a network cable?
A network cable is a physical transmission medium (a wire or fibre) used to connect two or more computers or network devices together so that data signals can pass between them.
(a)(ii) Three types of cable used in networking
(b) VSAT in full
VSAT means Very Small Aperture Terminal.
(c)(i) Difference between an e-mail address and a website address
An e-mail address is the unique identifier of an electronic mailbox to which messages are sent; it always contains the @ symbol in the form username@domain. A website address (URL) is the unique location of a web page or site on the Internet that you type into a browser to view the site; it has no @ symbol and usually begins with a protocol such as http:// or www.
(c)(ii) Examples of each
Question 3 Report
(a) Convert the following numbers to be stated bases:
(i) 125.625, to binary
(ii) 111011.10011, to octal
(b) Given that X = 11011011 and Y= 11101101,
(i) construct the truth table for X, Y Z;
(ii) draw the logic gate of X, Y = Z
(iii) state the value of Z.
(a) Number-base conversions
(i) Convert \(125.625_{10}\) to base 2.
For the integer part, successive division by 2 gives:
| Division | Quotient | Remainder |
|---|---|---|
| 125 ÷ 2 | 62 | 1 |
| 62 ÷ 2 | 31 | 0 |
| 31 ÷ 2 | 15 | 1 |
| 15 ÷ 2 | 7 | 1 |
| 7 ÷ 2 | 3 | 1 |
| 3 ÷ 2 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 |
Reading the remainders from bottom to top gives \(125_{10}=1111101_2\).
For the fractional part:
\(0.625\times2=1.25\), giving 1;
\(0.25\times2=0.5\), giving 0;
\(0.5\times2=1.0\), giving 1.
Therefore, \(0.625_{10}=0.101_2\), and hence
\[\boxed{125.625_{10}=1111101.101_2}\]
(ii) Convert \(111011.10011_2\) to base 8. Group the binary digits in threes from the binary point:
\[111\;011.100\;110_2=73.46_8\]
\[\boxed{111011.10011_2=73.46_8}\]
(b) \(X=11011011\), \(Y=11101101\), and \(Z=X\mathbin{\text{ AND }}Y\).
(i) Truth table for the AND operation
| X | Y | Z = X AND Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
(ii) Logic-gate diagram
(iii) Value of \(Z\)
| Bit position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| X | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| Y | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 |
| Z | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
\[\boxed{Z=11001001_2}\]
Answer Details
(a) Number-base conversions
(i) Convert \(125.625_{10}\) to base 2.
For the integer part, successive division by 2 gives:
| Division | Quotient | Remainder |
|---|---|---|
| 125 ÷ 2 | 62 | 1 |
| 62 ÷ 2 | 31 | 0 |
| 31 ÷ 2 | 15 | 1 |
| 15 ÷ 2 | 7 | 1 |
| 7 ÷ 2 | 3 | 1 |
| 3 ÷ 2 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 |
Reading the remainders from bottom to top gives \(125_{10}=1111101_2\).
For the fractional part:
\(0.625\times2=1.25\), giving 1;
\(0.25\times2=0.5\), giving 0;
\(0.5\times2=1.0\), giving 1.
Therefore, \(0.625_{10}=0.101_2\), and hence
\[\boxed{125.625_{10}=1111101.101_2}\]
(ii) Convert \(111011.10011_2\) to base 8. Group the binary digits in threes from the binary point:
\[111\;011.100\;110_2=73.46_8\]
\[\boxed{111011.10011_2=73.46_8}\]
(b) \(X=11011011\), \(Y=11101101\), and \(Z=X\mathbin{\text{ AND }}Y\).
(i) Truth table for the AND operation
| X | Y | Z = X AND Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
(ii) Logic-gate diagram
(iii) Value of \(Z\)
| Bit position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| X | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| Y | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 |
| Z | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
\[\boxed{Z=11001001_2}\]
Question 4 Report
(a) State two BASIC programming keywords.
(b) Write the following equations in BASIC format:
(i) I = PRT/100: (ii) Y= AX\(_2\) +BX+C; (iii) Y= MX + C
(c) Write a BASIC program that accepts mass and volume of a liquid as input from the User.
The program should compute.and display the density of the liquid. Density = mass volume
Question 5 Report
(a) An electronic book is made up of 500 pages and there are 16 lines of 24,000 bits line on each page.
Complete the space it will occupy on a computer in megabytes.
(b) Give an example of auxiliary storage.
(c) In the table below:
(i) List the two main types of computer software.
(ii) Give two examples under each in 4(c)i)
| main type | example |
| i | A |
| B | |
| ii | A B |
(a) Space occupied by the electronic book.
Given: 500 pages; 16 lines per page; 24,000 bits per line.
Step 1 - total number of bits:
\[ \text{Total bits} = 500 \times 16 \times 24000 = 192{,}000{,}000 \text{ bits} \]
Step 2 - convert bits to bytes (8 bits = 1 byte):
\[ \text{Bytes} = \dfrac{192{,}000{,}000}{8} = 24{,}000{,}000 \text{ bytes} \]
Step 3 - convert bytes to megabytes:
Using \(1\,\text{MB} = 1{,}000{,}000\) bytes:
\[ \dfrac{24{,}000{,}000}{1{,}000{,}000} = 24 \text{ MB} \]
The book therefore occupies about 24 MB. (If the binary convention \(1\,\text{MB} = 1{,}048{,}576\) bytes is used instead, the answer is \(24{,}000{,}000 \div 1{,}048{,}576 \approx 22.89\,\text{MB}\).)
(b) An example of auxiliary (secondary) storage: hard disk drive (also acceptable: flash/USB drive, CD-ROM, DVD, or magnetic tape).
(c) The two main types of computer software with examples:
| Main type | Examples |
|---|---|
| (i) System software | A - Operating system (e.g. Windows) B - Utility program / device driver |
| (ii) Application software | A - Word processor (e.g. Microsoft Word) B - Spreadsheet (e.g. Microsoft Excel) |
Answer Details
(a) Space occupied by the electronic book.
Given: 500 pages; 16 lines per page; 24,000 bits per line.
Step 1 - total number of bits:
\[ \text{Total bits} = 500 \times 16 \times 24000 = 192{,}000{,}000 \text{ bits} \]
Step 2 - convert bits to bytes (8 bits = 1 byte):
\[ \text{Bytes} = \dfrac{192{,}000{,}000}{8} = 24{,}000{,}000 \text{ bytes} \]
Step 3 - convert bytes to megabytes:
Using \(1\,\text{MB} = 1{,}000{,}000\) bytes:
\[ \dfrac{24{,}000{,}000}{1{,}000{,}000} = 24 \text{ MB} \]
The book therefore occupies about 24 MB. (If the binary convention \(1\,\text{MB} = 1{,}048{,}576\) bytes is used instead, the answer is \(24{,}000{,}000 \div 1{,}048{,}576 \approx 22.89\,\text{MB}\).)
(b) An example of auxiliary (secondary) storage: hard disk drive (also acceptable: flash/USB drive, CD-ROM, DVD, or magnetic tape).
(c) The two main types of computer software with examples:
| Main type | Examples |
|---|---|
| (i) System software | A - Operating system (e.g. Windows) B - Utility program / device driver |
| (ii) Application software | A - Word processor (e.g. Microsoft Word) B - Spreadsheet (e.g. Microsoft Excel) |
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