9th Class Computer Science — MCQs & Practice Questions

RAM is volatile, fast and comparatively small; it holds the programs and data currently in use and loses everything when the power goes. Secondary storage such as an SSD is non-volatile, slower and much larger; it keeps data permanently.

The control unit, which fetches and decodes instructions and sends control signals to the rest of the system; and the arithmetic and logic unit, which performs calculations and logical comparisons.

It holds the address of the next instruction to be fetched. It increments during the fetch stage so that the following cycle picks up the next instruction rather than repeating the current one.

1. Fetch: the address in the program counter is placed on the address bus and the instruction is brought back along the data bus into the instruction registermention of a bus expected[1]
2. Decode: the control unit interprets the instruction to determine the operation and its operands[1]
3. Execute: the ALU carries out the operation[1]
4. Store: the result is written back to a register or to memory, and the cycle repeatsthe repetition should be stated[1]

Fetch, decode, execute, store — repeated for every instruction.

1. The address bus carries the memory address being accessed[1]
2. The data bus carries the actual data or instruction[1]
3. The control bus carries signals such as read and write[1]
4. The address bus is unidirectional — addresses only ever travel from the CPU to memory, whereas data travels both waysthe reason is required[1]

Address = where (one-way), data = what (two-way), control = read or write.

1. Laptop B[1]
2. Four cores can execute four instruction streams at once, so it handles several programs far better than one faster core[1]
3. 16 GB of RAM means far less swapping to disk, and the SSD has no moving parts so it reads much faster than a hard diskeither point accepted[1]
4. Laptop A has 1 TB of storage against 512 GB, so it holds roughly twice as many filescapacity is the only advantage here[1]
5. Clock speed measures cycles per second on ONE core, so it says nothing about how many cores there are[1]
6. It also ignores cache size and the amount of RAM, both of which decide how often the CPU has to wait for data[1]

(a) Laptop B — more cores, far more RAM, and an SSD (b) twice the storage capacity (c) clock speed ignores cores, cache and RAM

A computer is built from electronic switches, each of which has only two reliable states — on and off, or high and low voltage. Binary has exactly two digits, so each digit maps directly onto one switch, making the hardware simple and resistant to electrical noise.

Place values 16, 8, 4, 2, 1: 16 + 8 + 0 + 2 + 0 = 26.

It is far shorter and easier to read — one hex digit replaces four bits, so a 32-bit address takes 8 hex characters instead of 32 binary ones — while converting between the two needs no arithmetic at all.

1. 2 in hex is 0010 in binaryfour bits per hex digit[1]
2. C is 12 in decimal, which is 1100A=10, B=11, C=12, D=13, E=14, F=15[1]
3. Together: 00101100₂concatenate the groups, do not add them[1]
4. Decimal: 32 + 8 + 4 = 44or 2×16 + 12 = 44 directly from hex[1]

2C₁₆ = 00101100₂ = 44₁₀

1. Adds column by column from the right, carrying where a column totals 2[1]
2. Result within 8 bits: 01000010 with a carry out of the leftmost column[1]
3. In decimal 202 + 115 = 317, which exceeds the 8-bit maximum of 255[1]
4. Overflow occurs — the true answer needs 9 bits, so the stored 8-bit result is incorrectthe consequence must be stated[1]

Carry out of the final column; overflow occurs because 317 > 255.

1. 2⁷ = 128 characters[1]
2. Urdu uses a different script with many additional letterforms, and 128 slots are already taken by English letters, digits and punctuation — so Unicode is used insteadnaming Unicode expected[1]
3. 2²⁴ = 16 777 216 coloursaccept "about 16.7 million"[1]
4. Total pixels = 100 × 200 = 20 000[1]
5. Each pixel takes 24 bits = 3 bytes, so 20 000 × 3 = 60 000 bytesconverting bits to bytes is the step most often missed[1]
6. 60 000 / 1024 = 58.6 KBaccept 58.6 or 60 KB if 1000 bytes per KB is stated[1]

(a) 128, too few for a second script (b) 16 777 216 (c) about 58.6 KB

Inputs 00 → 1, 01 → 1, 10 → 1, 11 → 0. NAND is the inverse of AND, so it outputs 0 only when both inputs are 1.

(A · B)′ = A′ + B′ and (A + B)′ = A′ · B′. Complementing a whole expression swaps AND for OR and complements each variable.

Because any logic function whatsoever can be built using copies of just one of them. NOT, AND, OR and every larger circuit can be constructed from NAND alone, or from NOR alone, which is why chip manufacturers can make a whole processor from one repeated gate design.

1. Four rows for two inputs, with columns for A′ and B′[1]
2. A=0,B=0 → 0; A=0,B=1 → 1A′·B is 1 in the second row[1]
3. A=1,B=0 → 1; A=1,B=1 → 0[1]
4. Output is 1 exactly when the inputs differ, so this is an XOR gatethe identification is the point of the question[1]

The table matches XOR: output 1 when A and B differ.

1. Expand: A·A + A·B′ + B·A + B·B′distributive law[1]
2. A·A = A (idempotent) and B·B′ = 0 (complement)[1]
3. X = A + A·B′ + A·Bthe zero term disappears[1]
4. By absorption A + A·anything = A, so X = Anaming the law is required by the question[1]

X = A

1. 0+0 → Sum 0, Carry 0; 0+1 → Sum 1, Carry 0[1]
2. 1+0 → Sum 1, Carry 0; 1+1 → Sum 0, Carry 1the last row is the whole point — 1+1 in binary is 10[1]
3. Sum is produced by an XOR gateoutput 1 when the inputs differ[1]
4. Carry is produced by an AND gateoutput 1 only when both are 1[1]
5. A half adder has only two inputs, so it has nowhere to accept the carry produced by the previous column[1]
6. A full adder is needed instead, taking three inputs — A, B and carry-innaming the full adder is expected[1]

(b) XOR for Sum, AND for Carry (c) it has no carry-in input, so a full adder is required

So that the outcome identifies the cause. If several things are changed together and the fault disappears, you do not know which change fixed it — and if it does not, you have introduced new variables that must themselves be undone.

Any two of: installing operating system and application updates; running antivirus scans; taking regular backups; uninstalling unused software; clearing temporary files.

Overheating is the most likely cause, since the fault appears under load. Check whether the fans are running and whether the vents are blocked with dust, and monitor the CPU temperature.

1. Identify the problem — establish exactly what happens and what changed recently[1]
2. Establish a theory of the probable cause, starting with the simplest and most likely[1]
3. Test the theory by changing one thing, and implement the fix if confirmed[1]
4. Verify that the fault is gone and nothing else has broken, then document the fault and the solutiondocumentation is regularly omitted and regularly marked[1]

Identify, theorise, test, fix, verify, document.

1. Boot the machine from alternative media, such as a USB drive with a live operating system[1]
2. If the fault disappears in that environment, the installed software is at fault; if it persists, the hardware isthe interpretation is the mark, not the action[1]
3. Test the suspect component in a known-good machine, or a known-good component in this machinecomponent swapping[1]
4. Check whether the same fault occurs in more than one application — a fault in only one program points to that program rather than the hardware[1]

Boot from other media, swap components, and check whether the fault is confined to one application.

1. Accumulated temporary files, downloads and unused installed software filling the disk[1]
2. Too many programs configured to start automatically, consuming RAM, and dust build-up causing the CPU to slow itself to avoid overheatingany second sensible cause[1]
3. Follow the 3-2-1 rule: three copies of the data[1]
4. On two different types of media, with one copy kept off-site or in cloud storage[1]
5. Backups should be automatic and scheduled rather than left to the user, and should be tested by restoring from theman untested backup is not known to work[1]
6. Hardware maintenance: clean dust from fans, vents and heatsinks each term, and check that all fans still spinaccept surge protection or checking disk health[1]

(a) accumulated files and startup programs, plus dust causing thermal throttling (b) 3-2-1, automated and tested (c) scheduled cleaning of fans and vents

System software manages and maintains the computer itself — for example Windows, an operating system. Application software performs tasks for the user — for example a word processor. The test is who the software serves: the machine, or the person using it.

Any two of: process management (scheduling which program uses the CPU), memory management (allocating RAM and keeping programs separate), file management, device management through drivers, and providing a user interface.

Utility software is system software that maintains, analyses or protects the computer rather than doing the user's own work. Examples: antivirus, backup software, disk defragmenter, file compression.

1. A compiler translates the entire program at once, producing an executable file; an interpreter translates and runs one line at a time[1]
2. A compiler reports all errors together at the end of compilation; an interpreter halts at the first error it reaches[1]
3. A compiler is preferable for finished software being distributed, since it runs faster and the source code need not be shipped[1]
4. An interpreter is preferable while learning or testing, because a change can be run immediately with no compilation step[1]

Whole program vs line by line; all errors vs first error; compiled for distribution, interpreted for development.

1. Proprietary: professional technical support is guaranteed, with someone responsible for fixing faults[1]
2. Proprietary: releases are formally tested, and compatibility with other schools and offices is more predictable[1]
3. Open source: no licence cost, which for a school with many machines is a substantial saving[1]
4. Open source: the source code can be inspected and modified, and the school is not locked into one supplier[1]

Proprietary: support and tested compatibility. Open source: no cost and freedom to modify.

1. Applications cannot access hardware directly — they make requests to the operating system, which carries them out[1]
2. Without an OS there is nothing to load the application into memory, allocate it CPU time, or give it access to files[1]
3. A device driver is software that translates general operating system requests into the specific commands one particular device understands[1]
4. So the OS can say "print this page" without knowing anything about that printer model; installing a new printer means installing its driver[1]
5. A CLI uses far fewer system resources than a graphical desktop, leaving more memory and CPU for the server's actual work[1]
6. Commands can be written into scripts and repeated automatically, and a CLI works well over a slow remote connectioneither reason accepted[1]

(a) applications reach hardware only through the OS (b) a driver translates OS requests for one specific device (c) lighter on resources and scriptable

A LAN covers a small geographical area such as one building and its hardware is usually owned by the organisation. A WAN covers a large area and typically uses third-party communication links.

An IP address identifies a device on a network and can change when the device moves to a different network. A MAC address is fixed in the hardware and uniquely identifies the network adapter itself.

The data is divided into equal-sized blocks. Each packet carries a header with the source and destination addresses and its sequence number, and packets may travel by different routes and are reassembled in order at the destination.

1. A LAN[1]
2. The computers are all within a single building and the school owns the cabling[1]
3. A switch connects devices within the LAN[1]
4. It forwards each frame only to the port of the intended recipient, using MAC addresses[1]
5. A router connects the LAN to another network, usually the internet[1]
6. It forwards packets between networks using IP addresses[1]
7. Advantage: files and printers can be shared[1]
8. Advantage: software and backups can be managed centrally[1]

1. The browser sends the domain name to a DNS server[1]
2. The server looks up the matching IP address[1]
3. The IP address is returned and the browser requests the page from that server[1]
4. HTTPS encrypts the data in transit, HTTP sends it in plain text[1]
5. So HTTPS protects information such as passwords from being read if intercepted[1]
6. To block access to unsuitable or unauthorised sites[1]
7. To inspect incoming traffic and block unauthorised access to the network[1]

1. Faster and more reliable, with less interference[1]
2. Devices can be moved freely and no cabling is needed[1]
3. Use WPA encryption on the wireless network[1]
4. Data sent over the air is scrambled[1]
5. So anyone intercepting the signal cannot read it without the keyaccept MAC filtering or a strong passphrase, explained[1]

A finite sequence of unambiguous instructions that solves a problem or completes a task, with clearly defined inputs and outputs.

Abstraction is removing the details that are not relevant to the problem being solved, so that the essential structure is easier to work with. A metro map shows the order of the stations and the connections between lines, but omits real distances and street layout — which makes it more useful for planning a journey, not less.

A flowchart is graphical, using standard shapes and arrows, and shows the flow of control visually. Pseudocode is textual, resembles program code and is quicker to write and to translate into a real language. Flowcharts become unwieldy for long algorithms, where pseudocode stays readable.

1. INPUT mark[1]
2. Validation: WHILE mark < 0 OR mark > 100 → OUTPUT "Invalid" and INPUT mark againa loop, not a single if — the user may enter a bad value twice[1]
3. IF mark >= 40 THEN OUTPUT "Pass"the >= is required; > alone fails a mark of exactly 40[1]
4. ELSE OUTPUT "Fail" ENDIF[1]

Input with a validation loop, then IF mark >= 40 output Pass ELSE output Fail.

1. Columns for n, i, total, and outputone column per variable[1]
2. After pass 1: i = 2, total = 1; after pass 2: i = 3, total = 3[1]
3. After pass 3: i = 4, total = 6; after pass 4: i = 5, total = 10[1]
4. The loop ends because 5 > 4; output is 10the algorithm sums 1 to n[1]

Output 10 — the algorithm computes 1 + 2 + 3 + 4.

1. Break the system into separate sub-problems, such as: record one student's attendance; store a day's records; calculate one student's monthly totalat least two named sub-problems[1]
2. And: produce the report; handle login. Each is small enough to be designed and tested on its ownthe reason for decomposing is part of the answer[1]
3. Abstraction: store only what the task needs — a student's roll number and present/absent flag — and ignore irrelevant details such as their address or hobbies[1]
4. The unnecessary detail would make the system larger and slower without improving the report[1]
5. Boundary data tests the values at the edges of what is acceptable, which is where most errors occur[1]
6. Example: a month with 31 days, or a class with exactly one student, or an attendance percentage of exactly 75% if that is the pass thresholdany sensible boundary value[1]

(a) split into recording, storing, calculating and reporting (b) store only roll number and present/absent (c) edges are where errors hide — e.g. exactly 75% attendance