This set covers the whole of hardware and processing: the instruction cycle, registers, memory and storage, input and output choices, and keeping claims inside the model. The questions go from easy to harder.
Attempt each one on paper first. Use the restricted pseudocode trace trainer or the safe Python reasoning sandbox only to check a trace after you have written your own.
Questions
Question 1 (easy)
State what the program counter (PC) holds and when its value changes in the simplified cycle.
Show answer
The PC holds the address of the next instruction to fetch. It increases by 1 during every fetch. A jump instruction overwrites it with a new address during the execute stage.
Question 2 (easy)
Name the register for each description: (a) holds an address being read, (b) holds the instruction being decoded, (c) holds the result of an addition.
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(a) MAR. (b) CIR. (c) ACC.
Question 3 (easy)
Which of RAM, ROM and secondary storage (a) loses its contents when the power is off, (b) holds the start-up instructions, (c) keeps a saved homework file?
Show answer
(a) RAM, because it is volatile. (b) ROM, which is non-volatile. (c) Secondary storage, which keeps files without power.
Question 4 (medium)
Complete a trace table for this program. LDA 90 loads the value at address 90, SUB 91 subtracts the value at address 91 from the ACC, STO 92 stores the ACC, and END stops. The PC starts at 50.
| Address | Contents |
|---|---|
| 50 | LDA 90 |
| 51 | SUB 91 |
| 52 | STO 92 |
| 53 | END |
| 90 | 15 |
| 91 | 6 |
| 92 | 0 |
Show answer
| Step | PC | MAR | MDR | CIR | ACC |
|---|---|---|---|---|---|
| Start | 50 | - | - | - | - |
| Fetch 1 | 51 | 50 | LDA 90 | LDA 90 | - |
| Execute 1 | 51 | 90 | 15 | LDA 90 | 15 |
| Fetch 2 | 52 | 51 | SUB 91 | SUB 91 | 15 |
| Execute 2 | 52 | 91 | 6 | SUB 91 | 9 |
| Fetch 3 | 53 | 52 | STO 92 | STO 92 | 9 |
| Execute 3 | 53 | 92 | 9 | STO 92 | 9 |
Check: 15 − 6 = 9. Address 92 now holds 9.
Question 5 (medium)
A laptop has 4 GB of RAM.
Three programs are open, needing 1.2 GB, 2.5 GB and 0.8 GB. (a) How much RAM do they need in total? (b) What happens, and why does it matter?
Show answer
(a) 1.2 + 2.5 + 0.8 = 4.5 GB.
(b) This is 0.5 GB more than the 4 GB of RAM. The computer uses part of secondary storage as virtual memory, swapping data in and out. This is slower than RAM, so the computer slows down.
Question 6 (medium)
A student photographs homework with a phone, edits it on a laptop and saves it. Say where the file is (a) while it is being edited and (b) after the laptop is switched off.
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(a) The open file is loaded into RAM so the processor can work on it quickly. (b) The saved file stays in secondary storage. RAM would have lost it when the power went off.
Question 7 (medium)
A library lets members borrow books by scanning a card and a book. Choose one input device and one output device and justify each in a sentence.
Show answer
Input: a barcode scanner, because the card and book each carry a code, so scanning records them quickly without typing errors.
Output: a receipt printer (or screen message) confirming the loan and the return date, so the member has proof of what was borrowed.
Question 8 (harder)
Trace this program and state the final value in address 32. JMP 13 sets the PC to 13. The PC starts at 10.
| Address | Contents |
|---|---|
| 10 | LDA 30 |
| 11 | JMP 13 |
| 12 | ADD 31 |
| 13 | STO 32 |
| 14 | END |
| 30 | 4 |
| 31 | 100 |
| 32 | 0 |
Show answer
| Step | PC | MAR | MDR | CIR | ACC |
|---|---|---|---|---|---|
| Start | 10 | - | - | - | - |
| Fetch 1 | 11 | 10 | LDA 30 | LDA 30 | - |
| Execute 1 | 11 | 30 | 4 | LDA 30 | 4 |
| Fetch 2 | 12 | 11 | JMP 13 | JMP 13 | 4 |
| Execute 2 | 13 | 11 | JMP 13 | JMP 13 | 4 |
| Fetch 3 | 14 | 13 | STO 32 | STO 32 | 4 |
| Execute 3 | 14 | 32 | 4 | STO 32 | 4 |
| Fetch 4 | 15 | 14 | END | END | 4 |
The jump overwrote the PC (12 became 13), so ADD 31 at address 12 never ran. Address 32 holds 4, not 104.
Question 9 (harder)
A student writes: “This phone has an 8 core processor at 2.8 GHz, so it runs at 22.4 GHz and will always be faster than a 3.0 GHz laptop.” Identify two errors and rewrite the claim.
Show answer
Error 1: cores do not add their clock speeds. 8 × 2.8 = 22.4 is not a valid speed. The clock speed is 2.8 GHz, which is 2.8 billion cycles per second.
Error 2: “always faster” is not supported. Speed also depends on cores, cache, and whether the software can use the cores.
Rewrite: “The phone’s processor runs at 2.8 GHz with 8 cores. The laptop has a higher clock speed (3.0 GHz), but the cores and the type of task also affect performance, so the data given does not show which is faster.”
Question 10 (harder)
Predict the full output of this Python program, which uses the same model as the lessons.
memory = {50: "LDA 90", 51: "ADD 91", 52: "STO 92", 53: "END",
90: 15, 91: 6, 92: 0}
pc = 50
acc = 0
while True:
mar = pc
pc = pc + 1
mdr = memory[mar]
cir = mdr
parts = cir.split()
op = parts[0]
if op == "END":
break
addr = int(parts[1])
if op == "LDA":
acc = memory[addr]
elif op == "ADD":
acc = acc + memory[addr]
elif op == "STO":
memory[addr] = acc
print(cir, "ACC =", acc, "PC =", pc)
print(memory[92])Show answer
The loop prints after each instruction except END:
LDA 90 ACC = 15 PC = 51
ADD 91 ACC = 21 PC = 52
STO 92 ACC = 21 PC = 53
21Check: 15 + 6 = 21. The PC has already increased during the fetch, so it prints 51, 52 and 53. The final line prints the value stored at address 92, which is 21. The loop stops when END is fetched, before anything is printed for it.
If you got these wrong
| What went wrong | Revisit |
|---|---|
| PC value off by one, or MAR and PC mixed up (Questions 1, 4, 8, 10) | Trace a simplified instruction cycle |
| Wrong register named (Questions 2, 4) | Relate a register to its role |
| RAM, ROM or storage confused, or virtual memory missed (Questions 3, 5, 6) | Compare memory and storage using a task |
| Device named without a reason from the task (Question 7) | Explain an input-output choice |
| Added a claim the question did not give (Question 9) | Avoid claiming a brand specification from a generic model |
Keep a note of each repeated slip in the mistake log and retest queue, and try a fresh question on that skill a few days later.
If a pattern keeps returning, our teachers in online one-to-one Computer Science tuition can trace with you and find where it begins. Return to the module overview to see the study order.