Advanced level 2025 Holy Infant High School mock guide Computer Science

Advanced level 2025 Holy Infant High School mock guide Computer Science

Advanced level 2025 Holy Infant High School mock guide Computer Science

D. 3CD. The other options do not correctly convert the decimal
number to hexadecimal.
2. C/D. The expression 𝒛 = 𝒙(𝒙̅+ 𝒚̅) = 𝒙𝒙̅+ 𝒙𝒚̅= 𝒙𝒚̅(𝑪) 𝒐𝒓 𝒛 =𝒙(𝒙̅+
𝒚̅) =𝒙(𝒙 ̅̅̅𝒚̅) (𝑫)𝒖𝒔𝒊𝒏g 𝑫𝒆 𝑴𝒐𝒓g𝒂𝒏′𝒔 𝒕𝒉𝒆𝒐𝒓𝒆𝒎. A is incorrect
because it represents a different expression. B is incorrect
because it represents a different expression.
3. D. Flip-Flop. A flip-flop is a basic memory element that can store
one bit of data. A. Register, B. Encoder, C. Cell: While these
components relate to data storage, only a flip-flop specifically stores
a single bit.
4. C. Bubble sort. Bubble sort works by repeatedly stepping through
the list, comparing adjacent elements and swapping them if they are
in the wrong order. A. Quick sort, B. Merge sort, D. Selection
sort: These algorithms do not focus on adjacent elements; Quick
sort uses partitioning, Merge sort combines sorted arrays, and
Selection sort selects the smallest element iteratively.
5. A. Divide-and-conquer approach. Recursive algorithms often
break a problem down into smaller subproblems, which is a divideand-conquer strategy. B. Modular approach, C. Structural
approach, D. Hierarchical approach: These do not accurately
describe the nature of recursion, which inherently involves breaking
down problems.
6. A. O(n). This function performs a linear number of additions, which
leads to a time complexity of O(n). B. O(log n), C. O(n²), D. O(n!):
These complexities do not reflect the linear recursive nature of the
function, which sums up n steps.
7. B. DRAM. Dynamic RAM (DRAM) requires periodic refreshing to
maintain the stored data. A. SRAM, C. ROM, D. Flash memory:
SRAM is static and does not require refreshing, ROM is nonvolatile and retains data without power, while Flash memory is also
non-volatile and doesn’t need refreshing.
8. B. Shortest Job First. The Shortest Job First (SJF) scheduling
algorithm minimizes the average waiting time. A. First Come First
Served, C. Round Robin, D. Priority Job First: These algorithms
can lead to longer average wait times compared to SJF, especially in
the presence of varying job lengths.
9. B. More than one program in memory. A time sharing system
implies that more than one program is in memory and multiple
programs are executed concurrently by sharing the CPU’s time. The
other options are incorrect because they do not accurately describe
the concept of a time sharing system.
10. C. Reduce costs. Banks deploy web self-services mainly to cut
operational expenses (e.g., staffing, infrastructure) tied to traditional
support channels. Reducing costs directly improves
profitability. A is incorrect because self-service enhances customer
value. B (risk management) is unrelated. D (opportunities) is
secondary.
11. B. Documentation. Documentation includes comments, guides, and
explanations to clarify code functionality for
developers. A (commentary) is a subset of
documentation. C (implementation) refers to coding itself, not
explanations. D (programming) is the act of writing code.
12. B. Systems programs. Systems software (e.g., OS, drivers) sets up
the foundational environment for machines to
operate. A (applications) run on top of systems. C (libraries) are
code modules. D (utilities) perform specific tasks but aren’t
foundational.
13. C. Object code. Object code is machine-readable output from a
compiler. It requires linking to libraries to form
executables. A (source) is human-readable. B (executable) is final
linked code. D (byte code) is intermediate (e.g., Java).
14. C. UPDATE USER SET City = ‘DC’ WHERE City = ‘Yde’;
SQL uses UPDATE (not MODIFY) to alter records. A/B use invalid
syntax. D swaps ‘Yde’ and ‘DC’, reversing the intent.
15. D. 0–iii, 1–I, 2–ii, 3–iv
– 0NF: Repeating values (iii) are allowed.
– 1NF: Single-valued attributes (i), no repeating groups.
–

– 16. D. All of the above. In the design phase of program development,
the term “module” refers to functions, procedures, and subprograms.
All of these options are correct as they represent different ways of
organizing and structuring code in a modular manner.
17. A. 00 101 000. Correct binary subtraction. B: Incorrect due to wrong
result and sign bit.; C: Incorrect value after subtraction.; D: Incorrect
result and sign bit.
18. D. All. NAND & NOR gates will resolve to 𝐶 = 𝐴. 𝐵 which gives
C = 0 when A = 0 & B = 1. XOR will resolves to 𝐶 = 0.
19. D. 613. Final value via indirect addressing. A: Initial address, not
value; B: Intermediate address; C: Intermediate address.
20. C. 0101110100 000101. The normalized floating point
representation of 23.25 is 0101110100 000101. Option A has
incorrect bits in the exponent. Option B has incorrect bits in both the
mantissa and the exponent. Option D has incorrect bits in the
mantissa.
21. C. 64K memory locations. With a 16-bit program counter, the CPU
can address 216 = 64K memory locations. A is incorrect because it
represents a smaller memory capacity. B is incorrect because it also
represents a smaller memory capacity. D is incorrect because it
represents a larger memory capacity.
22. C. Memory parity error has the highest priority as it indicates a
critical error in memory integrity. A is a software exception, B is a
lower priority interrupt, and D is also a lower priority interrupt.
23. B. Optical fiber provides the highest data rate. A has a lower data
rate, C has a lower data rate, and D have a lower data rate compared
to optical fiber.
24. B. 110110. To find the 2’s complement of 1010 in 6 bits, we invert
the bits and add 1. Options A, C, and D do not correctly represent
the calculation of the 2’s complement.
25. A. 𝒏(𝒏-𝟏)
𝟐
. B. Incorrect because (𝑛+1)(𝑛-1)
2
is not the formula for
calculating connections in a full mesh topology. C. Incorrect
because 𝑛(𝑛+1)
2
is not the formula for calculating connections in a
full mesh topology. D. Incorrect because (𝑛-1)(𝑛-2)
2
is not the
formula for calculating connections in a full mesh topology.
26. D: traversal. Traversal refers to the operation of processing each
element in a list or data structure. Sorting (A) arranges elements in a
specific order, searching (B) finds a specific element, and indexing
(C) refers to accessing elements by their position.
27. B. Central point of failure. A star topology’s central hub or switch
is crucial; if it fails, the entire network goes down, affecting all
connected devices. A. High costs of cabling is a financial issue but
not as immediate as a single point of failure. C. Difficulty in
troubleshooting is incorrect because star topology makes fault
detection easier due to its centralized structure D. Limited
scalability, while true that performance can degrade with many
devices, isn’t the primary disadvantage compared to a central failure