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Sample PEC EPE Computer & Allied Practice Questions

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1A divide-and-conquer algorithm solves a problem of size n by recursively solving 4 subproblems of size n/2, with the partitioning and combination step taking cn^2 time (for some constant c > 0). According to the Master Theorem, what is the asymptotic runtime complexity T(n)?
A.Θ(n^2 log n)
B.Θ(n^2)
C.Θ(n^(log_2 4)) = Θ(n^3)
D.Θ(n log n)
Explanation: For recurrence T(n) = aT(n/b) + f(n), we have a = 4, b = 2, and f(n) = Θ(n^2). The critical exponent is log_b(a) = log_2(4) = 2. Since f(n) = Θ(n^{log_b a}) = Θ(n^2), Case 2 of the Master Theorem applies, yielding T(n) = Θ(n^{log_b a} log n) = Θ(n^2 log n).
2In an undirected graph G = (V, E) representing a telecommunications mesh, there are exactly 15 network nodes (vertices), and each node is connected to exactly 4 distinct transmission links (edges). How many total transmission links exist in this network?
A.30 links
B.60 links
C.15 links
D.120 links
Explanation: According to the Euler Handshaking Lemma for any undirected graph, the sum of vertex degrees equals twice the number of edges: ∑_{v∈V} deg(v) = 2|E|. With 15 vertices each having degree 4, the degree sum is 15 × 4 = 60. Therefore, |E| = 60 / 2 = 30 links.
3Which of the following Boolean expressions represents the minimal sum-of-products (SOP) form for the function F(A, B, C) = ∑ m(0, 1, 4, 5, 6)?
A.B' + A C'
B.A' B' + A C'
C.B' C + A B
D.A' C' + B
Explanation: Group minterms on a 3-variable K-map: m(0, 1, 4, 5) corresponds to cells where B=0, giving prime implicant B' (since A and C vary across all combinations). Minterm m(6) (110) can be grouped with m(4) (100) along the bottom row where A=1 and C=0, giving prime implicant A C'. Combining these yields the minimal SOP expression F = B' + A C'.
4A communications channel has an analog bandwidth of 4 kHz and a signal-to-noise power ratio (SNR) of 31 (approximately 15 dB). According to the Shannon-Hartley theorem, what is the theoretical maximum channel capacity?
A.20 kbps
B.16 kbps
C.32 kbps
D.124 kbps
Explanation: The Shannon channel capacity formula is C = B × log_2(1 + SNR). Substituting B = 4,000 Hz and SNR = 31 yields C = 4000 × log_2(1 + 31) = 4000 × log_2(32). Since 32 = 2^5, log_2(32) = 5. Therefore, C = 4,000 × 5 = 20,000 bps = 20 kbps.
5What is the total number of structurally distinct binary search trees that can be formed using 4 distinct keys (or 4 unlabeled nodes)?
A.14
B.24
C.42
D.16
Explanation: The number of structurally unique binary trees with n nodes is given by the n-th Catalan number: C_n = (1 / (n + 1)) × (2n choose n). For n = 4, C_4 = (1 / 5) × (8 choose 4) = (1 / 5) × (8 × 7 × 6 × 5) / (4 × 3 × 2 × 1) = 70 / 5 = 14 distinct structures.
6Consider the conditional proposition: 'If process P holds the resource lock and queue Q is empty, then transaction T will successfully commit.' What is the logically equivalent contrapositive of this proposition?
A.If transaction T fails to commit, then process P does not hold the resource lock or queue Q is not empty.
B.If transaction T commits, then process P holds the resource lock and queue Q is empty.
C.If process P does not hold the resource lock or queue Q is not empty, then transaction T will not commit.
D.If transaction T fails to commit, then process P does not hold the resource lock and queue Q is not empty.
Explanation: The original statement has form (A ∧ B) → C. The contrapositive is ¬C → ¬(A ∧ B). By De Morgan's Law, ¬(A ∧ B) ≡ (¬A ∨ ¬B). Therefore, the contrapositive is: 'If transaction T does not commit (¬C), then process P does not hold the lock (¬A) OR queue Q is not empty (¬B)'.
7A discrete-time Markov chain modeling a cloud server state has two states: 0 (Idle) and 1 (Busy). The transition probability matrix is P = [[0.8, 0.2], [0.3, 0.7]], where P[i][j] is the probability of transitioning from state i to state j. What is the long-run steady-state probability π_1 of the server being in the Busy state?
A.0.40
B.0.60
C.0.20
D.0.50
Explanation: The stationary distribution satisfies [π_0, π_1] P = [π_0, π_1] with π_0 + π_1 = 1. From the system: π_0 = 0.8 π_0 + 0.3 π_1 => 0.2 π_0 = 0.3 π_1 => π_0 = 1.5 π_1. Substituting into π_0 + π_1 = 1 gives 1.5 π_1 + π_1 = 1 => 2.5 π_1 = 1 => π_1 = 1 / 2.5 = 0.40 (and π_0 = 0.60).
8In a dynamically resizable array initialized to capacity 1, the buffer size doubles whenever an insertion exceeds capacity. Which is the tightest listed bound on the amortized cost per append across n insertions starting from empty?
A.O(1)
B.O(log n)
C.O(n)
D.O(n log n)
Explanation: Using aggregate analysis, resizing occurs at powers of two: 1, 2, 4, 8, ..., 2^k ≤ n. The total element copy operations during reallocations sum to 1 + 2 + 4 + ... + 2^k < 2n. Across n appends, the total work is n (for individual insertions) + < 2n (for copies) < 3n. Dividing total cost by n operations yields an amortized time of O(1) per operation.
9What is the modular multiplicative inverse of 7 modulo 31 (that is, an integer x such that (7 × x) ≡ 1 mod 31)?
A.9
B.13
C.4
D.22
Explanation: Using the Extended Euclidean Algorithm or testing multiples: 7 × 9 = 63. Dividing 63 by 31 yields 63 = 2 × 31 + 1, which means 63 ≡ 1 (mod 31). Hence, the modular inverse of 7 modulo 31 is 9.
10Why is the information-theoretic lower bound for comparison-based sorting of n arbitrary elements Ω(n log n) in the worst case?
A.A decision tree must have at least n! leaves to distinguish all permutations, requiring a height of at least log_2(n!) = Ω(n log n).
B.Every comparison reduces the remaining search space by at most a factor of n, yielding n comparisons.
C.Any comparison-based algorithm must perform at least n(n-1)/2 comparisons to ensure complete pairwise transitivity.
D.Cache line invalidations and branch mispredictions impose an inherent Ω(n log n) overhead on modern hardware.
Explanation: Any comparison sort can be modeled as a binary decision tree where each leaf represents one of the n! possible permutations of the input. A binary tree of height h has at most 2^h leaves. Therefore, 2^h ≥ n!, which implies h ≥ log_2(n!). By Stirling's approximation, log_2(n!) = n log_2(n) - n log_2(e) + O(log n) = Ω(n log n).

About the PEC EPE Computer & Allied Exam

The current PEC syllabus names Computer and Allied Engineering (Computer/Computer Systems/Software). Registered Engineers take the EPE toward Professional Engineer registration. OpenExamPrep offers independent English-language practice across Computer Systems and Software topics. This survey does not replace the selected Depth syllabus and is not a calibrated examination simulation.

Exam sponsor: Pakistan Engineering Council (PEC). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Two MCQ parts: Part-I closed-book, 2 hours; a 90-minute break; Part-II open-book, 3 hours. Choose Computer Systems or Software Engineering Depth. No additional mandatory EPE assignment, oral, practical, or case-study component is listed.

Time Limit

3 hours (Part-II); 2 hours (Part-I), with a 90-minute break

Passing Score

60% in each part independently

Exam / Certification Fees

Rs. 5,000 new candidate; Rs. 2,500 single-part reappearance, plus bank charges

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

15% + 15% of official Breadth

Mathematics and programming

Engineering mathematics and basic computer programming.

10% + 10% of official Breadth

Electrical engineering and architecture

Basic electrical engineering; computer architecture and organization.

10% each of official Breadth

Data structures, software engineering, and databases

Data structures and algorithms; basic software engineering; database systems.

10% each of official Breadth

Networks and operating systems

Computer networks and operating systems. Breadth totals 25 MCQs.

35 of 60 official Part-II MCQs

Chosen Depth area

Choose Computer Systems Engineering or Software Engineering. Advanced topics differ by option; the full official document includes areas beyond this practice survey.

Preparing for the PEC EPE Computer & Allied Exam

What You Need to Know

  • Passing score: 60% in each part independently
  • Assessment: Two MCQ parts: Part-I closed-book, 2 hours; a 90-minute break; Part-II open-book, 3 hours. Choose Computer Systems or Software Engineering Depth. No additional mandatory EPE assignment, oral, practical, or case-study component is listed.
  • Time limit: 3 hours (Part-II); 2 hours (Part-I), with a 90-minute break
  • Exam / certification fees: Rs. 5,000 new candidate; Rs. 2,500 single-part reappearance, plus bank charges Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

PEC EPE Computer & Allied: Suggested Study Strategy

1Choose the correct Depth option before prioritizing revision.
2State the language, database engine, and protocol version when applying implementation-specific rules.
3Work complexity, cache, subnetting, and scheduling calculations rather than memorizing answer positions.

Frequently Asked Questions

How is the practice bank distributed?

The 100 questions comprise 12 mathematics/algorithms, 15 architecture, 15 OS/networks, 20 software engineering, 15 databases, 13 security/distributed systems, and 10 embedded/IoT items. These are practice inventory counts, not official subdomain weights. Use PEC's selected Depth outline to address gaps such as electrical foundations, compilers, VLSI, AI/ML, or other option-specific topics.

Which languages are officially available?

The official syllabus and sample material reviewed are in English, but PEC's current sources do not separately state assessment-language options. This bank is English-language independent study material and is not an official translation.

What may I bring to Part-II?

PEC permits bound textbooks, reference books, and standards. Follow the current candidate guidelines; loose notes, laptops, and prohibited electronic devices are not allowed.

Are experience and CPD practical exam components?

They are eligibility requirements. The published EPE assessment consists of two MCQ parts, with no additional mandatory assignment, oral, practical, or separate case-study assessment listed.