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Sample Matura Computer Science Practice Questions

Try these sample questions to review concepts for the Matura Computer Science exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A conventional binary search probes one midpoint per iteration in a sorted array of 1,024 elements. Counting one three-way key comparison per probe, what is the maximum number of probes?
A.10
B.11
C.512
D.1,024
Explanation: Binary search halves the search space at each iteration. For an array of size n = 1,024, the maximum number of comparisons in the worst case is floor(log2(n)) + 1 = 10 + 1 = 11 comparisons (or ceil(log2(n + 1)) = 11). Each comparison either finds the target or eliminates at least half of the remaining candidates.
2An algorithm satisfies T(n) = 2T(n/2) + Theta(n), with constant-size base cases. Which is the smallest of the listed valid Big-O bounds for its running time?
A.O(n)
B.O(n log n)
C.O(n^2)
D.O(log n)
Explanation: Here a = 2, b = 2, and f(n) = Theta(n). Since n^(log_b(a)) = n, Case 2 of the Master Theorem gives T(n) = Theta(n log n). Thus O(n log n) is the smallest listed valid upper bound. Each recursion level does Theta(n) work, and there are Theta(log n) levels.
3Trace the Euclidean algorithm using the modulo operator to compute the greatest common divisor gcd(198, 84). What is the resulting GCD?
A.2
B.6
C.12
D.14
Explanation: Applying the Euclidean algorithm: gcd(198, 84) -> 198 mod 84 = 30; next step gcd(84, 30) -> 84 mod 30 = 24; next step gcd(30, 24) -> 30 mod 24 = 6; next step gcd(24, 6) -> 24 mod 6 = 0. When the remainder is 0, the divisor 6 is the greatest common divisor.
4Using the fast modular exponentiation algorithm (repeated squaring), compute the value of (3^13) mod 7.
A.1
B.3
C.5
D.6
Explanation: The binary representation of 13 is 1101_2 (powers 8, 4, 1). Successive squarings modulo 7 yield: 3^1 mod 7 = 3; 3^2 mod 7 = 9 mod 7 = 2; 3^4 mod 7 = 2^2 mod 7 = 4; 3^8 mod 7 = 4^2 mod 7 = 16 mod 7 = 2. Now multiply powers corresponding to set bits: (3^8 * 3^4 * 3^1) mod 7 = (2 * 4 * 3) mod 7 = 24 mod 7 = 3.
5Which expression gives the tightest listed asymptotic running-time bound for the standard Sieve of Eratosthenes finding all primes up to n?
A.O(n)
B.O(n log log n)
C.O(n log n)
D.O(n^2)
Explanation: The Sieve of Eratosthenes marks multiples of each prime p <= sqrt(n). The total number of marking steps is proportional to n * sum_{p <= n}(1/p). According to Mertens' theorem, sum_{p <= n}(1/p) = ln(ln n) + M, giving an asymptotic time complexity of O(n log log n).
6Consider an array of n distinct integers that is already sorted in ascending order. What are the best-case time complexities of Bubble Sort (with an early exit swapped flag) and Insertion Sort?
A.Both are O(n)
B.Bubble Sort is O(n^2), Insertion Sort is O(n)
C.Bubble Sort is O(n), Insertion Sort is O(n log n)
D.Both are O(n log n)
Explanation: When an array is already sorted, an optimized Bubble Sort completes a single pass of n - 1 comparisons, detects zero swaps via its boolean flag, and terminates in O(n) time. Similarly, Insertion Sort checks each element against its immediate predecessor, finds it already in order, and takes exactly 1 comparison per element, running in O(n) linear time.
7What is the tightest listed worst-case time bound for standard QuickSort with a first-element pivot on an array of n distinct elements already sorted in ascending order?
A.O(n)
B.O(n log n)
C.O(n^2)
D.O(2^n)
Explanation: If the first element is chosen as pivot on an already sorted array, the pivot is always the minimum element. The partition produces one empty subarray and one subarray of size n - 1. This yields the recurrence T(n) = T(n - 1) + O(n) = O(n^2), resulting in quadratic execution time.
8To find both the minimum and maximum elements in an unsorted array of n elements (where n is even), what is the minimum number of comparisons required using the tournament / pairwise comparison method?
A.n - 1
B.3n/2 - 2
C.2n - 2
D.n log2(n)
Explanation: The pairwise comparison method compares elements in pairs (n/2 comparisons), sending the larger to a candidate max pool and the smaller to a candidate min pool. Finding the max among n/2 candidates takes (n/2) - 1 comparisons, and finding the min among n/2 candidates takes (n/2) - 1 comparisons. Total comparisons: n/2 + (n/2 - 1) + (n/2 - 1) = 3n/2 - 2.
9In the classical 0/1 Knapsack problem with n items and capacity W, let dp[i][w] denote the maximum value achievable using a subset of the first i items with total weight at most w. Which recurrence relation correctly defines dp[i][w] when w >= weight[i]?
A.dp[i][w] = dp[i - 1][w] + value[i]
B.dp[i][w] = max(dp[i - 1][w], dp[i - 1][w - weight[i]] + value[i])
C.dp[i][w] = max(dp[i - 1][w], dp[i][w - weight[i]] + value[i])
D.dp[i][w] = dp[i - 1][w - weight[i]] + dp[i - 1][w]
Explanation: For each item i, we decide either to exclude it (yielding value dp[i - 1][w]) or include it (yielding value dp[i - 1][w - weight[i]] + value[i]). The 0/1 constraint allows each item to be chosen at most once, which requires referencing row i - 1 when including item i.
10Evaluate the polynomial W(x) = 2x^3 - 4x^2 + 5x - 7 at x = 3 using Horner's method (schemat Hornera). What is the value of W(3)?
A.14
B.26
C.38
D.44
Explanation: Horner's arrangement is W(x) = ((2 * x - 4) * x + 5) * x - 7. Tracing with x = 3: Step 1 starts with coefficient 2; Step 2: 2 * 3 + (-4) = 6 - 4 = 2; Step 3: 2 * 3 + 5 = 6 + 5 = 11; Step 4: 11 * 3 + (-7) = 33 - 7 = 26. The result is 26.

About the Matura Computer Science Exam

Independent practice for Polish Matura Computer Science. This is an English-language MCQ study adaptation, not an official translation, an official-format simulation, or a simulation of the official language environment. It provides selected topic revision and does not substitute for writing full responses in the official assessment language. It does not replace practical work on a computer.

Exam sponsor: Centralna Komisja Egzaminacyjna (CKE) and the regional Okręgowe Komisje Egzaminacyjne (OKE), Poland. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

One continuous 210-minute session with computer access, including programming, spreadsheet, database, and written tasks. This is not the older two-part arrangement.

Time Limit

210 minutes

Passing Score

No minimum score for an additional extended-level subject in 2026. Certificate rules separately require 30% in each compulsory written and oral exam and attendance at an additional subject unless exempt.

Exam / Certification Fees

Generally free. In 2026, PLN 50 per subject, part, and level applies to third and subsequent sittings and certain previously declared additional-exam no-shows. OKE fee exemptions may apply; check the individual payment obligation.

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.

27% of this study bank

Algorithms & Computational Complexity

Algorithm design, Big-O asymptotic analysis, recursive relations, divide-and-conquer, greedy techniques, dynamic programming, and numerical algorithms such as fast exponentiation and Euclidean GCD.

25% of this study bank

Programming Concepts & Implementation

Implementation in C++, Python, or Java; modular structure, control flow, functions, recursion, input/output text file parsing, string parsing, and debugging.

13% of this study bank

Data Structures & Abstract Data Types

Linear structures (arrays, dynamic lists, stacks, queues), linked representations, binary trees, binary search trees, heaps, hash tables, and graph representations.

15% of this study bank

Relational Databases & SQL

Relational database concepts, primary and foreign keys, normalization, multi-table joins, subqueries, aggregation (GROUP BY, HAVING), sorting, and data filtering.

10% of this study bank

Computer Architecture, Networking & Security

Positional numeral systems (binary, octal, hexadecimal, two's complement), data representations (ASCII, UTF-8, floating-point IEEE 754), IP networking (IPv4 addressing, subnet masks), and cryptography and cybersecurity fundamentals.

10% of this study bank

Spreadsheet Data Analysis

Relative and absolute references, conditional formulas, aggregation, data import, sorting, charts, and missing values.

Preparing for the Matura Computer Science Exam

What You Need to Know

  • Passing score: No minimum score for an additional extended-level subject in 2026. Certificate rules separately require 30% in each compulsory written and oral exam and attendance at an additional subject unless exempt.
  • Assessment: One continuous 210-minute session with computer access, including programming, spreadsheet, database, and written tasks. This is not the older two-part arrangement.
  • Time limit: 210 minutes
  • Exam / certification fees: Generally free. In 2026, PLN 50 per subject, part, and level applies to third and subsequent sittings and certain previously declared additional-exam no-shows. OKE fee exemptions may apply; check the individual payment obligation. Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
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Matura Computer Science: Suggested Study Strategy

1Solve complete tasks on a computer, including file processing, spreadsheet analysis, and database queries.
2Use the CKE informator and released papers for the complete scope and scoring criteria. These MCQs cover selected topics and skills rather than every assessed requirement.

Frequently Asked Questions

What does this practice bank cover?

Independent practice for Polish Matura Computer Science. This is an English-language MCQ study adaptation, not an official translation, an official-format simulation, or a simulation of the official language environment. It provides selected topic revision and does not substitute for writing full responses in the official assessment language. It does not replace practical work on a computer. The section percentages describe this study bank, not official topic weightings. Consult the CKE informator for the complete requirements.

Are there separate assignments or performance requirements for this subject paper?

The subject assessment described here is the examination itself; no separate coursework portfolio or advance assignment is specified. One continuous 210-minute session with computer access, including programming, spreadsheet, database, and written tasks. This is not the older two-part arrangement. Whole-certificate requirements and accommodations are governed by the 2026 CKE instructions.

Who can sit the examination, and what does it cost?

Eligible secondary-school graduates must submit the examination declaration. The applicable formula and subject requirements depend on school type, graduation cohort, and prior examination history; consult the 2026 CKE instructions. Generally free. In 2026, PLN 50 per subject, part, and level applies to third and subsequent sittings and certain previously declared additional-exam no-shows. OKE fee exemptions may apply; check the individual payment obligation.