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Key Facts: FEI STU Math Entrance Exam Exam

The FEI STU entrance test covers secondary-school mathematics and is written in Slovak.

The test is scored out of 100 points and passed with at least 30 points.

It is required for applied informatics, digital technologies, information and communication technologies, and robotics and cybernetics.

The dean waives the test for at least the 70th percentile in the external maturita test in mathematics or at least 50 competition points.

The published 2025 papers had 15 single-answer questions with six options and an 80-minute limit.

Only correct answers were counted in the 2025 papers, with no penalty for wrong answers.

FEI publishes nine topic areas, from propositional logic to plane analytic geometry.

For 2026/2027 entry the processing fee was 30 EUR and the test dates were 29–30 April and 27 July 2026.

The FEI STU mathematics entrance test is a written Slovak-language test of secondary-school mathematics scored out of 100 points and passed with 30; the 2025 papers had 15 six-option questions in 80 minutes, and strong maturita or competition results can waive it.

Sample FEI STU Math Entrance Exam Practice Questions

Try these sample questions to review concepts for the FEI STU Math Entrance Exam exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which of the following compound propositions is logically equivalent to the implication p => q?
A.not p or q
B.p and not q
C.not p and q
D.p or not q
Explanation: By definition of material implication in classical logic, p => q is false if and only if p is true and q is false. Therefore, p => q is equivalent to not p or q.
2What is the logical negation of the statement: 'Every student in the seminar passed the examination'?
A.No student in the seminar passed the examination.
B.At least one student in the seminar did not pass the examination.
C.Every student in the seminar failed the examination.
D.At least one student in the seminar passed the examination.
Explanation: The statement has the form of a universally quantified proposition: for all x in S, P(x). Its formal negation is the existentially quantified proposition: there exists x in S such that not P(x), meaning at least one student did not pass.
3According to De Morgan's laws, which statement is logically equivalent to not (A and not B)?
A.not A and B
B.A and not B
C.not A or B
D.A or not B
Explanation: Applying De Morgan's law to the conjunction yields not (A and not B) <=> not A or not (not B). Using double negation elimination on not (not B) gives not A or B.
4Consider propositions p: 'x > 2' and q: 'x^2 > 4'. Which logical relationship correctly describes them for all real numbers x?
A.p <=> q
B.p => q, but q does not imply p
C.q => p, but p does not imply q
D.Neither p => q nor q => p holds
Explanation: If x > 2, then squaring both sides preserves the inequality since both sides are positive, yielding x^2 > 4. Thus p => q is true. Conversely, if x = -3, then x^2 = 9 > 4 is true, but -3 > 2 is false, so q does not imply p.
5What is the truth value of the formula ((p => q) and p) => q for all possible truth assignments of p and q?
A.It is a tautology (always true).
B.It is a contradiction (always false).
C.It is true only when both p and q are true.
D.It is false only when p is false and q is true.
Explanation: This formula represents the classical inference rule Modus Ponens. If p is true and p => q is true, then q must be true, so the implication ((p => q) and p) => q is true under all truth assignments, making it a tautology.
6What is the contrapositive (obmena) of the statement: 'If an integer n is divisible by 6, then n is divisible by 2 and by 3'?
A.If an integer n is not divisible by 6, then n is not divisible by 2 or not divisible by 3.
B.If an integer n is not divisible by 2 or not divisible by 3, then n is not divisible by 6.
C.If an integer n is divisible by 2 and by 3, then n is divisible by 6.
D.If an integer n is not divisible by 2 and not divisible by 3, then n is not divisible by 6.
Explanation: The contrapositive of p => q is not q => not p. Here q is 'divisible by 2 and divisible by 3'. By De Morgan's law, not q is 'not divisible by 2 or not divisible by 3'. Thus the contrapositive is 'If n is not divisible by 2 or not divisible by 3, then n is not divisible by 6'.
7For how many of the 4 possible truth assignments of variables (p, q) is the expression (p or q) => (p and q) true?
A.1
B.2
C.3
D.4
Explanation: Evaluate the 4 truth assignments: (T, T) gives T => T (true); (T, F) gives T => F (false); (F, T) gives T => F (false); (F, F) gives F => F (true). The formula is true for exactly 2 assignments: (T, T) and (F, F).
8Which of the following propositional formulas is a contradiction (false under all truth assignments)?
A.(p => q) and (p and not q)
B.(p or q) and (not p or not q)
C.(p => q) or (q => p)
D.not (p and q) <=> (not p or not q)
Explanation: The formula (p => q) and (p and not q) conjoins an implication with its exact negation, since not (p => q) is logically equivalent to p and not q. A proposition conjoined with its negation A and not A is always false, making it a contradiction.
9Consider the statement: 'For every real number x, there exists a real number y such that x + y = 0'. What is its negation, and what is the truth value of the original statement in the real numbers?
A.Negation: 'There exists a real x such that for all real y, x + y != 0'; original statement is true.
B.Negation: 'For all real x and all real y, x + y != 0'; original statement is true.
C.Negation: 'There exists a real x and a real y such that x + y != 0'; original statement is false.
D.Negation: 'For every real x, there exists a real y such that x + y != 0'; original statement is false.
Explanation: The original statement is the additive inverse axiom for the real field: for every x in R, setting y = -x gives x + (-x) = 0, so it is true. Negating the alternating quantifiers changes 'for all x exists y' into 'there exists x such that for all y, x + y != 0'.
10If proposition (p and not q) => (r or not s) is false, what are the truth values of p, q, r, and s respectively?
A.p = true, q = false, r = false, s = true
B.p = true, q = true, r = false, s = false
C.p = false, q = false, r = true, s = true
D.p = true, q = false, r = true, s = false
Explanation: An implication A => B is false if and only if A is true and B is false. For A = (p and not q) to be true, we must have p = true and not q = true (so q = false). For B = (r or not s) to be false, both disjuncts must be false: r = false and not s = false (so s = true). Thus (p, q, r, s) = (true, false, false, true).

About the FEI STU Math Entrance Exam Exam

Independent study practice for the mathematics entrance examination of the Faculty of Electrical Engineering and Information Technology of the Slovak University of Technology in Bratislava (FEI STU). Under Article 5 of FEI's further admission conditions for 2026/2027, applicants to the bachelor programmes applied informatics, digital technologies, information and communication technologies, and robotics and cybernetics are admitted on the basis of a written test in Slovak covering secondary-school mathematics. The test is scored out of 100 points and passed with at least 30 points; successful applicants are ranked by points. The dean waives the test for applicants with at least the 70th percentile in the external maturita test in mathematics or at least 50 points from secondary-school competitions under FEI's published table, and such applicants receive 100 points. The published 2025 papers had 15 single-answer questions with six options in 80 minutes. For 2026/2027 entry the test dates were 29–30 April 2026 and 27 July 2026, and the processing fee was 30 EUR. The 100 practice questions cover FEI's nine published topic areas. This bank is an English-language, four-option multiple-choice study adaptation; the real paper is in Slovak with six options including 'none of the other answers is correct'.

Exam sponsor: Fakulta elektrotechniky a informatiky Slovenskej technickej univerzity v Bratislave (FEI STU). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Questions

15 questions

Time Limit

80 minutes (2025 papers)

Passing Score

At least 30 of 100 points; a waived test counts as 100 points.

Exam / Certification Fees

30 EUR admission-processing fee for 2026/2027 entry

Exam sponsor website

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

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.

No official weighting; 10 of 100 practice questions

Výroková logika (Propositional Logic)

Basic logical operations and rules: negation, conjunction, disjunction, implication and equivalence, quantifiers, and truth tables.

No official weighting; 10 of 100 practice questions

Množiny (Sets)

Sets and set operations, number sets and intervals, Venn diagrams and inclusion-exclusion.

No official weighting; 12 of 100 practice questions

Kombinatorika (Combinatorics)

Permutations, variations and combinations with and without repetition, and combination numbers.

No official weighting; 12 of 100 practice questions

Úpravy algebraických výrazov (Algebraic Expressions)

Order of operations, brackets, fractions, powers and roots, factoring, and simplifying algebraic expressions with their domains.

No official weighting; 14 of 100 practice questions

Elementárne funkcie (Elementary Functions)

Linear, quadratic, polynomial and linear fractional functions, power, exponential and logarithmic functions, and trigonometric functions and identities.

No official weighting; 16 of 100 practice questions

Rovnice a nerovnice (Equations and Inequalities)

Linear, quadratic, trigonometric, exponential and logarithmic equations and inequalities, linear fractional expressions, and equations and inequalities with absolute value.

No official weighting; 10 of 100 practice questions

Postupnosti (Sequences)

Arithmetic and geometric sequences, their terms and sums, including infinite geometric series.

No official weighting; 8 of 100 practice questions

Planimetria (Plane Geometry)

Triangles and quadrilaterals, the Pythagorean theorem, the sine and cosine rules, circles and their chords, areas and perimeters.

No official weighting; 8 of 100 practice questions

Analytická geometria v rovine (Plane Analytic Geometry)

Vectors, equations of a line in the plane, and the relative position of lines (parallel, intersecting, perpendicular).

Preparing for the FEI STU Math Entrance Exam Exam

What You Need to Know

  • Passing score: At least 30 of 100 points; a waived test counts as 100 points.
  • Exam length: 15 questions
  • Time limit: 80 minutes (2025 papers)
  • Exam / certification fees: 30 EUR admission-processing fee for 2026/2027 entry 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

FEI STU Math Entrance Exam: Suggested Study Strategy

1Work through FEI's published sample test and the 2024 and 2025 papers under timed conditions; they are the best guide to the style and difficulty.
2Because the real paper includes the option 'none of the other answers is correct', always finish the calculation instead of guessing from the listed options.
3Check domains carefully in rational, radical, logarithmic and absolute-value problems, where extraneous solutions are a common trap.
4Practise combinatorics and sequences with short calculations you can do quickly without a calculator.
5Review propositional logic and set notation in Slovak (negácia, implikácia, prienik, zjednotenie) so the wording of the real test is familiar.

Frequently Asked Questions

Which FEI STU programmes require the mathematics entrance test?

Under Article 5 of FEI's further admission conditions for 2026/2027, the test is required for the bachelor programmes applied informatics, digital technologies, information and communication technologies, and robotics and cybernetics. Electronic systems and chip design has its own admission test and waiver rules under Article 6.

What does the test look like and how is it scored?

It is a written test in Slovak covering secondary-school mathematics, scored out of 100 points; an applicant passes with at least 30 points and successful applicants are ranked by points. The published 2025 papers had 15 single-answer questions with six options (a–f), including 'none of the other answers is correct', in 80 minutes, and only correct answers were counted.

Can the test be waived?

Yes. The dean waives the test for applicants who reached at least the 70th percentile in the external maturita test in mathematics, or who earned at least 50 points from secondary-school competitions under the table published by FEI STU. A waived test counts as the full 100 points.

Is this practice bank an official FEI STU test?

No. This is independent practice by OpenExamPrep: an English-language, four-option multiple-choice study adaptation of FEI's nine published topic areas. It is not an official test or translation; the real paper is in Slovak and offers six options per question.