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100+ Free NIMCET Practice Questions

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Key Facts: NIMCET Exam

120 MCQs

Number of questions on the NIMCET exam

NIMCET Information Brochure

2 hours

Total exam duration

NIMCET Information Brochure

INR 2,500

Application fee for OPEN/OBC candidates

NIMCET Information Brochure

Mathematics

Highest weighted section with 50 questions (500 marks maximum)

NIMCET Information Brochure

NIMCET is the national computer-based test for admission to the MCA program at participating NITs. The 2-hour exam has 120 MCQs across 4 sections with 25% negative marking and is scaled out of 1000 total marks.

Sample NIMCET Practice Questions

Try these sample questions to test your NIMCET exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1If A and B are two sets such that n(A) = 35, n(B) = 30, and n(A ∪ B) = 55, what is the value of n(A ∩ B)?
A.5
B.10
C.15
D.20
Explanation: According to the principle of inclusion-exclusion, n(A ∪ B) = n(A) + n(B) - n(A ∩ B). Substituting the given values, we get 55 = 35 + 30 - n(A ∩ B), which simplifies to n(A ∩ B) = 65 - 55 = 10.
2If log_10(x^2 - 6x + 45) = 2, what are the possible real values of x?
A.9 and -5
B.5 and -11
C.11 and -5
D.5 and -9
Explanation: Rewriting the logarithmic equation in exponential form gives x^2 - 6x + 45 = 10^2 = 100. This simplifies to the quadratic equation x^2 - 6x - 55 = 0. Factoring this yields (x - 11)(x + 5) = 0, which gives the roots x = 11 and x = -5.
3What is the sum of the infinite geometric series: 1 + 1/3 + 1/9 + 1/27 + ...?
A.5/4
B.4/3
C.2
D.3/2
Explanation: This is an infinite geometric series with first term a = 1 and common ratio r = 1/3. Since |r| < 1, the sum S is given by the formula S = a / (1 - r). Substituting the values, S = 1 / (1 - 1/3) = 1 / (2/3) = 3/2.
4For what value of k will the system of equations x + y + z = 1, x + 2y + 4z = k, and x + 4y + 10z = k^2 have a solution?
A.0 or 1
B.1 or 2
C.2 or 3
D.1 or 3
Explanation: We can write the augmented matrix and apply row operations. R2 -> R2 - R1 gives R2: [0, 1, 3 | k - 1]. R3 -> R3 - R1 gives R3: [0, 3, 9 | k^2 - 1]. Now R3 -> R3 - 3*R2 gives [0, 0, 0 | k^2 - 3k + 2]. For the system to have a solution, we must have k^2 - 3k + 2 = 0. Solving this quadratic equation yields (k - 1)(k - 2) = 0, so k = 1 or k = 2.
5If the matrix A = [[1, 2], [3, 4]], what is the determinant of A^3?
A.-2
B.-8
C.8
D.-6
Explanation: The determinant of A is det(A) = 1*4 - 2*3 = 4 - 6 = -2. Using the property det(A^n) = (det(A))^n, the determinant of A^3 is (det(A))^3 = (-2)^3 = -8.
6Find the number of ways in which the letters of the word 'NIMCET' can be arranged such that the vowels always occur together.
A.720
B.120
C.240
D.144
Explanation: The word 'NIMCET' has 6 letters: N, I, M, C, E, T. The vowels are I and E. Treat the group of vowels (I, E) as a single entity. This leaves 5 entities to arrange: {IE}, N, M, C, T. These can be arranged in 5! = 120 ways. Within the vowel group, I and E can be arranged in 2! = 2 ways. The total number of arrangements is 120 * 2 = 240.
7What is the limit of (sin(3x) / tan(5x)) as x approaches 0?
A.0
B.5/3
C.1
D.3/5
Explanation: We can rewrite the expression as (sin(3x) / (3x)) * ((5x) / tan(5x)) * (3x / 5x). As x approaches 0, sin(3x)/(3x) approaches 1, and (5x)/tan(5x) approaches 1. Therefore, the limit is 1 * 1 * (3/5) = 3/5. Alternatively, using L'Hopital's rule, the derivative of sin(3x) is 3cos(3x) and the derivative of tan(5x) is 5sec^2(5x). Evaluating at x = 0 gives 3cos(0) / 5sec^2(0) = 3/5.
8What is the derivative of x^x with respect to x?
A.x * x^(x-1)
B.x^x (1 + ln(x))
C.x^x * ln(x)
D.x^x
Explanation: Let y = x^x. Taking the natural logarithm of both sides gives ln(y) = x * ln(x). Differentiating implicitly with respect to x: (1/y) * dy/dx = ln(x) + x * (1/x) = ln(x) + 1. Multiplying by y, we get dy/dx = y(1 + ln(x)) = x^x (1 + ln(x)).
9What is the maximum value of the function f(x) = x * e^(-x) for x > 0?
A.e
B.1/e
C.1
D.1/e^2
Explanation: To find the maximum, we find the first derivative of f(x) = x * e^(-x): f'(x) = 1*e^(-x) - x*e^(-x) = e^(-x)(1 - x). Setting f'(x) = 0 gives x = 1. The second derivative f''(x) = -e^(-x)(1 - x) - e^(-x) = e^(-x)(x - 2). At x = 1, f''(1) = -1/e < 0, indicating a local maximum. The maximum value is f(1) = 1 * e^(-1) = 1/e.
10What is the value of the definite integral of sin^2(x) from 0 to pi/2?
A.1/2
B.pi/2
C.pi/4
D.pi
Explanation: Let I = ∫[0 to pi/2] sin^2(x) dx. Using the property ∫[a to b] f(x) dx = ∫[a to b] f(a + b - x) dx, we get I = ∫[0 to pi/2] sin^2(pi/2 - x) dx = ∫[0 to pi/2] cos^2(x) dx. Adding these two expressions: 2I = ∫[0 to pi/2] (sin^2(x) + cos^2(x)) dx = ∫[0 to pi/2] 1 dx = pi/2. Hence, I = pi/4.

About the NIMCET Exam

NIMCET is the national-level entrance examination conducted for admission to the Master of Computer Applications (MCA) programme at participating NITs. The 120-minute exam consists of 120 multiple choice questions covering Mathematics, Logical Reasoning, Computer Awareness, and General English, with scores scaled to 1000 total marks.

Assessment

Four sections with a total of 120 multiple choice questions: Mathematics (50 questions), Analytical Ability & Logical Reasoning (40 questions), Computer Awareness (20 questions), and General English (10 questions).

Time Limit

120 minutes

Passing Score

No minimum passing score; admission is based on the rank in the merit list during counseling

Exam Fee

INR 2,500 for OPEN/OBC, INR 1,250 for SC/ST/PwD (National Institutes of Technology (NITs))

NIMCET Exam Content Outline

50%

Mathematics

Calculus, vectors, set theory, coordinate geometry, trigonometry, algebra, quadratic equations, determinants, matrices, and basic probability

33.3%

Analytical Ability & Logical Reasoning

Logical patterns, sequences, coding-decoding, blood relations, circular and linear arrangements, direction sense, and puzzles

16.7%

Computer Awareness

CPU/memory organization, integer and floating-point representations, binary/hexadecimal arithmetic, Boolean algebra, and logic gates

8.3%

General English

Reading comprehension, English grammar, synonyms and antonyms, tenses, prepositions, and sentence correction

How to Pass the NIMCET Exam

What You Need to Know

  • Passing score: No minimum passing score; admission is based on the rank in the merit list during counseling
  • Assessment: Four sections with a total of 120 multiple choice questions: Mathematics (50 questions), Analytical Ability & Logical Reasoning (40 questions), Computer Awareness (20 questions), and General English (10 questions).
  • Time limit: 120 minutes
  • Exam fee: INR 2,500 for OPEN/OBC, INR 1,250 for SC/ST/PwD

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

NIMCET Study Tips from Top Performers

1Focus heavily on Mathematics as it accounts for 50% of the questions and is the major score differentiator.
2Practice solving coordinate geometry, calculus, and matrix algebra under timed conditions.
3Brush up on computer basics including Boolean algebra simplification, binary logic, and memory structures.
4Solve logical reasoning puzzles and tenses/preposition usage regularly to secure high accuracy on the English and Reasoning sections.

Frequently Asked Questions

How many questions are on the NIMCET exam?

The NIMCET exam has 120 multiple-choice questions: 50 in Mathematics, 40 in Analytical Ability & Logical Reasoning, 20 in Computer Awareness, and 10 in General English.

What is the duration of the NIMCET exam?

The total duration of the NIMCET exam is 120 minutes (2 hours).

Is there negative marking in NIMCET?

Yes. NIMCET has 25% negative marking per section: Mathematics awards 12 marks per correct answer and deducts 3 for wrong; Reasoning and Computer Awareness award 6 and deduct 1.5; General English awards 4 and deducts 1. Unattempted questions receive zero.

What is the fee for NIMCET registration?

The registration fee is INR 2,500 for OPEN, OPEN-EWS, and OBC candidates, and INR 1,250 for SC, ST, and PwD candidates.