9 Practice Problems with Solutions for GATE, UGC NET, ISRO, and NIELIT Exams

Competitive exams like GATE, UGC NET, ISRO, and NIELIT require not only theoretical knowledge but also rigorous practice to excel. Solving a variety of logical reasoning and mathematical problems is crucial to mastering the concepts and boosting your confidence.This article presents 9 practice problems, carefully selected to test your understanding of critical topics while preparing for these exams. Each problem is accompanied by detailed solutions, ensuring that you fully grasp the underlying concepts.

Why Are Practice Problems Essential?

• Reinforce Concepts: Practical application helps cement your understanding of theoretical principles.
• Develop Problem-Solving Skills: Exposure to diverse problems prepares you for the unexpected in exams.
• Time Management: Practice allows you to solve questions quickly and efficiently during exams.

By working through these problems, you’ll gain insights into commonly tested areas and learn strategies to solve them effectively.

Topics Covered

These practice problems focus on the following key areas:

• Logical reasoning with quantifiers and implications.
• Graph theory concepts like cyclic cardinality and degree sequences.
• Recurrence relations and their applications in binary strings.
• Set theory operations like union, intersection, and their counterexamples.
• Functional analysis involving inverse functions.
• Divisibility rules for numbers.
• Relation matrices and their higher-order computations.

Practice Problem Highlights

Here’s a brief overview of what each problem entails:

• Problem 1: Analyze logical statements using quantifiers, implications, and counterexamples.
• Problem 2: Calculate the cyclic cardinality of a tree using combinations in graph theory.
• Problem 3: Test degree sequences to determine their graphical validity.
• Problem 4: Solve recurrence relations for binary strings without consecutive 1s.
• Problem 5: Explore counterintuitive examples in set theory operations.
• Problem 6: Form 3-digit numbers under divisibility constraints.
• Problem 7: Deduce logical conclusions from cricket-based scenarios.
• Problem 8: Compute and validate inverse functions through algebraic manipulations.
• Problem 9: Decode adjacency matrices and compute relations with paths of length three.

How to Use These Problems?

Follow these steps to make the most of these practice problems:

1. Solve the Problems: Attempt each problem on your own before reviewing the solutions.
2. Understand the Solutions: Carefully read the explanations to grasp the underlying concepts.
3. Practice More: Use these problems as a foundation to explore similar questions and solidify your knowledge.

Why These Problems Are Unique

• Relevance: These problems cover topics frequently asked in competitive exams like GATE, UGC NET, ISRO, and NIELIT.
• Practical Application: Each problem is designed to improve your problem-solving skills with real-world relevance.
• Detailed Explanations: The solutions break down complex concepts into simple, easy-to-follow steps.

Conclusion

These 9 practice problems are a comprehensive resource to boost your preparation for GATE, UGC NET, ISRO, and NIELIT exams. They are designed to enhance your logical reasoning, mathematical aptitude, and exam performance. By solving these problems and understanding their solutions, you’ll gain the confidence needed to excel.

Stay consistent with your practice, and you’ll be well-prepared for your exams. Bookmark this page for future reference and share it with fellow aspirants. Good luck with your preparation!

Question:

Consider the following statements:

• P₁: ∀x ∀y ((x < 0) ∧ (y < 0) ⇒ (x * y > 0))
• P₂: ∀x ∃y (x + y = 0)
• P₃: ∀x ∀y ((x < 0) ∧ (y ≥ 0) ⇒ (x * y < 0))

Assume the domain to be the set of all integers. Which of the following statements is/are correct?

Options:

1. P₁ and P₂ only
2. P₂ and P₃ only
3. P₁ and P₃ only
4. P₁, P₂, and P₃

Solution:

Correct Option: A (P₁ and P₂ only)

Explanation:

• P₁: True. The product of two negative integers is always positive, which satisfies the given implication.
• P₂: True. For any integer x, there exists an integer y such that their sum is zero (y = -x, the additive inverse of x).
• P₃: False. If y = 0, then the product of x and y is zero, which violates the condition that the product must always be negative as stated in the implication.

Question:

Consider a tree T with n vertices and (n − 1) edges. The cyclic cardinality of a tree is defined as the number of cycles created when any two vertices of T are joined by an edge. Given a tree with 10 vertices, what is the cyclic cardinality of this tree?

Options:

1. 10
2. 100
3. 45
4. 90

Solution:

Correct Option: C (45)

Explanation:

For a tree with n vertices, the cyclic cardinality is equal to the number of ways to select two vertices from the n vertices, which is computed as ⁿC₂.

Given n = 10,

10 × (10 − 1)
2

= 45

Thus, the cyclic cardinality of the tree is 45.

Question:

A degree sequence (d₁, d₂, …, d_n) is graphical if there exists a simple undirected graph with n vertices having degrees d₁, d₂, …, d_n. The sequence is in non-increasing order (d₁ ≥ d₂ ≥ … ≥ d_n).

Consider the following sequences:

• Q₁: (2^8, 2^7, 2^6, 2^5, 2^4, 2^3, 2^2, 2^1)
• Q₂: (8^0, 7^0, 6^0, 5^0, 4^0, 3^0, 2^0, 1^0)

Which of the sequences given above is graphical?

Options:

1. Q₁ and Q₂ only
2. Only Q₁
3. Only Q₂
4. None of these

Solution:

Correct Option: C (Only Q₂)

Explanation:

• In Q₁, all the degrees are distinct, which makes it impossible to construct a simple undirected graph. Hence, Q₁ is not graphical.
• In Q₂, the degrees of all vertices satisfy the graphical sequence conditions:
  • The degree of each vertex is less than the total number of vertices.
  • The number of vertices with odd degrees is even.

  Therefore, Q₂ is graphical.

An example graph corresponding to Q₂:

Vertices: A, B, C, D, E, F, G, H

Edges:

• A – B
• C – D
• E – F
• G – H

This forms a valid graphical representation for Q₂.

Question:

Let M(n) represent the number of n-bit binary strings in which no two consecutive bits are 1. Which of the following correctly represents the recurrence relation for M(n)?

Options:

1. M(n) = 2M(n − 1) + M(n − 2); M(1) = 2, M(2) = 3
2. M(n) = M(n − 1) + M(n − 2); M(1) = 2, M(2) = 3
3. M(n) = (M(n − 1) + M(n − 2)) / 2; M(1) = 2, M(2) = 3
4. M(n) = M(n − 1) − M(n − 2); M(1) = 2, M(2) = 3

Solution:

Correct Option: B (M(n) = M(n − 1) + M(n − 2); M(1) = 2, M(2) = 3)

Explanation:

To determine the recurrence relation, we compute the number of n-bit binary strings without consecutive 1s:

• For n = 2, the possible strings are: 00, 01, and 10. Thus, M(2) = 3.
• For n = 3, the possible strings are: 000, 001, 010, 100, and 101. Thus, M(3) = 5.
• For n = 4, the possible strings are: 0000, 0001, 0010, 0100, 0101, 1000, 1001, and 1010. Thus, M(4) = 8.

From this pattern, it is clear that:

M(n) = M(n − 1) + M(n − 2)

This recurrence relation matches the Fibonacci sequence, where each term is the sum of the two preceding terms.

Hence, the correct recurrence relation is:

M(1) = 2, M(2) = 3, M(n) = M(n − 1) + M(n − 2)

Question:

Consider three sets A, B, and C. Now, consider the following statements:

• P₁: If A ∪ C = B ∪ C, then A = B.
• P₂: If A ∩ C = B ∩ C, then A = B.

Which of the above statements is/are true?

Options:

1. P₁ and P₂ only
2. Only P₁
3. Only P₂
4. None of these

Solution:

Correct Option: D (None of these)

Explanation:

• • P₁ is false:

Consider the counterexample:

• • • A = {1}, B = {1, 2}, C = {2, 3}
    • A ∪ C = {1, 2, 3}
    • B ∪ C = {1, 2, 3}

Here, A ≠ B, so P₁ is false.

• • P₂ is false:

Let C = ∅ (empty set), A = {1}, and B = {2}. Then:

• • • A ∩ C = ∅
    • B ∩ C = ∅

But A ≠ B, so P₂ is false.

Hence, both statements are false, and the correct answer is None of these.

Question:

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 such that the numbers are divisible by 6 and no digit is repeated?

Options:

1. 6
2. 8
3. 10
4. 12

Solution:

Correct Option: B (8)

Explanation:

To form a number divisible by 6, it must satisfy two conditions:

• Divisible by 2: The last digit (units place) must be even (2 or 4).
• Divisible by 3: The sum of all three digits must be divisible by 3.

Steps:

1. • Possible pairs of digits whose sum is divisible by 3:
     • {1, 5}: Sum = 6 (divisible by 3)
     • {3, 4}: Sum = 7 (not divisible by 3, invalid)
       • For digits {1, 3}, their sum is 4, not divisible by 3. Hence, this combination is invalid.
       • Valid combinations: ….If the last digit is 2, the remaining two digits must form a sum that is divisible by 3:

       Question:

       How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 such that the numbers are divisible by 6 and no digit is repeated?

       Options:

       1. 6
       2. 8
       3. 10
       4. 12

       Solution:

       Correct Option: B (8)

       Explanation:

       To form a 3-digit number divisible by 6, it must satisfy two conditions:

       • Divisibility by 2: The last digit (units place) must be even. The available even digits are 2 and 4.
       • Divisibility by 3: The sum of all three digits must be divisible by 3.

       We will calculate the valid combinations for each case:

       Case 1: Last digit = 2

       If the last digit is 2, the sum of the other two digits must be divisible by 3. The remaining digits are {1, 3, 4, 5}:

   Question:

   How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 such that the numbers are divisible by 6 and no digit is repeated?

   Options:

   1. 6
   2. 8
   3. 10
   4. 12

   Solution:

   Correct Option: B (8)

   Explanation:

   To form a 3-digit number divisible by 6, it must satisfy two conditions:

   • Divisibility by 2: The last digit (units place) must be even. The available even digits are 2 and 4.
   • Divisibility by 3: The sum of all three digits must be divisible by 3.

   Case 1: Last digit = 2

   If the last digit is 2, the sum of the other two digits must be divisible by 3. The remaining digits are {1, 3, 4, 5}:

   • Possible pairs of digits whose sum is divisible by 3:
     • {1, 5}: Sum = 6 (divisible by 3)
     • {3, 4}: Sum = 7 (not divisible by 3, invalid)
   • Valid numbers: 152, 512

   Case 2: Last digit = 4

   If the last digit is 4, the sum of the other two digits must be divisible by 3. The remaining digits are {1, 2, 3, 5}:

   • Possible pairs of digits whose sum is divisible by 3:
     • {1, 2}: Sum = 3 (divisible by 3)
     • {3, 5}: Sum = 8 (not divisible by 3, invalid)
   • Valid numbers: 124, 214, 324, 234

   Total valid numbers:

   • From Case 1: 2 numbers
   • From Case 2: 6 numbers

   Total = 2 + 6 = 8 numbers.

   Question:

   Consider the following statements:

   • P₁: Sachin Tendulkar gets out before the tea break only if Ishant Sharma comes out to bat.
   • P₂: Ishant Sharma won’t come out to bat if Lasith Malinga is not called to bowl.
   • P₃: Sachin Tendulkar got out before the tea break.

   Which of the following does not follow from P₁, P₂, and P₃?

   Options:

   1. Lasith Malinga is called to bowl.
   2. Ishant Sharma comes out to bat.
   3. Sachin Tendulkar got out after the tea break.
   4. None of these

   Solution:

   Correct Option: C (Sachin Tendulkar got out after the tea break)

   Explanation:

   • P₁: Sachin Tendulkar getting out before the tea break implies Ishant Sharma must have come out to bat. This statement does not conflict with the given conditions.
   • P₂: If Lasith Malinga is not called to bowl, then Ishant Sharma won’t come out to bat. However, since Sachin Tendulkar is already out before the tea break (from P₃), this implies Lasith Malinga was called to bowl (to satisfy P₂).
   • P₃: It is explicitly stated that Sachin Tendulkar got out before the tea break. Therefore, the statement “Sachin Tendulkar got out after the tea break” directly contradicts P₃.

   Since statement C contradicts the given information, it does not follow from P₁, P₂, and P₃. Hence, the correct answer is C.

   Question:

   Let f(x, y) = (x + y, x − y). What is f⁻¹(x, y)?

   Options:

   1. (x − y, x + y)
   2. (x − 2y, x + 2y)
   3. (x − y)/2, (x + y)/2
   4. (x + y)/2, (x − y)/2

   Solution:

   Correct Option: D ((x + y)/2, (x − y)/2)

   Explanation:

   To find f⁻¹(x, y), we solve the equations for x and y:

   • Given f(x, y) = (u, v) where u = x + y and v = x − y, we can express x and y in terms of u and v:
     • Add the two equations: u + v = (x + y) + (x − y) = 2x ⇒ x = (u + v)/2
     • Subtract the two equations: u − v = (x + y) − (x − y) = 2y ⇒ y = (u − v)/2

   Thus, f⁻¹(u, v) = ((u + v)/2, (u − v)/2).

   Verification:

   Let’s verify with an example:

   • Let x = 1 and y = 2. Then f(1, 2) = (1 + 2, 1 − 2) = (3, −1).
   • Now, applying f⁻¹(3, −1):
     • x = (3 + (−1))/2 = 2/2 = 1
     • y = (3 − (−1))/2 = 4/2 = 2
   • This matches the original values of x and y.

   Options A, B, and C fail to satisfy this condition. Hence, the correct option is D.

   Question:

   Given below is the matrix representation (M_R) of a relation R with 4 elements: {1, 2, 3, 4} respectively.

       1 2 3 4
   1 | 1 1 0 1
   2 | 0 1 0 0
   3 | 0 1 1 0
   4 | 0 0 0 0
   

   Which of the following correctly represents R³ in set-builder notation?

   Options:

   1. {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 3), (3, 3)}
   2. {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4)}
   3. {(1, 1), (1, 2), (1, 4), (2, 2), (2, 4), (3, 3), (3, 3)}
   4. {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 3), (3, 4)}

   Solution:

   Correct Option: B

   Explanation:

   The matrix M_R represents the adjacency matrix of a directed graph for relation R. Each entry M_R[i][j] is 1 if there is an edge from node i to node j, and 0 otherwise.

   Step 1: Convert the matrix to a directed graph.

   • From the matrix, we derive the edges:
     • (1, 1), (1, 2), (1, 4)
     • (2, 2)
     • (3, 2), (3, 3)

   Step 2: Find R³.

   R³ represents all pairs (a, b) such that there exists a path of length 3 from node a to node b.

   • From the graph, compute R³ by chaining paths of length 3:
     • For example, (1 → 2 → 3 → 4) is part of R³.
     • Similarly, paths like (1 → 1 → 1 → 1) and (1 → 2 → 2 → 4) are included.

   After computation, the set representing R³ is:

   {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4)}

   Hence, the correct option is B.