Engineering Mathematics Practice Questions for GATE CS

GATE CS Engineering Mathematics: 4 units and 19 chapters, weightage from 105 previous year questions across 10 papers, a study order by exam weight and 1656 practice questions.

About Engineering Mathematics Practice Questions

1656 practice questions for Engineering Mathematics in GATE CS, sorted chapter by chapter and graded from basic to exam level, each with a full solution.

GATE CS Engineering Mathematics Unit-wise Weightage from Past Papers

We counted every GATE CS Engineering Mathematics previous year question in our bank (105 questions from 10 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Discrete Mathematics	7	47	45%	4.7
Linear Algebra	4	20	19%	2
Calculus	3	13	12%	1.3
Probability and Statistics	5	25	24%	2.5

Suggested Engineering Mathematics Study Order for GATE CS

Discrete Mathematics: 45% of past Engineering Mathematics questions, about 4.7 per paper.
Probability and Statistics: 24% of past Engineering Mathematics questions, about 2.5 per paper.
Linear Algebra: 19% of past Engineering Mathematics questions, about 2 per paper.
Calculus: 12% of past Engineering Mathematics questions, about 1.3 per paper.

Start where the marks are. Units at the top of this list have appeared most often in past GATE CS papers.

Units in GATE CS Engineering Mathematics

All Engineering Mathematics chapters

One Solved Question from Each Engineering Mathematics Chapter

Question 1 · Logic and Proof Techniques MCQ

Consider the following two statements regarding nested quantifier conjectures:

Assertion (A): The conjecture

\forall x (P (x) ⟹ \exists y Q (x, y))

implies that for every

x

satisfying

P (x)

, there is a specific

y

(which may depend on

x

) satisfying

Q (x, y)

.

Reason (R): The existential quantifier

\exists y

is inside the scope of the universal quantifier

\forall x

, meaning the choice of

y

is independent of

x

.

Which of the following options is correct?

A.
Both A and R are true and R is the correct explanation of A.
B.
Both A and R are true but R is not the correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: The order of nested quantifiers dictates dependency. An existential quantifier inside a universal quantifier means the existential variable depends on the universal variable.

Exam route: Assertion A correctly states that $y$ may depend on $x$ because $\exists y$ is inside $\forall x$ . Reason R claims that being inside the scope means $y$ is independent of $x$ , which is the exact opposite of the truth. Thus, A is true, but R is false.

Learning route:

Step 1: Evaluate Assertion (A). The statement $\forall x \exists y$ means "for every $x$ , there is a $y$ ". The $y$ can be chosen differently for each $x$ . Thus, $y$ depends on $x$ . A is True.

Step 2: Evaluate Reason (R). The reason claims that $\exists y$ being inside $\forall x$ means $y$ is independent of $x$ . This is false. Scope dependency means the inner variable depends on the outer variable. R is False.

Step 3: Combine the evaluations. A is true, but R is false.

Answer: A is true but R is false (Option C).

Tempting wrong path: A student might confuse the scope rule, thinking that being "inside" the scope isolates the variable, making it independent (Option A). This breaks at misunderstanding the direction of quantifier dependency.

Generalization: Inner quantifiers depend on outer quantifiers. $\forall x \exists y$ means $y$ depends on $x$ .

Verification: If $y$ were independent, the statement would be written $\exists y \forall x$ , meaning a single $y$ works for all $x$ .

Question 2 · Sets, Relations, Functions and Order Structures MCQ

Let $g : A \to B$ and $f : B \to C$ be functions such that their composition $f \circ g : A \to C$ is injective. Which of the following properties MUST hold for the individual functions?

A.
$f$ must be injective
B.
$g$ must be injective
C.
$f$ must be surjective
D.
$g$ must be surjective

Correct Answer:

Step-by-Step Solution

Key idea: This is a composition properties question, recognizable by the given injectivity of a composite function $f \circ g$ .

Step 1: Recall the definition of injectivity for $f \circ g$ . If $(f \circ g) (x_{1}) = (f \circ g) (x_{2})$ , then $x_{1} = x_{2}$ .

Step 2: This means $f (g (x_{1})) = f (g (x_{2})) ⟹ x_{1} = x_{2}$ . For this to hold, $g$ must map distinct $x_{1}, x_{2}$ to distinct values $g (x_{1}), g (x_{2})$ . Thus, $g$ must be injective.

Step 3: Does $f$ need to be injective? No. $f$ only needs to be injective on the range of $g$ . It can map other elements of $B$ to the same values.

Step 4: Does $g$ need to be surjective? No. $g$ can leave some elements of $B$ unmapped, as long as the elements it does map are distinct.

Answer: Option B.

Question 3 · Combinatorics and Counting MCQ

The formula for the number of non-negative integer solutions to the equation $x_{1} + x_{2} + \dots + x_{k} = n$ is:

A.
$inom{n+k-1}{k-1}$
B.
$inom{n-1}{k-1}$
C.
$inom{n+k}{k}$
D.
$r a c n! k! (n - k)!$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct formula recall question for the standard Stars and Bars method with non-negative integer solutions.

Step 1: Recall the setup. We are distributing $n$ identical items into $k$ distinct bins, allowing empty bins.

Step 2: Recall the formula. The number of ways is the number of ways to arrange $n$ stars and $k - 1$ bars in $n + k - 1$ total positions.

Step 3: This is given by the combination $(k - 1 n + k - 1)$ .

Answer: A

Question 4 · Recurrences and Generating Functions MCQ

Consider the recurrence $a_{n} = 5 a_{n - 1} - 6 a_{n - 2}$ . Which of the following statements about its characteristic equation is TRUE?

A.
The characteristic equation is $r^{2} - 5 r - 6 = 0$ .
B.
The characteristic roots are $2$ and $3$ .
C.
The characteristic equation is $r^{2} + 5 r + 6 = 0$ .
D.
The characteristic roots are $- 2$ and $- 3$ .

Correct Answer:

Step-by-Step Solution

Key idea: For a recurrence $a_{n} = c_{1} a_{n - 1} + c_{2} a_{n - 2}$ , the characteristic equation is $r^{2} - c_{1} r - c_{2} = 0$ .

Step 1: Identify coefficients from $a_{n} = 5 a_{n - 1} - 6 a_{n - 2}$ . Here $c_{1} = 5$ and $c_{2} = - 6$ .

Step 2: Form the characteristic equation: $r^{2} - 5 r - (- 6) = 0 ⟹ r^{2} - 5 r + 6 = 0$ .

Step 3: Solve the quadratic equation. Factor it: $(r - 2) (r - 3) = 0$ .

Step 4: The roots are $r = 2$ and $r = 3$ .

Answer: The characteristic roots are 2 and 3.

Question 5 · Algebraic Structures and Groups MCQ

Assertion (A): The number of self-inverse elements in $Z_{2} \times Z_{3} \times Z_{4}$ is 4.

Reason (R): An element $(x, y, z)$ is self-inverse if $2 x \equiv 0 (mod 2)$ , $2 y \equiv 0 (mod 3)$ , and $2 z \equiv 0 (mod 4)$ .

A.
A is true but R is false.
B.
Both A and R are true but R is NOT the correct explanation of A.
C.
A is false but R is true.
D.
Both A and R are true and R is the correct explanation of A.

Correct Answer:

Step-by-Step Solution

Key idea: In a direct product, an element is self-inverse iff each component is self-inverse.

Step 1: In $Z_{n}$ under addition, the self-inverse condition is $2 x \equiv 0 (mod n)$ . This makes R true.

Step 2: For $Z_{2}$ , $2 x \equiv 0 (mod 2) ⟹ x \in {0, 1}$ (2 solutions).

Step 3: For $Z_{3}$ , $2 y \equiv 0 (mod 3) ⟹ y = 0$ (1 solution).

Step 4: For $Z_{4}$ , $2 z \equiv 0 (mod 4) ⟹ z \in {0, 2}$ (2 solutions).

Step 5: Total self-inverse elements = $2 \times 1 \times 2 = 4$ . This makes A true.

Step 6: R correctly provides the method to compute A, so R is the correct explanation.

Answer: D

Question 6 · Graph Fundamentals, Trees and Connectivity MCQ

Consider a connected simple graph with 10 vertices. If we partition its edges into a spanning tree and a set of extra edges, and the total number of edges is 15, how many edges are in the extra set?

A.
6
B.
5
C.
4
D.
9

Correct Answer:

Step-by-Step Solution

Insight: A spanning tree of an $n$ -vertex graph always contains exactly $n - 1$ edges.

Exam route: The spanning tree has $10 - 1 = 9$ edges. The extra edges are the total minus the tree edges: $15 - 9 = 6$ .

Learning route:

Identify the total number of vertices $n = 10$ and total edges $E = 15$ .
Recall that any spanning tree of a graph with $n$ vertices must have exactly $n - 1$ edges.
Calculate the number of edges in the spanning tree: $10 - 1 = 9$ .
The extra edges are those not in the spanning tree. Subtract the tree edges from the total: $15 - 9 = 6$ .

Answer: 6

Question 7 · Graph Coloring, Covers, Matrices and Special Graphs MCQ

Let $A$ be the adjacency matrix of a simple undirected graph. Which of the following statements is ALWAYS TRUE regarding the matrix $A^{k}$ for any integer $k \geq 1$ ?

A.
The entry $(A^{k})_{ij}$ gives the number of simple paths of length $k$ from $v_{i}$ to $v_{j}$ .
B.
The trace of $A^{k}$ gives the total number of closed walks of length $k$ in the graph.
C.
The sum of all elements in $A^{k}$ is equal to the total number of cycles of length $k$ .
D.
The diagonal entries of $A^{k}$ are always zero for all $k$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding/statement truth question that tests the fundamental properties of matrix powers and the constraints of graph traversals.

Step 1: Evaluate Option A. $A^{k}$ counts all walks, not just simple paths. Walks allow repeated vertices. Thus, A is false.

Step 2: Evaluate Option B. The trace of a matrix is the sum of its diagonal entries. The diagonal entry $(A^{k})_{ii}$ is the number of closed walks of length $k$ starting and ending at $v_{i}$ . Summing over all $i$ gives the total number of closed walks of length $k$ in the graph. Thus, B is true.

Step 3: Evaluate Option C. The sum of all elements counts all walks of length $k$ , not just cycles. Cycles are simple closed paths. Thus, C is false.

Step 4: Evaluate Option D. For even $k$ , there are closed walks (e.g., $v_{i} \to v_{j} \to v_{i}$ for $k = 2$ ). Thus, diagonal entries are non-zero for even $k$ . D is false.

Answer: B

Question 8 · Systems of Linear Equations and LU Decomposition MCQ

For a system of linear equations $A X = B$ with $n$ variables, which of the following conditions guarantees that the system has infinitely many solutions?

A.
$e x t r ank (A) < e x t r ank ([A ∣ B])$
B.
$e x t r ank (A) = e x t r ank ([A ∣ B]) < n$
C.
$e x t r ank (A) = e x t r ank ([A ∣ B]) = n$
D.
$e x t r ank (A) > e x t r ank ([A ∣ B])$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct definition question about the Rouché-Capelli theorem, recognisable because it asks for the condition of infinite solutions in terms of matrix ranks.

Step 1: Recall the consistency condition.

A system $A X = B$ is consistent if and only if $rank (A) = rank ([A ∣ B])$ .

Step 2: Distinguish between unique and infinite solutions.

If the common rank equals the number of variables $n$ , the system has exactly one unique solution.

If the common rank is strictly less than $n$ , there are $n - rank (A)$ free variables, leading to infinitely many solutions.

Step 3: Evaluate the options.

Option A implies inconsistency (no solution).

Option B implies consistency with free variables (infinite solutions).

Option C implies consistency with no free variables (unique solution).

Option D is mathematically impossible.

Answer: B

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Engineering Mathematics Practice Questions for GATE CS

About Engineering Mathematics Practice Questions

GATE CS Engineering Mathematics Unit-wise Weightage from Past Papers

Suggested Engineering Mathematics Study Order for GATE CS

Units in GATE CS Engineering Mathematics

All Engineering Mathematics chapters

One Solved Question from Each Engineering Mathematics Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution