Jamia Millia Islamia MCA Sets and Relations Previous Year Question Paper and Solution with PDF

Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

Jamia Millia Islamia MCA
Previous Year Questions

Jamia Millia Islamia MCA
Previous Year Questions
Jamia Millia Islamia MCA

Jamia Millia Islamia MCA Previous Year Questions (PYQs)

Jamia Millia Islamia MCA Sets And Relations PYQ

Qus : 1 Jamia Millia Islamia MCA PYQ

Points within a set are connected by a line segment must follow the condition that points are

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

In a convex set, any line joining two points lies entirely within the set.

Qus : 2 Jamia Millia Islamia MCA PYQ

Order of the power set $ P(A) $ of a set $ A $ of order $ n $ is:

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

$ |A| = n $

Then

$ |P(A)| = 2^n $

Number of subsets $ = 2^n $

Qus : 3 Jamia Millia Islamia MCA PYQ

$\text{The set }(A\cap B)'\ \cup\ (B\cap C)\ \text{ is equal to:},$

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

$ (A\cap B)'=A'\cup B' ,$ (De Morgan) $\Rightarrow (A\cap B)'\cup(B\cap C)=(A'\cup B')\cup(B\cap C)=A'\cup\big(B'\cup(B\cap C)\big)$ $B'\cup(B\cap C)=(B'\cup B)\cap(B'\cup C)=U\cap(B'\cup C)=B'\cup C$ $\Rightarrow A'\cup(B'\cup C)=A'\cup B'\cup C$

Qus : 4 Jamia Millia Islamia MCA PYQ

In the group G = {2, 4, 6, 8} under multiplication modulo 10, the identity element is

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

$(a × e) \bmod 10 = a.$ For $e=6$, $2×6=12\equiv2,\ 4×6=24\equiv4,\ 8×6=48\equiv8.$

Qus : 5 Jamia Millia Islamia MCA PYQ

A partition of {1, 2, 3, 4, 5} is the family

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

A partition divides a set into disjoint non-empty subsets covering all elements. {{1,2},{3,4,5}} satisfies it.

Qus : 6 Jamia Millia Islamia MCA PYQ

Let P(S) denote the power set of set S. Which of the following is always TRUE?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

A partition divides a set into disjoint non-empty subsets covering all elements. {{1,2},{3,4,5}} satisfies it.

Qus : 7 Jamia Millia Islamia MCA PYQ

G = {a, b, c} is an abelian group with ‘e’ as identity element. The order of other elements is A. 2, 2, 3 B. 3, 3, 3 C. 2, 2, 4 D. 2, 2, 2

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

With 3 elements, it must be cyclic of order 3.

Qus : 8 Jamia Millia Islamia MCA PYQ

The maximum number of equivalence relations on the set $A = \{1,2,3\}$ is:

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

Number of equivalence relations = Number of partitions of set $A$. For 3 elements, Bell number $B_3 = 5$. $\boxed{\text{Answer: (D) 5}}$

Qus : 9 Jamia Millia Islamia MCA PYQ

How many elements does the set $P\big({\varnothing, a, {a}, {{a}}}\big)$ has; where $a$ and $b$ are distinct elements and $P$ denotes the power set?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

The given set ${\varnothing, a, {a}, {{a}}}$ has $4$ elements.
Therefore,

|P(S)| = 2^{|S|} = 2^4 = 16

Qus : 10 Jamia Millia Islamia MCA PYQ

What is the cardinality of these sets in the order of their serial number? (i) ${a}$ (ii) ${{a}}$ (iii) ${a, {a}}$ (iv) ${a, {a}, {{a}}}$

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

(i) ${a}$ → 1 element
(ii) ${{a}}$ → 1 element
(iii) ${a, {a}}$ → 2 elements
(iv) ${a, {a}, {{a}}}$ → 3 elements

Qus : 11 Jamia Millia Islamia MCA PYQ

Suppose that $A_i = {1, 2, 3, \ldots, i}$ for $i = 1, 2, 3, \ldots$ Then find $\displaystyle \bigcup_{i=1}^{\infty} A_i = ?$ Here $Z$ denotes the set of integers.

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

$A_{i} = {1, 2, 3, \dots, i} (i = 1, 2, \dots)$ $\bigcup_{i=1}^{\infty}A_i=\{1,2,3,\dots\}= \mathbb Z^{+}$ $\boxed{\mathbb Z^{+}}$

Qus : 12 Jamia Millia Islamia MCA PYQ

$,\text{Find } \displaystyle \bigcup_{i=1}^{\infty} A_i \text{ and } \bigcap_{i=1}^{\infty} A_i,\ \text{for every positive integer } i \text{ where } A_i={-i,i}.$ (Here $\mathbb Z$ denotes the set of integers.)

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

$A_1={-1,1},\ A_2={-2,2},\ldots$ so $\displaystyle \bigcup_{i=1}^{\infty}A_i={\ldots,-3,-2,-1,1,2,3,\ldots}=\mathbb Z\setminus{0}$, $\displaystyle \bigcap_{i=1}^{\infty}A_i=\varnothing$.

Qus : 13 Jamia Millia Islamia MCA PYQ

Which of the following relations are functions? (i) ${(1,(a,b)),\ (2,(b,c)),\ (3,(c,a)),\ (4,(a,b))}$ (ii) ${(1,(a,b)),\ (2,(b,a)),\ (3,(c,a)),\ (1,(a,c))}$ (iii) ${(1,(a,b)),\ (2,(a,b)),\ (3,(a,b))}$ (iv) ${(1,(a,b)),\ (2,(b,c)),\ (1,(c,a))}$

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

A relation $R$ is a function iff every first component occurs exactly once. (i) first components $1,2,3,4$ appear once each $\Rightarrow$ function. (ii) $1$ appears twice $\Rightarrow$ not a function. (iii) $1,2,3$ appear once each $\Rightarrow$ function. (iv) $1$ appears twice $\Rightarrow$ not a function.

Qus : 14 Jamia Millia Islamia MCA PYQ

$(A\cap B')\ \cup\ (A'\cap B)\ \cup\ (A'\cap B')$ is equal to

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

$(A'\cap B)\cup(A'\cap B')=A'\cap(B\cup B')=A'$. Then $A'\cup(A\cap B')=(A'\cup A)\cap(A'\cup B')=\mathbf U\cap(A'\cup B')=A'\cup B'$.

Qus : 15 Jamia Millia Islamia MCA PYQ

The number of students who take both the subjects’ mathematics and chemistry are 30. This represents 10% of the enrolment in mathematics and 12% of enrolment in chemistry. How many students take at least one of these two subjects?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

Let total mathematics students be $M$, chemistry students be $C$, and both be $30$. \[ 0.1M=30 \Rightarrow M=300,\qquad 0.12C=30 \Rightarrow C=250. \] At least one $= M+C-\text{both}=300+250-30=520$.

Qus : 16 Jamia Millia Islamia MCA PYQ

Let $A$ and $B$ be two disjoint subsets of a universal set $E$. Then $(A \cup B)\cap B'$ is equal to

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

Since $A$ and $B$ are disjoint, elements common with $B'$ are only from $A$. Hence $(A \cup B)\cap B' = A.$

Qus : 17 Jamia Millia Islamia MCA PYQ

$(A - B) - A$ is equal to

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

$(A-B)$ means elements in $A$ but not in $B$. Subtracting $A$ again gives an empty set.

Qus : 18 Jamia Millia Islamia MCA PYQ

Let 10 is the cardinality of set A. The number of bijective mappings from set A to itself is

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

Bijections on a 10-element set = number of permutations = 10. So, 10 = 3628800.

Qus : 19 Jamia Millia Islamia MCA PYQ

Let cardinality of set A and B are 2 and 5 respectively. The number of relations from A to B is

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

$|A\times B| = 2\cdot5=10$. A relation is any subset of $A\times B$. Count = $2^{10}=1024$.

Qus : 20 Jamia Millia Islamia MCA PYQ

Let cardinality of A and B are 3 and 10 respectively. The number of one-to-one functions from A to B is

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

Injective maps from 3 to 10 = permutations $P(10,3)=10\cdot9\cdot8=720$.

Qus : 21 Jamia Millia Islamia MCA PYQ

Let $A=\{1,2,3,4\}$ and $B=\{a,b\}$. The number of surjective mappings from A to B is

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

Onto functions to a 2-element codomain: $2^{4}-\binom{2}{1}1^{4}=16-2=14$ (I.E. principle).

Qus : 22 Jamia Millia Islamia MCA PYQ

Let R be a relation on the set of ordered pairs of positive integers such that ((p, q), (r, s)) ∈ R if and only if p − s = q − r. Which one of the following is true about R?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

Let R be a relation on ordered pairs of positive integers such that ((p,q),(r,s)) R iff p-s=q-r Check reflexive: For reflexive, take (p,q)=(r,s) Then condition becomes p-q=q-p 2p=2q p=q This is not true for every ordered pair, so R is not reflexive. Check symmetric: Given p-s=q-r Rearrange: p+r=q+s Now for symmetry we need r-q=s-p This also gives r+p=s+q, same condition. So R is symmetric.

Qus : 23 Jamia Millia Islamia MCA PYQ

The relation represented by $R = \{(1,1), (2,2), (3,3), (1,3), (3,2), (1,2)\}$ on the set $A = \{1, 2, 3\}$ is what kind of relation?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

R includes $(1,1), (2,2), (3,3)$ → Reflexive. Check symmetry: $(1,2)$ exists but $(2,1)$ does not → Not symmetric. Check transitivity: $(1,2)$ and $(2,2)$ imply $(1,2)$ already → transitive holds. Hence, relation is reflexive and transitive but not symmetric.

Qus : 24 Jamia Millia Islamia MCA PYQ

For $a, b \in \mathbb{R}$ define $a = b$ to mean that $|x| = |y|$. If $[x]$ is an equivalence relation in $R$, then the equivalence relation for $[17]$ is...

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Given relation $a = b \iff |a| = |b|$. Then equivalence class of $17$ is $[17] = \{x \in \mathbb{R} : |x| = |17|\} = \{-17, 17\}.$

Qus : 25 Jamia Millia Islamia MCA PYQ

The sets A and B have the same cardinality if and only if there is a ..... correspondence from A to B.

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Two sets have the same cardinality if there exists a one-to-one and onto (bijective) correspondence between them.

Qus : 26 Jamia Millia Islamia MCA PYQ

The relation $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$ is:

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

Reflexive — yes (all $(a,a)$ are present). Symmetric — no, since $(1,2)\in R$ but $(2,1)\notin R$. Transitive — yes, because $(1,2)$ and $(2,3)$ imply $(1,3)$ (which exists).

Qus : 27 Jamia Millia Islamia MCA PYQ

For real numbers $x,y$, define $x\,R\,y$ iff $x-y+\sqrt{2}$ is irrational. Then the relation $R$ is:

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

For reflexivity: $xRx$ means $x-x+\sqrt{2}=\sqrt{2}$ (irrational) ⇒ true. For symmetry: $xRy⇒x-y+\sqrt{2}$ irrational, but $y-x+\sqrt{2}=-(x-y)+\sqrt{2}$ may be rational. Not always true ⇒ not symmetric. For transitivity: fails similarly.

Qus : 28 Jamia Millia Islamia MCA PYQ

If $A$ be a set of cardinality $n$, then number of one-to-one onto functions from set $A$ to $A$ is …

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

Number of bijective (one-one and onto) functions from a set of $n$ elements to itself $= n!$.

Qus : 29 Jamia Millia Islamia MCA PYQ

If $R$ is the largest equivalence relation on a set $A$ and $S$ is any relation on $A$, then:

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

Largest equivalence on $A$ is $A\times A$ (universal relation), so every relation $S$ on $A$ satisfies $S\subseteq A\times A=R$.

Qus : 30 Jamia Millia Islamia MCA PYQ

If $f$ is a function from a finite set $A$ having $10$ elements to a finite set $B$ having $5$ elements, then number of functions from $A$ to $B$ is …

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

Each of $10$ elements in $A$ can be mapped to any of $5$ elements in $B$. Hence total functions $= 5^{10}$.

Qus : 31 Jamia Millia Islamia MCA PYQ

If $A$ and $B$ are two sets, then $(A \cup B)' \cap B$ is equal to …

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

$(A \cup B)'$ means elements not in $A$ or $B$. Hence, $(A \cup B)' \cap B$ contains elements that are both in $B$ and not in $B$, i.e., none. So result is $\phi$.

Qus : 32 Jamia Millia Islamia MCA PYQ

If $A = \{a, b, c\}$ and $B = \{a, b, d, e, f\}$ are two sets, then number of elements in $(A - B) \times (A \cap B)$ is …

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

$A - B = \{c\}$ (only $c$ not in $B$). $A \cap B = \{a, b\}$. Then $(A - B) \times (A \cap B) = \{(c, a), (c, b)\}$ → 2 elements.

Qus : 33 Jamia Millia Islamia MCA PYQ

If $A$ and $B$ are two disjoint sets having $3$ and $5$ elements respectively, then power-set of $A \times (B - A)$ contains … elements.

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

Since $A$ and $B$ are disjoint, $B - A = B$. Then $A \times B$ has $3 \times 5 = 15$ ordered pairs. Power-set of it will have $2^{15}$ elements.

Qus : 34 Jamia Millia Islamia MCA PYQ

Let A={1,2,3} and R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. Then R is

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

$A={1,2,3},; R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}$

Reflexive: $(1,1),(2,2),(3,3)\in R\Rightarrow$ reflexive.

Symmetric: $(1,2)\in R$ but $(2,1)\notin R\Rightarrow$ not symmetric.

Transitive: check the only nontrivial chain: $(1,2)$ and $(2,3)\Rightarrow (1,3)$, which is in $R$. Pairs with $(x,x)$ keep the second pair; $(1,3)$ can only compose with $(3,3)$ giving $(1,3)$; $(2,3)$ with $(3,3)$ gives $(2,3)$. All required compositions are in $R$. $\boxed{\text{$R$ is reflexive and transitive, but not symmetric.}}$

Qus : 35 Jamia Millia Islamia MCA PYQ

On A={1,2,3} let R={(1,2)}. Then R is

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

Not reflexive (missing $(1,1),(2,2),(3,3)$). Not symmetric (since $(2,1)\notin R$). Transitive holds vacuously because there is no pair starting at $2$ to trigger $(1,2)$∘$(2,\cdot)$.

Qus : 36 Jamia Millia Islamia MCA PYQ

A={1,2,3}. Which $f:A\to A$ does not have an inverse?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

A, C, and D are bijections (permutations), hence invertible. In B, both $2$ and $3$ map to $1$ (not one-to-one), so not bijective $\Rightarrow$ no inverse.

Qus : 37 Jamia Millia Islamia MCA PYQ

The maximum number of equivalence relations on the set $A = {1, 2, 3}$ are

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

Number of equivalence relations = Number of partitions of the set = Bell number $B_3 = 5$.

Qus : 38 Jamia Millia Islamia MCA PYQ

A relation $R$ in a set $A$ is called ________, if $(a_1, a_2) \in R$ implies $(a_2, a_1) \in R$, for all $a_1, a_2 \in A$.

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

Qus : 39 Jamia Millia Islamia MCA PYQ

If A={1,2,3} and B={4,5}. What is the number of elements in A×B ?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

The number of elements in a Cartesian product is:

A× B= A× B

Here,

|A| = 3

|B| = 2

So,

3 2 = 6

Qus : 40 Jamia Millia Islamia MCA PYQ

Power set of empty set has exactly _____ subset.

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Power set of $\phi$ = ${\phi}$ has 1 subset.

Qus : 41 Jamia Millia Islamia MCA PYQ

What is the cardinality of the power set of ${0,1,2}$?

1
2
3
4
Go to Discussion

Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

For a set of $n$ elements, power set size $= 2^n = 2^3 = 8.$

Jamia Millia Islamia MCA

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Jamia Millia Islamia MCA

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

Please provide more information - at least 10 characters

Feedback

Info.

Information

Aspire's Library

Jamia Millia Islamia MCA Previous Year Questions (PYQs)

Jamia Millia Islamia MCA Sets And Relations PYQ

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Jamia Millia Islamia MCA

Jamia Millia Islamia MCA

Game Changer NIMCET Test Series 2026

Info.

Confirm Delete

Edit