Jamia Millia Islamia 2023 Previous Year Question Paper and Solution with PDF

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Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia 2023 PYQ

Qus : 1 Jamia Millia Islamia PYQ 2023

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?

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Solution

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

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

Qus : 2 Jamia Millia Islamia PYQ 2023

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

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Solution

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

Qus : 3 Jamia Millia Islamia PYQ 2023

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.

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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 : 4 Jamia Millia Islamia PYQ 2023

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

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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 : 5 Jamia Millia Islamia PYQ 2023

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) ${(1,(a,b)),\ (2,(b,c)),\ (1,(c,a))}$

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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 : 6 Jamia Millia Islamia PYQ 2023

There is a direct flight from Trichy to New Delhi and 2 direct trains. There are 6 trains from Trichy to Chennai and 4 trains from Chennai to Delhi. Also, there are 2 trains from Trichy to Mumbai and 8 flights from Mumbai to New Delhi. In how many ways can a person travel from Trichy to New Delhi?

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Solution

Routes and counts: Direct flight: $1$; direct trains: $2$; via Chennai: $6\times4=24$; via Mumbai: $2\times8=16$. Total $=1+2+24+16=43$.

Qus : 7 Jamia Millia Islamia PYQ 2023

If $P,Q,R$ have truth values $T,T,F$, then the truth values of $\Big(P\to(Q\to R)\Big)\to\Big((P\to Q)\to(P\to R)\Big)$ and $P\to(Q\vee R)$ are:

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Solution

$Q\to R = T\to F = F$, so $P\to(Q\to R)=T\to F=F$. $P\to Q = T\to T = T$, $P\to R = T\to F = F$, hence $(P\to Q)\to(P\to R)=T\to F=F$. Therefore $(F)\to(F)=T$. Also $Q\vee R = T\vee F = T$, so $P\to(Q\vee R)=T\to T=T$.

Qus : 8 Jamia Millia Islamia PYQ 2023

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

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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 : 9 Jamia Millia Islamia PYQ 2023

The floor function $[x]$ is

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Solution

The floor function $f(x) = [x]$ maps every real number to the greatest integer less than or equal to $x$. Different real numbers can have the same floor value, so it is not one-to-one, but for every integer $n \in \mathbb{Z}$, there exists an $x \in \mathbb{R}$ such that $[x]=n$. Hence, it is onto but not one-to-one.

Qus : 10 Jamia Millia Islamia PYQ 2023

The domain of the real-valued function $f(x) = \sqrt{x - 3} + \sqrt{x - 4}$ is the set of all values of $x$ satisfying

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Solution

For $f(x)$ to be real, both radicals must be defined: $x - 3 \ge 0 \quad \text{and} \quad x - 4 \ge 0$ $\Rightarrow x \ge 4$ So the domain is $[4, \infty)$.

Qus : 11 Jamia Millia Islamia PYQ 2023

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?

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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 : 12 Jamia Millia Islamia PYQ 2023

Find the value of $ \dfrac{1}{\sin 10^\circ} - \dfrac{\sqrt{3}}{\cos 10^\circ} = ? $

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Solution

For this expression, $\dfrac{1}{\sin 10^\circ} - \dfrac{\sqrt{3}}{\cos 10^\circ} = \dfrac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}.$ Now, $\cos 10^\circ - \sqrt{3}\sin 10^\circ = 2\cos(60^\circ + 10^\circ) = 2\cos 70^\circ.$ Also, $\sin 10^\circ \cos 10^\circ = \dfrac{1}{2}\sin 20^\circ.$ Hence, $\dfrac{2\cos70^\circ}{\frac{1}{2}\sin20^\circ} = \dfrac{2\sin20^\circ}{\sin20^\circ} = 2.$

Qus : 13 Jamia Millia Islamia PYQ 2023

The value of $\sin\dfrac{\pi}{16} \sin\dfrac{3\pi}{16} \sin\dfrac{5\pi}{16} \sin\dfrac{7\pi}{16}$ is —

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Solution

We know that $\sin x \sin\!\left(\dfrac{\pi}{4} - x\right) \sin\!\left(\dfrac{\pi}{4} + x\right) = \dfrac{1}{4}\sin(4x).$ Let $x = \dfrac{\pi}{16}$. Then, $\sin\dfrac{\pi}{16} \sin\dfrac{3\pi}{16} \sin\dfrac{5\pi}{16} \sin\dfrac{7\pi}{16} = \dfrac{\sqrt{2}}{32}.$

Qus : 14 Jamia Millia Islamia PYQ 2023

Number of unimodular complex numbers which satisfy the locus $\arg\!\left(\dfrac{z - 1}{z + i}\right) = \dfrac{\pi}{2}$ is —

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Solution

Let $z = x + iy$ and $|z| = 1$. The given condition means $(z - 1)$ is perpendicular to $(z + i)$. \[ (x - 1, y) \cdot (x, y + 1) = 0 \Rightarrow x^2 - x + y^2 + y = 0 \] Since $x^2 + y^2 = 1$ (unimodular), \[ 1 - x + y = 0 \Rightarrow y = x - 1. \] Substitute in $x^2 + y^2 = 1$: \[ x^2 + (x - 1)^2 = 1 \Rightarrow 2x^2 - 2x = 0 \Rightarrow x(x - 1) = 0. \] Hence $x = 0$ or $x = 1$. For $x = 0$, $z = -i$ (denominator $z+i=0$). For $x = 1$, $z = 1$ (numerator $z-1=0$). Both make $\arg$ undefined. Therefore, no unimodular complex number satisfies the condition.

Qus : 15 Jamia Millia Islamia PYQ 2023

The values of the parameter $a$ such that the roots $\alpha, \beta$ of $2x^2 + 6x + a = 0$ satisfy the inequality $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} < 2$ are —

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Solution

For the quadratic $2x^2 + 6x + a = 0$: $\alpha + \beta = -\dfrac{6}{2} = -3$, and $\alpha\beta = \dfrac{a}{2}$. Now, $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \dfrac{9 - 2\cdot \frac{a}{2}}{\frac{a}{2}} = \dfrac{18 - 2a}{a}.$ Given $\dfrac{18 - 2a}{a} < 2 \Rightarrow 18 - 2a < 2a \Rightarrow 18 < 4a \Rightarrow a > \dfrac{9}{2}.$ For real roots, discriminant $36 - 8a \ge 0 \Rightarrow a \le \dfrac{9}{2}.$ Hence, no real $a$ satisfies both.

Qus : 16 Jamia Millia Islamia PYQ 2023

The 120 permutations of “MAHES” are arranged in dictionary order, as if each were an ordinary 5-letter word. The last letter of the $86^{th}$ word in the list is —

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Solution

Total letters = 5 distinct → $5! = 120$ words. Fixing first letter in alphabetical order: A, E, H, M, S. Each block = $4! = 24$ words. After A (24), E (24), H (24) → total = 72. So the $86^{th}$ word lies in M-block (73–96). Within M-block, order remaining letters: A, E, H, S. Each gives $3! = 6$ words. $73$–$78$: MA… $79$–$84$: ME… $85$–$90$: MH… Hence, the $86^{th}$ word lies in MH-series. So the last letter = H.

Qus : 17 Jamia Millia Islamia PYQ 2023

A person writes letters to 6 friends and addresses the corresponding envelopes. Let $x$ be the number of ways so that **at least 2 letters** are in wrong envelopes and $y$ be the number of ways so that **all letters** are in wrong envelopes. Then find $x - y$.

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Solution

Total ways to arrange 6 letters = $6! = 720$. Let $D_6$ = number of derangements (no letter in correct envelope): $D_6 = 6! \left(1 - \dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} - \dfrac{1}{5!} + \dfrac{1}{6!}\right) = 265.$ Now, $x =$ number of ways with at least 2 letters wrong $= 6! - [\text{exactly 0 or 1 correct letters}]$ For exactly 0 correct = $D_6 = 265$. For exactly 1 correct: choose 1 correct letter $(6C1)$ × derange remaining 5 $(D_5)$. $D_5 = 44$, so ways = $6 \times 44 = 264.$ Hence, $x = 720 - (1 + 264) = 455.$ $\therefore x - y = 455 - 265 = 190.$

Qus : 18 Jamia Millia Islamia PYQ 2023

In how many ways can the following diagram be colored, subject to: (i) Each of the smaller triangles is to be painted with one of three colors — red, blue, or green. (ii) No two adjacent regions have the same color.

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Solution

The figure has 4 small triangles (1 top, 3 at the base). Let top be colored in 3 ways. The 3 base triangles are adjacent, so they must all be different and also different from the top if adjacent. ⇒ Choose color for top: 3 ways ⇒ Choose 3 distinct colors for bottom row (arranged 3! = 6 ways). ⇒ But bottom middle and top are adjacent → exclude same color combinations. Valid colorings = $3 \times (3! - 3! / 3) = 3 \times 8 = 24.$

Qus : 19 Jamia Millia Islamia PYQ 2023

The tens digit of $1! + 2! + 3! + 4! + \dots + 49!$ is —

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Solution

From $10!$ onward, every factorial ends with at least two zeros. So, only the sum of first nine factorials affects the tens digit. $1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 = 409113.$ Tens digit = $\boxed{1}.$

Qus : 20 Jamia Millia Islamia PYQ 2023

The middle term in the expansion of $\left(1 + \dfrac{1}{x^2}\right)\!\left(1 + x^2\right)^n$ is —

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Solution

Expand $\left(1 + x^2\right)^n = \sum_{k=0}^{n} {}^{n}C_{k}x^{2k}.$ Multiply by $\left(1 + \dfrac{1}{x^2}\right)$: $= \sum_{k=0}^{n} {}^{n}C_{k}x^{2k} + \sum_{k=0}^{n} {}^{n}C_{k}x^{2k-2}.$ To find middle term → powers of $x$ that are equal when $2k = 2n - (2k-2)$. Simplifying gives $k = n.$ So middle term = ${}^{2n}C_{n}.$

Qus : 21 Jamia Millia Islamia PYQ 2023

The sum of the infinite series $\dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dfrac{2^8}{8!} + \dots$ is equal to —

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Solution

We know that $e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots$ Let $S = \dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dots$ Write using even terms of $e^x$: $e^2 = 1 + 2 + \dfrac{2^2}{2!} + \dfrac{2^3}{3!} + \dfrac{2^4}{4!} + \dots$ and $e^{-2} = 1 - 2 + \dfrac{2^2}{2!} - \dfrac{2^3}{3!} + \dfrac{2^4}{4!} - \dots$ Adding both, $e^2 + e^{-2} = 2\left(1 + \dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dots\right)$ So, $S = \dfrac{1}{2}\left(e^2 + e^{-2} - 2\right) = \dfrac{1}{2}\left(\dfrac{e^4 + 1 - 2e^2}{e^2}\right) = \dfrac{(e^2 - 1)^2}{2e^2}.$

Qus : 22 Jamia Millia Islamia PYQ 2023

If ‘a’ is the arithmetic mean of ‘b’ and ‘c’, and $G_1$ and $G_2$ are the two geometric means between them, then $G_1^3 + G_2^3$ is equal to —

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Solution

Given $a = \dfrac{b + c}{2}$ and $G_1, G_2$ are geometric means between $b$ and $c$. So $b, G_1, G_2, c$ are in G.P. Let common ratio = $r$. Then $G_1 = br$ and $G_2 = br^2$, $c = br^3$. Hence $a = \dfrac{b + c}{2} = \dfrac{b + br^3}{2} = \dfrac{b(1 + r^3)}{2}.$ Now, $G_1^3 + G_2^3 = b^3(r^3 + r^6) = b^3r^3(1 + r^3).$ But $c = br^3$ and $(1 + r^3) = \dfrac{2a}{b}$. So, $G_1^3 + G_2^3 = b^3r^3 \times \dfrac{2a}{b} = 2ab^2r^3 = 2abc.$

Qus : 23 Jamia Millia Islamia PYQ 2023

For $x \in \mathbb{R}$, find $\displaystyle \lim_{x \to \infty} \left(\dfrac{x - 3}{x + 2}\right)^x = ?$

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Solution

Let $L = \left(\dfrac{x - 3}{x + 2}\right)^x = \left(1 - \dfrac{5}{x + 2}\right)^x.$ Take $\ln L = x \ln\!\left(1 - \dfrac{5}{x + 2}\right).$ For large $x$, use $\ln(1 - y) \approx -y$. So, $\ln L \approx x \left(-\dfrac{5}{x + 2}\right) \to -5.$ Hence, $L = e^{-5}.$

Qus : 24 Jamia Millia Islamia PYQ 2023

The contrapositive of $p \to (\neg q \to \neg r)$ is —

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Solution

Given statement: $p \to (\neg q \to \neg r)$ Contrapositive = $\neg(\neg q \to \neg r) \to \neg p$ Now, $\neg(\neg q \to \neg r) \equiv (\neg q \land r)$ Hence, contrapositive = $(\neg q \land r) \to \neg p.$

Qus : 25 Jamia Millia Islamia PYQ 2023

The mean of 100 observations is 50 and their standard deviation is 5. The sum of squares of all observations is —

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Solution

Let $\bar{x} = 50$, $\sigma = 5$, and $n = 100.$ We know, $\sigma^2 = \dfrac{\sum x^2}{n} - \bar{x}^2$ $\Rightarrow \sum x^2 = n(\sigma^2 + \bar{x}^2)$ $= 100(5^2 + 50^2) = 100(25 + 2500) = 100 \times 2525 = 2,52,500.$

Qus : 26 Jamia Millia Islamia PYQ 2023

A card is drawn from a pack of 52 cards. A gambler bets that it is a spade or an ace. What are the odds against his winning this bet?

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Solution

Total cards = 52 Favorable cards = 13 spades + 3 other aces = 16 Odds **against** = $\dfrac{\text{unfavorable}}{\text{favorable}} = \dfrac{52 - 16}{16} = \dfrac{36}{16} = 9:4.$

Qus : 27 Jamia Millia Islamia PYQ 2023

If $Z$ is an idempotent matrix, then $(I + Z)^n$ is —

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Solution

Since $Z$ is idempotent, $Z^2 = Z.$ $(I + Z)^2 = I + 2Z + Z^2 = I + 3Z$ $(I + Z)^3 = (I + Z)(I + 3Z) = I + 4Z$ Hence, by induction: $(I + Z)^n = I + (2^n - 1)Z.$

Qus : 28 Jamia Millia Islamia PYQ 2023

If $A^2 - A = 3I$, then $A^{-1}$ is —

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Solution

Given $A^2 - A = 3I$ $\Rightarrow A(A - I) = 3I$ Multiply both sides by $A^{-1}$: $(A - I) = 3A^{-1}$ $\Rightarrow A^{-1} = \dfrac{1}{3}(A - I).$

Qus : 29 Jamia Millia Islamia PYQ 2023

The system of linear equations is: $a + 2b + 3c = 7$ $2a + 4b + c = 12$ $3a + 6b + 4c = 20$

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Solution

We can write the augmented matrix as: $\begin{bmatrix} 1 & 2 & 3 & | & 7 \\ 2 & 4 & 1 & | & 12 \\ 3 & 6 & 4 & | & 20 \end{bmatrix}$ Perform the following operations: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$ $\Rightarrow \begin{bmatrix} 1 & 2 & 3 & | & 7 \\ 0 & 0 & -5 & | & -2 \\ 0 & 0 & -5 & | & -1 \end{bmatrix}$ Now subtract $R_3 - R_2$: $\Rightarrow \begin{bmatrix} 0 & 0 & 0 & | & 1 \end{bmatrix}$ This represents an inconsistent equation $0 = 1$. Hence, the system **has no solution.**

Qus : 30 Jamia Millia Islamia PYQ 2023

$Q30.$ If the rank of the matrix \[ \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \] is $2$, then find the correct condition.

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Solution

For a diagonal matrix, the rank equals the number of non-zero diagonal elements. If the rank is $2$, exactly two of $a, b, c$ must be non-zero and one must be zero. Thus, the possible condition is $ab \neq 0,\; c = 0$.

Qus : 31 Jamia Millia Islamia PYQ 2023

The general solution of differential equation $(\tan^{-1}y - x)dy = (1 + y^2)dx$ is:

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Solution

Let $x$ be a function of $y.$ $\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1 + y^2}$ $\Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}$ This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y),$ where $P(y) = \frac{1}{1 + y^2}.$ Integrating factor, $\mu = e^{\int P(y) dy} = e^{\tan^{-1}y}.$ $\Rightarrow \frac{d}{dy}(xe^{\tan^{-1}y}) = e^{\tan^{-1}y} \frac{\tan^{-1}y}{1 + y^2}$ Let $t = \tan^{-1}y \Rightarrow dt = \frac{dy}{1 + y^2}$ $xe^{t} = \int t e^{t} dt = e^{t}(t - 1) + C$ $\Rightarrow x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$

Qus : 32 Jamia Millia Islamia PYQ 2023

A pair of fair dice is thrown independently 3 times. The probability of getting a score of exactly 9 twice is

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Solution

$P(\text{sum} = 9) = \dfrac{4}{36} = \dfrac{1}{9}$ $\Rightarrow P(\text{exactly 2 times}) = \binom{3}{2}\left(\dfrac{1}{9}\right)^2\left(\dfrac{8}{9}\right) = 3 \times \dfrac{1}{81} \times \dfrac{8}{9} = \dfrac{8}{243}$

Qus : 33 Jamia Millia Islamia PYQ 2023

Solution of the differential equation $\displaystyle \frac{dx}{dy}-\frac{x\log x}{1+\log x}=\frac{e^{y}}{1+\log x},\ \text{ if } y(1)=0,$ is –

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Solution

Treat $x$ as a function of $y$ and set $u=1+\log x\ (\Rightarrow x=e^{u-1},\ \frac{dx}{dy}=x\frac{du}{dy})$. The DE becomes $u\frac{du}{dy}=u-1+e^{\,1+y-u}.$ This transforms to a first-order non-linear equation in $u(y)$ whose implicit integral does **not** reduce to any of the listed closed forms (A)–(C). With the initial condition $y(1)=0$ (i.e., $u=1$ at $y=0$), the solution is an implicit relation not matching (A)–(C).

Qus : 34 Jamia Millia Islamia PYQ 2023

Every gram of wheat provides $0.1$ gram of proteins and $0.25$ gram of carbohydrates. The corresponding values of rice are $0.05$ gram and $0.5$ gram respectively. The minimum daily requirements of proteins and carbohydrates for an average child are $50$ gram and $200$ gram respectively. Then in what quantities wheat and rice be mixed in daily diet to provide minimum daily requirement of proteins and carbohydrates at minimum cost?

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Solution

Let $x =$ grams of wheat and $y =$ grams of rice. Protein constraint: $0.1x + 0.05y = 50$ Carbohydrate constraint: $0.25x + 0.5y = 200$ Simplify: $2x + y = 1000$ $x + 2y = 800$ Solving the two equations gives: $x = 400,\ y = 200.$ Hence, wheat = $400$ g, rice = $200$ g.

Qus : 35 Jamia Millia Islamia PYQ 2023

$Z = 7x + y$, subject to constraints: $5x + y \ge 5,\ x + y \ge 3,\ x \ge 0,\ y \ge 0.$ Then minimum value of $Z$ occurs at –

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Solution

Solution: Corner points by solving the constraints: $(0,5),\ (3,0),\ \left(\frac{1}{2},\frac{5}{2}\right)$ Now compute $Z = 7x + y$: At $(0,5): Z = 5$ At $(3,0): Z = 21$ At $\left(\frac{1}{2},\frac{5}{2}\right): Z = \frac{7}{2} + \frac{5}{2} = 6$ Minimum $Z = 5$ at $(0,5)$

Qus : 36 Jamia Millia Islamia PYQ 2023

The point of inflection for $f(x) = 3x^4 - 4x^3$ are –

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Solution

Solution: $f'(x) = 12x^3 - 12x^2$ $f''(x) = 36x^2 - 24x = 12x(3x - 2)$ Set $f''(x) = 0 \Rightarrow x = 0,\ x = \dfrac{2}{3}$ Hence, points of inflection are $x = 0$ and $x = \dfrac{2}{3}.$

Qus : 37 Jamia Millia Islamia PYQ 2023

$\displaystyle \int_{0}^{1000} e^{\,x-[x]}\,dx$ is –

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Solution

Solution: For $x \in [n,n+1)$, we have $[x]=n$. $\displaystyle \int_n^{n+1} e^{\,x-[x]}dx = e^{-n}\!\int_n^{n+1} e^{x}dx = e^{-n}(e^{n+1}-e^{n})=e-1.$ The interval $[0,1000)$ has $1000$ such unit pieces, so the total integral is $1000(e-1)$.

Qus : 38 Jamia Millia Islamia PYQ 2023

Let the equation of a curve passing through $(0,1)$ be $y=\displaystyle\int x^{2}e^{x^{3}}\,dx$. If the curve is written as $x=f(y)$, then $f(y)$ is –

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Solution

Solution: Let $t=x^3 \Rightarrow dt=3x^2dx$, so $y=\dfrac{1}{3}e^{x^3}+C$. Using $(0,1)$: $1=\dfrac{1}{3}+C \Rightarrow C=\dfrac{2}{3}$. Hence $3y-2=e^{x^3} \Rightarrow x^3=\log_e(3y-2)$, so $x=\sqrt[3]{\log_e(3y-2)}$.

Qus : 39 Jamia Millia Islamia PYQ 2023

The value of $\displaystyle \int_{0}^{\pi/2}\sin^{4}x\,\cos^{4}x\,dx$ is –

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Solution

Solution: $\sin^{4}x\cos^{4}x=\big(\sin^{2}x\cos^{2}x\big)^2 =\left(\dfrac{\sin 2x}{2}\right)^{4} =\dfrac{1}{16}\sin^{4}2x.$ Thus $J=\displaystyle\int_{0}^{\pi/2}\sin^{4}x\cos^{4}x\,dx =\dfrac{1}{16}\!\int_{0}^{\pi/2}\!\sin^{4}2x\,dx =\dfrac{1}{32}\!\int_{0}^{\pi}\!\sin^{4}u\,du.$ Using $\int_{0}^{\pi}\sin^{4}u\,du=\dfrac{3\pi}{8}$, we get $J=\dfrac{1}{32}\cdot\dfrac{3\pi}{8}=\dfrac{3\pi}{256}$.

Qus : 40 Jamia Millia Islamia PYQ 2023

If $49^{n}+16n+\lambda$ is divisible by $64$ for all $n\in\mathbb{N}$, then the least negative value of $\lambda$ is –

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Solution

Solution: Work modulo $64$. $49\equiv -15\pmod{64}$ and $49^{1}\equiv49,\ 49^{2}\equiv33,\ 49^{3}\equiv17,\ 49^{4}\equiv1$; hence $49^{n}$ is periodic with period $4$. Also $16n\equiv 0,16,32,48\ (\bmod\ 64)$ for $n\equiv 0,1,2,3$. For each residue class: $n\equiv0$: $1+0+\lambda\equiv0\Rightarrow \lambda\equiv -1$ $n\equiv1$: $49+16+\lambda\equiv1+\lambda\equiv0$ $n\equiv2$: $33+32+\lambda\equiv1+\lambda\equiv0$ $n\equiv3$: $17+48+\lambda\equiv1+\lambda\equiv0$ All give $\lambda\equiv -1\pmod{64}$. The least negative representative is $\boxed{-1}$.

Qus : 41 Jamia Millia Islamia PYQ 2023

A Disk Defragmenter is an example of –

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Solution

Solution: A Disk Defragmenter is a **system utility** that rearranges fragmented data on a disk so that files are stored in contiguous sectors, improving access speed and efficiency. Hence, it belongs to the category of **Utility Programs**.

Qus : 42 Jamia Millia Islamia PYQ 2023

Convert the following decimal number to a number system with radix 3: $(106)_{10} = (?)_{3}$

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Solution

Solution: Convert $106$ to base $3$: $106 \div 3 = 35$ remainder $1$ $35 \div 3 = 11$ remainder $2$ $11 \div 3 = 3$ remainder $2$ $3 \div 3 = 1$ remainder $0$ $1 \div 3 = 0$ remainder $1$ Reading remainders from bottom to top: $(106)_{10} = (10221)_{3}$

Qus : 43 Jamia Millia Islamia PYQ 2023

Which of the following is an encoding scheme created for Indian scripts?

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Solution

Solution: **ISCII (Indian Standard Code for Information Interchange)** is the encoding scheme developed specifically for representing Indian scripts such as Devanagari, Tamil, Bengali, etc.

Qus : 44 Jamia Millia Islamia PYQ 2023

Convert $(10025)_{10}$ to hexadecimal $(?)_{16}$

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Solution

Solution: Convert $10025$ to base $16$: $10025 \div 16 = 626$ remainder $9$ $626 \div 16 = 39$ remainder $2$ $39 \div 16 = 2$ remainder $7$ $2 \div 16 = 0$ remainder $2$ Reading remainders bottom to top: $(10025)_{10} = (2729)_{16}$ Since none of the given options match,

Qus : 45 Jamia Millia Islamia PYQ 2023

Consider the following C language declarations & statements. Which statement is erroneous? float f1 = 9.9; float f2 = 66; const float *ptrF1; float * const ptrF2 = &f2; ptrF1 = &f1; ptrF2++; ptrF1++;

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Solution

- `const float *ptrF1;` → pointer to constant float, can move pointer, not modify value. - `float * const ptrF2 = &f2;` → constant pointer to float, cannot change address, but can modify value. - Statement `ptrF2++` tries to move a **constant pointer**, which is **not allowed**. Hence, `ptrF2++` is **erroneous**.

Qus : 46 Jamia Millia Islamia PYQ 2023

What will be output of following statements? int n1 = 3, n2 = 6, a; printf("(n1 ^ n2) + (a ^ a) = %d", (n1 ^ n2) + (a ^ a));

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Solution

Solution: Operator `^` is **bitwise XOR**. $n1 = 3 \Rightarrow 011_2$ $n2 = 6 \Rightarrow 110_2$ $n1 \oplus n2 = 101_2 = 5$ Variable `a` is declared but **not initialized**. Using an uninitialized variable in expression `(a ^ a)` causes **undefined behavior**. Hence, the program compiles but produces a **run-time error or unpredictable result**.

Qus : 47 Jamia Millia Islamia PYQ 2023

What will be output of following statements? char ch; ch = 130; printf("\nvalue of ch=%d", ch);

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Solution

Solution: In C, the range of `char` is from $-128$ to $127$. When we assign $130$ to a `char`, it overflows: $130 - 256 = -126$ Thus the value stored in `ch` is $-126$. Output: value of ch = -126

Qus : 48 Jamia Millia Islamia PYQ 2023

What is the output of the following C code segment?

int i;

for(i = 0; i <= 2; i++)

{

switch(i)

{

case 1: printf("%2d", i);

case 2: printf("%2d", i); continue;

default: printf("%2d", i);

}

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Solution

Solution:

Let’s analyze iteration by iteration —

**Iteration 1:** `i = 0`

→ `switch(0)` → goes to `default:` → prints `0`.

**Iteration 2:** `i = 1`

→ executes `case 1:` → prints `1`

→ no `break`, so it falls through to `case 2:` → prints `1`

→ encounters `continue;` → skips remaining statements in the `for` loop body and proceeds to next iteration.

**Iteration 3:** `i = 2`

→ executes `case 2:` → prints `2`

→ `continue;` moves control to the next iteration, but loop ends because `i <= 2` condition fails.

Final output:

`0 1 1 2`

Qus : 49 Jamia Millia Islamia PYQ 2023

What is the output of the following C program?

int main()

{

char ch = 'A';

int x = 97;

int y = sizeof(++x);

printf("\nx is %d", x);

while (ch <= 'F')

{

switch (ch)

{

case 'A':

case 'B':

case 'C':

case 'D': ch++; break;

case 'E':

case 'F': ch++;

}

putchar(ch);

}

return 0;

}

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Solution

Solution:

1️⃣ `int y = sizeof(++x);`

→ `sizeof` is a compile-time operator, so `++x` is **not evaluated**.

Hence, `x` remains unchanged (`x = 97`).

2️⃣ `printf("\nx is %d", x);`

→ prints: `x is 97`

3️⃣ Loop execution:

|------|----------------------|--------------------|------------------|

| 'A' | falls to case 'D' → ch++ → break | prints `'B'` | 'B' |

| 'B' | falls to case 'D' → ch++ → break | prints `'C'` | 'C' |

| 'C' | falls to case 'D' → ch++ → break | prints `'D'` | 'D' |

| 'D' | case 'D' → ch++ → break | prints `'E'` | 'E' |

| 'E' | case 'E' → fall through to case 'F' → ch++ | prints `'F'` | 'F' |

Output sequence = `x is 97 BCDEFG`

Qus : 50 Jamia Millia Islamia PYQ 2023

What is the output of following C program?

void e(int x)

{

if (x > 0)

{

e(- -x);

printf("%2d", x);

e(- -x);

}

int main()

{

e(3);

return 0;

}

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Solution

Solution:

In C, the tokens `- -x` are parsed as two unary minuses with a space, i.e. $-(-x)=x$.

So both recursive calls are `e(x)`, not `e(--x)`.

For `x=3`, the function calls itself **with the same positive value** forever:

`e(3) → e(3) → e(3) → ...` and never reaches a base case.

This causes infinite recursion (stack overflow) at run time.

Qus : 51 Jamia Millia Islamia PYQ 2023

How will this call assign values when the user input is `29␠w` (a space between 29 and w)? scanf("%i%c", &i, &c);

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Solution

Solution: `%i` reads an integer and **skips leading whitespace**, so it consumes `29`. `%c` reads the **next character exactly as-is** (does **not** skip whitespace). The next character after `29` is the space. Therefore: $i=29$, $c=$ space `' '`.

Qus : 52 Jamia Millia Islamia PYQ 2023

Minimum & Maximum range of values for ‘float’ data type in C is:

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Solution

In standard C (IEEE-754 single precision), the approximate range of a `float` is from $1.17 \times 10^{-37}$ to $3.4 \times 10^{38}$.

Qus : 53 Jamia Millia Islamia PYQ 2023

Which out of these is NOT valid for C language?

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Solution

Solution: When a local variable has the same name as a global variable, the **local variable overrides** (takes precedence over) the global one inside its block. Thus (C) is **invalid** for C.

Qus : 54 Jamia Millia Islamia PYQ 2023

C was originally developed in the 1970s by Dennis Ritchie at Bell Telephone Laboratories, Inc., which is an outgrowth of two earlier languages, called:

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Solution

Solution: The C language was developed as an evolution of **BCPL** and its successor **B**. C inherited features from both.

Qus : 55 Jamia Millia Islamia PYQ 2023

Multiply 1101 by 1011 (binary multiplication):

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Solution

$1101_2 = 13_{10}$ $1011_2 = 11_{10}$ $13 \times 11 = 143_{10}$ Convert $143_{10}$ to binary: $143_{10} = 10001111_2$

Qus : 56 Jamia Millia Islamia PYQ 2023

Subtract $(2761)_8$ from $(6357)_8$ :

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Solution

Solution: Convert to decimal: $(6357)_8 = 6×512 + 3×64 + 5×8 + 7 = 3327$ $(2761)_8 = 2×512 + 7×64 + 6×8 + 1 = 1505$ Now subtract: $3327 - 1505 = 1822$ Convert $1822$ to octal: $1822 ÷ 8 = 227$ R6 $227 ÷ 8 = 28$ R3 $28 ÷ 8 = 3$ R4 $3 ÷ 8 = 0$ R3 $\Rightarrow (1822)_{10} = (3436)_8$ None of the given options matches exactly, but the **closest correct result** (likely typo) is $(3376)_8$.

Qus : 57 Jamia Millia Islamia PYQ 2023

Which out of these is NOT correct pairing?

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Solution

Solution: - BCD (Binary Coded Decimal) uses **4 bits**, not 7 bits. - EBCDIC (Extended Binary Coded Decimal Interchange Code) uses **8 bits**. - ASCII (American Standard Code for Information Interchange) is **7-bit**, extended ASCII is 8-bit. Hence, “BCD – 7 bit” is incorrect.

Qus : 58 Jamia Millia Islamia PYQ 2023

Which out of these does NOT support VoIP?

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Solution

Solution: VoIP = Voice over Internet Protocol — used for calling over the internet. All three — WhatsApp, FaceTime, and IMO — support VoIP. So none of these options is correct.

Qus : 59 Jamia Millia Islamia PYQ 2023

By using .............. addition or subtraction of signed numbers are performed.

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Solution

Solution: Signed number arithmetic (addition/subtraction) is performed using **2’s complement representation** in digital systems.

Qus : 60 Jamia Millia Islamia PYQ 2023

Which statement out of these is NOT correct about multiprocessor systems?

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Solution

Solution: Statement (B) is **incorrect** — tightly coupled systems are **not** necessarily more energy-efficient than clusters; clusters can be optimized for energy efficiency.

Qus : 61 Jamia Millia Islamia PYQ 2023

Which file format is NOT suitable for SD card in Android phone?

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Solution

Android devices natively support **FAT32** and **exFAT** file systems. However, **NTFS** is not fully supported without third-party drivers or root access.

Qus : 62 Jamia Millia Islamia PYQ 2023

Which out of these is NOT a type of ROM?

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Solution

Masked ROM, EEPROM, and Flash BIOS are all types of **Read Only Memory (ROM)**. - **Flash drive** is a type of **secondary storage (non-volatile)** device, not ROM.

Qus : 63 Jamia Millia Islamia PYQ 2023

Select the next to smallest memory size from the given options:

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Solution

Solution: Memory size order: Kilobyte < Megabyte < Gigabyte < Terabyte < **Petabyte** < **Exabyte** < **Zettabyte** < **Yottabyte** Next to smallest → after Petabyte → **Exabyte**

Qus : 64 Jamia Millia Islamia PYQ 2023

When you simplify algebraically the given expression to a minimum sum of products, how many terms do you get? (A + B' + C + E') (A + B' + D' + E) (B' + C' + D' + E')

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Solution

Solution: Let’s analyze: We can simplify using Boolean algebra rules. After simplification (by K-map or expansion reduction), the minimum sum of products results in **4 terms**.

Qus : 65 Jamia Millia Islamia PYQ 2023

The simplified form of the given Boolean expression is: A'CD'E + A'B'D' + ABCE + ABD

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Solution

Solution: Simplify step-by-step: $A'CD'E + A'B'D' + ABCE + ABD$ → Combine using absorption and distributive laws. $A'B'D' + ABD + ACD'E$ Hence, the final simplified expression is:

Qus : 66 Jamia Millia Islamia PYQ 2023

Example of 5th generation language is:

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Solution

Solution: 5th Generation Languages (5GL) are logic-based, AI-oriented languages that use **constraints** and **declarative problem-solving**, not step-by-step coding. Examples: Prolog, Mercury, OPS5, etc. - ASP and JavaScript → 4th generation or scripting languages - SQL → 4th generation language Hence, none of these belong to 5GL.

Qus : 67 Jamia Millia Islamia PYQ 2023

The output of following C language statement is: printf("\nhello" + 3);

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Solution

Solution: String: "\nhello" Character positions: 0:'\n' 1:'h' 2:'e' 3:'l' 4:'l' 5:'o' → "\nhello" + 3 points to index 3, i.e., `"llo"` Hence it prints “llo”.

Qus : 68 Jamia Millia Islamia PYQ 2023

Give output of following C code:

int count(unsigned x)

{

int b;

for (b = 0; x != 0; x >>= 1)

if (x & 1)

b++;

return b;

}

int main()

{

unsigned int a = 3;

printf("%d", count(a));

return 0;

}

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Solution

```ruby Solution: The function `count()` counts number of 1-bits in the binary representation of `x`. For `a = 3` Binary of 3 = `11` → Number of 1 bits = 2 Hence output = 2.

Qus : 69 Jamia Millia Islamia PYQ 2023

What is the data type of the following expression:

expr₁ ? expr₂ : expr₃

if expr₁ is of type float & expr₂ is of type int.

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Solution

Solution: In the **ternary operator** `(expr₁ ? expr₂ : expr₃)`, the **data type of result** is the **common type** of `expr₂` and `expr₃`. Here, - `expr₁` → condition (float type, irrelevant for result type) - `expr₂` → int - `expr₃` → (not specified but implied same as expr₂ type logic) When int and float are combined → **result type = float** (implicit type conversion).

Qus : 70 Jamia Millia Islamia PYQ 2023

Which operator out of these has got the highest precedence?

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Solution

Solution: Operator precedence (from high to low in C): `[]` > `<` > `?:` > `,` Therefore, the highest precedence among these is **array subscript `[ ]`**.

Qus : 71 Jamia Millia Islamia PYQ 2023

Which operator out of these has left to right associativity?

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Solution

Operator associativity in C: - Most unary operators (!, ++, --, sizeof, etc.) → **Right to Left** - Conditional (`?:`) → **Right to Left** - Comma operator (`,`) → **Left to Right** Hence, only the **comma operator** has **left-to-right** associativity.

Qus : 72 Jamia Millia Islamia PYQ 2023

Consider the following code segment:

if (n > 0)

for (i = 0; i < 3; i++)

if (array[i] > 0)

printf("%d\n", array[i]);

else

printf("\n n is negative\n");

Here, ‘else’ is paired with which ‘if’?

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Solution

Solution: In C, **“else” is always paired with the nearest unmatched “if”** (no braces rule). Here: - The inner `if (array[i] > 0)` is the **nearest unmatched `if`** before `else`. So, the `else` is paired with the **second (inner)** `if`.

Qus : 73 Jamia Millia Islamia PYQ 2023

For this kind of declaration of main() function in a C program ‘copy.C’:

int main(int argc, char *argv[]) { }

and this call of main function at command prompt:

C:\tc\bin>copy file1 file2 file3

What will be the value passed in parameter argc?

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Solution

Solution: The **argc (argument count)** includes the program name itself. So the arguments are: 1️⃣ "copy" (program name) 2️⃣ "file1" 3️⃣ "file2" 4️⃣ "file3" Thus, total arguments = 4.

Qus : 74 Jamia Millia Islamia PYQ 2023

What is the correct file mode that opens preexisting file in read and write mode?

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

Solution

Solution: - `"r+b"` or `"rb+"` → open **existing file** for both reading and writing (no truncation). - `"w+b"` or `"wb+"` → open file for reading and writing but **creates/truncates** file. - `"ab"` → append binary mode. Hence, the correct mode for **existing file in read-write** is `"r+b"`.

Qus : 75 Jamia Millia Islamia PYQ 2023

Which C expression correctly represents this statement: “It decrements pointer p before fetching the character that p points to.”

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

Solution

Solution: - `p--` → post-decrement (use then decrement) - `--p` → pre-decrement (decrement then use) Here, we need to **decrement p first**, then fetch the value it points to. Hence, correct expression: `*--p`

Qus : 76 Jamia Millia Islamia PYQ 2023

How many times this statement will execute: for (; *s == *t && *t != '\0'; s++, t++) if both character pointers ‘s’ and ‘t’ point to the same string “abc”.

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

Solution

Solution: String = "abc" → characters: a, b, c, '\0' Loop condition: - Iteration 1: *s = 'a', *t = 'a' → true - Iteration 2: *s = 'b', *t = 'b' → true - Iteration 3: *s = 'c', *t = 'c' → true - Iteration 4: *s = '\0', *t = '\0' → fails because *t != '\0' → false Hence, loop executes **3 times**.

Qus : 77 Jamia Millia Islamia PYQ 2023

Which out of these statements is NOT true:

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

Solution

Solution: All three statements (A), (B), and (C) are correct according to C language behavior. Hence, there is **no incorrect statement**.

Qus : 78 Jamia Millia Islamia PYQ 2023

Which out of these is NOT the keyword C99 has added in addition to 32 keywords defined by ANSI C?

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

Solution

Solution: C99 added these new keywords: - `_Bool` - `inline` - `restrict` - `_Complex` - `_Imaginary` `register` is an **old ANSI C keyword**, not added in C99.

Qus : 79 Jamia Millia Islamia PYQ 2023

Which out of these is NOT a valid C version?

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

Solution

Solution: Valid C versions are: - C89 (ANSI standard, 1989) - C90 (ISO standard, 1990) - C99, C11, C17, and C23 There is **no version called C2007 or CIX**.

Qus : 80 Jamia Millia Islamia PYQ 2023

Who developed World Wide Web version 3 which is known as “Semantic Web”?

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

Solution

Solution: The **Semantic Web (Web 3.0)** concept was introduced and developed by **Tim Berners-Lee**, the original inventor of the World Wide Web.

Qus : 81 Jamia Millia Islamia PYQ 2023

His speech was full of affectation.

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

Solution

Solution: “Affectation” means **artificial behavior or pretence** intended to impress others.

Qus : 82 Jamia Millia Islamia PYQ 2023

Reading of poetry is not *congenial* to her taste.

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

Solution

Solution: “Congenial” means **suited, agreeable, or pleasing** to one’s taste or nature. So, “not congenial” means **not suited**.

Qus : 83 Jamia Millia Islamia PYQ 2023

Select the phrase which means nearly the same as this idiomatic phrase: “General act of forgiveness on a national occasion.”

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

Solution

Solution: “Amnesty” means a **general pardon or forgiveness**, usually granted by a government.

Qus : 84 Jamia Millia Islamia PYQ 2023

Pick the antonym of *vacillating.*

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

Solution

“Vacillating” means **indecisive or wavering**. Its opposite (antonym) is **resolute**, meaning **firm and determined**.

Qus : 85 Jamia Millia Islamia PYQ 2023

Pick the synonym of “patronage”

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

Solution

Solution: “Patronage” means **support, sponsorship, or backing**, especially in the sense of helping an individual, organization, or activity. Hence, synonym is **support**.

Qus : 86 Jamia Millia Islamia PYQ 2023

Select the closest meaning of idiom “stick to one’s guns”.

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

Solution

Solution: Idiom “stick to one’s guns” means **to maintain one’s position or opinion even in the face of opposition**.

Qus : 87 Jamia Millia Islamia PYQ 2023

Did you think you …………….. (see) me somewhere before?

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

Solution

Solution: Sentence is in **past tense** (“Did you think”), so the action that happened **before** that must be in **past perfect tense**. Correct form: “Did you think you **had seen** me somewhere before?”

Qus : 88 Jamia Millia Islamia PYQ 2023

Having placed …………….. proposals before you, I now …………….. your decision.

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

Solution

Solution: Correct phrasing: “Having placed **many** proposals before you, I now **am waiting** your decision.”

Qus : 89 Jamia Millia Islamia PYQ 2023

Choose the correct preposition: He was of a charitable disposition, but did not like a number of his relatives trying to live …………….. him without trying to earn their living.

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

Solution

Solution: Phrase “to live off someone” means **to depend on someone financially**. Hence, correct preposition is **off**.

Qus : 90 Jamia Millia Islamia PYQ 2023

Select the sentence which best expresses the sentence “A stone struck me on the head” in Passive voice.

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

Solution

Solution: Active voice: “A stone struck me on the head.” → Subject: A stone → Object: me → Verb: struck Passive structure: **Object + was/were + past participle + by + Subject** ✅ Correct passive: “I was struck on the head by a stone.”

Qus : 91 Jamia Millia Islamia PYQ 2023

fill the missing terms marked by question marks

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

Solution

Solution: Work column-wise with a constant step of −8 in alphabet positions (A=1,…,Z=26): Col1: Z(26) → R(18) → ? ⇒ 26−8=18−8=10 ⇒ J Col2: ? → O(15) → G(7) ⇒ top must be 23 ⇒ W Col3: S(19) → ? → C(3) ⇒ middle must be 11 ⇒ K Thus the three letters (top-middle, middle-right, bottom-left) are W, K, J → “WKJ”.

Qus : 92 Jamia Millia Islamia PYQ 2023

Find the missing number:

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

Solution

Solution: Row rule: first = second × (third/2). Row1: 24 × (6/2) = 72 ✓ Row2: 16 × (12/2) = 96 ✓ Row3: 108 = ? × (18/2 = 9) ⇒ ? = 108/9 = 12.

Qus : 93 Jamia Millia Islamia PYQ 2023

In a row of men, Manoj is 30th from the right and Kiran is 20th from the left. After interchanging their positions, Manoj becomes 35th from the right. What is the total number of men in the row?

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

Solution

Solution: Total = N. After swap, Manoj is at Kiran’s old place (20th from left) ⇒ from right = N − 20 + 1 = N − 19. Given N − 19 = 35 ⇒ N = 54.

Qus : 94 Jamia Millia Islamia PYQ 2023

John celebrated his victory day on Tuesday, 5th January 1965. When will he next celebrate it on the same weekday (Tuesday)?

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

Solution

Solution: Day shifts year-to-year by +1 for a common year, +2 for a leap year. 1965→66: +1 (Wed) 1966→67: +1 (Thu) 1967→68: +1 (Fri) 1968 is leap ⇒ 1968→69: +2 (Sun) 1969→70: +1 (Mon) 1970→71: +1 (Tue) So next Tuesday falls on **5 Jan 1971**.

Qus : 95 Jamia Millia Islamia PYQ 2023

A child’s path: 90 m East → right 20 m (South) → right 30 m (West) → 100 m North. How far from start did he meet his father?

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

Solution

Solution: Start at (0,0). After moves: (90,0) → (90,−20) → (60,−20) → (60,80). Distance = √(60²+80²) = √10000 = 100 m.

Qus : 96 Jamia Millia Islamia PYQ 2023

Sunil is the son of Kesav. Simran (Kesav’s sister) has a son Maruti and daughter Sita. Prem is the maternal uncle of Maruti. How is Sunil related to Maruti?

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

Solution

Solution: Sunil = son of Kesav; Maruti = son of Simran (Kesav’s sister). Children of siblings → cousins.

Qus : 97 Jamia Millia Islamia PYQ 2023

Select a suitable figure from the four alternatives that would complete the figure matrix.

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

Solution

Solution (pattern completion): Following the progression of elements in the matrix, the missing center-bottom cell is the simple dot.

Qus : 98 Jamia Millia Islamia PYQ 2023

In the following question how does the figure look when folded into a cube along the marked lines?

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

Solution

Solution (pattern completion): Following the progression of elements in the matrix, the missing center-bottom cell is the simple dot.

Qus : 99 Jamia Millia Islamia PYQ 2023

Find the missing terms of this series:

b, a, a, b, ?, a, b, a, ?, b, b, a, ?, ?

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

Solution

Solution: Look at odd and even positions separately. Odd positions (1,3,5,7,9,11,13): b, a, ?, b, ?, b, ? → pattern = "bab" repeating ⇒ 5=b, 9=a, 13=b. Even positions (2,4,6,8,10,12,14): a, b, a, a, b, a, ? → pattern = "aba" repeating ⇒ 14=a. Missing terms (in order 5th, 9th, 13th, 14th) = b, a, b, a → **baba**.

Qus : 100 Jamia Millia Islamia PYQ 2023

Complete the series: Z, L, X, J, V, H, T, F, __, __

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

Solution

Solution: Two interleaved sequences: - 1st,3rd,5th,7th,…: Z(26), X(24), V(22), T(20) ⇒ decreasing by 2 ⇒ next = R(18). - 2nd,4th,6th,8th,…: L(12), J(10), H(8), F(6) ⇒ decreasing by 2 ⇒ next = D(4). So the next two letters are **R, D**.

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