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

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Jamia Millia Islamia
Previous Year Questions

Jamia Millia Islamia
Previous Year Questions
Jamia Millia Islamia

2. P4 Q. Microsoft Corporation

3. Java R. IBM

4. PC S. Intel Corporation

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

Solution

lution: - DOS was developed by **Microsoft Corporation** → Q - P4 (Pentium 4) is a processor by **Intel Corporation** → S - Java was developed by **Sun Microsystems** → P - PC (Personal Computer) was introduced by **IBM** → R Hence the correct matching: $(1, Q), (2, S), (3, P), (4, R)$

Qus : 39 Jamia Millia Islamia PYQ

Which of the following languages is case sensitive?

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

Solution

C distinguishes uppercase/lowercase identifiers.

Qus : 40 Jamia Millia Islamia PYQ

Kernel is:

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Solution

The kernel is the core providing fundamental (primitive) services on which the rest of the OS runs.

Qus : 41 Jamia Millia Islamia PYQ

If $(123)_5=(A3)_B$, then the number of possible values of $A$ is:

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

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 : 63 Jamia Millia Islamia PYQ

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

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 : 64 Jamia Millia Islamia PYQ

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

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 : 65 Jamia Millia Islamia PYQ

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

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.

$(1056)_x = x^3 + 5x + 6$. So, $x^3 + 5x + 6 = 780 \Rightarrow x = 9$.

Qus : 90 Jamia Millia Islamia PYQ

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

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 : 91 Jamia Millia Islamia PYQ

If $(2?1)_7 = (120)_{10}$, then the missing digit is

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

Solution

$2\times7^2 + ?\times7 + 1 = 120$ $\Rightarrow 99 + 7? = 120 \Rightarrow ? = 3$.

Qus : 92 Jamia Millia Islamia PYQ

$A + B = \text{NAND}(\text{NAND}(A,A), \text{NAND}(B,B))$. Hence, $3$ NAND gates required. $\boxed{3}$

Qus : 98 Jamia Millia Islamia PYQ

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 : 99 Jamia Millia Islamia PYQ

The time required for an electronic circuit to change its state is called

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

Solution

Output delay between input and output change = propagation time. $\boxed{\text{Propagation time}}$

Qus : 100 Jamia Millia Islamia PYQ

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

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

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

Solution

i-- uses the old value 1, then i becomes 0. Expression is 1 - (-2) = 3.

Qus : 122 Jamia Millia Islamia PYQ

What will be output of following statements?

int i = 1, j;

j = --i - 2;

printf("%d", j);

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

Solution

i = 1 → --i = 0 → j = 0 - 2 = -2

Qus : 123 Jamia Millia Islamia PYQ

What is the output of following C Program?

#include <stdio.h>

int main()

{

char grade[] = {'A','B','C'};

printf("GRADE=%c, ", *grade);

Far pointer can access ____

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

QWERTY, AZERTY, and DVORAK are keyboard layouts.

SULTRY is an English word, not a layout.

Qus : 161 Jamia Millia Islamia PYQ

Which of the following describes the characteristics of SRAM?

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

Solution

All of them let you define new data types

Qus : 162 Jamia Millia Islamia PYQ

Match List–I and List–II and select the correct group of matching.

List–I

P. RAM

Q. CPU Speed

R. Monitor

S. CD–ROM Speed

List–II

1. Hertz

(231)_4 = 2·4^2 + 3·4 + 1 = 45_(10).

45 to base 3 → 1·3^3 + 2·3^2 + 0·3 + 0 = 1200_3.

Qus : 176 Jamia Millia Islamia PYQ

Consider the following lists, and then select the correct option after matching them. $\begin{array}{|c|l|c|l|} \hline \textbf{List–I} & & \textbf{List–II} & \\ \hline 1. & \text{Procedural Oriented Language} & P. & \text{COBOL} \\ \hline 2. & \text{Object Oriented Language} & Q. & \text{HTML} \\ \hline 3. & \text{Business Oriented Language} & R. & \text{C++} \\ \hline 4. & \text{Web Page} & S. & \text{Pascal} \\ \hline \end{array}$

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Procedural Oriented → Pascal $(1,S)$ Object Oriented → C++ $(2,R)$ Business Oriented → COBOL $(3,P)$ Web Page → HTML $(4,Q)$ Correct match: $(1,S), (2,R), (3,P), (4,Q)$

Qus : 177 Jamia Millia Islamia PYQ

Convert (1278)_12 into (____)_4.

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

Solution

(1278)_12 = 1·12^3 + 2·12^2 + 7·12 + 8 = 2108_(10).

2108 to base 4:

Group binary digits in groups of 3 from right: $(00\ 100\ 101\ 00)_2 = (010\ 010\ 100)_2 = (224)_8$

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