2020

11. Which one of the following is *not* a vector image format ?

Which one of the following is *not* a vector image format ?

[amp_mcq option1=”CGM” option2=”PNG” option3=”SVG” option4=”SWF” correct=”option2″]

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PNG (Portable Network Graphics) is a raster image format, meaning it represents images as a grid of pixels.
Vector image formats use mathematical equations to describe shapes, lines, and curves, making them scalable without loss of quality. Raster formats store images as a fixed grid of pixels, which can lose quality when scaled.
CGM (Computer Graphics Metafile), SVG (Scalable Vector Graphics), and SWF (Shockwave Flash, often used for vector animations) are examples of vector image formats. Other common raster formats include JPEG, GIF, and BMP.

12. Which one of the following languages is interpreter ?

Which one of the following languages is interpreter ?

[amp_mcq option1=”FORTRAN” option2=”Pascal” option3=”Python” option4=”C++” correct=”option3″]

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Python is an interpreted language, unlike FORTRAN, Pascal, and C++, which are primarily compiled languages.
Interpreted languages execute code line by line using an interpreter, while compiled languages translate the entire code into machine code before execution.
FORTRAN (Formula Translation), Pascal, and C++ are widely considered compiled languages. Python, while sometimes using compilation steps (like bytecode generation), is fundamentally executed through an interpreter. Other interpreted languages include Ruby, JavaScript, PHP, etc.

13. Convert F(A, B, C) = (A + $\bar{B}$) ($\bar{B}$ + C) into canonical Pr

Convert F(A, B, C) = (A + $\bar{B}$) ($\bar{B}$ + C) into canonical Product of Sum form.

[amp_mcq option1=”(A + B + C) (A + B + $\bar{C}$) (A + $\bar{B}$ + C)” option2=”(A + B + C) ($\bar{A}$ + B + $\bar{C}$) ($\bar{A}$ + $\bar{B}$ + $\bar{C}$)” option3=”(A + $\bar{B}$ + C) (A + $\bar{B}$ + $\bar{C}$) ($\bar{A}$ + B + $\bar{C}$)” option4=”(A + B + $\bar{C}$) ($\bar{A}$ + $\bar{B}$ + $\bar{C}$) (A + $\bar{B}$ + C)” correct=”option3″]

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The canonical Product of Sums (POS) form is a product of maxterms, where a maxterm is a sum containing every variable in either complemented or uncomplemented form. The maxterms included in the canonical POS form are those for which the function evaluates to 0.
The given function is $F(A, B, C) = (A + \bar{B}) (\bar{B} + C)$.
We can find the minterms where F is 0 by finding where $(A + \bar{B})=0$ OR $(\bar{B} + C)=0$.
$(A + \bar{B}) = 0$ if and only if $A=0$ AND $\bar{B}=0$, which means $A=0$ and $B=1$. For three variables, this corresponds to minterms 010 ($m_2$) and 011 ($m_3$).
$(\bar{B} + C) = 0$ if and only if $\bar{B}=0$ AND $C=0$, which means $B=1$ and $C=0$. For three variables, this corresponds to minterms 010 ($m_2$) and 110 ($m_6$).
So, F=0 for the union of these minterms: $\{m_2, m_3\} \cup \{m_2, m_6\} = \{m_2, m_3, m_6\}$.
The canonical POS form is the product of the corresponding maxterms $M_2, M_3, M_6$.
The maxterm $M_i$ corresponds to the binary representation of $i$, where a 0 corresponds to the uncomplemented variable and a 1 corresponds to the complemented variable in the sum term.
$M_2$ from 010: $(A + \bar{B} + C)$
$M_3$ from 011: $(A + \bar{B} + \bar{C})$
$M_6$ from 110: $(\bar{A} + \bar{B} + C)$
The canonical POS form is $(A + \bar{B} + C)(A + \bar{B} + \bar{C})(\bar{A} + \bar{B} + C)$.
Comparing this derived form to the options, Option C is $(A + \bar{B} + C) (A + \bar{B} + \bar{C}) (\bar{A} + B + \bar{C})$.
Option C has the first two terms correct (M2 and M3). However, the third term in Option C is $(\bar{A} + B + \bar{C})$, which is $M_5$ (from 101). The correct third term should be $(\bar{A} + \bar{B} + C)$, which is $M_6$ (from 110).
There appears to be an error in the provided options as none exactly matches the derived canonical POS form. However, option C contains two out of the three correct maxterms and is the closest match structurally. Assuming a likely typo in the third term of Option C, it is the most probable intended answer.
– Canonical POS is a product of maxterms.
– Maxterms correspond to the minterms where the function is 0.
– For a variable in a maxterm, it is uncomplemented if its value is 0 in the corresponding minterm binary representation, and complemented if its value is 1.
The original function simplifies to $\bar{B} + AC$. Its minterms are $\sum m(0, 1, 4, 5, 7)$, and maxterms are $\prod M(2, 3, 6)$. This confirms the derived maxterms.

14. Which OSI layer is *not* part of TCP/IP model ?

Which OSI layer is *not* part of TCP/IP model ?

[amp_mcq option1=”Application layer” option2=”Session layer” option3=”Network layer” option4=”Physical layer” correct=”option2″]

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The OSI (Open Systems Interconnection) model has seven layers: Physical, Data Link, Network, Transport, Session, Presentation, and Application.
The TCP/IP model is typically described with four layers: Network Interface (or Link), Internet, Transport, and Application. Sometimes a five-layer model is used, splitting the Network Interface layer into Physical and Data Link.
Comparing the models:
– OSI Physical and Data Link correspond to TCP/IP Network Interface/Link (or Physical and Data Link).
– OSI Network corresponds to TCP/IP Internet.
– OSI Transport corresponds to TCP/IP Transport.
– OSI Application corresponds to TCP/IP Application, but the TCP/IP Application layer encompasses the functions of the OSI Session, Presentation, and Application layers.
The OSI Session layer and Presentation layer do not have distinct corresponding layers in the standard TCP/IP model; their functions are integrated into other layers (primarily the Application layer). Among the options provided, the Session layer is the one from the OSI model that is not a separate layer in the TCP/IP model.
– OSI has 7 layers, TCP/IP has 4 or 5 layers.
– TCP/IP combines functions of some OSI layers into its own layers.
– OSI Session and Presentation layers are not distinct layers in the TCP/IP model.
While TCP/IP is the dominant model for the internet, the OSI model is still valuable for understanding networking concepts and relationships between different functions.

15. On which storage device is track pattern spiral ?

On which storage device is track pattern spiral ?

[amp_mcq option1=”Magnetic disk” option2=”Optical disk” option3=”Floppy disk” option4=”Pendrive” correct=”option2″]

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Optical storage devices such as CDs, DVDs, and Blu-ray discs store data along a single, continuous track that spirals outwards from the center of the disc. The read head follows this spiral track to access data sequentially or jump to specific locations.
Magnetic disks (like HDDs and floppy disks) use multiple concentric circular tracks on the surface of the platter(s).
Flash memory (like pendrives or SSDs) uses semiconductor memory chips and does not have a physical track pattern in the same sense as disk media.
– Optical disks use a single spiral track.
– Magnetic disks use concentric circular tracks.
– Flash memory uses solid-state storage without physical tracks.
The spiral track allows optical drives to read data in a continuous stream, which is efficient for audio or video playback. The concentric tracks on magnetic disks allow for faster random access using read/write heads that move radially across the platters.

16. Exclusive-OR binary operation can be represented as

Exclusive-OR binary operation can be represented as

[amp_mcq option1=”$\bar{A} \cdot B + A \cdot \bar{B}$” option2=”$A \cdot \bar{B} + \bar{A} \cdot B$” option3=”$A \cdot B + \bar{A} \cdot \bar{B}$” option4=”$(\bar{A} + \bar{B}) \cdot (A + B)$” correct=”option1″]

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The Exclusive-OR (XOR) binary operation of two inputs A and B is true (1) if and only if the inputs are different. Its truth table is:
A | B | A XOR B
–|—|——–
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
The Sum of Products (SOP) representation for this function includes minterms where the output is 1. These are when (A=0 and B=1) or (A=1 and B=0).
– A=0 and B=1 is represented as $\bar{A} \cdot B$.
– A=1 and B=0 is represented as $A \cdot \bar{B}$.
Combining these with an OR operator gives the SOP form: $\bar{A} \cdot B + A \cdot \bar{B}$.
Option A is $\bar{A} \cdot B + A \cdot \bar{B}$, which directly matches the standard SOP form of XOR.
– XOR outputs 1 when inputs are different.
– The standard SOP form of XOR(A, B) is $\bar{A}B + A\bar{B}$.
– Boolean algebra allows representing logic functions using AND (`.`), OR (`+`), and NOT (`bar` or prime).
Option B is identical to A due to the commutativity of addition. Option C represents XNOR ($A \cdot B + \bar{A} \cdot \bar{B}$), which is the complement of XOR. Option D, $(A+B)(\bar{A}+\bar{B})$, is the Product of Sums (POS) canonical form for XOR, also a correct representation. However, option A is the standard SOP form.

17. In case of ASCII representation of characters, in which order will a c

In case of ASCII representation of characters, in which order will a computer sort the strings 23, A1, 1A, a2, 2a, aA and Aa ?

[amp_mcq option1=”1A<23<2a

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Computers typically sort strings based on the numerical values assigned to characters in a character encoding standard like ASCII. In ASCII:
– Digits (‘0’-‘9’) have values 48-57.
– Uppercase letters (‘A’-‘Z’) have values 65-90.
– Lowercase letters (‘a’-‘z’) have values 97-122.
Thus, digits come before uppercase letters, which come before lowercase letters. Sorting is done character by character from left to right.
Let’s list the strings and their first characters’ ASCII values:
– 23: ‘2’ (50)
– A1: ‘A’ (65)
– 1A: ‘1’ (49)
– a2: ‘a’ (97)
– 2a: ‘2’ (50)
– aA: ‘a’ (97)
– Aa: ‘A’ (65)
Based on the first character, the order is 1A (49), then 23 and 2a (both 50), then A1 and Aa (both 65), then a2 and aA (both 97).
Now compare within groups with the same first character:
– 23 vs 2a: ‘3’ (51) vs ‘a’ (97). ‘3’ < 'a', so 23 comes before 2a. Order: 23, 2a. - A1 vs Aa: '1' (49) vs 'a' (97). '1' < 'a', so A1 comes before Aa. Order: A1, Aa. - a2 vs aA: '2' (50) vs 'A' (65). '2' < 'A', so a2 comes before aA. Order: a2, aA. Combining the ordered groups based on the first character's order: 1A (starts with '1') 23, 2a (start with '2') A1, Aa (start with 'A') a2, aA (start with 'a') The final sorted order is 1A, 23, 2a, A1, Aa, a2, aA. This matches option A.
– ASCII sorting is based on the numerical value of each character.
– Sorting is lexicographical, comparing characters from left to right.
– Digit characters have lower ASCII values than uppercase letters, which have lower values than lowercase letters.
Different character encodings (like EBCDIC) have different sorting orders. However, ASCII and its extensions are prevalent, especially in contexts like file systems and databases, making this sorting order common.

18. Convert 1AC in hexadecimal into a decimal number.

Convert 1AC in hexadecimal into a decimal number.

[amp_mcq option1=”428″ option2=”424″ option3=”408″ option4=”420″ correct=”option1″]

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To convert a hexadecimal number (base 16) to a decimal number (base 10), we multiply each digit by its corresponding power of 16 and sum the results. The hexadecimal number is 1AC.
Starting from the rightmost digit, the powers of 16 are $16^0, 16^1, 16^2$, and so on.
In hexadecimal, A represents 10, B represents 11, C represents 12.
$1AC_{16} = 1 \times 16^2 + A \times 16^1 + C \times 16^0$
$ = 1 \times 256 + 10 \times 16 + 12 \times 1$
$ = 256 + 160 + 12$
$ = 416 + 12 = 428$.
The decimal equivalent is 428.
– Hexadecimal uses digits 0-9 and letters A-F to represent values 0-15.
– Conversion to decimal involves summing products of digits and powers of the base (16).
– The position of the digit determines the power of 16.
Hexadecimal is often used in computing as a compact representation of binary data because $16 = 2^4$, meaning each hexadecimal digit represents exactly four binary bits.

19. Which of the following is/are bit manipulation operators ? 1. || 2.

Which of the following is/are bit manipulation operators ?

  • 1. ||
  • 2. ^
  • 3. >>
  • 4. <<

Select the correct answer using the code given below :

[amp_mcq option1=”1 only” option2=”1, 2 and 3″ option3=”2 and 4 only” option4=”1, 2 and 4″ correct=”option3″]

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Bit manipulation operators (or bitwise operators) perform operations on individual bits of integers. From the given list:
1. `||` is the logical OR operator, which operates on boolean values (or treats non-zero as true). It is not a bit manipulation operator.
2. `^` is the bitwise XOR operator, which compares corresponding bits and returns 1 if they are different, and 0 if they are the same. This is a bit manipulation operator.
3. `>>` is the bitwise right shift operator, which shifts all bits in an operand to the right by a specified number of positions. This is a bit manipulation operator.
4. `<<` is the bitwise left shift operator, which shifts all bits in an operand to the left by a specified number of positions. This is a bit manipulation operator. Operators 2 (`^`), 3 (`>>`), and 4 (`<<`) are bit manipulation operators. Option C lists 2 and 4 only. While 3 is also correct, Option C contains only correct operators from the list, unlike options B and D which incorrectly include 1 (`||`). Given the options, C is the most plausible correct answer, assuming it lists a subset of the correct operators from the provided list.
– Bitwise operators work on the binary representation of numbers.
– Common bitwise operators include AND (`&`), OR (`|`), XOR (`^`), NOT (`~`), left shift (`<<`), and right shift (`>>`).
– Logical operators (`&&`, `||`, `!`) work on boolean values.
The options provided seem incomplete as `>>` (3) is also a bit manipulation operator. However, option C correctly identifies two bit manipulation operators from the list without including any incorrect ones.

20. Find the decimal equivalent of the binary number 110.101.

Find the decimal equivalent of the binary number 110.101.

[amp_mcq option1=”6.5″ option2=”6.625″ option3=”6.5″ option4=”6.25″ correct=”option2″]

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To convert a binary number with a fractional part to decimal, we sum the products of each digit and its corresponding power of 2. For the number 110.101:
Integer part (110): $1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 1 \times 4 + 1 \times 2 + 0 \times 1 = 4 + 2 + 0 = 6$.
Fractional part (.101): $1 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} = 1 \times 0.5 + 0 \times 0.25 + 1 \times 0.125 = 0.5 + 0 + 0.125 = 0.625$.
The decimal equivalent is the sum of the integer and fractional parts: $6 + 0.625 = 6.625$.
– The position of each binary digit corresponds to a power of 2.
– Digits to the left of the decimal point have positive powers ($2^0, 2^1, …$).
– Digits to the right of the decimal point have negative powers ($2^{-1}, 2^{-2}, …$).
$2^0 = 1$, $2^1 = 2$, $2^2 = 4$, $2^{-1} = 0.5$, $2^{-2} = 0.25$, $2^{-3} = 0.125$. This method applies to converting any base-N number to decimal.
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