1. |
## Any signed negative binary number is recognised by its |

A. | msb |

B. | lsb |

C. | byte |

D. | nibble |

Answer» A. msb | |

Explanation: any negative number is recognized by its msb (most significant bit). |

2. |
## The parameter through which 16 distinct values can be represented is known as |

A. | bit |

B. | byte |

C. | word |

D. | nibble |

Answer» C. word | |

Explanation: it can be represented up to 16 different values with the help of a word. nibble is a combination of four bits and byte is a combination of 8 bits. it is “word” which is said to be a collection of 16-bits on most of the systems. |

3. |
## If the decimal number is a fraction then its binary equivalent is obtained by the number continuously by 2. |

A. | dividing |

B. | multiplying |

C. | adding |

D. | subtracting |

Answer» B. multiplying | |

Explanation: on multiplying the decimal number continuously by 2, the binary equivalent is obtained by the collection of the integer part. however, if it’s an integer, then it’s binary equivalent is determined by dividing the number by 2 and collecting the remainders. |

4. |
## The representation of octal number (532.2)8 in decimal is |

A. | (346.25)10 |

B. | (532.864)10 |

C. | (340.67)10 |

D. | (531.668)10 |

Answer» A. (346.25)10 | |

Explanation: octal to decimal conversion is obtained by multiplying 8 to the power of base index along with the value at that index position. |

5. |
## The decimal equivalent of the binary number (1011.011)2 is |

A. | (11.375)10 |

B. | (10.123)10 |

C. | (11.175)10 |

D. | (9.23)10 |

Answer» A. (11.375)10 | |

Explanation: binary to decimal conversion is obtained by multiplying 2 to the power of base index along with the value at that index position. |

6. |
## An important drawback of binary system is |

A. | it requires very large string of 1’s and 0’s to represent a decimal number |

B. | it requires sparingly small string of 1’s and 0’s to represent a decimal number |

C. | it requires large string of 1’s and small string of 0’s to represent a decimal number |

D. | it requires small string of 1’s and large string of 0’s to represent a decimal number |

Answer» A. it requires very large string of 1’s and 0’s to represent a decimal number | |

Explanation: the most vital drawback of binary system is that it requires very large string of 1’s and 0’s |

7. |
## The decimal equivalent of the octal number (645)8 is |

A. | (450)10 |

B. | (451)10 |

C. | (421)10 |

D. | (501)10 |

Answer» C. (421)10 | |

Explanation: octal to decimal conversion is obtained by multiplying 8 to the power of base index along with the value at that index position. |

8. |
## The largest two digit hexadecimal number is |

A. | (fe)16 |

B. | (fd)16 |

C. | (ff)16 |

D. | (ef)16 |

Answer» C. (ff)16 | |

Explanation: (fe)16 is 254 in decimal system, while (fd)16 is 253. (ef)16 is 239 in decimal system. and, |

9. |
## What is the addition of the binary numbers 11011011010 and 010100101? |

A. | 0111001000 |

B. | 1100110110 |

C. | 11101111111 |

D. | 10011010011 |

Answer» C. 11101111111 | |

Explanation: the rules for binary addition are : 0 + 0 = 0 |

10. |
## Representation of hexadecimal number (6DE)H in decimal: |

A. | 6 * 162 + 13 * 161 + 14 * 160 |

B. | 6 * 162 + 12 * 161 + 13 * 160 |

C. | 6 * 162 + 11 * 161 + 14 * 160 |

D. | 6 * 162 + 14 * 161 + 15 * 160 |

Answer» A. 6 * 162 + 13 * 161 + 14 * 160 | |

Explanation: hexadecimal to decimal conversion is obtained by multiplying 16 to the power of base index along with the value at that index position. |

11. |
## 100101 × 0110 = ? |

A. | 1011001111 |

B. | 0100110011 |

C. | 101111110 |

D. | 0110100101 |

Answer» C. 101111110 | |

Explanation: the rules for binary multiplication are: 0 * 0 = 0 |

12. |
## Perform multiplication of the binary numbers: 01001 × 01011 = ? |

A. | 001100011 |

B. | 110011100 |

C. | 010100110 |

D. | 101010111 |

Answer» A. 001100011 | |

Explanation: the rules for binary multiplication are: 0 * 0 = 0 |

13. |
## Divide the binary numbers: 111101 ÷ 1001 and find the remainder |

A. | 0010 |

B. | 1010 |

C. | 1100 |

D. | 0011 |

Answer» D. 0011 | |

Explanation: binary division is accomplished using long division method. |

14. |
## Divide the binary number (011010000) by (0101) and find the quotient |

A. | 100011 |

B. | 101001 |

C. | 110010 |

D. | 010001 |

Answer» B. 101001 |

15. |
## Binary coded decimal is a combination of |

A. | two binary digits |

B. | three binary digits |

C. | four binary digits |

D. | five binary digits |

Answer» C. four binary digits | |

Explanation: binary coded decimal is a combination of 4 binary digits. for example-8421. |

16. |
## The decimal number 10 is represented in its BCD form as |

A. | 10100000 |

B. | 01010111 |

C. | 00010000 |

D. | 00101011 |

Answer» C. 00010000 | |

Explanation: the decimal number 10 is represented in its bcd form as 0001 0000, in accordance to 8421 for each of the two digits. |

17. |
## Carry out BCD subtraction for (68) – (61) using 10’s complement method. |

A. | 00000111 |

B. | 01110000 |

C. | 100000111 |

D. | 011111000 |

Answer» A. 00000111 | |

Explanation: first the two numbers are converted into their respective bcd form using 8421 sequence. then binary subtraction is carried out. |

18. |
## When numbers, letters or words are represented by a special group of symbols, this process is called |

A. | decoding |

B. | encoding |

C. | digitizing |

D. | inverting |

Answer» B. encoding | |

Explanation: when numbers, letters or words are represented by a special group of symbols, this process is called encoding. encoding in the sense of fetching the codes or words in a computer. it is done to secure the transmission of information. |

19. |
## A three digit decimal number requires for representation in the conventional BCD format. |

A. | 3 bits |

B. | 6 bits |

C. | 12 bits |

D. | 24 bits |

Answer» C. 12 bits | |

Explanation: the number of bits needed to represent a given decimal number is always greater than the number of bits required for a straight binary encoding of the same. hence, a three digit decimal number requires 12 bits for representation in bcd format. |

20. |
## How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes? |

A. | 12 |

B. | 14 |

C. | 16 |

D. | 18 |

Answer» B. 14 | |

Explanation: total number of decimals to be represented = 10000 = 104 = 2n (where n is the number of bits required) = 213.29. therefore, the number of bits required for straight binary encoding = 14. |

21. |
## The excess-3 code for 597 is given by |

A. | 100011001010 |

B. | 100010100111 |

C. | 010110010111 |

D. | 010110101101 |

Answer» A. 100011001010 | |

Explanation: the addition of ‘3’ to each digit yields the three new digits ‘8’, ’12’ and ’10’. hence, the corresponding four-bit binary equivalents are 100011001010, in accordance to 8421 format. |

22. |
## The decimal equivalent of the excess-3 number 110010100011.01110101 is |

A. | 970.42 |

B. | 1253.75 |

C. | 861.75 |

D. | 1132.87 |

Answer» A. 970.42 | |

Explanation: the conversion of binary numbers into digits ‘1100’, ‘1010’, ‘0011’, ‘0111’ and ‘0101’ gives |

23. |
## In boolean algebra, the OR operation is performed by which properties? |

A. | associative properties |

B. | commutative properties |

C. | distributive properties |

D. | all of the mentioned |

Answer» D. all of the mentioned | |

Explanation: the expression for associative property is given by a+(b+c) = (a+b)+c & a*(b*c) = (a*b)*c. |

24. |
## The expression for Absorption law is given by |

A. | a + ab = a |

B. | a + ab = b |

C. | ab + aa’ = a |

D. | a + b = b + a |

Answer» A. a + ab = a | |

Explanation: the expression for absorption law is given by: a+ab = a. |

25. |
## DeMorgan’s theorem states that |

A. | (ab)’ = a’ + b’ |

B. | (a + b)’ = a’ * b |

C. | a’ + b’ = a’b’ |

D. | (ab)’ = a’ + b |

Answer» A. (ab)’ = a’ + b’ | |

Explanation: the demorgan’s law states that (ab)’ = a’ + b’ & (a + b)’ = a’ * b’, as per the dual property. |

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