Important MCQ questions for Class 11 Computer Science Chapter-Nature and Purpose of the Business
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MCQ Questions set-1 for chapter-Nature and Purpose of the Business class 11 Computer Science
Computer Science - MCQ on Nature and Purpose of the Business
Class XI
Q.1. ASCII stands for
a. Authorised Standard for Coding and Information Interchanging.
b. American Scientific Code For Information Exchange.
c. American Standard Code for Information Interchange.
d. American Standard Compression with Information Interchange.
Answer:
(c)
Explanation: It is the most widely used alphanumeric code, which was proposed by the American Standard for Information Interchange.
Q.2. 43 in binary number can be represented as
a. 100110.
b. 110101.
c. 010110.
d. 101011.
Answer:
(d)
Explanation: Divide the given number by two and write down the remainder from down to up.
Q.3. In hexadecimal number system F represents the decimal number
a. 10.
b. 12.
c. 15.
d. 14.
Answer:
(c)
Explanation: In hexadecimal number systems 15 is represented as F.
Q.4. One’s compliment of –13 is
a. 00001101.
b. 00010111.
c. 11010010.
d) 11110010.
Answer:
(d)
Explanation: Find the binary equivalent of 13 and then replace 0’s by 1 and 1’s by 0’s.
Q.5. 10110 in decimal form can be represented as
a. 26.
b. 25.
c. 22.
d. 19.
Answer:
(c)
Explanation: Multiply each one by 2^(n-1) where n is its position.
Q.6. (38.21) can be represented in binary form as
a. 100110.0011010111.
b. 110001.1100010110.
c. 100011.0011010111.
d. 110100.0110100101.
Answer:
(a)
Explanation: Convert 38 to its binary equivalent and then .21
Q.7. MSB stands for
a. Memory Standard Bit.
b. Memory Stored Bit.
c. Most significant Bit.
d. Most Standard Bit.
Answer:
(c)
Explanation: The rightmost bit is known as the most significant bit.
Q.8. Numbers that can be represented by N bit word is
a. 2^n.
b. 2n.
c. 2n-1.
d. 2^n-1.
Answer:
(d)
Explanation: An n-bit word can represent 2^n-1 words.
Q.9. LSB stands for
a. low standard bit.
b. least standard bit.
c. least significant bit.
d. leftmost significant bit.
Answer:
(c)
Explanation: The leftmost bit in a binary number is referred to as Least Significant Bit.
Q.10. 4A8C hex in binary form would be equivalent to
a. 0100101010001100.
b. 0100110010001010.
c. 1100010110101110.
d. 0101110110001101.
Answer:
(a)
Explanation: Convert each digit to its decimal equivalent.
Q.11. FACE is a hex. It can be represented in binary as
a. 1011101000111110.
b. 1100100011101110.
c. 0000010100110001.
d. 1111101011001110.
Answer:
(d)
Explanation: Convert each digit to its decimal equivalent.
Q.12. The 1’s complement of 1010111001010 is
a. 0010110110001.
b. 0101000110101.
c. 0101011000110.
d. 0101000010110.
Answer:
(b)
Explanation: Replace 1 by 0 and 0 by 1.
Q.13. The binary equivalent of hex F4B is
a. 110110101001.
b. 101010100110.
c. 111110010110.
d. 111101001011.
Answer:
(d)
Explanation: Convert each digit to its decimal equivalent.
Q.14. Unicode Stands for
a. universal coding.
b. unique coding.
c. uppercase coding.
d. unique code.
Answer:
(a)
Explanation: It is a new universal code standard, which was adopted for all newer platforms by the Unicode Consortium.
Q.15. ASCII code is a
a. 8-bit code.
b. 2-bit code.
c. 7-bit code.
d. 9-bit code.
Answer:
(c)
Explanation: It is a 7-bit parity code and can code up to 2^7-1=128 characters.
Q.16. ISCII code is a
a. 7-bit code.
b. 5-bit code.
c. 8-bit code.
d. 16-bit code.
Answer:
(c)
Explanation: ISCII is a 8 bit code capable of coding 256 characters.
Q.17. The binary equivalent of 22.25 is
a. 10110.01.
b. 01011.11.
c. 11001.01.
d. 10101.10.
Answer:
(a)
Explanation: First convert 22 into binary and then .25.
Q.18. 11011110101110 in hexadecimal form will be represented as
a. 4B3E.
b. 37AE.
c. EA73.
d. E3B4.
Answer:
(b)
Explanation: Group into four and write their hex equivalent.
Q.19. The number of digits in octal system is
a. 6.
b. 7.
c. 8.
d. 15.
Answer:
(c)
Explanation: An octal number system has 2^3=8 digits.
Q.20. Numbers are stored and transmitted inside a computer in
a. decimal number system.
b. binary number system.
c. hexadecimal number system.
d. octal number system.
Answer:
(b)
Explanation: The machine language coverts the numbers in binary format.
Q.21. The base of octal number system is
a. 2.
b. 16.
c. 8.
d. 4.
Answer:
(d)
Explanation: Octal number system has base 8.
Q.22. A byte corresponds to
a. 12 bits.
b. 4 bits.
c. 16 bits.
d. 8 bits.
Answer:
(d)
Explanation: 1 byte=8 bits.
Q.23. Hexadecimal number F is equal to octal number
a. 15.
b. 16.
c. 17.
d. 18.
Answer:
(b)
Explanation: In hexadecimal number system F represents 15.
Q.24. The ASCII code of ‘A’ is
a. 66D.
b. 41H.
c. 100 0001.
d. 110 0001.
Answer:
(c)
Explanation: ASCII code is a 7-bit code.
25. The minimum number of bits required to store FF is
a. 2.
b. 7.
c. 16.
d. 8.
Answer:
(c)
Explanation: FF is a hexadecimal representation.
MCQ Questions set-2 for chapter-Nature and Purpose of the Business class 11 Computer Science
Q.26. The code used by the java language to represent its characters
a. ASCII.
b. ISCII.
c. Unicode.
d. extended ASCII.
Answer:
(c)
Explanation: Java code is written in Unicode to make it a portable language.
Q.27. The integer range of ASCII codes is
a. –127 to 127.
b. 0 to 127.
c. 0 to 270.
d. –32768 to 32767.
Answer:
(b)
Explanation: ASCII code is a 7-bit code. It codes 2^7 bits ranging from 0 to 127.
Q.28. The two's complement of an integer is formed by
a. reversing (inverting) the bits and adding 1.
b. adding 1 and reversing the bits.
c. calculating the integer's additive inverse.
d. changing the highest bit to a 1.
Answer:
(a)
Explanation: To obtain the 2’s compliment of a number first reverse the bits and then add 1 to it.
Q.29. The largest signed integer that may be stored in 32 bits is
a. 2^31.
b. 2^32-1.
c. 2^31-1.
d. 2^32.
Answer:
(b)
Explanation: In n bits format one can store 2^n-1 bits.
Q.30. The ASCII representation of 0 is
a. 0110010.
b. 0110011.
c. 0110000.
d. 0001011.
Answer:
(c)
Explanation: Decimal numbers are represented in ASCII starting from 011.
Q.31. The binary equivalent of 225.35 is
a. 11100101.01011.
b. 10100101.01011.
c. 11000101.01011.
d. 10110100.01011.
Answer:
(a)
Explanation: Convert 225 to binary and then .35 to its binary equivalent.
Q.32. 146 in binary form can be represented as
a. 10001010.
b. 10010010.
c. 01010101.
d. 11011100.
Answer:
(b)
Explanation: Divide successively by 2 and write the remainders from top to down.
Q.33. 106 is a decimal number, when converted to binary it becomes
a. 1010101.
b. 0101001.
c. 1101010.
d. 1110100.
Answer:
(c)
Explanation: Divide successively by 2 and write the remainders from top to down.
Q.34. The decimal equivalent of 101010 is
a. 39.
b. 40.
c. 41.
d. 42.
Answer:
(d)
Explanation: 0x(2^0) +1x(2^1)+0x (2^2) + 1x (2^3) + 0x(2^4) + 1x(2^5)=42.
Q.35. The octal representation of 1011110100011 is
a. 1,364,307.
b. 3,647,107.
c. 1,364,073.
d. 1,647,307.
Answer:
(a)
Explanation: Divide into groups of three and express the groups in decimal.
Q.36. The hexadecimal representation of 101011101 is
a. AB2.
b. 2AB.
c. 2BA.
d. BA2.
Answer:
(c)
Explanation: Group into 4 and express the groups in hexadecimal.
Q.37. The hexadecimal representation of 10101110.010111 is
a. 5E.AC.
b. EA.C5.
c. 5E.AC.
d. CE.5A.
Answer:
(c)
Explanation: Group into 4 and express the groups in hexadecimal.
Q.38. 10101100.0100 in hexadecimal form becomes
a. AC.4.
b. CA.3.
c. AC.5.
d. CA.4.
Answer:
(a)
Explanation: Group into 4 and express the groups in hexadecimal.
Q.39. 72905 on converting to hexadecimal number system becomes
a. CC901.
b. 9CC01.
c. 11CC9.
d. 1C1C9.
Answer:
(c)
Explanation: Divide successively by 16 and write the remainders from bottom to up.
Q.40. The two’s compliment of –4 is
a. 11011111.
b. 01100101.
c. 00110101.
d. 11111100.
Answer:
(d)
Explanation: Positive expression of the number is 0000 0100.
Compliment 1111 1011.
Add 1 1111 1100.
Therefore –4 equals 1111 1100.
Q.41. The decimal equivalent EB4A16 is
a. 61230.
b. 60231.
c. 60234.
d. 61234.
Answer:
(c)
Explanation: EB4A16 =14x163 +11x162+4x161+10x160.
=6023410.
Q.42. The hexadecimal equivalent of 72905 would be
a. 9AB1.
b. CC9B.
c. CC91.
d. 11CC9.
Answer:
(d)
Explanation: 16 72905
16 4556 9 9
16 284 12 C
16 17 12 C
16 1 1 1
0 1 1
Q.43. The way by which computers manipulate data into information is called
a. programming.
b. processing.
c. storing.
d. organizing.
Answer:
(b)
Explanation: the extraction of information from raw facts i.e. data is called processing.
Q.44. Computers process data into information by working exclusively with
a. multimedia.
b. words.
c. characters.
d. numbers.
Answer:
(d)
Explanation: Data is converted into ASCII format for processing. In ASCII format each and every character is represented as a number.
Q.45. In the binary language each letter of the alphabet, each number and each special character is made up of a unique combination of
a. eight bytes.
b. eight kilobytes.
c. eight characters.
d. eight bits.
Answer:
(d)
Explanation: Each character is stored in one byte of information, which is eight bits of memory.
Q.46. A string of eight 0s and 1s is called a
a. megabyte.
b. byte.
c. kilobyte.
d. gigabyte.
Answer:
(a)
Explanation: 1 byte when represented in binary form becomes a string of eight 0’s and 1’s.
Q.47. A million bytes would be approximately equal to
a. gigabyte.
b. kilobyte.
c. megabyte.
d. terabyte.
Answer:
(c)
Explanation: Megabyte is a SI-multiple (see prefix mega-) of the unit byte for digital information storage or transmission and is equal to 106 (1,000,000) bytes.
Q.48. One billion bytes is approximately equal to
a. kilobyte.
b. bit.
c. gigabyte.
d. megabyte.
Answer:
(c)
Explanation: Gigabyte is an SI-multiple of the unit byte for digital information storage. Since the giga- prefix means 109, gigabyte means 1,000,000,000 bytes (10003, 109). However, this term is also often used meaning 1,073,741,824 bytes (10243, 230).
Q.49. A byte can hold one
a. bit.
b. binary digit.
c. character.
d. kilobyte.
Answer:
(c)
Explanation: Each of the byte can hold a single character of information in it.
Q.50. 345 in binary format would be equivalent to
a. 101011001.
b. 100101100.
c. 101010011.
d. 101011010.
Answer:
(a)
Explanation: Follow successive division method by two and write the remainders from bottom to top.