Question 1

Question 1Convert decimal number into binary number.(a) 52(b) 44(c) 25.80(d) 62.325

Accepted Answer

(a) 52252Remainder2260 (LSB)2130261230211 01 (MSB)Therefore, (52)10 = (110100)2.(b) 44244Remainder2220 (LSB)2110251221210 01 (MSB)Therefore, (44)10 = (101100)2.(c) 25.80Let us first convert 25 into binary as shown below:225Remainder2121 (LSB)260230211 01 (MSB)The binary equivalent of integer part (25)10 = (11001)2.Now let us convert (0.80)10 into binary as shown below:Multiply=IntegerFraction0.80 x 2=10.6 (MSB)0.6 x 2=10.20.2 x 2=00.40.4 x 2=00.80.8 x 2=10.6 (LSB)The binary equivalent of fractional part (0.8)10 = (11001)2.Therefore, (25.80)10 = (11001.11001)2.(d) 62.325Let us first convert 62 into binary as shown below:262Remainder2310 (LSB)2151271231211 01 (MSB)The binary equivalent of integer part (62)10 = (111110)2.Now let us convert (0.325)10 into binary as shown below:Multiply=IntegerFraction0.325 x 2=00.65 (MSB)0.65 x 2=10.30.3 x 2=00.60.6 x 2=10.20.2 x 2=00.40.4 x 2=00.80.8 x 2=10.6 (LSB)The binary equivalent of fractional part (0.325)10 = (0101001)2.Therefore, (62.325)10 = (111110.0101001)2.

Question 2

Question 2Convert decimal number into octal number.(a) 911(b) 540(c) 61(d) 132

Accepted Answer

(a) 9118911Remainder81137 (LSB)8141816 01 (MSB)Therefore, (911)10 = (1617)8.(b) 5408540Remainder8674 (LSB)883810 01 (MSB)Therefore, (540)10 = (1034)8.(c) 61861Remainder875 (LSB) 07 (MSB)Therefore, (61)10 = (75)8.(d) 1328132Remainder8164 (LSB)820 02 (MSB)Therefore, (132)10 = (204)8.

Question 3

Question 3Convert decimal number into hexadecimal number.(a) 132(b) 3619(c) 206(d) 752

Accepted Answer

(a) 13216132Remainder1684 (LSB) 08 (MSB)Therefore, (132)10 = (84)16.(b) 3619163619Remainder162263 (LSB)16142 0E (14) (MSB)Therefore, (3619)10 = (E23)16.(c) 20616206Remainder1612E (14) (LSB) 0C (12) (MSB)Therefore, (206)10 = (CE)16.(d) 75216752Remainder16470 (LSB)162F (15) 02 (MSB)Therefore, (752)10 = (2F0)16

Question 4

Question 4Convert binary number into decimal number.(a) 10111(b) 111101(c) 101010.011(d) 101101

Accepted Answer

(a) 10111BinaryNoPowerValueResult1 (LSB)2011x1=112121x2=212241x4=402380x8=01 (MSB)24161x16=16Equivalent decimal number = 1 + 2 + 4 + 16 = 23Therefore, (10111)2 = (23)10.(b) 111101BinaryNoPowerValueResult1 (LSB)2011x1=102120x2=012241x4=412381x8=8124161x16=161 (MSB)25321x32=32Equivalent decimal number = 1 + 4 + 8 + 16 + 32 = 61Therefore, (111101)2 = (61)10.(c) 101010.011For the integer part (101010):BinaryNoPowerValueResult0 (LSB)2010x1=012121x2=202240x4=012381x8=8024160x16=01 (MSB)25321x32=32Equivalent decimal number = 2 + 8 + 32 = 42For the fractional part (0.011):BinaryNoPowerValueResult0 (LSB)2-10.50x0.5=012-20.251x0.25=0.2512-30.1251x0.125=0.125Equivalent decimal number = 0.25 + 0.125 = 0.375Therefore, (101010.011)2 = (42.375)10.(d) 101101BinaryNoPowerValueResult1 (LSB)2011x1=102120x2=012241x4=412381x8=8024160x16=01 (MSB)25321x32=32Equivalent decimal number = 1 + 4 + 8 + 32 = 45Therefore, (101101)2 = (45)10.

Question 5

Question 5Convert octal number into decimal number.(a) 75(b) 321(c) 142(d) 205

Accepted Answer

(a) 75OctalNoPowerValueResult5 (LSB)8015x1=57 (MSB)8187x8=56Equivalent decimal number = 5 + 56 = 61Therefore, (75)8 = (61)10.(b) 321OctalNoPowerValueResult1 (LSB)8011x1=128182x8=163 (MSB)82643x64=192Equivalent decimal number = 1 + 16 + 192 = 209Therefore, (321)8 = (209)10.(c) 142OctalNoPowerValueResult2 (LSB)8012x1=248184x8=321 (MSB)82641x64=64Equivalent decimal number = 2 + 32 + 64 = 98Therefore, (142)8 = (98)10.(d) 205OctalNoPowerValueResult5 (LSB)8015x1=508180x8=562 (MSB)82642x64=128Equivalent decimal number = 5 + 128 = 133Therefore, (205)8 = (133)10.

Question 6

Question 6Convert hexadecimal number into decimal number.(a) A2(b) 13B(c) 271(d) 132

Accepted Answer

(a) A2HexadecimalNumberPowerValueResult216012x1=2A (10)1611610x16=160Equivalent decimal number = 2 + 160 = 162Therefore, (A2)16 = (162)10.(b) 13BHexadecimalNumberPowerValueResultB (11)160111x1=113161163x16=4811622561x256=256Equivalent decimal number = 11 + 48 + 256 = 315Therefore, (13B)16 = (315)10.(c) 271HexadecimalNumberPowerValueResult116011x1=17161167x16=11221622562x256=512Equivalent decimal number = 1 + 112 + 512 = 625Therefore, (271)16 = (625)10.(d) 132HexadecimalNumberPowerValueResult216012x1=23161163x16=4811622561x256=256Equivalent decimal number = 2 + 48 + 256 = 306Therefore, (132)16 = (306)10.

Question 7

Question 7Discuss UTF-8 encoding scheme. How is it different from UTF-32 encoding scheme?

Accepted Answer

UTF-8 is a variable-width encoding that can represent every character in the Unicode character set. The code unit of UTF-8 is 8 bits (an octet). It uses 1 to 4 octets to represent code points, depending on their size. For example, it uses 1 byte for ASCII characters and up to 4 bytes for others. This makes UTF-8 a multi-byte encoding that efficiently handles different character sets.UTF-32, in contrast, is a fixed-length encoding scheme that always uses exactly 4 bytes (32 bits) to represent all Unicode code points, regardless of their size, making it less space-efficient than UTF-8.

Question 8

Question 9What are Boolean operators? Explain AND, OR, NOT operators.

Accepted Answer

Operators used in Boolean algebra are known as Boolean/Logical operators.1. AND Operator — AND operator is a binary operator that operates on two variables and the operation performed by AND operator is known as logical multiplication. The symbol used for logical multiplication is dot(.) operator. The truth table for AND operator is as follows:Truth TableABOutput A.B000010100111The AND operation will result in true value (1) when both inputs are 1 (true/high) and for all other values it results in 0 (false/low).2. OR Operator — The OR operator is a binary operator that operates on two variables and the operation performed by OR operator is known as logical addition. The symbol used for logical addition is plus (+) operator. The truth table for OR operator is as follows:Truth TableABOutput A + B000011101111The OR operation results in true value (1) when either of the inputs is 1 (true) or both the inputs are 1 (true/high), and for all other values of inputs it results in 0 (false/low).3. NOT Operator — The NOT operator is a unary operator that operates on one variable and the operation performed by NOT operator is known as negation or complementation. The truth table for NOT operator is as follows:Truth TableAOutput A'0110It means that the logical statements A and A' are opposite to each other. If the value of a variable A is 0, its complement would be 1, and if the value of the variable A is 1, its complement would be 0.

Question 9

Question 10What is the use of Hexadecimal number system in computers?

Accepted Answer

The use of the hexadecimal number system in computers includes the following:The Hexadecimal number system is used in computers to specify memory addresses (which are 16-bit or 32-bit long). For example, a memory address 1101011010101111 is a big binary address but with hex it is D6AF which is easier to remember.The Hexadecimal number system is used to represent color codes. For example, FFFFFF represents White, FF0000 represents Red, etc.

Question 10

Question 11Verify the following using truth table:(i) a.(a+b) = a(ii) X.(Y+Z) = X.Y + X.Z

Accepted Answer

(i) a.(a+b) = aTruth Tableaba + ba.(a+b)a00000011001011111111As columns "a.(a+b)" and "a" have same values, hence the expression is proved.(ii) X.(Y+Z) = X.Y + X.ZTruth TableXYZY+ZX.(Y+Z)X.YX.ZX.Y+X.Z0000000000110000010100000111000010000000101110111101110111111111As columns X.(Y+Z) and X.Y+X.Z have same values, hence the expression is proved.

Question 11

Question 12Obtain the Boolean expression for the logic circuit shown below:

Accepted Answer

The boolean expression for the above logic circuit is xn.y‾+x.yn‾‾\overline{\overline{x^n.y} + \overline{x .y^n}}xn.y​+x.yn​​.

Question 12

Question 13State De Morgan's law and prove it with a truth table.

Accepted Answer

The first law states that when two (or more) input variables are OR'ed and negated, they are equivalent to the AND of the complements of the individual variables.A+B‾\overline{A+B}A+B​ = A‾.B‾\overline{A}.\overline{B}A.BProof using truth table:ABA'B'A + B(A+B)'A'.B'0011011011010010011001100100The second law states that when two (or more) input variables are AND'ed and negated, they are equivalent to the OR of the complements of the individual variables.A.B‾\overline{A.B}A.B = A‾+B‾\overline{A} + \overline{B}A+BProof using truth table:ABA'B'A.B(A.B)'A' + B'0011011011001110010111100100

Question 13

Question 14What is the advantage of preparing a digital content in Indian language using UNICODE font?

Accepted Answer

Following are the advantages of preparing a digital content in Indian language using UNICODE font:Unicode provides a universal character encoding standard, ensuring consistent display and compatibility across different platforms, devices, and applications.It supports a wide range of Indian scripts, allowing seamless text exchange and readability without font issues.Content prepared using Unicode can be easily indexed by search engines, improving accessibility and searchability.

Question 14

Question 15Encode the word 'COMPUTER' using ASCII and convert the encoded value into binary values.

Accepted Answer

Representing the word 'COMPUTER' in ASCII values of its characters:COMPUTER → 67 79 77 80 85 84 69 82Binary Equivalent → 01000011 01001111 01001101 01010000 01010101 01010100 01000101 01010010ExplanationASCII value of C is 67 and its equivalent 7-bit binary code = 01000011ASCII value of O is 79 and its equivalent 7-bit binary code = 01001111ASCII value of M is 77 and its equivalent 7-bit binary code = 01001101ASCII value of P is 80 and its equivalent 7-bit binary code = 01010000ASCII value of U is 85 and its equivalent 7-bit binary code = 01010101ASCII value of T is 84 and its equivalent 7-bit binary code = 01010100ASCII value of E is 69 and its equivalent 7-bit binary code = 01000101ASCII value of R is 82 and its equivalent 7-bit binary code = 01010010Hence, binary value for the word 'COMPUTER' is 01000011 01001111 01001101 01010000 01010101 01010100 01000101 01010010.

Question 15

Question 16Give full form of ASCII and ISCII.

Accepted Answer

ASCII — American Standard Code for Information InterchangeISCII — Indian Standard Code for Information Interchange

Source: This question is from Data Representation and Boolean Logic, Computer Science &mdash; Class 11, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Data Representation and Boolean Logic: Question, Full, Form, Ascii, Iscii, Data Representation and Boolean Logic. These are fundamental topics in Computer Science that students are expected to master as part of the CBSE Class 11 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Computer Science papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberWrite programs with proper indentation and comments.
Trace through your code with sample inputs to verify correctness.
Explain the logic behind each step of your solution.
Familiarise yourself with common library functions and methods.
Related Questions from Data Representation and Boolean LogicQ1: Question 1 Convert decimal number into binary number. (a) 52 (b) 44 (c) 25.80 (d) 62.325
Q10: Question 10 What is the use of Hexadecimal number system in computers?
Q11: Question 11 Verify the following using truth table: (i) a.(a+b) = a (ii) X.(Y+Z) = X.Y + X.Z
Q12: Question 12 Obtain the Boolean expression for the logic circuit shown below:
Q13: Question 13 State De Morgan&#x27;s law and prove it with a truth table.
Practice more questions from Data Representation and Boolean Logic &mdash; Computer Science, Class 11 CBSE

Question 16

Question 17Draw a logic circuit of the following Boolean expression:(i) (A.B‾\overline{B}B) + (C + D‾\overline{D}D) + (B.D‾\overline{D}D)(ii) (A+B).(BC+D‾\overline{D}D)

Accepted Answer

(i)(ii)

Question 17

Question 18Write binary equivalent of the following octal numbers:(a) 2306(b) 5610(c) 742(d) 65.203

Accepted Answer

(a) 2306OctalNumberBinaryEquivalent6110000030112010Therefore, (2306)8 = (010undefined 011undefined 000undefined 110undefined\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{110}}010​011​000​110​)2.(b) 5610OctalNumberBinaryEquivalent0000100161105101Therefore, (5610)8 = (101undefined 110undefined 001undefined 000undefined\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{000}}101​110​001​000​)2.(c) 742OctalNumberBinaryEquivalent201041007111Therefore, (742)8 = (111undefined 100undefined 010undefined\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{010}}111​100​010​)2.(d) 65.203OctalNumberBinaryEquivalent30110000201051016110Therefore, (65.203)8 = (110undefined 101undefined. 010undefined 000undefined 011undefined\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{101}}.\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{011}}110​101​.010​000​011​)2.

Question 18

Question 19Write binary representation of the following hexadecimal numbers:(a) 4026(b) BCA1(c) 98E(d) 132.45

Accepted Answer

(a) 4026HexadecimalNumberBinaryEquivalent60110200100000040100Therefore, (4026)16 = (0100000000100110)2.(b) BCA1HexadecimalNumberBinaryEquivalent10001A (10)1010C (12)1100B (11)1011B → 1011C → 1100A → 10101 → 0001Therefore, (BCA1)16 = (1011110010100001)2.(c) 98EHexadecimalNumberBinaryEquivalentE11108100091001Therefore, (98E)16 = (100110001110)2.(d) 132.45HexadecimalNumberBinaryEquivalent5010140100200103001110001Therefore, (132.45)16 = (000100110010.01000101)2.

Question 19

Question 20Perform the following conversions from decimal number to other number systems:(a) (54)10 = (?)2(b) (120)10 = (?)2(c) (76)10 = (?)8(d) (889)10 = (?)8(e) (789)10 = (?)16(f) (108)10 = (?)16

Accepted Answer

(a) (54)10254Remainder2270 (LSB)2131261230211 01 (MSB)Therefore, (54)10 = (110110)2.(b) (120)102120Remainder2600 (LSB)23002150271231211 01 (MSB)Therefore, (120)10 = (1111000)2.(c) (76)10876Remainder894 (LSB)811 01 (MSB)Therefore, (76)10 = (114)8.(d) (889)108889Remainder81111 (LSB)8137815 01 (MSB)Therefore, (889)10 = (1571)8.(e) (789)1016789Remainder16495 (LSB)1631 03 (MSB)Therefore, (789)10 = (315)16.(f) (108)1016108Remainder16612 (C) (LSB) 06 (MSB)Therefore, (108)10 = (6C)16.

Question 20

Question 21Express the following binary numbers into their equivalent octal numbers:(a) 100101111(b) 111011010(c) 1010011(d) 10011101

Accepted Answer

(a) 100101111Grouping in bits of 3:100undefined101undefined111undefined\underlinesegment{100} \quad \underlinesegment{101} \quad \underlinesegment{111}100​101​111​BinaryNumberEquivalentOctal100410151117Therefore, (100101111)2 = (457)8(b) 111011010Grouping in bits of 3:111undefined011undefined010undefined\underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{010}111​011​010​BinaryNumberEquivalentOctal111701130102Therefore, (111011010)2 = (732)8(c) 1010011Grouping in bits of 3:001undefined010undefined011undefined\underlinesegment{001} \quad \underlinesegment{010} \quad \underlinesegment{011}001​010​011​BinaryNumberEquivalentOctal001101020113Therefore, (1010011)2 = (123)8(d) 10011101Grouping in bits of 3:010undefined011undefined101undefined\underlinesegment{010} \quad \underlinesegment{011} \quad \underlinesegment{101}010​011​101​BinaryNumberEquivalentOctal010201131015Therefore, (10011101)2 = (235)8

Data Representation and Boolean Logic

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