Question 1

Question 6Convert the following binary numbers to decimal and hexadecimal:(a) 1010(b) 111010(c) 101011111(d) 1100(e) 10010101(f) 11011100

Accepted Answer

(a) 1010Converting to decimal:BinaryNoPowerValueResult0 (LSB)2010x1=012121x2=202240x4=01 (MSB)2381x8=8Equivalent decimal number = 8 + 2 = 10Therefore, (1010)2 = (10)10Converting to hexadecimal:Grouping in bits of 4:1010undefined\underlinesegment{1010}1010​BinaryNumberEquivalentHexadecimal1010A (10)Therefore, (1010)2 = (A)16(b) 111010Converting to decimal:BinaryNoPowerValueResult0 (LSB)2010x1=012121x2=202240x4=012381x8=8124161x16=161 (MSB)25321x32=32Equivalent decimal number = 32 + 16 + 8 + 2 = 58Therefore, (111010)2 = (58)10Converting to hexadecimal:Grouping in bits of 4:0011undefined1010undefined\underlinesegment{0011} \quad \underlinesegment{1010}0011​1010​BinaryNumberEquivalentHexadecimal1010A (10)00113Therefore, (111010)2 = (3A)16(c) 101011111Converting to decimal:BinaryNoPowerValueResult1 (LSB)2011x1=112121x2=212241x4=412381x8=8124161x16=16025320x32=0126641x64=640271280x128=01 (MSB)282561x256=256Equivalent decimal number = 256 + 64 + 16 + 8 + 4 + 2 + 1 = 351Therefore, (101011111)2 = (351)10Converting to hexadecimal:Grouping in bits of 4:0001undefined0101undefined1111undefined\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1111}0001​0101​1111​BinaryNumberEquivalentHexadecimal1111F (15)0101500011Therefore, (101011111)2 = (15F)16(d) 1100Converting to decimal:BinaryNoPowerValueResult0 (LSB)2010x1=002120x2=012241x4=41 (MSB)2381x8=8Equivalent decimal number = 8 + 4 = 12Therefore, (1100)2 = (12)10Converting to hexadecimal:Grouping in bits of 4:1100undefined\underlinesegment{1100}1100​BinaryNumberEquivalentHexadecimal1100C (12)Therefore, (1100)2 = (C)16(e) 10010101Converting to decimal:BinaryNoPowerValueResult1 (LSB)2011x1=102120x2=012241x4=402380x8=0124161x16=16025320x32=0026640x64=01 (MSB)271281x128=128Equivalent decimal number = 1 + 4 + 16 + 128 = 149Therefore, (10010101)2 = (149)10Converting to hexadecimal:Grouping in bits of 4:1001undefined0101undefined\underlinesegment{1001} \quad \underlinesegment{0101}1001​0101​BinaryNumberEquivalentHexadecimal0101510019Therefore, (101011111)2 = (95)16(f) 11011100Converting to decimal:BinaryNoPowerValueResult0 (LSB)2010x1=002120x2=012241x4=412381x8=8124161x16=16025320x32=0126641x64=641 (MSB)271281x128=128Equivalent decimal number = 4 + 8 + 16 + 64 + 128 = 220Therefore, (11011100)2 = (220)10Converting to hexadecimal:Grouping in bits of 4:1101undefined1100undefined\underlinesegment{1101} \quad \underlinesegment{1100}1101​1100​BinaryNumberEquivalentHexadecimal1100C (12)1101D (13)Therefore, (11011100)2 = (DC)16

Question 2

Question 9Convert the following binary numbers to hexadecimal and octal :(a) 10011011101(b) 1111011101011011(c) 11010111010111(d) 1010110110111(e) 10110111011011(f) 1111101110101111

Accepted Answer

(a) 10011011101Converting to hexadecimal:Grouping in bits of 4:0100undefined1101undefined1101undefined\underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101}0100​1101​1101​BinaryNumberEquivalentHexadecimal1101D (13)1101D (13)01004Therefore, (10011011101)2 = (4DD)16Converting to Octal:Grouping in bits of 3:010undefined011undefined011undefined101undefined\underlinesegment{010} \quad \underlinesegment{011} \quad \underlinesegment{011} \quad \underlinesegment{101}010​011​011​101​BinaryNumberEquivalentOctal1015011301130102Therefore, (10011011101)2 = (2335)8(b) 1111011101011011Converting to hexadecimal:Grouping in bits of 4:1111undefined0111undefined0101undefined1011undefined\underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011}1111​0111​0101​1011​BinaryNumberEquivalentHexadecimal1011B (11)01015011171111F (15)Therefore, (1111011101011011)2 = (F75B)16Converting to Octal:Grouping in bits of 3:001undefined111undefined011undefined101undefined011undefined011undefined\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{101} \quad \underlinesegment{011} \quad \underlinesegment{011}001​111​011​101​011​011​BinaryNumberEquivalentOctal011301131015011311170011Therefore, (1111011101011011)2 = (173533)8(c) 11010111010111Converting to hexadecimal:Grouping in bits of 4:0011undefined0101undefined1101undefined0111undefined\underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111}0011​0101​1101​0111​BinaryNumberEquivalentHexadecimal011171101D (13)0101500113Therefore, (11010111010111)2 = (35D7)16Converting to Octal:Grouping in bits of 3:011undefined010undefined111undefined010undefined111undefined\underlinesegment{011} \quad \underlinesegment{010} \quad \underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111}011​010​111​010​111​BinaryNumberEquivalentOctal11170102111701020113Therefore, (11010111010111)2 = (32727)8(d) 1010110110111Converting to hexadecimal:Grouping in bits of 4:0001undefined0101undefined1011undefined0111undefined\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1011} \quad \underlinesegment{0111}0001​0101​1011​0111​BinaryNumberEquivalentHexadecimal011171011B (11)0101500011Therefore, (1010110110111)2 = (15B7)16Converting to Octal:Grouping in bits of 3:001undefined010undefined110undefined110undefined111undefined\underlinesegment{001} \quad \underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{111}001​010​110​110​111​BinaryNumberEquivalentOctal11171106110601020011Therefore, (1010110110111)2 = (12667)8(e) 10110111011011Converting to hexadecimal:Grouping in bits of 4:0010undefined1101undefined1101undefined1011undefined\underlinesegment{0010} \quad \underlinesegment{1101} \quad \underlinesegment{1101} \quad \underlinesegment{1011}0010​1101​1101​1011​BinaryNumberEquivalentHexadecimal1011B (11)1101D (13)1101D (13)00102Therefore, (10110111011011)2 = (2DDB)16Converting to Octal:Grouping in bits of 3:010undefined110undefined111undefined011undefined011undefined\underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{011}010​110​111​011​011​BinaryNumberEquivalentOctal01130113111711060102Therefore, (10110111011011)2 = (26733)8(f) 1111101110101111Converting to hexadecimal:Grouping in bits of 4:1111undefined1011undefined1010undefined1111undefined\underlinesegment{1111} \quad \underlinesegment{1011} \quad \underlinesegment{1010} \quad \underlinesegment{1111}1111​1011​1010​1111​BinaryNumberEquivalentHexadecimal1111F (15)1010A (10)1011B (11)1111F (15)Therefore, (1111101110101111)2 = (FBAF)16Converting to Octal:Grouping in bits of 3:001undefined111undefined101undefined110undefined101undefined111undefined\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{101} \quad \underlinesegment{110} \quad \underlinesegment{101} \quad \underlinesegment{111}001​111​101​110​101​111​BinaryNumberEquivalentOctal111710151106101511170011Therefore, (1111101110101111)2 = (175657)8

Question 3

Question 7What is the most significant bit and the least significant bit in a binary code ?

Accepted Answer

In a binary code, the leftmost bit is called the most significant bit or MSB. It carries the largest weight. The rightmost bit is called the least significant bit or LSB. It carries the smallest weight. For example:1MSB0110110LSB\begin{matrix} \underset{\bold{MSB}}{1} & 0 & 1 & 1 & 0 & 1 & 1 & \underset{\bold{LSB}}{0} \end{matrix}MSB1​​0​1​1​0​1​1​LSB0​​

Question 4

Question 1Convert the following binary numbers to decimal:(a) 1101

Accepted Answer

BinaryNoPowerValueResult1 (LSB)2011x1=102120x2=012241x4=41 (MSB)2381x8=8Equivalent decimal number = 1 + 4 + 8 = 13Therefore, (1101)2 = (13)10(b) 111010BinaryNoPowerValueResult0 (LSB)2010x1=012121x2=202240x4=012381x8=8124161x16=161 (MSB)25321x32=32Equivalent decimal number = 2 + 8 + 16 + 32 = 58Therefore, (111010)2 = (58)10(c) 101011111BinaryNoPowerValueResult1 (LSB)2011x1=112121x2=212241x4=412381x8=8124161x16=16025320x32=0126641x64=640271280x128=01 (MSB)282561x256=256Equivalent decimal number = 1 + 2 + 4 + 8 + 16 + 64 + 256 = 351Therefore, (101011111)2 = (351)10

Question 5

Question 2Convert the following binary numbers to decimal :(a) 1100

Accepted Answer

BinaryNoPowerValueResult0 (LSB)2010x1=002120x2=012241x4=41 (MSB)2381x8=8Equivalent decimal number = 4 + 8 = 12Therefore, (1100)2 = (12)10(b) 10010101BinaryNoPowerValueResult1 (LSB)2011x1=102120x2=012241x4=402380x8=0124161x16=16025320x32=0026640x64=01 (MSB)271281x128=128Equivalent decimal number = 1 + 4 + 16 + 128 = 149Therefore, (10010101)2 = (149)10(c) 11011100BinaryNoPowerValueResult0 (LSB)2010x1=002120x2=012241x4=412381x8=8124161x16=16025320x32=0126641x64=641 (MSB)271281x128=128Equivalent decimal number = 4 + 8 + 16 + 64 + 128 = 220Therefore, (11011100)2 = (220)10

Question 6

Question 3Convert the following decimal numbers to binary:(a) 23

Accepted Answer

2QuotientRemainder2231 (LSB)2111251220211 (MSB) 0 Therefore, (23)10 = (10111)2(b) 1002QuotientRemainder21000 (LSB)250022512120260231211 (MSB) 0 Therefore, (100)10 = (1100100)2(c) 1452QuotientRemainder21451 (LSB)272023602180291240220211 (MSB) 0 Therefore, (145)10 = (10010001)2(d) 0.25Multiply=ResultantCarry0.25 x 2=0.500.5 x 2=01Therefore, (0.25)10 = (0.01)2

Question 7

Question 4Convert the following decimal numbers to binary:(a) 19

Accepted Answer

2QuotientRemainder2191 (LSB)291240220211 (MSB) 0 Therefore, (19)10 = (10011)2(b) 1222QuotientRemainder21220 (LSB)261123002151271231211 (MSB) 0 Therefore, (122)10 = (1111010)2(c) 1612QuotientRemainder21611 (LSB)2800240022002100251220211 (MSB) 0 Therefore, (161)10 = (10100001)2(d) 0.675Multiply=ResultantCarry0.675 x 2=0.3510.35 x 2=0.700.7 x 2=0.410.4 x 2=0.800.8 x 2=0.61(We stop after 5 iterations if fractional part doesn't become 0)Therefore, (0.675)10 = (0.10101)2

Question 8

Question 5Convert the following decimal numbers to octal:(a) 19

Accepted Answer

8QuotientRemainder8193 (LSB)822 (MSB) 0 Therefore, (19)10 = (23)8(b) 1228QuotientRemainder81222 (LSB)8157811 (MSB) 0 Therefore, (122)10 = (172)8(c) 1618QuotientRemainder81611 (LSB)8204822 (MSB) 0 Therefore, (161)10 = (241)8(d) 0.675Multiply=ResultantCarry0.675 x 8=0.450.4 x 8=0.230.2 x 8=0.610.6 x 8=0.840.8 x 8=0.46Therefore, (0.675)10 = (0.53146)8

Question 9

Question 8Convert the following binary numbers to hexadecimal:(a) 10011011101

Accepted Answer

Grouping in bits of 4:0100undefined1101undefined1101undefined\underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101}0100​1101​1101​BinaryNumberEquivalentHexadecimal1101D (13)1101D (13)01004Therefore, (10011011101)2 = (4DD)16(b) 1111011101011011Grouping in bits of 4:1111undefined0111undefined0101undefined1011undefined\underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011}1111​0111​0101​1011​BinaryNumberEquivalentHexadecimal1011B (11)01015011171111F (15)Therefore, (1111011101011011)2 = (F75B)16(c) 11010111010111Grouping in bits of 4:0011undefined0101undefined1101undefined0111undefined\underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111}0011​0101​1101​0111​BinaryNumberEquivalentHexadecimal011171101D (13)0101500113Therefore, (11010111010111)2 = (35D7)16

Question 10

Question 10Convert the following octal numbers to decimal:(a) 257

Accepted Answer

OctalNoPowerValueResult7 (LSB)8017x1=758185x8=402 (MSB)82642x64=128Equivalent decimal number = 7 + 40 + 128 = 175Therefore, (257)8 = (175)10(b) 3527OctalNoPowerValueResult7 (LSB)8017x1=728182x8=16582645x64=3203 (MSB)835123x512=1536Equivalent decimal number = 7 + 16 + 320 + 1536 = 1879Therefore, (3527)8 = (1879)10(c) 123OctalNoPowerValueResult3 (LSB)8013x1=328182x8=161 (MSB)82641x64=64Equivalent decimal number = 3 + 16 + 64 = 83Therefore, (123)8 = (83)10(d) 605.12Integral partOctalNoPowerValueResult58015x1=508180x8=0682646x64=384Fractional partOctalNoPowerValueResult18-10.1251x0.125=0.12528-20.01562x0.0156=0.0312Equivalent decimal number = 5 + 384 + 0.125 + 0.0312 = 389.1562Therefore, (605.12)8 = (389.1562)10

Question 11

Question 11Convert the following hexadecimal numbers to decimal:(a) A6

Accepted Answer

HexadecimalNumberPowerValueResult616016x1=6A (10)1611610x16=160Equivalent decimal number = 6 + 160 = 166Therefore, (A6)16 = (166)10(b) A13BHexadecimalNumberPowerValueResultB (11)160111x1=113161163x16=4811622561x256=256A (10)163409610x4096=40960Equivalent decimal number = 11 + 48 + 256 + 40960 = 41275Therefore, (A13B)16 = (41275)10(c) 3A5HexadecimalNumberPowerValueResult516015x1=5A (10)1611610x16=16031622563x256=768Equivalent decimal number = 5 + 160 + 768 = 933Therefore, (3A5)16 = (933)10

Question 12

Question 12Convert the following hexadecimal numbers to decimal:(a) E9

Accepted Answer

HexadecimalNumberPowerValueResult916019x1=9E (14)1611614x16=224Equivalent decimal number = 9 + 224 = 233Therefore, (E9)16 = (233)10(b) 7CA3HexadecimalNumberPowerValueResult3 (11)16013x1=3A (10)1611610x16=160C (12)16225612x256=3072716340967x4096=28672Equivalent decimal number = 3 + 160 + 3072 + 28672 = 31907Therefore, (7CA3)16 = (31907)10

Question 13

Question 13Convert the following decimal numbers to hexadecimal:(a) 132

Accepted Answer

16QuotientRemainder1613241688 0 Therefore, (132)10 = (84)16(b) 235216QuotientRemainder16235201614731699 0 Therefore, (2352)10 = (930)16(c) 12216QuotientRemainder16122A (10)1677 0 Therefore, (122)10 = (7A)16(d) 0.675Multiply=ResultantCarry0.675 x 16=0.8A (10)0.8 x 16=0.8C (12)0.8 x 16=0.8C (12)0.8 x 16=0.8C (12)0.8 x 16=0.8C (12)(We stop after 5 iterations if fractional part doesn't become 0)Therefore, (0.675)10 = (0.ACCCC)16

Question 14

Question 14Convert the following decimal numbers to hexadecimal:(a) 206

Accepted Answer

16QuotientRemainder16206E (14)1612C (12) 0 Therefore, (206)10 = (CE)16(b) 361916QuotientRemainder16361931622621614E (14) 0 Therefore, (3619)10 = (E23)16

Question 15

Question 15Convert the following hexadecimal numbers to octal:(a) 38AC

Accepted Answer

HexadecimalNumberBinaryEquivalentC (12)1100A (10)10108100030011(38AC)16 = (11100010101100)2Grouping in bits of 3:011undefined 100undefined 010undefined 101undefined 100undefined\underlinesegment{011}\medspace\underlinesegment{100}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{100}011​100​010​101​100​BinaryNumberEquivalentOctal10041015010210040113(38AC)16 = (34254)8(b) 7FD6HexadecimalNumberBinaryEquivalent60110D (13)1101F (15)111170111(7FD6)16 = (111111111010110)2Grouping in bits of 3:111undefined 111undefined 111undefined 010undefined 110undefined\underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{010}\medspace\underlinesegment{110}111​111​111​010​110​BinaryNumberEquivalentOctal11060102111711171117(7FD6)16 = (77726)8(c) ABCDHexadecimalNumberBinaryEquivalentD (13)1101C (12)1100B (11)1011A (10)1010(ABCD)16 = (1010101111001101)2Grouping in bits of 3:001undefined 010undefined 101undefined 111undefined 001undefined 101undefined\underlinesegment{001}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{111}\medspace\underlinesegment{001}\medspace\underlinesegment{101}001​010​101​111​001​101​BinaryNumberEquivalentOctal101500111117101501020011(ABCD)16 = (125715)8

Question 16

Question 16Convert the following octal numbers to binary:(a) 123

Accepted Answer

OctalNumberBinaryEquivalent301120101001Therefore, (123)8 = (001undefined 010undefined 011undefined\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{011}}001​010​011​)2(b) 3527OctalNumberBinaryEquivalent7111201051013011Therefore, (3527)8 = (011undefined 101undefined 010undefined 111undefined\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{111}}011​101​010​111​)2(c) 705OctalNumberBinaryEquivalent510100007111Therefore, (705)8 = (111undefined 000undefined 101undefined\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{101}}111​000​101​)2

Question 17

Question 17Convert the following octal numbers to binary:(a) 7642

Accepted Answer

OctalNumberBinaryEquivalent2010410061107111Therefore, (7642)8 = (111undefined 110undefined 100undefined 010undefined\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{010}}111​110​100​010​)2(b) 7015OctalNumberBinaryEquivalent5101100100007111Therefore, (7015)8 = (111undefined 000undefined 001undefined 101undefined\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{101}}111​000​001​101​)2(c) 3576OctalNumberBinaryEquivalent6110711151013011Therefore, (3576)8 = (011undefined 101undefined 111undefined 110undefined\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{110}}011​101​111​110​)2(d) 705OctalNumberBinaryEquivalent510100007111Therefore, (705)8 = (111undefined 000undefined 101undefined\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{101}}111​000​101​)2

Question 18

Question 18Convert the following binary numbers to octal(a) 111010

Accepted Answer

Grouping in bits of 3:111undefined010undefined\underlinesegment{111} \quad \underlinesegment{010}111​010​BinaryNumberEquivalentOctal01021117Therefore, (111010)2 = (72)8(b) 110110101Grouping in bits of 3:110undefined110undefined101undefined\underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{101}110​110​101​BinaryNumberEquivalentOctal101511061106Therefore, (110110101)2 = (665)8(c) 1101100001Grouping in bits of 3:001undefined101undefined100undefined001undefined\underlinesegment{001} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \underlinesegment{001}001​101​100​001​BinaryNumberEquivalentOctal0011100410150011Therefore, (1101100001)2 = (1541)8

Question 19

Question 19Convert the following binary numbers to octal(a) 11001

Accepted Answer

Grouping in bits of 3:011undefined001undefined\underlinesegment{011} \quad \underlinesegment{001}011​001​BinaryNumberEquivalentOctal00110113Therefore, (11001)2 = (31)8(b) 10101100Grouping in bits of 3:010undefined101undefined100undefined\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100}010​101​100​BinaryNumberEquivalentOctal100410150102Therefore, (10101100)2 = (254)8(c) 111010111Grouping in bits of 3:111undefined010undefined111undefined\underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111}111​010​111​BinaryNumberEquivalentOctal111701021117Therefore, (111010111)2 = (727)8

Question 20

Question 20Add the following binary numbers:(i) 10110111 and 1100101

Accepted Answer

1101110111111+1100101100011100\begin{matrix} & & \overset{1}{1} & \overset{1}{0} & 1 & 1 & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 \ + & & & 1 & 1 & 0 & 0 & 1 & 0 & 1 \ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{0} & \bold{1} & \bold{1} & \bold{1} & \bold{0} & \bold{0} \end{matrix}+​1​110​0110​110​101​0101​1111​1100​110​​Therefore, (10110111)2 + (1100101)2 = (100011100)2(ii) 110101 and 10111111110111011+1011111100100\begin{matrix} & & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{0} & 1 \ + & & 1 & 0 & 1 & 1 & 1 & 1 \ \hline & \bold{1} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{0} \end{matrix}+​1​1111​1100​0110​1111​0110​110​​Therefore, (110101)2 + (101111)2 = (1100100)2(iii) 110111.110 and 11011101.0100101111101111111.1110+11011101.010100010101.000\begin{matrix} & & \overset{1}{0} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{1} & . & \overset{1}{1} & 1 & 0 \ + & & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & . & 0 & 1 & 0 \ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{.} & \bold{0} & \bold{0} & \bold{0} \end{matrix}+​1​0110​0110​1100​1111​0110​1111​1100​1111​...​1100​110​000​​Therefore, (110111.110)2 + (11011101.010)2 = (100010101)2(iv) 1110.110 and 11010.011011111101.1110+11010.011101001.001\begin{matrix} & & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 & \overset{1}{0} & . & \overset{1}{1} & 1 & 0 \ + & & 1 & 1 & 0 & 1 & 0 & . & 0 & 1 & 1 \ \hline & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{.} & \bold{0} & \bold{0} & \bold{1} \end{matrix}+​1​0110​1111​1100​110​0101​...​1100​110​011​​Therefore, (1110.110)2 + (11010.011)2 = (101001.001)2

Data Representation

Checkpoint 2.1

Type A: Short Answer Questions

Type B: Application Based Questions