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CBSE Class 11 Computer Science Question 91 of 91

Data Representation — Question 22

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Question 22

Convert the following binary numbers to decimal, octal and hexadecimal numbers.

(i) 100101.101

Answer

Decimal Conversion of integral part:

Binary
No		PowerValueResult
12011x1=1
02120x2=0
12241x4=4
02380x8=0
024160x16=0
125321x32=32

Decimal Conversion of fractional part:

Binary
No		PowerValueResult
12-10.51x0.5=0.5
02-20.250x0.25=0
12-30.1251x0.125=0.125

Equivalent decimal number = 1 + 4 + 32 + 0.5 + 0.125 = 37.625

Therefore, (100101.101)2 = (37.625)10

Octal Conversion

Grouping in bits of 3:

100undefined101undefined.101undefined\underlinesegment{100} \quad \underlinesegment{101} \quad \bold{.} \quad \underlinesegment{101}

Binary
Number		Equivalent
Octal
1015
1004
..
1015

Therefore, (100101.101)2 = (45.5)8

Hexadecimal Conversion

Grouping in bits of 4:

0010undefined0101undefined . 1010undefined\underlinesegment{0010} \quad \underlinesegment{0101} \medspace . \medspace \underlinesegment{1010}

Binary
Number		Equivalent
Hexadecimal
01015
00102
. 
1010A (10)

Therefore, (100101.101)2 = (25.A)16

(ii) 10101100.01011

Decimal Conversion of integral part:

Binary
No		PowerValueResult
02010x1=0
02120x2=0
12241x4=4
12381x8=8
024160x16=0
125321x32=32
026640x64=0
1271281x128=128

Decimal Conversion of fractional part:

Binary
No		PowerValueResult
02-10.50x0.5=0
12-20.251x0.25=0.25
02-30.1250x0.125=0
12-40.06251x0.0625=0.0625
12-50.031251x0.03125=0.03125

Equivalent decimal number = 4 + 8 + 32 + 128 + 0.25 + 0.0625 + 0.03125 = 172.34375

Therefore, (10101100.01011)2 = (172.34375)10

Octal Conversion

Grouping in bits of 3:

010undefined101undefined100undefined.010undefined110undefined\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{110}

Binary
Number		Equivalent
Octal
1004
1015
0102
..
0102
1106

Therefore, (10101100.01011)2 = (254.26)8

Hexadecimal Conversion

Grouping in bits of 4:

1010undefined1100undefined . 0101undefined 1000undefined\underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1000}

Binary
Number		Equivalent
Hexadecimal
1100C (12)
1010A (10)
. 
01015
10008

Therefore, (10101100.01011)2 = (AC.58)16

(iii) 1010

Decimal Conversion:

Binary
No		PowerValueResult
02010x1=0
12121x2=2
02240x4=0
12381x8=8

Equivalent decimal number = 2 + 8 = 10

Therefore, (1010)2 = (10)10

Octal Conversion

Grouping in bits of 3:

001undefined010undefined\underlinesegment{001} \quad \underlinesegment{010}

Binary
Number		Equivalent
Octal
0102
0011

Therefore, (1010)2 = (12)8

Hexadecimal Conversion

Grouping in bits of 4:

1010undefined\underlinesegment{1010}

Binary
Number		Equivalent
Hexadecimal
1010A (10)

Therefore, (1010)2 = (A)16

(iv) 10101100.010111

Decimal Conversion of integral part:

Binary
No		PowerValueResult
02010x1=0
02120x2=0
12241x4=4
12381x8=8
024160x16=0
125321x32=32
026640x64=0
1271281x128=128

Decimal Conversion of fractional part:

Binary
No		PowerValueResult
02-10.50x0.5=0
12-20.251x0.25=0.25
02-30.1250x0.125=0
12-40.06251x0.0625=0.0625
12-50.031251x0.03125=0.03125
12-60.0156251x0.015625=0.015625

Equivalent decimal number = 4 + 8 + 32 + 128 + 0.25 + 0.0625 + 0.03125 + 0.015625 = 172.359375

Therefore, (10101100.010111)2 = (172.359375)10

Octal Conversion

Grouping in bits of 3:

010undefined101undefined100undefined.010undefined111undefined\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{111}

Binary
Number		Equivalent
Octal
1004
1015
0102
..
0102
1117

Therefore, (10101100.010111)2 = (254.27)8

Hexadecimal Conversion

Grouping in bits of 4:

1010undefined1100undefined . 0101undefined 1100undefined\underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1100}

Binary
Number		Equivalent
Hexadecimal
1100C (12)
1010A (10)
. 
01015
1100C (12)

Therefore, (10101100.010111)2 = (AC.5C)16