No
Equivalent decimal number = 1 + 4 + 8 = 13
Therefore, (1101) 2 = (13) 10
Binary No | Power | Value | Result |
---|---|---|---|
0 | 2 | 1 | 0x1=0 |
1 | 2 | 2 | 1×2=2 |
0 | 2 | 4 | 0x4=0 |
1 | 2 | 8 | 1×8=8 |
1 | 2 | 16 | 1×16=16 |
1 | 2 | 32 | 1×32=32 |
Equivalent decimal number = 2 + 8 + 16 + 32 = 58
Therefore, (111010) 2 = (58) 10
(c) 101011111
Binary No | Power | Value | Result |
---|---|---|---|
1 | 2 | 1 | 1×1=1 |
1 | 2 | 2 | 1×2=2 |
1 | 2 | 4 | 1×4=4 |
1 | 2 | 8 | 1×8=8 |
1 | 2 | 16 | 1×16=16 |
0 | 2 | 32 | 0x32=0 |
1 | 2 | 64 | 1×64=64 |
0 | 2 | 128 | 0x128=0 |
1 | 2 | 256 | 1×256=256 |
Equivalent decimal number = 1 + 2 + 4 + 8 + 16 + 64 + 256 = 351
Therefore, (101011111) 2 = (351) 10
Convert the following binary numbers to decimal :
Binary No | Power | Value | Result |
---|---|---|---|
0 | 2 | 1 | 0x1=0 |
0 | 2 | 2 | 0x2=0 |
1 | 2 | 4 | 1×4=4 |
1 | 2 | 8 | 1×8=8 |
Equivalent decimal number = 4 + 8 = 12
Therefore, (1100) 2 = (12) 10
(b) 10010101
Binary No | Power | Value | Result |
---|---|---|---|
1 | 2 | 1 | 1×1=1 |
0 | 2 | 2 | 0x2=0 |
1 | 2 | 4 | 1×4=4 |
0 | 2 | 8 | 0x8=0 |
1 | 2 | 16 | 1×16=16 |
0 | 2 | 32 | 0x32=0 |
0 | 2 | 64 | 0x64=0 |
1 | 2 | 128 | 1×128=128 |
Equivalent decimal number = 1 + 4 + 16 + 128 = 149
Therefore, (10010101) 2 = (149) 10
(c) 11011100
Binary No | Power | Value | Result |
---|---|---|---|
0 | 2 | 1 | 0x1=0 |
0 | 2 | 2 | 0x2=0 |
1 | 2 | 4 | 1×4=4 |
1 | 2 | 8 | 1×8=8 |
1 | 2 | 16 | 1×16=16 |
0 | 2 | 32 | 0x32=0 |
1 | 2 | 64 | 1×64=64 |
1 | 2 | 128 | 1×128=128 |
Equivalent decimal number = 4 + 8 + 16 + 64 + 128 = 220
Therefore, (11011100) 2 = (220) 10
Convert the following decimal numbers to binary:
2 | Quotient | Remainder |
---|---|---|
2 | 23 | 1 (LSB) |
2 | 11 | 1 |
2 | 5 | 1 |
2 | 2 | 0 |
2 | 1 | 1 (MSB) |
0 |
Therefore, (23) 10 = (10111) 2
2 | Quotient | Remainder |
---|---|---|
2 | 100 | 0 (LSB) |
2 | 50 | 0 |
2 | 25 | 1 |
2 | 12 | 0 |
2 | 6 | 0 |
2 | 3 | 1 |
2 | 1 | 1 (MSB) |
0 |
Therefore, (100) 10 = (1100100) 2
2 | Quotient | Remainder |
---|---|---|
2 | 145 | 1 (LSB) |
2 | 72 | 0 |
2 | 36 | 0 |
2 | 18 | 0 |
2 | 9 | 1 |
2 | 4 | 0 |
2 | 2 | 0 |
2 | 1 | 1 (MSB) |
0 |
Therefore, (145) 10 = (10010001) 2
Multiply | = | Resultant | Carry |
---|---|---|---|
0.25 x 2 | = | 0.5 | 0 |
0.5 x 2 | = | 0 | 1 |
Therefore, (0.25) 10 = (0.01) 2
2 | Quotient | Remainder |
---|---|---|
2 | 19 | 1 (LSB) |
2 | 9 | 1 |
2 | 4 | 0 |
2 | 2 | 0 |
2 | 1 | 1 (MSB) |
0 |
Therefore, (19) 10 = (10011) 2
2 | Quotient | Remainder |
---|---|---|
2 | 122 | 0 (LSB) |
2 | 61 | 1 |
2 | 30 | 0 |
2 | 15 | 1 |
2 | 7 | 1 |
2 | 3 | 1 |
2 | 1 | 1 (MSB) |
0 |
Therefore, (122) 10 = (1111010) 2
2 | Quotient | Remainder |
---|---|---|
2 | 161 | 1 (LSB) |
2 | 80 | 0 |
2 | 40 | 0 |
2 | 20 | 0 |
2 | 10 | 0 |
2 | 5 | 1 |
2 | 2 | 0 |
2 | 1 | 1 (MSB) |
0 |
Therefore, (161) 10 = (10100001) 2
Multiply | = | Resultant | Carry |
---|---|---|---|
0.675 x 2 | = | 0.35 | 1 |
0.35 x 2 | = | 0.7 | 0 |
0.7 x 2 | = | 0.4 | 1 |
0.4 x 2 | = | 0.8 | 0 |
0.8 x 2 | = | 0.6 | 1 |
(We stop after 5 iterations if fractional part doesn’t become 0)
Therefore, (0.675) 10 = (0.10101) 2
Convert the following decimal numbers to octal:
8 | Quotient | Remainder |
---|---|---|
8 | 19 | 3 (LSB) |
8 | 2 | 2 (MSB) |
0 |
Therefore, (19) 10 = (23) 8
8 | Quotient | Remainder |
---|---|---|
8 | 122 | 2 (LSB) |
8 | 15 | 7 |
8 | 1 | 1 (MSB) |
0 |
Therefore, (122) 10 = (172) 8
8 | Quotient | Remainder |
---|---|---|
8 | 161 | 1 (LSB) |
8 | 20 | 4 |
8 | 2 | 2 (MSB) |
0 |
Therefore, (161) 10 = (241) 8
Multiply | = | Resultant | Carry |
---|---|---|---|
0.675 x 8 | = | 0.4 | 5 |
0.4 x 8 | = | 0.2 | 3 |
0.2 x 8 | = | 0.6 | 1 |
0.6 x 8 | = | 0.8 | 4 |
0.8 x 8 | = | 0.4 | 6 |
Therefore, (0.675) 10 = (0.53146) 8
Convert the following hexadecimal numbers to binary:
Hexadecimal Number | Binary Equivalent |
---|---|
6 | 0110 |
A (10) | 1010 |
(A6) 16 = (10100110) 2
Hexadecimal Number | Binary Equivalent |
---|---|
7 | 0111 |
0 | 0000 |
A (10) | 1010 |
(A07) 16 = (101000000111) 2
Hexadecimal Number | Binary Equivalent |
---|---|
4 | 0100 |
B (11) | 1011 |
A (10) | 1010 |
7 | 0111 |
(7AB4) 16 = (111101010110100) 2
Hexadecimal Number | Binary Equivalent |
---|---|
D (13) | 1101 |
3 | 0011 |
2 | 0010 |
(23D) 16 = (1000111101) 2
Hexadecimal Number | Binary Equivalent |
---|---|
9 | 1001 |
C (12) | 1100 |
B (11) | 1011 |
(BC9) 16 = (101111001001) 2
Hexadecimal Number | Binary Equivalent |
---|---|
8 | 1000 |
C (12) | 1100 |
B (11) | 1011 |
9 | 1001 |
(9BC8) 16 = (1001101111001000) 2
Convert the following binary numbers to hexadecimal:
(a) 10011011101
Grouping in bits of 4:
0100undefined 0100 1101 1101
Binary Number | Equivalent Hexadecimal |
---|---|
1101 | D (13) |
1101 | D (13) |
0100 | 4 |
Therefore, (10011011101) 2 = (4DD) 16
(b) 1111011101011011
1111undefined 1111 0111 0101 1011
Binary Number | Equivalent Hexadecimal |
---|---|
1011 | B (11) |
0101 | 5 |
0111 | 7 |
1111 | F (15) |
Therefore, (1111011101011011) 2 = (F75B) 16
(c) 11010111010111
0011undefined 0011 0101 1101 0111
Binary Number | Equivalent Hexadecimal |
---|---|
0111 | 7 |
1101 | D (13) |
0101 | 5 |
0011 | 3 |
Therefore, (11010111010111) 2 = (35D7) 16
(a) 1010110110111
0001undefined 0001 0101 1011 0111
Binary Number | Equivalent Hexadecimal |
---|---|
0111 | 7 |
1011 | B (11) |
0101 | 5 |
0001 | 1 |
Therefore, (1010110110111) 2 = (15B7) 16
(b) 10110111011011
0010undefined 0010 1101 1101 1011
Binary Number | Equivalent Hexadecimal |
---|---|
1011 | B (11) |
1101 | D (13) |
1101 | D (13) |
0010 | 2 |
Therefore, (10110111011011) 2 = (2DDB) 16
(c) 0110101100
0001undefined 0001 1010 1100
Binary Number | Equivalent Hexadecimal |
---|---|
1100 | C (12) |
1010 | A (10) |
0001 | 1 |
Therefore, (0110101100) 2 = (1AC) 16
Convert the following octal numbers to decimal:
Octal No | Power | Value | Result |
---|---|---|---|
7 | 8 | 1 | 7×1=7 |
5 | 8 | 8 | 5×8=40 |
2 | 8 | 64 | 2×64=128 |
Equivalent decimal number = 7 + 40 + 128 = 175
Therefore, (257) 8 = (175) 10
Octal No | Power | Value | Result |
---|---|---|---|
7 | 8 | 1 | 7×1=7 |
2 | 8 | 8 | 2×8=16 |
5 | 8 | 64 | 5×64=320 |
3 | 8 | 512 | 3×512=1536 |
Equivalent decimal number = 7 + 16 + 320 + 1536 = 1879
Therefore, (3527) 8 = (1879) 10
Octal No | Power | Value | Result |
---|---|---|---|
3 | 8 | 1 | 3×1=3 |
2 | 8 | 8 | 2×8=16 |
1 | 8 | 64 | 1×64=64 |
Equivalent decimal number = 3 + 16 + 64 = 83
Therefore, (123) 8 = (83) 10
Octal No | Power | Value | Result |
---|---|---|---|
5 | 8 | 1 | 5×1=5 |
0 | 8 | 8 | 0x8=0 |
6 | 8 | 64 | 6×64=384 |
Octal No | Power | Value | Result |
---|---|---|---|
1 | 8 | 0.125 | 1×0.125=0.125 |
2 | 8 | 0.0156 | 2×0.0156=0.0312 |
Equivalent decimal number = 5 + 384 + 0.125 + 0.0312 = 389.1562
Therefore, (605.12) 8 = (389.1562) 10
Convert the following hexadecimal numbers to decimal:
Hexadecimal Number | Power | Value | Result |
---|---|---|---|
6 | 16 | 1 | 6×1=6 |
A (10) | 16 | 16 | 10×16=160 |
Equivalent decimal number = 6 + 160 = 166
Therefore, (A6) 16 = (166) 10
Hexadecimal Number | Power | Value | Result |
---|---|---|---|
B (11) | 16 | 1 | 11×1=11 |
3 | 16 | 16 | 3×16=48 |
1 | 16 | 256 | 1×256=256 |
A (10) | 16 | 4096 | 10×4096=40960 |
Equivalent decimal number = 11 + 48 + 256 + 40960 = 41275
Therefore, (A13B) 16 = (41275) 10
Hexadecimal Number | Power | Value | Result |
---|---|---|---|
5 | 16 | 1 | 5×1=5 |
A (10) | 16 | 16 | 10×16=160 |
3 | 16 | 256 | 3×256=768 |
Equivalent decimal number = 5 + 160 + 768 = 933
Therefore, (3A5) 16 = (933) 10
Hexadecimal Number | Power | Value | Result |
---|---|---|---|
9 | 16 | 1 | 9×1=9 |
E (14) | 16 | 16 | 14×16=224 |
Equivalent decimal number = 9 + 224 = 233
Therefore, (E9) 16 = (233) 10
Hexadecimal Number | Power | Value | Result |
---|---|---|---|
3 (11) | 16 | 1 | 3×1=3 |
A (10) | 16 | 16 | 10×16=160 |
C (12) | 16 | 256 | 12×256=3072 |
7 | 16 | 4096 | 7×4096=28672 |
Equivalent decimal number = 3 + 160 + 3072 + 28672 = 31907
Therefore, (7CA3) 16 = (31907) 10
Convert the following decimal numbers to hexadecimal:
16 | Quotient | Remainder |
---|---|---|
16 | 132 | 4 |
16 | 8 | 8 |
0 |
Therefore, (132) 10 = (84) 16
16 | Quotient | Remainder |
---|---|---|
16 | 2352 | 0 |
16 | 147 | 3 |
16 | 9 | 9 |
0 |
Therefore, (2352) 10 = (930) 16
16 | Quotient | Remainder |
---|---|---|
16 | 122 | A (10) |
16 | 7 | 7 |
0 |
Therefore, (122) 10 = (7A) 16
Multiply | = | Resultant | Carry |
---|---|---|---|
0.675 x 16 | = | 0.8 | A (10) |
0.8 x 16 | = | 0.8 | C (12) |
0.8 x 16 | = | 0.8 | C (12) |
0.8 x 16 | = | 0.8 | C (12) |
0.8 x 16 | = | 0.8 | C (12) |
Therefore, (0.675) 10 = (0.ACCCC) 16
16 | Quotient | Remainder |
---|---|---|
16 | 206 | E (14) |
16 | 12 | C (12) |
0 |
Therefore, (206) 10 = (CE) 16
16 | Quotient | Remainder |
---|---|---|
16 | 3619 | 3 |
16 | 226 | 2 |
16 | 14 | E (14) |
0 |
Therefore, (3619) 10 = (E23) 16
Convert the following hexadecimal numbers to octal:
Hexadecimal Number | Binary Equivalent |
---|---|
C (12) | 1100 |
A (10) | 1010 |
8 | 1000 |
3 | 0011 |
(38AC) 16 = (11100010101100) 2
Grouping in bits of 3:
011undefined 100undefined 010undefined 101undefined 100undefined 011 100 010 101 100
Binary Number | Equivalent Octal |
---|---|
100 | 4 |
101 | 5 |
010 | 2 |
100 | 4 |
011 | 3 |
(38AC) 16 = (34254) 8
Hexadecimal Number | Binary Equivalent |
---|---|
6 | 0110 |
D (13) | 1101 |
F (15) | 1111 |
7 | 0111 |
(7FD6) 16 = (111111111010110) 2
111undefined 111undefined 111undefined 010undefined 110undefined 111 111 111 010 110
Binary Number | Equivalent Octal |
---|---|
110 | 6 |
010 | 2 |
111 | 7 |
111 | 7 |
111 | 7 |
(7FD6) 16 = (77726) 8
Hexadecimal Number | Binary Equivalent |
---|---|
D (13) | 1101 |
C (12) | 1100 |
B (11) | 1011 |
A (10) | 1010 |
(ABCD) 16 = (1010101111001101) 2
001undefined 010undefined 101undefined 111undefined 001undefined 101undefined 001 010 101 111 001 101
Binary Number | Equivalent Octal |
---|---|
101 | 5 |
001 | 1 |
111 | 7 |
101 | 5 |
010 | 2 |
001 | 1 |
(ABCD) 16 = (125715) 8
Convert the following octal numbers to binary:
Octal Number | Binary Equivalent |
---|---|
3 | 011 |
2 | 010 |
1 | 001 |
Therefore, (123) 8 = ( 001undefined 010undefined 011undefined 001 010 011 ) 2
Octal Number | Binary Equivalent |
---|---|
7 | 111 |
2 | 010 |
5 | 101 |
3 | 011 |
Therefore, (3527) 8 = ( 011undefined 101undefined 010undefined 111undefined 011 101 010 111 ) 2
Octal Number | Binary Equivalent |
---|---|
5 | 101 |
0 | 000 |
7 | 111 |
Therefore, (705) 8 = ( 111undefined 000undefined 101undefined 111 000 101 ) 2
Octal Number | Binary Equivalent |
---|---|
2 | 010 |
4 | 100 |
6 | 110 |
7 | 111 |
Therefore, (7642) 8 = ( 111undefined 110undefined 100undefined 010undefined 111 110 100 010 ) 2
Octal Number | Binary Equivalent |
---|---|
5 | 101 |
1 | 001 |
0 | 000 |
7 | 111 |
Therefore, (7015) 8 = ( 111undefined 000undefined 001undefined 101undefined 111 000 001 101 ) 2
Octal Number | Binary Equivalent |
---|---|
6 | 110 |
7 | 111 |
5 | 101 |
3 | 011 |
Therefore, (3576) 8 = ( 011undefined 101undefined 111undefined 110undefined 011 101 111 110 ) 2
Convert the following binary numbers to octal
111undefined 111 010
Binary Number | Equivalent Octal |
---|---|
010 | 2 |
111 | 7 |
Therefore, (111010) 2 = (72) 8
(b) 110110101
110undefined 110 110 101
Binary Number | Equivalent Octal |
---|---|
101 | 5 |
110 | 6 |
110 | 6 |
Therefore, (110110101) 2 = (665) 8
(c) 1101100001
001undefined 001 101 100 001
Binary Number | Equivalent Octal |
---|---|
001 | 1 |
100 | 4 |
101 | 5 |
001 | 1 |
Therefore, (1101100001) 2 = (1541) 8
011undefined 011 001
Binary Number | Equivalent Octal |
---|---|
001 | 1 |
011 | 3 |
Therefore, (11001) 2 = (31) 8
(b) 10101100
010undefined 010 101 100
Binary Number | Equivalent Octal |
---|---|
100 | 4 |
101 | 5 |
010 | 2 |
Therefore, (10101100) 2 = (254) 8
(c) 111010111
111undefined 111 010 111
Binary Number | Equivalent Octal |
---|---|
111 | 7 |
010 | 2 |
111 | 7 |
Therefore, (111010111) 2 = (727) 8
Add the following binary numbers:
(i) 10110111 and 1100101
1101110111111+1100101100011100 + 1 1 1 0 0 1 1 0 11 0 10 1 0 1 0 1 1 1 1 1 1 1 0 0 11 0
Therefore, (10110111) 2 + (1100101) 2 = (100011100) 2
(ii) 110101 and 101111
11110111011+1011111100100 + 1 1 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 0 1 1 0 11 0
Therefore, (110101) 2 + (101111) 2 = (1100100) 2
(iii) 110111.110 and 11011101.010
0101111101111111.1110+11011101.010100010101.000 + 1 0 1 1 0 0 1 1 0 1 1 0 0 1 1 1 1 0 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 .. . 1 1 0 0 11 0 00 0
Therefore, (110111.110) 2 + (11011101.010) 2 = (100010101) 2
(iv) 1110.110 and 11010.011
011111101.1110+11010.011101001.001 + 1 0 1 1 0 1 1 1 1 1 1 0 0 11 0 0 1 0 1 .. . 1 1 0 0 11 0 01 1
Therefore, (1110.110) 2 + (11010.011) 2 = (101001.001) 2
Given that A’s code point in ASCII is 65, and a’s code point is 97. What is the binary representation of ‘A’ in ASCII ? (and what’s its hexadecimal representation). What is the binary representation of ‘a’ in ASCII ?
Binary representation of ‘A’ in ASCII will be binary representation of its code point 65.
Converting 65 to binary:
2 | Quotient | Remainder |
---|---|---|
2 | 65 | 1 (LSB) |
2 | 32 | 0 |
2 | 16 | 0 |
2 | 8 | 0 |
2 | 4 | 0 |
2 | 2 | 0 |
2 | 1 | 1 (MSB) |
0 |
Therefore, binary representation of ‘A’ in ASCII is 1000001.
Converting 65 to Hexadecimal:
16 | Quotient | Remainder |
---|---|---|
16 | 65 | 1 |
16 | 4 | 4 |
0 |
Therefore, hexadecimal representation of ‘A’ in ASCII is (41) 16 .
Similarly, converting 97 to binary:
2 | Quotient | Remainder |
---|---|---|
2 | 97 | 1 (LSB) |
2 | 48 | 0 |
2 | 24 | 0 |
2 | 12 | 0 |
2 | 6 | 0 |
2 | 3 | 1 |
2 | 1 | 1 (MSB) |
0 |
Therefore, binary representation of ‘a’ in ASCII is 1100001.
Convert the following binary numbers to decimal, octal and hexadecimal numbers.
(i) 100101.101
Decimal Conversion of integral part:
Binary No | Power | Value | Result |
---|---|---|---|
1 | 2 | 1 | 1×1=1 |
0 | 2 | 2 | 0x2=0 |
1 | 2 | 4 | 1×4=4 |
0 | 2 | 8 | 0x8=0 |
0 | 2 | 16 | 0x16=0 |
1 | 2 | 32 | 1×32=32 |
Decimal Conversion of fractional part:
Binary No | Power | Value | Result |
---|---|---|---|
1 | 2 | 0.5 | 1×0.5=0.5 |
0 | 2 | 0.25 | 0x0.25=0 |
1 | 2 | 0.125 | 1×0.125=0.125 |
Equivalent decimal number = 1 + 4 + 32 + 0.5 + 0.125 = 37.625
Therefore, (100101.101) 2 = (37.625) 10
Octal Conversion
100undefined 100 101 . 101
Binary Number | Equivalent Octal |
---|---|
101 | 5 |
100 | 4 |
101 | 5 |
Therefore, (100101.101) 2 = (45.5) 8
Hexadecimal Conversion
0010undefined 0010 0101 . 1010
Binary Number | Equivalent Hexadecimal |
---|---|
0101 | 5 |
0010 | 2 |
. | |
1010 | A (10) |
Therefore, (100101.101) 2 = (25.A) 16
(ii) 10101100.01011
Binary No | Power | Value | Result |
---|---|---|---|
0 | 2 | 1 | 0x1=0 |
0 | 2 | 2 | 0x2=0 |
1 | 2 | 4 | 1×4=4 |
1 | 2 | 8 | 1×8=8 |
0 | 2 | 16 | 0x16=0 |
1 | 2 | 32 | 1×32=32 |
0 | 2 | 64 | 0x64=0 |
1 | 2 | 128 | 1×128=128 |
Binary No | Power | Value | Result |
---|---|---|---|
0 | 2 | 0.5 | 0x0.5=0 |
1 | 2 | 0.25 | 1×0.25=0.25 |
0 | 2 | 0.125 | 0x0.125=0 |
1 | 2 | 0.0625 | 1×0.0625=0.0625 |
1 | 2 | 0.03125 | 1×0.03125=0.03125 |
Equivalent decimal number = 4 + 8 + 32 + 128 + 0.25 + 0.0625 + 0.03125 = 172.34375
Therefore, (10101100.01011) 2 = (172.34375) 10
010undefined 010 101 100 . 010 110
Binary Number | Equivalent Octal |
---|---|
100 | 4 |
101 | 5 |
010 | 2 |
010 | 2 |
110 | 6 |
Therefore, (10101100.01011) 2 = (254.26) 8
1010undefined 1010 1100 . 0101 1000
Binary Number | Equivalent Hexadecimal |
---|---|
1100 | C (12) |
1010 | A (10) |
. | |
0101 | 5 |
1000 | 8 |
Therefore, (10101100.01011) 2 = (AC.58) 16
Decimal Conversion:
Binary No | Power | Value | Result |
---|---|---|---|
0 | 2 | 1 | 0x1=0 |
1 | 2 | 2 | 1×2=2 |
0 | 2 | 4 | 0x4=0 |
1 | 2 | 8 | 1×8=8 |
Equivalent decimal number = 2 + 8 = 10
Therefore, (1010) 2 = (10) 10
001undefined 001 010
Binary Number | Equivalent Octal |
---|---|
010 | 2 |
001 | 1 |
Therefore, (1010) 2 = (12) 8
1010undefined 1010
Binary Number | Equivalent Hexadecimal |
---|---|
1010 | A (10) |
Therefore, (1010) 2 = (A) 16
(iv) 10101100.010111
Binary No | Power | Value | Result |
---|---|---|---|
0 | 2 | 0.5 | 0x0.5=0 |
1 | 2 | 0.25 | 1×0.25=0.25 |
0 | 2 | 0.125 | 0x0.125=0 |
1 | 2 | 0.0625 | 1×0.0625=0.0625 |
1 | 2 | 0.03125 | 1×0.03125=0.03125 |
1 | 2 | 0.015625 | 1×0.015625=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
010undefined 010 101 100 . 010 111
Binary Number | Equivalent Octal |
---|---|
100 | 4 |
101 | 5 |
010 | 2 |
010 | 2 |
111 | 7 |
Therefore, (10101100.010111) 2 = (254.27) 8
1010undefined 1010 1100 . 0101 1100
Binary Number | Equivalent Hexadecimal |
---|---|
1100 | C (12) |
1010 | A (10) |
. | |
0101 | 5 |
1100 | C (12) |
Therefore, (10101100.010111) 2 = (AC.5C) 16
print() takes a few additional arguments that provide modest control over the format of the output. Each of these is a special type of argument called a keyword argument.
Point to remember:
Keyword arguments have the form <keyword>=<value>.
Any keyword arguments passed to print() must come at the end, after the list of objects to display.
Adding the keyword argument sep=<str> causes objects to be separated by the string <str> instead of the default single space:
>>> print(‘foo’, 42, ‘bar’)
>>> print(‘foo’, 42, ‘bar’, sep=’/’)
>>> print(‘foo’, 42, ‘bar’, sep=’ ‘)
>>> d = {‘foo’: 1, ‘bar’: 2, ‘baz’: 3}
>>> for k, v in d.items():
print(k, v, sep=’ -> ‘)
foo -> 1
bar -> 2
baz -> 3
The keyword argument end=<str> causes output to be terminated by <str> instead of the default newline:
print(‘foo’, end=’/’)
print(42, end=’/’)
print(‘bar’)
Computer science, 2023-24 syllabus.
Data representation in computer MCQ . Questions and answers with PDF for all Computer Related Entrance & Competitive Exams Preparation. Helpful for Class 11, GATE, IBPS, SBI (Bank PO & Clerk), SSC, Railway etc.
1. To perform calculation on stored data computer, uses ……… number system. [SBI Clerk 2009]
(1) decimal
(2) hexadecimal
2. The number system based on ‘0’ and ‘1’ only, is known as
(1) binary system
(2) barter system
(3) number system
(4) hexadecimal system
3. Decimal number system is the group of ………… numbers.
(4) 0 to 9 and A to F
4. A hexadecimal number is represented by
(1) three digits
(2) four binary digits
(3) four digits
(4) All of these
5. A hexadigit can be represented by [IBPS Clerk 2012]
(1) three binary (consecutive) bits
(2) four binary (consecutive) bits
(3) eight binary (consecutive) bits
(4) sixteen binary (consecutive) bits
(5) None of the above
6. What type of information system would be recognised by digital circuits?
(1) Hexadecimal system
(2) Binary system
(3) Both ‘1’ and ‘2’
(4) Only roman system
7. The binary equivalent of decimal number 98 is [IBPS Clerk 2012]
(1) 1110001
(2) 1110100
(3) 1100010
(4) 1111001
(5) None of these
8. What is the value of the binary number 101?
9. The binary number 10101 is equivalent to decimal number ………….
10. To convert binary number to decimal, multiply the all binary digits by power of
11. Which of the following is a hexadecimal number equal to 3431 octal number?
12. LSD stands for
(1) Long Significant Digit
(2) Least Significant Digit
(3) Large Significant Digit
(4) Longer Significant Decimal
13. How many values can be represented by a single byte?
14. Which of the following is not a computer code?
(4) UNICODE
15. MSD refers as
(1) Most Significant Digit
(2) Many Significant Digit
(3) Multiple Significant Digit
(4) Most Significant Decimal
16. The most widely used code that represents each character as a unique 8-bit code is [IBPS Clerk 2011]
(2) UNICODE
17. Today’s mostly used coding system is/are
(4) Both ‘1’ and ‘2’
18. In EBCDIC code, maximum possible characters set size is
19. Code ‘EBCDIC’ that is used in computing stands for
(1) Extension BCD Information Code
(2) Extended BCD Information Code
(3) Extension BCD Interchange Conduct
(4) Extended BCD Interchange Conduct
20. Most commonly used codes for representing bits are
21. The coding system allows non-english characters and special characters to be represented
22. Which of the following is invalid hexadecimal number?
23. Gate having output 1 only when one of its input is 1 is called
24. Gate is also known as inverter.
25. The only function of NOT gate is to ……..
(1) Stop signal
(2) Invert input signal
(3) Act as a universal gate
(4) Double input signal
26. Which of following are known as universal gates?
(1) NAND & NOR
(2) AND & OR
(3) XOR & OR
27. Gate whose output is 0 only when inputs are different is called
28. In the binary language, each letter of the alphabet, each number and each special character is made up of a unique combination of [BOB Clerk 2010]
c) 8 character
29. Decimal equivalent of (1111) 2 is [IBPS Clerk 2012]
30. ASCII code for letter A is
a) 1100011
b) 1000001
c) 1111111
31. Which of the following is not a binary number? [IBPS Clerk 2011]
32. Which of the following is an example of binary number? [IBPS Clerk 2011]
33. Numbers that are written with base 10 are classified as
(1) decimal number
(2) whole number
(3) hexadecimal number
(4) exponential integers
34. The octal system [IBPS Clerk 2011]
(1) needs less digits to represent a number than in the binary system
(2) needs more digits to represent a number than in the binary system
(3) needs the same number of digits to represent a number as in the binary system
(4) needs the same number of digits to represent a number as in the decimal system
35. Hexadecimal number system has ………. base.
36. ASCII stands for [IBPS Clerk 2011,2014]
(1) American Special Computer for Information Interaction
(2) American Standard Computer for Information Interchange
(3) American Special Code for Information Interchange
(4) American Special Computer for Information Interchange
(5) American Standard Code for Information Interchange
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1. What is data representation in computer science? |
2. How is data represented in Python? |
3. What are the advantages of using binary data representation in computers? |
4. How does data representation affect the performance of a computer system? |
5. What are some common data representation errors in Python programming? |
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Data can be anything, including a number, a name, musical notes, or the colour of an image. The way that we stored, processed, and transmitted data is referred to as data representation. We can use any device, including computers, smartphones, and iPads, to store data in digital format. The stored data is handled by electronic circuitry. A bit is a 0 or 1 used in digital data representation.
Data Representation Techniques
Computer scans are classified broadly based on their speed and computing power.
1. Microcomputers or PCs (Personal Computers): It is a single-user computer system with a medium-power microprocessor. It is referred to as a computer with a microprocessor as its central processing unit.
Microcomputer
2. Mini-Computer: It is a multi-user computer system that can support hundreds of users at the same time.
Types of Mini Computers
3. Mainframe Computer: It is a multi-user computer system that can support hundreds of users at the same time. Software technology is distinct from minicomputer technology.
Mainframe Computer
4. Super-Computer: With the ability to process hundreds of millions of instructions per second, it is a very quick computer. They are used for specialised applications requiring enormous amounts of mathematical computations, but they are very expensive.
Supercomputer
Every value saved to or obtained from computer memory uses a specific number system, which is the method used to represent numbers in the computer system architecture. One needs to be familiar with number systems in order to read computer language or interact with the system.
Types of Number System
There are only two digits in a binary number system: 0 and 1. In this number system, 0 and 1 stand in for every number (value). Because the binary number system only has two digits, its base is 2.
A bit is another name for each binary digit. The binary number system is also a positional value system, where each digit's value is expressed in powers of 2.
The following are the primary characteristics of the binary system:
It only has two digits, zero and one.
Depending on its position, each digit has a different value.
Each position has the same value as a base power of two.
Because computers work with internal voltage drops, it is used in all types of computers.
Binary Number System
The decimal number system is a base ten number system with ten digits ranging from 0 to 9. This means that these ten digits can represent any numerical quantity. A positional value system is also a decimal number system. This means that the value of digits will be determined by their position.
Ten units of a given order equal one unit of the higher order, making it a decimal system.
The number 10 serves as the foundation for the decimal number system.
The value of each digit or number will depend on where it is located within the numeric figure because it is a positional system.
The value of this number results from multiplying all the digits by each power.
Decimal Number System
Decimal | Binary |
0 | 0000 |
1 | 0001 |
2 | 0010 |
3 | 0011 |
4 | 0100 |
5 | 0101 |
6 | 0110 |
7 | 0111 |
8 | 1000 |
9 | 1001 |
10 | 1010 |
11 | 1011 |
12 | 1100 |
13 | 1101 |
14 | 1110 |
15 | 1111 |
There are only eight (8) digits in the octal number system, from 0 to 7. In this number system, each number (value) is represented by the digits 0, 1, 2, 3,4,5,6, and 7. Since the octal number system only has 8 digits, its base is 8.
Contains eight digits: 0,1,2,3,4,5,6,7.
Also known as the base 8 number system.
Each octal number position represents a 0 power of the base (8).
An octal number's last position corresponds to an x power of the base (8).
Octal Number System
There are sixteen (16) alphanumeric values in the hexadecimal number system, ranging from 0 to 9 and A to F. In this number system, each number (value) is represented by 0, 1, 2, 3, 5, 6, 7, 8, 9, A, B, C, D, E, and F. Because the hexadecimal number system has 16 alphanumeric values, its base is 16. Here, the numbers are A = 10, B = 11, C = 12, D = 13, E = 14, and F = 15.
A system of positional numbers.
Has 16 symbols or digits overall (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F). Its base is, therefore, 16.
Decimal values 10, 11, 12, 13, 14, and 15 are represented by the letters A, B, C, D, E, and F, respectively.
A single digit may have a maximum value of 15.
Each digit position corresponds to a different base power (16).
Since there are only 16 digits, any hexadecimal number can be represented in binary with 4 bits.
Hexadecimal Number System
So, we've seen how to convert decimals and use the Number System to communicate with a computer. The full character set of the English language, which includes all alphabets, punctuation marks, mathematical operators, special symbols, etc., must be supported by the computer in addition to numerical data.
Choose the correct answer:.
1. Which computer is the largest in terms of size?
Minicomputer
Micro Computer
2. The binary number 11011001 is converted to what decimal value?
1. Give some examples where Supercomputers are used.
Ans: Weather Prediction, Scientific simulations, graphics, fluid dynamic calculations, Nuclear energy research, electronic engineering and analysis of geological data.
2. Which of these is the most costly?
Mainframe computer
Ans: C) Supercomputer
1. What is the distinction between the Hexadecimal and Octal Number System?
The octal number system is a base-8 number system in which the digits 0 through 7 are used to represent numbers. The hexadecimal number system is a base-16 number system that employs the digits 0 through 9 as well as the letters A through F to represent numbers.
2. What is the smallest data representation?
The smallest data storage unit in a computer's memory is called a BYTE, which comprises 8 BITS.
3. What is the largest data unit?
The largest commonly available data storage unit is a terabyte or TB. A terabyte equals 1,000 gigabytes, while a tebibyte equals 1,024 gibibytes.
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Get answers to all exercises of Chapter 2: Data Representation Sumita Arora Computer Science with Python CBSE Class 11 book. Clear your computer doubts instantly & get more marks in computers exam easily. Master the concepts with our detailed explanations & solutions.
Contents NCERT Solutions for Class 11 Computer Science (Python) - Data Representation TOPIC 1 Number System and Its Conversion Short Answer Type Questions-II Question 1: Explain octal and hexadecimal number. Answer: Octal (base 8) was previously a popular choice for representing digital circuit numbers in a form that is more compact than binary. Octal is […]
During the Class: Students can utilise the Data Representation questions from the Class 11 Computer Science NCERT Solutions during the class as an additional supplement. By using it as an additional supplement, students can understand the concepts of Data Representation in a better way.
Download Button will appear in 25 Seconds. NCERT Solutions Class 11 18. Computer Science Data Representation. Data Representation is a chapter in 18. Computer Science Solutions of Class 11 which is issued by NCERT, These Solutionss are considered to be the best for preparing for competitive exam and having clear understanding of concept.
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Chapterwise Question Bank CBSE Class 11 Computer Science (Python) Unit 1 : Computer Fundamentals 1. Computer overview and its Basics 2. Software Concepts 3. Data Representation 4. Microprocessor and Memory Concepts Unit 2 : Programming Methodology 5. Programming Methodology 6. Algorithms and Flowcharts Unit 3 : Introduction to Python 7. Introduction to Python 8. Getting […]
This page contains the CBSE class 11 Computer Science with Python chapter 3, Data Representation in Computers .You can find the questions/answers/solutions for the chapter 3 of Unit 1 of CBSE class 11 Computer Science with Python in this page.
Data Representation (Part 1) | Number System | Class 11 Computer ScienceIn this video, you will understand the Introduction of Digital Number System and how...
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Chapter 3-Data Representation I PUC, MDRPUC, Hassan 5 | P a g e. 2. Steps to convert decimal fraction number to binary number: Step 1: Multiply the given decimal fraction number by 2. Step 2: Note the carry and the product. Step 3: Repeat the Step 1 and Step 2 until the decimal number cannot be divided further. Step 4: The first carry will be ...
Videos. Free Notes for 11th Class Computers Science Data Representation.
The way that we stored, processed, and transmitted data is referred to as data representation. We can use any device, including computers, smartphones, and iPads, to store data in digital format. The stored data is handled by electronic circuitry. A bit is a 0 or 1 used in digital data representation. Data Representation Techniques.
NCERT Class 11 Computer Science Data Representation in Computers - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The document discusses various number systems used in computers like binary, octal, decimal, and hexadecimal. It provides methods to convert between these number systems. Key points covered include: - Binary, octal, decimal, and hexadecimal number ...
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