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IT Theory IEB Grade 10 2020 Unit 1

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A comprehensive summary of IT Theory on a Grade 10 Level. It contains all the necessary information in a colourful, table format. All information has been typed for improved legibility. These notes have been used by hundreds of learners throughout my school (including myself) and is guaranteed to get you an A. All information is relevant to the current (2nd) edition of my textbook, named below **This section covers Unit 1 of (EXPLORING IT: THEORY GRADE 10 2ND ED ISBN 3620)**

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Zeyn Mahomed Grade 10 2020




LU1 - Data Representation and Storage
1. Decimal, Binary and Hexadecimal Number Systems

Decimal (Base-10) Binary (Base-2) Hexadecimal (Base-16)

Binary Decimal Decimal
● Continuously divide by 2 ● Order the binary number in ● Order the hex number in
● Write down the remainder columns with increasing columns with increasing
● Repeat until the quotient is 0 powers of 2 powers of 16
● Read the remainders from ● Going from right to left, ● Going from right to left,
bottom to top starting at 0 starting at 0
● Multiply the bit by the ● Multiply the integer value
Base 140 Remainder
power of 2 in its column by the power of 16 in its
2 70 0 column
2 35 0 27 26 25 24 23 22 21 20 161 160
2 17 1 1 0 0 0 1 1 0 0
8 C
2 8 1
2 x 1 = 128 +
7
2 x0=0 6
8 x 161 12 x 160
2 4 0 25 x 0 = 0 + 24 x 0 = 0
2 x1=8
3
+ 22 x 1 = 4
128 12
2 2 0
2 x0=0
1
+ 20 x 0 = 0
1000 11002 = 14010
2 1 0 128 + 12 = 140
2 0 1
8C16 = 14010
14010 = 1000 11002

Hexadecimal Hexadecimal Binary
● Continuously divide by 16 ● Number of bits must be a ● First convert to decimal as
● Write down the remainder multiple of 4 shown above
● Repeat until the quotient is 0 ● Leading zeros must be ● Then use the decimal
● Read the remainders from added method to convert to
bottom to top ● Split the binary number into binary
groups of 4 bits
0 1 2 3 4 5 6 7 8 9
● Multiply each bit by the
0 1 2 3 4 5 6 7 8 9 multiple of 2
● Match the Hex value
10 11 12 13 14
1 0 0 0 1 1 0 0
A B C D E
x x x x x x x x
23 22 21 20 23 22 21 20
8C16 = 1000 11002
Base 140 Remainder
8 0 0 0 8 4 0 0
16 8 12
8=8 12 = c
16 0 8
161 162
8=8 12 = C 8=8
8 C
14010 = 8C16
1000 11002 = 8C16

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