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Prepare for your exams
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In these notes we provide the imp questions from ch 2 to prepare for your exam these questions works as a notes also
Typology: Exercises
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(i) 13: Repeated division by 2. 13 ÷ 2 = 6 (R1), 6 ÷ 2 = 3 (R0), 3 ÷ 2 = 1 (R1), 1 ÷ 2 = 0 (R1) Answer: (13)₁₀ = (1101)₂ (ii) 106: 106 →53(R0)→26(R1)→13(R0)→6(R1)→3(R0)→1(R1)→0(R1) Answer: (106)₁₀ = (1101010)₂ (iii) 3.25: Int: 3 = 11₂. Frac: 0.25×2 = 0.5(0), 0.5×2 = 1.0(1) Answer: (3.25)₁₀ = (11.01)₂
Answer: (423)₁₀ = (1A7)₁₆ 72905: 72905÷16=4556(R9), 4556÷16=284(RC), 284÷16=17(RC), 17÷16=1(R1) Answer: (72905)₁₀ = (11CC9)₁₆
Q.2 (i) 10010₂: 1×2⁴ + 0×2³ + 0×2² + 1×2¹ + 0×2⁰ = 16 + 2 = 18
Q.2 (iii) 101010.01₂: 42 + (0×2⁻¹ + 1×2⁻²) = 42 + 0.25 = 42. Q.5 (2C9)₁₆: 2×16² + 12×16¹ + 9×16⁰ = 512 + 192 + 9 = 713 Q.7 (EB4A)₁₆: 14×16³ + 11×16² + 4×16¹ + 10 = 57344 + 2816 + 64 + 10 = 60234
3. Grouping Conversions (Hex/Octal)
Q.3 Binary to Hex (Group 4): (1101110101110)₂ → 0001 | 1011 | 1010 | 1110 = 1BAE₁₆ Q.10 Binary to Octal (Group 3): (11111011110101)₂ → 011 | 110 | 111 | 101 | 011 = 36753 ₈ Q.12 Octal to Binary: 3(011) 6(110) 7(111) 4(100) = 011110111100 ₂ Q.13 Hex to Octal: (B2F)₁₆ → Bin (1011 0010 1111) → Oct (101 | 100 | 101 | 111) = 5457 ₈
4. Binary Arithmetic
(a) Addition:
(111) 110101 <- Carry
(b) Addition: (111) 10110 <- Carry