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AppendixAUnit8_1Sol

# AppendixAUnit8_1Sol - 8 Appendix B Sample Problem Solutions...

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293 8. Appendix B: Sample Problem Solutions The following are solutions to the problems presented in Appendix A. Use these only after attempting the problems on your own.

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294 8.1. Unit 1 - Number Systems, Conversions, Signed Representations and Arithmetic 1. a. 1110110.010111 2 = 1*2 6 + 1*2 5 +1*2 4 + 1*2 2 + 1*2 1 + 1*2 -2 + 1*2 -4 + 1*2 -5 + 1*2 -6 = 118.359375 10 = 001 | 110 | 110.010 | 111 2 = 166.27 8 = 0111 | 0110.0101 | 1100 2 = 76.5C 16 b. 15B.35 16 = 1*16 2 + 5*16 1 +11*16 0 + 3*16 -1 + 5*16 -2 = 347.20703125 10 = 0001 | 0101 | 1011.0011 | 0101 2 = 101 | 011 | 011.001 | 101 | 010 2 = 533.152 8 2. Following Generalized Integer Method and Fraction Method, we have: 923.625 10 = 1110011011.101 2 Integer Quotient Rem. Coeff. Int. Frac. Coeff. 923 ÷ 2 = 461 1 a 0 = 1 461 ÷ 2 = 230 1 a 1 = 1 230 ÷ 2 = 115 0 a 2 = 0 115 ÷ 2 = 57 1 a 3 = 1 57 ÷ 2 = 28 1 a 4 = 1 28 ÷ 2 = 14 0 a 5 = 0 14 ÷ 2 = 7 0 a 6 = 0 7 ÷ 2 = 3 1 a 7 = 1 3 ÷ 2 = 1 1 a 8 = 1 1 ÷ 2 = 0 1 a 9 = 1 0.625 × 2 = 1 0.25 a -1 = 1 0.25 × 2 = 0 0.5 a -2 = 0 0.5 × 2 = 1 0 a -3 = 1 (923) 10 = (1110011011) 2 (0.625) 10 = (0.101) 2 3. a. 4530.152 8 = 100 101 011 000.001 101 01 2 = 1001 | 0101 | 1000.0011 | 0101 2 = 958.35 16 4. a. DABB.AD00 16 = 1101 1010 1011 1011.1010 1101 2 = 001 | 101 | 101 | 010 | 111 | 011.101 | 011 | 010 2 = 155273.532 8 b. BAD.A 16 = 11*16 2 + 10*16 1 +13*16 0 + 10*16 -1 = 2989.625 10 5.
295 a. Following Generalized Integer Method and Fraction Method by using base 5, we have 2143.64 10 = 32033.31 5 Integer Quotient Rem. Coeff. Int.

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AppendixAUnit8_1Sol - 8 Appendix B Sample Problem Solutions...

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