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Number conversions (122)10 to BCD etc.

Short answer

a) 122 BCD: 0001 0010 0010. b) 423.25 octal to binary: 100010011.010101. c) 179=10110011, .03125=2⁻⁵ → 10110011.00001. d) 49-25 by 2’s complement = 24.

Perform the following: a) (122)10 = (?)BCD b) (423.25)8 = (?)2 c) (179.03125)10 = (?)2 d) Subtract (25)10 from (49)10 using 2's complement method.

a) (122)₁₀ to BCD

1 = 0001, 2 = 0010, 2 = 0010

(122)₁₀ = (0001 0010 0010)BCD

b) (423.25)₈ to binary

4 = 100, 2 = 010, 3 = 011 . 2 = 010, 5 = 101

(423.25)₈ = (100010011.010101)₂

c) (179.03125)₁₀ to binary

  • 179 = 128 + 32 + 16 + 2 + 1 = 10110011
  • 0.03125 = 2⁻⁵ = .00001

(179.03125)₁₀ = (10110011.00001)₂

d) 49 − 25 by 2's complement (8 bits)

  • 25 = 00011001; 2's complement = 11100111
  • 00110001 + 11100111 = 1 00011000
  • End carry is discarded, so the result is positive: 00011000

Answer: 49 − 25 = 24 = (00011000)₂

✓ Answer checked by Claude Opus 5.5 (high effort)

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