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Number conversions (375.37)8 etc.

Short answer

a) Octal 375.37 to hex via binary: FD.7C. b) 169=10101001, fraction 1/32=.00001 → 10101001.00001. c) 905 BCD: 1001 0000 0101. d) 2’s complement of 25 is 11100111; 49+(-25)=24 with end carry discarded.

Perform the following: a) (375.37)8 = (?)16 b) (169.03125)10 = (?)2 c) (905)10 = (?)BCD d) Subtract (25)10 from (49)10 using 2's complement method.

a) (375.37)₈ to hexadecimal

Octal to binary (3 bits each), then regroup in 4 bits.

375  = 011 111 101
.37  = 011 111
Integer: 0 1111 1101 -> 0 F D
Fraction: 0111 1100  (padded with zeros) -> 7 C

(375.37)₈ = (FD.7C)₁₆

b) (169.03125)₁₀ to binary

  • 169: 128 + 32 + 8 + 1 = 10101001
  • 0.03125 = 1/32 = 2⁻⁵ = .00001

(169.03125)₁₀ = (10101001.00001)₂

c) (905)₁₀ to BCD

9 = 1001, 0 = 0000, 5 = 0101

(905)₁₀ = (1001 0000 0101)BCD

d) 49 − 25 using 2's complement (8 bits)

  • 49 = 00110001, 25 = 00011001
  • 2's complement of 25 = 11100110 + 1 = 11100111
   00110001
 + 11100111
 ----------
 1 00011000   discard the end carry

End carry present, so the result is positive: 00011000 = 24.

Answer: 49 − 25 = 24

✓ Answer checked by Claude Opus 5.5 (high effort)

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